Question

A man covers a certain distance on bike. Had he moved $4kmph$ faster, he would have taken $72$ minutes less. If he moves slow by $4kmph$, he will take $2$ hours more to cover the distance. What is the distance traveled by him?

• A. $96$ km
• B. $40$ km
• C. $50$ km
• D. $25$ km

Answer 1

Answer: (A)

$96$ km

Let the distance travelled be $dkm$ and speed be be $skm/hr$

Time taken if travelled at normal speed $=\frac{d}{s}hr$

Had he moved $4kmph$ faster, he would have taken $72$ minutes less

$\Rightarrow \frac{d}{s+4}=\frac{d}{s}-\frac{72}{60}$

$\Rightarrow \frac{d}{s+4}=\frac{d}{s}-\frac{6}{5}$

$\Rightarrow d(\frac{1}{s+4}-\frac{1}{s})=-\frac{6}{5}$

$\Rightarrow d\left(\frac{s-s-4}{s(s+4)}\right)=-\frac{6}{5}$

$\Rightarrow d\left(\frac{-4}{s(s+4)}\right)=-\frac{6}{5}$

$\Rightarrow d\left(\frac{2}{s(s+4)}\right)=\frac{3}{5}$

$\Rightarrow 10d=3s(s+4)$ ............$\left(1\right)$

If he moves slow by $4kmph$, he will take $2$ hours more to cover the distance

$\Rightarrow \frac{d}{s-4}=\frac{d}{s}+2$

$\Rightarrow \frac{d}{s-4}-\frac{d}{s}=2$

$\Rightarrow d(\frac{1}{s-4}-\frac{1}{s})=2$

$\Rightarrow d\left(\frac{s-s+4}{s(s-4)}\right)=2$

$\Rightarrow d\left(\frac{2}{s^2-4s}\right)=1$

$\Rightarrow 2d=s^2-4s$

or $\Rightarrow 2d\times 5=(s^2-4s)\times 5$

$\Rightarrow 10d=5s^2-20s$ ..........$\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow 3s(s+4)=5s^2-20s$

$\Rightarrow 3s^2+12s=5s^2-20s$

$\Rightarrow 3s^2+12s-5s^2+20s=0$

$\Rightarrow -2s^2+32s=0$

$\Rightarrow -2s(s-16)=0$

$\therefore s=16kmph$ and $s\ne 0$

Substituting $s=16kmph$ in $\left(1\right)$ we get

$10d=3\times 16(16+4)$

$\Rightarrow 10d=3\times 16\times 20$

$\Rightarrow d=3\times 16\times 2=96km$

Hence he travelled $96km$