A and B start at the same time to reach the same destination. B travelled at 5/7 of A's speed and reached the destination 1 hour 20 minutes after A. What was the time taken by B to reach the destination?
- (a)4 hours 40 minutes
- (b)4 hours 55 minutes
- (c)5 hours 5 minutes
- (d)5 hours 15 minutes
Correct — A, 4 hours 40 minutes. Over a fixed distance, time varies inversely with speed. B travels at 5/7 of A's speed, so B takes 7/5 of A's time. The difference between them is 7/5 − 1 = 2/5 of A's time, and that difference is given as 1 hour 20 minutes, or 80 minutes. So (2/5) of A's time = 80 minutes, making A's time 200 minutes. B's time is 200 + 80 = 280 minutes, which is 4 hours 40 minutes. The same figure follows from 7/5 × 200 = 280.
- (b)4 hours 55 minutes — 295 minutes would make A's time 215 minutes, and 215 × 7/5 is 301, not 295. The ratio does not hold.
- (c)5 hours 5 minutes — 305 minutes implies A took 225 minutes, but 225 × 7/5 = 315. The 80-minute gap would then be 90.
- (d)5 hours 15 minutes — 315 minutes is 7/5 of 225 minutes, so it fits the ratio but forces a gap of 90 minutes rather than the 80 the question gives.
For a fixed distance, speed and time are inversely proportional: if speeds are in the ratio m : n, the times are in the ratio n : m. Turning the given speed ratio into a time ratio converts a rate problem into simple proportion, and any stated difference in arrival times becomes the difference between the two parts of that ratio.
Working in minutes avoids the mixed units that cost marks here. A speed factor of 5/7 gives a time factor of 7/5, and the useful quantity is the surplus 2/5, which is what the 80 minutes measures. Two checks are worth making at the end: B's time must exceed A's, and the ratio of the two times must come back to 7 : 5, since 280 : 200 reduces to 7 : 5.
- Over the same distance, time is inversely proportional to speed.
- A speed ratio of 5 : 7 gives a time ratio of 7 : 5.
- The difference of 2 parts in that ratio equals 1 hour 20 minutes, so one part is 40 minutes.
- A's time is 5 parts, 200 minutes; B's time is 7 parts, 280 minutes, or 4 hours 40 minutes.
- Converting hours and minutes into minutes before dividing removes most of the arithmetic risk.
A takes 5 parts, that is 200 minutes, and the 80-minute difference is the two extra parts B needs.
- Multiplying A's time by 5/7 instead of 7/5.
- Treating 1 hour 20 minutes as 1.20 hours instead of 80 minutes.
- Reporting A's time of 200 minutes when the question asks for B's.
A proportion item dressed as a journey — no distance is given because none is needed once the speed ratio is turned upside down.
A person travelled from one place to another at an average speed of 40 kilometres/hour and back to the original place at an average speed of 50 kilometres/hour. What is his average speed in kilometres/hour during the entire roundtrip?
- (a) 45
- (b) 20
- (c) 400/9
- (d) Impossible to find out unless the distance between the two places is known
Answer(c) 400/9
The same principle that the distance cancels. There it lets two speeds be combined without knowing the route length; here it lets a ratio of speeds be turned into a ratio of times.
- practice — not a real PYQ
A covers a distance in 6 hours. B travels at 3/4 of A's speed. How long does B take?
- (a)7 hours
- (b)7 hours 30 minutes
- (c)8 hours
- (d)9 hours
Answer(c) 8 hours — B's time is 4/3 of A's, and 4/3 × 6 = 8.
- practice — not a real PYQ
Two runners cover the same track, one at 12 km/hr and the other at 15 km/hr. The slower takes 12 minutes longer. What is the length of the track?
- (a)8 km
- (b)10 km
- (c)12 km
- (d)15 km
Answer(c) 12 km — d/12 − d/15 = 0.2 hour gives d/60 = 0.2, so d = 12 km.