A box contains 420 coins in rupee. 50 paisa and 20 paisa coins, the ratio of their rupee values being 13: 11 : 7. The number of 50 paisa coins is,

Options

A. 42
B. 78
C. 66
D. 132

Answer

To solve the problem, we need to find the number of 50 paisa coins in a box that contains a total of 420 coins consisting of rupee coins, 50 paisa coins, and 20 paisa coins. The rupee values of these coins are in the ratio of 13:11:7.

Let's denote:
- The number of rupee coins as \( x \)
- The number of 50 paisa coins as \( y \)
- The number of 20 paisa coins as \( z \)

From the problem, we know:
1. The total number of coins:
\[ x + y + z = 420 \quad \text{(1)} \]

2. The rupee values of the coins:
- The value of \( x \) rupee coins is \( x \) rupees.
- The value of \( y \) 50 paisa coins is \( \frac{y}{2} \) rupees (since 50 paisa = 0.5 rupee).
- The value of \( z \) 20 paisa coins is \( \frac{z}{5} \) rupees (since 20 paisa = 0.2 rupee).

The ratio of their rupee values is given as 13:11:7. Therefore, we can write:
\[ \frac{x}{13} = \frac{\frac{y}{2}}{11} = \frac{\frac{z}{5}}{7} \quad \text{(2)} \]

Let’s denote the common ratio as \( k \). Then we can express \( x \), \( y \), and \( z \) in terms of \( k \):
- From \( \frac{x}{13} = k \), we have \( x = 13k \).
- From \( \frac{\frac{y}{2}}{11} = k \), we have \( y = 22k \).
- From \( \frac{\frac{z}{5}}{7} = k \), we have \( z = 35k \).

Now, substituting \( x \), \( y \), and \( z \) back into equation (1):
\[ 13k + 22k + 35k = 420 \]
\[ 70k = 420 \]
\[ k = \frac{420}{70} = 6 \]

Now we can find the number of each type of coin:
- \( x = 13k = 13 \times 6 = 78 \) (number of rupee coins)
- \( y = 22k = 22 \times 6 = 132 \) (number of 50 paisa coins)
- \( z = 35k = 35 \times 6 = 210 \) (number of 20 paisa coins)

Thus, the number of 50 paisa coins is \( y = 132 \).

### Final Answer:
The number of 50 paisa coins is **132**.