Question 1

A train crosses a 100 m platform in 30 s at 45 km/h. How long will the same train take to cross a single electric pole? (Find train length first.)

Accepted Answer

Introduction / Context: Crossing a platform requires the train to cover its own length plus the platform length. Crossing a pole requires only the train’s own length. We compute the train length from the platform scenario and then find the time for the pole case. Given Data / Assumptions: Platform length = 100 m. Speed v = 45 km/h = 12.5 m/s. Time over platform = 30 s. Concept / Approach: Total distance over platform = (train length + platform length) = v * t. Then train length = (v * t − platform length). For a pole, time = (train length) / v. Step-by-Step Solution: v = 45 km/h = 45 * 5/18 = 12.5 m/s.Distance over platform = 12.5 * 30 = 375 m.Train length = 375 − 100 = 275 m.Time to cross pole = 275 / 12.5 = 22 s. Verification / Alternative check: Check platform case: (275 + 100)/12.5 = 375/12.5 = 30 s, consistent. Why Other Options Are Wrong: 8 s, 30 s, and 52 s do not match the computed 22 s. Therefore “None of these” is correct here. Common Pitfalls: Forgetting that platform adds extra length, or failing to convert km/h to m/s correctly. Final Answer: None of these

Question 2

A 440 m long train runs at 240 km/h. In the opposite direction, a man runs at 24 km/h. In what time will the train completely pass the man?

Accepted Answer

Introduction / Context: When two objects approach each other, relative speed is the sum of their speeds. For a train crossing a man, the distance to be covered is just the train’s length. Compute relative speed, convert units, then apply time = distance / speed. Given Data / Assumptions: Train length L = 440 m. Train speed = 240 km/h. Man speed (opposite) = 24 km/h. Concept / Approach: v_rel = (240 + 24) km/h = 264 km/h. Convert to m/s via 5/18; then t = L / v_rel. Step-by-Step Solution: v_rel = 264 km/h = 264 * 5/18 = 73.333... m/s.t = 440 / 73.333... = 6 s. Verification / Alternative check: In 6 s at 73.333... m/s, the train covers 440 m exactly, so it clears the man. Why Other Options Are Wrong: 4 s needs 110 m/s; 9 s implies 48.9 m/s; 12 s means 36.7 m/s. None match 73.333... m/s. Common Pitfalls: Using speed difference instead of sum for opposite directions or skipping unit conversion. Final Answer: 6 s

Question 3

A 280 m long train crosses a platform that is three times its own length in 6 min 40 s. What is the speed of the train (in m/s)?

Accepted Answer

Introduction / Context: Crossing a platform requires the train to travel the sum of its own length and the platform length. The speed follows directly from total distance over total time, provided all units are consistent. Given Data / Assumptions: Train length = 280 m. Platform length = 3 * 280 = 840 m. Total time = 6 min 40 s = 6 * 60 + 40 = 400 s. Concept / Approach: Total distance to clear platform = 280 + 840 = 1120 m. Speed v = distance / time in m/s because distance is in meters and time in seconds. Step-by-Step Solution: Total distance = 1120 m.Total time = 400 s.Speed v = 1120 / 400 = 2.8 m/s. Verification / Alternative check: If v = 2.8 m/s, time required for 1120 m is 1120/2.8 = 400 s, which matches 6 min 40 s. Why Other Options Are Wrong: 1.4 m/s or 3.2 m/s produce times inconsistent with 400 s; “Cannot be determined” is false because sufficient data are given. Common Pitfalls: Forgetting to multiply platform length correctly or converting the time from minutes and seconds. Final Answer: 2.8 m/s

Question 4

Two trains of lengths 512 m and 528 m run towards each other on parallel lines at 84 km/h and 60 km/h, respectively. From the instant they meet, in how many seconds will they be completely clear of each other?

Accepted Answer

Introduction / Context: When two trains approach each other, their relative speed is the sum of their speeds. From the moment their fronts meet, they must collectively cover the sum of their lengths to be completely clear of one another. Given Data / Assumptions: L1 = 512 m, L2 = 528 m. v1 = 84 km/h, v2 = 60 km/h. Uniform motion on parallel tracks. Concept / Approach: Total length to clear = L1 + L2. Relative speed v_rel = (v1 + v2) converted to m/s. Then time = (L1 + L2) / v_rel. Step-by-Step Solution: Total length = 512 + 528 = 1040 m.v_rel = (84 + 60) km/h = 144 km/h = 144 * 5/18 = 40 m/s.t = 1040 / 40 = 26 s. Verification / Alternative check: At 40 m/s, covering 1040 m needs 26 s—consistent. Why Other Options Are Wrong: 25 s or 27 s would imply non-integer effective speeds; 15 s is far too short. Common Pitfalls: Using difference of speeds (only for same-direction) or forgetting to convert km/h to m/s. Final Answer: 26 s

Question 5

A train 100 m in length passes a milestone (point) in 10 s and another train of the same length coming from the opposite direction in 8 s. Find the speed of the second train (in km/h).

Accepted Answer

Introduction / Context: This problem combines a point-crossing time to determine the first train’s speed and a full-train crossing time (opposite directions) to determine relative speed, from which the second train’s speed follows by subtraction. Given Data / Assumptions: Each train length = 100 m. First train passes a point in 10 s ⇒ speed1 = 100/10 = 10 m/s. Opposite-direction cross time for both trains together = 8 s. Concept / Approach: For opposite directions, relative speed = (sum of individual speeds). To fully cross, required distance = sum of lengths = 200 m. Hence v_rel = 200/8 = 25 m/s. Therefore speed2 = v_rel − speed1. Step-by-Step Solution: speed1 = 10 m/s = 36 km/h.v_rel = 200 / 8 = 25 m/s.speed2 = 25 − 10 = 15 m/s = 15 * 18/5 = 54 km/h. Verification / Alternative check: Check: 36 + 54 = 90 km/h; in m/s that is 25 m/s; 200/25 = 8 s, consistent. Why Other Options Are Wrong: 36 km/h is the first train; 48 and 60 km/h lead to wrong crossing times. Common Pitfalls: Forgetting to add lengths for full crossing or mixing up km/h and m/s conversions. Final Answer: 54 kmph

Question 6

Trains A and B start simultaneously from stations X and Y towards each other. After meeting, A takes 4 h 48 min to reach Y and B takes 3 h 20 min to reach X. If A's speed is 45 km/h, find the speed of B.

Accepted Answer

Introduction / Context: For two trains starting from opposite ends and meeting once, the distances remaining after the meeting are proportional to their speeds, hence the times to finish are inversely proportional to their speeds. This yields a simple ratio to relate speeds using the post-meeting times. Given Data / Assumptions: Time after meeting: t_A = 4 h 48 min = 4.8 h; t_B = 3 h 20 min = 3.333... h. Speed of A: v_A = 45 km/h. Straight track, constant speeds. Concept / Approach: Speeds are inversely proportional to the post-meeting times: v_A / v_B = t_B / t_A. Solve for v_B, then check consistency. Step-by-Step Solution: t_A = 4.8 h; t_B = 3.333... h = 10/3 h.v_A / v_B = t_B / t_A = (10/3) / 4.8 = (10/3) / (24/5) = 50/72 = 25/36.Thus v_B = v_A * (36/25) = 45 * 36/25 = 64.8 km/h. Verification / Alternative check: v_A : v_B = 45 : 64.8 = 25 : 36, matching t_B : t_A = 25 : 36 inverted. Why Other Options Are Wrong: 60 km/h and 54 km/h do not satisfy the inverse-time ratio; 37.5 km/h would make B slower than A, contradicting the times. Common Pitfalls: Using direct (not inverse) proportionality or mis-converting minutes to hours. Final Answer: 64.8 km/hr

Question 7

Ajanta and Taj express start simultaneously from Meerut towards Nagpur (distance 390 km) on parallel tracks. Their initial speed ratio is 6:7. After Taj has covered a distance D km, the trains exchange speeds instantly and continue so that both reac…

Accepted Answer

Introduction / Context: This is a speed-exchange puzzle. Two trains start together over the same origin–destination span. After the faster train has covered D km, both instantly swap speeds and continue, arriving simultaneously. We form total-time expressions before and after the swap and equate them. Given Data / Assumptions: Total distance L = 390 km. Initial speeds: Ajanta = 6u, Taj = 7u. Speed swap occurs at the same instant for both trains (not necessarily at the same location). Concept / Approach: Let t1 = time until swap. Then D = 7u * t1 (distance Taj covers), while Ajanta covers 6u * t1. After swapping, Taj's remaining distance = L − D at speed 6u; Ajanta's remaining = L − 6u t1 at speed 7u. Equal arrival times ⇒ post-swap travel times are equal. Step-by-Step Solution: t1 = D / (7u); Ajanta has covered 6D/7 by then.Set (L − D)/(6u) = (L − 6D/7)/(7u).Multiply by 42u: 7(L − D) = 6(L − 6D/7) ⇒ 7L − 7D = 6L − 36D/7.Rearrange: L = (13/7)D ⇒ D = (7/13)L = (7/13)*390 = 210 km. Verification / Alternative check: Plugging D = 210 satisfies symmetry of post-swap times; any other D breaks equality. Why Other Options Are Wrong: 150 km and 190 km do not make the post-swap times equal; “can’t be determined” is incorrect because algebra yields a unique D. Common Pitfalls: Assuming they must meet at the swap point; the swap is instantaneous and simultaneous, not collocated. Final Answer: 210 km

Question 8

Train P leaves Mokama at 5:00 pm and reaches Hazipur at 6:00 pm (1 hour). Train Q leaves Hazipur at 5:00 pm and reaches Mokama at 6:30 pm (1.5 hours). Assuming constant speeds and simultaneous start at 5:00 pm, at what time do they meet?

Accepted Answer

Introduction / Context: Two trains start simultaneously from opposite stations and move towards each other with constant speeds. The ratio of their speeds equals the inverse ratio of their individual travel times over the same fixed distance. Meeting time is obtained from the sum of distances covered until meeting equaling the full inter-station distance. Given Data / Assumptions: P takes 1 h for the full route ⇒ v_P = D / 1 h. Q takes 1.5 h ⇒ v_Q = D / 1.5 h. They start at 5:00 pm from opposite stations and move towards each other. Concept / Approach: At meeting, v_P * t + v_Q * t = D. Express each speed as a fraction of D per hour; solve for t in hours, then convert to minutes to get the wall-clock time after 5:00 pm. Step-by-Step Solution: v_P = 1 * D per hour; v_Q = (2/3) * D per hour.(1 + 2/3) * D * t = D ⇒ (5/3) * t = 1 ⇒ t = 3/5 h = 36 min.Meeting time = 5:00 pm + 36 min = 5:36 pm. Verification / Alternative check: In 36 minutes, P covers 0.6D and Q covers 0.4D; together they span D. Why Other Options Are Wrong: Other times do not correspond to t = 36 minutes after a simultaneous 5:00 pm start. Common Pitfalls: Mistaking absolute times for durations; always compute elapsed time from the simultaneous start. Final Answer: 5 : 36 pm

Question 9

A train’s average speed on the onward (outbound) journey is 25% higher than on the return journey. After reaching the destination, the train halts for 2 hours. The total time for the complete to-and-fro trip is 32 hours, and the total distance cover…

Accepted Answer

Introduction / Context: This problem tests weighted time calculations with different speeds on two legs of a round trip and a fixed halt. Translating words into equations and keeping careful units is the key. Given Data / Assumptions: Total distance (to and fro) = 1600 km ⇒ 800 km each way. Onward speed = 1.25 * return speed. Halt at destination = 2 h. Total elapsed time (travel + halt) = 32 h. Concept / Approach: If return speed is v, onward speed is 1.25v. Travel time = 800/(1.25v) + 800/v. Add the 2 h halt and set equal to 32 h. Step-by-Step Solution: Onward time = 800/(1.25v) = 640/v.Return time = 800/v.Total travel time = 640/v + 800/v = 1440/v.Include halt: 1440/v + 2 = 32 ⇒ 1440/v = 30 ⇒ v = 48 km/h.Onward speed = 1.25 * 48 = 60 km/h. Verification / Alternative check: At v = 48, times are 640/48 = 13.333… h and 800/48 = 16.666… h. Sum = 30 h; plus 2 h halt = 32 h as required. Why Other Options Are Wrong: Values like 56.25, 66.50, or 67 km/h violate the time equation after including the halt; they do not total 32 h. Common Pitfalls: Forgetting the halt, or assuming 1600 km is one-way. Also mixing up “25% more speed” with “25% less time” can lead to algebra errors. Final Answer: 60 km/h

Question 10

A train overtakes two people walking along the track in the same direction at 4.5 km/h and 5.4 km/h, respectively. It takes 8.4 s to overtake the first and 8.5 s to overtake the second. Assuming uniform speed and length for the train, find the train…

Accepted Answer

Introduction / Context: In same-direction overtaking, relative speed equals (train speed − walker speed). The time to overtake equals train length divided by relative speed. Equal train length in both cases lets us equate two expressions and solve for the train’s speed. Given Data / Assumptions: Walker speeds: 4.5 km/h and 5.4 km/h. Overtaking times: 8.4 s and 8.5 s. Train length = constant; uniform speed. Concept / Approach: Let train speed be V (km/h) and length be L (m). Convert to m/s when multiplying by seconds. For same direction,L = (V − 4.5) * 5/18 * 8.4 = (V − 5.4) * 5/18 * 8.5.Cancel 5/18 and solve for V. Step-by-Step Solution: (V − 4.5) * 8.4 = (V − 5.4) * 8.58.4V − 37.8 = 8.5V − 45.9 ⇒ 8.1 = 0.1V ⇒ V = 81 km/h. Verification / Alternative check: Compute implied length: L = (81 − 4.5) * 5/18 * 8.4 ≈ 178.5 m. Using the other case with 8.5 s gives the same L, confirming consistency. Why Other Options Are Wrong: 66, 72, 78, 60 km/h fail the equality of lengths or produce inconsistent train lengths across the two overtakings. Common Pitfalls: Adding speeds (opposite direction rule) or leaving speeds in km/h when multiplying by seconds (unit mismatch). Final Answer: 81 km/h

Question 11

Two trains start at the same time from Howrah and Patna toward each other. After passing, they take 4 h 48 min and 3 h 20 min, respectively, to reach their destinations. If the Howrah train runs at 45 km/h, find the speed of the other train (km/h).

Accepted Answer

Introduction / Context: When two trains start together and meet, the times they take after meeting to reach the endpoints are inversely related to their speeds (with a known relation). This classic result avoids needing the total distance. Given Data / Assumptions: Time after meeting: t_Howrah→Patna = 4 h 48 min = 4.8 h. Time after meeting: t_Patna→Howrah = 3 h 20 min = 3.333… h. Speed of Howrah train v_H = 45 km/h. Concept / Approach: Result: v_H / v_P = sqrt(t_P / t_H). Equivalently, v_P = v_H / sqrt(t_P / t_H). This follows from t_H = (v_P / v_H) * t_meet and t_P = (v_H / v_P) * t_meet. Step-by-Step Solution: Compute ratio: t_P / t_H = (3.333…)/(4.8) = 1/1.44 ≈ 0.694444…v_H / v_P = sqrt(0.694444…) ≈ 0.833333… ⇒ v_P ≈ 45 / 0.833333… = 54 km/h. Verification / Alternative check: Check symmetric relation: t_H * t_P stays proportional to 1/(v_H * v_P); numbers are consistent and realistic. Why Other Options Are Wrong: 60 km/h assumes simple inverse without the square-root relation; 45 km/h means equal speeds despite unequal post-meeting times; 35 km/h is too low for the computed ratio. Common Pitfalls: Using direct inverse (v ∝ 1/t) here; the correct formula involves a square root derived from relative-motion geometry. Final Answer: 54 km/h

Question 12

A train travelling at 36 km/h completely crosses another train that has half its length and is coming from the opposite direction at 54 km/h in 12 seconds. The same train passes a railway platform in 1.5 minutes. Find the length of the platform (in …

Accepted Answer

Introduction / Context: This mixes two standard train problems: first find the train’s own length using an “opposite direction” crossing, then use that length and known speed to compute platform length from a platform crossing time. Given Data / Assumptions: Our train speed v = 36 km/h = 10 m/s. Other train speed = 54 km/h. Other train length = (1/2)*our train length. Crossing time (trains) = 12 s; platform time = 1.5 min = 90 s. Concept / Approach: Opposite direction ⇒ relative speed = 36 + 54 = 90 km/h = 25 m/s. If our train length is L, other is L/2; total distance to clear each other = 1.5L. Step-by-Step Solution: 1.5L / 25 = 12 ⇒ 1.5L = 300 ⇒ L = 200 m.Platform crossing: distance = L + P; time = 90 s at 10 m/s ⇒ 10 * 90 = 200 + P ⇒ P = 700 m. Verification / Alternative check: Units are consistent (metres and seconds). Magnitudes are realistic for a moderate-speed train and a long platform segment. Why Other Options Are Wrong: 560, 620, 650, 750 m do not satisfy both crossing conditions simultaneously. Common Pitfalls: Using only our train’s length in the first crossing; forgetting to convert km/h to m/s; misreading 1.5 minutes as 1.5 seconds. Final Answer: 700 metres

Question 13

A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?

Accepted Answer

Relative speed =280/9 m / sec = (280/9*18/5) kmph = 112 kmph. Speed of goods train = (112 - 50) kmph = 62 kmph.

Question 14

A train of length 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

Accepted Answer

6

Question 15

Two trains started at the same time, one from A to B and the other from B to A . If they arrived at B and A respectively 4 hours and 9 hours after they passed each other the ratio of the speeds of the two trains was

Accepted Answer

Note : If two 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) Therefore, Ratio of the speeds of two trains = 9 : 4 = 3 : 2

Question 16

A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is :

Accepted Answer

Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr. Therefore, Man's rate against the current = (12.5 - 2.5) = 10 km/hr.

Question 17

Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds?

Accepted Answer

24

Question 18

Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is

Accepted Answer

Distance covered = 120+120 = 240 m Time = 12 s Let the speed of each train = v. Then relative speed = v + v = 2v 2v = distance/time = 240/12 = 20 m/s Speed of each train = v = 20/2 = 10 m/s = 10 × 36/10 km/hr = 36 km/hr

Question 19

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

Accepted Answer

79.2 km/hr

Question 20

A train moves with a speed of 108 kmph. Its speed in metres per second is :

Accepted Answer

108 kmph = 108*[5/18] m/sec = 30 m / s.

Problems on Trains Questions