Box A contained 200 fifty-cent coins and 90 one-dollar coins. Box B contained 80 fifty-cent coins and 550 one-dollar coins. Some coins were transferred from Box A to Box B such tha... Box A contained 200 fifty-cent coins and 90 one-dollar coins. Box B contained 80 fifty-cent coins and 550 one-dollar coins. Some coins were transferred from Box A to Box B such that 1/2 of the coins in Box A and 3/10 of the coins in Box B were fifty-cent coins. Find the total value of the fifty-cent coins in Box B in the end. Read more

Steps to Solve

1. Calculate the number of coins remaining in Box A after the transfer

Let $x$ be the number of coins transferred from Box A to Box B. After the transfer, the number of coins remaining in Box A is $200 + 90 - x = 290 - x$. Since $\frac{1}{2}$ of the coins in Box A are fifty-cent coins, the number of fifty-cent coins remaining in Box A is $\frac{1}{2}(290-x)$. The number of fifty-cent coins initially in Box A is 200. Therefore, the number of fifty-cent coins transferred from Box A to Box B is $200 - \frac{1}{2}(290-x)$.

2. Calculate the number of coins in Box B after the transfer

The total number of coins in Box B after the transfer is $80 + 550 + x = 630 + x$. Since $\frac{3}{10}$ of the coins in Box B are fifty-cent coins, the number of fifty-cent coins in Box B is $\frac{3}{10}(630+x)$. The number of fifty-cent coins initially in Box B is 80. Therefore, the number of fifty-cent coins transferred from Box A to Box B is $\frac{3}{10}(630+x) - 80$.

3. Set up an equation and solve for x

Since the number of fifty-cent coins transferred from Box A to Box B must be the same regardless of whether we calculate it from Box A's perspective or Box B's perspective, we can set up the following equation: $$200 - \frac{1}{2}(290-x) = \frac{3}{10}(630+x) - 80$$ Now, solve for $x$: $$200 - 145 + \frac{1}{2}x = 189 + \frac{3}{10}x - 80$$ $$55 + \frac{1}{2}x = 109 + \frac{3}{10}x$$ $$\frac{1}{2}x - \frac{3}{10}x = 109 - 55$$ $$\frac{5}{10}x - \frac{3}{10}x = 54$$ $$\frac{2}{10}x = 54$$ $$\frac{1}{5}x = 54$$ $$x = 54 \times 5 = 270$$

4. Calculate the number of fifty-cent coins in Box B after the transfer

The number of fifty-cent coins in Box B after the transfer is $\frac{3}{10}(630+x) = \frac{3}{10}(630+270) = \frac{3}{10}(900) = 3 \times 90 = 270$.

5. Calculate the total value of the fifty-cent coins in Box B

Since each fifty-cent coin is worth $0.50, the total value of the fifty-cent coins in Box B is $270 \times 0.50 = 135$.