Question

A man covered a certain distance at some speed had he moved 3 km/hr faster he would have taken 40 minutes less If he had moved 2 km/hr slower he would have taken 40 minutes more The distance ( in km) is

• A. 35
• B. $37\frac{1}{2}$
• C. $36\frac{2}{3}$
• D. 40

Answer 1

Answer: (D)

Let the distance = x km and usual rate = y km/hr Then $\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\Rightarrow \frac{x(y+3)-xy}{y(y+3)}=\frac{2}{3}\Rightarrow \frac{3x}{y(y+3)}=\frac{2}{3}$
$\Rightarrow 2y(y+3)=9x...............\left(i\right)$
and $\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\Rightarrow \frac{xy-x(y-2)}{y(y-2)}=\frac{2}{3}$ $\Rightarrow \frac{2x}{y(y-2)}=\frac{2}{3}$
$\Rightarrow y(y-2)=3x...............\left(ii\right)$
On dividing eqn (i) by eqn (ii) we get
$\frac{2(y+3)}{y-2}=3\Rightarrow 2y+6=3y-6\Rightarrow y=12$
$\therefore$ Putting the value of y in (i) we get 2 x 12 x 15 = 9x $\Rightarrow$ x = 40 km/hr