Caroline by walking at (8/9) of her usual speed covers a certain distance in 12 minutes more than the time taken by her at normal speed. Calculate the usual time taken by her to cover the certain distance?

Question

Question

This question was previously asked in

RPF SI (2018) Official Paper (Held On: 24 Dec, 2018 Shift 1)

Answer 1

Given:

Caroline walks at $\frac{8}{9}$ of her usual speed and takes 12 minutes more than her usual time to cover a certain distance.

Formula Used:

Time = Distance / Speed

Calculation:

Let the usual time taken by Caroline be $t$ minutes.

Let the usual speed be $S$ and the distance be $D$ .

Usual time: $t = \frac{D}{S}$

New speed = $\frac{8}{9}S$

New time: $t + 12 = \frac{D}{\frac{8}{9}S} = \frac{9D}{8S}$

Using the usual time equation:

$t + 12 = \frac{9}{8}t$

⇒ $8(t + 12) = 9t$

⇒ $8t + 96 = 9t$

⇒ $96 = 9t - 8t$

⇒ $96 = t$

The usual time taken by Caroline to cover the certain distance is 96 minutes.