A train starts from station A at 9 AM and reaches station B at 2 PM. Another train starts from station B at 11 AM and reaches station A at 3 PM. On their way, they meet at a point P. The distances from station A to P and from station B to P bear the ratio :
- (a)1 : 2
- (b)2 : 1
- (c)2 : 3
- (d)3 : 2
Answer
Why
Correct — B, (b) 2 : 1. Distance A to P is twice distance B to P.
Set the distance between the stations as D. Train 1 runs A to B from 9 AM to 2 PM, that is 5 hours, so its speed is D/5. Train 2 runs B to A from 11 AM to 3 PM, that is 4 hours, so its speed is D/4.
The head start. Train 2 leaves two hours after train 1. In those two hours train 1 covers 2 × D/5 = 2D/5, so when train 2 starts, the gap between them is D − 2D/5 = 3D/5.
Closing the gap. Moving towards each other, they close the gap at the sum of their speeds: D/5 + D/4 = 4D/20 + 5D/20 = 9D/20 per hour. Time to meet after 11 AM = (3D/5) ÷ (9D/20) = (3D/5) × (20/9D) = 60/45 = 4/3 hours, that is 1 hour 20 minutes. They meet at 12:20 PM. Note that D cancels — the answer cannot depend on the actual distance.
The two distances. Train 1 has been running from 9 AM to 12:20 PM, which is 2 + 4/3 = 10/3 hours, so AP = (D/5) × (10/3) = 2D/3. Train 2 has been running from 11 AM to 12:20 PM, which is 4/3 hours, so BP = (D/4) × (4/3) = D/3.
The ratio AP : BP = 2D/3 : D/3 = 2 : 1.
Two checks. First, the two parts must add up to the whole: 2D/3 + D/3 = D, which they do. Second, a sense check — train 1 is both slower and earlier, and the earliness dominates here, so it should have covered the larger share of the route by the time they meet. It has: two-thirds of it.
The same answer by a shorter route. Once the meeting time is known, the two distances are just speed multiplied by time, so their ratio is (D/5 × 10/3) ÷ (D/4 × 4/3) — and rather than evaluating both, note that it is (10/3 ÷ 4/3) × (D/5 ÷ D/4) = (10/4) × (4/5) = 2. The ratio of running times multiplied by the ratio of speeds gives the ratio of distances, and each factor is easy on its own.
Why the others are wrong
- (a)1 : 2 — The correct pair of numbers with the two stations interchanged. The stem asks for the distances 'from station A to P and from station B to P', in that order, so the first term must be the part covered by the train that started at A. That train had a two-hour start and covered the larger share. Where an option set contains a ratio and its reverse, the order in the stem is the whole of the question.
- (c)2 : 3 — This is close to the ratio of the two trains' SPEEDS rather than of the distances they covered — the speeds are D/5 and D/4, that is in the ratio 4 : 5. A candidate who compares journey times or speeds without allowing for the two-hour head start produces a ratio of this shape. The head start is worth 2D/5 of the route and cannot be left out.
- (d)3 : 2 — Right in direction — the first train has gone further — but wrong in size, and it typically comes from taking the meeting time as the midpoint of some interval rather than computing it. The meeting is at 12:20 PM, which is 3 hours 20 minutes into train 1's journey and 1 hour 20 minutes into train 2's; those two times, weighted by the respective speeds, give 2 : 1 and not 3 : 2.
Concept
Two bodies approaching each other close the distance between them at the SUM of their speeds; two moving in the same direction close it at the difference. That is the whole of relative-speed reasoning. When one starts earlier, the standard method is to advance the clock to the moment the second one starts, compute how much of the route the first has already covered, and treat the remainder as a fresh approach problem. Where the distance is not given, setting it as D and expressing both speeds in terms of D lets it cancel — which it must, since a ratio of two parts of the same route cannot depend on how long the route is. This family also includes overtaking problems, trains crossing platforms and poles, and boats in streams, all built on the same addition or subtraction of speeds.
The examiner has given the timings in a form that makes the durations easy — five hours and four hours — but has staggered the departures, which is where the work lies. Note also the stem's construction: it ends in a colon and the options complete the sentence, and two of the four options are the same ratio reversed. Both are standard features of this paper, and both reward reading the stem to its last word before choosing.
Two of the four options are the same ratio reversed, and so are the other two — which means the order in which the stem names the distances is doing as much work as the arithmetic. That construction appears repeatedly on this paper's quantitative items, and it punishes a candidate who computes correctly and then writes the terms in the order that feels natural rather than the order asked for. Note also that the stem ends in a colon and the options complete the sentence; reading stem and option together as one sentence is the way to keep the order straight.
Key facts
- Train 1 takes 5 hours for the route (9 AM to 2 PM); train 2 takes 4 hours (11 AM to 3 PM).
- Speeds are therefore D/5 and D/4 for a route of length D.
- In its two-hour head start train 1 covers 2D/5, leaving a gap of 3D/5 at 11 AM.
- Approaching each other, they close that gap at D/5 + D/4 = 9D/20 per hour.
- Time to meet after 11 AM = (3D/5) ÷ (9D/20) = 4/3 hours, so they meet at 12:20 PM.
- AP = (D/5) × (10/3) = 2D/3 and BP = (D/4) × (4/3) = D/3, giving 2 : 1.
- The distances add back to D, which is the natural check on this kind of answer.
- Bodies moving towards each other close at the sum of their speeds; in the same direction, at the difference.
Study next
Common traps
- Reversing the order of the ratio; the stem names A to P first.
- Ignoring the two-hour head start.
- Comparing the speeds instead of the distances actually covered.
- Assuming the trains meet midway or at the midpoint in time.
Meeting-point problems are a fixture of the quantitative block. Solve them by fixing the distance as one symbol, converting every timing into a speed, and advancing the clock to the later departure; the symbol always cancels when the answer is a ratio. Expect one meeting-point or overtaking problem per paper, and expect the option list to contain the reversed ratio; the arithmetic and the reading are examined together.
Related PYQs
EPFO_EOAO_2020_Q82If the ratio of speeds of ‘A’ and ‘B’ is 5 : 6 and ‘B’ allows ‘A’ a start of 70 metres in a 1·2 km race, who will win the race and by what distance ?
- (a) ‘A’ wins by 30 m.
- (b) ‘B’ wins by 200 m.
- (c) ‘B’ wins by 130 m.
- (d) The race finishes in a dead heat.
Answer(c) ‘B’ wins by 130 m.
The other relative-speed item in these papers — two runners whose speeds are in a given ratio with one allowed a start, decided by the same comparison of rates over a fixed distance.
Practice
- practice — not a real PYQ
A train leaves station A at 8 AM and reaches station B at 12 noon. Another train leaves station B at 8 AM and reaches station A at 2 PM. At what time do they meet ?
- (a)9 : 24 AM
- (b)10 : 00 AM
- (c)10 : 24 AM
- (d)11 : 00 AM
Answer(c) 10 : 24 AM