Question

$X$ takes $3$ hours more than $Y$ to walk $30km$. But, if $X$ doubles his pace, he is ahead of $Y$ by $1\frac{1}{2}$ hours. Find their speed of walking.

Answer 1

Let the speed of X and Y be x km/hr and y km/hr respectively. Then,

Time taken by X to cover $30$km $=\frac{30}{x}$hrs,

and, Time taken by Y to cover $30$km $=\frac{30}{y}$hrs

By the given conditions, we have

$\frac{30}{x}-\frac{30}{y}=3\Rightarrow \frac{10}{x}-\frac{10}{y}=1$ .(i)

If X doubles his pace, then speed of X is $2x$ km/hr

Times taken by X to cover $30$ km $=\frac{30}{2x}$hrs

Times taken by Y to cover $30$km $=\frac{30}{y}$hrs

According to the given conditions, we have

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

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

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

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

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, in equations (i) and (ii) we get

$10u-10v=1\Rightarrow 10u-10v-1=0$ (iii)

$-10u+20v=1\Rightarrow -10u+20v-1=0$ (iv)

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

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

Now, $u=\frac{3}{10}\Rightarrow \frac{1}{x}=\frac{3}{10}\Rightarrow x=\frac{10}{3}$ and, $v=\frac{1}{5}\Rightarrow \frac{1}{y}=\frac{1}{5}\Rightarrow y=5$

Hence, X's speed $=\frac{10}{3}$ km/hr and, Y's speed $=5$ km/hr.