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

135
236 ²/₃
337 ¹/₂
440

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let distance = x k m and usual rate = y k m p h \\ Then, \frac{x}{y} - \frac{x}{y + 3} = \frac{40}{60} \\ \Rightarrow 2 y (y + 3) = 9 x . . . . . . . . . . (i) \\ and, \frac{x}{y - 2} - \frac{x}{y} = \frac{40}{60} \\ \Rightarrow y (y - 2) = 3 x . . . . . . . . . . . . . (i i) \\ On dividing (i) by (ii), we get x = 40 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{distance}} = x\,km\,{\text{and}}\,{\text{usual}}\,{\text{rate}} = y\,kmph \cr & {\text{Then}},\,\frac{x}{y} - \frac{x}{{y + 3}} = \frac{{40}}{{60}} \cr & \Rightarrow 2y\left( {y + 3} \right) = 9x\,..........\left( i \right) \cr & {\text{and}},\,\frac{x}{{y - 2}} - \frac{x}{y} = \frac{{40}}{{60}} \cr & \Rightarrow y(y - 2) = 3x\,.............\left( {ii} \right) \cr & {\text{On}}\,{\text{dividing}}\,\left( {\text{i}} \right)\,{\text{by}}\,\left( {{\text{ii}}} \right){\text{,}}\,{\text{we}}\,{\text{get}}\,x = 40 \cr}$