🔬 Model Lab

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad

Problem 1:
We need paths from (0,0) to (3,2) using only East and North steps.
To go 3 east and 2 north needs 5 blocks in total.

Without restriction, any order of 3 E's and 2 N's works.
Number = choose positions of the 2 N's among 5 steps:
C(5,2) = 10. This is choice E, the tempting total.

We must avoid (1,1), one east and one north of start.

Count paths that DO go through (1,1) to subtract them.

From (0,0) to (1,1): need 1 E, 1 N.
Number = C(2,1) = 2.

From (1,1) to (3,2): need 2 more E, 1 more N.
Number = C(3,1) = 3, choosing position of the N.

So paths through forbidden point = 2 * 3 = 6.

Allowed paths = all paths - bad paths = 10 - 6 = 4.

Check: 8 would be double counting, 5,6 ignore correct subtraction.
ANSWER 1: A

Problem 2:
Each child boy/girl with probability 1/2. There are 2^4 = 16 equally likely boy-girl sequences.

Count sequences for each type:

A. all 4 are boys: only BBBB. 1 sequence. Prob = 1/16.

B. all 4 are girls: only GGGG. 1 sequence. Prob = 1/16.

C. 2 girls and 2 boys: number = C(4,2) = 6 sequences. Prob = 6/16.

D. 3 of one gender and 1 of the other: this includes TWO cases:
 3 boys 1 girl: C(4,3)=4 sequences,
 3 girls 1 boy: C(4,3)=4 sequences.
 Total = 8 sequences. Prob = 8/16 = 1/2.

So D > C > A = B. E is false because probabilities differ.

The trap is thinking the balanced 2-2 is most likely, but D lumps both 3-1 and 1-3 together.
ANSWER 2: D

Problem 3:
Let total marbles = T.
Blue = T/3, Red = T/4 must be whole numbers.
So T is divisible by 3 and 4, hence by 12. Write T = 12k for integer k >=1.

Then blue + red = T/3 + T/4 = 7T/12 = 7k.
Green = 6.
Yellow = T - blue - red - green = T - 7T/12 - 6 = 5T/12 - 6 = 5k - 6.

Yellow cannot be negative, so 5k - 6 >= 0.
k=1 gives 5-6 = -1, impossible since total would be 12 but 4+3+6=13 >12.
So k >= 2.

For k=2: T=24, blue=8, red=6, green=6, yellow = 5*2-6=4.
This works: 8+6+6+4=24.

Larger k gives larger yellow, so smallest possible yellow is 4.

Choices 1,2,3 are too small to make the total work.
ANSWER 3: D

Problem 4:
Let the number be n.
Given: 5 times a number is 2, i.e. 5n = 2.

So n = 2/5.
Reciprocal of n = 1/n = 5/2 = 2.5.

100 times the reciprocal = 100 * (5/2) = 500/2 = 250.

Note: 2.5 is just the reciprocal, not times 100. 40,50,500 are arithmetic errors.
ANSWER 4: D

Problem 5:
We need n divisible by its units digit. Units digit 0 is impossible because division by 0 is undefined, so all numbers ending in 0 fail.

Check 10-49 (10 and 50 ending in 0 fail anyway, so endpoints don't matter):

10s: 11,12,15 work. 13,14,16,17,18,19 fail. 10 fails. => 3
20s: 21=21/1, 22/2=11, 24/4=6, 25/5=5 work. => 4
30s: 31/1, 32/2, 33/3=11, 35/5, 36/6=6 work. => 5
40s: 41/1, 42/2, 44/4=11, 45/5, 48/8=6 work. => 5

Total = 3+4+5+5 = 17.

E.g. 15 works but 16 does not since 16 is not divisible by 6, etc.
ANSWER 5: C

Problem 6:
Suppose the correct score has tens digit t and units digit u, e.g. value ...10t+u.
Reversed it becomes ...10u+t.

Difference = (10t+u) - (10u+t) = 9t - 9u = 9(t-u).
Absolute error = 9 times |t-u|.

Hence the error must be a multiple of 9.

Also |t-u| <= 9, so error is 0,9,18,27,36,45,54,63,72,81.

Of the choices:
45 is divisible by 9,
46,47,48,49 are not.

So only 45 could occur, e.g. reversing 61 to 16 differs by 45.
ANSWER 6: A

Problem 7:
With n teams, each pair plays once.
Number of games = C(n,2) = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try n=7: 7*6 = 42. Works.
n=6 gives 15 games, n=8 gives 28 games.

So 7 teams.
ANSWER 7: B

Problem 8:
Let sides be integers l,w >0.
Perimeter = 2(l+w) = 50, so l+w = 25, w = 25-l.

Area = l*w = l(25-l).

To maximize with fixed sum, make sides as equal as possible. Since 25 is odd, closest integers are 12 and 13.
Max area = 12*13 = 156.

To minimize, make sides as unequal as possible: smallest positive integer is 1, so 1 and 24.
Min area = 1*24 = 24. Sides cannot be 0.

Difference = 156 - 24 = 132.

76,120,128,136 come from wrong extremes.
ANSWER 8: D

Problem 9:
Mean of 4 numbers is 85, so sum of 4 = 4*85 = 340.

Largest is 97. Sum of remaining 3 = 340 - 97 = 243.

Mean of remaining 3 = 243/3 = 81.0.

Choices around 82-84 come from subtraction or division errors.
ANSWER 9: A

Problem 10:
We need ceiling of 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5.

Integer parts: 2+3+4+5 = 14.
Fraction parts: 1/2 + 1/3 + 1/4 + 1/5.

Common denominator 60: = 30/60 + 20/60 + 15/60 + 12/60 = 77/60 = 1 + 17/60.

So total = 14 + 1 + 17/60 = 15 + 17/60, about 15.28.

Whole numbers larger than this start at 16. 15 is smaller, 14 is far smaller.

So smallest whole number larger is 16.
ANSWER 10: C

Problem 1:
Jack wants to bike from his house at coordinates $(0,0)$ to Jill's house at $(3,2)$, biking along the grid for a total of $3 + 2 = 5$ blocks (3 blocks East and 2 blocks North). He must avoid the dangerous intersection at $(1,1)$.

First, we determine the total number of 5-block paths from $(0,0)$ to $(3,2)$ with no restrictions:
$$\binom{3+2}{2} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \text{ paths.}$$

Next, we count how many of these paths pass through the dangerous intersection at $(1,1)$:
- Paths from $(0,0)$ to $(1,1)$: moving 1 block East and 1 block North gives $\binom{1+1}{1} = \binom{2}{1} = 2$ ways.
- Paths from $(1,1)$ to $(3,2)$: moving $3 - 1 = 2$ blocks East and $2 - 1 = 1$ block North gives $\binom{2+1}{1} = \binom{3}{1} = 3$ ways.

By the multiplication principle, the number of paths that pass through $(1,1)$ is:
$$2 \times 3 = 6 \text{ paths.}$$

Subtracting the forbidden paths from the total gives the number of safe paths:
$$10 - 6 = 4 \text{ paths.}$$

ANSWER 1: A

---

Problem 2:
We are considering all possible gender outcomes for four children. Since each child is equally likely to be a boy (B) or a girl (G), there are $2^4 = 16$ equally likely possible sequences of births.

Let us find the probability of each outcome listed in the choices:
- Outcome A (all 4 are boys): There is only 1 sequence (BBBB), so the probability is $\frac{1}{16}$.
- Outcome B (all 4 are girls): There is only 1 sequence (GGGG), so the probability is $\frac{1}{16}$.
- Outcome C (2 are girls and 2 are boys): The number of sequences is $\binom{4}{2} = 6$, so the probability is $\frac{6}{16}$.
- Outcome D (3 are of one gender and 1 is of the other): This can happen as 3 boys and 1 girl ($\binom{4}{3} = 4$ ways) or 3 girls and 1 boy ($\binom{4}{1} = 4$ ways), giving a total of $4 + 4 = 8$ sequences. The probability is $\frac{8}{16} = \frac{1}{2}$.

Comparing the counts, Outcome D has 8 favorable outcomes out of 16, which is strictly greater than Outcome C (6 outcomes) and Outcomes A and B (1 outcome each). Therefore, Outcome D is the most likely.

ANSWER 2: D

---

Problem 3:
Let $N$ be the total number of marbles Marcy has.
- Blue marbles: $B = \frac{1}{3}N$
- Red marbles: $R = \frac{1}{4}N$
- Green marbles: $G = 6$
- Yellow marbles: $Y$

Since the count of blue and red marbles must be integers, $N$ must be a multiple of $\text{lcm}(3, 4) = 12$.

The total number of marbles is:
$$N = B + R + G + Y = \frac{1}{3}N + \frac{1}{4}N + 6 + Y$$
$$N = \frac{7}{12}N + 6 + Y$$
$$Y = \frac{5}{12}N - 6$$

Since $N$ is a positive multiple of 12:
- If $N = 12$, then $Y = \frac{5}{12}(12) - 6 = 5 - 6 = -1$, which is impossible because the number of marbles cannot be negative.
- If $N = 24$, then $Y = \frac{5}{12}(24) - 6 = 10 - 6 = 4$, which is a valid number of marbles.
- If $N = 36$, then $Y = \frac{5}{12}(36) - 6 = 15 - 6 = 9$.

The smallest possible number of yellow marbles Marcy could have is 4.

ANSWER 3: D

---

Problem 4:
Let the unknown number be $x$.
We are given that:
$$5x = 2 \implies x = \frac{2}{5}$$

The reciprocal of $x$ is:
$$\frac{1}{x} = \frac{5}{2}$$

We are asked to find 100 times the reciprocal of the number:
$$100 \times \frac{1}{x} = 100 \times \frac{5}{2} = 50 \times 5 = 250$$

ANSWER 4: D

---

Problem 5:
We need to count how many whole numbers strictly between 10 and 50 (i.e., from 11 to 49) are divisible by their units digit. Note that division by 0 is undefined, so numbers with a units digit of 0 (20, 30, 40) do not have this property.

We test each decade:
- **Teens (11 to 19):**
  - $11 \div 1 = 11$ (Yes)
  - $12 \div 2 = 6$ (Yes)
  - $15 \div 5 = 3$ (Yes)
  - Other numbers (13, 14, 16, 17, 18, 19) are not divisible by their units digits.
  - Count: 3 numbers.

- **Twenties (21 to 29):**
  - $21 \div 1 = 21$ (Yes)
  - $22 \div 2 = 11$ (Yes)
  - $24 \div 4 = 6$ (Yes)
  - $25 \div 5 = 5$ (Yes)
  - Others (23, 26, 27, 28, 29) are not.
  - Count: 4 numbers.

- **Thirties (31 to 39):**
  - $31 \div 1 = 31$ (Yes)
  - $32 \div 2 = 16$ (Yes)
  - $33 \div 3 = 11$ (Yes)
  - $35 \div 5 = 7$ (Yes)
  - $36 \div 6 = 6$ (Yes)
  - Others (34, 37, 38, 39) are not.
  - Count: 5 numbers.

- **Forties (41 to 49):**
  - $41 \div 1 = 41$ (Yes)
  - $42 \div 2 = 21$ (Yes)
  - $44 \div 4 = 11$ (Yes)
  - $45 \div 5 = 9$ (Yes)
  - $48 \div 8 = 6$ (Yes)
  - Others (43, 46, 47, 49) are not.
  - Count: 5 numbers.

Adding these up:
$$3 + 4 + 5 + 5 = 17 \text{ numbers.}$$

ANSWER 5: C

---

Problem 6:
Let the original two-digit score have tens digit $t$ and units digit $u$.
The value of the original score is:
$$10t + u$$

When Clara reversed the digits, the score became:
$$10u + t$$

The difference between the incorrect sum and the correct sum is equal to the difference between these two values:
$$|(10t + u) - (10u + t)| = |9t - 9u| = 9|t - u|$$

Since $t$ and $u$ are single-digit integers, the difference must be a multiple of 9.
Looking at the choices:
- A: $45 = 9 \times 5$ (a multiple of 9)
- B: 46 (not divisible by 9)
- C: 47 (not divisible by 9)
- D: 48 (not divisible by 9)
- E: 49 (not divisible by 9)

Only 45 is a multiple of 9.

ANSWER 6: A

---

Problem 7:
Let $n$ be the number of teams in the BIG N conference.
Since every team plays each other team exactly once, the total number of games played is given by the combination formula:
$$\binom{n}{2} = \frac{n(n - 1)}{2}$$

We are given that 21 conference games were played:
$$\frac{n(n - 1)}{2} = 21$$
$$n(n - 1) = 42$$

Since $n$ must be a positive integer, we look for two consecutive integers whose product is 42:
$$7 \times 6 = 42 \implies n = 7$$

There were 7 teams in the BIG N conference.

ANSWER 7: B

---

Problem 8:
Let the length and width of the rectangle be integers $L$ and $W$.
The perimeter is given by:
$$2(L + W) = 50 \implies L + W = 25$$

The area of the rectangle is:
$$A = L \times W = W(25 - W)$$

Since $L$ and $W$ are positive integers, without loss of generality we can assume $1 \le W \le 12$:
- To **minimize** the are

## Problem 1:
Jack must travel 3 blocks east and 2 blocks north — 5 blocks total.

**Total paths (ignoring the dangerous intersection):**
This is a path of 5 moves where we choose which 2 are north:
$$\binom{5}{2} = 10$$

**Paths through the bad intersection (1 east, 1 north from home):**
- Ways to reach it: $\binom{2}{1} = 2$
- Ways to go from there to Jill's house (remaining 2 east, 1 north): $\binom{3}{1} = 3$
- Paths to subtract: $2 \times 3 = 6$

**Valid paths:** $10 - 6 = 4$

Check: This rules out the tempting answer E (10), which forgets the forbidden intersection.

ANSWER 1: A

## Problem 2:
Each child is independently a boy or girl with probability 1/2, so there are $2^4 = 16$ equally likely gender sequences.

**Count each outcome:**
- All 4 boys: 1 way → probability 1/16
- All 4 girls: 1 way → probability 1/16
- 2 girls and 2 boys: $\binom{4}{2} = 6$ ways → probability 6/16
- 3 of one gender, 1 of the other: $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ ways → probability 8/16

The outcome "3 of one gender and 1 of the other" has 8 favorable sequences, the most likely.

ANSWER 2: D

## Problem 3:
Let the total number of marbles be $N$. Blue = $N/3$, Red = $N/4$, so $N$ must be divisible by 3 and 4, i.e., by 12. Write $N = 12k$.

**Green + Yellow marbles:** These make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total:
$$\text{Green} + \text{Yellow} = \frac{5}{12}(12k) = 5k$$

Since Green = 6:
$$\text{Yellow} = 5k - 6$$

**Find smallest valid k:** We need Yellow ≥ 1 (at least one yellow, since "smallest number" implies yellows exist; also Yellow must be a non-negative integer).
- $k = 1$: Yellow = $5 - 6 = -1$ (impossible)
- $k = 2$: Yellow = $10 - 6 = 4$ ✓ (check: $N = 24$, blue = 8, red = 6, green = 6, yellow = 4 ✓)

ANSWER 3: D

## Problem 4:
Let the number be $x$. We're told $5x = 2$, so:
$$x = \frac{2}{5}$$

The reciprocal of the number is:
$$\frac{1}{x} = \frac{5}{2}$$

Then:
$$100 \times \frac{5}{2} = \frac{500}{2} = 250$$

(Check: a common error is computing $100 \times \frac{2}{5} = 40$, which is answer B — but the problem asks for the reciprocal.)

ANSWER 4: D

## Problem 5:
A number is divisible by its units digit (the units digit can't be 0, since division by 0 is undefined). Check each possible units digit for numbers 10–50:

- **Units 1** (divide by 1, always true): 11, 21, 31, 41 → 4
- **Units 2** (must be even, always true): 12, 22, 32, 42 → 4
- **Units 3** (must be divisible by 3): only 33 → 1
- **Units 4** (must be divisible by 4): 24, 44 → 2
- **Units 5** (must end in 5, always true): 15, 25, 35, 45 → 4
- **Units 6** (must be even AND divisible by 3): 36 → 1
- **Units 7** (must be divisible by 7): 49 → 1
- **Units 8** (must be divisible by 8): 48 → 1
- **Units 9** (must be divisible by 9): multiples of 9 are 18, 27, 36, 45 — none end in 9 → 0
- **Units 0**: undefined → 0

**Total:** $4 + 4 + 1 + 2 + 4 + 1 + 1 + 1 = 18$

ANSWER 5: D

## Problem 6:
Suppose the score had tens digit $a$ and units digit $b$, so Clara wrote $10b + a$ instead of $10a + b$.

**The error in the sum:**
$$(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$$

The difference must be a multiple of 9.

**Check the options:**
- 45 = 9 × 5 ✓
- 46: not divisible by 9 ✗
- 47: not divisible by 9 ✗
- 48: not divisible by 9 ✗
- 49: not divisible by 9 ✗

Only 45 can result from reversing digits (e.g., 94 vs. 49).

ANSWER 6: A

## Problem 7:
With $n$ teams where each pair plays exactly once, the total number of games is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$

**Set up the equation:**
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42 \implies n^2 - n - 42 = 0 \implies (n-7)(n+6) = 0$$

So $n = 7$ (rejecting the negative root).

**Check:** $\binom{7}{2} = \frac{7 \cdot 6}{2} = 21$ ✓

ANSWER 7: B

## Problem 8:
Perimeter 50 means:
$$2(l + w) = 50 \implies l + w = 25$$

With integer sides, the dimensions are $1 \times 24, 2 \times 23, \ldots, 12 \times 13$.

**Largest area** (as close to a square as possible): $12 \times 13 = 156$

**Smallest area** (most elongated): $1 \times 24 = 24$

**Difference:**
$$156 - 24 = 132$$

ANSWER 8: D

## Problem 9:
The mean of four numbers is 85, so their sum is:
$$4 \times 85 = 340$$

The largest number is 97, so the sum of the remaining three numbers is:
$$340 - 97 = 243$$

**Mean of the remaining three:**
$$\frac{243}{3} = 81.0$$

ANSWER 9: A

## Problem 10:
Each term is a mixed number: $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

**Separate whole parts and fractions:**
- Whole parts: $2 + 3 + 4 + 5 = 14$
- Fraction parts: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$

**Compute the fractions (common denominator 60):**
$$\frac{30 + 20 + 15 + 12}{60} = \frac{77}{60} = 1\frac{17}{60}$$

**Total sum:**
$$14 + \frac{77}{60} = 14 + 1.283\ldots = 15.283\ldots$$

The smallest whole number **larger** than this sum is 16.

(Note: A tempting trap is answer B (15), but 15 is *less than* the sum, and we need a whole number strictly greater.)

ANSWER 10: C

Problem 1:
Jack must go 3 blocks east and 2 blocks north, for a total of 5 blocks. Without any restrictions, the number of shortest paths is the number of ways to arrange 3 E’s and 2 N’s:

\[
\binom{5}{2}=10.
\]

He must avoid the intersection one block east and one block north, which is point \((1,1)\). Count paths that pass through \((1,1)\).

From Jack’s house to \((1,1)\), he needs 1 E and 1 N:

\[
\binom{2}{1}=2
\]

ways.

From \((1,1)\) to Jill’s house at \((3,2)\), he needs 2 E’s and 1 N:

\[
\binom{3}{1}=3
\]

ways.

So the number of forbidden paths is

\[
2\cdot 3=6.
\]

Therefore, the number of safe paths is

\[
10-6=4.
\]

ANSWER 1: A

Problem 2:
Each child is independently equally likely to be a boy or a girl, so there are

\[
2^4=16
\]

equally likely gender sequences.

Now count the sequences in each listed outcome.

- A. all 4 are boys: exactly 1 sequence.
- B. all 4 are girls: exactly 1 sequence.
- C. 2 girls and 2 boys:

\[
\binom{4}{2}=6
\]

sequences.
- D. 3 of one gender and 1 of the other: this includes 3 boys and 1 girl, or 3 girls and 1 boy.

\[
\binom{4}{3}+\binom{4}{1}=4+4=8
\]

sequences.

So the probabilities are:

\[
A=\frac1{16},\quad B=\frac1{16},\quad C=\frac6{16},\quad D=\frac8{16}.
\]

Thus the most likely outcome is D.

ANSWER 2: D

Problem 3:
Let the total number of marbles be \(N\). Since one third are blue and one fourth are red, \(N\) must be divisible by both 3 and 4, so \(N\) is a multiple of 12.

Write

\[
N=12k
\]

for some positive integer \(k\).

Then the number of blue marbles is

\[
\frac{12k}{3}=4k,
\]

the number of red marbles is

\[
\frac{12k}{4}=3k,
\]

and the number of green marbles is 6. Therefore the number of yellow marbles is

\[
12k-4k-3k-6=5k-6.
\]

This must be at least 0:

\[
5k-6\ge 0 \implies k\ge 2.
\]

The smallest possible value occurs at \(k=2\):

\[
5(2)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4.

ANSWER 3: D

Problem 4:
Let the number be \(x\). We are told

\[
5x=2.
\]

So

\[
x=\frac25.
\]

The reciprocal of \(x\) is

\[
\frac1x=\frac52.
\]

Then 100 times the reciprocal is

\[
100\cdot \frac52=50\cdot 5=250.
\]

ANSWER 4: D

Problem 5:
We need whole numbers between 10 and 50 that are divisible by their units digit. A number with units digit 0 cannot be used, since division by 0 is undefined.

A two-digit number with tens digit \(a\) and units digit \(b\) is \(10a+b\). It is divisible by \(b\) exactly when

\[
10a+b \equiv 10a \pmod b,
\]

so we need

\[
b\mid 10a.
\]

Check each tens digit.

- 10s: \(a=1\). Need \(b\mid 10\). Possible \(b=1,2,5\), giving 11, 12, 15: 3 numbers.
- 20s: \(a=2\). Need \(b\mid 20\). Possible \(b=1,2,4,5\), giving 21, 22, 24, 25: 4 numbers.
- 30s: \(a=3\). Need \(b\mid 30\). Possible \(b=1,2,3,5,6\), giving 31, 32, 33, 35, 36: 5 numbers.
- 40s: \(a=4\). Need \(b\mid 40\). Possible \(b=1,2,4,5,8\), giving 41, 42, 44, 45, 48: 5 numbers.
- 50: units digit is 0, so it does not count.

Total:

\[
3+4+5+5=17.
\]

ANSWER 5: C

Problem 6:
Suppose the original score has tens digit \(a\) and units digit \(b\). The original score is

\[
10a+b.
\]

After reversing the tens and units digits, the incorrect score is

\[
10b+a.
\]

The difference is

\[
(10b+a)-(10a+b)=9b-9a=9(b-a).
\]

So any possible difference must be a multiple of 9.

Among the choices,

\[
45,\ 46,\ 47,\ 48,\ 49,
\]

only 45 is a multiple of 9. It could occur, for example, if a score of 16 was written as 61.

ANSWER 6: A

Problem 7:
If there are \(n\) teams, and every pair of teams plays exactly once, the number of games is the number of ways to choose 2 teams from \(n\):

\[
\binom{n}{2}=\frac{n(n-1)}2.
\]

We are told this equals 21:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Now

\[
7\cdot 6=42,
\]

so

\[
n=7.
\]

ANSWER 7: B

Problem 8:
Let the rectangle have side lengths \(x\) and \(y\), where \(x\) and \(y\) are positive integers. The perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
A=xy=x(25-x).
\]

To find the largest possible area, choose integer side lengths as close as possible to each other. Since \(x+y=25\), the closest pair is 12 and 13:

\[
A_{\max}=12\cdot 13=156.
\]

To find the smallest possible area, choose the most unbalanced positive integer side lengths: 1 and 24:

\[
A_{\min}=1\cdot 24=24.
\]

The difference is

\[
156-24=132.
\]

ANSWER 8: D

Problem 9:
The average of four numbers is 85, so their total sum is

\[
4\cdot 85=340.
\]

The largest of the four numbers is 97. Therefore, the sum of the remaining three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

ANSWER 9: A

Problem 10:
The expression is the mixed-number sum

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Separate the whole-number parts and fractional parts:

\[
(2+3+4+5)+\left(\frac12+\frac13+\frac14+\frac15\right).
\]

The whole-number sum is

\[
2+3+4+5=14.
\]

Now compute the fractions. Use denominator 60:

\[
\frac12=\frac{30}{60},\quad
\frac13=\frac{20}{60},\quad
\frac14=\frac{15}{60},\quad
\frac15=\frac{12}{60}.
\]

So

\[
\frac12+\frac13+\frac14+\frac15
=\frac{30+20+15+12}{60}
=\frac{77}{60}
=1+\frac{17}{60}.
\]

Therefore the total is

\[
14+1+\frac{17}{60}=15+\frac{17}{60}.
\]

This is between 15 and 16. The smallest whole number larger than the sum is 16.

ANSWER 10: C

# Problem 1:

Jack must go 3 blocks east (E) and 2 blocks north (N), i.e. arrange 5 letters: EEENN.

**Total paths:** C(5,2) = 10.

**Bad paths** (through the intersection 1 east, 1 north):
- House → (1,1): sequences EN or NE → 2 ways
- (1,1) → (3,2): needs 2 E's and 1 N → C(3,1) = 3 ways
- Total bad: 2 × 3 = 6

**Safe paths:** 10 − 6 = 4.

(Choice E is the total ignoring the danger; C counts the bad paths instead.)

ANSWER 1: A

# Problem 2:

There are 2⁴ = 16 equally likely gender sequences.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: C(4,2) = 6 ways → 6/16
- 3 of one gender, 1 of the other: C(4,3) = 4 (3 boys, 1 girl) + 4 (3 girls, 1 boy) = 8 ways → 8/16

The largest probability is 8/16 for "3 of one gender and 1 of the other." (Choice C, 6/16, is the tempting answer, but it loses because option D combines two cases.)

ANSWER 2: D

# Problem 3:

Let the total be $n$. Blue = n/3, red = n/4, so blue + red = 7n/12, and green + yellow = 5n/12. Since n/3 and n/4 must be whole numbers, n is a multiple of 12.

Green + yellow = 5n/12 must be at least 6 + 1 (need green = 6 and yellow ≥ 1... actually yellow ≥ 0, but we need 5n/12 ≥ 6):

- n = 12: 5n/12 = 5 < 6 — impossible.
- n = 24: 5n/12 = 10, so yellow = 10 − 6 = 4. ✓ (blue 8, red 6, green 6, yellow 4 — total 24)

Smallest possible yellow count is 4.

ANSWER 3: D

# Problem 4:

5x = 2 → x = 2/5.

Reciprocal: 1/x = 5/2 = 2.5.

100 × 2.5 = 250.

(Choice A is just the reciprocal; E would come from using 5 instead of 5/2.)

ANSWER 4: D

# Problem 5:

Check numbers 11–49; a number "n" with units digit d works if d divides n (d = 0 is impossible).

- Units 1: 11, 21, 31, 41 → all work (4)
- Units 2: 12, 22, 32, 42 → all even (4)
- Units 3: 33 only (1)
- Units 4: 24, 44 (2)
- Units 5: 15, 25, 35, 45 (4)
- Units 6: 36 only (1)
- Units 7: none (7, 77 are the only such multiples) (0)
- Units 8: 48 only (1)
- Units 9: none (9, 99) (0)

Total: 4+4+1+2+4+1+0+1+0 = **17**

ANSWER 5: C

# Problem 6:

A two-digit score 10a + b written as 10b + a differs by
(10a + b) − (10b + a) = 9(a − b).

So the error must be a multiple of 9. Among 45, 46, 47, 48, 49, only **45** = 9 × 5 is a multiple of 9 (and it's achievable, e.g. 61 → 16).

ANSWER 6: A

# Problem 7:

With n teams each playing every other once, the number of games is C(n,2) = n(n−1)/2.

n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7 × 6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 — neither is 21.

ANSWER 7: B

# Problem 8:

Perimeter 50 → length + width = 25, with positive integers.

- Largest area: sides as close as possible: 12 × 13 = 156
- Smallest area: sides as far apart as possible: 1 × 24 = 24

Difference: 156 − 24 = **132**.

(Choice E, 136, comes from wrongly using 12.5² or 2×23; 128 comes from mis-taking min as 28.)

ANSWER 8: D

# Problem 9:

Sum of the four numbers = 4 × 85 = 340.

Remove the largest: 340 − 97 = 243.

Mean of remaining three: 243 ÷ 3 = 81.0.

ANSWER 9: A

# Problem 10:

The sum is 2½ + 3⅓ + 4¼ + 5⅕.

Whole parts: 2 + 3 + 4 + 5 = 14.

Fractions: 1/2 + 1/3 + 1/4 + 1/5 = (30 + 20 + 15 + 12)/60 = 77/60 ≈ 1.28.

Total ≈ 15.28. The smallest whole number larger than this is **16**.

(Choice B, 15, fails since the fractions add to more than 1.)

ANSWER 10: C

Problem 1:

Jack must travel 3 blocks east and 2 blocks north, for a total of 5 blocks.

Without the dangerous intersection, the number of routes would be the number of ways to arrange 3 E's and 2 N's:

\[
\binom{5}{2}=10.
\]

However, he must avoid the intersection 1 block east and 1 block north, namely \((1,1)\).

To reach that dangerous intersection, Jack must travel one east and one north:

\[
EN \text{ or } NE,
\]

so there are 2 ways to get there.

From \((1,1)\), he must still travel 2 east and 1 north. The number of ways to do that is

\[
\binom{3}{1}=3.
\]

Thus, the number of bad routes is

\[
2\cdot 3=6.
\]

Therefore, the number of safe routes is

\[
10-6=4.
\]

Choices B, C, D, and E do not account correctly for all six routes that pass through the dangerous intersection.

ANSWER 1: A

Problem 2:

Each of the four children can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so probability is \(\frac{1}{16}\).
- All 4 girls: only \(GGGG\), so probability is \(\frac{1}{16}\).
- Exactly 2 girls and 2 boys: choose which 2 of the 4 are girls:

\[
\binom{4}{2}=6,
\]

so probability is \(\frac{6}{16}\).

- Exactly 3 of one gender and 1 of the other:
  - 3 boys and 1 girl: \(\binom41=4\) ways.
  - 3 girls and 1 boy: \(\binom41=4\) ways.

So there are

\[
4+4=8
\]

ways, with probability

\[
\frac{8}{16}.
\]

Since \(\frac{8}{16}\) is greater than \(\frac{6}{16}\) and \(\frac{1}{16}\), choice D is most likely. Choice E is false because these outcomes do not have equal probabilities.

ANSWER 2: D

Problem 3:

Let the total number of marbles be \(N\).

Blue marbles:

\[
\frac13N
\]

Red marbles:

\[
\frac14N
\]

Since these must both be whole numbers, \(N\) must be divisible by both 3 and 4. Therefore, \(N\) must be a multiple of 12.

The number of yellow marbles is

\[
N-\frac13N-\frac14N-6.
\]

Combine the blue and red fractions:

\[
\frac13+\frac14=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}.
\]

Thus yellow marbles equal

\[
N-\frac{7}{12}N-6=\frac{5}{12}N-6.
\]

Try the smallest possible multiple of 12:

\[
N=12.
\]

Then yellow marbles would be

\[
\frac{5}{12}(12)-6=5-6=-1,
\]

which is impossible.

The next multiple of 12 is \(24\). Then:

\[
\frac{5}{12}(24)-6=10-6=4.
\]

So the smallest possible number of yellow marbles is 4. Choices A, B, and C are too small, while E is possible only with more marbles than necessary.

ANSWER 3: D

Problem 4:

Let the number be \(x\).

We are told:

\[
5x=2.
\]

Divide by 5:

\[
x=\frac25.
\]

Its reciprocal is

\[
\frac{1}{x}=\frac52.
\]

Therefore,

\[
100\left(\frac52\right)=50\cdot 5=250.
\]

Choices A, B, and C result from incorrect arithmetic or using the original number instead of its reciprocal. Choice E is twice the correct value.

ANSWER 4: D

Problem 5:

A number must be divisible by its units digit. Its units digit cannot be 0, since division by 0 is impossible.

Consider numbers from 10 through 49.

- Units digit 1: Every number ending in 1 is divisible by 1.

\[
11,21,31,41
\]

There are 4.

- Units digit 2: Every even number ending in 2 is divisible by 2.

\[
12,22,32,42
\]

There are 4.

- Units digit 3:

\[
33
\]

is divisible by 3. There is 1.

- Units digit 4:

\[
24,44
\]

are divisible by 4. There are 2.

- Units digit 5: Every number ending in 5 is divisible by 5.

\[
15,25,35,45
\]

There are 4.

- Units digit 6:

\[
36
\]

is divisible by 6. There is 1.

- Units digit 7: None of \(17,27,37,47\) is divisible by 7. There are 0.

- Units digit 8:

\[
48
\]

is divisible by 8. There is 1.

- Units digit 9: None of \(19,29,39,49\) is divisible by 9. There are 0.

The total is

\[
4+4+1+2+4+1+0+1+0=17.
\]

Thus choices A and B are too small, and D and E count too many numbers.

ANSWER 5: C

Problem 6:

Suppose the tens digit is \(a\) and the units digit is \(b\).

The correct score is

\[
10a+b.
\]

After reversing the digits, Clara uses

\[
10b+a.
\]

The difference is

\[
(10a+b)-(10b+a)=9a-9b=9(a-b).
\]

Therefore, the difference must be a multiple of 9.

Check the choices:

\[
45=9\cdot 5
\]

is a multiple of 9.

But

\[
46,47,48,49
\]

are not multiples of 9.

So the incorrect sum might differ from the correct sum by 45.

ANSWER 6: A

Problem 7:

If there are \(n\) teams and every pair of teams plays exactly once, then the number of games is

\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

For comparison, 6 teams would give only \(\binom62=15\) games, while 8 teams would give \(\binom82=28\) games.

ANSWER 7: B

Problem 8:

Let the integer side lengths of the rectangle be \(x\) and \(y\).

Its perimeter is 50, so

\[
2x+2y=50.
\]

Divide by 2:

\[
x+y=25.
\]

The area is

\[
xy.
\]

To get the largest area with a fixed sum, the side lengths should be as close together as possible. The closest integers adding to 25 are 12 and 13:

\[
12\cdot 13=156.
\]

To get the smallest area, use the most unequal positive integer side lengths: 1 and 24.

\[
1\cdot 24=24.
\]

The requested difference is

\[
156-24=132.
\]

The other choices do not equal the difference between the maximum possible area and minimum possible area.

ANSWER 8: D

Problem 9:

The average of four numbers is 85, so their total is

\[
4\cdot 85=340.
\]

The largest number is 97. Therefore, the sum of the other three numbers is

\[
340-97=243.
\]

Their average is

\[
\frac{243}{3}=81.
\]

Thus the mean of the remaining three numbers is \(81.0\). The other choices would give a total other than 243 for the three remaining numbers.

ANSWER 9: A

Problem 10:

The sum is

\[
2\frac12+3\frac13+4\frac14+5\frac15.
\]

Rewrite each mixed number:

\[
2\frac12=\frac52,\qquad
3\frac13=\frac{10}{3},\qquad
4\frac14=\frac{17}{4},\qquad
5\frac15=\frac{26}{5}.
\]

Using denominator 60,

\[
\frac52=\frac{150}{60},
\quad
\frac{10}{3}=\frac{200}{60},
\quad
\frac{17}{4}=\frac{255}{60},
\quad