PPS Sample Puzzle 14 — Marcy Mathworks - Free Printable

Let's solve each problem step by step:

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Problem 1: How many squares are in the figure at right?

The figure is a 3x3 grid of smaller squares. To count all the squares, we need to consider squares of different sizes:

- 1x1 squares: There are 9 individual squares.
- 2x2 squares: There are 4 squares (top-left, top-right, bottom-left, bottom-right).
- 3x3 squares: There is 1 large square that encompasses the entire grid.

Total number of squares = \( 9 + 4 + 1 = 14 \).

Answer: 14

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Problem 2: Twelve pennies are arranged in a row on a table. Every other coin is replaced with a nickel. Then every third coin is replaced with a dime. Finally, every fourth coin is replaced with a quarter. What is the total value of the coins on the table?

Let's follow the steps:

1. Initial arrangement: 12 pennies (P).
- Value: \( 12 \times 0.01 = \$0.12 \).

2. Replace every other coin with a nickel (N):
- Positions: 2, 4, 6, 8, 10, 12.
- New arrangement: P, N, P, N, P, N, P, N, P, N, P, N.
- Value: \( 6 \times 0.01 + 6 \times 0.05 = \$0.12 + \$0.30 = \$0.42 \).

3. Replace every third coin with a dime (D):
- Positions: 3, 6, 9, 12.
- New arrangement: P, N, D, N, P, D, P, N, D, N, P, D.
- Value: \( 4 \times 0.01 + 4 \times 0.05 + 4 \times 0.10 = \$0.04 + \$0.20 + \$0.40 = \$0.64 \).

4. Replace every fourth coin with a quarter (Q):
- Positions: 4, 8, 12.
- New arrangement: P, N, D, Q, P, D, P, Q, D, N, P, Q.
- Value: \( 5 \times 0.01 + 3 \times 0.05 + 3 \times 0.10 + 3 \times 0.25 = \$0.05 + \$0.15 + \$0.30 + \$0.75 = \$1.25 \).

Answer: \$1.25

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Problem 3: If eggs are $2.00 a dozen and cost $1.00 per pound, how much does an egg weigh?

1. Cost per egg: Since a dozen eggs cost \$2.00, the cost per egg is:
\[
\frac{2.00}{12} = \$0.1667 \text{ (approximately)}.
\]

2. Weight of a dozen eggs: Since a dozen eggs cost \$1.00 per pound, the weight of a dozen eggs is 1 pound.

3. Weight per egg: Since a dozen eggs weigh 1 pound, the weight of one egg is:
\[
\frac{1 \text{ pound}}{12} = \frac{1}{12} \text{ pound}.
\]

Answer: \(\frac{1}{12}\) pound

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Problem 4: There are 4 black, 4 red, and 4 white marbles in a box. How many marbles do you have to take out (without looking) to be sure you have at least two marbles of the same color?

To ensure at least two marbles of the same color, consider the worst-case scenario:
- You could pick one marble of each color first (black, red, white).
- After picking 3 marbles, you still don't have two of the same color.
- The next marble you pick (the 4th one) must match one of the colors already picked.

Answer: 4

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Problem 5: If it takes 2 bricklayers an hour and a half to lay a total of 90 bricks, how many bricklayers would it take to lay 210 bricks in an hour?

1. Rate of work for 2 bricklayers: They lay 90 bricks in 1.5 hours.
- Bricks laid per hour by 2 bricklayers:
\[
\frac{90}{1.5} = 60 \text{ bricks/hour}.
\]

2. Rate of work per bricklayer: Since 2 bricklayers lay 60 bricks per hour, one bricklayer lays:
\[
\frac{60}{2} = 30 \text{ bricks/hour}.
\]

3. Bricks needed per hour: We need to lay 210 bricks in 1 hour.
- Number of bricklayers required:
\[
\frac{210}{30} = 7.
\]

Answer: 7

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Problem 6: In the kingdom of Up, royal license plates have 4 letters. Each letter is either a U or a P, but there can never be two P's next to each other. How many different license plates are possible?

We need to count all valid sequences of length 4 using the letters U and P, where no two P's are adjacent.

#### Step 1: Define the problem recursively
Let \( a_n \) be the number of valid sequences of length \( n \):
- If the sequence ends in U, the preceding \( n-1 \) letters can be any valid sequence of length \( n-1 \).
- If the sequence ends in P, the preceding \( n-1 \) letters must end in U (to avoid two consecutive P's), so the preceding \( n-2 \) letters can be any valid sequence of length \( n-2 \).

Thus, the recurrence relation is:
\[
a_n = a_{n-1} + a_{n-2}
\]
with initial conditions:
\[
a_1 = 2 \quad (\text{U, P}), \quad a_2 = 3 \quad (\text{UU, UP, PU}).
\]

#### Step 2: Compute \( a_4 \)
Using the recurrence relation:
\[
a_3 = a_2 + a_1 = 3 + 2 = 5,
\]
\[
a_4 = a_3 + a_2 = 5 + 3 = 8.
\]

Answer: 8

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Problem 7: If it were one hour later, it would be half as long to midnight as it is now. If it were two hours earlier, it would be twice as long to midnight as it is now. What time is it?

Let the current time be \( x \) hours before midnight.

1. One hour later: The time to midnight would be \( x - 1 \). According to the problem:
\[
x - 1 = \frac{x}{2}.
\]
Solving for \( x \):
\[
2(x - 1) = x \implies 2x - 2 = x \implies x = 2.
\]

2. Two hours earlier: The time to midnight would be \( x + 2 \). According to the problem:
\[
x + 2 = 2x.
\]
Solving for \( x \):
\[
x + 2 = 2x \implies 2 = x.
\]

Both conditions are satisfied when \( x = 2 \). Therefore, the current time is 2 hours before midnight, which is 10:00 PM.

Answer: 10:00 PM

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Problem 8: The sum of the ages of three children is 32. The age of the oldest is twice the age of the youngest. The ages of the two older children differ by three years. What are the ages of the three children?

Let the ages of the children be \( y \) (youngest), \( m \) (middle), and \( o \) (oldest).

1. Sum of ages:
\[
y + m + o = 32.
\]

2. Oldest is twice the youngest:
\[
o = 2y.
\]

3. Difference between the two older children:
\[
o - m = 3 \quad \text{or} \quad m - o = 3.
\]
Since \( o = 2y \), we use \( o - m = 3 \):
\[
2y - m = 3 \implies m = 2y - 3.
\]

4. Substitute into the sum equation:
\[
y + (2y - 3) + 2y = 32.
\]
Simplify:
\[
y + 2y - 3 + 2y = 32 \implies 5y - 3 = 32 \implies 5y = 35 \implies y = 7.
\]

5. Find \( m \) and \( o \):
\[
m = 2y - 3 = 2(7) - 3 = 14 - 3 = 11,
\]
\[
o = 2y = 2(7) = 14.
\]

Answer: 7, 11, 14

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Problem 9: Write a number in each blank to make the following statement true:

"In this sentence, the number of occurrences of 1 is __, of 2 is __, of 3 is __, of 4 is __, and of 5 is __."

This is a self-referential puzzle. Let's denote the blanks as \( a, b, c, d, e \) for the numbers of occurrences of 1, 2, 3, 4, and 5, respectively.

1. Count the digits in the sentence:
- The sentence has 5 blanks, so the total number of digits mentioned must be 5.
- The sentence itself contains the digits 1, 2, 3, 4, and 5 exactly once in the blanks.

2. Trial and error:
- If \( a = 1 \), \( b = 1 \), \( c = 2 \), \( d = 1 \), \( e = 0 \), the sentence becomes:
"In this sentence, the number of occurrences of 1 is 1, of 2 is 1, of 3 is 2, of 4 is 1, and of 5 is 0."
- Count the digits:
- 1 appears 3 times (in "1, 1, 1").
- 2 appears 1 time (in "2").
- 3 appears 1 time (in "3").
- 4 appears 1 time (in "4").
- 5 appears 0 times.
- This does not match.

3. Correct solution:
- If \( a = 1 \), \( b = 2 \), \( c = 1 \), \( d = 1 \), \( e = 1 \), the sentence becomes:
"In this sentence, the number of occurrences of 1 is 1, of 2 is 2, of 3 is 1, of 4 is 1, and of 5 is 1."
- Count the digits:
- 1 appears 3 times (in "1, 1, 1").
- 2 appears 2 times (in "2, 2").
- 3 appears 1 time (in "3").
- 4 appears 1 time (in "4").
- 5 appears 1 time (in "5").
- This matches.

Answer: 1, 2, 1, 1, 1

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Problem 10: A man leaves home in a big hurry. He makes three left turns. Then he returns home and finds himself face-to-face with two masked men. Who are the masked men?

The man makes three left turns and returns home. This implies he is in a place where making three left turns brings him back to his starting point. The most logical place is a cubicle or a room with walls forming a closed loop.

When he returns home, he sees two masked men. These are likely the postmen or delivery people who were waiting for him outside his house.

Answer: Postmen/Delivery people

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Final Answers:

1. 14
2. \$1.25
3. \(\frac{1}{12}\) pound
4. 4
5. 7
6. 8
7. 10:00 PM
8. 7, 11, 14
9. 1, 2, 1, 1, 1
10. Postmen/Delivery people

\boxed{14, 1.25, \frac{1}{12}, 4, 7, 8, 10:00 \text{ PM}, 7, 11, 14, 1, 2, 1, 1, 1, \text{Postmen/Delivery people}}