A man running at the speed of 3 km/hr reaches his destination 10 minutes later than the normal time. If he increases his speed by 1 km/hr, then he reaches his destination 15 minutes before. Find the distance of his destination.

Correct option is A

Given:$$\frac{d}{3} - \frac{d}{4} = \frac{5}{12}$$​​

1. Speed at first: 3 kmph
2. Increased speed: 3 + 1 = 4 kmph
3. Time difference between the two speeds: 10 + 15 = 25 minutes = $$\frac{25}{60} = \frac{5}{12} \, \text{hours}$$ ​
4. Let the distance to the destination be d km

Formula Used:
1. Time for a journey:

​$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$​​


2. Difference in time for two speeds:


Step-by-Step Solution:

Step 1: Set up the equation:

$$\frac{d}{3} - \frac{d}{4} = \frac{5}{12}$$​​

Take the LCM of 3 and 4, which is 12:

$$\frac{4d}{12} - \frac{3d}{12} = \frac{5}{12}$$​​


Simplify:
​$$\frac{d}{12} = \frac{5}{12}$$​​


Step 2: Solve for d:

d = 5 km