Time, Speed and Work — Practice Questions
CSAT-pattern practice on time, speed and work — meeting trains, combined work, pipes with a leak, late-arrival ratios, average speed and efficiency, fully worked.
Questions
These are original CSAT-pattern practice questions written for this site — not past UPSC papers. No calculator; target time about 9 minutes.
Two stations A and B are 330 km apart. A train leaves A for B at 8:00 a.m. at 60 km/h. Another train leaves B for A at 9:00 a.m. at 75 km/h. At what time will the two trains meet?
Solution
By 9:00 a.m. the first train has covered 60 km → gap $= 330 - 60 = 270$ km.
From 9:00, relative speed $= 60 + 75 = 135$ km/h → time to meet $= \dfrac{270}{135} = 2$ h → 11:00 a.m.
Check: first train $60 \times 3 = 180$ km, second $75 \times 2 = 150$ km; $180 + 150 = 330$ ✓.
Answer: B
A can do a piece of work in 10 days, B in 15 days and C in 30 days. In how many days will they finish the work if all three work together?
Solution
Total work $= \text{LCM}(10, 15, 30) = 30$ units. Efficiencies: A $= 3$, B $= 2$, C $= 1$ unit/day → $6$ units/day.
Time $= \dfrac{30}{6} = 5$ days.
Answer: C
Pipe A can fill a tank in 12 hours, while pipe B can empty the full tank in 18 hours. If both pipes are opened together when the tank is empty, in how many hours will the tank be full?
Solution
Capacity $= \text{LCM}(12, 18) = 36$ units. A: $+3$/h; B: $-2$/h → net $+1$/h.
Time $= 36$ hours.
Answer: D
Walking at three-fourths of his usual speed, a man reaches his office 20 minutes late. What is his usual time to reach the office?
Solution
Distance is fixed, so time ∝ $\dfrac{1}{\text{speed}}$. Speed $\times \tfrac{3}{4}$ → time $\times \tfrac{4}{3}$.
Extra time $= \tfrac{4}{3}T - T = \tfrac{1}{3}T = 20$ min → $T = 60$ minutes.
(80 minutes is the new time — a trap.)
Answer: A
A car covers the first one-third of a journey at 20 km/h, the second one-third at 30 km/h and the last one-third at 60 km/h. What is its average speed for the whole journey?
Solution
Take the distance as 180 km (LCM of the speeds), so each third is 60 km.
Times: $\dfrac{60}{20} = 3$ h, $\dfrac{60}{30} = 2$ h, $\dfrac{60}{60} = 1$ h → total $6$ h.
Average speed $= \dfrac{180}{6} = 30$ km/h. (The simple mean, $\approx 36.7$, is the trap.)
Answer: C
A is twice as efficient as B. Working together, they complete a job in 12 days. In how many days can A alone complete the job?
Solution
Let B $= 1$ unit/day, A $= 2$ units/day. Together $3$ units/day for 12 days → job $= 36$ units.
A alone $= \dfrac{36}{2} = 18$ days. (B alone would take 36 days.)
Answer: B