If a person travels in a car at a speed of 36 km/hour, then he will reach his destination on time. He covers half the journey in (4/5)th time. What should be his speed for the remaining part of the journey so that he reaches his destination on time?

Correct option is C

Given:

Speed for the first part of the journey = 36 km/h

The person covers half of the journey in $$\frac{4}{5}$$​th of the total time.

Formula Used:

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

Solution:

Let total distance = D km.

Let total time = T hours.

Original speed = 36 km/h.

Original condition:

D = $$36 \times T \quad \text{(Distance = Speed × Time)}$$​​

Distance covered = $$\frac{D}{2}$$​ km

Time taken = $$\frac{4}{5}T$$​ hours

Speed for first half

Speed = $$\frac{\frac{D}{2}}{\frac{4}{5}T} = \frac{5D}{8T}$$​​

Speed = $$\frac{5 \times 36T}{8T} = \frac{180}{8}$$​ = 22.5 km/h

This is the speed for the first half, but the problem states he travels at 36 km/h to reach on time. This suggests the first half was slower.

Remaining distance = $$\frac{D}{2}$$​ km

Remaining time = $$T - \frac{4}{5}T = \frac{1}{5}T$$​ hours

Speed = $$\frac{\frac{D}{2}}{\frac{1}{5}T} = \frac{5D}{2T}$$​​

Substituting D = 36T

Speed = $$\frac{5 \times 36T}{2T} = \frac{180}{2}$$​ = 90 km/h