If a man walks at 5 km/hr he reaches the bus stop 5 minutes too late and if he walks at 6 km/hr he reaches the bus stop 5 minutes too early. The bus stop is at a distance of-

Options

A. 3 km
B. 5 km
C. 6 km
D. 10 km

Answer

To solve the problem step by step, let's denote:

- Distance to the bus stop = D km
- Time taken to reach the bus stop at speed 5 km/hr = T1 hours
- Time taken to reach the bus stop at speed 6 km/hr = T2 hours

### Step 1: Set up the equations based on the information given.

1. When the man walks at 5 km/hr, he is 5 minutes late:
\[ T1 = \frac{D}{5} \quad \text{(time taken at 5 km/hr)} \]
Since he is 5 minutes late, we can express this as:
\[ T1 = T + \frac{5}{60} \quad \text{(where T is the actual time he should take)} \]

2. When the man walks at 6 km/hr, he is 5 minutes early:
\[ T2 = \frac{D}{6} \quad \text{(time taken at 6 km/hr)} \]
Since he is 5 minutes early, we can express this as:
\[ T2 = T - \frac{5}{60} \]

### Step 2: Equate the two expressions for T1 and T2.

From the above equations, we can write:
\[ \frac{D}{5} = T + \frac{5}{60} \quad \text{(1)} \]
\[ \frac{D}{6} = T - \frac{5}{60} \quad \text{(2)} \]

### Step 3: Rearrange both equations to isolate T.

From equation (1):
\[ T = \frac{D}{5} - \frac{5}{60} \]

From equation (2):
\[ T = \frac{D}{6} + \frac{5}{60} \]

### Step 4: Set the two expressions for T equal to each other.

\[ \frac{D}{5} - \frac{5}{60} = \frac{D}{6} + \frac{5}{60} \]

### Step 5: Clear the fractions by multiplying through by 60 (the LCM of the denominators).

\[ 60 \left(\frac{D}{5}\right) - 5 = 60 \left(\frac{D}{6}\right) + 5 \]
This simplifies to:
\[ 12D - 5 = 10D + 5 \]

### Step 6: Solve for D.

Rearranging gives:
\[ 12D - 10D = 5 + 5 \]
\[ 2D = 10 \]
\[ D = 5 \text{ km} \]

### Conclusion

The distance to the bus stop is **5 km**.

---