A, B and C start walking from same point at same time is same direction with a speed of `2 (1)/(2)` km./hr., 3 km./hr. and 2 km./hr. respectively on a circular track of 5 km. long. In how many hours will they meet at starting point again ?

Options

A. 30 hour
B. 6 hour
C. 10 hour
D. 15 hour

Answer

To solve the problem, we need to determine how long it will take for A, B, and C to meet at the starting point again after starting from the same point on a circular track of 5 km.

### Step-by-Step Solution:

1. **Identify the speeds of A, B, and C:**
- Speed of A (S1) = \(2 \frac{1}{2}\) km/hr = \( \frac{5}{2} \) km/hr
- Speed of B (S2) = 3 km/hr
- Speed of C (S3) = 2 km/hr

2. **Calculate the time taken by each person to complete one lap around the circular track:**
- Time taken by A (T1):
\[ T1 = \frac{\text{Distance}}{\text{Speed}} = \frac{5 \text{ km}}{\frac{5}{2} \text{ km/hr}} = \frac{5 \times 2}{5} = 2 \text{ hours} \]
- Time taken by B (T2):
\[ T2 = \frac{5 \text{ km}}{3 \text{ km/hr}} = \frac{5}{3} \text{ hours} \]
- Time taken by C (T3):
\[ T3 = \frac{5 \text{ km}}{2 \text{ km/hr}} = \frac{5}{2} \text{ hours} \]

3. **Find the least common multiple (LCM) of the times T1, T2, and T3:**
- Convert T2 and T3 to a common fraction format:
- T1 = 2 hours = \( \frac{2}{1} \)
- T2 = \( \frac{5}{3} \)
- T3 = \( \frac{5}{2} \)

- To find the LCM of \( \frac{2}{1}, \frac{5}{3}, \frac{5}{2} \):
- LCM of the numerators (2, 5, 5) = 10
- GCD of the denominators (1, 3, 2) = 1
- Therefore, LCM:
\[ \text{LCM} = \frac{\text{LCM of numerators}}{\text{GCD of denominators}} = \frac{10}{1} = 10 \text{ hours} \]

4. **Conclusion:**
- A, B, and C will meet at the starting point again after **10 hours**.