Direction : Read the data given below carefully and answer the questions. A…
2020
Direction : Read the data given below carefully and answer the questions.
A train running between four stations i.e. A, B, C & D on Monday and Tuesday. The average speed of train on Monday during whole journey is 50 kmph and average speed of train on Tuesday during whole journey is 62.5 kmph. On Monday train takes one hour less to cover distance between A and B as compare to that of on Tuesday. Train takes equal time to cover distance between B to C on both the given days, while train takes three hours more to cover distance between C and D as compare to that of on Tuesday. The distance between A and B is 40% less than that of between C and D. The distance between B and C is 50% more than that of between A and B.
Note - There is not any halt or stoppage from the station A to D.
On Monday speed of train between stations B to C is 50% more than speed of train between C to D and time taken to cover distance between B to C is 2 hours less than that of time taken between C and D. Find the speed of train between C to D on Monday?
- A.
20 kmph
- B.
60 kmph
- C.
30 kmph
- D.
50 kmph
- E.
40 kmph
Attempted by 1 students.
Show answer & explanation
Correct answer: E
Concept
Average speed over a multi-leg journey is total distance ÷ total time, NOT the average of the leg speeds. For any leg, distance = speed × time. When two relations are given on one day — "one leg is k% faster than another" and "its time is some hours less" — combine them as time = distance ÷ speed to pin the unknown leg times before touching distances.
Set up the distances
Let the C–D distance be d. Then:
A–B = 40% less than C–D = 0.6d
B–C = 50% more than A–B = 1.5 × 0.6d = 0.9d
Total distance A→D = 0.6d + 0.9d + d = 2.5d
Application — Monday's C–D and B–C times
On Monday the B–C speed is 50% more than the C–D speed, so for the same kind of split, B–C time = (0.9d) ÷ (1.5·v) and C–D time = d ÷ v, where v is the C–D speed. The B–C time is 2 hours less than the C–D time:
0.9d ÷ (1.5v) = 0.6 × (d ÷ v), so B–C time = 0.6 × C–D time.
0.6 × (d/v) = (d/v) − 2 ⇒ −0.4 × (d/v) = −2 ⇒ d/v = 5.
Hence Monday C–D time = 5 hours and Monday B–C time = 5 − 2 = 3 hours.
Link the two days
B–C time is equal on both days, so Tuesday B–C time = 3 h. Monday C–D takes 3 h more than Tuesday, so Tuesday C–D time = 5 − 3 = 2 h. Monday A–B takes 1 h less than Tuesday, so Monday A–B time = (Tuesday A–B time) − 1.
Use total distance = average speed × total time on each day. With total distance = 2.5d and Tuesday A–B time = x:
Monday: 2.5d = 50 × ((x − 1) + 3 + 5) = 50 × (x + 7).
Tuesday: 2.5d = 62.5 × (x + 3 + 2) = 62.5 × (x + 5).
Equate: 50(x + 7) = 62.5(x + 5) ⇒ 350 − 312.5 = 12.5x ⇒ x = 3 h.
Result
Tuesday total time = 3 + 5 = 8 h, so 2.5d = 62.5 × 8 = 500 ⇒ d = 200 km. Monday C–D speed = d ÷ (Monday C–D time) = 200 ÷ 5 = 40 kmph.
Cross-check
Distances become A–B = 120, B–C = 180, C–D = 200 (total 500 km). Monday times 2 + 3 + 5 = 10 h give 500 ÷ 10 = 50 kmph; Tuesday times 3 + 3 + 2 = 8 h give 500 ÷ 8 = 62.5 kmph — both averages match, and Monday B–C speed 180 ÷ 3 = 60 is exactly 50% more than the C–D speed of 40.