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?

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.