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Question

A man walks a certain distance with certain speed. If he walks $$\dfrac {1}{2}$$ 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.


Solution

Let the original speed by $$y\ km/hr$$ and total time taken be $$x$$ $$hrs$$. Then,
$$Distance=(xy)\ km$$.

Since distance covered in each case in same.

$$\therefore \left( y+\dfrac { 1 }{ 2 }  \right) \left( x-1 \right) =xy$$
and, $$(y-1)(x+3)=xy$$
$$\Rightarrow \dfrac { x }{ 2 } -y-\dfrac { 1 }{ 2 } =0$$ and $$-x+3y-3=0$$

Solving equations $$x-2y-1=0$$,$$x-3y+3=0$$
we get,$$x=9$$ and $$y=4$$

Therefore, Distance=xy=36 km
Original rate of walking is $$ 4\,km/hr$$ 

Mathematics

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