SSC CPO SI and ASI Model Paper 4

(2) Let the speed of the bus bex kmph and distance be y.
∴ Original time required to reachdestination $=\frac{\mathit{y}}{\mathit{x}}$h
Time required at the speed of20 kmph $=\frac{\mathit{y}}{20}\mathit{h}$
Time required at the speed of25 kmph $=\frac{\mathit{y}}{25}\mathit{h}$
∴ According to the question,
$\frac{\mathit{y}}{20}-1=\frac{\mathit{y}}{25}$
⇒ 5y – 4y = 200
⇒ y = 200 km