CAT 2025 Slot 3 · Quant · Question 6

Let the original speed be v km/h. The scheduled journey runs 5 pm to 11 pm, so 6 hours.

What matters is only the part of the trip after the stop. Call the distance still to cover R and the time still available at the original speed T, so R=vT.

With a 20-minute stop, the remaining time shrinks by \frac{1}{3} hour and the speed becomes v+3, so \frac{R}{v+3}=T-\frac{1}{3}.

Substituting R=vT and cross-multiplying: vT=\left( v+3 \right)\left( T-\frac{1}{3}\right), which simplifies to 3T=\frac{v}{3}+1.

With a 30-minute stop and speed v+5: \frac{R}{v+5}=T-\frac{1}{2}, which in the same way simplifies to 5T=\frac{v}{2}+\frac{5}{2}.

From these, T=\frac{v}{9}+\frac{1}{3} and T=\frac{v}{10}+\frac{1}{2}.

Equating, \frac{v}{9}-\frac{v}{10}=\frac{1}{2}-\frac{1}{3}, so \frac{v}{90}=\frac{1}{6} and v=15 km/h.

Shortcut. The 6-hour total and the point of the stop never enter the working — only the leftover distance and leftover time do, and they cancel. Naming R and T for the remainder instead of the whole journey removes two unknowns before you start.