Question

A person walks from his house at a speed of $4km/hr$ and reaches his school $5$ minutes late. If his speed had been $5km/hr$ , he would have reached $10$ minutes earlier. The distance between the school from his house is
(A) $5km$
(B) $6km$
(C) $7km$
(D) $8km$

Answer 1

This problem can be solved by equating the difference between the time taken by the two speeds, which can be calculated using the formula for speed, to the total time difference mentioned in the question $(5 + 10)\min$ .

Formula used:
Speed, $v = \dfrac{x}{T}$
where, $x$ is the total distance travelled and $T$ is the total time taken.

Complete step by step answer:
Let $x$ be the exact distance between the person’s house and school.
In the first case, he travels at a speed of $4km/hr$ .
Therefore the time taken say ${t_1}$ for him to travel to his school from his house can be given as,
${t_1} = \dfrac{x}{4}$
In the second case, he travels at a speed of $5km/hr$ .
Therefore the time taken say ${t_2}$ for him to travel to his school from his house can be given as,
${t_2} = \dfrac{x}{5}$
Now, in the first case, he was $5$ minutes late to reach his school, while in the second case he was $10$ minutes earlier to reach his school.
Therefore we can say that the difference between the two times taken with two different speeds is,
${t_1} - {t_2} = (5 + 10)\min$
Also, we know that $1\min = \dfrac{1}{{60}}hr$ .
Therefore upon substituting these values we get,
$\dfrac{x}{4} - \dfrac{x}{5} = \dfrac{{5 + 10}}{{60}}$
Upon further solving this equation we get,
$\dfrac{x}{4} - \dfrac{x}{5} = \dfrac{{15}}{{60}}$
$\Rightarrow \dfrac{{5x - 4x}}{{20}} = \dfrac{1}{4}$
Multiplying with $4$ on both sides we get,
$\dfrac{x}{5} = 1$
or, $x = 5$
Hence, the distance of the school from this person’s house is $5km$ .
Therefore option (A) is the correct answer.

Note:
Keep in mind of the unit in this question. The unit of speed is given as $km/hr$ but the time is given in minutes. Hence it is necessary to convert the unit of time from minutes to hour in order to solve this question. Conversion from minutes to seconds can also be done, but in that case, the speed needs to be converted to $m/s$ as well.