Question

If a man walks $1km/h$ faster then his usual speed. Then he covers a distance of $3km$ in $15$ minutes less time. Find the usual speed.

Answer 1

Let usual speed $=x$

time $t_1=\frac{3}{x}$ [when he walks in usual speed]

when he walks 1 km/h as for

$t_2=\frac{3}{x+1}$

given that

$t_1-t_2=15min=\frac{15}{60}hr$

$\frac{3}{x}-\frac{3}{x+1}=\frac{15}{60}=\frac{1}{4}$

$3[\frac{1}{x}-\frac{1}{x+1}]=\frac{1}{4}$

$12\left[\frac{x+1-x}{x(x+1)}\right]=1$

$x^2+x-12=0$

$x^2+4x-3x-12=0$

$x(x+4)-3(x+4)=0$

$x=3,-4$

$x$ cant be negative hence $x=3$