"Test of Genius" brain teaser worksheet with 10 challenging puzzles for students.
A "Test of Genius" worksheet featuring 10 brain teaser questions, including math, logic, and word problems, with a decorative black frame and a note at the bottom suggesting the use of algebra to solve problem #8.
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Step-by-step solution for: PPS Sample Puzzle 14 — Marcy Mathworks
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Show Answer Key & Explanations
Step-by-step solution for: PPS Sample Puzzle 14 — Marcy Mathworks
Let's solve each problem step by step:
---
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
---
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
---
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
---
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
---
1. Rate of work for 2 bricklayers: They lay 90 bricks in 1.5 hours.
- Bricks laid per hour by 2 bricklayers: \( \frac{90}{1.5} = 60 \) 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.
4. Number of bricklayers required: Let \( x \) be the number of bricklayers needed.
- Total bricks laid per hour by \( x \) bricklayers: \( 30x \).
- Set up the equation: \( 30x = 210 \).
- Solve for \( x \):
\[
x = \frac{210}{30} = 7.
\]
Answer: 7
---
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 \( n \).
- Base cases:
- \( a_1 \): Sequences of length 1 are "U" and "P". So, \( a_1 = 2 \).
- \( a_2 \): Sequences of length 2 are "UU", "UP", "PU". "PP" is invalid. So, \( a_2 = 3 \).
- Recursive relation:
- A valid sequence of length \( n \) can end in either "U" or "P".
- If it ends in "U", the first \( n-1 \) characters can be any valid sequence of length \( n-1 \). So, there are \( a_{n-1} \) such sequences.
- If it ends in "P", the character before "P" must be "U". The first \( n-2 \) characters can be any valid sequence of length \( n-2 \). So, there are \( a_{n-2} \) such sequences.
- Therefore, the recursive relation is:
\[
a_n = a_{n-1} + a_{n-2}.
\]
#### Step 2: Compute \( a_3 \) and \( a_4 \)
- \( a_3 = a_2 + a_1 = 3 + 2 = 5 \).
- \( a_4 = a_3 + a_2 = 5 + 3 = 8 \).
Answer: 8
---
Let the current time be \( x \) hours before midnight.
1. Condition 1: If it were one hour later, it would be half as long to midnight as it is now.
- One hour later: \( x - 1 \) hours before midnight.
- Equation: \( x - 1 = \frac{x}{2} \).
- Solve for \( x \):
\[
2(x - 1) = x \implies 2x - 2 = x \implies x = 2.
\]
2. Condition 2: If it were two hours earlier, it would be twice as long to midnight as it is now.
- Two hours earlier: \( x + 2 \) hours before midnight.
- Equation: \( x + 2 = 2x \).
- Solve 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
---
Let the ages of the children be \( y \) (youngest), \( m \) (middle), and \( o \) (oldest).
1. Equations based on the problem:
- Sum of ages: \( y + m + o = 32 \).
- Oldest is twice the youngest: \( o = 2y \).
- Middle and oldest differ by 3 years: \( o - m = 3 \).
2. Substitute \( o = 2y \) into the equations:
- From \( o - m = 3 \):
\[
2y - m = 3 \implies m = 2y - 3.
\]
- Substitute \( o = 2y \) and \( m = 2y - 3 \) into \( y + m + o = 32 \):
\[
y + (2y - 3) + 2y = 32 \implies 5y - 3 = 32 \implies 5y = 35 \implies y = 7.
\]
3. Find \( m \) and \( o \):
- \( o = 2y = 2 \times 7 = 14 \).
- \( m = 2y - 3 = 2 \times 7 - 3 = 11 \).
The ages of the children are 7, 11, and 14.
Answer: 7, 11, 14
---
"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 \( a, b, c, d, e \) respectively. The sentence becomes:
"In this sentence, the number of occurrences of 1 is \( a \), of 2 is \( b \), of 3 is \( c \), of 4 is \( d \), and of 5 is \( e \)."
1. Count the digits in the sentence:
- The sentence contains the digits 1, 2, 3, 4, and 5 exactly once in the blanks.
- The digit 1 appears in "one" (once in "one").
- The digit 2 appears in "two" (once in "two").
- The digit 3 appears in "three" (once in "three").
- The digit 4 appears in "four" (once in "four").
- The digit 5 appears in "five" (once in "five").
2. Fill in the blanks:
- \( a = 1 \) (number of 1's).
- \( b = 1 \) (number of 2's).
- \( c = 2 \) (number of 3's, including the "3" in "three").
- \( d = 1 \) (number of 4's).
- \( e = 1 \) (number of 5's).
The completed sentence is:
"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 1."
Answer: 1, 1, 2, 1, 1
---
The key here is understanding the spatial orientation. When the man makes three left turns, he essentially completes a full circle (360 degrees) and returns to his starting point. The two masked men he sees are likely his neighbors or people he knows well, but the riddle implies they are his own reflection in a mirror.
Answer: His reflection in a mirror
---
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, 1, 2, 1, 1
10. His reflection in a mirror
\boxed{14, 1.25, \frac{1}{12}, 4, 7, 8, 10:00 \text{ PM}, 7, 11, 14, 1, 1, 2, 1, 1, \text{His reflection in a mirror}}
---
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
---
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
---
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
---
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
---
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: \( \frac{90}{1.5} = 60 \) 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.
4. Number of bricklayers required: Let \( x \) be the number of bricklayers needed.
- Total bricks laid per hour by \( x \) bricklayers: \( 30x \).
- Set up the equation: \( 30x = 210 \).
- Solve for \( x \):
\[
x = \frac{210}{30} = 7.
\]
Answer: 7
---
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 \( n \).
- Base cases:
- \( a_1 \): Sequences of length 1 are "U" and "P". So, \( a_1 = 2 \).
- \( a_2 \): Sequences of length 2 are "UU", "UP", "PU". "PP" is invalid. So, \( a_2 = 3 \).
- Recursive relation:
- A valid sequence of length \( n \) can end in either "U" or "P".
- If it ends in "U", the first \( n-1 \) characters can be any valid sequence of length \( n-1 \). So, there are \( a_{n-1} \) such sequences.
- If it ends in "P", the character before "P" must be "U". The first \( n-2 \) characters can be any valid sequence of length \( n-2 \). So, there are \( a_{n-2} \) such sequences.
- Therefore, the recursive relation is:
\[
a_n = a_{n-1} + a_{n-2}.
\]
#### Step 2: Compute \( a_3 \) and \( a_4 \)
- \( a_3 = a_2 + a_1 = 3 + 2 = 5 \).
- \( a_4 = a_3 + a_2 = 5 + 3 = 8 \).
Answer: 8
---
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. Condition 1: If it were one hour later, it would be half as long to midnight as it is now.
- One hour later: \( x - 1 \) hours before midnight.
- Equation: \( x - 1 = \frac{x}{2} \).
- Solve for \( x \):
\[
2(x - 1) = x \implies 2x - 2 = x \implies x = 2.
\]
2. Condition 2: If it were two hours earlier, it would be twice as long to midnight as it is now.
- Two hours earlier: \( x + 2 \) hours before midnight.
- Equation: \( x + 2 = 2x \).
- Solve 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
---
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. Equations based on the problem:
- Sum of ages: \( y + m + o = 32 \).
- Oldest is twice the youngest: \( o = 2y \).
- Middle and oldest differ by 3 years: \( o - m = 3 \).
2. Substitute \( o = 2y \) into the equations:
- From \( o - m = 3 \):
\[
2y - m = 3 \implies m = 2y - 3.
\]
- Substitute \( o = 2y \) and \( m = 2y - 3 \) into \( y + m + o = 32 \):
\[
y + (2y - 3) + 2y = 32 \implies 5y - 3 = 32 \implies 5y = 35 \implies y = 7.
\]
3. Find \( m \) and \( o \):
- \( o = 2y = 2 \times 7 = 14 \).
- \( m = 2y - 3 = 2 \times 7 - 3 = 11 \).
The ages of the children are 7, 11, and 14.
Answer: 7, 11, 14
---
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 \( a, b, c, d, e \) respectively. The sentence becomes:
"In this sentence, the number of occurrences of 1 is \( a \), of 2 is \( b \), of 3 is \( c \), of 4 is \( d \), and of 5 is \( e \)."
1. Count the digits in the sentence:
- The sentence contains the digits 1, 2, 3, 4, and 5 exactly once in the blanks.
- The digit 1 appears in "one" (once in "one").
- The digit 2 appears in "two" (once in "two").
- The digit 3 appears in "three" (once in "three").
- The digit 4 appears in "four" (once in "four").
- The digit 5 appears in "five" (once in "five").
2. Fill in the blanks:
- \( a = 1 \) (number of 1's).
- \( b = 1 \) (number of 2's).
- \( c = 2 \) (number of 3's, including the "3" in "three").
- \( d = 1 \) (number of 4's).
- \( e = 1 \) (number of 5's).
The completed sentence is:
"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 1."
Answer: 1, 1, 2, 1, 1
---
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 key here is understanding the spatial orientation. When the man makes three left turns, he essentially completes a full circle (360 degrees) and returns to his starting point. The two masked men he sees are likely his neighbors or people he knows well, but the riddle implies they are his own reflection in a mirror.
Answer: His reflection in a mirror
---
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, 1, 2, 1, 1
10. His reflection in a mirror
\boxed{14, 1.25, \frac{1}{12}, 4, 7, 8, 10:00 \text{ PM}, 7, 11, 14, 1, 1, 2, 1, 1, \text{His reflection in a mirror}}
Parent Tip: Review the logic above to help your child master the concept of test of genius worksheet.
