Speed, Time & Distance: previous-year questions

Answer: (B) 24 m/s

When equal distances are covered at speeds $v_1$ and $v_2$, the average speed is the harmonic mean $\frac{2v_1v_2}{v_1 + v_2} = \frac{2 \times 20 \times 30}{20 + 30} = 24$ m/s. Option C (25 m/s) is the simple average, which would be correct only if equal times, not equal distances, were spent at each speed; the true value is lower because the body spends longer on the slower half. Option A confuses average speed with average velocity in a round trip. Check with a distance of 120 m: time $= \frac{60}{20} + \frac{60}{30} = 5$ s, so speed $= \frac{120}{5} = 24$ m/s.

Answer: (A) is 16 m/s.

For two equal distances the average speed is the harmonic mean: $\frac{2 \times 10 \times 40}{10 + 40} = \frac{800}{50} = 16$ m/s. It does not depend on the total distance (option D), because the distance cancels out: for a total distance $2d$, time $= \frac{d}{10} + \frac{d}{40} = \frac{d}{8}$, and speed $= \frac{2d}{d/8} = 16$. Option C (25 m/s) is the arithmetic mean, valid only when equal times are spent at each speed. Notice how strongly the slow half pulls the average down: the answer is much nearer 10 than 40.

Answer: (A) 10:10 am

The cars move towards each other, so their relative speed is $50 + 30 = 80$ km/h. Time to meet $= \frac{160}{80} = 2$ hours, and 2 hours after 8:10 am is 10:10 am. They meet $50 \times 2 = 100$ km from A and $30 \times 2 = 60$ km from B. Rule: for bodies moving in opposite directions add the speeds; for the same direction subtract them. If the two cars were travelling in the same direction, the faster one would close a 160 km gap in $\frac{160}{20} = 8$ hours.

Answer: (B) 500 m

60 km/h is 1 km per minute, so in 2 minutes the train travels 2 km. To pass completely through a tunnel a train must cover the length of the tunnel plus its own length, so its length is $2 - 1.5 = 0.5$ km $= 500$ m. The same rule applies to bridges and platforms: distance covered equals the length of the object plus the length of the train, whereas for passing a pole or a standing man it is the train's length alone. Useful conversion: 1 km/h $= \frac{5}{18}$ m/s, so 60 km/h $= 16\frac{2}{3}$ m/s.

Answer: (B) 28 minutes

For the same distance time is inversely proportional to speed; speed × 1·25 means time ÷ 1·25. So train B takes 35 ÷ 1·25 = 28 minutes. Simply cutting 25% of the time would wrongly give about 26 minutes, and 30 or 36 minutes are too slow.

Answer: (B) 4 hours

Call each of AB and BC one stretch. The round trip A→B→A is one stretch with the stream and one against it, taking 6 hours; A to C (two stretches) taking 8 hours means A→C is upstream at 4 hours a stretch. So one stretch downstream takes 6 − 4 = 2 hours, and C to A (two stretches downstream) takes 4 hours.

Answer: (B) 4 hours

A's speed is (12 + s) and B's is (15 − s) for stream speed s, so their closing speed is 27 km/hr whatever s is. They meet after 108 ÷ 27 = 4 hours. 4·5 and 6 hours would need a slower closing speed.

Answer: (B) 9 kmph

Taking three times as long upstream means the downstream speed is three times the upstream speed: 18 + s = 3(18 − s). So 4s = 36 and s = 9 kmph. With s = 6 the downstream speed would be only twice the upstream speed.

Answer: (B) 3 : 4

To pass a pole the train covers its own length, 240 m; to cross the bridge it covers 240 + 80 = 320 m. At the same speed the times are in the ratio 240 : 320 = 3 : 4. Forgetting the train's own length on the bridge gives a wrong ratio such as 3 : 1.

Answer: (A) 230 m

In opposite directions the relative speed is 200 km/h = $\frac{500}{9}$ m/s, so in 9 seconds they cover 500 m together. That is the sum of the two lengths, so the other train is 500 − 270 = 230 m.

Answer: (B) 4 : 3

For two trains starting together towards each other, speed ratio = $\sqrt{t_2} : \sqrt{t_1}$, where $t_1$, $t_2$ are their times after meeting. Here that is $\sqrt{16} : \sqrt{9}$ = 4 : 3. 9 : 16 is the ratio of the times, not of the speeds.

Answer: (A) 12 km/hr

For equal distances the average speed is $\frac{2xy}{x + y} = \frac{2 \times 10 \times 15}{25} = 12$ km/hr. 12·5 km/hr is the simple mean of the speeds, which is wrong because more time is spent at the slower speed.

No option is correct as printed — this question is disputed. WBPSC awarded the mark to all candidates: no correct option — the journey works out to 70 km, which is not listed (140 km would need a 30-minute delay).

The extra time is $\frac{d}{35} - \frac{d}{40} = \frac{d}{280}$ hours, and setting this to $\frac{1}{4}$ hour gives d = 70 km. 70 km is not among the options; 140 km would make the train 30 minutes late. The Commission therefore awarded the mark to all candidates.

Answer: (A) 50 km

In the same time, the faster speed covers 4 km more each hour, so 20 km extra takes 5 hours. The actual distance at 10 km/hr is 10 × 5 = 50 km. 70 km is the distance at the faster speed.

Answer: (A) 72 km/hr

The extra 60 m of bridge takes 3 extra seconds, so the speed is 20 m/s. 20 m/s × $\frac{18}{5}$ = 72 km/hr (the train is 120 m long). The other speeds do not fit both timings.

Answer: (D) 5 km/hr

Downstream speed = 24 ÷ 4 = 6 km/hr and upstream speed = 24 ÷ 6 = 4 km/hr. The boat's speed in still water is their average, (6 + 4) ÷ 2 = 5 km/hr. 6 km/hr is the downstream speed.

Answer: (B) 60 minutes

At $\frac{5}{4}$ of the speed the time is $\frac{4}{5}$ of the usual time, a saving of $\frac{1}{5}$ of it. So $\frac{1}{5}$ of the usual time = 12 minutes, and the usual time is 60 minutes. 48 minutes is the time at the faster speed.

Answer: (C) 3·5 km/hr

Downstream speed = 15 ÷ 3 = 5 km/hr and upstream speed = 15 ÷ 7·5 = 2 km/hr. Speed in still water = (5 + 2) ÷ 2 = 3·5 km/hr. 1·5 km/hr is the speed of the stream.

Answer: (A) 50 km

In the same time the faster pace covers 4 km more per hour, so the 20 km extra means 5 hours of walking. At the actual speed of 10 km/hr he covered 50 km. 70 km is what he would have covered at 14 km/hr.

Answer: (A) 1 : 12

The hour hand completes one rotation ($360^{\circ}$) in 12 hours, while the minute hand does so in 1 hour. Since angular speed $\omega=\frac{2\pi}{T}$, we get $\omega_h:\omega_m=\frac{1}{12}:1=1:12$. Option (B) is the inverse (minute hand : hour hand). For comparison, minute hand : second hand $=1:60$ and hour hand : second hand $=1:720$; the hour hand moves $0.5^{\circ}$ per minute and the minute hand $6^{\circ}$ per minute.