IAS 2012 — CSAT Maths

Correct Option (B)

Let the total distance to the office be d km. The average speed of Mr. Kumar is 48 km/h.

The first 60% of the distance is 0.6d km, and the remaining distance is 0.4d km.

The time taken to cover the first 60% of the distance is $\frac{0.6d}{48}$480.6d​ hours.

The time taken to cover the remaining distance is $\frac{0.4d}{48}$480.4d​ hours.

According to the problem statement, the time taken for the first 60% of the distance is 10 minutes more than the time taken for the remaining distance. Converting 10 minutes to hours, we get $\frac{10}{60} = \frac{1}{6}$6010​=61​ hours.

Formulating the equation based on the given information:

$$\frac{0.6d}{48} - \frac{0.4d}{48} = \frac{10}{60}$$480.6d​−480.4d​=6010​

Simplifying the equation:

$$\frac{0.2d}{48} = \frac{1}{6}$$480.2d​=61​

Multiplying both sides by 48:

$$0.2d = \frac{48}{6}$$0.2d=648​

$$0.2d = 8$$0.2d=8

Solving for d:

$$d = \frac{8}{0.2}$$d=0.28​

$$d = 40$$d=40

Therefore, the total distance to the office is 40 km.

Incorrect Options:

Options A (30 km), C (45 km), and D (48 km) are incorrect. These values do not satisfy the conditions specified in the problem statement when subjected to the correct mathematical formulation and calculation. The distance of 40 km is uniquely determined by the given average speed and the time difference between covering the two segments of the journey.