Let’s break down the puzzle

Puzzle Recap:

• You have a box with 100 balls in total. The balls are either red or blue.
• Scenario 1: If you take out one red ball and replace it with a blue ball, the number of red balls becomes half the number of blue balls.
• Scenario 2: If you take out one blue ball and replace it with a red ball, the number of red balls becomes twice the number of blue balls.

Let the number of red balls be $r$ and the number of blue balls be $b$. The total number of balls is 100, so:

$$r+b=100$$

Now, let’s use the information in the puzzle to create two equations.

1. First Scenario:

• When you replace one red ball with a blue ball, the number of red balls becomes $r-1$ and the number of blue balls becomes $b+1$.
• According to the puzzle, after this change, the number of red balls is half the number of blue balls. This gives us the equation:

r−1=12×(b+1)r – 1 = \frac{1}{2} \times (b + 1)r−1=21​×(b+1)

2. Second Scenario:

• When you replace one blue ball with a red ball, the number of red balls becomes $r+1$ and the number of blue balls becomes $b-1$.
• According to the puzzle, after this change, the number of red balls becomes twice the number of blue balls. This gives us the equation:

$$r+1=2\times (b-1)$$

Solving the Equations:

We now have two equations:

1. r−1=12×(b+1)r – 1 = \frac{1}{2} \times (b + 1)r−1=21​×(b+1)
2. $r+1=2\times (b-1)$

Let’s solve these equations step by step.

Step 1: Solve for $r$ in terms of $b$ from the first equation.

From the first equation:

r−1=12×(b+1)r – 1 = \frac{1}{2} \times (b + 1)r−1=21​×(b+1)

Multiply both sides by 2 to eliminate the fraction:

$$2(r-1)=b+1$$ $$2r-2=b+1$$ $$2r=b+3$$

So, r=b+32r = \frac{b + 3}{2}r=2b+3​.

Step 2: Solve for $r$ in terms of $b$ from the second equation.

From the second equation:

$$r+1=2\times (b-1)$$

Simplify the right side:

$$r+1=2b-2$$ $$r=2b-3$$

Step 3: Set the two expressions for $r$ equal to each other.

We have:

b+32=2b−3\frac{b + 3}{2} = 2b – 32b+3​=2b−3

Multiply both sides by 2 to eliminate the fraction:

$$b+3=4b-6$$

Now, solve for $b$:

$$b+3=4b-6$$ $$3+6=4b-b$$ $$9=3b$$ $$b=3$$

Step 4: Find $r$.

Substitute $b=3$ into $r=2b-3$:

$$r=2(3)-3=6-3=3$$

Final Answer:

• The number of red balls is 3.
• The number of blue balls is 97.

Thus, there are 3 red balls and 97 blue balls in the box.