A, B and C start to walk around a circular track. A completes one round in 91/6 seconds, B completes in 65/4 seconds and C completes in 56/3 seconds respectively. After how many seconds will they meet to each other at the starting point again?

Options

A. 3460 s
B. 1820 s
C. 3640 s
D. none of these

Answer

To solve the problem of when A, B, and C will meet at the starting point again after starting to walk around a circular track, we need to find the least common multiple (LCM) of the times taken by each of them to complete one round.

### Step-by-Step Solution:

1. **Identify the times taken by A, B, and C:**
- A takes \( \frac{91}{6} \) seconds.
- B takes \( \frac{65}{4} \) seconds.
- C takes \( \frac{56}{3} \) seconds.

2. **Convert the times into improper fractions:**
- A: \( \frac{91}{6} \)
- B: \( \frac{65}{4} \)
- C: \( \frac{56}{3} \)

3. **Find the LCM of the numerators (91, 65, 56):**
- Prime factorization:
- \( 91 = 7 \times 13 \)
- \( 65 = 5 \times 13 \)
- \( 56 = 2^3 \times 7 \)
- LCM is found by taking the highest power of each prime:
- \( 2^3 \) from 56,
- \( 5^1 \) from 65,
- \( 7^1 \) from 91 and 56,
- \( 13^1 \) from 91 and 65.
- Thus, LCM = \( 2^3 \times 5^1 \times 7^1 \times 13^1 = 8 \times 5 \times 7 \times 13 \).

4. **Calculate the LCM:**
- \( 8 \times 5 = 40 \)
- \( 40 \times 7 = 280 \)
- \( 280 \times 13 = 3640 \)
- Therefore, LCM of the numerators is 3640.

5. **Find the GCD of the denominators (6, 4, 3):**
- The GCD of 6, 4, and 3 is 1 (since they have no common factors other than 1).

6. **Calculate the LCM of the fractions:**
- LCM of the times = \( \frac{\text{LCM of numerators}}{\text{GCD of denominators}} = \frac{3640}{1} = 3640 \) seconds.

7. **Conclusion:**
- A, B, and C will all meet at the starting point again after **3640 seconds**.