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

• A. Distance = 45 km, Original speed = 5 km/hr
• B. Distance = 36 km, Original speed = 4 km/hr
• C. Distance = 63 km, Original speed = 7 km/hr
• D. Distance = 72 km, Original speed = 8 km/hr

Answer 1

Answer: (B)

Distance = 36 km, Original speed = 4 km/hr

Let the original speed of the man$=x$ km/hr
The schedule time of walking$=y$ hrs.
$\therefore$ $distance=yx$ km ..... $Distance=Time\times Speed$

According to the first condition
$increasedspeed=(x+\frac{1}{2})$ km/hr

$reducedtime=(y-1)hrs.$

$\therefore distance=(x+\frac{1}{2})(y-1)$
$\Rightarrow xy+\frac{y}{2}-x-\frac{1}{2}$

In both case distance are equal

$\Rightarrow xy=xy+\frac{y}{2}-x-\frac{1}{2}$
$\Rightarrow y-2x-1\iff 2x-y=-1$ .......eq1

According to the second condition
$reducedspeed=(x-1)$ km/hr
$increasedtime=(y+3)$ hrs.
$\therefore distance=(x-1)(y+3)$
$\Rightarrow xy-y+3x-3$

In both case distance are equal
$\Rightarrow xy=xy-y+3x-3\Rightarrow 3x-y=3$ ........eq2

Subtract eq 1 and eq2
$\Rightarrow x=4$

put $x=4$ in eq1
$\Rightarrow 2\times 4-y=-1$
$\Rightarrow -y=-9\Rightarrow y=9$

Hence,The speed of the man$=4$ km/hr
The time of walking$=9$ hrs.
The distance covered by man $4\times 9=36$ km.