SSC Sample Paper Mock Test-8 SSC CGL Tear-II Paper-1

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

A) 20

B) 35

C) \[36\frac{2}{3}\]

D) 40

Correct Answer: D

Solution :

Let the distance be x km and initial speed y km/h.

According to question, \[\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\] ...(i)

and \[\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\] ?(ii)

From Eqs. (i) and (ii), we get

\[\frac{x}{y}-\frac{x}{y+3}=\frac{x}{y-2}-\frac{x}{y}\]

\[\Rightarrow \] \[\frac{1}{y}-\frac{1}{y+3}=\frac{1}{y-2}-\frac{1}{y}\]

\[\Rightarrow \] \[\frac{y+3-y}{y\,\,(y+3)}=\frac{y-y+2}{y\,\,(y-2)}\]

\[\Rightarrow \] \[3\,\,(y-2)=2(y+3)\]

\[\Rightarrow \] \[3y-6=2y+6\]

\[\Rightarrow \] \[y=12\]

From Eq. (i) \[\frac{x}{12}-\frac{x}{15}=\frac{40}{60}\]\[\Rightarrow \]\[\frac{5x-4x}{60}=\frac{2}{3}\]

\[\Rightarrow \] \[x=\frac{2}{3}\times 60=40\]

\[\therefore \] Distance = 40 km