A man walks a certain distance with certain speed. If he walks 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.
Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,
Distance covered= speed $\times$ dis tance
$=x \times y$
$=x y \ldots(i)$
If the speed is increased by $\frac{1}{2} \mathrm{Km} / \mathrm{hr}$, then time of journey is reduced by 1 hour i.e., when speed is $\left(x+\frac{1}{2}\right) \mathrm{km} / \mathrm{hr}$, time of journey is $(y-1)$ hours
Distance covered $=x y \mathrm{~km}$
$x y=(x-1)(y+3)$
$x y=(x-1)(y+3)$
$x y=x y-1 y+3 x-3$
Thus we obtain the following equations
$-2 x+1 y-1=0$
$3 x-1 y-3=0$
By using elimination method, we have
Putting the value $x=4$ in equation (iii) we get
$3 x-1 y-3=0$
$3 \times 4-1 y-3=0$
$12-1 y-3=0$
$12-3-1 y=0$
$9-1 y=0$
Putting the value of x and y in equations (i) we get
Distance covered $=x y$
$=4 \times 9$
$=36 \mathrm{~km}$
Hence, the distance is $36 \mathrm{~km}$,
The speed of walking is $4 \mathrm{~km} / \mathrm{hr}$.