Question

A man walks a certain distance with 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.

Answer 1

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

Since distance covered in each case in same.

$\therefore (y+\frac{1}{2})(x-1)=xy$
and, $(y-1)(x+3)=xy$
$\Rightarrow \frac{x}{2}-y-\frac{1}{2}=0$ 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 $4km/hr$