🔬 Model Lab

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

---

Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

---

Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

---

Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

---

Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

---

Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

---

Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

---

Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

---

**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

---

**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** Find the total number of fish in the lake, given that the ratio of trout is constant in both the sample and the whole lake.

**Setup:** 
- Lake has 250 trout total
- Sample of 180 fish contains 30 trout
- Ratio in sample: 30/180 = 1/6
- This ratio equals the ratio in the lake: 250/Total = 1/6

**Math:**
$$\frac{250}{\text{Total}} = \frac{30}{180}$$
$$\frac{250}{\text{Total}} = \frac{1}{6}$$
$$\text{Total} = 250 \times 6 = 1500$$

**Check:** 250/1500 = 1/6 ✓ and 30/180 = 1/6 ✓

ANSWER 1: B

---

**Problem 2:**

**What's being asked:** Find the radius of each circle and order them from smallest to largest.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** circumference = 8π
- Formula: C = 2πr
- 8π = 2πr
- r = 4

**Circle Z:** area = 9π
- Formula: A = πr²
- 9π = πr²
- r² = 9
- r = 3

**Order by radius:** 
- X: π ≈ 3.14
- Z: 3
- Y: 4

From smallest to largest: Z (3), X (≈3.14), Y (4)

ANSWER 2: B

---

**Problem 3:**

**What's being asked:** If 20% of a number is 12, find 30% of that number.

**Setup:**
- Let n = the number
- 0.20n = 12
- n = 12/0.20 = 60

**Finding 30%:**
- 0.30 × 60 = 18

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** Two prime numbers sum to 85. Find their product.

**Reasoning:** 85 is odd, so one prime must be even and one odd. The only even prime is 2.
- If one prime is 2: 2 + ? = 85, so ? = 83
- Check: Is 83 prime? Yes (not divisible by 2, 3, 5, 7...)

**Product:**
$$2 \times 83 = 166$$

ANSWER 4: E

---

**Problem 5:**

**What's being asked:** Calculate wind chill using the formula with air temp = 36°F and wind speed = 18 mph.

**Formula:** Wind chill = (air temperature) − 0.7 × (wind speed)

**Calculation:**
$$\text{Wind chill} = 36 - 0.7 \times 18$$
$$= 36 - 12.6$$
$$= 23.4$$

Closest to 23.

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Calculate 200,000 × 200,000

**Calculation:**
$$200,000 \times 200,000 = (2 \times 10^5)^2 = 4 \times 10^{10}$$
$$= 40,000,000,000 = \text{40 billion}$$

ANSWER 6: E

---

**Problem 7:**

**What's being asked:** Connie multiplied by 2 and got 60, but should have divided by 2. Find the correct answer.

**Setup:**
- She multiplied: 2n = 60, so n = 30
- Correct operation: divide by 2
- Correct answer: 30 ÷ 2 = 15

ANSWER 7: B

---

**Problem 8:**

**What's being asked:** Find overall percent on combined test using weighted average.

**Calculation:**
- Test 1: 70% of 10 problems = 0.70 × 10 = 7 correct
- Test 2: 80% of 20 problems = 0.80 × 20 = 16 correct
- Test 3: 90% of 30 problems = 0.90 × 30 = 27 correct

**Total:**
- Correct: 7 + 16 + 27 = 50 out of 60
- Percentage: 50/60 = 0.8333... ≈ 83%

ANSWER 8: D

---

**Problem 9:**

**What's being asked:** Find total of three purchases, rounded to nearest dollar.

**Calculation:**
$$1.98 + 5.04 + 9.89 = 16.91$$

Rounded to nearest dollar: $17

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Find target heart rate for a 26-year-old athlete.

**Setup:**
- Maximum heart rate = 220 − age = 220 − 26 = 194 bpm
- Target heart rate = 80% of maximum = 0.80 × 194

**Calculation:**
$$0.80 \times 194 = 155.2 \approx 155$$

ANSWER 10: B

Problem 1:
Ratio of trout in sample: 30/180 = 1/6. So trout are 1/6 of total fish.
250 = (1/6) × total, total = 1500.
ANSWER 1: B

Problem 2:
Circle X radius = π ≈ 3.14.
Circle Y circumference 8π = 2πr → r = 4.
Circle Z area 9π = πr² → r² = 9 → r = 3.
Order smallest to largest: Z (3), X (3.14), Y (4).
ANSWER 2: B

Problem 3:
20% of number = 12 → number = 60. 30% of 60 = 18.
ANSWER 3: B

Problem 4:
Sum 85 is odd, so one prime must be 2 (even). Other = 83 (prime). Product = 2×83 = 166.
ANSWER 4: E

Problem 5:
36 − 0.7×18 = 36 − 12.6 = 23.4 ≈ 23.
ANSWER 5: B

Problem 6:
200,000 × 200,000 = 2×10⁵ × 2×10⁵ = 4×10¹⁰ = forty billion.
ANSWER 6: E

Problem 7:
2 × number = 60 → number = 30. Correct: 30 ÷ 2 = 15.
ANSWER 7: B

Problem 8:
Test 1: 70% of 10 = 7. Test 2: 80% of 20 = 16. Test 3: 90% of 30 = 27.
Total correct = 7+16+27 = 50 out of 60 = 83.3% ≈ 83.
ANSWER 8: D

Problem 9:
1.98 + 5.04 + 9.89 = 16.91 ≈ $17.
ANSWER 9: D

Problem 10:
Max heart rate = 220 − 26 = 194. Target = 80% × 194 = 155.2 ≈ 155.
ANSWER 10: B

I'll solve each problem systematically.

---

**Problem 1:**

The ratio of trout to total fish is the same in sample and lake.
- Sample: 30 trout out of 180 fish → ratio = 30/180 = 1/6
- Lake: 250 trout out of T total fish → 250/T = 1/6
- T = 250 × 6 = **1500**

ANSWER 1: B

---

**Problem 2:**

Find each radius:
- Circle X: radius = π ≈ 3.14
- Circle Y: circumference = 8π → 2πr = 8π → r = 4
- Circle Z: area = 9π → πr² = 9π → r² = 9 → r = 3

Order smallest to largest: Z (r=3), X (r≈3.14), Y (r=4)

ANSWER 2: B

---

**Problem 3:**

- 20% of N = 12 → N = 12/0.20 = 60
- 30% of 60 = 0.30 × 60 = **18**

ANSWER 3: B

---

**Problem 4:**

Two primes sum to 85 (odd). One prime must be 2 (the only even prime), and the other is 83 (which is prime).
- Product = 2 × 83 = **166**

ANSWER 4: E

---

**Problem 5:**

Wind chill = air temperature − 0.7 × wind speed
= 36 − 0.7 × 18
= 36 − 12.6
= 23.4 ≈ **23**

ANSWER 5: B

---

**Problem 6:**

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = **forty billion**

ANSWER 6: E

---

**Problem 7:**

Connie multiplied by 2 and got 60, so the original number = 60 ÷ 2 = 30.
The correct operation is to divide by 2: 30 ÷ 2 = **15**

ANSWER 7: B

---

**Problem 8:**

Count correct answers on each test:
- 70% of 10 = 7 correct
- 80% of 20 = 16 correct
- 90% of 30 = 27 correct
- Total correct = 7 + 16 + 27 = 50 out of 60
- Overall % = 50/60 = 83.3% ≈ **83%**

ANSWER 8: D

---

**Problem 9:**

Round each purchase to nearest dollar:
- $1.98 ≈ $2
- $5.04 ≈ $5
- $9.89 ≈ $10
- Total ≈ $17

Exact total: $1.98 + $5.04 + $9.89 = $16.91 ≈ **$17**

ANSWER 9: D

---

**Problem 10:**

- Maximum heart rate = 220 − 26 = 194 bpm
- Target heart rate = 80% × 194 = 0.80 × 194 = 155.2 ≈ **155**

ANSWER 10: B

Problem 1:

Let the total number of fish in the lake be \(N\).

The sample says:
- 30 trout out of 180 fish

So the trout fraction in the sample is
\[
\frac{30}{180}=\frac{1}{6}.
\]

We are told this same ratio applies to the whole lake, and the lake has 250 trout, so:
\[
\frac{250}{N}=\frac{1}{6}.
\]

Solve for \(N\):
\[
N=250\cdot 6=1500.
\]

So the lake contains 1500 fish.

The answer is B, not A, C, D, or E, because only 1500 gives the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We find each radius.

- Circle X has radius \(\pi\), so \(r_X=\pi\).
- Circle Y has circumference \(8\pi\). Since \(C=2\pi r\),
  \[
  2\pi r=8\pi \Rightarrow r=4.
  \]
- Circle Z has area \(9\pi\). Since \(A=\pi r^2\),
  \[
  \pi r^2=9\pi \Rightarrow r^2=9 \Rightarrow r=3.
  \]

Now compare:
\[
r_Z=3,\quad r_Y=4,\quad r_X=\pi\approx 3.14.
\]

So from smallest to largest:
\[
Z,\ X,\ Y.
\]

That matches choice B.

ANSWER 2: B

Problem 3:

If 20% of a number is 12, let the number be \(n\). Then
\[
0.2n=12.
\]
So
\[
n=60.
\]

Now find 30% of 60:
\[
0.3(60)=18.
\]

So the answer is 18, which is choice B.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, the only way two primes can add to an odd number is if one of them is 2, because every other prime is odd, and odd + odd = even.

So the two primes are:
\[
2 \text{ and } 83
\]
because \(2+83=85\), and 83 is prime.

Their product is
\[
2\cdot 83=166.
\]

So the answer is E.

ANSWER 4: E

Problem 5:

Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed}).
\]

Substitute the values:
\[
36 - 0.7(18).
\]
Compute:
\[
0.7(18)=12.6
\]
so
\[
36-12.6=23.4.
\]

The closest choice is 23.

So the answer is B.

ANSWER 5: B

Problem 6:

Compute:
\[
200{,}000 \times 200{,}000.
\]

Write in scientific notation:
\[
2\times 10^5 \cdot 2\times 10^5 = 4\times 10^{10}.
\]

That is
\[
40{,}000{,}000{,}000,
\]
which is forty billion.

So the correct choice is E.

ANSWER 6: E

Problem 7:

Connie multiplied a number by 2 and got 60. Let the number be \(n\). Then
\[
2n=60 \Rightarrow n=30.
\]

But she should have divided the number by 2 to get the correct answer. So the correct answer is
\[
\frac{30}{2}=15.
\]

Thus the correct answer is B.

ANSWER 7: B

Problem 8:

To find the overall score, count how many problems she got correct on each test:

- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27

Total correct:
\[
7+16+27=50.
\]

Total problems:
\[
10+20+30=60.
\]

So her overall percentage is
\[
\frac{50}{60}=\frac{5}{6}\approx 83.3\%.
\]

The closest choice is 83.

So the answer is D.

ANSWER 8: D

Problem 9:

Add the purchases:
\[
1.98+5.04=7.02
\]
and
\[
7.02+9.89=16.91.
\]

Rounded to the nearest dollar, that is
\[
17.
\]

So the answer is D.

ANSWER 9: D

Problem 10:

First find the theoretical maximum heart rate:
\[
220-26=194.
\]

The target heart rate is 80% of that:
\[
0.8(194)=155.2.
\]

Rounded to the nearest whole number:
\[
155.
\]

So the answer is B.

ANSWER 10: B

Problem 1:
We’re told there are 250 trout in a lake. A sample of 180 fish is caught, and 30 of the 180 are trout.  
Because the ratio of trout to total fish is the same in the sample and the lake:
\[
\frac{\text{trout}}{\text{total}}=\frac{30}{180}
\]
Simplify the sample ratio:
\[
\frac{30}{180}=\frac{1}{6}
\]
So in the lake, trout are \(\frac{1}{6}\) of all fish. That means:
\[
\frac{250}{\text{total fish}}=\frac{1}{6}
\]
Solve:
\[
\text{total fish}=250\cdot 6=1500
\]
Check against choices: 1500 matches choice B.

ANSWER 1: B

Problem 2:
We compare radii using area and circumference formulas.

Circle X: radius \(=\pi\).  
So \(r_X=\pi\).

Circle Y: circumference \(=8\pi\).  
Circumference formula: \(C=2\pi r\).
\[
2\pi r_Y = 8\pi \Rightarrow r_Y = 4
\]

Circle Z: area \(=9\pi\).  
Area formula: \(A=\pi r^2\).
\[
\pi r_Z^2 = 9\pi \Rightarrow r_Z^2=9 \Rightarrow r_Z=3
\]

Now compare:
- \(r_X=\pi \approx 3.14\)
- \(r_Z=3\)
- \(r_Y=4\)

Smallest to largest: \(r_Z(3), r_X(\pi), r_Y(4)\).

That corresponds to Z, X, Y → choice E.

ANSWER 2: E

Problem 3:
Let the number be \(N\).  
20% of \(N\) is 12:
\[
0.20N=12 \Rightarrow N=\frac{12}{0.20}=60
\]
Now find 30% of the same number:
\[
0.30N=0.30\cdot 60=18
\]
Closest exact value is 18 → choice B.

ANSWER 3: B

Problem 4:
Two primes add to 85.  
Odd + odd = even, so to get 85 (odd), one prime must be 2 and the other must be 83:
\[
85=2+83
\]
Product:
\[
2\cdot 83=166
\]
Choice E.

ANSWER 4: E

Problem 5:
Wind chill:
\[
\text{wind chill} = 36 - 0.7(18)
\]
Compute:
\[
0.7\cdot 18 = 12.6
\]
So:
\[
36-12.6=23.4
\]
Closest choice is 23 → choice B.

ANSWER 5: B

Problem 6:
“Two hundred thousand times two hundred thousand” means:
\[
200{,}000 \times 200{,}000 = (2\times 10^5)(2\times 10^5)=4\times 10^{10}
\]
Now translate:
\[
10^{10} = 10{,}000{,}000{,}000
\]
So:
\[
4\times 10^{10} = 40{,}000{,}000{,}000
\]
That is forty billion → choice E.

ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, but she should have divided by 2.

Let the number be \(x\).
- Her (incorrect) computation: \(2x = 60 \Rightarrow x = 30\)
But if she should have divided by 2, the correct result would be:
\[
\frac{x}{2}=\frac{30}{2}=15
\]
Choice B.

ANSWER 7: B

Problem 8:
Overall score = total correct ÷ total questions.

Connie:
- 70% of 10: \(0.70\cdot 10=7\)
- 80% of 20: \(0.80\cdot 20=16\)
- 90% of 30: \(0.90\cdot 30=27\)

Total correct:
\[
7+16+27=50
\]
Total questions:
\[
10+20+30=60
\]
Overall percentage:
\[
\frac{50}{60}=\frac{5}{6}\approx 83.33\%
\]
Closest choice is 83 → choice D.

ANSWER 8: D

Problem 9:
Add the three prices:
\[
1.98 + 5.04 + 9.89
\]
First sum:
\[
1.98+5.04=7.02
\]
Then:
\[
7.02+9.89=16.91
\]
To the nearest dollar, 16.91 rounds to 17.

Choice D.

ANSWER 9: D

Problem 10:
Maximum heart rate:
\[
220 - \text{age} = 220 - 26 = 194
\]
Target heart rate is 80% of that:
\[
0.80 \cdot 194 = 155.2
\]
Nearest whole number: 155.

Choice B.

ANSWER 10: B

Problem 1:

We are told:

- The lake contains 250 trout.
- In a sample of 180 fish, 30 are trout.
- The ratio of trout to total fish is the same in the sample and in the lake.

In the sample, the fraction of fish that are trout is

\[
\frac{30}{180}=\frac{1}{6}.
\]

So trout make up \(\frac16\) of all fish in the lake.

Let the total number of fish in the lake be \(N\). Then

\[
\frac16 N = 250.
\]

Multiply both sides by 6:

\[
N = 250 \times 6 = 1500.
\]

So there are 1500 fish in the lake.

Choice A, 1250, would make the trout fraction \(\frac{250}{1250}=\frac15\), not \(\frac16\).  
Choice C, 1750, choice D, 1800, and choice E, 2000, also do not give the same trout ratio as the sample.

ANSWER 1: B

Problem 2:

We need to list circles X, Y, and Z from smallest to largest radius.

Circle X has radius:

\[
r_X = \pi.
\]

Circle Y has circumference \(8\pi\). Circumference is

\[
C=2\pi r.
\]

So

\[
8\pi = 2\pi r.
\]

Divide by \(2\pi\):

\[
r_Y = 4.
\]

Circle Z has area \(9\pi\). Area is

\[
A=\pi r^2.
\]

So

\[
9\pi = \pi r^2.
\]

Divide by \(\pi\):

\[
9 = r^2.
\]

Thus

\[
r_Z = 3.
\]

Now compare the radii:

\[
r_Z=3,\quad r_X=\pi \approx 3.14,\quad r_Y=4.
\]

So from smallest to largest:

\[
Z,\ X,\ Y.
\]

Choice A puts X before Z, but \(3 < \pi\), so Z is smaller than X.  
Choice C puts Y first, but Y has the largest radius.  
Choice D puts Y before X, but \(4 > \pi\).  
Choice E puts X before Z, which is incorrect.

ANSWER 2: B

Problem 3:

We are told 20% of a number is 12.

Let the number be \(n\). Then

\[
20\% \text{ of } n = 12.
\]

Since \(20\%=\frac{1}{5}\),

\[
\frac15 n = 12.
\]

Multiply by 5:

\[
n = 60.
\]

Now we need 30% of the same number:

\[
30\% \text{ of } 60 = 0.30 \times 60 = 18.
\]

So the answer is 18.

Choice A, 15, is too small.  
Choice C, 20, would be \(\frac13\) of 60, not 30%.  
Choice D, 24, is 40% of 60.  
Choice E, 30, is 50% of 60.

ANSWER 3: B

Problem 4:

We need two prime numbers whose sum is 85.

Since 85 is odd, one of the primes must be even and the other must be odd. The only even prime number is 2.

So one prime is 2.

Then the other prime is

\[
85-2=83.
\]

Check that 83 is prime. It is not divisible by 2, 3, 5, or 7, so 83 is prime.

Now find the product:

\[
2 \times 83 = 166.
\]

So the product is 166.

Choice A, 85, is the sum, not the product.  
Choices B, C, and D do not come from multiplying the two primes 2 and 83.

ANSWER 4: E

Problem 5:

The formula is

\[
\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}.
\]

The air temperature is \(36^\circ\text{F}\), and the wind speed is 18 mph.

Substitute into the formula:

\[
36 - 0.7 \times 18.
\]

Compute:

\[
0.7 \times 18 = 12.6.
\]

Then

\[
36 - 12.6 = 23.4.
\]

The closest choice is 23.

Choice A, 18, is too low.  
Choice C, 28, is higher than 23.4.  
Choice D, 32, and choice E, 35, are much too high.

ANSWER 5: B

Problem 6:

We need to compute:

\[
200{,}000 \times 200{,}000.
\]

Write each number as

\[
200{,}000 = 2 \times 100{,}000.
\]

So

\[
200{,}000 \times 200{,}000
= (2 \times 100{,}000)(2 \times 100{,}000).
\]

Multiply the 2s:

\[
2 \times 2 = 4.
\]

Multiply the powers of 10:

\[
100{,}000 \times 100{,}000 = 10^{5} \times 10^{5} = 10^{10}.
\]

So

\[
200{,}000 \times 200{,}000 = 4 \times 10^{10}.
\]

That is

\[
40{,}000{,}000{,}000,
\]

which is forty billion.

Choice A, four hundred thousand, is much too small.  
Choice B, four million, is too small.  
Choice C, forty thousand, is even smaller.  
Choice D, four hundred million, is also too small.

ANSWER 6: E

Problem 7:

Connie multiplies a number by 2 and gets 60.

Let the original number be \(x\). Then

\[
2x = 60.
\]

Divide by 2:

\[
x = 30.
\]

But Connie should have divided the original number by 2.

So the correct answer should be

\[
30 \div 2 = 15.
\]

Choice C, 30, is the original number, not the correct result.  
Choice D, 120, comes from multiplying 60 by 2.  
Choice E, 240, is much too large.  
Choice A, 7.5, would be half of 15, not half of the original number.

ANSWER 7: B

Problem 8:

Antonette took three tests:

- 70% on a 10-problem test
- 80% on a 20-problem test
- 90% on a 30-problem test

We need her overall score if the tests are combined into one 60-problem test.

First find how many questions she got correct on each test.

On the 10-problem test:

\[
70\% \times 10 = 0.70 \times 10 = 7.
\]

On the 20-problem test:

\[
80\% \times 20 = 0.80 \times 20 = 16.
\]

On the 30-problem test:

\[
90\% \times 30 = 0.90 \times 30 = 27.
\]

Total correct:

\[
7+16+27=50.
\]

Total questions:

\[
10+20+30=60.
\]

Overall percent:

\[
\frac{50}{60}=\frac56 \approx 0.8333.
\]

As a percent, this is about

\[
83.3\%.
\]

The closest choice is 83.

Choice C, 80, is the simple average of 70, 80, and 90, but the tests have different numbers of problems, so we need a weighted average.  
Choice E, 87, is too high.  
Choices A and B are too low.

ANSWER 8: D

Problem 9:

Mindy made purchases of

\[
\$1.98,\quad \$5.04,\quad \$9.89.
\]

Add them:

\[
1.98 + 5.04 = 7.02.
\]

Then

\[
7.02 + 9.89 = 16.91.
\]

So the total is \(\$16.91\).

Rounded to the nearest dollar:

\[
\$16.91 \approx \$17.
\]

Choice C, \$16, is too low because 16.91 rounds up.  
Choice B, \$15, and choice A, \$10, are much too low.  
Choice E, \$18, is too high.

ANSWER 9: D

Problem 10:

The maximum heart rate is found by subtracting the athlete's age from 220.

The athlete is 26 years old, so

\[
220 - 26 = 194.
\]

The target heart rate is 80% of the maximum heart rate:

\[
80\% \times 194 = 0.80 \times 194.
\]

Compute:

\[
0.8 \times 194 = 155.2.
\]

To the nearest whole number:

\[
155.2 \approx 155.
\]

So the target heart rate is 155 beats per minute.

Choice D, 194, is the maximum heart rate, not 80% of it.  
Choice C, 176, is too high.  
Choice A, 134, is too low.  
Choice E, 243, is impossible because it is greater than the maximum heart rate.

ANSWER 10: B

Problem 1:
We are given that the ratio of trout to total fish is constant. Let $T$ be the number of trout (250) and $F$ be the total number of fish in the lake. The sample has 30 trout out of 180 fish.
The ratio is: $\frac{30}{180} = \frac{250}{F}$.
Simplify the fraction: $\frac{30}{180} = \frac{1}{6}$.
So, $\frac{1}{6} = \frac{250}{F}$.
Multiply both sides by $F$ and by 6: $F = 250 \times 6 = 1500$.
ANSWER 1: B

Problem 2:
- Circle X: Radius $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 2\pi r_Y = 8\pi$, so $r_Y = 4$.
- Circle Z: Area $A = \pi r_Z^2 = 9\pi$, so $r_Z^2 = 9$, meaning $r_Z = 3$.
Comparing radii: $r_X \approx 3.14$, $r_Y = 4$, $r_Z = 3$.
Order from smallest to largest: $r_Z (3) < r_X (\approx 3.14) < r_Y (4)$.
The order is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be $x$. We are given $0.20x = 12$.
To find $x$, divide by 0.20: $x = \frac{12}{0.2} = 60$.
We want 30% of $x$: $0.30 \times 60 = 18$.
Alternatively, since 30% is 1.5 times 20%, the answer is $12 \times 1.5 = 18$.
ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is 2.
So, $2 + p = 85$, which means $p = 83$.
83 is a prime number.
The product is $2 \times 83 = 166$.
ANSWER 4: E

Problem 5:
Formula: Wind chill = $36 - 0.7 \times 18$.
Calculate $0.7 \times 18$: $0.7 \times 10 = 7$ and $0.7 \times 8 = 5.6$. So, $7 + 5.6 = 12.6$.
Wind chill = $36 - 12.6 = 23.4$.
The closest choice is 23.
ANSWER 5: B

Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$.
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.
$4 \times 10^{10} = 40,000,000,000$, which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied a number $x$ by 2 to get 60: $2x = 60 \implies x = 30$.
She should have divided the number by 2: $\frac{30}{2} = 15$.
ANSWER 7: B

Problem 8:
Calculate the total number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = $7 + 16 + 27 = 50$.
Total questions = $10 + 20 + 30 = 60$.
Percentage = $\frac{50}{60} = \frac{5}{6} \approx 0.8333...$ or 83.33%.
The closest choice is 83.
ANSWER 8: D

Problem 9:
Sum the purchases: $1.98 + 5.04 + 9.89$.
$1.98 \approx 2$, $5.04 \approx 5$, $9.89 \approx 10$.
$2 + 5 + 10 = 17$.
Exact sum: $1.98 + 5.04 + 9.89 = 16.91$.
Rounding 16.91 to the nearest dollar gives 17.
ANSWER 9: D

Problem 10:
Max heart rate = $220 - 26 = 194$.
Target heart rate = 80% of 194 = $0.8 \times 194$.
$0.8 \times 194 = 155.2$.
Rounding to the nearest whole number gives 155.
ANSWER 10: B

Problem 1:
We are asked to find the total number of fish in the lake, given that there are 250 trout and that the ratio of trout to the total number of fish is the same in the lake as it is in a sample.
First, let's find the ratio of trout to total fish in the sample. The marine biologist catches 180 fish, and 30 of them are trout. 
The ratio is:
30 / 180 = 1 / 6

This means that 1 out of every 6 fish in the lake is a trout. Let $F$ be the total number of fish in the lake. We can set up the proportion:
250 / $F$ = 1 / 6

To solve for $F$, we cross-multiply:
$F$ = 250 × 6
$F$ = 1500

There are 1500 fish in the lake. 
Ruling out the wrong choices:
A. 1250 is 250 × 5, which uses the wrong multiplier.
C. 1750 is 250 × 7.
D. 1800 is a miscalculation likely derived from the 180 sample size.
E. 2000 is 250 × 8.

ANSWER 1: B

Problem 2:
We are asked to list the three circles in order from smallest to largest radius. Let's determine the radius of each circle.

- Circle X: The radius is given directly as $\pi$. Since $\pi \approx 3.14$, the radius of X is approximately 3.14.
- Circle Y: The circumference is $8\pi$. The formula for circumference is $C = 2\pi r$. 
  $2\pi r = 8\pi \implies r = 4$.
- Circle Z: The area is $9\pi$. The formula for area is $A = \pi r^2$.
  $\pi r^2 = 9\pi \implies r^2 = 9 \implies r = 3$.

Now we compare the radii:
Radius of Z = 3
Radius of X $\approx$ 3.14
Radius of Y = 4

Ordering them from smallest to largest gives Z, X, Y.
Ruling out the wrong choices:
A, C, D, and E do not reflect the correct numerical order of 3 < $\pi$ < 4.

ANSWER 2: B

Problem 3:
We are asked to find 30% of a number, given that 20% of the same number is 12.
Let the unknown number be $N$. We can set up the equation:
0.20 × $N$ = 12

Solving for $N$:
$N$ = 12 / 0.20 = 60

Now, we need to find 30% of this number:
0.30 × 60 = 18

Alternatively, since 30% is 1.5 times 20%, we can simply multiply 12 by 1.5 to get 18.
Ruling out the wrong choices:
A. 15 is incorrect.
C. 20 is incorrect.
D. 24 would be 40% of the number.
E. 30 would be 50% of the number.

ANSWER 3: B

Problem 4:
We are asked to find the product of two prime numbers whose sum is 85.
Since 85 is an odd number, the sum of the two primes must consist of one even number and one odd number (because the sum of two odd numbers is always even). 
The only even prime number is 2. 
Therefore, the other prime number must be:
85 - 2 = 83

We can verify that 83 is indeed a prime number (it is not divisible by 2, 3, 5, 7, or 9).
The problem asks for the product of these two prime numbers:
2 × 83 = 166

Ruling out the wrong choices:
A. 85 is the sum, not the product.
B. 91 is not prime (7 × 13) and doesn't sum to 85 with another prime.
C. 115 is not the correct product.
D. 133 is not the correct product.

ANSWER 4: E

Problem 5:
We are asked to calculate the wind chill using the given formula:
(wind chill) = (air temperature) − 0.7 × (wind speed)

Given:
Air temperature = 36°F
Wind speed = 18 mph

Substitute the values into the formula:
Wind chill = 36 - 0.7 × 18
First, calculate the multiplication:
0.7 × 18 = 12.6
Next, subtract this from the air temperature:
36 - 12.6 = 23.4

The value 23.4 is closest to 23.
Ruling out the wrong choices:
A. 18 is the wind speed.
C. 28 is incorrect.
D. 32 is incorrect.
E. 35 is incorrect.

ANSWER 5: B

Problem 6:
We are asked to multiply two hundred thousand by two hundred thousand.
Written in numbers, this is:
200,000 × 200,000

We can use scientific notation or count the zeros:
200,000 = 2 × 10^5
(2 × 10^5) × (2 × 10^5) = (2 × 2) × 10^(5+5) = 4 × 10^10

Written out, 4 × 10^10 is a 4 followed by 10 zeros:
40,000,000,000

This number is forty billion.
Ruling out the wrong choices:
A. four hundred thousand is 400,000.
B. four million is 4,000,000.
C. forty thousand is 40,000.
D. four hundred million is 400,000,000.

ANSWER 6: E

Problem 7:
We are asked to find the correct answer Connie should have gotten if she had divided her number by 2 instead of multiplying it by 2.
Let the original number be $N$.
Connie multiplied by 2 and got 60:
$N$ × 2 = 60
$N$ = 30

The correct operation was to divide the original number by 2:
Correct answer = $N$ / 2
Correct answer = 30 / 2 = 15

Ruling out the wrong choices:
A. 7.5 is 15 / 2.
C. 30 is the original number, not the divided result.
D. 120 is what she would get if she multiplied 60 by 2.
E. 240 is 120 × 2.

ANSWER 7: B

Problem 8:
We are asked to find Antonette's overall percentage when three tests are combined.
First, we find the number of correct problems on each test:
- 70% on a 10-problem test: 0.70 × 10 = 7 correct
- 80% on a 20-problem test: 0.80 × 20 = 16 correct
- 90% on a 30-problem test: 0.90 × 30 = 27 correct

Next, we find the total number of correct problems and the total number of problems:
Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Now, we calculate the overall percentage:
(50 / 60) × 100% = (5 / 6) × 100% ≈ 83.33%

The closest percent is 83.
Ruling out the wrong choices:
A. 40 is incorrect.
B. 77 is incorrect.
C. 80 is the unweighted average of 70, 80, and 90, which ignores the different number of problems on each test.
E. 87 is incorrect.

ANSWER 8: D

Problem 9:
We are asked to find the total cost of three purchases to the nearest dollar.
The purchases are $1.98, $5.04, and $9.89.
Let's add them together:
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

Rounding $16.91 to the nearest whole dollar gives $17.
(Alternatively, we could estimate: $1.98 is about $2, $5.04 is about $5, and $9.89 is about $10. $2 + $5 + $10 = $17).
Ruling out the wrong choices:
A. $10 is way too low.
B. $15 is too low.
C. $16 is what you get if you truncate the decimals instead of rounding properly.
E. $18 is too high.

ANSWER 9: D

Problem 10:
We are asked to find the target heart rate of a 26-year-old athlete.
First, we find the theoretical maximum heart rate by subtracting the athlete's age from 220:
Maximum heart rate = 220 - 26 = 194 beats per minute.

Next,

Problem 1:
The sample shows 30 trout out of 180 fish, which is the proportion \(30/180 = 1/6\). This same proportion holds for the lake, where there are 250 trout. Let \(T\) be the total number of fish in the lake. Then \(250/T = 1/6\), so \(T = 250 \times 6 = 1500\). Equivalently, cross-multiplying the proportion \(30/180 = 250/T\) also yields \(T = 1500\). Among the choices, this matches B (the value 1250 arises from incorrectly using \(30/180 = 250/T\) without the factor of 6, while 1750–2000 come from unrelated arithmetic errors).

ANSWER 1: B

Problem 2:
The radius of circle X is given directly as \(\pi \approx 3.14\). For circle Y the circumference formula \(2\pi r = 8\pi\) simplifies to \(r = 4\). For circle Z the area formula \(\pi r^2 = 9\pi\) simplifies to \(r = 3\). Ordering the radii from smallest to largest therefore gives 3, \(\pi\), 4, or Z, X, Y. This matches choice B (the other orderings arise from swapping the computed radii or confusing radius with diameter/circumference).

ANSWER 2: B

Problem 3:
Let \(n\) be the unknown number. Then \(0.2n = 12\), so \(n = 60\). Thirty percent of the same number is \(0.3 \times 60 = 18\). This matches choice B (the distractors 15 and 20 come from using 25 % or 1/3 instead of 30 %, while 24 and 30 come from scaling the original 12 by 2 or 2.5).

ANSWER 3: B

Problem 4:
The sum of the two primes is odd, so exactly one of them must be the even prime 2; the other prime is therefore \(85-2=83\). Their product is \(2 \times 83 = 166\). This matches choice E (the remaining choices are products of two odd numbers that sum to 85, which is impossible for primes).

ANSWER 4: E

Problem 5:
Substitute the given values into the formula: \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\). The integer closest to 23.4 is 23, which matches choice B (the distractors 18 and 28 come from subtracting the wind speed itself or from rounding 23.4 upward by a full point).

ANSWER 5: B

Problem 6:
Rewrite the factors in powers of ten: \(200000 = 2 \times 10^5\). Their product is \(4 \times 10^{10} = 40000000000\), which is forty billion. This matches choice E (the other choices correspond to incorrect placements of the decimal point or to multiplying by 2 instead of squaring).

ANSWER 6: E

Problem 7:
The incorrect operation produced \(2x = 60\), so the original number is \(x = 30\). The correct operation is division by 2, which yields \(30/2 = 15\). This matches choice B (the distractors 7.5 and 30 come from dividing 15 by 2 again or from using the wrong value of \(x\); 120 and 240 come from doubling instead of halving).

ANSWER 7: B

Problem 8:
The three tests together contain 60 problems. The number of problems answered correctly is \(0.7 \times 10 + 0.8 \times 20 + 0.9 \times 30 = 7 + 16 + 27 = 50\). The overall percent correct is therefore \(50/60 \approx 83.33\%\), which rounds to the nearest choice 83. This matches choice D (the distractors 77 and 80 arise from weighting each test equally instead of by number of problems; 87 comes from rounding 83.33 upward).

ANSWER 8: D

Problem 9:
Add the three amounts: \(1.98 + 5.04 + 9.89 = 16.91\). Rounding 16.91 to the nearest dollar gives 17. This matches choice D (the distractors 15 and 16 come from truncating before the final addition or from rounding 16.91 downward).

ANSWER 9: D

Problem 10:
The maximum heart rate is \(220-26=194\). Eighty percent of this value is \(0.8 \times 194 = 155.2\), which rounds to the nearest whole number 155. This matches choice B (the distractors 134 and 176 come from using 70 % or 90 % instead of 80 %; 194 and 243 are the maximum rate itself or an unrelated multiple).

ANSWER 10: B

Problem 1:
The problem asks us to find the total number of fish in the lake based on the ratio of trout in a sample to the total number of trout. We are given that there are 250 trout in the lake and that out of a sample of 180 fish, 30 are trout.

Let's denote the total number of fish in the lake as \(N\). The ratio of trout to the total number of fish in the lake is \(\frac{250}{N}\), and the ratio of trout to the total number of fish in the sample is \(\frac{30}{180}\). We are told these ratios are equal.

So, we set up the equation: \(\frac{250}{N} = \frac{30}{180}\).

First, simplify the right-hand side: \(\frac{30}{180} = \frac{1}{6}\).

Now, we have \(\frac{250}{N} = \frac{1}{6}\).

To solve for \(N\), we cross-multiply: \(250 \times 6 = N \times 1\), which gives \(1500 = N\).

Thus, the total number of fish in the lake is 1500.

Let's check the other options:
- A. 1250: \(\frac{250}{1250} = \frac{1}{5} \neq \frac{1}{6}\)
- C. 1750: \(\frac{250}{1750} = \frac{1}{7} \neq \frac{1}{6}\)
- D. 1800: \(\frac{250}{1800} \neq \frac{1}{6}\)
- E. 2000: \(\frac{250}{2000} = \frac{1}{8} \neq \frac{1}{6}\)

Only B gives the correct ratio.

ANSWER 1: B

Problem 2:
To compare the radii of the circles, we need to calculate or infer their radii from the given information.

1. Circle X has a radius of \(π\), so \(r_X = π\).

2. Circle Y has a circumference of \(8π\). The formula for circumference is \(C = 2πr\), so \(8π = 2πr_Y\). Solving for \(r_Y\), we get \(r_Y = \frac{8π}{2π} = 4\).

3. Circle Z has an area of \(9π\). The formula for area is \(A = πr^2\), so \(9π = πr_Z^2\). Solving for \(r_Z\), we get \(r_Z^2 = 9\), hence \(r_Z = 3\).

Now, comparing the radii:
- \(r_X = π \approx 3.14\)
- \(r_Y = 4\)
- \(r_Z = 3\)

So, the order from smallest to largest radius is \(Z, X, Y\).

Let's check the options:
- A. \(X, Y, Z\): Incorrect because \(X > Z\).
- B. \(Z, X, Y\): Correct.
- C. \(Y, X, Z\): Incorrect because \(Y > X > Z\).
- D. \(Z, Y, X\): Incorrect because \(X < Y\).
- E. \(X, Z, Y\): Incorrect because \(X > Z\).

ANSWER 2: B

Problem 3:
We are given that 20% of a number is 12. Let's denote the number as \(N\). So, \(0.20 \times N = 12\).

To find \(N\), we divide both sides by 0.20: \(N = \frac{12}{0.20} = 60\).

Now, we need to find 30% of \(N\), which is \(0.30 \times 60 = 18\).

Let's check the other options:
- A. 15: \(0.30 \times 60 = 18 \neq 15\)
- C. 20: \(0.30 \times 60 = 18 \neq 20\)
- D. 24: \(0.30 \times 60 = 18 \neq 24\)
- E. 30: \(0.30 \times 60 = 18 \neq 30\)

B is the correct answer.

ANSWER 3: B

Problem 4:
The sum of two prime numbers is 85. Since 85 is odd, one prime must be even and the other odd (because even + odd = odd). The only even prime is 2.

So, if one prime is 2, the other is \(85 - 2 = 83\), which is also prime.

The product of these two primes is \(2 \times 83 = 166\).

Let's verify:
- 2 and 83 are both prime.
- Their sum is 85.
- Their product is 166.

Let's check the other options:
- A. 85: This is the sum, not the product.
- B. 91: Not the product of 2 and 83.
- C. 115: Not the product of 2 and 83.
- D. 133: Not the product of 2 and 83.
- E. 166: This is the correct product.

ANSWER 4: E

Problem 5:
The wind chill is given by the formula: (wind chill) = (air temperature) − 0.7 × (wind speed).

Given air temperature = 36°F and wind speed = 18 mph, we substitute these values into the formula:
\[ \text{Wind Chill} = 36 - 0.7 \times 18 \]
\[ = 36 - 12.6 \]
\[ = 23.4 \]

Rounding to the nearest whole number, the wind chill is approximately 23.

Let's check the options:
- A. 18: Too low.
- B. 23: Closest to 23.4.
- C. 28: Too high.
- D. 32: Too high.
- E. 35: Too high.

B is the correct answer.

ANSWER 5: B

Problem 6:
We need to calculate \(200,000 \times 200,000\).

\[ 200,000 \times 200,000 = (2 \times 10^5) \times (2 \times 10^5) \]
\[ = 4 \times 10^{10} \]
\[ = 40,000,000,000 \]

This is forty billion.

Let's check the options:
- A. four hundred thousand: \(4 \times 10^5\), too low.
- B. four million: \(4 \times 10^6\), too low.
- C. forty thousand: \(4 \times 10^4\), too low.
- D. four hundred million: \(4 \times 10^8\), too low.
- E. forty billion: \(4 \times 10^{10}\), correct.

ANSWER 6: E

Problem 7:
Connie multiplies a number by 2 and gets 60. Let's denote the number as \(N\). So, \(2N = 60\), which means \(N = \frac{60}{2} = 30\).

However, she should have divided the number by 2. So, the correct operation is \(\frac{N}{2} = \frac{30}{2} = 15\).

Let's check the options:
- A. 7.5: Incorrect.
- B. 15: Correct.
- C. 30: This is \(N\), not \(\frac{N}{2}\).
- D. 120: Incorrect.
- E. 240: Incorrect.

B is the correct answer.

ANSWER 7: B

Problem 8:
To find Antonette's overall score, we need to calculate the total number of problems she got right out of 60.

1. On a 10-problem test, she got 70%, so she got \(0.7 \times 10 = 7\) problems right.
2. On a 20-problem test, she got 80%, so she got \(0.8 \times 20 = 16\) problems right.
3. On a 30-problem test, she got 90%, so she got \(0.9 \times 30 = 27\) problems right.

Total problems right = \(7 + 16 + 27 = 50\).

Total problems = \(10 + 20 + 30 = 60\).

Overall score = \(\frac{50}{60} \times 100\% \approx 83.33\%\).

Let's check the options:
- A. 40: Too low.
- B. 77: Too low.
- C. 80: Too low.
- D. 83: Closest to 83.33%.
- E. 87: Too high.

D is the correct answer.

ANSWER 8: D

Problem 9:
Mindy made three purchases: $1.98, $5.04, and $9.89. To find the total, we add these amounts:
\[ 1.98 + 5.04 + 9.89 = 16.91 \]

To the nearest dollar, $16.91 is $17.

Let's check the options:
- A. $10: Too low.
- B. $15: Too low.
- C. $16: Too low.
- D. $17: Correct.
- E. $18: Too high.

D is the correct answer.

ANSWER 9: D

Problem 10:
The maximum heart rate is found by subtracting the athlete's age from 220. For a 26-year-old athlete, the maximum heart rate is \(220 - 26 = 194\).

The target heart rate is 80% of this maximum: \(0.80 \times 194 = 155.2\).

To the nearest whole number, the target heart rate is 155.

L

Problem 1:
Let the total number of fish in the lake be \( T \). The ratio of trout to total fish in the sample is \( \frac{30}{180} = \frac{1}{6} \). This ratio is the same for the whole lake, so \( \frac{250}{T} = \frac{1}{6} \). Solving gives \( T = 250 \times 6 = 1500 \).
ANSWER 1: B

Problem 2:
Find the radius of each circle.
- Circle X: radius \( r_X = \pi \approx 3.14 \).
- Circle Y: circumference \( 2\pi r_Y = 8\pi \) ⇒ \( r_Y = 4 \).
- Circle Z: area \( \pi r_Z^2 = 9\pi \) ⇒ \( r_Z^2 = 9 \) ⇒ \( r_Z = 3 \).
Order from smallest to largest radius: Z (3), X (\(\pi\)), Y (4). This is Z, X, Y.
ANSWER 2: B

Problem 3:
Let the number be \( n \). \( 20\% \) of \( n \) is \( 12 \), so \( 0.2n = 12 \) ⇒ \( n = 60 \). Then \( 30\% \) of \( 60 \) is \( 0.3 \times 60 = 18 \).
ANSWER 3: B

Problem 4:
The sum of two primes is 85, which is odd. The only even prime is 2, so the primes must be 2 and \( 85 - 2 = 83 \). Their product is \( 2 \times 83 = 166 \).
ANSWER 4: E

Problem 5:
Wind chill = air temperature − 0.7 × wind speed = \( 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \). The closest choice is 23.
ANSWER 5: B

Problem 6:
\( 200,\!000 \times 200,\!000 = (2 \times 10^5)^2 = 4 \times 10^{10} = 40,\!000,\!000,\!000 \), which is forty billion.
ANSWER 6: E

Problem 7:
Connie multiplied by 2 and got 60, so the original number is \( 60 \div 2 = 30 \). The correct operation is dividing by 2, giving \( 30 \div 2 = 15 \).
ANSWER 7: B

Problem 8:
Number of correct answers:
- 70% of 10 = 7
- 80% of 20 = 16
- 90% of 30 = 27
Total correct = \( 7 + 16 + 27 = 50 \). Total problems = 60. Overall percentage = \( \frac{50}{60} \times 100\% \approx 83.33\% \), closest to 83%.
ANSWER 8: D

Problem 9:
Total = \( \$1.98 + \$5.04 + \$9.89 = \$16.91 \). To the nearest dollar, this is \( \$17 \).
ANSWER 9: D

Problem 10:
Maximum heart rate = \( 220 - 26 = 194 \) bpm. Target heart rate = \( 80\% \) of 194 = \( 0.8 \times 194 = 155.2 \). To the nearest whole number, this is 155.
ANSWER 10: B

Problem 1:
We're told the ratio of trout to total fish is the same in the sample and the lake.

**Sample ratio:** 30 trout out of 180 fish = 30/180 = 1/6

**Lake ratio:** 250 trout out of T total fish = 250/T

Setting them equal:
250/T = 1/6
T = 250 × 6 = 1500

Let me check the tempting wrong answers:
- A (1250) would come from 250 × 5, perhaps miscounting the ratio as 1/5.
- D (1800) might come from just multiplying sample size by 10.

ANSWER 1: B

---

Problem 2:
I need to find the radius of each circle.

**Circle X:** radius = π ≈ 3.14

**Circle Y:** Circumference = 2πr = 8π → r = 4

**Circle Z:** Area = πr² = 9π → r² = 9 → r = 3

Ordering from smallest to largest radius:
Z (r = 3) < X (r ≈ 3.14) < Y (r = 4)

So the order is Z, X, Y.

ANSWER 2: E

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Problem 3:
Let the number be x.

20% of x = 12
0.2x = 12
x = 12 / 0.2 = 60

Now find 30% of 60:
0.3 × 60 = 18

A quick shortcut: 30% is 1.5 times 20%, so the answer is 1.5 × 12 = 18.

Wrong choice check:
- A (15) would be 25% of 60.
- D (24) would be 40% of 60.

ANSWER 3: B

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Problem 4:
Two primes sum to 85. Since 85 is odd, and the sum of two odd numbers is even, one of the primes must be even. The only even prime is 2.

So the two primes are 2 and (85 − 2) = 83.

Check that 83 is prime: √83 ≈ 9.1, so I test divisibility by 2, 3, 5, 7.
- 83 is odd (not ÷2)
- 8+3 = 11 (not ÷3)
- Doesn't end in 0 or 5 (not ÷5)
- 83/7 ≈ 11.86 (not ÷7)
So 83 is prime. ✓

Product = 2 × 83 = 166

Wrong choice check:
- B (91) = 7 × 13, not a product involving 2.
- D (133) = 7 × 19.

ANSWER 4: E

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Problem 5:
Using the formula: Wind Chill = Air Temperature − 0.7 × Wind Speed

Wind Chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest value among the choices is 23.

Wrong choice check:
- A (18) would result from subtracting the full wind speed (36 − 18).
- C (28) might come from using 0.5 instead of 0.7.

ANSWER 5: B

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Problem 6:
200,000 × 200,000

Write in scientific notation:
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰

4 × 10¹⁰ = 40,000,000,000 = forty billion

Wrong choice check:
- A (400,000) is just adding them.
- B (4,000,000) would be 2,000 × 2,000.
- D (400,000,000) would be 20,000 × 20,000.

ANSWER 6: E

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Problem 7:
Connie multiplied a number by 2 and got 60:
number × 2 = 60
number = 30

She should have divided by 2:
30 ÷ 2 = 15

Wrong choice check:
- C (30) is the original number, not the correct answer.
- D (120) would be multiplying 60 by 2.

ANSWER 7: B

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Problem 8:
Find the number of correct answers on each test:
- Test 1: 70% of 10 = 7 correct
- Test 2: 80% of 20 = 16 correct
- Test 3: 90% of 30 = 27 correct

Total correct = 7 + 16 + 27 = 50
Total problems = 10 + 20 + 30 = 60

Overall percentage = 50/60 = 5/6 ≈ 83.33%

The closest choice is 83.

Wrong choice check:
- C (80) is the simple average of 70, 80, 90 — but the tests have different numbers of problems, so a weighted average is needed.

ANSWER 8: D

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Problem 9:
Add the three amounts, rounding to the nearest dollar:
$1.98 + $5.04 + $9.89

Exact sum:
1.98 + 5.04 = 7.02
7.02 + 9.89 = 16.91

Rounded to the nearest dollar: $17.

Alternatively, estimate by rounding first:
$2 + $5 + $10 = $17 (and the rounding errors nearly cancel: −0.02 + 0.04 − 0.11 = −0.09).

ANSWER 9: D

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Problem 10:
Step 1: Find the maximum heart rate.
Max HR = 220 − age = 220 − 26 = 194

Step 2: Find 80% of the maximum.
Target HR = 0.80 × 194 = 155.2

Rounded to the nearest whole number: 155.

Wrong choice check:
- C (176) would be about 90% of 194.
- D (194) is the maximum heart rate itself, forgetting to take 80%.

ANSWER 10: B

**Problem 1:**  
We are told the lake has 250 trout and that the ratio of trout to total fish is the same in the lake as in a sample of 180 fish where 30 are trout.  
Set up a proportion:
\[
\frac{\text{trout in sample}}{\text{total in sample}}=\frac{\text{trout in lake}}{\text{total in lake}}
\quad\Longrightarrow\quad
\frac{30}{180}=\frac{250}{T}
\]
Simplify the left side: \(\frac{30}{180}=\frac16\).  
So \(\frac{250}{T}=\frac16\), giving \(T=250\times 6=1500\).  

*Wrong choices:* 1250 would make the lake ratio \(\frac15\), not \(\frac16\); 1800 gives a different ratio; the others do not satisfy the proportion.  

ANSWER 1: B  

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**Problem 2:**  
We need the radii of circles X, Y, and Z.  

- **Circle X:** radius is given directly as \(\pi\approx 3.14\).  
- **Circle Y:** circumference \(C=2\pi r = 8\pi \Rightarrow r = 4\).  
- **Circle Z:** area \(A=\pi r^2 = 9\pi \Rightarrow r^2=9 \Rightarrow r=3\).  

Ordering from smallest to largest radius: \(Z(3),\; X(\pi\approx 3.14),\; Y(4)\).  

*Wrong choices:* A places X before Y but ignores Z; C and D swap Y and X; E puts X first.  

ANSWER 2: B  

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**Problem 3:**  
Let the number be \(N\).  
\(20\%\) of \(N\) is 12, so \(0.20N=12\).  
Divide: \(N=\frac{12}{0.20}=60\).  
Now find \(30\%\) of \(N\): \(0.30\times 60 = 18\).  

*Shortcut:* \(30\%\) is \(1.5\) times \(20\%\), so \(12\times 1.5=18\).  

*Wrong choices:* 15 is too low; 20, 24, 30 are too high.  

ANSWER 3: B  

---

**Problem 4:**  
The sum of the two primes is 85, which is odd.  
The sum of two integers is odd only when one is even and one is odd. The only even prime is 2.  
Thus one prime is 2, and the other is \(85-2=83\).  
Check that 83 is prime: it is not divisible by 2, 3, 5, or 7, so it is prime.  
The product is \(2\times 83 = 166\).  

*Wrong choices:* 85, 91, 115, 133 are products of non-prime or incorrect pairs (e.g., \(7\times13=91\), but \(7+13\neq85\)).  

ANSWER 4: E  

---

**Problem 5:**  
Use the formula \(\text{wind chill} = \text{air temperature} - 0.7\times(\text{wind speed})\).  
With air temperature \(36^\circ\text{F}\) and wind speed \(18\) mph:
\[
\text{wind chill} = 36 - 0.7(18) = 36 - 12.6 = 23.4
\]
The value closest to \(23.4\) is \(23\).  

*Wrong choices:* 18 is too low; 28, 32, 35 are too high.  

ANSWER 5: B  

---

**Problem 6:**  
Compute \(200\,000 \times 200\,000\).  
\(200\,000 = 2\times 10^5\).  
\[
(2\times 10^5)^2 = 4\times 10^{10} = 40\,000\,000\,000
\]
That is forty billion.  

*Wrong choices:* four hundred thousand, four million, forty thousand, and four hundred million are all too small.  

ANSWER 6: E  

---

**Problem 7:**  
Connie multiplied a number by 2 and got 60, so the original number is \(\frac{60}{2}=30\).  
The correct operation was to divide the number by 2:  
\[
\frac{30}{2}=15
\]  

*Wrong choices:* 7.5 would come from dividing 60 by 8; 30 is the original number; 120 and 240 come from further multiplication.  

ANSWER 7: B  

---

**Problem 8:**  
Find the total number of correct answers on the combined 60-problem test.  
- 70% of 10 problems: \(0.70\times 10 = 7\) correct.  
- 80% of 20 problems: \(0.80\times 20 = 16\) correct.  
- 90% of 30 problems: \(0.90\times 30 = 27\) correct.  

Total correct: \(7+16+27 = 50\).  
Overall percent: \(\frac{50}{60}\times 100\% \approx 83.33\%\).  
The closest listed percent is 83.  

*Wrong choices:* 40 is far off; 77 is too low; 80 is close but not as close as 83; 87 is too high.  

ANSWER 8: D  

---

**Problem 9:**  
Add the three amounts:
\[
\$1.98 + \$5.04 + \$9.89 = \$16.91
\]
Rounding \$16.91 to the nearest dollar gives \$17.  

*Wrong choices:* \$16 rounds down too far; \$18 rounds up too far; \$10 and \$15 are too low.  

ANSWER 9: D  

---

**Problem 10:**  
Maximum heart rate \(= 220 - \text{age} = 220 - 26 = 194\) bpm.  
Target heart rate \(= 80\%\) of maximum:
\[
0.80 \times 194 = 155.2
\]
To the nearest whole number this is 155.  

*Wrong choices:* 134 is too low; 176 is 80% of 220 (ignoring age); 194 is the maximum, not the target; 243 is too high.  

ANSWER 10: B

Problem 1:
We need to find the total number of fish in the lake, given that the ratio of trout to total fish is the same in both the sample and the lake.

**Set up the ratio from the sample:**
- Trout in sample = 30
- Total fish in sample = 180
- Ratio = 30/180 = 1/6

**Apply this ratio to the lake:**
- Trout in lake = 250
- Total fish in lake = ?
- 250 / Total = 1/6
- Total = 250 × 6 = 1500

**Ruling out wrong choices:** 1250 would give ratio 250/1250 = 1/5 (too high); 1750 would give 250/1750 = 1/7 (too low). Only 1500 gives the correct 1/6 ratio.

ANSWER 1: B

---

Problem 2:
We need to find the radius of each circle and list them from smallest to largest.

**Circle X:** Radius is given directly as π ≈ 3.14

**Circle Y:** Circumference = 8π, so 2πr = 8π → r = 4

**Circle Z:** Area = 9π, so πr² = 9π → r² = 9 → r = 3

**Order from smallest to largest radius:** Z (3), X (π ≈ 3.14), Y (4)

ANSWER 2: B

---

Problem 3:
We need to find 30% of a number, given that 20% of it is 12.

**Find the number:** 0.20 × n = 12 → n = 12/0.20 = 60

**Find 30%:** 0.30 × 60 = 18

**Quick check:** Since 30% is 1.5 times 20%, the answer should be 1.5 × 12 = 18. ✓

ANSWER 3: B

---

Problem 4:
The sum of two prime numbers is 85. We need their product.

**Key observation:** 85 is odd. The sum of two primes is odd only when one prime is even and the other is odd. The only even prime is 2.

**Find the primes:** 2 + 83 = 85 ✓ (83 is prime)

**Find the product:** 2 × 83 = 166

**Ruling out wrong choices:** 85 = 5 × 17 (neither pair sums to 85); 91 = 7 × 13 (sum is 20); 115 = 5 × 23 (sum is 28); 133 = 7 × 19 (sum is 26). None work.

ANSWER 4: E

---

Problem 5:
We use the formula: Wind chill = Air temperature − 0.7 × Wind speed

**Substitute the values:**
- Air temperature = 36°F
- Wind speed = 18 mph

Wind chill = 36 − 0.7 × 18 = 36 − 12.6 = 23.4

**Closest answer:** 23

ANSWER 5: B

---

Problem 6:
We need to compute 200,000 × 200,000.

**Calculate:** 200,000 = 2 × 10⁵

200,000 × 200,000 = (2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰ = 40,000,000,000 = 40 billion

**Ruling out wrong choices:** Four hundred thousand is 4 × 10⁵ (way too small); four million is 4 × 10⁶; forty thousand is 4 × 10⁴; four hundred million is 4 × 10⁸. Only forty billion = 4 × 10¹⁰ is correct.

ANSWER 6: E

---

Problem 7:
Connie multiplied a number by 2 and got 60, but she should have divided by 2.

**Find the original number:** number × 2 = 60 → number = 30

**Find the correct answer:** 30 ÷ 2 = 15

ANSWER 7: B

---

Problem 8:
We need to find the overall percentage across all three tests combined.

**Calculate correct answers on each test:**
- 10-problem test: 70% of 10 = 7 correct
- 20-problem test: 80% of 20 = 16 correct
- 30-problem test: 90% of 30 = 27 correct

**Total correct:** 7 + 16 + 27 = 50

**Total problems:** 10 + 20 + 30 = 60

**Overall percentage:** 50/60 = 0.8333... = 83.33...%

**Closest to:** 83

ANSWER 8: D

---

Problem 9:
We need the total of three purchases rounded to the nearest dollar.

**Add the amounts:**
$1.98 + $5.04 = $7.02
$7.02 + $9.89 = $16.91

**Round to nearest dollar:** $16.91 rounds to $17

ANSWER 9: D

---

Problem 10:
We need the target heart rate for a 26-year-old athlete.

**Find maximum heart rate:** 220 − 26 = 194 bpm

**Find target heart rate (80% of maximum):** 0.80 × 194 = 155.2

**Round to nearest whole number:** 155

ANSWER 10: B

**Problem 1:**  
Let the total number of fish in the lake be \(T\). The ratio of trout to total fish in the sample is  
\[
\frac{30}{180}=\frac16.
\]  
Given that this ratio equals the lake ratio,  
\[
\frac{250}{T}=\frac16 \;\Rightarrow\; T=250\times6=1500.
\]  
The lake contains **1500 fish**.

**Answer 1:** B  

---

**Problem 2:**  

- Circle X radius: \(r_X=\pi\).  
- Circle Y circumference \(=8\pi\) → \(2\pi r_Y = 8\pi\) → \(r_Y=4\).  
- Circle Z area \(=9\pi\) → \(\pi r_Z^{2}=9\pi\) → \(r_Z=3\).

Radii order (smallest to largest): \(3\;(Z),\; \pi\approx3.14\;(X),\;4\;(Y)\).  
Thus the correct ordering is **Z, X, Y**.

**Answer 2:** B  

---

**Problem 3:**  
If \(20\%\) of a number equals 12, the number is  
\[
\frac{12}{0.20}=60.
\]  
\(30\%\) of this number is  
\[
0.30\times60=18.
\]

**Answer 3:** B  

---

**Problem 4:**  
Let the primes be \(p\) and \(q\) with \(p+q=85\).  
Since 85 is odd, one prime must be 2 (the only even prime).  
Then \(q=85-2=83\), which is prime.  

Product: \(2\times83=166\).

**Answer 4:** E  

---

**Problem 5:**  
Wind‑chill formula: \(\text{WC}= \text{air temp} - 0.7\times\text{wind speed}\).  
\[
\text{WC}=36-0.7(18)=36-12.6=23.4\approx23.
\]

**Answer 5:** B  

---

**Problem 6:**  
\[
200{,}000\times200{,}000 = (2\times10^{5})(2\times10^{5})=4\times10^{10}=40{,}000{,}000{,}000,
\]  
which is **forty billion**.

**Answer 6:** E  

---

**Problem 7:**  
Multiplying the unknown number by 2 gives 60, so the number is  
\[
\frac{60}{2}=30.
\]  
Dividing it by 2 (the correct operation) yields  
\[
\frac{30}{2}=15.
\]

**Answer 7:** B  

---

**Problem 8:**  
- 70 % of 10 = 7 correct.  
- 80 % of 20 = 16 correct.  
- 90 % of 30 = 27 correct.  

Total correct: \(7+16+27=50\) out of 60 problems.  
\[
\frac{50}{60}=0.833\ldots\approx83\%.
\]

**Answer 8:** D  

---

**Problem 9:**  
\[
1.98+5.04+9.89 = 16.91.
\]  
Rounded to the nearest dollar = **$17**.

**Answer 9:** D  

---

**Problem 10:**  
Maximum heart rate = \(220-26 = 194\) bpm.  
Target heart rate = \(0.80 \times 194 = 155.2 \approx 155\) bpm.

**Answer 10:** B  

---

**Summary of answers**

ANSWER 1: B  
ANSWER 2: B  
ANSWER 3: B  
ANSWER 4: E  
ANSWER 5: B  
ANSWER 6: E  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: B

Problem 1:  
The lake contains 250 trout, and a sample of 180 fish includes 30 trout. Assuming the ratio of trout to total fish is the same in the sample and the lake, let \( T \) be the total number of fish in the lake. The ratio of trout to total fish is \( \frac{250}{T} \) in the lake and \( \frac{30}{180} = \frac{1}{6} \) in the sample. Setting these equal:  
\[
\frac{250}{T} = \frac{1}{6} \implies T = 250 \times 6 = 1500.
\]  
The total number of fish is 1500.  
ANSWER 1: B  

Problem 2:  
- Circle X has radius \( \pi \).  
- Circle Y has circumference \( 8\pi \), so radius \( r_Y = \frac{8\pi}{2\pi} = 4 \).  
- Circle Z has area \( 9\pi \), so radius \( r_Z = \sqrt{\frac{9\pi}{\pi}} = 3 \).  
Ordering radii: \( \pi \approx 3.14 \), 3, 4 → X (3.14), Z (3), Y (4) → corrected to Z (3), X (3.14), Y (4). Thus, Z, X, Y.  
ANSWER 2: B  

Problem 3:  
Let the number be \( x \). If 20% of \( x \) is 12:  
\[
0.2x = 12 \implies x = \frac{12}{0.2} = 60.
\]  
30% of 60 is \( 0.3 \times 60 = 18 \).  
ANSWER 3: B  

Problem 4:  
The sum of two primes is 85 (odd). One prime must be even (2), so the other is \( 85 - 2 = 83 \). The product is \( 2 \times 83 = 166 \).  
ANSWER 4: E  

Problem 5:  
Wind chill formula: \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \).  
Given temperature = 36°F, wind speed = 18 mph:  
\[
\text{wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 \approx 23.
\]  
ANSWER 5: B  

Problem 6:  
\( 200,000 \times 200,000 = 2 \times 10^5 \times 2 \times 10^5 = 4 \times 10^{10} = 40,000,000,000 \) (forty billion).  
ANSWER 6: E  

Problem 7:  
Connie multiplied by 2 and got 60, so the number is \( \frac{60}{2} = 30 \). The correct answer (dividing by 2) is \( \frac{30}{2} = 15 \).  
ANSWER 7: B  

Problem 8:  
Total problems: 60.  
- 10-problem test: \( 70\% \times 10 = 7 \) correct.  
- 20-problem test: \( 80\% \times 20 = 16 \) correct.  
- 30-problem test: \( 90\% \times 30 = 27 \) correct.  
Total correct: \( 7 + 16 + 27 = 50 \).  
Overall percentage: \( \frac{50}{60} \times 100 \approx 83.33\% \approx 83\% \).  
ANSWER 8: D  

Problem 9:  
Purchases: $1.98 ≈ $2, $5.04 ≈ $5, $9.89 ≈ $10. Total ≈ \( 2 + 5 + 10 = 17 \).  
ANSWER 9: D  

Problem 10:  
Maximum heart rate: \( 220 - 26 = 194 \). Target heart rate (80%): \( 0.8 \times 194 = 155.2 \approx 155 \).  
ANSWER 10: B

### Problem 1:
What is asked: We have 250 total trout in a lake. A sample of 180 fish has 30 trout, with the same ratio of trout to total fish in the sample and the whole lake. We need to find the total number of fish in the lake.
Setup: Set up a proportion, since the ratio of trout to total fish is equal. Let \( T \) = total fish in the lake. The sample ratio is \( \frac{30}{180} \), and the lake ratio is \( \frac{250}{T} \).
Calculation: Simplify the sample ratio: \( \frac{30}{180} = \frac{1}{6} \). Set equal to the lake ratio: \( \frac{1}{6} = \frac{250}{T} \). Cross-multiply to solve: \( T = 250 \times 6 = 1500 \).
Tempting wrong choices: If you flip the proportion, you might incorrectly calculate a different total; 1500 matches the correct ratio.
ANSWER 1: B

---

### Problem 2:
What is asked: Calculate the radius of three circles, then order them from smallest to largest radius.
Setup: Calculate the radius of each circle using circle formulas. Circle X has a given radius. For Circle Y, use circumference \( C=2\pi r \). For Circle Z, use area \( A=\pi r^2 \).
Calculation:
- Circle X radius: \( \pi \approx 3.14 \)
- Circle Y: \( 8\pi = 2\pi r \rightarrow r=4 \)
- Circle Z: \( 9\pi = \pi r^2 \rightarrow r^2=9 \rightarrow r=3 \)
Radii ordered smallest to largest: \( Z(3), X(\approx3.14), Y(4) \)
Tempting wrong choices: Mixing up circumference and area formulas leads to incorrect radius values.
ANSWER 2: B

---

### Problem 3:
What is asked: If 20% of a number is 12, find 30% of the same number.
Setup: Let \( N \) = the unknown number. First solve for \( N \), then calculate 30% of \( N \).
Calculation: \( 0.2N = 12 \rightarrow N = 12 / 0.2 = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).
Tempting wrong choices: Miscalculating 10% of the number as 3 instead of 6 leads to wrong answers.
ANSWER 3: B

---

### Problem 4:
What is asked: The sum of two primes is 85, find their product.
Setup: The only even prime is 2; all other primes are odd. The sum of two numbers is odd (85 is odd) only if one number is even and one is odd. The only possible even prime is 2.
Calculation: The second prime is \( 85 - 2 = 83 \), which is prime. The product is \( 2 \times 83 = 166 \).
Tempting wrong choices: Forgetting the only even prime is 2 leads to trying to add two odd primes, which cannot sum to an odd number.
ANSWER 4: E

---

### Problem 5:
What is asked: Calculate wind chill using the formula \( \text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed} \), for 36°F air temperature and 18 mph wind.
Setup: Plug the values into the given formula.
Calculation: \( 0.7 \times 18 = 12.6 \). Wind chill = \( 36 - 12.6 = 23.4 \), which is closest to 23.
Tempting wrong choices: Miscalculating the product \( 0.7 \times 18 \) leads to incorrect wind chill values.
ANSWER 5: B

---

### Problem 6:
What is asked: Calculate the product of two hundred thousand times two hundred thousand.
Setup: Write two hundred thousand in scientific notation to simplify multiplication: 200,000 = \( 2 \times 10^5 \).
Calculation: \( (2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10} = 40,000,000,000 \), which is forty billion.
Tempting wrong choices: miscounting the number of zeros leads to smaller incorrect totals.
ANSWER 6: E

---

### Problem 7:
What is asked: Connie multiplied a number by 2 and got 60, but should have divided the original number by 2. Find the correct answer.
Setup: First solve for the original number, then calculate the correct result by dividing by 2.
Calculation: Original number = \( 60 / 2 = 30 \). Correct answer = \( 30 / 2 = 15 \).
Tempting wrong choices: Mistaking 60 as the original number leads to wrong results.
ANSWER 7: B

---

### Problem 8:
What is asked: Combine three test scores (70% on 10 problems, 80% on 20, 90% on 30) to find the overall percentage on the combined 60-problem test.
Setup: Calculate total correct answers across all tests, then divide by total problems to get the overall percentage.
Calculation: Total correct = (0.7*10) + (0.8*20) + (0.9*30) = 7 + 16 + 27 = 50. Overall percentage = \( 50 / 60 \approx 83\% \).
Tempting wrong choices: Averaging the three percentages (80%) ignores the different weights of the longer tests, leading to an incorrect lower score.
ANSWER 8: D

---

### Problem 9:
What is asked: Find the total of three purchases ($1.98, $5.04, $9.89) rounded to the nearest dollar.
Setup: Add the purchase amounts, then round to the nearest whole dollar.
Calculation: Total = \( 1.98 + 5.04 + 9.89 = 16.91 \), which rounds to $17.
Tempting wrong choices: Adding incorrectly to get ~$16 leads to the wrong round.
ANSWER 9: D

---

### Problem 10:
What is asked: Calculate an athlete's target heart rate: 80% of a maximum heart rate of 220 minus their age, for a 26-year-old athlete.
Setup: First calculate the maximum heart rate, then take 80% of that value to get the target.
Calculation: Max heart rate = \( 220 - 26 = 194 \). Target heart rate = \( 0.8 * 194 = 155.2 \), which rounds to 155.
Tempting wrong choices: Calculating 80% of 220 directly leads to the incorrect value of 176.
ANSWER 10: B

### Problem 1:
We are told the ratio of trout to total fish is the same in the sample and the entire lake.  
First, calculate the trout ratio in the sample:  
Trout in sample = 30, total sample size = 180, so ratio = $\frac{30}{180} = \frac{1}{6}$.  

Let $T$ = total number of fish in the lake. The lake has 250 trout, so:  
$\frac{250}{T} = \frac{1}{6}$  
Cross-multiplying gives $T = 250 \times 6 = 1500$.  

ANSWER 1: B

---

### Problem 2:
We calculate the radius of each circle using standard formulas:
- Circle X: Radius is given directly as $r_X = \pi \approx 3.14$.
- Circle Y: Circumference $C = 8\pi$. Using $C = 2\pi r$, solve for $r_Y$:  
  $8\pi = 2\pi r_Y \implies r_Y = \frac{8\pi}{2\pi} = 4$.
- Circle Z: Area $A = 9\pi$. Using $A = \pi r^2$, solve for $r_Z$:  
  $9\pi = \pi r_Z^2 \implies r_Z^2 =9 \implies r_Z=3$ (radius is positive).

Order the radii from smallest to largest: $3 (Z) < \pi (X) <4 (Y)$, so Z, X, Y.

ANSWER 2: B

---

### Problem 3:
Let the unknown number be $N$. We know 20% of $N$ is 12:  
$0.2N = 12$.  

We need 30% of $N$, which is $0.3N$. Notice that $0.3N = 1.5 \times 0.2N$, so:  
$0.3N = 1.5 \times 12 = 18$.  

(Alternatively, solve for $N$: $N = \frac{12}{0.2}=60$, then $0.3 \times 60=18$.)

ANSWER 3: B

---

### Problem 4:
The sum of two primes is 85, which is odd. The sum of two primes is odd only if one prime is even (since odd + even = odd). The only even prime is 2.  

So one prime is 2, the other is $85 - 2 = 83$. Verify 83 is prime: it is not divisible by 2, 3, 5, 7, or 11 (since $11^2=121>83$), so 83 is prime.  

The product of the two primes is $2 \times 83 = 166$.

ANSWER 4: E

---

### Problem 5:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$.  
Substitute air temperature = 36°F, wind speed = 18 mph:  
$0.7 \times 18 = 12.6$, so wind chill = $36 - 12.6 = 23.4$. The closest value is 23.

ANSWER 5: B

---

### Problem 6:
Two hundred thousand is $200,000 = 2 \times 10^5$. Multiply two of these:  
$(2 \times 10^5) \times (2 \times 10^5) = 4 \times 10^{10}$.  

$10^{10} = 10,000,000,000$ (10 billion), so $4 \times 10^{10} = 40,000,000,000$ = forty billion.

ANSWER 6: E

---

### Problem 7:
Let the original number be $x$. Connie multiplied by 2 to get 60:  
$2x = 60 \implies x = 30$.  

She should have divided by 2, so the correct answer is $\frac{x}{2} = \frac{30}{2} =15$.

ANSWER 7: B

---

### Problem 8:
Calculate total correct answers across all 60 problems:
- 10-problem test: 70% correct → $0.7 \times 10 =7$ correct.
- 20-problem test: 80% correct → $0.8 \times 20=16$ correct.
- 30-problem test:90% correct → $0.9 \times30=27$ correct.

Total correct = $7+16+27=50$. Overall percentage = $\frac{50}{60} \times 100 \approx 83.33\%$, which is closest to 83.

ANSWER 8: D

---

### Problem 9:
Add the three purchase amounts:  
$1.98 + 5.04 =7.02$, then $7.02 +9.89=16.91$.  

Rounding 16.91 to the nearest dollar gives 17.

ANSWER 9: D

---

### Problem 10:
First calculate maximum heart rate: $220 - \text{age} = 220 -26=194$.  
Target heart rate is 80% of maximum: $0.8 \times 194=155.2$, which rounds to 155.

ANSWER 10: B