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 is sitting in a train that is moving at 50 km/h. He observes that a goods train, 280 meters long, passes him completely in 9 seconds while moving in the opposite direction. What is the speed of the goods train in km/h?

Accepted Answer

Introduction / Context:This is a classic relative speed question involving two trains moving in opposite directions. One train carries an observer and the other is a goods train of known length. The goal is to find the speed of the goods train using the time taken to pass the observer. Given Data / Assumptions:- Speed of the passenger train with the man inside is 50 km/h.- The goods train is 280 meters long.- The goods train passes the man completely in 9 seconds.- The two trains move in opposite directions in a straight line.- We must find the speed of the goods train in km/h. Concept / Approach:When two objects move in opposite directions, their relative speed is the sum of their individual speeds. The time taken for the goods train to pass the man equals the time it takes for its entire length to cover the relative distance at the relative speed. We convert everything to consistent units, compute the relative speed, and then find the unknown speed of the goods train. Step-by-Step Solution:Step 1: Let the speed of the goods train be v km/h.Step 2: Since the trains are moving in opposite directions, relative speed = 50 + v km/h.Step 3: Convert the relative speed into meters per second by multiplying by 5 / 18.Step 4: The time taken to pass completely is 9 seconds and the distance covered relative to the man is the length of the goods train, which is 280 meters.Step 5: Relative speed in m/s = distance / time = 280 / 9 m/s.Step 6: Equate the two expressions for relative speed: (50 + v) * 5 / 18 = 280 / 9.Step 7: Multiply both sides by 18 to get 5(50 + v) = 560.Step 8: Simplify: 5(50 + v) = 560 leads to 50 + v = 112, so v = 62 km/h. Verification / Alternative check:Compute the relative speed in km/h from 280 / 9 m/s. Multiply 280 / 9 by 18 / 5 to convert, which gives (280 * 18) / (9 * 5) = 5040 / 45 = 112 km/h. Since the passenger train moves at 50 km/h, the goods train must move at 112 - 50 = 62 km/h. This matches the earlier result. Why Other Options Are Wrong:Options 60, 64, 65, and 58 km/h are values that do not produce the correct relative speed. For example, if the goods train had speed 60 km/h, relative speed would be 110 km/h, which in meters per second is about 30.56 and would give a different passing time. Only 62 km/h gives a relative speed of 112 km/h that matches the observed time and length. Common Pitfalls:Students often forget to add the speeds for opposite directions or they subtract them instead. Another common error is to skip unit conversion and mix km/h with meters and seconds. Correctly applying the conversion factor 5 / 18 and carefully equating the relative speed to distance over time is crucial. Final Answer:The speed of the goods train is 62 km/h.

Question 14

A train is 110 meters long and runs at 60 km/h. In how many seconds will it completely pass a man who is running at 6 km/h in the direction opposite to that of the train?

Accepted Answer

Introduction / Context:This question involves relative speed between a moving train and a man on the track. The train and the man are moving in opposite directions, and we need to determine the time taken for the train to pass the man completely. Given Data / Assumptions:- Length of the train is 110 meters.- Speed of the train is 60 km/h.- Speed of the man is 6 km/h.- The man runs in the direction opposite to the direction of the train.- We must find the time taken in seconds for the train to pass the man. Concept / Approach:When two objects move in opposite directions, their relative speed is the sum of their speeds. The time required for the train to completely pass the man equals the length of the train divided by the relative speed, with all units consistent. So we convert relative speed from km/h to m/s and then apply the formula time = distance / speed. Step-by-Step Solution:Step 1: Compute the relative speed since directions are opposite: 60 km/h + 6 km/h = 66 km/h.Step 2: Convert 66 km/h to meters per second by multiplying by 5 / 18.Step 3: Relative speed in m/s = 66 * 5 / 18 = 330 / 18 = 55 / 3 m/s.Step 4: The distance to be covered for complete passing is the length of the train, which is 110 meters.Step 5: Time = distance / speed = 110 / (55 / 3) seconds.Step 6: Simplify the fraction: 110 * 3 / 55 = 330 / 55 = 6 seconds.Step 7: Thus, the train passes the man completely in 6 seconds. Verification / Alternative check:We can approximate the relative speed numerically. 66 km/h is approximately 18.33 m/s. Dividing the train length 110 meters by 18.33 m/s gives about 6 seconds. This confirms the exact calculation of 6 seconds. Why Other Options Are Wrong:10 or 8 seconds would arise if one incorrectly used only the train speed or subtracted the man's speed instead of adding. 4 seconds would require a much higher relative speed than actually present. 12 seconds comes from treating the man as stationary with half the actual relative speed. Only 6 seconds correctly matches the relative speed calculation. Common Pitfalls:Typical mistakes include subtracting speeds when directions are opposite, failing to convert km/h to m/s, or using the sum of speeds but not converting units properly. Forgetting that only the train length is relevant for passing a point or a person is another frequent error. Final Answer:The train will completely pass the man in 6 seconds.

Question 15

Two trains start at the same time from stations A and B and travel towards each other. After they meet, one train reaches B in 4 hours and the other reaches A in 9 hours. What is the ratio of their speeds?

Accepted Answer

Introduction / Context:This question is a well known relative motion problem involving two trains starting from opposite ends of a route. After meeting, they take different times to reach the opposite stations. The task is to determine the ratio of their speeds using time taken after the meeting point. Given Data / Assumptions:- Train 1 starts from station A and Train 2 starts from station B at the same time.- They travel towards each other and meet at some point between A and B.- After meeting, Train 1 takes 4 hours to reach B.- Train 2 takes 9 hours to reach A.- Both trains move at constant speeds. Concept / Approach:When two trains travel towards each other, the distances they cover before and after meeting are related to their speeds. After the meeting, Train 1 covers the distance originally travelled by Train 2, and Train 2 covers the distance originally travelled by Train 1. The key fact is that the ratio of speeds is inversely proportional to the time taken to cover the same distances. Step-by-Step Solution:Step 1: Let v1 be the speed of the train starting from A and v2 be the speed of the train starting from B.Step 2: Let d1 be the distance from A to the meeting point and d2 be the distance from B to the meeting point.Step 3: Before meeting, Train 1 covers d1 in some time t and Train 2 covers d2 in the same time t, so v1 = d1 / t and v2 = d2 / t.Step 4: After meeting, Train 1 covers distance d2 in 4 hours so v1 = d2 / 4. Train 2 covers distance d1 in 9 hours so v2 = d1 / 9.Step 5: From the before meeting relations, v1 / v2 = d1 / d2.Step 6: From the after meeting relations, v1 / v2 = (d2 / 4) / (d1 / 9) = (d2 * 9) / (4 * d1) = (9 / 4) * (d2 / d1).Step 7: Equate the two expressions for v1 / v2: d1 / d2 = (9 / 4) * (d2 / d1).Step 8: Multiply both sides by d1 / d2 to get (d1 / d2)^2 = 9 / 4, so d1 / d2 = 3 / 2.Step 9: Since v1 / v2 = d1 / d2, the ratio of speeds is v1 : v2 = 3 : 2. Verification / Alternative check:A shortcut formula for such problems states that if trains take t1 and t2 hours respectively to reach destinations after meeting, then their speed ratio is proportional to the square root of t2 / t1 or can be derived as above. However, using the detailed distance logic is more transparent and confirms the ratio 3 : 2. Why Other Options Are Wrong:Ratios 2 : 1, 4 : 3, 5 : 4, or 9 : 4 do not satisfy the relationship between distances and times derived from the motion before and after meeting. They would produce inconsistent distances when back substituted. Only the ratio 3 : 2 matches the distance and time conditions simultaneously. Common Pitfalls:One common mistake is to assume speed ratio equal to time ratio 4 : 9 instead of understanding the inverse relationship. Others forget that the trains swap the remaining distances after meeting. Ignoring this leads to incorrect proportional reasoning. Final Answer:The ratio of the speeds of the two trains is 3 : 2.

Question 16

A man can travel at 15 km/h with the current of a river and the speed of the current alone is 2.5 km/h. What is the speed of the man in still water against the current?

Accepted Answer

Introduction / Context:This is a standard boats and streams type problem. It uses the relationship between speeds in still water, speeds of the current, and effective speeds with and against the current. Understanding these relationships is important for many motion based aptitude questions. Given Data / Assumptions:- Speed of the man with the current (downstream speed) is 15 km/h.- Speed of the current alone is 2.5 km/h.- The man rows in a straight river with uniform current.- We are asked to find the speed of the man against the current (upstream speed). Concept / Approach:For motion in a stream, let v be the speed of the man in still water and c be the speed of the current. The downstream speed is v + c and the upstream speed is v - c. Given downstream speed and c, we can find v and then compute upstream speed. Algebraic manipulation is straightforward once the relationships are remembered. Step-by-Step Solution:Step 1: Let v be the speed of the man in still water and c be the speed of the current.Step 2: We are given c = 2.5 km/h.Step 3: Downstream speed is v + c and is given as 15 km/h, so v + 2.5 = 15.Step 4: Solve for v: v = 15 - 2.5 = 12.5 km/h.Step 5: Upstream speed is v - c = 12.5 - 2.5 = 10 km/h.Step 6: Therefore, the man's speed against the current is 10 km/h. Verification / Alternative check:If the man rows in still water at 12.5 km/h and the current flows at 2.5 km/h, then downstream he moves at 12.5 + 2.5 = 15 km/h, which matches the given downstream speed. Upstream, the current opposes his motion so the effective speed is 12.5 - 2.5 = 10 km/h. This confirms that the calculation is consistent. Why Other Options Are Wrong:9.5 km/h or 10.5 km/h would imply incorrect values for the still water speed when added to or subtracted from 2.5 km/h and would not reproduce the downstream speed of 15 km/h. 11 km/h also fails this consistency check. 12.5 km/h is the still water speed, not the upstream speed, so it is not the correct answer to the question asked. Common Pitfalls:Some learners confuse the meanings of downstream and upstream and may add when they should subtract or vice versa. Others set up the equation as 15 - 2.5 = v - c, which mixes given and unknown speeds incorrectly. Remembering the basic formulas v + c for downstream and v - c for upstream helps avoid these mistakes. Final Answer:The speed of the man against the current is 10 km/h.

Question 17

In a system of two meshed cogged wheels, one wheel has 32 cogs and the other wheel has 54 cogs. The wheels work into each other without slipping. If the wheel with 54 cogs makes 80 complete revolutions in three-quarters of a minute, how many complet…

Accepted Answer

Introduction / Context:This aptitude problem on cogged wheels (gears) tests the concept of relative motion and how rotational speed changes when two wheels with different numbers of cogs mesh with each other. Such questions are common in quantitative aptitude exams because they combine proportional reasoning with unit conversions involving minutes and seconds. Given Data / Assumptions: First wheel has 32 cogs. Second wheel has 54 cogs. The 54-cog wheel makes 80 revolutions in three-quarters of a minute, that is in 45 seconds. The wheels are perfectly meshed without slipping, so the linear motion of the teeth is the same for both wheels. We must find how many revolutions the 32-cog wheel makes in 8 seconds. Concept / Approach:When two cogged wheels mesh, the number of teeth passing the point of contact per unit time is the same for both wheels. Therefore, the speed of rotation is inversely proportional to the number of cogs. If one wheel has more cogs, it turns more slowly, and if it has fewer cogs, it turns more quickly. We first find the angular speed of the 54-cog wheel, then use the tooth ratio to obtain the speed of the 32-cog wheel, and finally convert that result to revolutions in 8 seconds. Step-by-Step Solution:Step 1: Time for 80 revolutions of the 54-cog wheel is 45 seconds, so its revolutions per second are 80 / 45.Step 2: For meshed gears, speed of 32-cog wheel / speed of 54-cog wheel = number of cogs on 54-cog wheel / number of cogs on 32-cog wheel = 54 / 32.Step 3: Revolutions per second of 32-cog wheel = (80 / 45) * (54 / 32).Step 4: Simplify: 80 * 54 / (45 * 32) = 135 / 45 = 3 revolutions per second.Step 5: In 8 seconds, revolutions of 32-cog wheel = 3 * 8 = 24. Verification / Alternative check:If the smaller wheel has fewer cogs, it must rotate faster than the larger wheel.The ratio of speeds should be 54 : 32, which reduces to 27 : 16, a value greater than 1, confirming that the 32-cog wheel is faster.Using 3 revolutions per second for the smaller wheel is reasonable because in 45 seconds it would complete 135 revolutions, which matches 80 revolutions of the larger wheel times 54 / 32. Why Other Options Are Wrong:Values like 36, 38 or 48 assume incorrect proportionality or mishandled time conversion. For example, 48 would correspond to using a wrong ratio or doubling the correct value. The option 30 is also incorrect because it ignores the exact 45 second to 8 second proportion combined with the tooth ratio. Common Pitfalls:Candidates often reverse the ratio of cogs and incorrectly assume that speed is directly proportional to the number of cogs. Another frequent error is to forget that three-quarters of a minute is 45 seconds, not 30 seconds, which leads to wrong revolutions per second. Some learners also multiply by the time interval in minutes rather than converting everything into seconds before calculating. Final Answer:The 32-cog wheel makes 24 complete revolutions in 8 seconds.

Question 18

Two trains of equal length are running in opposite directions at the same uniform speed. The length of each train is 120 metres. If they completely cross each other in 12 seconds, what is the speed of each train in km/h?

Accepted Answer

Introduction / Context:This question is a classic trains problem involving relative speed when two trains move in opposite directions. It checks whether a student can correctly apply the idea that when bodies move towards each other, their relative speed is the sum of their individual speeds, and then convert from metres per second to kilometres per hour. Given Data / Assumptions: Each train is 120 metres long. The trains are moving in opposite directions with the same speed. They completely cross each other in 12 seconds. The motion is along the same straight track, and speeds are constant. We must find the speed of each train in km/h. Concept / Approach:When two trains move in opposite directions, the distance to be covered when they cross each other is the sum of their lengths. Since both lengths are equal here, the total crossing distance is 120 + 120 metres. The relative speed when moving in opposite directions is the sum of their individual speeds. Once the relative speed in metres per second is found from distance divided by time, we divide by 2 to get the speed of each train, and finally convert this speed into km/h using the standard factor 18 / 5. Step-by-Step Solution:Step 1: Total length to be covered while crossing each other = 120 + 120 = 240 metres.Step 2: Time taken to cross each other = 12 seconds, so relative speed = 240 / 12 = 20 metres per second.Step 3: Since the trains have equal speed and move in opposite directions, each train has speed = relative speed / 2 = 20 / 2 = 10 metres per second.Step 4: Convert this speed into km/h using 1 metre per second = 18 / 5 km/h.Step 5: Speed in km/h = 10 * 18 / 5 = 10 * 3.6 = 36 km/h. Verification / Alternative check:If each train runs at 36 km/h, then in metres per second the speed is 36 * 5 / 18 = 10 metres per second.Relative speed in opposite directions = 10 + 10 = 20 metres per second.Time = distance / speed = 240 / 20 = 12 seconds, which matches the given crossing time. Why Other Options Are Wrong:20 km/h and 28 km/h are too low; if each train had such a speed, the time required to cross would be higher than 12 seconds. The option 42 km/h is higher than the correct value and would lead to a shorter crossing time. The option 32 km/h does not satisfy the exact condition when checked back using distance and time. Common Pitfalls:Many learners forget to add the lengths of both trains when they cross each other and instead use only one length. Another common mistake is to forget that the speeds add when trains move in opposite directions. Finally, some students perform the km/h to m/s conversion incorrectly, using 5 / 18 instead of 18 / 5 or vice versa in the wrong place. Careful unit handling avoids such errors. Final Answer:Each train is moving at a speed of 36 km/h.

Question 19

A train passes a telegraph post in 8 seconds and then passes completely over a bridge that is 264 metres long in 20 seconds. What is the speed of the train in km/h?

Accepted Answer

Introduction / Context:This problem combines basic distance–speed–time relationships with the idea that a moving train must cover different effective distances when it passes a single point versus when it passes a bridge. It tests the ability to form and solve simple equations to find both the length of the train and its speed in km/h. Given Data / Assumptions: Time to pass a telegraph post completely = 8 seconds. Time to cross a bridge that is 264 metres long = 20 seconds. The train runs at uniform speed in a straight line. We assume no delays or changes in speed between the two situations. We must find the speed of the train in km/h. Concept / Approach:When a train passes a single point such as a post, the distance covered is equal to the length of the train. When it passes a bridge, the distance covered is the sum of the length of the train and the length of the bridge. Let the length of the train be L metres and the speed be v metres per second. We use the relationship speed = distance / time in both scenarios, set up equations, solve for L, then compute v and convert to km/h using the factor 18 / 5. Step-by-Step Solution:Step 1: Let the train length be L metres and its speed be v metres per second.Step 2: From passing the post, L / v = 8, so v = L / 8.Step 3: When crossing the 264 metre bridge, the distance is L + 264 and time is 20 seconds, so (L + 264) / v = 20.Step 4: Substitute v = L / 8 into the second equation to get (L + 264) / (L / 8) = 20.Step 5: Simplify: 8 * (L + 264) / L = 20, so 8L + 2112 = 20L, giving 12L = 2112 and L = 176 metres.Step 6: Speed v = L / 8 = 176 / 8 = 22 metres per second.Step 7: Convert to km/h: speed in km/h = 22 * 18 / 5 = 79.2 km/h. Verification / Alternative check:Check with L = 176 metres and v = 22 metres per second.Time to pass post = 176 / 22 = 8 seconds, which agrees with the data.Time to cross bridge = (176 + 264) / 22 = 440 / 22 = 20 seconds, which also matches the given information. Why Other Options Are Wrong:Speeds such as 69.5 km/h, 70 km/h, 72 km/h or 79 km/h are all close but incorrect. They arise if one approximates during intermediate steps or uses a wrong conversion factor between metres per second and km/h. Only 79.2 km/h satisfies both time conditions exactly when back-substituted. Common Pitfalls:Candidates sometimes forget that the distance covered while crossing the bridge includes both the train length and the bridge length. Another typical mistake is incorrect algebra while solving for L and v, such as cross multiplying incorrectly. Errors in unit conversion, such as using 5 / 18 instead of 18 / 5 to convert metres per second to km/h, are also frequent. Working carefully through each step and checking dimensions helps avoid these issues. Final Answer:The speed of the train is 79.2 km/h.

Question 20

An express train is moving at a constant speed of 108 km/h. What is this speed expressed in metres per second?

Accepted Answer

Introduction / Context:Unit conversion is a fundamental skill tested in almost every quantitative aptitude exam. This question asks you to convert a speed given in kilometres per hour into an equivalent value in metres per second, which is the unit often used in physics and many train or motion based problems. Given Data / Assumptions: Speed of the train = 108 km/h. We need to convert this speed into metres per second. We assume uniform motion without acceleration. Concept / Approach:The key relationship here is the numerical conversion between kilometres per hour and metres per second. One kilometre is 1000 metres and one hour is 3600 seconds. Therefore, 1 km/h equals 1000 / 3600 metres per second, which simplifies to 5 / 18 metres per second. Conversely, to convert from metres per second to km/h, we use the factor 18 / 5. In this problem, we multiply 108 km/h by 5 / 18 to obtain the equivalent speed in metres per second. Step-by-Step Solution:Step 1: Recall that 1 km/h = 5 / 18 metres per second.Step 2: Speed in metres per second = 108 * 5 / 18.Step 3: Simplify by dividing 108 by 18 to get 6.Step 4: Now compute 6 * 5 = 30.Step 5: Thus, the speed is 30 metres per second. Verification / Alternative check:Convert 30 metres per second back into km/h to verify.Using the reverse factor, multiply 30 by 18 / 5 to get 30 * 18 / 5 = 30 * 3.6 = 108 km/h.Since we arrive at the original speed given in the question, the conversion is correct. Why Other Options Are Wrong:10.8 metres per second and 18 metres per second correspond to incomplete or incorrect use of the conversion factor. The value 38.8 metres per second is too high and could come from mixing up 5 / 18 and 18 / 5. The value 24 metres per second ignores the exact factor and leads to an incorrect back-converted speed in km/h. Only 30 metres per second accurately represents 108 km/h. Common Pitfalls:A common mistake is to use 18 / 5 instead of 5 / 18 when converting from km/h to metres per second. Another frequent error is to forget either the 1000 metres in a kilometre or the 3600 seconds in an hour, leading to incorrect fractions such as 10 / 36 instead of 5 / 18. To avoid confusion, remember: to go from km/h to metres per second, multiply by 5 / 18; to go from metres per second to km/h, multiply by 18 / 5. Final Answer:The speed of 108 km/h is equal to 30 metres per second.

Problems on Trains Questions

Questions

Practice smarter