Question 1

Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.

Accepted Answer

Given Two 500 m trains run in opposite directions. Speeds: slower = 30 km/h, faster driver = 45 km/h. Find time taken by the slower train to pass the driver (a point) of the faster train. ComputationRelative speed = 30 + 45 = 75 km/h = 75 × (5/18) = 20.833... m/sDistance to be covered = length of slower train = 500 mTime = Distance / Relative speed = 500 / 20.833... = 24 s

Question 2

Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

Accepted Answer

Given Each train length = 100 m. They move in opposite directions and cross in 8 s. One train is twice as fast as the other. Let slower speed = u m/s, faster speed = 2u m/sRelative speed = u + 2u = 3uTotal distance to cross = 100 + 100 = 200 m200 = 3u × 8 ⇒ u = 25/3 m/sFaster speed = 2u = 50/3 m/s = (50/3) × 3.6 = 60 km/h

Question 3

A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:

Accepted Answer

Given Train A: length 108 m, speed 50 km/h. Train B: length 112 m, opposite direction, unknown speed v km/h. They cross in 6 s. ComputeTotal length = 108 + 112 = 220 mRelative speed = 220 / 6 = 36.666... m/s = 36.666... × 18/5 = 132 km/hSo, 50 + v = 132 ⇒ v = 82 km/h

Question 4

A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons ar…

Accepted Answer

Given Walkers: 4.5 km/h and 5.4 km/h (same direction as train). Overtake times: 8.4 s and 8.5 s respectively. Let train speed be v km/h, length L m.L = (v − 4.5) × (5/18) × 8.4L = (v − 5.4) × (5/18) × 8.5Equate and solve ⇒ v = 81 km/h Check (optional)L = (81 − 4.5) × (5/18) × 8.4 = 178.5 m (consistent)

Question 5

A train runs at 200 km/h. Convert this speed to metres per second (m/s).

Accepted Answer

Introduction / Context:Unit conversion between km/h and m/s is routine in motion problems. The factor is 1000/3600 = 5/18 to go from km/h to m/s. Accuracy to two decimal places is commonly acceptable for train-speed conversions in aptitude questions. Given Data / Assumptions: Speed = 200 km/h. 1 km = 1000 m; 1 h = 3600 s. Concept / Approach:Use v(m/s) = v(km/h) * (1000/3600) = v(km/h) * (5/18). Keep sufficient decimal precision when required by options. Step-by-Step Solution: v = 200 * (5/18) = 1000/18 ≈ 55.555… m/s.Rounded to two decimals: 55.55 m/s. Verification / Alternative check:Reverse conversion: 55.55 * (18/5) ≈ 199.98 km/h ≈ 200 km/h (rounding). Why Other Options Are Wrong:50 or 48 m/s are too low; 66.66 m/s would correspond to ~240 km/h, not 200 km/h. Common Pitfalls:Using 18/5 instead of 5/18 when converting km/h to m/s; truncation rather than rounding. Final Answer:55.55 m/sec

Question 6

A train crosses a pole in 10 s and a 200 m platform in 20 s. What is the length of the train (in metres)?

Accepted Answer

Introduction / Context:Let the train length be L and speed be v (m/s). Passing a pole takes L/v seconds, while passing a platform of length P takes (L + P)/v seconds at the same speed. Using the two given times allows solving for L directly without computing v numerically first. Given Data / Assumptions: Pole time = 10 s ⇒ L = 10v. Platform length P = 200 m; platform time = 20 s ⇒ L + 200 = 20v. Concept / Approach:Eliminate v by substituting L = 10v into the platform equation and solve for L. This exploits the same-speed condition across both scenarios. Step-by-Step Solution: L + 200 = 20v and L = 10v ⇒ 10v + 200 = 20v.200 = 10v ⇒ v = 20 m/s.Then L = 10v = 200 m. Verification / Alternative check:Check: time across platform = (L + 200)/v = (200 + 200)/20 = 20 s—consistent. Why Other Options Are Wrong:100/250/300 m contradict either the pole time or platform time at a single consistent speed. Common Pitfalls:Assuming 20 s implies double the length only; it is double total distance L + P, not L itself unless computed properly. Final Answer:200m

Question 7

Two trains of lengths 200 m and 300 m run on parallel tracks in opposite directions at 36 km/h and 54 km/h, respectively. How many seconds will they take to completely cross each other?

Accepted Answer

Introduction / Context:For opposite-direction motion, relative speed is the sum of speeds. The distance to be covered for a complete pass is the sum of train lengths. Converting km/h to m/s ensures compatible units with metres and seconds for time calculation. Given Data / Assumptions: Lengths: 200 m and 300 m ⇒ total = 500 m. Speeds: 36 km/h and 54 km/h. Concept / Approach:Relative speed = 36 + 54 = 90 km/h = 90 * (1000/3600) = 25 m/s. Time = distance / relative speed in SI units. Step-by-Step Solution: Relative speed = 25 m/s.Time = 500 / 25 = 20 s. Verification / Alternative check:Unit check: metres divided by metres/second yields seconds—correct dimensionally. Why Other Options Are Wrong:25/30/35 s correspond to lower relative speeds than 25 m/s, inconsistent with 36+54 km/h. Common Pitfalls:Forgetting to add speeds for opposite directions; mixing km/h directly with metres without conversion. Final Answer:20 sec

Question 8

A 200 m long train runs at 60 km/h. How many seconds does it take to pass a man standing still on a platform?

Accepted Answer

Introduction / Context:When a train passes a stationary point-like observer, the distance to be covered is its own length. Time equals length divided by speed, but speed must be in m/s if length is in metres and time is in seconds. So unit conversion is crucial. Given Data / Assumptions: Train length L = 200 m. Speed v = 60 km/h = 60 * (1000/3600) m/s. Observer is stationary (relative speed equals the train’s speed). Concept / Approach:Compute v in m/s, then apply t = L / v. No platform length is involved because the person is a point reference. Step-by-Step Solution: v = 60 * (1000/3600) = 16.666… m/s.t = 200 / 16.666… = 12 s. Verification / Alternative check:Check with fraction: 60 km/h = 5/3 m/s * 10 = 50/3? No: correctly 60 * (5/18) = 50/3 ≈ 16.666…; 200 * 3 / 50 = 12 s. Why Other Options Are Wrong:10/15/18 s imply speeds different from 60 km/h for a 200 m train. Common Pitfalls:Using km/h directly with metres; or mistakenly adding extra distance as if passing a platform. Final Answer:12 sec

Question 9

Problems on Trains – Train + platform crossing time (speed and lengths given): A train is running at 264 km/h. If the train length is 330 m and the platform length is 330 m, how long (in seconds) will the train take to completely cross the platform?

Accepted Answer

Introduction / Context:In platform-crossing problems, the train must cover its own length plus the platform length. With a given speed, time is distance divided by speed after converting all units consistently to SI (metres and seconds). Given Data / Assumptions: Train speed = 264 km/h. Train length = 330 m. Platform length = 330 m. Motion is uniform and straight, with no acceleration phases. Concept / Approach:Total distance to clear the platform = L_train + L_platform. Convert speed from km/h to m/s using 1 km/h = 5/18 m/s. Then compute time = distance / speed in seconds. Step-by-Step Solution:Total distance = 330 + 330 = 660 m.Speed (m/s) = 264 * (5/18) = 73.333… m/s.Time = 660 / 73.333… = 9 s. Verification / Alternative check:Quick ratio check: 73.333 m each second; in 9 s, distance covered = 660 m, exactly the required sum of lengths. Why Other Options Are Wrong:11 s, 12 s, and 15 s overestimate the time (would imply lower speeds than 264 km/h). 8 s would cover only ~586.7 m, insufficient to clear both lengths. Common Pitfalls:Forgetting to add train and platform lengths; mixing units by using km and m together; converting 264/3.6 incorrectly. Final Answer:9 sec

Question 10

Problems on Trains – Find train speed from pole-crossing time: A 264 m long train passes a pole in 12 seconds. What is the speed of the train in km/h?

Accepted Answer

Introduction / Context:When a train passes a pole, the distance covered during the crossing equals the train’s own length. With distance and time known, the speed is computed and then converted to km/h. Given Data / Assumptions: Train length = 264 m. Time to pass pole = 12 s. Pole has negligible length (treated as a point). Concept / Approach:Speed (m/s) = distance / time. To convert m/s to km/h, multiply by 3.6. Alternatively, compute directly in m/s and then scale. Step-by-Step Solution:Speed = 264 / 12 = 22 m/s.Speed in km/h = 22 * 3.6 = 79.2 km/h. Verification / Alternative check:If the speed were 80 km/h, in 12 s the distance would be 80*(1000/3600)*12 ≈ 266.7 m, slightly more than 264 m; hence 79.2 km/h fits exactly. Why Other Options Are Wrong:80, 80.6, and 82.2 km/h yield distances larger than 264 m in 12 s; 76.8 km/h is smaller than required. Common Pitfalls:Using the wrong conversion factor (3.6), or dividing time by distance inadvertently. Final Answer:79.2 km/hr

Question 11

Problems on Trains – Bridge length from crossing time: A 200 m long train runs at 36 km/h and crosses a bridge in 30 seconds. What is the length of the bridge?

Accepted Answer

Introduction / Context:When a train crosses a bridge, the total distance traveled during the crossing equals train length + bridge length. With speed and total time known, compute the total crossing distance and subtract the train length to get the bridge length. Given Data / Assumptions: L_train = 200 m. Speed = 36 km/h = 10 m/s. Crossing time = 30 s. Concept / Approach:Total distance = speed * time. Then L_bridge = total distance − L_train. Step-by-Step Solution:Speed = 36 * (1000/3600) = 10 m/s.Total distance = 10 * 30 = 300 m.Bridge length = 300 − 200 = 100 m. Verification / Alternative check:Check: At 10 m/s, 100 m are covered in 10 s. The remaining 200 m (train length) take 20 s. Sum = 30 s, consistent. Why Other Options Are Wrong:150 m, 200 m, 125 m, 180 m do not satisfy total distance = 300 m with the given train length. Common Pitfalls:Forgetting to convert speed to m/s or failing to add train and bridge distances when forming the total crossing distance. Final Answer:100 m

Question 12

Problems on Trains – Platform-crossing time from pole time: A 360 m long train takes 36 seconds to cross a telegraph pole. How long will it take to cross a 600 m long platform?

Accepted Answer

Introduction / Context:The time to cross a pole gives the speed because the train covers only its own length. Once the speed is known, platform-crossing time follows by using total distance = train length + platform length. Given Data / Assumptions: L_train = 360 m. Time to pass pole = 36 s. L_platform = 600 m. Concept / Approach:First find speed v = L_train / t_pole in m/s. Then compute total distance through the platform D = L_train + L_platform, and time t = D / v. Step-by-Step Solution:Speed v = 360 / 36 = 10 m/s.Total distance = 360 + 600 = 960 m.Time = 960 / 10 = 96 s. Verification / Alternative check:If the platform were 600 m and speed 10 m/s, every second adds 10 m; 96 s accounts exactly for 960 m. Why Other Options Are Wrong:60, 66, 50, 80 s do not match D/v with v fixed by the pole-crossing time. Common Pitfalls:Using only platform length instead of train + platform; misreading the pole time as the platform time. Final Answer:96 sec

Question 13

Problems on Trains – Find train length from platform-crossing data: A passenger train runs at 72 km/h and crosses a 400 m long platform in 30 seconds. What is the length of the train?

Accepted Answer

Introduction / Context:Crossing a platform requires the train to cover its own length plus the platform length. With speed and time known, total distance gives the sum, from which we deduce the train’s length. Given Data / Assumptions: Speed = 72 km/h = 20 m/s. Time = 30 s. Platform length = 400 m. Concept / Approach:Total distance = speed * time = (L_train + L_platform). Solve for L_train. Step-by-Step Solution:Speed (m/s) = 72 * (1000/3600) = 20 m/s.Total distance = 20 * 30 = 600 m.Train length = 600 − 400 = 200 m. Verification / Alternative check:At 20 m/s, the train covers 200 m in 10 s and the platform in 20 s; total 30 s, matching the data. Why Other Options Are Wrong:150, 180, 220, 240 m cannot satisfy total distance 600 m with a 400 m platform. Common Pitfalls:Forgetting unit conversion or subtracting in the wrong order. Final Answer:200 m

Question 14

Problems on Trains – Train length equals platform length: A train travels at 72 km/h and takes 2 minutes to cross a platform whose length equals the length of the train. What is the length of the train?

Accepted Answer

Introduction / Context:When the platform length equals the train length L, the total distance to fully clear the platform is L + L = 2L. With the crossing time and speed, we can compute 2L and hence L. Given Data / Assumptions: Speed = 72 km/h = 20 m/s. Crossing time = 2 minutes = 120 s. Platform length = train length = L. Concept / Approach:Distance covered in crossing = 2L = speed * time. Then solve for L. Step-by-Step Solution:Speed (m/s) = 20 m/s.Distance in 120 s = 20 * 120 = 2400 m.Hence 2L = 2400 ⇒ L = 1200 m. Verification / Alternative check:If L = 1200 m, the platform is 1200 m too, so total 2400 m. At 20 m/s, time = 2400/20 = 120 s (2 minutes), consistent. Why Other Options Are Wrong:600, 800, 900, 1000 m would not yield a 2-minute crossing at 72 km/h when the platform equals the train length. Common Pitfalls:Using only one L instead of 2L; converting 2 minutes incorrectly; misreading km/h to m/s. Final Answer:1200 m

Question 15

Problems on Trains – Find train length from two platform times: A train takes 24 s to cross a 480 m platform and 36 s to cross a 900 m platform. What is the length of the train?

Accepted Answer

Introduction / Context:Two crossing scenarios with different platform lengths allow us to eliminate speed and solve for train length directly. Represent total distance as (L_train + L_platform) = v * t in each case and subtract the equations. Given Data / Assumptions: (L + 480) = v * 24. (L + 900) = v * 36. Uniform speed v. Concept / Approach:Subtracting the two equations removes L: (900 − 480) = v(36 − 24), yielding v. Substitute back to find L. Step-by-Step Solution:420 = v * 12 ⇒ v = 35 m/s.From L + 480 = 35 * 24 = 840 ⇒ L = 360 m. Verification / Alternative check:For the 900 m platform: L + 900 = 360 + 900 = 1260 m; time = 1260 / 35 = 36 s, valid. Why Other Options Are Wrong:400, 390, 430, 320 m will not satisfy both time equations simultaneously with one constant speed. Common Pitfalls:Adding times or lengths incorrectly; not recognizing that subtracting the equations isolates v immediately. Final Answer:360 m

Question 16

Problems on Trains – Two trains on parallel tracks, same direction: Train-1 is 200 m long at 54 km/h; Train-2 is 300 m long at 72 km/h. On parallel tracks and same direction, how long will they take to completely cross each other?

Accepted Answer

Introduction / Context:For trains in the same direction, the relative speed equals the difference of their speeds. The distance to be covered for complete crossing is the sum of their lengths. Given Data / Assumptions: L1 = 200 m at 54 km/h. L2 = 300 m at 72 km/h. Same direction on parallel tracks. Concept / Approach:Relative speed (m/s) = (72 − 54) km/h converted to m/s. Crossing distance = L1 + L2. Then time = distance / relative speed. Step-by-Step Solution:Relative speed = 18 km/h = 18 * (1000/3600) = 5 m/s.Distance = 200 + 300 = 500 m.Time = 500 / 5 = 100 s. Verification / Alternative check:Opposite-direction check would be much faster (sum of speeds); since here it is much slower, a large time like 100 s is reasonable. Why Other Options Are Wrong:90, 120, 150, 80 s do not match the computed distance/relative-speed ratio. Common Pitfalls:Adding speeds instead of subtracting for same direction; failing to convert km/h to m/s. Final Answer:100 sec

Question 17

Problems on Trains – Tunnel length from total crossing time: A 400 m long train running at 54 km/h (constant speed) crosses a tunnel in 2 minutes. What is the tunnel’s length?

Accepted Answer

Introduction / Context:Passing completely through a tunnel requires the train to cover its own length plus the tunnel length. With speed and total time known, compute the total distance; subtract the train length to obtain the tunnel length. Given Data / Assumptions: L_train = 400 m. Speed = 54 km/h = 15 m/s. Time = 2 minutes = 120 s. Concept / Approach:Total distance = speed * time; then L_tunnel = total distance − L_train. Step-by-Step Solution:Total distance = 15 * 120 = 1800 m.Tunnel length = 1800 − 400 = 1400 m. Verification / Alternative check:At 15 m/s, 1400 m takes 93.33 s; with 400 m more (train length), additional 26.67 s; sum 120 s, confirming the calculation. Why Other Options Are Wrong:1200, 1600, 1800, 1000 m do not satisfy the total distance constraint with the given speed and time. Common Pitfalls:Using tunnel length alone for distance; not converting time to seconds or speed to m/s. Final Answer:1400 m

Question 18

Problems on Trains – Average speed for equal distances (find return speed): A train goes from A to B at 64 km/h and returns from B to A at a slower speed. If the average speed for the round trip is 56 km/h, what is the return speed?

Accepted Answer

Introduction / Context:For equal distances each way, the average speed is the harmonic mean: v_avg = 2uv/(u + v), where u and v are the onward and return speeds. With average and one leg’s speed known, solve for the other speed. Given Data / Assumptions: Onward speed u = 64 km/h. Average speed v_avg = 56 km/h. Distances A→B and B→A are equal. Concept / Approach:Use v_avg = 2uv/(u + v). Rearranging for v gives: 56(64 + v) = 128v ⇒ solve linearly. Step-by-Step Solution:56(64 + v) = 128v.3584 + 56v = 128v ⇒ 3584 = 72v ⇒ v = 3584/72 = 49.777… km/h. Verification / Alternative check:Harmonic mean with u = 64 and v ≈ 49.777 gives v_avg ≈ 56, confirming correctness. Why Other Options Are Wrong:48, 52, 47.46, 50 km/h do not satisfy the harmonic-mean equation with 64 km/h producing 56 km/h average. Common Pitfalls:Using arithmetic mean instead of harmonic mean; assuming average = (64 + v)/2, which is incorrect for equal distances at different speeds. Final Answer:49.77 km/hr.

Question 19

Problems on Trains – Crossing a moving man (opposite direction): A 270 m train moves at 25 km/h and meets a man walking toward it at 2 km/h. In how many seconds will the train completely pass the man?

Accepted Answer

Introduction / Context:When objects move toward each other, their relative speed is the sum of their individual speeds (with consistent units). To pass a man, the train must cover only its own length relative to the man. Given Data / Assumptions: L_train = 270 m. Train speed = 25 km/h. Man speed (opposite) = 2 km/h. Concept / Approach:Relative speed = (25 + 2) km/h = 27 km/h. Convert to m/s by multiplying by 1000/3600 = 5/18. Step-by-Step Solution:Relative speed = 27 * (5/18) = 7.5 m/s.Time = distance / relative speed = 270 / 7.5 = 36 s. Verification / Alternative check:At 7.5 m/s, 36 s produces 270 m, which matches the train length and thus completes the pass. Why Other Options Are Wrong:32, 28, 24, 30 s correspond to incorrect relative speeds or arithmetic errors. Common Pitfalls:Subtracting speeds instead of adding; forgetting to convert km/h to m/s consistently. Final Answer:36 seconds

Question 20

Problems on Trains – Speed from time to pass a standing man: A train 300 m long passes a stationary man in 15 seconds. What is the speed of the train?

Accepted Answer

Introduction / Context:The distance covered while passing a standing man is simply the train’s own length. Use speed = distance / time, then convert to km/h if required. Given Data / Assumptions: L_train = 300 m. Time = 15 s. Man is stationary (speed = 0). Concept / Approach:Speed (m/s) = 300 / 15 = 20 m/s. Convert to km/h by multiplying by 3.6. Step-by-Step Solution:Speed = 20 m/s = 20 * 3.6 = 72 km/h. Verification / Alternative check:At 72 km/h (20 m/s), 15 s * 20 m/s = 300 m, consistent with the train length. Why Other Options Are Wrong:40, 50, 54, 60 km/h correspond to speeds that would not cover 300 m in 15 s. Common Pitfalls:Using 3.8 or 3.2 as the conversion factor; inverting the conversion accidentally. Final Answer:72 km/hr

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