Question

Ravi takes 3 hours more than Abhishek to walk 30 km. But if Ravi doubles his pace, he is ahead of Abhishek by $1\frac{1}{2}$ hours. What was Ravi's original speed?

• A. $5km/hr$
• B. $9km/hr$
• C. $\frac{9}{7}km/hr$
• D. $\frac{10}{3}km/hr$

Answer 1

The correct option is D

$\frac{10}{3}km/hr$

Let the speed of Ravi and Abhishek be x km/hr and y km/hr respectively.

Then, Time taken by Ravi to cover 30 km = $\frac{30}{x}hr$

and time taken by Abhishek to cover 30 km = $\frac{30}{y}hr$

By the given conditions, we have

$\frac{30}{x}-\frac{30}{y}=3$

$\Rightarrow \frac{10}{x}-\frac{10}{y}=1.....\left(i\right)$

If Ravi doubles his pace then Ravi's speed is 2x km/hr.

Time taken by Ravi to cover 30 km $=\frac{30}{2x}hr$

We have,

$\frac{30}{y}-\frac{30}{2x}=1\frac{1}{2}$

$\Rightarrow \frac{30}{y}-\frac{30}{2x}=\frac{3}{2}$

$\Rightarrow \frac{10}{y}-\frac{5}{x}=\frac{1}{2}$

$\Rightarrow \frac{-10}{x}+\frac{20}{y}=\mathrm{1......}\left(ii\right)$

Let $\frac{1}{x}=uand\frac{1}{y}=v$.

In equation (i) and (ii), we get:

$10u-10v=1\Rightarrow 10u-10v-1=\mathrm{0......}\left(iii\right)$

$and-10u+20v=1\Rightarrow -10u+20v-1=\mathrm{0.....}\left(iv\right)$

Adding equations (iii) and (iv), we get

$10v-2=0\Rightarrow v=\frac{1}{5}$

Putting $v=\frac{1}{5}$ in equation (iii), we get

$\Rightarrow 10u-10\frac{1}{5}-1=0$

$\Rightarrow 10u-3=0\Rightarrow u=\frac{3}{10}$

$Now,u=\frac{3}{10}$

$\Rightarrow \frac{1}{x}=\frac{3}{10}$

$\Rightarrow x=\frac{10}{3}$

$\therefore$ Ravi's speed is $\frac{10}{3}km/hr$