SSC CHSL 6 Mar 2018 Shift 1 Paper

Let the distance between A and B be = $3\mathit{d}$ km
Distance covered at the speed of 4 km/hr = $\frac{1}{3}\times 3\mathit{d}=\mathit{d}$ km
Thus, remaining distance covered at the speed of 5 km/hr = $\left(3\mathit{d}-\mathit{d}\right)=2\mathit{d}$ km
Using, time = distance/speed
⇒ $\frac{\mathit{d}}{4}+\frac{2\mathit{d}}{5}=26$
⇒ $\frac{\left(5\mathit{d}+8\mathit{d}\right)}{20}=26$
⇒ $\mathit{d}=26\times \frac{20}{13}$
⇒ $\mathit{d}=2\times 20=40$
∴ Distence between A and B (in km ) = $3\times 40=120$