Question

A man covers a distance by certain speed. If he increases his speed by 3 km/hr, he reaches 20 min before the actual time. If he reduces speed by 2 km/hr, then he will be late by 20 min. Find distance?

Answer 1

Let n ve the event speed

Let T ve the time taken as speed

Let D ve the distance covered in T & x

Given

$20minutes=\frac{1}{3}hourD=xT$

$\frac{D}{(x+3)}=T-\frac{1}{3}$

$\frac{D}{x-2}=T+1/3$

Adding the two equations we get

$\frac{D(2x+1)}{(x+3)(x-2)}=2T$

$\frac{x(2x+1)}{(x+3)(x-2)}=2$

$2x^2+x=2x^2+2x-12$

$12=x$

$\therefore$ Speed 12 km/hr

$\therefore 15T=D$

$18(T-1/3)=D$

$18T-6=D$

$3T-6=0$

$T=2hours$

$\therefore$ Distance $=24cm$