Question

Walking ${\frac{6}{7}}^{\mathrm{th}}$ of his usual speed, a man is $12\mathrm{minutes}$ too late. What is the usual time taken by him to cover that distance?

Answer 1

Step 1: Finding new speed and new time:

We are given that,

Walking ${\frac{6}{7}}^{\mathrm{th}}$ of his usual speed, a man is $12\mathrm{minutes}$ too late.

We need to find the usual time taken by him to cover that distance.

Let the distance covered be $x$.

Let the original speed be $y$.

We know that, $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

So, the original time taken is $\frac{x}{y}$.

Walking ${\frac{6}{7}}^{\mathrm{th}}$ of his usual speed.

i.e. new speed is $\frac{6}{7}y$

Therefore, New time is $\frac{x}{{\displaystyle \frac{6}{7}}y}$.

Hence, $\mathrm{Time}=\frac{7\mathrm{x}}{6\mathrm{y}}$

Step 2: Finding usual time taken to cover the distance

We know, A man is $12\mathrm{minutes}$ too late.

Hence,

$$\begin{gathered} \mathrm{New}\mathrm{time}–\mathrm{Original}\mathrm{time}=12 \\ \Rightarrow \frac{7\mathrm{x}}{6\mathrm{y}}-\frac{\mathrm{x}}{\mathrm{y}}=12 \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}(\frac{7}{6}-1)=12 \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=12\times 6 \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=72 \end{gathered}$$

Hence, the usual time taken by him to cover the distance is$72\mathrm{minutes}\mathrm{or}1\mathrm{hour}12\mathrm{minutes}.$