Question

Walking $6/{7^{th}}$ of his usual speed, Sanchit is 12 minutes late. The usual time taken by him to cover that distance is:
A. 2 hr
B. 1 hr 12 mins
C. 1 hr 15 mins
D. 45 mins

Answer 1

Speed of any object is given by the distance covered by that object in a particular interval of time. Therefore, speed is equal to the distance divided by the time. In the problem, it is given that Sanchit is 12 minutes late when he walks with $6/{7^{th}}$ of his usual speed. The usual time taken by him to cover that distance can be calculated by applying the formula of speed-distance -time.

Formula Used:
The formula for speed is given as: $s = \dfrac{d}{t}$
where, $s$ is the speed , $d$ is the distance and $t$ is the time

Complete step by step answer:
Let the usual time taken by Sanchit to cover the given distance $d$ be $t$. Then his speed $s$ is given by
$s = \dfrac{d}{t}$
Rearranging the equation in terms of $t$
$t = \dfrac{d}{s}$ $\to (1)$
This is the usual time taken.
If he walks with $6/{7^{th}}$ of the original speed then the new speed will be;
$s' = \dfrac{6}{7}s$
where, $s'$ is the new speed and $s$ is the usual speed.
Then, the new time will be

\Rightarrow t' = \dfrac{{7d}}{{6s}}$$ $$ \to (2)$$ With this new time, Sanchit is 12minutes late. $${\text{New time - Usual Time = 12}}$$ $$\dfrac{{7d}}{{6s}} - \dfrac{d}{s} = 12 \\\ \Rightarrow \dfrac{{7d - 6d}}{{6s}} = 12 \\\ \Rightarrow \dfrac{d}{{6s}} = 12 \\\ \therefore \dfrac{d}{s} = 72$$ But from equation (1),$$\dfrac{d}{s} = t$$. Therefore, $$t = 72{\text{ minutes}}$$. Therefore, the usual time taken by him to cover that distance is 72 minutes or 1 hour 12 minutes. **Hence, option B is the correct answer.** **Note:** In the given problem, Sanchit is walking $$6/{7^{th}}$$ of his usual speed.Therefore, he is 12 minutes late. This is because, speed is inversely proportional to time.When he is walking $$6/{7^{th}}$$ of his usual speed, his speed decreases and therefore the time taken to cover the distance increases. Thus, he needed an extra 12 minutes to cover the distance.