🔬 Model Lab

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E

Problem 1:
We are asked to add $1/10 + 9/100 + 9/1000 + 7/10000$.

Convert each fraction to a decimal:
$1/10 = 0.1$
$9/100 = 0.09$
$9/1000 = 0.009$
$7/10000 = 0.0007$

Add:
$0.1 + 0.09 = 0.19$
$0.19 + 0.009 = 0.199$
$0.199 + 0.0007 = 0.1997$

Check choices: 0.0026 is far too small, 0.0197 misplaces the decimal, 0.26 would be $1/10+9/100+...$ rounded incorrectly, 1.997 is about 10 times too big. The correct sum is 0.1997.
ANSWER 1: C

Problem 2:
We need final savings as a percent of the original after a 20% decrease then a 50% increase.

Let original be 100 dollars for ease.
July: decreased by 20%: keep $100\% - 20\% = 80\%$.
$100 \times 0.80 = 80$.

August: increased by 50% of the new amount: keep $100\% + 50\% = 150\%$ of July amount.
$80 \times 1.50 = 80 + 40 = 120$.

120 compared to original 100 is $120/100 = 120\%$.
So not 80% (forgetting the increase), not 90-110% (mis-adding percents like $-20+50=30$). Multiplication gives 120.
ANSWER 2: E

Problem 3:
We need total correct divided by total problems.

Test 1: 80% of 25: $0.80 \times 25 = 20$ correct.
Test 2: 90% of 40: $0.90 \times 40 = 36$ correct.
Test 3: 70% of 10: $0.70 \times 10 = 7$ correct.

Total correct $= 20+36+7 = 63$.
Total problems $= 25+40+10 = 75$.

Percent correct $= 63/75 \times 100\%$.
$63/75 = 21/25 = 84/100 = 84\%$.

We cannot just average 80,90,70 because tests have different numbers of problems. Simple average $(80+90+70)/3 = 80$ is wrong choice C.
ANSWER 3: D

Problem 4:
Compute $8 \times 0.25 \times 2 \times 0.125$.

$8 \times 0.25 = 2$ since a quarter of 8 is 2.
Then $2 \times 2 = 4$.
Then $4 \times 0.125 = 4 \times 1/8 = 1/2$ since $0.125 = 1/8$.

Alternatively with fractions: $8 \times 1/4 \times 2 \times 1/8 = (8/4)\times(2/8)=2\times1/4=1/2$.

So 1/8, 1/4, 1, 2 are incorrect.
ANSWER 4: C

Problem 5:
With $30.00, sandwiches $4.50 each, maximize sandwiches, then spend rest on $1.00 drinks. Count total items.

Max number of sandwiches: $30 \div 4.50 = 300 \div 45 = 6.66...$ so at most 6 whole sandwiches.
Cost of 6 sandwiches: $6 \times 4.50 = 27.00$.
Money left: $30.00-27.00 = 3.00$.
Drinks at $1.00 each: 3 drinks.

Total items $= 6+3 = 9$.

7 sandwiches would cost $31.50 > 30$, impossible. 6 items would be ignoring drinks. So 9 is correct.
ANSWER 5: D

Problem 6:
Estimate $(1.8)(40.3+0.07)$.

First add: $40.3+0.07 = 40.37$.
So product $= 1.8 \times 40.37$.

Approximate: $1.8 \approx 2$, $40.37 \approx 40$, so $\approx 2\times40 = 80$.
Exact: $40.37 \times 1.8 = 40.37 \times 18 \div 10 = 726.66 \div 10 = 72.666$.

Compare to choices: 7 is 10x too small, 42 is about half, 84 is about 11 away, 737 is 10x too big. 74 is only about 1.3 away from 72.7, so closest.
ANSWER 6: C

Problem 7:
Original price P, half-price sale, then extra 20% off sale price. What % off original?

Sale price $= 1/2 \times P = 0.5P$.
Coupon gives 20% discount, so you pay $100\%-20\% = 80\%$ of sale price:
pay $= 0.80 \times 0.5P = 0.40P$.

So you pay 40% of original, meaning you save $100\%-40\% = 60\%$ off.
10% would be just the coupon confusion, 40% is the amount paid not off, 50% forgets coupon.
ANSWER 7: D

Problem 8:
We interpret as $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}$.

Numerator $= 8! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 = 40320$.
Check: $1\cdot2\cdot3\cdot4\cdot5 =120$, $\times6=720$, $\times7=5040$, $\times8=40320$.

Denominator $= 1+2+3+4+5+6+7+8 = 36$. Indeed sum $1$ to $n$ is $n(n+1)/2 = 8\cdot9/2=36$.

Divide: $40320 \div 36 = 40320 \div 12 \div 3 = 3360 \div 3 = 1120$.

Other choices are arithmetic errors.
ANSWER 8: B

Problem 9:
Start $100, -15% first year, +20% second year. Net change?

After year 1: keep $85\%$: $100 \times 0.85 = 85$.
After year 2: gain 20% on 85: $85 \times 1.20 = 85 + 17 = 102$.

Net: from 100 to 102 is +2, i.e. 2% gain.
Simply doing $-15\%+20\%=+5\%$ is wrong because the 20% applies to the smaller 85, not the original 100. So E is tempting but wrong.
ANSWER 9: D

Problem 10:
Compute $(1+11+21+31+41)+(9+19+29+39+49)$.

Direct: first sum $=1+11=12$, $+21=33$, $+31=64$, $+41=105$.
Second sum $=9+19=28$, $+29=57$, $+39=96$, $+49=145$.
Total $=105+145=250$.

Clever pairing: match ones to make 50:
$1+49=50$, $11+39=50$, $21+29=50$, $31+19=50$, $41+9=50$. Five 50s $=250$.

150,199,200,249 are too low.
ANSWER 10: E

Problem 1:
We are asked to compute the sum of the four fractions:
$$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$$

We convert each fraction into its decimal representation:
* $\frac{1}{10} = 0.1$
* $\frac{9}{100} = 0.09$
* $\frac{9}{1000} = 0.009$
* $\frac{7}{10000} = 0.0007$

Adding these decimals together:
$$0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$$

Thus, the sum is $0.1997$.

ANSWER 1: C

---

Problem 2:
We are asked to find Brynn's final savings as a percentage of her original savings after a decrease of $20\%$ followed by an increase of $50\%$.

Let Brynn's original savings be $100$ dollars.
1. In July, her savings decreased by $20\%$:
   $$\text{July savings} = 100 - (0.20 \times 100) = 80$$
2. In August, her savings increased by $50\%$ of the new amount:
   $$\text{August savings} = 80 + (0.50 \times 80) = 80 + 40 = 120$$

Since she now has $120$ dollars compared to the original $100$ dollars, her current savings are $\frac{120}{100} \times 100\% = 120\%$ of the original amount.

ANSWER 2: E

---

Problem 3:
We need to find the overall percentage of questions Ryan answered correctly across three different tests.

First, determine the number of correct questions for each test:
* Test 1: $80\%$ of $25 = 0.80 \times 25 = 20$
* Test 2: $90\%$ of $40 = 0.90 \times 40 = 36$
* Test 3: $70\%$ of $10 = 0.70 \times 10 = 7$

Next, find the total number of correct answers:
$$20 + 36 + 7 = 63$$

Find the total number of problems:
$$25 + 40 + 10 = 75$$

Finally, calculate the overall percentage:
$$\frac{63}{75} = \frac{21}{25} = \frac{84}{100} = 84\%$$

ANSWER 3: D

---

Problem 4:
We are asked to evaluate the product:
$$8 \times 0.25 \times 2 \times 0.125$$

It is easiest to convert the decimals to fractions or rearrange using compatible numbers:
$$0.25 = \frac{1}{4} \quad \text{and} \quad 0.125 = \frac{1}{8}$$

Now group the terms:
$$(8 \times 0.125) \times (2 \times 0.25) = \left(8 \times \frac{1}{8}\right) \times \left(2 \times \frac{1}{4}\right) = 1 \times \frac{2}{4} = \frac{1}{2}$$

ANSWER 4: C

---

Problem 5:
Ike and Mike have $\$30.00$ to spend. They want to buy as many sandwiches as possible at $\$4.50$ each, and then use the remaining money on soft drinks at $\$1.00$ each.

First, calculate the maximum number of sandwiches they can buy:
$$30.00 \div 4.50 = \frac{30}{\frac{9}{2}} = \frac{60}{9} = 6\frac{2}{3}$$
So, the maximum number of whole sandwiches they can buy is $6$.

The cost for $6$ sandwiches is:
$$6 \times \$4.50 = \$27.00$$

The remaining money is:
$$\$30.00 - \$27.00 = \$3.00$$

With $\$3.00$, they can buy:
$$\$3.00 \div \$1.00 = 3 \text{ soft drinks}$$

Counting both sandwiches and soft drinks, the total number of items they will buy is:
$$6 + 3 = 9$$

ANSWER 5: D

---

Problem 6:
We want to find which choice is closest to the product:
$$(1.8)(40.3 + 0.07)$$

First, simplify the expression inside the parentheses:
$$40.3 + 0.07 = 40.37$$

Now, multiply by $1.8$:
$$1.8 \times 40.37 = 72.666$$

Looking at the choices:
* $|7 - 72.666| = 65.666$
* $|42 - 72.666| = 30.666$
* $|74 - 72.666| = 1.334$
* $|84 - 72.666| = 11.334$
* $|737 - 72.666| = 664.334$

The value $72.666$ is clearly closest to $74$.

ANSWER 6: C

---

Problem 7:
Let the original price of an item be $P$.
1. The "half price" sale reduces the price to:
   $$0.50P$$
2. The coupon gives an additional $20\%$ discount on the sale price:
   $$\text{Final Price} = 0.50P \times (1 - 0.20) = 0.50P \times 0.80 = 0.40P$$

The final price is $40\%$ of the original price. Therefore, the discount (percentage off the original price) is:
$$100\% - 40\% = 60\%$$

ANSWER 7: D

---

Problem 8:
We are asked to evaluate the fraction:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}$$

Calculate the numerator:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8! = 40320$$

Calculate the denominator using the formula for the sum of the first $n$ integers, $\frac{n(n+1)}{2}$:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \frac{8 \times 9}{2} = 36$$

Now, divide the numerator by the denominator:
$$\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \frac{(4 \cdot 3 \cdot 3) \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36} = \frac{36 \times (1 \cdot 2 \cdot 5 \cdot 2 \cdot 7 \cdot 8)}{36}$$
Alternatively, simply compute:
$$\frac{40320}{36} = 1120$$

ANSWER 8: B

---

Problem 9:
Tammy starts with an investment of $\$100$.
1. During the first year, her investment suffers a $15\%$ loss:
   $$\text{Value after Year 1} = 100 - (0.15 \times 100) = \$85$$
2. During the second year, the remaining investment shows a $20\%$ gain:
   $$\text{Value after Year 2} = 85 + (0.20 \times 85) = 85 + 17 = \$102$$

The net change over the two-year period is:
$$\$102 - \$100 = +\$2$$
As a percentage of the initial $\$100$, this is a $2\%$ gain.

ANSWER 9: D

---

Problem 10:
We need to calculate:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

We can pair each term from the first group with the corresponding term from the second group:
* $1 + 9 = 10$
* $11 + 19 = 30$
* $21 + 29 = 50$
* $31 + 39 = 70$
* $41 + 49 = 90$

Now, add these sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

ANSWER 10: E

# Problem 1:
We need to convert each fraction to a decimal and add:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Adding: 0.1 + 0.09 + 0.009 + 0.0007 = 0.1997

Check the tempting wrong answers: 0.0026 (choice A) would require all four terms to be tiny, but 1/10 = 0.1 alone is larger than that. 0.26 (choice D) mistakenly uses 6/100 instead of 7/10000 style place values. 1.997 (choice E) is about ten times too big.

ANSWER 1: C

# Problem 2:
Let the original savings be S.

**July (decrease by 20%):** Savings become S − 0.20S = 0.80S.

**August (increase by 50% of the new amount):** Savings become 0.80S + 0.50(0.80S) = 0.80S × 1.50 = 1.20S.

So the savings are 1.20S = **120%** of the original amount.

Common error to rule out: answering 80 + 50 = 130 or thinking the percent changes cancel out (choice C, 100) — but percentages apply to different base amounts, so they don't cancel.

ANSWER 2: E

# Problem 3:
Compute the number of correct problems on each test:
- 80% of 25 = 20 correct
- 90% of 40 = 36 correct
- 70% of 10 = 7 correct

Total correct = 20 + 36 + 7 = 63.
Total problems = 25 + 40 + 10 = 75.

Percent correct = 63/75 = 0.84 = **84%**.

Rule out the tempting shortcut: averaging the percentages, (80 + 90 + 70)/3 = 80 (choice C), is wrong because the tests have different numbers of problems. The correct weighted answer is 84.

ANSWER 3: D

# Problem 4:
Convert the decimals to fractions and multiply step by step:
- 8 × 0.25 = 8 × 1/4 = 2
- 2 × 2 = 4
- 4 × 0.125 = 4 × 1/8 = 4/8 = 1/2

So the product equals **1/2**.

Rule out D: if you mistakenly compute 4 × 0.125 as 4 × 1/4 you'd get 1, but 0.125 = 1/8, not 1/4.

ANSWER 4: C

# Problem 5:
Sandwiches cost $4.50. With $30.00:
$30.00 ÷ $4.50 = 6.67, so they can buy at most **6 sandwiches**.

Cost: 6 × $4.50 = $27.00. Remaining money: $30.00 − $27.00 = $3.00.
Soft drinks at $1.00 each: they buy **3 soft drinks**.

Total items = 6 + 3 = **9**.

Rule out E (10): buying 7 sandwiches would cost $31.50, which exceeds $30. Rule out A/B/C: these don't account for spending the leftover money on drinks.

ANSWER 5: D

# Problem 6:
First compute inside the parentheses:
40.3 + 0.07 = 40.37

Then multiply:
1.8 × 40.37 ≈ 1.8 × 40 = 72, and more precisely 1.8 × 40.37 = 72.666

Compare to the choices: 72.666 is about 1.3 away from 74, but about 30 away from 42 and very far from 7 or 737. So the closest value is **74**.

ANSWER 6: C

# Problem 7:
Let the original price be P.

**Half price:** sale price = 0.50P.
**Coupon takes 20% off the sale price:** final price = 0.50P × 0.80 = 0.40P.

So the customer pays 40% of the original price, meaning the discount is 100% − 40% = **60% off**.

Rule out common errors: adding 50% + 20% = 70% (choice E) is wrong because the 20% applies to the reduced price, not the original. Getting 40 (choice C) is the amount *paid*, not the amount *off*.

ANSWER 7: D

# Problem 8:
Compute the numerator and denominator separately.

**Denominator (sum):** 1 + 2 + ... + 8 = (8 × 9)/2 = 36

**Numerator (product):** 8! = 40320

Divide: 40320 ÷ 36 = **1120**.

Quick check of other choices: 1020 would come from an arithmetic slip; 2240 and 3360 are 2× and 3× the correct value.

ANSWER 8: B

# Problem 9:
Start: $100.

**Year 1 (15% loss):** $100 × 0.85 = $85.

**Year 2 (20% gain on $85):** $85 × 1.20 = $102.

Final amount: $102, which is a change of +$2 on $100, a **2% gain**.

Rule out A/E: the changes don't simply cancel to 5% loss or gain, since the 20% gain is applied to the smaller $85 base, not $100.

ANSWER 9: D

# Problem 10:
Pair each term from the first group with a term from the second group — they conveniently sum to round numbers:
- 1 + 9 = 10
- 11 + 19 = 30
- 21 + 29 = 50
- 31 + 39 = 70
- 41 + 49 = 90

Total = 10 + 30 + 50 + 70 + 90 = 250.

(Verification by direct sums: 1+11+21+31+41 = 105 and 9+19+29+39+49 = 145; 105 + 145 = 250.) ✓

Rule out C (200): that would be the answer if the pairs summed to 40 each, but they increase by 20 each time.

ANSWER 10: E

Problem 1:
The expression is the sum of fractions:
\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}
\]
Convert each term to ten-thousandths:
\[
\frac{1}{10}=\frac{1000}{10000},\quad
\frac{9}{100}=\frac{900}{10000},\quad
\frac{9}{1000}=\frac{90}{10000},\quad
\frac{7}{10000}=\frac{7}{10000}
\]
Now add:
\[
\frac{1000+900+90+7}{10000}=\frac{1997}{10000}=0.1997
\]
The tempting choice 0.26 would come from misplacing the decimal places, but the exact sum is 0.1997.

ANSWER 1: C

Problem 2:
Let the original savings be \(S\).

A 20% decrease in July leaves:
\[
S-0.20S=0.80S
\]

In August, the savings increase by 50% of the new amount:
\[
0.80S+0.50(0.80S)=0.80S+0.40S=1.20S
\]

So the final amount is \(120\%\) of the original.

ANSWER 2: E

Problem 3:
Find the number of correct answers on each test.

First test:
\[
25(0.80)=20
\]

Second test:
\[
40(0.90)=36
\]

Third test:
\[
10(0.70)=7
\]

Total correct answers:
\[
20+36+7=63
\]

Total number of problems:
\[
25+40+10=75
\]

Percent correct:
\[
\frac{63}{75}\times 100=84\%
\]

A tempting but wrong approach is to average the three percentages, but the tests have different numbers of problems.

ANSWER 3: D

Problem 4:
Rewrite the decimals as fractions:
\[
.25=\frac14,\qquad .125=\frac18
\]

So the product is:
\[
8\cdot \frac14\cdot 2\cdot \frac18
\]

Compute step by step:
\[
8\cdot \frac14=2
\]
\[
2\cdot 2=4
\]
\[
4\cdot \frac18=\frac48=\frac12
\]

Thus the product is \(\frac12\).

ANSWER 4: C

Problem 5:
Ike and Mike have $30.00.

Sandwiches cost $4.50 each. The maximum number of sandwiches they can buy is:
\[
\left\lfloor \frac{30}{4.50} \right\rfloor = 6
\]

Cost of 6 sandwiches:
\[
6(4.50)=27.00
\]

Remaining money:
\[
30.00-27.00=3.00
\]

Soft drinks cost $1.00 each, so they can buy:
\[
3
\]
soft drinks.

Total items bought:
\[
6+3=9
\]

ANSWER 5: D

Problem 6:
First add inside the parentheses:
\[
40.3+.07=40.37
\]

Now multiply:
\[
(1.8)(40.37)
\]

Compute:
\[
1.8(40.37)=40.37(2-0.2)=80.74-8.074=72.666
\]

Compare with the choices:
- 7 is far away
- 42 is far away
- 74 is close
- 84 is farther away
- 737 is extremely far away

The closest choice is 74.

ANSWER 6: C

Problem 7:
Let the original price be \(P\).

“Half price” means the sale price is:
\[
0.50P
\]

The coupon gives an additional 20% discount on the sale price, so they pay 80% of the sale price:
\[
0.80(0.50P)=0.40P
\]

So the final price is 40% of the original price. That means the total discount is:
\[
100\%-40\%=60\%
\]

A common mistake is to add 50% and 20% to get 70%, but the discounts are applied to different amounts.

ANSWER 7: D

Problem 8:
The expression is the product of the numbers 1 through 8 divided by their sum:
\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{1+2+3+4+5+6+7+8}
\]

The numerator is:
\[
8! = 40320
\]

The denominator is the sum from 1 to 8:
\[
1+2+\cdots+8=\frac{8(9)}{2}=36
\]

Now divide:
\[
\frac{40320}{36}=1120
\]

ANSWER 8: B

Problem 9:
Start with $100.

After a 15% loss:
\[
100-0.15(100)=100-15=85
\]

Then a 20% gain on the remaining $85:
\[
85+0.20(85)=85+17=102
\]

The investment went from $100 to $102, a $2 increase. As a percent of the original $100:
\[
\frac{2}{100}\times 100=2\%
\]

So the two-year change is a 2% gain.

ANSWER 9: D

Problem 10:
Add each group separately.

First group:
\[
1+11+21+31+41=105
\]

Second group:
\[
9+19+29+39+49=145
\]

Total:
\[
105+145=250
\]

Alternatively, pair terms:
\[
(1+9)+(11+19)+(21+29)+(31+39)+(41+49)
\]
\[
=10+30+50+70+90=250
\]

ANSWER 10: E

Problem 1:
We need 1/10 + 9/100 + 9/1000 + 7/10000.

Convert each to a decimal:
- 1/10 = 0.1
- 9/100 = 0.09
- 9/1000 = 0.009
- 7/10000 = 0.0007

Add: 0.1 + 0.09 = 0.19; 0.19 + 0.009 = 0.199; 0.199 + 0.0007 = 0.1997.

(A tempting error is misplacing the decimal, giving 0.0197 or 1.997 — but the largest term is 0.1, so the sum must be between 0.1 and 0.2.)

ANSWER 1: C

Problem 2:
Let the original savings be 100.
- July: decreased 20% → 100 × 0.80 = 80.
- August: increased by 50% of the *new* amount → 80 × 1.50 = 120.

So savings are 120/100 = 120% of the original.

(Choice C, 100, would be right only if percentages "cancelled," but a 50% increase on the smaller base 80 adds 40, more than the 20 lost.)

ANSWER 2: D

Problem 3:
Count correct answers on each test:
- 80% of 25 = 20
- 90% of 40 = 36
- 70% of 10 = 7

Total correct = 20 + 36 + 7 = 63. Total problems = 25 + 40 + 10 = 75.

Percent = 63/75 = 0.84 = 84%.

(Averaging 80, 90, 70 gives 80 — choice C — but that ignores the different test sizes; the 40-problem test with 90% carries more weight.)

ANSWER 3: D

Problem 4:
8 × 0.25 × 2 × 0.125

Group conveniently: 8 × 0.25 = 2, and 2 × 0.125 = 0.25.
Then 2 × 0.25 = 0.5 = 1/2.

(Check another way: 0.25 = 1/4, 0.125 = 1/8, so 8 · (1/4) · 2 · (1/8) = 16/32 = 1/2.)

ANSWER 4: C

Problem 5:
They buy as many $4.50 sandwiches as possible from $30.00:
- 6 sandwiches cost 6 × 4.50 = $27.00
- 7 sandwiches would cost $31.50 > $30.00, too much.

So 6 sandwiches, leaving 30.00 − 27.00 = $3.00, which buys 3 soft drinks at $1.00 each.

Total items = 6 + 3 = 9.

(Choice A counts only sandwiches; choice E would require money they don't have.)

ANSWER 5: D

Problem 6:
First, 40.3 + 0.07 = 40.37.
Then 1.8 × 40.37 ≈ 1.8 × 40 = 72, more precisely 1.8 × 40.37 = 72.666.

Among the choices, 74 is closest (|74 − 72.67| ≈ 1.3, while |42 − 72.67| ≈ 30.7 and |84 − 72.67| ≈ 11.3).

ANSWER 6: C

Problem 7:
Let the original price be 100.
- Half price: 100 × 0.50 = 50.
- Coupon takes 20% off the sale price: 50 × 0.80 = 40.

Final price = 40, i.e., 40% of the original, so the discount is 100 − 40 = 60% off.

(Choice C, 40, is the *percent paid*, not the percent off; choice E, 70, comes from wrongly adding 50% + 20%.)

ANSWER 7: D

Problem 8:
The expression is (1·2·3·4·5·6·7·8) / (1+2+3+4+5+6+7+8).

Numerator: 8! = 40320.
Denominator: 1+2+…+8 = 8·9/2 = 36.

40320 / 36 = 1120.

ANSWER 8: B

Problem 9:
Start with $100.
- Year 1: 15% loss → 100 × 0.85 = 85.
- Year 2: 20% gain → 85 × 1.20 = 102.

Final value $102 vs. original $100, a 2% gain.

(Simply subtracting 20% − 15% = 5% gain, choice E, is wrong because the gain applies to the reduced amount $85.)

ANSWER 9: D

Problem 10:
First group: 1 + 11 + 21 + 31 + 41 = 105.
Second group: 9 + 19 + 29 + 39 + 49 = 145.

Total: 105 + 145 = 250.

(Nice check: pair terms as (1+9) + (11+19) + (21+29) + (31+39) + (41+49) = 10 + 30 + 50 + 70 + 90 = 250.)

ANSWER 10: E

Problem 1:

We are asked to evaluate

\[
\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{7}{10000}.
\]

Write each fraction as a decimal:

\[
\frac{1}{10}=0.1,\qquad \frac{9}{100}=0.09,\qquad \frac{9}{1000}=0.009,\qquad \frac{7}{10000}=0.0007.
\]

Now add:

\[
0.1+0.09+0.009+0.0007=0.1997.
\]

Choice C is correct. A and B have decimal places too far to the left, while D and E are too large.

ANSWER 1: C

Problem 2:

We are asked what percent of the original savings remains after a 20% decrease and then a 50% increase of the new amount.

Start with an original amount of \(100\) dollars.

After a 20% decrease:

\[
100-20=80.
\]

In August, the savings increase by 50% of \(80\):

\[
50\%\text{ of }80=40.
\]

So the new amount is

\[
80+40=120.
\]

Since the original amount was \(100\), the final amount is \(120\%\) of the original.

A common wrong thought is that \(-20\%+50\%=30\%\), but the 50% increase is based on the reduced amount, not the original amount.

ANSWER 2: E

Problem 3:

We need the percent correct over all three tests, so we must count the total number correct rather than average the three percentages.

On the 25-problem test, Ryan got

\[
80\%\text{ of }25=0.8(25)=20
\]

correct.

On the 40-problem test, he got

\[
90\%\text{ of }40=0.9(40)=36
\]

correct.

On the 10-problem test, he got

\[
70\%\text{ of }10=0.7(10)=7
\]

correct.

Thus, he got

\[
20+36+7=63
\]

problems correct out of

\[
25+40+10=75
\]

total problems.

Therefore, his overall percentage is

\[
\frac{63}{75}\times100\%=84\%.
\]

Simply averaging \(80\%,90\%,70\%\) would be incorrect because the tests have different numbers of problems.

ANSWER 3: D

Problem 4:

We are asked to compute

\[
8\times 0.25\times 2\times 0.125.
\]

Use fractions:

\[
0.25=\frac14,\qquad 0.125=\frac18.
\]

So

\[
8\cdot \frac14\cdot 2\cdot \frac18.
\]

Pair \(8\) with \(\frac18\):

\[
8\cdot \frac18=1.
\]

Then the product becomes

\[
1\cdot \frac14\cdot 2=\frac24=\frac12.
\]

ANSWER 4: C

Problem 5:

Ike and Mike have \(\$30.00\). Sandwiches cost \(\$4.50\) each, and they want as many sandwiches as possible.

Find the greatest whole number of sandwiches they can buy:

\[
30\div 4.50=6\frac{2}{3}.
\]

So they can buy \(6\) sandwiches, since \(7\) sandwiches would cost

\[
7(4.50)=31.50,
\]

which is too much.

The cost of \(6\) sandwiches is

\[
6(4.50)=27.00.
\]

Their remaining money is

\[
30.00-27.00=3.00.
\]

Soft drinks cost \(\$1.00\), so they can buy

\[
3
\]

soft drinks.

The total number of items is

\[
6+3=9.
\]

ANSWER 5: D

Problem 6:

We need to estimate

\[
(1.8)(40.3+0.07).
\]

First add inside the parentheses:

\[
40.3+0.07=40.37.
\]

Then estimate:

\[
1.8(40.37)\approx 1.8(40)=72.
\]

More accurately,

\[
1.8(40.37)=72.666.
\]

The closest choice to \(72.666\) is \(74\).

Choices such as \(84\) or \(737\) result from inaccurate place-value estimation.

ANSWER 6: C

Problem 7:

We are asked for the total percent discount from the original price.

Let the original price be \(\$100\).

The half-price sale reduces the price to

\[
\$50.
\]

Then the coupon takes 20% off the sale price. Twenty percent of \(\$50\) is

\[
0.20(50)=10.
\]

So the final price is

\[
50-10=40.
\]

The original price was \(\$100\), and the final price is \(\$40\), so the discount is

\[
100-40=60.
\]

Thus, the price is \(60\%\) off the original price.

It is not \(70\%\), because the 20% coupon applies to the already reduced sale price, not to the original price.

ANSWER 7: D

Problem 8:

We are asked to evaluate

\[
\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}
{1+2+3+4+5+6+7+8}.
\]

First compute the numerator:

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40320.
\]

Now compute the denominator:

\[
1+2+3+4+5+6+7+8=36.
\]

Therefore,

\[
\frac{40320}{36}=1120.
\]

ANSWER 8: B

Problem 9:

Tammy starts with \(\$100\).

After a 15% loss, she has

\[
100(1-0.15)=100(0.85)=85.
\]

During the second year, her investment gains 20% of the remaining \(\$85\):

\[
85(1+0.20)=85(1.20)=102.
\]

She started with \(\$100\) and ended with \(\$102\), so her investment increased by

\[
102-100=2.
\]

Since this is \(2\) out of the original \(100\),

\[
\frac{2}{100}=2\%.
\]

Thus, Tammy had a \(2\%\) gain.

A 15% loss followed by a 20% gain does not produce a 5% gain, because the gain is calculated from \(\$85\), not from the original \(\$100\).

ANSWER 9: D

Problem 10:

We need to find

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Compute the first group:

\[
1+11+21+31+41=105.
\]

Compute the second group:

\[
9+19+29+39+49=145.
\]

Now add:

\[
105+145=250.
\]

ANSWER 10: E