🔬 Model Lab

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B

Problem 1:
We are asked how long 60 pills last if the dose is one half pill every other day.

Restate: Every 2 days, 1/2 pill is used. So per day on average:
(1/2 pill) / 2 days = 1/4 pill per day.

Total pills = 60. Number of days = total / rate:
60 / (1/4) = 60 x 4 = 240 days.

A month is about 30 days. 240 / 30 = 8 months.

Check choices:
A. 1 month would be about 30 pills at 1 per day, far too short.
B. 4 months ~ 120 days ~ 30 pills used, too short.
C. 6 months ~ 180 days ~ 45 pills used, too short.
D. 8 months ~ 240 days ~ 60 pills used, matches.
E. 1 year ~ 365 days ~ 91 pills needed, too long.

So about 8 months.
ANSWER 1: D

Problem 2:
We are asked for p, r, s from exponential equations and their product.

Interpret the equations as:
3^p + 3^4 = 90,  2^r + 44 = 76,  5^3 + 6^s = 1421.

Solve first: 3^4 = 81.
So 3^p + 81 = 90 => 3^p = 90 - 81 = 9 = 3^2.
Thus p = 2.

Solve second: 2^r + 44 = 76 => 2^r = 76 - 44 = 32 = 2^5.
Thus r = 5.

Solve third: 5^3 = 125.
So 125 + 6^s = 1421 => 6^s = 1421 - 125 = 1296.
Check powers of 6: 6^1=6, 6^2=36, 6^3=216, 6^4=1296.
Thus s = 4.

Product p * r * s = 2 * 5 * 4 = 40.

Rule out: Linear solving would give non-integers; recognizing powers is key. 27=3^3, 50,70,90 do not equal 2*5*4.
ANSWER 2: B

Problem 3:
We need sets of three different numbers from 1 to 9 summing to 15, that contain a 5.

Let the set be {5, a, b} with a, b distinct, different from 5 and from each other, from 1-9.
Then 5 + a + b = 15 => a + b = 10.

List pairs of distinct numbers not 5 summing to 10:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 + 5 = 10 but numbers must be different and 5 already used, so reject.

That is 4 pairs, giving sets:
{5,1,9}, {5,2,8}, {5,3,7}, {5,4,6}.

No others. So 4 sets.

Choices 3,5,6,7 miscount pairs.
ANSWER 3: B

Problem 4:
We need total gumdrops, then new brown count after replacement.

Percentages: blue 30%, brown 20%, red 15%, yellow 10%.
Sum = 30+20+15+10 = 75%.
So remainder for green = 100% - 75% = 25%.

Given green = 30 gumdrops = 25% of total T.
So 0.25 T = 30 => T = 30 / 0.25 = 120.

Then:
Blue = 30% of 120 = 36.
Brown = 20% of 120 = 24.

Half of the blue are replaced with brown: half of 36 = 18.
Remove 18 blue, add 18 brown.

New brown = old brown + 18 = 24 + 18 = 42.

Check: A 35, B 36 confuses blue count with brown, D 48 doubles brown, E 64 too large.
ANSWER 4: C

Problem 5:
Ara and Shea started same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara now.

Let original height = H.
Shea now = H + 20% of H = 1.20 H = 60.
So H = 60 / 1.2 = 50 inches.

Shea's growth in inches = 60 - 50 = 10 inches.
Ara's growth = half as many inches = 10 / 2 = 5 inches.
Ara now = H + 5 = 50 + 5 = 55 inches.

Tempting error: half as much percent would be 10% growth => 55 also? Actually 50*1.1=55 same here, but reasoning must be inches. Choice 54 would be miscalculation, 48 would be original, etc.
ANSWER 5: E

Problem 6:
Rank S, M, L by value, i.e. lowest cost per amount of detergent is best.

Let small cost = Cs, amount = As.
Given:
Cm = 50% more than Cs => Cm = 1.5 Cs.
Al = twice As => Al = 2 As.
Am = 20% less than Al => Am = 0.8 Al = 0.8*2 As = 1.6 As.
Cl = 30% more than Cm => Cl = 1.3 Cm = 1.3*1.5 Cs = 1.95 Cs.

Unit price = cost / amount. Set Cs=1, As=1 for comparison:
S: 1 / 1 = 1.00
M: 1.5 / 1.6 = 0.9375
L: 1.95 / 2 = 0.975

Smallest = best: M (0.9375) < L (0.975) < S (1.00).
So best to worst: M L S.

This is choice E. Others reverse order.
ANSWER 6: E

Problem 7:
Start with number N. Multiply by 3/4 then divide by 3/5. What single operation is equivalent?

Compute: N * (3/4) / (3/5) = N * (3/4) * (5/3) because dividing by a fraction is multiplying by its reciprocal.
= N * (3*5)/(4*3) = N * 5/4.

So equivalent to multiplying by 5/4.

Check others:
A dividing by 4/3 = multiplying by 3/4, only first step.
B dividing by 9/20 = multiplying by 20/9, too big.
C multiplying by 9/20 = (3/4)*(3/5), that's multiply by both, not divide.
D dividing by 5/4 = multiplying by 4/5, reciprocal.
ANSWER 7: E

Problem 8:
x = 0.00...01 with 1996 zeros after decimal before the 1. So x is a tiny positive number, about 10^-1997. Compare 3+x, 3-x, 3x, 3/x, x/3.

A: 3 + x = just over 3.
B: 3 - x = just under 3, so less than A.
C: 3*x = tiny, near 0, much less than 3.
D: 3 / x = 3 divided by tiny = huge, about 3*10^1997, far larger than 3.
E: x / 3 = even tinier than x, near 0.

Largest is clearly 3/x because dividing by a number <1 magnifies.

Order: D >> A > B >> C > E.
ANSWER 8: D

Problem 9:
500 balls, 80% red. Remove some reds so reds are 75% of remainder.

Initial red = 0.80*500 = 400. Blue = 500-400 = 100. Blue count never changes.

Let removed = R. Remaining red = 400 - R. Remaining total = 500 - R.
Want (400 - R)/(500 - R) = 75% = 3/4.
Cross-multiply: 4(400 - R) = 3(500 - R)
1600 - 4R = 1500 - 3R
1600 - 1500 = 4R - 3R
100 = R.

So remove 100 red balls. Check: remaining 300 red of 400 total = 75%. Yes.

Smaller removals like 25,50 leave red % above 75%.
ANSWER 9: D

Problem 10:
Won 40 of first 50. Total season = 50 + 40 remaining = 90 games. Want exactly 70% wins overall.

Needed total wins = 70% of 90 = 0.7*90 = 63.
Already have 40. Need additional = 63 - 40 = 23.

So must win 23 of remaining 40.

Check: 40+23=63, 63/90=0.7. Yes.
20 would give 60/90=66.7%, 28 would give 68/90~75.6%.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
Jill's grandmother takes $\frac{1}{2}$ of a pill every other day (every 2 days). A full supply has 60 pills. We need to find approximately how long this supply will last.

**Step-by-step Solution:**
1. Determine the rate of pill consumption:
   Taking $\frac{1}{2}$ of a pill every 2 days means she consumes:
   $$\frac{1/2 \text{ pill}}{2 \text{ days}} = \frac{1}{4} \text{ pill per day}$$
   Equivalently, it takes 4 days to consume 1 full pill.

2. Calculate the total number of days the supply lasts:
   $$\text{Days} = 60 \text{ pills} \times 4 \text{ days/pill} = 240 \text{ days}$$

3. Convert days to months:
   Since an average month has approximately 30 days:
   $$\frac{240 \text{ days}}{30 \text{ days/month}} = 8 \text{ months}$$

Thus, the supply will last approximately 8 months.

ANSWER 1: D

---

Problem 2:

**Restatement of the Problem:**
Given the equations:
1. $3^p + 3^4 = 90$
2. $2^r + 44 = 76$
3. $5^3 + 6^s = 1421$

Find the product of $p$, $r$, and $s$.

**Step-by-step Solution:**
1. Solve for $p$:
   $$3^4 = 81$$
   $$3^p + 81 = 90 \implies 3^p = 9 \implies p = 2$$

2. Solve for $r$:
   $$2^r + 44 = 76 \implies 2^r = 76 - 44 = 32$$
   Since $32 = 2^5$, we have $r = 5$.

3. Solve for $s$:
   $$5^3 = 125$$
   $$125 + 6^s = 1421 \implies 6^s = 1421 - 125 = 1296$$
   Since $6^4 = 1296$, we have $s = 4$.

4. Compute the product $p \times r \times s$:
   $$p \times r \times s = 2 \times 5 \times 4 = 40$$

ANSWER 2: B

---

Problem 3:

**Restatement of the Problem:**
We want to choose sets of three distinct numbers $\{a, b, c\}$ from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that their sum is 15 and one of the numbers is 5. We need to find how many such sets exist.

**Step-by-step Solution:**
1. Since 5 must be in the set, let the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct numbers from the remaining set $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the three numbers is 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list all pairs $(a, b)$ of distinct numbers from the allowed set that sum to 10:
   - $a = 1, b = 9$ (valid: $\{1, 5, 9\}$)
   - $a = 2, b = 8$ (valid: $\{2, 5, 8\}$)
   - $a = 3, b = 7$ (valid: $\{3, 5, 7\}$)
   - $a = 4, b = 6$ (valid: $\{4, 5, 6\}$)
   - $a = 5, b = 5$ is not allowed because the three numbers must be distinct.

There are 4 such sets.

ANSWER 3: B

---

Problem 4:

**Restatement of the Problem:**
A jar contains gumdrops of five colors: 30% blue, 20% brown, 15% red, 10% yellow, and the remaining 30 gumdrops are green. If half of the blue gumdrops are replaced with brown ones, how many brown gumdrops will there be?

**Step-by-step Solution:**
1. Determine the percentage of green gumdrops:
   $$\text{Percentage of green} = 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\%$$

2. Find the total number of gumdrops:
   Since 25% corresponds to 30 gumdrops:
   $$\text{Total gumdrops} = \frac{30}{0.25} = 120$$

3. Find the initial count of each relevant color:
   - Blue gumdrops: $30\% \times 120 = 36$
   - Brown gumdrops: $20\% \times 120 = 24$

4. Replace half of the blue gumdrops with brown:
   - Half of the blue gumdrops is $\frac{36}{2} = 18$.
   - These 18 gumdrops become brown.
   - New total of brown gumdrops: $24 + 18 = 42$.

ANSWER 4: C

---

Problem 5:

**Restatement of the Problem:**
Ara and Shea originally had the same height. Shea grew 20% and is now 60 inches tall. Ara grew half as many inches as Shea. We want to find Ara's current height in inches.

**Step-by-step Solution:**
1. Find their original height, $h$:
   Shea grew by 20%, so her current height is:
   $$1.20 \times h = 60 \implies h = \frac{60}{1.20} = 50 \text{ inches}$$

2. Calculate how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10 \text{ inches}$$

3. Calculate how many inches Ara grew:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5 \text{ inches}$$

4. Determine Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55 \text{ inches}$$

ANSWER 5: E

---

Problem 6:

**Restatement of the Problem:**
A company sells detergent in small (S), medium (M), and large (L) boxes:
- Medium costs 50% more than Small and contains 20% less detergent than Large.
- Large contains twice as much detergent as Small and costs 30% more than Medium.

We need to rank the three sizes from best buy (lowest cost per unit of detergent) to worst buy (highest cost per unit of detergent).

**Step-by-step Solution:**
1. Let the amount of detergent in box S be $D_S = 10$ units:
   - Large contains twice as much as Small: $D_L = 2 \times 10 = 20$ units.
   - Medium contains 20% less than Large: $D_M = (1 - 0.20) \times 20 = 16$ units.

2. Let the cost of box S be $C_S = \$100$:
   - Medium costs 50% more than Small: $C_M = 1.50 \times 100 = \$150$.
   - Large costs 30% more than Medium: $C_L = 1.30 \times 150 = \$195$.

3. Calculate the unit price (cost per unit of detergent) for each size:
   - Small (S): $\frac{\$100}{10} = \$10.00$ per unit
   - Medium (M): $\frac{\$150}{16} = \$9.375$ per unit
   - Large (L): $\frac{\$195}{20} = \$9.75$ per unit

4. Compare and rank the sizes from best buy (cheapest unit cost) to worst buy:
   - Best: Medium (\$9.375)
   - Second: Large (\$9.75)
   - Worst: Small (\$10.00)

Thus, the ranking from best to worst buy is M, L, S.

ANSWER 6: E

---

Problem 7:

**Restatement of the Problem:**
Find which given operation has the same effect on a number $N$ as multiplying by $\frac{3}{4}$ and then dividing by $\frac{3}{5}$.

**Step-by-step Solution:**
1. Apply the two operations to an arbitrary number $N$:
   $$\left(N \times \frac{3}{4}\right) \div \frac{3}{5} = N \times \frac{3}{4} \times \frac{5}{3}$$

2. Simplify the multiplication:
   $$N \times \left(\frac{3}{4} \times \frac{5}{3}\right) = N \times \frac{5}{4}$$

3. This is identical to multiplying by $\frac{5}{4}$.

Looking at the options, choice E is "multiply

Problem 1:  
Jill’s grandmother takes one half of a pill every other day. “Every other day” means she takes 1 dose every 2 days. Each dose is \( \frac12 \) pill, so she uses

\[
\frac{1/2 \text{ pill}}{2 \text{ days}}=\frac{1 \text{ pill}}{4 \text{ days}}.
\]

A 60-pill supply lasts

\[
60 \times 4 = 240 \text{ days}.
\]

Approximating one month as 30 days,

\[
240 \div 30 = 8 \text{ months}.
\]

So the supply lasts approximately 8 months.

ANSWER 1: D

Problem 2:  
The intended equations are powers:

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First equation:

\[
3^4=81,
\]
so

\[
3^p+81=90 \implies 3^p=9=3^2 \implies p=2.
\]

Second equation:

\[
2^r+44=76 \implies 2^r=32=2^5 \implies r=5.
\]

Third equation:

\[
5^3=125,
\]
so

\[
125+6^s=1421 \implies 6^s=1296.
\]

Since

\[
6^4=1296,
\]

we have \(s=4\).

Thus the product is

\[
p r s = 2\cdot 5\cdot 4 = 40.
\]

ANSWER 2: B

Problem 3:  
We want sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) whose sum is 15 and which contain a 5.

Fix the 5. The other two numbers must be different from each other, different from 5, and add to

\[
15-5=10.
\]

The possible pairs are

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

They give the sets

\[
\{1,5,9\},\quad \{2,5,8\},\quad \{3,5,7\},\quad \{4,5,6\}.
\]

There are 4 such sets.

ANSWER 3: B

Problem 4:  
The gumdrops that are not blue, brown, red, or yellow are green. The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

So green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

There are 30 green gumdrops, so 25% of the total is 30. The total number of gumdrops is

\[
30 \div 0.25 = 120.
\]

Blue gumdrops:

\[
30\% \text{ of } 120 = 36.
\]

Brown gumdrops:

\[
20\% \text{ of } 120 = 24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac{36}{2}=18.
\]

So the number of brown gumdrops becomes

\[
24+18=42.
\]

ANSWER 4: C

Problem 5:  
Let their original height be \(h\) inches. Shea has now grown 20%, so her current height is

\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:

\[
1.20h=60 \implies h=50.
\]

Shea grew

\[
60-50=10 \text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5 \text{ inches}.
\]

Ara’s current height is

\[
50+5=55 \text{ inches}.
\]

ANSWER 5: E

Problem 6:  
“Best buy” means the most detergent per dollar. Let the small size cost 100 and contain 100 units of detergent.

Small:

\[
\text{cost}=100,\qquad \text{amount}=100.
\]

Medium costs 50% more than small:

\[
\text{medium cost}=150.
\]

Large contains twice as much detergent as small:

\[
\text{large amount}=200.
\]

Medium contains 20% less detergent than large:

\[
\text{medium amount}=0.80\times 200=160.
\]

Large costs 30% more than medium:

\[
\text{large cost}=1.30\times 150=195.
\]

Now compare detergent per dollar:

\[
S:\frac{100}{100}=1,
\]

\[
M:\frac{160}{150}\approx 1.067,
\]

\[
L:\frac{200}{195}\approx 1.026.
\]

The best buy is Medium, then Large, then Small:

\[
M,\ L,\ S.
\]

ANSWER 6: E

Problem 7:  
Multiplying by \(\frac34\) and then dividing by \(\frac35\) means multiplying by

\[
\frac34\cdot \frac{5}{3}.
\]

The 3’s cancel:

\[
\frac34\cdot \frac53=\frac{15}{12}=\frac54.
\]

So the operation is the same as multiplying by \(\frac54\).

ANSWER 7: E

Problem 8:  
The number \(x\) is

\[
0.\underbrace{00\cdots 0}_{1996\text{ zeros}}1,
\]

so \(x\) is extremely small and positive.

Compare the choices:

- \(3+x\) is slightly more than 3.
- \(3-x\) is slightly less than 3.
- \(3\cdot x\) is extremely small.
- \(\frac{x}{3}\) is also extremely small.
- \(\frac{3}{x}\) is extremely large because dividing 3 by a tiny positive number gives a huge number.

Thus the largest expression is

\[
\frac{3}{x}.
\]

ANSWER 8: D

Problem 9:  
There are 500 balls total. 80% are red:

\[
0.80\times 500=400
\]

red balls, so the other

\[
500-400=100
\]

balls are blue.

Only red balls are removed, so the number of blue balls remains 100. Let \(r\) be the number of red balls remaining. We want 75% of the remaining balls to be red:

\[
\frac{r}{r+100}=\frac34.
\]

Cross-multiply:

\[
4r=3(r+100)
\]

\[
4r=3r+300
\]

\[
r=300.
\]

Originally there were 400 red balls, so the number removed is

\[
400-300=100.
\]

ANSWER 9: D

Problem 10:  
The team has played 50 games and has 40 more games to play, so the season total is

\[
50+40=90
\]

games.

The team wants to win exactly 70% of all its games:

\[
0.70\times 90=63.
\]

It has already won 40 games, so it must win

\[
63-40=23
\]

more games.

ANSWER 10: B

# Problem 1:
Grandmother takes ½ pill every other day.

- One pill = 2 doses, and doses come 2 days apart, so **one pill lasts 4 days**.
- 60 pills × 4 days = 240 days.
- 240 days ÷ 30 days per month ≈ 8 months.

(Choice B, 4 months, comes from forgetting the "half pill"; 1 month or 1 year are far off.)

ANSWER 1: D

# Problem 2:
The equations are $3^p + 3^4 = 90$, $2^r + 44 = 76$, $5^3 + 6^s = 1421$.

- $3^p = 90 - 81 = 9 \Rightarrow p = 2$
- $2^r = 76 - 44 = 32 \Rightarrow r = 5$
- $6^s = 1421 - 125 = 1296 = 6^4 \Rightarrow s = 4$

Product: $2 \cdot 5 \cdot 4 = 40$.

ANSWER 2: B

# Problem 3:
We need three **different** numbers from 1–9 summing to 15, one of which is 5. The other two must sum to $15 - 5 = 10$, be different from each other and from 5:

- 1 + 9 ✓
- 2 + 8 ✓
- 3 + 7 ✓
- 4 + 6 ✓
- 5 + 5 ✗ (repeats 5)

That gives 4 sets: {1,5,9}, {2,5,8}, {3,5,7}, {4,5,6}.

ANSWER 3: B

# Problem 4:
Blue + brown + red + yellow = 30 + 20 + 15 + 10 = 75%, so green = 25%.

- 25% of total = 30 gumdrops ⇒ total = 120.
- Blue = 30% of 120 = 36; brown = 20% of 120 = 24.
- Half the blue = 18 are replaced by brown.
- New brown count = 24 + 18 = **42**.

(35 or 36 come from stopping at the blue count; 48 doubles brown incorrectly.)

ANSWER 4: C

# Problem 5:
Shea grew 20% to reach 60 inches: original height $h$ satisfies $1.2h = 60 \Rightarrow h = 50$.

- Shea's growth = 60 − 50 = 10 inches.
- Ara grew half as many inches: 5 inches.
- Ara's height = 50 + 5 = 55 inches.

(54 would come from mistakenly taking Shea's original height as 60·0.9 type errors; 51–52 come from misreading "half as many inches.")

ANSWER 5: E

# Problem 6:
Let small contain 1 unit and cost 1.

- Large contains twice small: 2 units.
- Medium contains 20% less than large: $0.8(2) = 1.6$ units.
- Medium costs 50% more than small: 1.5.
- Large costs 30% more than medium: $1.3(1.5) = 1.95$.

Cost per unit of detergent (smaller is better):
- S: $1/1 = 1.000$
- M: $1.5/1.6 = 0.9375$
- L: $1.95/2 = 0.975$

Best → worst: M, L, S.

ANSWER 6: E

# Problem 7:
Multiplying by $\frac34$ then dividing by $\frac35$:
$$n \cdot \frac34 \div \frac35 = n \cdot \frac34 \cdot \frac53 = n \cdot \frac{15}{12} = n \cdot \frac54.$$

So it's the same as multiplying by $\frac54$ (equivalently dividing by $\frac45$, which is not offered).

(Choice C, multiplying by 9/20, comes from multiplying by 3/5 instead of dividing; D would be dividing, not multiplying, by 5/4.)

ANSWER 7: E

# Problem 8:
Here $x = 10^{-1997}$, a tiny positive number.

- $3 + x \approx 3$
- $3 - x \approx 3$
- $3x \approx 0$
- $3/x = 3 \cdot 10^{1997}$ — enormous
- $x/3 \approx 0$

Dividing 3 by a tiny number gives the largest value.

ANSWER 8: D

# Problem 9:
Red = 80% of 500 = 400; blue = 100.

Removing only red balls keeps blue at 100. If red is 75% of the remainder, blue is 25%:
$$0.25 \cdot T = 100 \Rightarrow T = 400.$$
Remaining red = 400 − 100 = 300, so we removed $400 - 300 = 100$ red balls.

(Choice B, 50, comes from the wrong assumption that removing 5% of balls suffices.)

ANSWER 9: D

# Problem 10:
Total games = 50 + 40 = 90. To win exactly 70%: $0.70 \times 90 = 63$ wins needed.

Already has 40 wins, so it needs $63 - 40 = 23$ more wins (out of the remaining 40 — possible).

ANSWER 10: B

Problem 1:

Jill’s grandmother takes \(\frac12\) pill every other day.

- In 2 days, she uses \(\frac12\) pill.
- Therefore, 1 pill lasts \(4\) days.
- A supply of \(60\) pills lasts

\[
60\cdot 4=240\text{ days}.
\]

Using about 30 days per month,

\[
240\div 30=8\text{ months}.
\]

A tempting error is to say 60 pills lasts 120 days, but that would treat each half-pill as a whole pill. Since each pill provides two doses, and doses are every other day, each pill lasts 4 days.

ANSWER 1: D

Problem 2:

The equations are interpreted as

\[
3^p+3^4=90,\qquad 2^r+44=76,\qquad 5^3+6^s=1421.
\]

First, solve for \(p\):

\[
3^p+3^4=90
\]

\[
3^p+81=90
\]

\[
3^p=9=3^2.
\]

Thus,

\[
p=2.
\]

Next, solve for \(r\):

\[
2^r+44=76
\]

\[
2^r=32=2^5.
\]

Thus,

\[
r=5.
\]

Finally, solve for \(s\):

\[
5^3+6^s=1421
\]

\[
125+6^s=1421
\]

\[
6^s=1296=6^4.
\]

Thus,

\[
s=4.
\]

Therefore,

\[
prs=2\cdot 5\cdot 4=40.
\]

For example, \(2\cdot 5\cdot 3=30\) would incorrectly use \(s=3\), but \(6^3=216\), not \(1296\).

ANSWER 2: B

Problem 3:

We need sets of three different numbers from \(\{1,2,3,4,5,6,7,8,9\}\) that have sum 15 and contain 5.

If one number is 5, the other two numbers must add to

\[
15-5=10.
\]

The distinct pairs from the set that add to 10 are:

\[
1+9,\quad 2+8,\quad 3+7,\quad 4+6.
\]

So the possible sets are

\[
\{1,5,9\},\ \{2,5,8\},\ \{3,5,7\},\ \{4,5,6\}.
\]

There are \(4\) sets.

The pair \(5+5\) is not allowed because the three numbers must be different.

ANSWER 3: B

Problem 4:

The listed percentages are

\[
30\%+20\%+15\%+10\%=75\%.
\]

Therefore, green gumdrops make up the remaining

\[
100\%-75\%=25\%.
\]

We are told that 25% equals 30 green gumdrops, so the total number of gumdrops is

\[
30\div 0.25=120.
\]

Initially, the number of blue gumdrops is

\[
30\%\text{ of }120=36,
\]

and the number of brown gumdrops is

\[
20\%\text{ of }120=24.
\]

Half of the blue gumdrops are replaced with brown gumdrops:

\[
\frac12\cdot 36=18.
\]

So the new number of brown gumdrops is

\[
24+18=42.
\]

A common mistake is to take 30 green gumdrops as 30% of the jar, but green represents 25%, not 30%.

ANSWER 4: C

Problem 5:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew 20%, and Shea is now 60 inches tall:

\[
1.2h=60.
\]

Thus,

\[
h=60\div 1.2=50.
\]

So Shea grew

\[
60-50=10\text{ inches}.
\]

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5\text{ inches}.
\]

Therefore, Ara is now

\[
50+5=55\text{ inches tall}.
\]

It would be incorrect to make Ara grow 10% of her original height without first finding Shea’s actual growth in inches.

ANSWER 5: E

Problem 6:

Let the small box contain \(q\) units of detergent and cost \(p\) dollars.

The large box contains twice as much as the small box:

\[
L=2q.
\]

The medium box contains 20% less detergent than the large box:

\[
M=80\%\text{ of }2q=1.6q.
\]

For prices:

- Small costs \(p\).
- Medium costs 50% more than small:

\[
M_{\text{price}}=1.5p.
\]

- Large costs 30% more than medium:

\[
L_{\text{price}}=1.3(1.5p)=1.95p.
\]

Now compare cost per unit of detergent:

\[
\text{Small: }\frac{p}{q}=1\cdot\frac pq,
\]

\[
\text{Medium: }\frac{1.5p}{1.6q}=0.9375\cdot\frac pq,
\]

\[
\text{Large: }\frac{1.95p}{2q}=0.975\cdot\frac pq.
\]

Lower cost per unit means a better buy. Therefore:

\[
\text{Medium is best, Large is next, Small is worst.}
\]

So the ranking is

\[
M,\ L,\ S.
\]

Although large contains the most detergent, it is not the best value because its price per unit is higher than medium’s.

ANSWER 6: E

Problem 7:

We multiply by \(\frac34\), then divide by \(\frac35\).

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac34\div\frac35
=
\frac34\cdot\frac53.
\]

Then

\[
\frac34\cdot\frac53=\frac54.
\]

So the combined effect is multiplying by

\[
\frac54.
\]

A common error is to multiply by \(\frac45\), but that would reverse the reciprocal incorrectly.

ANSWER 7: E

Problem 8:

The number

\[
x=0.00\ldots 01
\]

is positive but extremely small.

Compare the choices:

- \(3+x\) is just a tiny amount bigger than 3.
- \(3-x\) is just a tiny amount smaller than 3.
- \(3x\) is extremely small.
- \(\frac{x}{3}\) is even smaller.
- \(\frac3x\) divides 3 by an extremely tiny positive number, making the result extremely large.

Therefore,

\[
\frac3x
\]

is the largest expression.

A tempting thought is that \(3+x\) should be largest because it adds something to 3, but dividing by such a tiny number produces a vastly larger number.

ANSWER 8: D

Problem 9:

Initially, 80% of 500 balls are red:

\[
0.80\cdot 500=400.
\]

So there are 400 red balls.

Let \(n\) red balls be removed. Then:

- remaining red balls: \(400-n\),
- remaining total balls: \(500-n\).

We want 75% of the remaining balls to be red:

\[
\frac{400-n}{500-n}=0.75.
\]

Solve:

\[
400-n=0.75(500-n)
\]

\[
400-n=375-0.75n
\]

\[
25=0.25n
\]

\[
n=100.
\]

Removing 100 red balls leaves 300 red balls out of 400 total balls:

\[
\frac{300}{400}=75\%.
\]

ANSWER 9: D

Problem 10:

The team plays

\[
50+40=90
\]

games total.

To win exactly 70% of 90 games, the team must win

\[
0.70\cdot 90=63
\]

games.

The team already won 40 games, so it must win

\[
63-40=23
\]

of the remaining games.

A tempting error is to find 70% of just the remaining 40 games, but the 70% goal is for all 90 games of the season.

ANSWER 10: B