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PROFIT AND LOSS PROBLEMS

Aptitude

PROBLEMS ON TRAINS -> IMPORTANT FORMULAE

1. a km/hr = [a * 5/18]m/s.
2. a m/s = [a * 18/5] km/hr.
3. Time taken by a trian of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.
4. Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l + b) metres.
5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relatives speed = (u - v) m/s.
6. Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s
7. If two trains of length a metres and b metres are moving in opposite directions at u
8. If two trains of length a metres and b metres are moving in the same direciton at u m/s and v m/s, then the time taken by the faster train to cross the
slower train = (a + b)/(u - v) sec.
9. If tow trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then
(A’s speed) : (B’s speed) = (√b : √a).

PROBLEMS ON TRAINS -> SOLVED EXAMPLES

1. Two tain 100 metres and 120 metres long are running in the same direction with speeds of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?
  Sol. Relative speed of the train = (72 - 54) km/hr = 18 km/hr
= [18 * 5/18] m/sec = 5 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (100 + 120) m at 5 m/sec = [220/5]sec = 44 sec.
2. A train 220 m long is running with a speed of 59 kmph. In what time will it pas a man who is running at 7 kmph in the direction opposite to that in which the tain is going?
  Sol. Speed of the train relative to man = (59 + 7) kmph
= [66 * 5/18] m/sec = [55/3] m/sec.
Time taken by the train to cross the man
= Time taken by it to cover 220m at [55/3] m/sec
= [220 * 3/55] sec = 12 sec.
3. A man siting in a trian which is travelling at 50 kmph observes that a goods trian, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.
  Sol.
Relative speed = [280/9] m/sec = [280/9 * 18/5] kmph = 112 kmph.
∴ Speed of goods train = (112 - 50) kmph = 62 kmph.

PROBLEMS ON TRAINS -> EXERCIES

25. Two trains 140 m and 160 m long run at the speed fo 60 k/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:
 
  • A. 7
  • B. 8.6
  • C. 10.8
  • D. 11
Ans: C.
Sol.
Relative speed = (60 + 40) km/hr = [100 * 5/18] m/sec = [250/9] m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [300 * 9/250] sec = 54/5 sec = 10.8 ec.
 
26. A train takes 18 seconds to pass completely through a statin 162 m long and 15 seconds through another station 120 m long. The length of the train is:
 
  • A. 30
  • B. 60
  • C. 90
  • D. 110
Ans: C.
Sol.
Let the length of the train be x metres.
∴ x + 162 / 18 = x + 120 / 15 ⇔ 15 (x + 162) = 18 (x + 120) ⇔ x = 90 m.
 
 
27. A train speeds past a pole in 15 seconds and a platfrom 100 m long in 25 seconds. Its length is :
 
  • A. 100 m
  • B. 125 m
  • C. 130 m
  • D. 150 m
Ans: D.
Sol.
Let the length of the train be x metres and its speed be y m/sec.
They, x / y = 15 ⇒ y = x/15
∴ x + 100 / 25 = x / 15 ⇔ x = 150 m.