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 kmOriginal rate of walking is $$4\,km/hr$$ Mathematics

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