A man walks a certain distance with a certain speed. If he walks $\frac{1}{2}$ km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.

Given:

A man walks a certain distance with a certain speed. If he walks $\frac{1}{2}$ km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours.

To do:

We have to find the distance covered by the man and his original rate of walking.

Solution:

Let the distance be $x$ km and the original speed be $y$ km/hr.

We know that,

Time $=$ Distance $\div$ Speed 

This implies,

Time taken to cover the distance of $x$ km at a speed of $y$ km/hr $=\frac{x}{y}$.

If he walks $\frac{1}{2}$ km an hour faster, he takes 1 hour less.

This implies,

Speed$_{1}$ $=y+\frac{1}{2}=\frac{2y+1}{2}$ km/hr.

Time $_{1}=\frac{x}{\frac{2y+1}{2}}=\frac{2x}{2y+1}$

According to the question,

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

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

$(x-y)(2y+1)=2x(y)$    (On cross multiplication)

$2xy+x-2y^2-y=2xy$

$x=2y^2+y$.....(i)

If he walks 1 km an hour slower, he takes 3 more hours.

This implies,

Speed$_{2}$ $=y-1$ km/hr.

Time $_{2}=\frac{x}{y-1}$ hr

According to the question,

$\frac{x}{y}+3=\frac{x}{y-1}$

$\frac{x+3y}{y}=\frac{x}{y-1}$

$(x+3y)(y-1)=x(y)$    (On cross multiplication)

$xy-x+3y^2-3y=xy$

$x=3y^2-3y$.....(ii)

From equations (i) and (ii), we get,

$2y^2+y=3y^2-3y$

$3y^2-2y^2-3y-y=0$

$y^2-4y=0$

$y(y-4)=0$

$y=0$ or $y=4$

But $y=0$ is not possible.

Therefore, $y=4$ 

$\Rightarrow x=2(4)^2+(4)$

$x=32+4$

$x=36$

Therefore, the distance covered by the man is $36\ km$ and his original rate of walking is $4\ km/hr$.

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Updated on: 10-Oct-2022

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