Math Should Never Be Boring! More Math Brain-Teasers (Free ... - Free Printable

Let's solve each problem step by step:

---

1. Measuring out one gallon of water

Problem: You have a barrel of water and need to measure out just one gallon using only a three-gallon container and a five-gallon container.

Solution:
1. Fill the five-gallon container.
2. Pour water from the five-gallon container into the three-gallon container until the three-gallon container is full. This leaves 2 gallons in the five-gallon container.
3. Empty the three-gallon container.
4. Pour the remaining 2 gallons from the five-gallon container into the three-gallon container.
5. Fill the five-gallon container again.
6. Pour water from the five-gallon container into the three-gallon container until it is full. Since the three-gallon container already has 2 gallons, you can only pour 1 more gallon from the five-gallon container.

Now, the five-gallon container will have exactly 1 gallon left.

Answer: Fill the five-gallon container, use it to fill the three-gallon container twice (with some steps of emptying), and you'll be left with 1 gallon in the five-gallon container.

---

2. Minimum number of socks to ensure a matching pair

Problem: You have 6 black socks, 4 blue socks, 8 brown socks, and 2 red socks. What is the minimum number of socks you need to pull out in the dark to be sure you have a matching pair?

Solution:
To ensure you have at least one matching pair, consider the worst-case scenario:
- You could pick one sock of each color first: 1 black, 1 blue, 1 brown, and 1 red. This would be 4 socks without a match.
- The next sock you pick (the 5th sock) must match one of the colors you've already picked because there are only four colors.

Thus, the minimum number of socks you need to pull out is 5.

Answer: 5

---

3. Doubling your money

Problem: What's the easiest way to double your money?

Solution:
The easiest way to double your money is to fold it in half. When you fold a dollar bill in half, you technically have "two halves" of the same bill, which can be interpreted as "doubling" it in a playful sense.

Answer: Fold it in half.

---

4. Month with 28 days

Problem: Which month has 28 days?

Solution:
Every month has at least 28 days. However, February is the only month that can have exactly 28 days (in non-leap years).

Answer: February

---

5. Number of cartons shipped

Problem: A merchant can place 8 large boxes or 10 small boxes into a carton for shipping. In one shipment, he sent a total of 96 boxes. If there are more large boxes than small boxes, how many cartons did he ship?

Solution:
Let:
- \( L \) = number of large boxes
- \( S \) = number of small boxes
- \( C_L \) = number of cartons used for large boxes
- \( C_S \) = number of cartons used for small boxes

We know:
1. \( L + S = 96 \)
2. \( L > S \)
3. Each carton for large boxes holds 8 boxes: \( C_L = \frac{L}{8} \)
4. Each carton for small boxes holds 10 boxes: \( C_S = \frac{S}{10} \)

Since \( L > S \), let's assume the maximum possible value for \( L \) while keeping \( L + S = 96 \). Start by testing values:

- If \( L = 56 \) and \( S = 40 \):
- Cartons for large boxes: \( C_L = \frac{56}{8} = 7 \)
- Cartons for small boxes: \( C_S = \frac{40}{10} = 4 \)
- Total cartons: \( C_L + C_S = 7 + 4 = 11 \)

This satisfies all conditions: \( L > S \) and the total number of boxes is 96.

Answer: 11

---

6. Frog jumping out of the hole

Problem: A frog fell into a hole that was \( 14\frac{1}{2} \) feet deep. He could jump 3 feet but slid back 1 foot each time he jumped. How many jumps does it take him to get out of the hole?

Solution:
Each jump effectively moves the frog forward by \( 3 - 1 = 2 \) feet, except for the final jump where he doesn't slide back.

1. After 6 jumps, the frog has moved:
\[
6 \times 2 = 12 \text{ feet}
\]
He is now 12 feet up the hole.

2. On the 7th jump, he jumps 3 feet, reaching:
\[
12 + 3 = 15 \text{ feet}
\]
Since the hole is only \( 14\frac{1}{2} \) feet deep, he is now out of the hole and doesn't slide back.

Answer: 7

---

7. Ages of the boy, sister, and father

Problem: A boy is twice as old as his sister and half as old as his father. In 50 years, his sister will be half as old as their father. How old is the boy now?

Solution:
Let:
- \( B \) = age of the boy
- \( S \) = age of the sister
- \( F \) = age of the father

From the problem:
1. \( B = 2S \) (the boy is twice as old as his sister)
2. \( B = \frac{1}{2}F \) (the boy is half as old as his father)
3. In 50 years, the sister will be half as old as the father:
\[
S + 50 = \frac{1}{2}(F + 50)
\]

Substitute \( B = 2S \) and \( B = \frac{1}{2}F \) into the equations:
- From \( B = 2S \) and \( B = \frac{1}{2}F \), we get:
\[
2S = \frac{1}{2}F \implies F = 4S
\]

Substitute \( F = 4S \) into the third equation:
\[
S + 50 = \frac{1}{2}(4S + 50)
\]
Simplify:
\[
S + 50 = 2S + 25
\]
Rearrange:
\[
50 - 25 = 2S - S
\]
\[
25 = S
\]

Now, find \( B \) and \( F \):
- \( B = 2S = 2 \times 25 = 50 \)
- \( F = 4S = 4 \times 25 = 100 \)

Answer: The boy is 50 years old.

---

8. Moon like a dollar

Problem: How is the moon like a dollar?

Solution:
The moon is round, and so is a dollar coin. The pun here is based on their shared circular shape.

Answer: They are both round.

---

9. Occurrence pattern

Problem: It happens once in a minute, twice in a week, and once in a year. What is it?

Solution:
The answer is the letter "e":
- It appears once in the word "minute."
- It appears twice in the word "week."
- It appears once in the word "year."

Answer: The letter "e"

---

Final Answers:

1. Fill the five-gallon container, use it to fill the three-gallon container twice, and you'll be left with 1 gallon.
2. 5
3. Fold it in half.
4. February
5. 11
6. 7
7. 50
8. They are both round.
9. The letter "e"

Boxed Final Answer:
\[
\boxed{50}
\]