SSC CHSL Model Paper 8

Let the distance between A and B be D km and the real speed of the car be S km/hr.
First time the car takes 90 minutes more and second time the car takes 75 minutes more than the scheduled time.
So, ${\mathit{T}}_1-{\mathit{T}}_2=\frac{15}{60}$
Difference in time taken during the two journeys:
$\frac{15}{60}=\left(\frac{100}{\mathit{s}}+\frac{\mathit{D}-100}{3\mathit{s}∕4}\right)$ $-\left(\frac{160}{\mathit{s}}+\frac{\mathit{D}-160}{3\mathit{s}∕4}\right)$
⇒ $\frac{15}{60}=\frac{100}{\mathit{s}}+\frac{4\mathit{D}}{3\mathit{s}}-\frac{400}{3\mathit{s}}-\frac{160}{\mathit{s}}$ $-\frac{4\mathit{D}}{3\mathit{s}}+\frac{640}{3\mathit{s}}$
⇒ $\frac{15}{60}==\frac{240}{3\mathit{s}}-\frac{60}{\mathit{s}}$
⇒ $\frac{15}{60}=\frac{20}{\mathit{s}}$
⇒ s = 80 km/hr