Two vehicles A and B travel with uniform speed 30 km per hour and 60 km per hour respectively in the same direction. They start at the same time and from the same place for a distance of 120 km. The faster vehicle B reaches the destination and travels back with the same speed. Assume that the loss of time to change the direction is negligible. If x is the distance travelled by the slower vehicle A before the two vehicles cross each other, then x is

Correct — C, (c) 80 km. The vehicles start together and travel the same way, so nothing happens until B reaches the destination. B covers 120 km at 60 km per hour, which takes 2 hours, and in those 2 hours A has covered 30 × 2 = 60 km. At that instant the two are 60 km apart, and B turns and comes back. From then on they are approaching each other, so the gap closes at the sum of the speeds, 30 + 60 = 90 km per hour, and it takes 60 ÷ 90 = 2/3 hour to disappear. In that time A covers a further 30 × 2/3 = 20 km. So A has travelled 60 + 20 = 80 km when the two cross, and x is 80 km — option (c).

There is a second route that is quicker once seen. Both vehicles are on the road for the same length of time, and by the moment they meet, A has covered x while B has gone out 120 km and come back to the same point, that is 120 + (120 − x). The two distances together therefore always add to 240 km, whatever the meeting point is. Since the vehicles travel for equal times, their distances stand in the ratio of their speeds, 30 : 60, that is 1 : 2. A therefore covers one third of the combined 240 km, which is 80 km, and B covers the other two thirds, 160 km. That B has indeed gone 160 km can be confirmed directly: 120 km out plus 40 km back leaves it at the 80 km post, exactly where A is.

The insight to carry away is that the turn converts a chase into a meeting. Before the turn the gap is widening at the difference of the speeds; after it, the gap closes at their sum. A candidate who models the journey as a single chase will find that the faster vehicle never meets the slower one, because it is always ahead.

Why the others are wrong

• (a)70 km — Seventy kilometres is too early, and the way to see that is to ask where B is at the moment A reaches that post. A covers 70 km in 70 ÷ 30 = 7/3 hours, which is 2 hours and 20 minutes. In that time B has travelled 60 × 7/3 = 140 km, which is 120 km out and 20 km back, leaving it at the 100 km post. B is therefore still 30 km ahead of A along the road and the two have not yet crossed. The general method here is worth more than the particular answer: on a meeting problem, every candidate value can be tested by converting it into a time and then locating the other body at that time. The two positions must coincide at the moment of crossing, and any option for which one vehicle is still ahead of the other is rejected without solving anything.
• (b)75 km — Seventy-five kilometres is the tempting round figure, because it is well past the halfway point and a candidate who reasons loosely — B turns at 120, A is somewhere behind, they must meet a little beyond the middle — will be satisfied by it. Tested properly it fails. A covers 75 km in 75 ÷ 30 = 2·5 hours, by which time B has travelled 60 × 2·5 = 150 km, that is 120 out and 30 back, standing at the 90 km post. B is still 15 km ahead. The meeting has not yet happened, so 75 km is short of the answer. The lesson is that on a two-phase journey the meeting point is decided by the arithmetic of the second phase and not by any intuition about the middle of the route; here the two-thirds and one-third split of the combined 240 km is the structure that fixes it.
• (d)90 km — Ninety kilometres is too late — the vehicles have already crossed by then. A reaches the 90 km post after 90 ÷ 30 = 3 hours, and in 3 hours B has travelled 180 km, which is 120 km out and 60 km back, leaving it at the 60 km post. B is now 30 km behind A, so the crossing took place earlier, and indeed it took place at 80 km after 8/3 hours. This option is what a candidate produces by allowing the closing phase to run too long, most often by dividing the 60 km gap by 30 rather than by the combined 90 and adding a full hour to A's 60 km. Dividing by the wrong speed is the standard error on the second phase of these problems: once the vehicles face each other, it is the sum of the speeds that governs, not either speed by itself.

This is a relative-speed problem in two phases, and the technique is to treat each phase separately and to join them at the moment the situation changes. In the first phase both vehicles travel the same way, so the gap between them widens at the difference of the speeds; the phase ends when the faster vehicle reaches the destination, at a time found by dividing the distance by its speed. At that instant one records two things: where the slower vehicle has reached, and therefore how large the gap is. In the second phase the vehicles face each other, so the gap closes at the sum of the speeds, and the time to meet is the gap divided by that sum. Adding the two times, or adding the distances the slower vehicle covers in each phase, answers the question. A second and often faster technique uses the fact that both vehicles are on the road for exactly the same length of time. Distances covered in equal times are in the ratio of the speeds, so if the total ground covered by both together can be written down — here 240 km, because the faster vehicle goes out and comes part of the way back to wherever the slower one has reached — then each vehicle's share follows from that ratio without any reference to time at all.

Key facts

• B covers the 120 km at 60 km per hour in 2 hours, by which time A has covered 60 km and the gap is 60 km.
• After B turns, the two approach each other, so the gap closes at the sum of the speeds, 30 + 60 = 90 km per hour.
• The gap of 60 km therefore closes in 60 ÷ 90 = 2/3 hour, during which A covers a further 20 km.
• A has travelled 60 + 20 = 80 km when the two cross, so x = 80 km.
• Alternative method: by the meeting, the two together have covered 240 km, and equal travel times split that in the ratio of the speeds, 1 : 2.
• One third of 240 km is 80 km for A and two thirds is 160 km for B, which is 120 km out and 40 km back to the same post.
• Before the turn the gap grows at the difference of the speeds; after the turn it shrinks at their sum.
• Any candidate answer can be tested by converting it into a time and locating the other vehicle at that instant.

Common traps

• Modelling the journey as a single chase, in which the faster vehicle stays ahead and no meeting ever occurs.
• Using one vehicle's speed rather than the sum of the speeds during the phase in which they approach each other.
• Forgetting to add the distance covered before the turn to the distance covered afterwards.
• Assuming the meeting point must lie near the middle of the route, which the speeds here do not support.
• Reading the question as asking where B is, when what is wanted is the distance travelled by the slower vehicle A.

• practice — not a real PYQ

  Two vehicles start at the same time from the same place towards a destination 60 km away, at uniform speeds of 20 km per hour and 40 km per hour. The faster vehicle turns back immediately on reaching the destination. At what distance from the starting place do the two vehicles cross each other?

  • (a)30 km
  • (b)35 km
  • (c)40 km
  • (d)45 km

  Answer(c) 40 km

• practice — not a real PYQ

  A cyclist travelling at a uniform 12 km per hour is overtaken by a motorcyclist who starts from the same place along the same route 2 hours later at a uniform 30 km per hour. How long after starting does the motorcyclist overtake the cyclist?

  • (a)1 hour
  • (b)1 hour 20 minutes
  • (c)1 hour 30 minutes
  • (d)2 hours

  Answer(b) 1 hour 20 minutes