A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace (speed) he is ahead of B by $1\frac{1}{2}$ hours. Find the speeds of A and B.

Given:

A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace (speed) he is ahead of B by $1\frac{1}{2}$ hours.
To do:

We have to find the speeds of A and B.

Solution:

Total distance $= 30\ km$

Let the speed of A be $x\ km/hr$ and the speed of B be $y\ km/hr$.

We know that,

$Time=\frac{Distance}{Speed}$

According to the question,

While covering a distance of 30 km, A takes 3 hours more than B.

This implies,

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

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

If A doubles his speed, he is ahead of B by $1\frac{1}{2}$ hours.

This implies,

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

$\frac{15}{x}-\frac{30}{y}=-1\frac{1}{2}$......(ii)

Subtracting equation (ii) from equation (i), we get,

$\frac{30}{x}-\frac{30}{y}-\frac{15}{x}+\frac{30}{y}=3-(-1\frac{1}{2})$

$\frac{30-15}{x}=3+\frac{2\times1+1}{2}$

$\frac{15}{x}=\frac{3\times2+3}{2}$

$x=\frac{15\times2}{9}$

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

Substituting the value of $x=\frac{10}{3}$ in equation (i), we get,

$\frac{30}{\frac{10}{3}}-\frac{30}{y}=3$

$\frac{30\times3}{10}-3=\frac{30}{y}$

$9-3=\frac{30}{y}$

$y=\frac{30}{6}$

$y=5$

The speed of A is $\frac{10}{3}\ km/hr$ and the speed of B is $5\ km/hr$. 

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Updated on: 10-Oct-2022

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