A and B start driving simultaneously from point X and go towards point Y. X and Y are 60 km apart. A's speed is 4 km/h less than that of B. B, after reaching Y, returns and meets A at a point 12 km away from Y. Find the speed of A.

Correct option is C

Given:

The distance between X and Y is 60 km.

A’s speed is 4 km/h less than B’s speed.

B reaches Y, returns, and meets A 12 km away from Y.

Formula Used:

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

Solution:

Let the speed of B be b km/h. Then, the speed of A is (b - 4) km/h.

B travels from X to Y (a distance of 60 km) and then returns.

B meets A at a point 12 km away from Y, which means A has traveled 60 - 12 = 48 km by the time they meet.

B travels 60 km to reach Y and then 12 km back, so B covers a total distance of 60 + 12 = 72 km.

The time taken by B to cover 72 km is:

Time taken by B =$$\frac{72}{b}$$​

A travels 48 km to the meeting point.

The time taken by A to cover 48 km is:

Time taken by A =$$\frac{48}{b - 4}$$​

Since both start simultaneously and meet after the same time, the times must be equal:

​$$\frac{72}{b} = \frac{48}{b - 4}$$​​

72(b - 4) = 48b

72b - 288 = 48b

24b = 288

b = 12

A’s speed is b - 4, so the speed of A is:

Speed of A = 12 – 4 = 8 km/h

Alternate Method:

A : B = (60 - 12) : (60 + 12) =  48 : 72 = 2 : 3  (Distance Ratio or Speed Ratio)

1 unit = 4 km/h

A’s speed = 2 $$\times$$ 4 km/h = 8 km/h​​