Question 1

Meeting-times trick – speed of the other train Trains from Delhi and Patna meet; after meeting, Delhi→Patna train takes 16 h, Patna→Delhi train takes 9 h. If Delhi-origin train speed is 90 km/h, find the other train’s speed.

Accepted Answer

Introduction / Context:A classic result: for two bodies starting toward each other and meeting, the ratio of their speeds equals the square root of the inverse ratio of their times to reach destinations after meeting. This avoids needing the total distance. Given Data / Assumptions: After meeting: time of Train D (from Delhi) = 16 h at 90 km/h. After meeting: time of Train P (from Patna) = 9 h at unknown speed v. Uniform speeds, same track distance both ways. Concept / Approach:Speed ratio identity: v_D / v_P = sqrt(t_P_after / t_D_after). Derivation comes from equal total distance and proportional segments before/after meeting. Step-by-Step Solution: v_D / v_P = sqrt(9 / 16) = 3 / 4Thus v_P = (4/3) * v_D = (4/3) * 90 = 120 km/h Verification / Alternative check:Remaining distances after meeting: D_rem_D = 90 * 16 = 1440 km; D_rem_P = 120 * 9 = 1080 km. Proportional relations reconcile with pre-meet segments so they meet once on the route. Why Other Options Are Wrong:125, 145, 190 km/h: Do not satisfy the square-root ratio with 16 h and 9 h. Common Pitfalls:Using direct time ratios (16:9) rather than the square-root relationship; missing the “after meeting” condition which is crucial to the identity. Final Answer:120

Question 2

Pole crossing – length from speed and time A train takes 9 s to cross a pole at 48 km/h. What is the train length?

Accepted Answer

Introduction / Context:When crossing a pole, the distance covered equals the train’s length. Convert speed to m/s and multiply by time to get length directly. Given Data / Assumptions: v = 48 km/h = 13.333... m/s. t = 9 s. Concept / Approach:L = v * t with consistent units. Step-by-Step Solution: L = (48 * 1000 / 3600) * 9 = 13.333... * 9 = 120 m Verification / Alternative check:At 13.333... m/s, 9 s gives exactly 120 m. Checks out. Why Other Options Are Wrong:150 m, 90 m, 80 m: Do not match the computed product. Common Pitfalls:Not converting km/h to m/s. Using 48 as m/s leads to large errors. Final Answer:120 m

Question 3

Pole crossing at 72 km/h – find length A train moving at 72 km/h crosses a pole in 15 s. What is its length?

Accepted Answer

Introduction / Context:The pole is a point obstacle; the entire train passes by, so distance equals the train length. Apply unit conversion and multiply speed by time to get the length. Given Data / Assumptions: v = 72 km/h = 20 m/s. t = 15 s. Concept / Approach:L = v * t after converting to m/s. Step-by-Step Solution: L = 20 * 15 = 300 m Verification / Alternative check:Reverse: 300 m / 20 m/s = 15 s, consistent. Why Other Options Are Wrong:150 m, 200 m, 350 m: Each contradicts the exact multiplication result. Common Pitfalls:Leaving units as km/h and seconds without conversion leads to wrong answers. Final Answer:300 m

Question 4

Bridge length from mixed time – fast train A 150 m train at 90 km/h crosses a bridge in 26 s. What is the bridge length?

Accepted Answer

Introduction / Context:Calculate total distance covered during the crossing time, then subtract train length to get the bridge length. Unit conversion remains essential. Given Data / Assumptions: L = 150 m. v = 90 km/h = 25 m/s. t = 26 s. Concept / Approach:D = v * t; bridge = D - L. Step-by-Step Solution: D = 25 * 26 = 650 mBridge = 650 - 150 = 500 m Verification / Alternative check:At 25 m/s, 26 s gives 650 m which equals 150 m train + 500 m bridge. Why Other Options Are Wrong:600 m and 550 m overshoot; 659 m is not founded on the given numbers. Common Pitfalls:Forgetting to subtract the train's own length from the total distance. Final Answer:500 m

Question 5

Train + platform – compute platform length A train at 60 km/h crosses a platform in 30 s. If the train length is 200 m, find the platform length.

Accepted Answer

Introduction / Context:We know speed and crossing duration; compute total traverse and subtract the known train length to isolate the platform length. Given Data / Assumptions: v = 60 km/h = 16.666... m/s. t = 30 s. L_train = 200 m. Concept / Approach:Total distance D = v * t = train + platform → platform = D - train. Step-by-Step Solution: D = (60 * 1000 / 3600) * 30 = 16.666... * 30 = 500 mPlatform = 500 - 200 = 300 m Verification / Alternative check:At 16.666... m/s, 30 s gives 500 m, consistent with 200 m + 300 m. Why Other Options Are Wrong:420 m, 500 m, 250 m: Do not satisfy D = v * t with the train length provided. Common Pitfalls:Rounding 16.666... too early; however, the final arithmetic resolves cleanly to 300 m. Final Answer:300 m

Question 6

Bridge and man timings – find train length A train crosses a 150 m bridge in 15 s and a man (standing) in 9 s at uniform speed. What is the length of the train?

Accepted Answer

Introduction / Context:Two timings allow solving for both the train length and speed. The man is a point obstacle (distance = train length). The bridge timing includes both train and bridge lengths. Given Data / Assumptions: Bridge length = 150 m. t_bridge = 15 s; t_man = 9 s. Uniform speed. Concept / Approach:Let v be speed in m/s and L the train length. From man: L = v * 9. From bridge: L + 150 = v * 15. Solve the two linear equations for L. Step-by-Step Solution: From man: L = 9vFrom bridge: 9v + 150 = 15v ⇒ 6v = 150 ⇒ v = 25 m/sTherefore L = 9 * 25 = 225 m Verification / Alternative check:Bridge check: 225 + 150 = 375 m; at 25 m/s, time = 15 s. Man check: 225 m at 25 m/s is 9 s. Both match. Why Other Options Are Wrong:200 m, 135 m, 90 m: None satisfies both timing equations simultaneously. Common Pitfalls:Mixing which timing corresponds to which distance; remember man → distance = L, bridge → distance = L + 150. Final Answer:225 m

Question 7

Pole and bridge timings – find train length A train crosses a pole in 2 s and a 250 m bridge in 7 s at uniform speed. What is the length of the train?

Accepted Answer

Introduction / Context:Two timings with two different obstacles (point and extended) allow solving for speed and length via simultaneous equations. Given Data / Assumptions: t_pole = 2 s (distance = L). t_bridge = 7 s (distance = L + 250 m). Uniform speed v. Concept / Approach:From pole: L = v * 2. From bridge: L + 250 = v * 7. Substitute L into the second equation to solve for v, then compute L. Step-by-Step Solution: Let L = 2v2v + 250 = 7v ⇒ 5v = 250 ⇒ v = 50 m/sL = 2 * 50 = 100 m Verification / Alternative check:Pole: 100 m / 50 m/s = 2 s. Bridge: (100 + 250) / 50 = 7 s. Both satisfied. Why Other Options Are Wrong:150 m, 120 m, 80 m: Fail one or both equations’ checks. Common Pitfalls:Using km/h inadvertently; keep everything in m and s to maintain consistency. Final Answer:100 m

Question 8

Equal train and platform lengths – find train length A train and platform have equal lengths. The train runs at 90 km/h and crosses the platform in 1 minute. Find the train length.

Accepted Answer

Introduction / Context:When train and platform lengths are equal, the total crossing distance is twice the train length. Convert the speed to m/s and multiply by time to get the total distance, then halve it. Given Data / Assumptions: Speed = 90 km/h = 25 m/s. Crossing time = 60 s. Train length = platform length. Concept / Approach:Total distance D = v * t. With equal lengths, D = L + L = 2L, so L = D / 2. Step-by-Step Solution: D = 25 * 60 = 1500 mL = D / 2 = 1500 / 2 = 750 m Verification / Alternative check:Reconstruct: Train 750 m + platform 750 m = 1500 m in 60 s at 25 m/s — consistent. Why Other Options Are Wrong:500 m, 550 m, 600 m: When doubled, these do not yield the 1500 m total distance implied by the given speed and time. Common Pitfalls:Forgetting to double the train length when platform length equals train length; or keeping speed in km/h with seconds. Final Answer:750 m

Question 9

A train travelling with uniform speed crosses two bridges of lengths 300 m and 240 m in 21 s and 18 s respectively. Find the speed of the train?

Accepted Answer

Let length of the train = L According to the question, (L + 300)/21 = (L + 240)/18 ⇒ (L + 300)/7 = (L + 240)/6 ⇒ 6L + 1800 = 7L + 1680 ∴ L = 120 mTaking the length of the 2nd bridge into consideration, Speed of train = (L+ 240)/18 = (120 + 240)/18 m/s = (360/18) x (18/5) km/h = 72 km/h

Question 10

Two train of length 70 m and 90 m are moving in opposite directions at 10 m/s and 6 m/s, respectively. Find the time taken by trains to cross each other .?

Accepted Answer

According to the formula . Required time = (x + y) / (u + v) Here, x = 70 m, y = 90 m, u = 10 m/s and v = 6 m/s ∴ Required time = (70 + 90) / (10 + 6) = 160/16 = 10 s

Question 11

A 400 m long train takes 36 s to cross a man walking at 20 km/h in the direction opposite to that of the train. What is the speed of the train?

Accepted Answer

Relative speed of train = 400/36 m/s = 400/36 x (18/5) = 40 km/h Relative speed of train = Speed of train + Speed of man ⇒ 40 = Speed of train + 20 ∴ Speed of train = 40 - 20 = 20 km/h

Question 12

Two trains of lengths 105 m and 90 m, respectively run at the speeds of 45 km/h and 72 km/h, respectively in opposite directions on parallel tracks. Find the time which they take to cross each other?

Accepted Answer

Total length of the train = 105 + 90 = 195 m Relative speed = 72 + 45 km/h = 117 km/h = 117 x (5/18) = 585/18 m/s ∴ Required time = (195 x (18/585) s = 6 s

Question 13

A 110 m long train is running at a speed of 60 km/h. How many seconds, does it take to cross an another train of length 170 m, which is standing on parallel track?

Accepted Answer

Speed = 60 km/h = 60 x (5/18) m/s = 50/3 m/s Required time = (110 + 170) / (50/3) = (280 x 3)/50 = 16.8 s

Question 14

Two train of equal length are running on parallel lines in the same direction at 46 km/h and 36 km/h. The faster train passes the slower train in 36 s. The length of each trains is?

Accepted Answer

Let the length of each train be L m. Relative speed = 46 - 36 = 10 km/h = 10 x 5/18 m/s = 25/9 m/s Sum of length of train / Relative speed of train = Time taken ∴ 2L/(25/9) = 36 ⇒ 2L = (36 x 25)/9 = 100 ∴ L = 50 m.

Question 15

Two trains of same length are running in parallel tracks in opposite directions with speeds 65 km/h and 85 km/h respectively. They cross each other in 6 s . The length of each train is?

Accepted Answer

Let length of each train be L m. Then. (65 + 85 ) x 5/18 = (L + L)/6 ⇒ (150 x 5 x 6)/18 = 2L ⇒ 2L = 250 ∴ L = 125 m

Question 16

Two train start at the same time from points x and y towards each other and after crossing, they take 9 h and 4 h in reaching points y and x, respectively. Find the ratio of speeds of the 1st train to that of the 2nd train?

Accepted Answer

Given that, T1 = 9 h and T 2 = 4 h According to the formula. (1st train's speed) : (2nd train's speed) = √4 : √9 = 2 : 3

Question 17

The ratio between the speeds of two train is 8 : 9 Second train covers 360 km in 4 h. Distance covered by first train in 3 h (in km ) is?

Accepted Answer

Let speed of 1st train be 8S and speed of 2nd train be 9S According to the question. Speed of second train = 360/4 = 90 km/h 9S = 90 ⇒ S = 10 So, speed of 1st train = 8 x 10 = 80 km/h ∴ Required distance = 80 x 3 = 240 km

Question 18

From station M and N, two trains start moving towards each other at speed 125 km/h and 75 km/h, respectively. When the two trains meet each other, it is founds that one train covers 50 km more than another. Find the distance between M and N.?

Accepted Answer

Let the trains meet after time t at a distance x from station N. Then another train coming from station M covers a distance of (x + 50) For station M, (x + 50) = 125t ⇒ x = 125t - 50 ..(i)For station N, x = 75t .....(ii)From Eqs. (i) and (ii), we get 75t = 125t - 50 ⇒ t = 1h Distance between station M and N = 125 + 75t = 200 x 1 = 200 km

Question 19

Excluding stoppages, a train runs at 108 km/h; including stoppages, its average speed is 90 km/h. For how many minutes does the train stop per hour?

Accepted Answer

Introduction / Context:A train’s “running speed” excludes stoppages, while the “average speed” includes them. Comparing distances covered in one hour gives the fraction of time actually spent moving versus stopping. Given Data / Assumptions: Running speed = 108 km/h Average speed (with stops) = 90 km/h Consider a 1-hour window. Concept / Approach:In 1 hour of clock time, distance covered = 90 km. If the train had run the whole hour, it would have covered 108 km. Therefore, the “running time” = distance/run speed = 90/108 hours. The remainder of the hour is stoppage time. Step-by-Step Solution: Running time = 90/108 = 5/6 hStoppage time per hour = 1 − 5/6 = 1/6 hConvert to minutes: (1/6) * 60 = 10 minutes Verification / Alternative check:In 50 minutes of running (5/6 h), distance = 108 * 5/6 = 90 km, matching the average of 90 km/h over the full hour. Why Other Options Are Wrong: 5, 6, 9 minutes do not produce 90 km in one clock hour at 108 km/h running speed. Common Pitfalls: Confusing average speed with running speed. Not standardizing to a 60-minute window. Final Answer:10

Question 20

Without stoppages, a train runs at 150 km/h; with stoppages, its average speed is 100 km/h. How many minutes per hour does the train stop?

Accepted Answer

Introduction / Context:The average speed over a clock hour falls below the running speed if the train stops. Using distance equivalence over one hour gives the stoppage duration per hour. Given Data / Assumptions: Running speed = 150 km/h Average speed = 100 km/h Concept / Approach:In one hour, actual distance = 100 km. Running time = distance / running speed = 100/150 h; stoppage = 1 − (100/150) h. Step-by-Step Solution: Running time = 100/150 = 2/3 h = 40 minutesStoppage per hour = 60 − 40 = 20 minutes Verification / Alternative check:Over 60 minutes: 40 minutes moving at 150 km/h gives 100 km average overall ✔ Why Other Options Are Wrong: 15, 25, 45 minutes would yield averages different from 100 km/h. Common Pitfalls: Dividing 150 by 100 instead of 100 by 150 to get running-time fraction. Final Answer:20

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