🔬 Model Lab

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A

Problem 1:
We are asked: toss a fair coin 3 times, find P(at least two consecutive heads).

Total outcomes: each toss has 2 possibilities, so $2^3 = 8$ equally likely outcomes.

List them:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

"At least two consecutive heads" means HH occurs as first two or last two, including three heads.
Favorable:
HHH - has HH
HHT - has HH at start
THH - has HH at end

That is 3 outcomes. HTH does NOT count because the heads are separated.

So probability $= 3/8$.

A $1/8$ is only HHH. B $1/4=2/8$ is too small. D $1/2=4/8$ and E $3/4=6/8$ are too big.
ANSWER 1: C

Problem 2:
We are asked the value of $\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$.

Work from the inside out.
Innermost: $\sqrt{4}=2$.
Then: $8 \cdot \sqrt{4} = 8 \cdot 2 = 16$.
Then: $\sqrt{8\cdot\sqrt{4}} = \sqrt{16}=4$.
Then: $16 \cdot \sqrt{8\cdot\sqrt{4}} = 16\cdot 4 = 64$.
Then: $\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{64}=8$.

So the value is 8. A is too small, E is forgetting the outer square root.
ANSWER 2: C

Problem 3:
We are asked: Abe picks at random from 1 green, 1 red. Bob picks at random from 1 green, 1 yellow, 2 red. Find P(same color shown).

Abe: $P(G)=1/2$, $P(R)=1/2$.
Bob: total $1+1+2=4$ beans, so $P(G)=1/4$, $P(Y)=1/4$, $P(R)=2/4=1/2$.

They can match only as both green or both red, since Abe has no yellow.
$P(\text{both green}) = \frac12 \cdot \frac14 = \frac18$.
$P(\text{both red}) = \frac12 \cdot \frac12 = \frac14 = \frac28$.

Add for mutually exclusive ways:
$P(\text{match}) = 1/8+2/8 = 3/8$.

A $1/4$ is only the red-red part, D $1/2$ overcounts.
ANSWER 3: C

Problem 4:
We are asked for % yellow after adding 5 liters yellow to 30 liters of 25% red, 30% yellow, 45% water.

Original amount of yellow:
$30\%$ of $30 = 0.30 \times 30 = 9$ liters.

Add 5 liters yellow:
new yellow $= 9+5 = 14$ liters.
new total $= 30+5 = 35$ liters.

New percent yellow $= 14/35 = 2/5 = 0.40 = 40\%$.

A 25, B 35 just add 5% incorrectly, D 45, E 50 are too high.
ANSWER 4: C

Problem 5:
We are asked: jumps are 1,2,4,... doubling each time. Which jump first exceeds 1000 meters?

Jump $n$ distance $=2^{\,n-1}$ meters:
1st:1, 2nd:2, 3rd:4, 4th:8, 5th:16, 6th:32, 7th:64, 8th:128, 9th:256, 10th:512, 11th:1024,...

$2^9 = 512 < 1000$, $2^{10}=1024 > 1000$.

So the 10th jump is still only 512 m. The 11th jump is first $>1000$.

B 10th is the tempting mistake of using $2^{10}$ but forgetting the 1st jump is $2^0$.
ANSWER 5: C

Problem 6:
We are asked in how many weeks Chandler can afford a $500$ bike.

Birthday money: $50+35+15 = 100$ dollars.
Still needed: $500-100 = 400$ dollars.

Paper route: $16$ dollars per week.
Weeks needed: $400 \div 16 = 25$.

Check: $25 \times 16 = 400$. Plus 100 = 500 exactly.

A 24 gives only $24\times16+100=484<500$.
ANSWER 6: B

Problem 7:
We are asked for length of lunch break in minutes. Total trip 3 hours, 100 miles, driving speed 40 mph.

Driving time only $=$ distance $\div$ speed:
$100 \div 40 = 2.5$ hours $= 150$ minutes.

Total elapsed $=3$ hours $=180$ minutes.
Break $=$ total $-$ driving $=180-150=30$ minutes.

A 15, C 40, D 45, E 60 do not match distance/speed calculation.
ANSWER 7: B

Problem 8:
We are asked to estimate total cars. 6 cars in 10 seconds, constant speed, total time 2 min 45 sec.

Convert: $2$ min $45$ sec $=2\times60+45=165$ seconds.

Rate: $6$ cars per $10$ sec $=0.6$ cars/sec.

Estimate: $165 \times 6/10 = 165 \times 0.6 = 99$ cars.

Nearest most likely choice is 100.

A 60 would be only about 100 sec, D 120, E 140 too many.
ANSWER 8: C

Problem 9:
We are asked: 3 pounds makes 8 hamburgers, how much for 24 hamburgers?

Ratio: $24 \div 8 = 3$ times as many hamburgers.
So need 3 times as much meat: $3 \times 3 = 9$ pounds.

Proportion: $3/8 = x/24$, so $x = 3\times24/8 =9$.

A 6 is doubling only, D 8 confuses counts.
ANSWER 9: E

Problem 10:
We are asked to compute $16+8\div4-2$.

By order of operations, multiply/divide before add/subtract.

First $8\div4 = 2$.
Then $16+2-2 = 16$.

Tempting wrong A 4 comes from doing left-to-right: $(16+8)\div4-2 =6-2=4$. That violates PEMDAS.
C 12 comes from $(16+8)\div(4-2)$.
ANSWER 10: D

Problem 1:
We are asked to find the probability of getting at least two consecutive heads when a fair coin is tossed 3 times.

First, list all possible equally likely outcomes when a coin is tossed 3 times. Since each toss has 2 outcomes (Heads, $H$, or Tails, $T$), there are $2^3 = 8$ total outcomes:
1. $HHH$
2. $HHT$
3. $HTH$
4. $HTT$
5. $THH$
6. $THT$
7. $TTH$
8. $TTT$

Next, identify the outcomes with at least two consecutive heads:
- $HHH$ (heads on tosses 1, 2, and 3)
- $HHT$ (heads on tosses 1 and 2)
- $THH$ (heads on tosses 2 and 3)

The outcome $HTH$ has two heads, but they are not consecutive. 

There are $3$ favorable outcomes out of $8$ total outcomes. Thus, the probability is:
$$\frac{3}{8}$$

ANSWER 1: C

---

Problem 2:
We are asked to evaluate the expression:
$$\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}$$

We evaluate the expression from the inside out:
1. The innermost square root is $\sqrt{4} = 2$.
2. Substitute this back into the next radical:
   $$\sqrt{8 \cdot \sqrt{4}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$$
3. Substitute this back into the outer radical:
   $$\sqrt{16 \cdot 4} = \sqrt{64} = 8$$

Thus, the value of the expression is $8$.

ANSWER 2: C

---

Problem 3:
We are asked for the probability that the colors of two randomly selected jelly beans match, where:
- Abe has 2 jelly beans: 1 green ($G$) and 1 red ($R$).
- Bob has 4 jelly beans: 1 green ($G$), 1 yellow ($Y$), and 2 red ($R$).

The total number of possible pairs of choices is $2 \times 4 = 8$ equally likely outcomes.

A match can occur in two disjoint cases:
1. Both pick Green:
   - Abe picks Green with probability $\frac{1}{2}$.
   - Bob picks Green with probability $\frac{1}{4}$.
   - Probability of both picking Green = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

2. Both pick Red:
   - Abe picks Red with probability $\frac{1}{2}$.
   - Bob picks Red with probability $\frac{2}{4} = \frac{1}{2}$.
   - Probability of both picking Red = $\frac{1}{2} \times \frac{2}{4} = \frac{2}{8} = \frac{1}{4}$.

Since yellow is only held by Bob, they cannot match on yellow.

Adding the probabilities of the two cases:
$$\text{P(match)} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 3: C

---

Problem 4:
We are given an initial mixture of $30$ liters of paint containing:
- $25\%$ red tint
- $30\%$ yellow tint
- $45\%$ water

First, calculate the volume of yellow tint in the original mixture:
$$\text{Original yellow tint} = 30\% \times 30 = 0.30 \times 30 = 9\text{ liters}$$

Next, $5$ liters of yellow tint are added to the mixture:
- New amount of yellow tint = $9 + 5 = 14$ liters.
- New total volume of the mixture = $30 + 5 = 35$ liters.

Now, compute the percentage of yellow tint in the new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

ANSWER 4: C

---

Problem 5:
We are given that the Hulk's jump distance doubles with each jump:
- Jump 1: $1\text{ m} = 2^0\text{ m}$
- Jump 2: $2\text{ m} = 2^1\text{ m}$
- Jump 3: $4\text{ m} = 2^2\text{ m}$
- Jump $n$: $2^{n-1}\text{ m}$

We want to find the first jump $n$ where the distance exceeds $1\text{ km} = 1000\text{ meters}$:
$$2^{n-1} > 1000$$

Powers of $2$ near $1000$:
- $2^9 = 512$
- $2^{10} = 1024$

Since $1024 > 1000$, we need:
$$n - 1 = 10 \implies n = 11$$

Thus, on the 11th jump, the Hulk will first jump more than $1000$ meters.

ANSWER 5: C

---

Problem 6:
We want to find the number of weeks Chandler needs to work to afford a $\$500$ mountain bike.

First, calculate the total birthday money Chandler received:
$$\$50 + \$35 + \$15 = \$100$$

Subtract this from the total cost of the bike to find the remaining amount needed:
$$\$500 - \$100 = \$400$$

Chandler earns $\$16$ per week from his paper route. The number of weeks required is:
$$\frac{400}{16} = 25\text{ weeks}$$

ANSWER 6: B

---

Problem 7:
Casey's road trip covered a total distance of $100$ miles at an average driving speed of $40$ miles per hour.

First, calculate the time Casey spent driving:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\text{ miles}}{40\text{ mph}} = 2.5\text{ hours}$$

Convert the driving time into minutes:
$$2.5\text{ hours} \times 60\text{ minutes/hour} = 150\text{ minutes}$$

The total trip took $3$ hours, which is:
$$3\text{ hours} \times 60\text{ minutes/hour} = 180\text{ minutes}$$

The duration of the lunch break is the difference between total trip time and driving time:
$$\text{Lunch break} = 180 - 150 = 30\text{ minutes}$$

ANSWER 7: B

---

Problem 8:
We are given:
- Trey counts $6$ cars passing in $10$ seconds.
- The train takes $2$ minutes and $45$ seconds to clear the crossing at a constant speed.

First, convert the total time into seconds:
$$2\text{ minutes and } 45\text{ seconds} = 2 \times 60 + 45 = 120 + 45 = 165\text{ seconds}$$

Find the rate at which cars pass the crossing:
$$\text{Rate} = \frac{6\text{ cars}}{10\text{ seconds}} = 0.6\text{ cars per second}$$

Calculate the total number of cars:
$$\text{Total cars} = 165\text{ seconds} \times 0.6\text{ cars/second} = 99\text{ cars}$$

Among the given choices (60, 80, 100, 120, 140), $99$ is closest to $100$.

ANSWER 8: C

---

Problem 9:
Rachelle uses $3$ pounds of meat to make $8$ hamburgers.
We need to find how many pounds of meat are required to make $24$ hamburgers.

Find the scaling factor for the number of hamburgers:
$$\frac{24}{8} = 3$$

Since the amount of meat is directly proportional to the number of hamburgers made, multiply the original amount of meat by $3$:
$$3\text{ pounds} \times 3 = 9\text{ pounds}$$

ANSWER 9: E

---

Problem 10:
We are asked to evaluate the arithmetic expression:
$$16 + 8 \div 4 - 2$$

According to the order of operations (PEMDAS):
1. Division comes before addition and subtraction:
   $$8 \div 4 = 2$$
2. Substitute back into the expression:
   $$16 + 2 - 2$$
3. Perform addition and subtraction from left to right:
   $$16 + 2 = 18$$
   $$18 - 2 = 16$$

Thus, the value is $16$.

ANSWER 10: D

Problem 1:
We need the probability that, in 3 coin tosses, there is at least one pair of consecutive heads.

There are \(2^3 = 8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes with at least two consecutive heads are:

- \(HHH\): has consecutive heads
- \(HHT\): first two tosses are heads
- \(THH\): last two tosses are heads

So there are 3 favorable outcomes out of 8 total outcomes.

\[
P=\frac{3}{8}
\]

ANSWER 1: C

Problem 2:
Evaluate from the inside out:

\[
\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}
\]

First,

\[
\sqrt{4}=2
\]

Then,

\[
8 \cdot 2 = 16
\]

So the inner square root becomes

\[
\sqrt{16}=4
\]

Now the expression is

\[
\sqrt{16 \cdot 4}
\]

\[
16 \cdot 4 = 64
\]

Finally,

\[
\sqrt{64}=8
\]

ANSWER 2: C

Problem 3:
Abe has 2 jelly beans: 1 green and 1 red.

Bob has 4 jelly beans: 1 green, 1 yellow, and 2 red.

The colors match only if both choose green or both choose red.

Probability both choose green:

\[
\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}
\]

Probability both choose red:

\[
\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}
\]

Add these probabilities:

\[
\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}
\]

ANSWER 3: C

Problem 4:
The original mixture is 30 liters, and 30% is yellow tint.

Yellow tint in the original mixture:

\[
0.30 \cdot 30 = 9 \text{ liters}
\]

Five liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14 \text{ liters}
\]

The new total mixture volume is

\[
30+5=35 \text{ liters}
\]

The percent yellow tint in the new mixture is

\[
\frac{14}{35}\cdot 100\% = 0.40 \cdot 100\% = 40\%
\]

ANSWER 4: C

Problem 5:
The jumps double each time:

\[
1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128,\ 256,\ 512,\ 1024,\dots
\]

The \(n\)-th jump has length

\[
2^{n-1}
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

\[
2^{10}=1024
\]

Since \(2^{10}\) corresponds to the 11th jump, the Hulk first jumps more than 1000 meters on the 11th jump.

ANSWER 5: C

Problem 6:
Chandler needs a total of 500 dollars.

Birthday money:

\[
50+35+15=100
\]

So he still needs

\[
500-100=400
\]

He earns 16 dollars per week. The number of weeks needed is

\[
\frac{400}{16}=25
\]

ANSWER 6: B

Problem 7:
Casey drove 100 miles at an average driving speed of 40 miles per hour.

Driving time:

\[
\frac{100}{40}=2.5 \text{ hours}
\]

Convert to minutes:

\[
2.5 \cdot 60 = 150 \text{ minutes}
\]

The total trip time, including lunch, was 3 hours:

\[
3 \cdot 60 = 180 \text{ minutes}
\]

Therefore, the lunch break lasted

\[
180-150=30 \text{ minutes}
\]

ANSWER 7: B

Problem 8:
Trey counted 6 cars in 10 seconds.

Total time for the train to clear the crossing:

\[
2 \text{ minutes } 45 \text{ seconds} = 2 \cdot 60 + 45 = 165 \text{ seconds}
\]

At a constant speed, the number of cars is proportional to time.

Since 10 seconds corresponds to 6 cars, 165 seconds corresponds to

\[
6 \cdot \frac{165}{10}
\]

\[
6 \cdot 16.5 = 99
\]

The train had about 99 cars, so the most likely answer among the choices is 100.

ANSWER 8: C

Problem 9:
Rachelle uses 3 pounds of meat for 8 hamburgers.

She needs to make 24 hamburgers, which is

\[
\frac{24}{8}=3
\]

times as many hamburgers.

So she needs 3 times as much meat:

\[
3 \cdot 3 = 9 \text{ pounds}
\]

ANSWER 9: E

Problem 10:
The expression as typed appears to be missing the division sign; the intended expression is

\[
16 + 8 \div 4 - 2
\]

Using order of operations, division comes before addition and subtraction:

\[
8 \div 4 = 2
\]

Now substitute:

\[
16 + 2 - 2 = 16
\]

ANSWER 10: D

# Problem 1:
We toss a fair coin 3 times and want the probability that **at least two heads occur in a row**.

All 8 equally likely outcomes:
- HHH — heads in positions 1,2 (and 2,3) → yes
- HHT — heads in positions 1,2 → yes
- HTH — heads are separated → no
- HTT — no
- THH — heads in positions 2,3 → yes
- THT — no
- TTH — no
- TTT — no

Favorable outcomes: HHH, HHT, THH = 3 out of 8.

Probability = **3/8**. (The tempting error is counting HTH as "two heads," but they are not *consecutive*, which would give 4/8 = 1/2.)

ANSWER 1: C

# Problem 2:
Work from the inside out.

√4 = 2

Next layer: 8 · 2 = 16, and √16 = 4

Outer layer: 16 · 4 = 64, and √64 = 8

(A common slip is stopping after an inner step and answering 4, or multiplying wrongly to get 8√2.)

ANSWER 2: C

# Problem 3:
Abe's hand: 1 green, 1 red → P(green) = 1/2, P(red) = 1/2.
Bob's hand: 1 green, 1 yellow, 2 red (4 beans) → P(green) = 1/4, P(yellow) = 1/4, P(red) = 2/4 = 1/2.

Colors match only if both show green or both show red (Abe has no yellow).

- Both green: (1/2)(1/4) = 1/8
- Both red: (1/2)(1/2) = 1/4 = 2/8

Total = 1/8 + 2/8 = **3/8**.

(Answering 1/2 comes from forgetting Bob has two red beans out of *four*, and 1/4 from counting only one matching case.)

ANSWER 3: C

# Problem 4:
Original 30 liters: yellow tint = 30% of 30 = 9 liters.

Add 5 liters of yellow tint:
- Yellow = 9 + 5 = 14 liters
- Total = 30 + 5 = 35 liters

Percent yellow = 14/35 = 0.40 = **40%**.

(Answer 35 comes from adding percentages incorrectly, forgetting the total volume also grew.)

ANSWER 4: C

# Problem 5:
Jump lengths: 1, 2, 4, 8, … so the nth jump is 2^(n−1) meters.

We need 2^(n−1) > 1000.
- 2^9 = 512 → that's the 10th jump (512 m, not enough)
- 2^10 = 1024 → that's the 11th jump (1024 m > 1000) ✓

So the **11th** jump is the first to exceed 1 km. (Choosing 10th mistakenly uses 2^n for the nth jump.)

ANSWER 5: C

# Problem 6:
Birthday money: 50 + 35 + 15 = 100 dollars.

Still needed: 500 − 100 = 400 dollars.

At 16 dollars per week: 400 ÷ 16 = 25 weeks exactly.

So he can buy the bike after **25** weeks. (24 weeks gives only 384 + 100 = 484 dollars — not enough.)

ANSWER 6: B

# Problem 7:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.

Total trip time = 3 hours, so lunch = 3 − 2.5 = 0.5 hour = **30 minutes**.

(40 or 45 minutes would come from mis-dividing; note 40 mph is the driving average, not the overall average.)

ANSWER 7: B

# Problem 8:
Rate of cars passing: 6 cars in 10 seconds = 0.6 cars per second.

Total time = 2 min 45 s = 120 + 45 = 165 seconds.

Number of cars ≈ 0.6 × 165 = 99 cars.

The closest choice is **100**.

ANSWER 8: C

# Problem 9:
Meat per hamburger = 3/8 pound.

For 24 hamburgers: (3/8) × 24 = 9 pounds.

Check with proportion: 24 hamburgers is 3 times 8 hamburgers, so 3 × 3 = **9** pounds.

(6 pounds would be doubling instead of tripling.)

ANSWER 9: E

# Problem 10:
The expression is the fraction (16 + 8)/(4 − 2).

Numerator: 16 + 8 = 24
Denominator: 4 − 2 = 2

24 ÷ 2 = **12**.

(Answer 16 comes from splitting the fraction wrongly, e.g., 16 + 8/4 − 2.)

ANSWER 10: C

Problem 1:

We want the probability of getting **at least two consecutive heads** in 3 tosses.

There are \(2^3=8\) equally likely outcomes:

\[
HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT
\]

The outcomes containing consecutive heads are:

\[
HHH,\ HHT,\ THH
\]

That is \(3\) favorable outcomes out of \(8\) total outcomes:

\[
\frac{3}{8}
\]

For example, \(HTH\) has two heads, but they are not consecutive, so it does not count.

ANSWER 1: C

Problem 2:

Evaluate

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

Work from the inside outward.

First,

\[
\sqrt4=2.
\]

So the middle square root becomes

\[
\sqrt{8\cdot 2}=\sqrt{16}=4.
\]

Now the whole expression is

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

Choices involving \(4\sqrt2\) or \(8\sqrt2\) result from stopping before fully simplifying a square root.

ANSWER 2: C

Problem 3:

Abe has:

- 1 green and 1 red jelly bean, so each color has probability \(\frac12\).

Bob has:

- 1 green, 1 yellow, and 2 red jelly beans, for 4 total beans.

The colors match if they both choose green or they both choose red.

Probability both choose green:

\[
\frac12\cdot\frac14=\frac18.
\]

Probability both choose red:

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

Add these mutually exclusive possibilities:

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot match because Abe has no yellow jelly bean.

ANSWER 3: C

Problem 4:

The original mixture has 30 liters, and \(30\%\) is yellow tint.

Amount of yellow tint originally:

\[
0.30(30)=9\text{ liters}.
\]

Then 5 liters of yellow tint are added, so the new amount of yellow tint is

\[
9+5=14\text{ liters}.
\]

The total amount of paint is now

\[
30+5=35\text{ liters}.
\]

Thus the new percentage of yellow tint is

\[
\frac{14}{35}=0.4=40\%.
\]

A common mistake is to add 5 directly to \(30\%\), but the total amount of mixture also changes.

ANSWER 4: C

Problem 5:

The distances form a doubling pattern:

\[
1,\ 2,\ 4,\ 8,\dots
\]

The \(n\)th jump has length

\[
2^{n-1}\text{ meters}.
\]

We need the first jump greater than 1000 meters.

\[
2^9=512
\]

and

\[
2^{10}=1024.
\]

Since \(1024>1000\), the first jump longer than 1 kilometer is the jump with distance \(2^{10}\).

Because the first jump is \(2^0\), the \(2^{10}\)-meter jump is the

\[
10+1=11\text{th jump}.
\]

ANSWER 5: C

Problem 6:

Chandler receives birthday money totaling

\[
50+35+15=100\text{ dollars}.
\]

The bike costs \$500, so he still needs

\[
500-100=400\text{ dollars}.
\]

He earns \$16 per week, so the number of weeks needed is

\[
\frac{400}{16}=25.
\]

He can buy the bike after 25 weeks. Adding the birthday money incorrectly to the weekly earnings without first finding the remaining cost could lead to an incorrect answer.

ANSWER 6: B

Problem 7:

Casey drove 100 miles at an average driving speed of 40 miles per hour.

Her driving time was

\[
\frac{100}{40}=2.5\text{ hours}.
\]

The total trip took 3 hours, including lunch. Therefore, the lunch break lasted

\[
3-2.5=0.5\text{ hour}.
\]

Convert \(0.5\) hour to minutes:

\[
0.5(60)=30\text{ minutes}.
\]

The 3 hours includes both driving and lunch, so it would be incorrect to treat all 3 hours as driving time.

ANSWER 7: B

Problem 8:

Trey counted 6 cars in 10 seconds. Thus the train passes at a rate of

\[
\frac{6}{10}=0.6
\]

cars per second.

The train takes 2 minutes and 45 seconds to clear the crossing:

\[
2(60)+45=165\text{ seconds}.
\]

At \(0.6\) cars per second, the estimated number of cars is

\[
0.6(165)=99.
\]

Since 99 is closest to 100, the most likely number of cars is 100.

The answer is not exactly 99 because the choices ask for the most likely choice, and counting cars over only the first 10 seconds is an estimate.

ANSWER 8: C

Problem 9:

Rachelle uses 3 pounds of meat for 8 hamburgers.

To make 24 hamburgers, she needs

\[
\frac{24}{8}=3
\]

times as many hamburgers.

Therefore, she needs 3 times as much meat:

\[
3\cdot 3=9
\]

pounds.

Choices such as 6 or \(7\frac12\) do not triple the original amount of meat.

ANSWER 9: E

Problem 10:

The expression is interpreted as

\[
\frac{16+8}{4}-2.
\]

First add in the numerator:

\[
16+8=24.
\]

Then divide:

\[
\frac{24}{4}=6.
\]

Finally subtract 2:

\[
6-2=4.
\]

It would be incorrect to subtract 2 before completing the division.

ANSWER 10: A