Question

A person walks from his house at a speed of $4{\text{ km/hr}}$ and reaches his school $5$ minutes late. If his speed would have been $5{\text{ km/hr}}$ then he would have reached $10$ minutes earlier. The distance of school from his house is:
A. $5{\text{km}}$
B. \[6{\text{km}}\]
C. $7{\text{km}}$
D. $8{\text{km}}$

Answer 1

Hint:
Let the distance of school from his house be $x{\text{km}}$and let as the speed $v$ he reaches the school in proper time $t$ and now at the speed of $4{\text{ km/hr}}$ he reaches the school in $(t + 5)$ minutes and at the speed $5{\text{ km/hr}}$ he reaches in the time $(t - 10)$ minutes and we know the formula that
${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$ and here the distance is same but speed and time changing in both the situations.

Complete step by step solution:
Let the distance of school from his house be $x{\text{km}}$and let as the speed $v$ he reaches the school in proper time $t$
According to the question, at the speed of $4{\text{ km/hr}}$ he reaches the school in $(t + 5)$ minutes
We know that ${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
Also we know that $1{\text{km}} = 1000{\text{m}}$
$1{\text{hr}} = {\text{60minutes}}$
$1{\text{kmph}} = \dfrac{{1{\text{km}}}}{{1{\text{hour}}}}$
Now we can convert the hours into minutes
$1{\text{kmph}} = \dfrac{{1{\text{km}}}}{{60{\text{minutes}}}}$
So ${\text{4kmph}} = \dfrac{{\text{4}}}{{60}}{\text{km/min}}$
As we are taking the distance in km and the time in minutes so we need to convert hours in minutes
So let us assume that the distance$ = x{\text{ km}}$
So speed we calculated$ = \dfrac{{\text{4}}}{{60}}{\text{km/min}}$
And we know that ${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
And the time taken is $t + 5{\text{ minutes}}$
So $\Rightarrow \dfrac{4}{{60}} = \dfrac{x}{{t + 5}}$
$\Rightarrow t + 5 = 15x - - - - (1)$
And it is also given that at the speed of $5{\text{ km/hr}}$ he reaches in the time $(t - 10)$ minutes
So again we can say that
$\Rightarrow 1{\text{kmph}} = \dfrac{{1{\text{km}}}}{{60{\text{minutes}}}}$
So $\Rightarrow {\text{5kmph}} = \dfrac{5}{{60}}{\text{km/min}}$
So we have ${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
$\Rightarrow \dfrac{5}{{60}} = \dfrac{x}{{t - 10}}$
$\Rightarrow t - 10 = 12x$$ - - - - (2)$
Now subtracting (1) and (2) we get that
$
  - 15 = - 3x \\ \Rightarrow
  3x = 15 \\ \Rightarrow
  x = 5{\text{km}} \\
 $

Hence the distance is $5{\text{km}}$

Note:
The given formula ${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$ is valid only when the speed is constant or when the acceleration is zero but is the acceleration is non-zero then ${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$ is not valid as the acceleration is the change of the speed’s rate and the speed is not constant.