SSC CHSL 15 Nov 2015 Shift 2 Paper

Let total distance travelled = 2d km and total time taken = 5 hr 15 min = 5$\frac{1}{4}$ hrs
Half of the distance (i.e. km) is travelled at 60 km/hr and remaining at 45 km/hr
Using, time = distance/speed
⇒ $\left(\frac{\mathit{d}}{60}\right)+\left(\frac{\mathit{d}}{45}\right)=5\frac{1}{4}$
⇒ $\frac{\mathit{d}}{15}\left(\frac{1}{4}+\frac{1}{3}\right)=\frac{21}{4}$
⇒ $\frac{\mathit{d}}{15}\times \left(\frac{7}{12}\right)=\frac{21}{4}$
⇒ $\frac{\mathit{d}}{15}\times \left(\frac{7}{12}\right)=\frac{21}{4}$
⇒ $\mathit{d}=\frac{21}{4}\times \frac{12}{7}\times 15$
⇒ $\mathit{d}=9\times 15=135$
∴ Total distence travelled = $2\times 1356=270$ km