Question 1

Two trains run in the same direction on parallel tracks. Lucknow Mail is 120 m long at 40 km/h; Punjab Mail is 80 m long at 50 km/h. How many seconds will they take to completely cross each other (overtaking, same direction)?

Accepted Answer

Introduction / Context: For trains in the same direction, relative speed equals the difference of speeds. The distance to be covered for a complete pass is the sum of their lengths. Given Data / Assumptions: Lengths: 120 m and 80 m ⇒ total = 200 m. Speeds: 40 km/h and 50 km/h (same direction). Concept / Approach: Relative speed = 50 − 40 = 10 km/h. Convert to m/s, then time = distance / speed. Step-by-Step Solution: Relative speed = 10 km/h = 10 * 1000 / 3600 = 2.777... m/s.Time = 200 / 2.777... = 72 s. Verification / Alternative check: 2.777... m/s * 72 s = 200 m, exactly the combined length. Why Other Options Are Wrong: 8/11.5/12.5 s require much larger relative speeds than 10 km/h. Common Pitfalls: Using sum of speeds instead of difference for same-direction motion. Final Answer: 72 sec.

Question 2

A train passes a pole in 10 seconds and a 120 m platform in 18 seconds. What is the length of the train (in metres)?

Accepted Answer

Introduction / Context: Let L be the train length and v its speed. Passing a pole takes time to cover just L; passing a platform of length P takes time to cover L + P at the same speed v. Given Data / Assumptions: Pole time = 10 s ⇒ L = v * 10. Platform = 120 m, platform time = 18 s ⇒ L + 120 = v * 18. Concept / Approach: Eliminate v by dividing equations or substituting v = L/10 into the platform equation to solve for L directly. Step-by-Step Solution: L + 120 = (L/10) * 18 = 1.8L.120 = 1.8L − L = 0.8L ⇒ L = 120 / 0.8 = 150 m. Verification / Alternative check: Speed v = L/10 = 15 m/s; platform pass distance = 270 m; 270/15 = 18 s, consistent. Why Other Options Are Wrong: 130/180/240 m violate either the pole or platform timing when checked with the same v. Common Pitfalls: Adding 120 m incorrectly or mixing seconds with metres without using speed. Final Answer: 150 m

Question 3

An engine wheel has circumference 300 cm (3 m) and makes 10 revolutions in 6 seconds. What is the linear speed of the wheel in km/h?

Accepted Answer

Introduction / Context: Linear speed = (revolutions per time) * circumference. Convert the resulting m/s to km/h by multiplying by 3.6. Given Data / Assumptions: Circumference = 300 cm = 3 m. 10 revolutions in 6 s. Concept / Approach: Distance traveled in 6 s = 10 * 3 = 30 m. So speed = 30/6 = 5 m/s. Convert to km/h: 5 * 3.6 = 18 km/h. Step-by-Step Solution: Distance per 6 s = 10 * 3 m = 30 m.Speed = 30 / 6 = 5 m/s.5 m/s = 18 km/h. Verification / Alternative check: Using unit factors: (5 m/s)*(18/5) = 18 km/h. Why Other Options Are Wrong: 20/27/36 km/h require different revolution counts or circumference. Common Pitfalls: Forgetting to convert cm to m or m/s to km/h. Final Answer: 18

Question 4

Anjali fires two shots 6 minutes apart from the same place. Bhagwat is in a car approaching the firing point and hears the second report 5 min 32 s after the first. Given speed of sound = 332 m/s, what is the car’s speed (in m/s)?

Accepted Answer

Introduction / Context: This is a moving-observer Doppler timing case. If the source emits with period T, an observer moving toward a stationary source perceives period T′ = T * (1 − v_o / v), where v_o is observer speed and v is sound speed. We compare the actual 6-minute interval with the heard 5 min 32 s. Given Data / Assumptions: T = 6 min = 360 s. T′ = 5 min 32 s = 332 s. v (sound) = 332 m/s. Concept / Approach: Use T′ = T * (1 − v_o / v). Solve for v_o = v * (1 − T′ / T). Round to the nearest option as practical measurement output. Step-by-Step Solution: v_o = 332 * (1 − 332/360) = 332 * (28/360) = 332 * (7/90) ≈ 25.82 m/s.Closest listed speed ≈ 28 m/s. Verification / Alternative check: Using 28 m/s back into T′ ≈ 360 * (1 − 28/332) ≈ 360 * 0.9157 ≈ 330 s (near 332 s), within typical rounding tolerance for such problems. Why Other Options Are Wrong: 56/102 m/s are unrealistically high for a car; 32 m/s overshoots the implied timing. Common Pitfalls: Using frequency formula directly without converting to period; mixing minutes and seconds. Final Answer: 28 m/s

Question 5

A car overtakes a man walking at 6 km/h on a foggy road. The man can see objects up to 450 m ahead in his direction of travel. He continues to see the car (moving in the same direction) for 4.5 minutes after being overtaken. What is the car’s speed …

Accepted Answer

Introduction / Context: After overtaking, the car pulls away. The man can see up to 450 m ahead; the car disappears when it is 450 m in front. The time to “disappear” equals distance / relative speed (car minus man). Given Data / Assumptions: Visibility ahead = 450 m. Man speed = 6 km/h. Visible duration after pass = 4.5 min = 270 s. Concept / Approach: Let v be car speed (km/h). Relative speed = v − 6 (km/h) = (v − 6) * 1000 / 3600 m/s. Time = 450 / relative_speed. Step-by-Step Solution: 270 = 450 / [ (v − 6)*1000/3600 ].(v − 6) = (450 / 270) * (3600/1000) = (5/3) * 3.6 = 6 km/h.v = 6 + 6 = 12 km/h. Verification / Alternative check: Relative 6 km/h covers 450 m in 270 s exactly. Why Other Options Are Wrong: 9/12.5/15 km/h yield different relative speeds and times. Common Pitfalls: Using 450 m as two-way visibility or using man speed instead of relative speed. Final Answer: 12 km/h

Question 6

X and Y start toward each other at 8:00 a.m. from points 17.5 km apart. Speeds: X = 3 km/h, Y = 4 km/h. At what time do they meet?

Accepted Answer

Introduction / Context: The classic meeting-time problem: when two objects move toward each other, add their speeds to get the closing rate. Time to meet = distance / (sum of speeds). Add this time to the start time to get the clock time. Given Data / Assumptions: Initial separation = 17.5 km. Speeds: 3 km/h and 4 km/h. Start time = 8:00 a.m. Concept / Approach: Closing speed = 3 + 4 = 7 km/h. Time in hours = 17.5 / 7 = 2.5 h. Step-by-Step Solution: Time to meet = 17.5 / 7 = 2.5 h = 2 hours 30 minutes.Meeting time = 8:00 a.m. + 2:30 = 10:30 a.m. Verification / Alternative check: Distances covered: X → 3 * 2.5 = 7.5 km; Y → 4 * 2.5 = 10 km; total 17.5 km. Why Other Options Are Wrong: 11:30 a.m./p.m. are an hour late; p.m. variants misplace time of day. Common Pitfalls: Forgetting to add speeds or mis-converting 0.5 h to 30 min. Final Answer: 10 : 30 am

Question 7

Two trains run between Manipur and Dispur. Train M: Manipur→Dispur, departs 6:00 a.m., arrives 10:00 a.m. (4 h). Train D: Dispur→Manipur, departs 8:00 a.m., arrives 11:00 a.m. (3 h). At what time do the two trains meet?

Accepted Answer

Introduction / Context: Assume constant speeds on the same route. Let the one-way distance be D. Speed of Train M = D/4 (h^−1), speed of Train D = D/3 (h^−1). The second train starts later, so first compute distance Train M covers by 8:00 a.m., then use relative speed to find the meeting time after 8:00 a.m. Given Data / Assumptions: Train M speed = D/4 per hour. Train D speed = D/3 per hour. By 8:00 a.m., Train M has run 2 h ⇒ distance = D/2. Concept / Approach: Remaining separation at 8:00 a.m. = D − D/2 = D/2. After 8:00 a.m., closing speed = D/4 + D/3 = 7D/12 per hour. Time after 8:00 a.m. to meet = (D/2) / (7D/12) = 6/7 h ≈ 0 h 51 m 26 s. Step-by-Step Solution: Meeting time ≈ 8:00 a.m. + 51 m 26 s ≈ 8:51 a.m. (to the nearest minute). Verification / Alternative check: Distances after 8:00 a.m.: M covers (D/4)*(6/7) = 3D/14; D covers (D/3)*(6/7) = 2D/7; sum = 3D/14 + 2D/7 = D/2, consistent. Why Other Options Are Wrong: 7:56 a.m. is too early (before Train D departs). 8:56 a.m. is several minutes late. p.m. options are irrelevant. Common Pitfalls: Taking an average of clock times or ignoring the 2 h head start. Final Answer: None of these (correct ≈ 8:51 a.m.)

Question 8

Anurag travels 45 km in 8 h 45 min (8.75 h) by walking part of the way at 3 km/h and cycling the rest at 8 km/h. How many kilometres did he travel by bicycle?

Accepted Answer

Introduction / Context: This is a two-speed mixture problem. Let x be distance cycled at 8 km/h; the remaining (45 − x) km is on foot at 3 km/h. Total time is the sum of segment times. Given Data / Assumptions: Total distance = 45 km. Total time = 8.75 h. Walking speed = 3 km/h; cycling speed = 8 km/h. Concept / Approach: Time equation: x/8 + (45 − x)/3 = 8.75. Solve for x. Since distance and time are positive, the solution must lie between 0 and 45 km. Step-by-Step Solution: x/8 + (45 − x)/3 = 8.75.Multiply by 24: 3x + 8(45 − x) = 210 ⇒ 3x + 360 − 8x = 210.−5x = −150 ⇒ x = 30 km. Verification / Alternative check: Time = 30/8 + 15/3 = 3.75 + 5 = 8.75 h, matches total. Why Other Options Are Wrong: 15/20/25 km lead to total times not equal to 8.75 h. Common Pitfalls: Mixing time/pace or converting 8 h 45 min incorrectly (8.75 h is exact). Final Answer: 30 km

Question 9

A train runs at 60 km/h and passes completely through a tunnel 1.5 km long in 2 minutes. What is the length of the train (in metres)?

Accepted Answer

Introduction / Context: While passing a tunnel, the train’s front travels train_length + tunnel_length. Use distance = speed * time (consistent units) to find the distance covered in the 2-minute interval, then subtract the tunnel length to get the train length. Given Data / Assumptions: Speed = 60 km/h. Tunnel length = 1.5 km. Time = 2 min = 1/30 h. Concept / Approach: Total distance covered in time T: S = v * T. Train length = S − tunnel_length (in km), then convert to metres. Step-by-Step Solution: S = 60 * (1/30) = 2 km.Train length = 2 − 1.5 = 0.5 km = 500 m. Verification / Alternative check: At 60 km/h, 2 minutes corresponds to 2 km—consistent. Why Other Options Are Wrong: 250/1000/1500 m imply tunnel pass times other than 2 minutes. Common Pitfalls: Using only tunnel length instead of tunnel + train while timing. Final Answer: 500 m

Question 10

A train’s onward speed is 25% more than its return speed. It halts 1 hour at the destination. Total time for a round trip of 800 km is 17 h (including the 1 h halt). What is the onward (faster) speed (in km/h)?

Accepted Answer

Introduction / Context: Let return speed be v; onward speed is 1.25v. One-way distance is 400 km. Exclude the 1 h halt first to get actual travel time, then solve for v from the total-time equation. Given Data / Assumptions: Total distance = 800 km. Halt = 1 h; total time = 17 h ⇒ travel time = 16 h. Onward speed = 1.25v; return speed = v. Concept / Approach: Travel time T = 400/(1.25v) + 400/v = 16. Solve for v, then compute 1.25v. Step-by-Step Solution: 400/(1.25v) = 320/v.320/v + 400/v = 720/v = 16 ⇒ v = 45 km/h.Onward speed = 1.25 * 45 = 56.25 km/h. Verification / Alternative check: Times: 400/56.25 = 7.111... h; 400/45 = 8.888... h; sum ≈ 16 h; +1 h halt = 17 h total. Why Other Options Are Wrong: 45 km/h is the return speed, not onward; others misfit the 16 h travel time. Common Pitfalls: Forgetting to subtract the 1 h halt before solving for speeds. Final Answer: 56.25 km/h

Question 11

Speeds of Meena and Tenna are in the ratio 2:3 for the same distance A→B. Meena takes 20 minutes more than Tenna for A→B. If Meena doubled her speed, how long would she take for A→B (in minutes)?

Accepted Answer

Introduction / Context: Let speeds be 2k and 3k for Meena and Tenna; let distance be D. Time is distance/speed. Use the 20-minute difference to find D/k, then compute Meena’s time if she doubles her speed to 4k. Given Data / Assumptions: Speed ratio 2:3 (Meena:Tenna). Time difference = 20 min = 1/3 h. Concept / Approach: t_M − t_T = D/(2k) − D/(3k) = D/(6k) = 1/3 ⇒ D/k = 2. Then doubled-speed time = D/(4k) = (D/k)/4. Step-by-Step Solution: D/k = 2.New time = (D/k)/4 = 2/4 = 1/2 h = 30 min. Verification / Alternative check: Original times: D/(2k) = 1 h; D/(3k) = 2/3 h; difference 1/3 h = 20 min (consistent when D/k = 2). Why Other Options Are Wrong: 60/45/110 min do not match the derived new-time expression. Common Pitfalls: Applying the 2:3 ratio to times instead of speeds (times are inversely proportional). Final Answer: 30 min

Question 12

A local train at 29 km/h and an express train at 65 km/h run in the same direction on parallel tracks. From the local’s cab, the faster train takes exactly 16 seconds to pass by. What is the length of the faster train (in metres)?

Accepted Answer

Introduction / Context: For same-direction trains, use relative speed (difference). The faster train’s length equals (relative speed) * (overtaking time), since we observe it pass a reference point in the local cab. Given Data / Assumptions: Speeds: 65 km/h and 29 km/h. Overtaking time = 16 s. Concept / Approach: Relative speed = (65 − 29) km/h = 36 km/h = 10 m/s. Length = 10 * 16 = 160 m. Step-by-Step Solution: Relative speed = 36 km/h = 10 m/s.Length = 10 m/s * 16 s = 160 m. Verification / Alternative check: Unit-consistent multiplication yields metres directly. Why Other Options Are Wrong: 60/120/240 m do not match the observed 16 s at 10 m/s. Common Pitfalls: Adding speeds instead of subtracting for same-direction motion. Final Answer: 160 m

Question 13

A man completes 2/25 of a journey by bus, 21/50 by car, and the remaining 2 km on foot. What is the total distance of the journey (in km)?

Accepted Answer

Introduction / Context: Part-by-part distance problems often provide fractional segments plus a fixed remainder. Sum the fractions to find what part of the journey the fixed remainder represents, then scale to get the total distance. Given Data / Assumptions: Bus fraction = 2/25. Car fraction = 21/50. Foot distance = 2 km (remaining). Concept / Approach: Total fractional part done by bus + car = 2/25 + 21/50 = 4/50 + 21/50 = 25/50 = 1/2. Hence remaining half equals 2 km ⇒ total is 4 km. Step-by-Step Solution: Fraction by bus+car = 1/2.Remaining fraction = 1/2 corresponds to 2 km.Total distance = 2 / (1/2) = 4 km. Verification / Alternative check: Half of 4 km is 2 km, matching the on-foot segment. Why Other Options Are Wrong: 2.7/3.4/3.8 km would not leave exactly 2 km as one half. Common Pitfalls: Failing to convert 2/25 to 4/50 before addition. Final Answer: 4 km

Question 14

A driver counts 25 telephone poles in exactly 1 minute along a straight road. Poles are 40 m apart (uniform spacing). At what speed is the car travelling (in km/h)?

Accepted Answer

Introduction / Context: When counting poles, the number of intervals is usually one less than the number of poles seen during the timing interval (starting at a pole). Thus, distance ≈ (poles − 1) * spacing. Convert metres per minute to km/h for the final speed. Given Data / Assumptions: Poles counted = 25 in 1 min. Spacing = 40 m. Assume counting starts at a pole and ends at the 25th pole. Concept / Approach: Intervals = 25 − 1 = 24; distance in 1 min = 24 * 40 = 960 m. Convert 960 m/min to km/h by multiplying by 60 and dividing by 1000. Step-by-Step Solution: Distance per minute = 960 m.Speed = 960 * 60 / 1000 = 57.6 km/h. Verification / Alternative check: Using 25 intervals would yield 60 km/h, but that assumes counting from mid-interval rather than pole-to-pole; the standard interpretation uses intervals. Why Other Options Are Wrong: 52.4/48.2/44.9 km/h do not align with either interpretation; 57.6 km/h matches the standard pole-interval method. Common Pitfalls: Taking 25 intervals instead of 24; mis-converting m/min to km/h. Final Answer: 57.6 km/h

Question 15

A man walks 7.5 km at 3 km/h. What speed must he maintain for the next 2 hours so that his overall average speed for the whole session is 4 km/h?

Accepted Answer

Introduction / Context: Average speed = total distance / total time. Compute required remaining distance to hit the target average over the total duration, then divide by the remaining time to get the needed speed for the second segment. Given Data / Assumptions: First segment: 7.5 km at 3 km/h ⇒ 2.5 h. Second segment duration: 2 h at unknown speed v. Target average = 4 km/h over total 4.5 h. Concept / Approach: Total distance needed = 4 * 4.5 = 18 km. Remaining distance = 18 − 7.5 = 10.5 km. Required v = 10.5 / 2. Step-by-Step Solution: v = 10.5 / 2 = 5.25 km/h. Verification / Alternative check: Total distance = 7.5 + 10.5 = 18 km; time = 2.5 + 2 = 4.5 h; average = 18/4.5 = 4 km/h. Why Other Options Are Wrong: 3.65/4.75/6.50 km/h would not produce the exact 18 km total in 4.5 h. Common Pitfalls: Using average of speeds instead of weighted-average based on time and distance. Final Answer: 5.25 km/h

Question 16

Ram and Sita walk toward each other along a 160 km road at 8 km/h and 2 km/h, respectively. How many hours until they meet?

Accepted Answer

Introduction / Context: Meeting time problems use the sum of approach speeds. Here, walking speeds are constant and the road is straight with fixed length 160 km. Given Data / Assumptions: Distance = 160 km. Speeds: 8 km/h (Ram) and 2 km/h (Sita). Concept / Approach: Closing speed = 8 + 2 = 10 km/h. Time = distance / closing speed. Step-by-Step Solution: Time = 160 / 10 = 16 h. Verification / Alternative check: Distances covered at meeting: 8 * 16 = 128 km and 2 * 16 = 32 km; sum = 160 km. Why Other Options Are Wrong: 8/10/20 h correspond to wrong closing speeds or distances. Common Pitfalls: Subtracting speeds instead of adding for opposite directions. Final Answer: 16 h

Question 17

Two trains leave New Delhi at the same time on straight tracks. One goes due North at 60 km/h; the other goes due South at 40 km/h. After how many hours will they be 150 km apart?

Accepted Answer

Introduction / Context: When two trains move in opposite directions from the same point, separation grows at the sum of their speeds. Time to reach a given separation = separation / (sum of speeds). Given Data / Assumptions: Speeds: 60 km/h northbound; 40 km/h southbound. Desired separation: 150 km. Concept / Approach: Sum speed = 60 + 40 = 100 km/h. Time = 150 / 100 = 1.5 h. Step-by-Step Solution: Time = 1.5 h = 3/2 h. Verification / Alternative check: After 1.5 h: distances 90 km and 60 km; separation 150 km. Why Other Options Are Wrong: 4/3 and 3/4 h yield separations ≠ 150 km; 15/2 h is far too long. Common Pitfalls: Subtracting speeds instead of adding for diverging paths. Final Answer: 3/2 h

Question 18

Walking at 5/7 of his usual speed, a boy reaches school 6 minutes late. What is his usual time to reach the school (in minutes)?

Accepted Answer

Introduction / Context: For a fixed distance, time is inversely proportional to speed. If speed is scaled by a factor, time is scaled by the reciprocal. Use the late time (excess) to back out the usual time. Given Data / Assumptions: New speed = (5/7) * usual speed. Delay = 6 minutes. Concept / Approach: Let T be usual time. New time = T / (5/7) = (7/5)T. Lateness = (7/5)T − T = (2/5)T = 6 ⇒ T = 15 min. Step-by-Step Solution: (2/5)T = 6 ⇒ T = 15. Verification / Alternative check: New time = (7/5)*15 = 21 min, which is 6 min more than the usual 15 min. Why Other Options Are Wrong: 10/12/18 min do not satisfy the proportionality equation. Common Pitfalls: Adding/subtracting speeds instead of using inverse proportionality for time. Final Answer: 15 min

Question 19

A car travels in three stages: • 30 min at 40 km/h, • 45 min at 60 km/h, • 2 h at 70 km/h. What is the car’s average speed over the entire journey (in km/h)?

Accepted Answer

Introduction / Context: Average speed over multiple segments with different speeds equals total distance divided by total time, not the simple average of speeds. Convert each stage to hours, compute distances, sum, then divide by the total time. Given Data / Assumptions: Stage 1: 0.5 h at 40 km/h. Stage 2: 0.75 h at 60 km/h. Stage 3: 2 h at 70 km/h. Concept / Approach: Total distance = Σ(v * t). Total time = 0.5 + 0.75 + 2 = 3.25 h. Average speed = total distance / 3.25. Step-by-Step Solution: Distances: 0.5*40 = 20 km; 0.75*60 = 45 km; 2*70 = 140 km.Total distance = 20 + 45 + 140 = 205 km.Average speed = 205 / 3.25 ≈ 63.08 km/h ≈ 63 km/h. Verification / Alternative check: Weighted average by time yields the same result; simple mean of speeds would be incorrect. Why Other Options Are Wrong: 58/67/71 km/h do not match total distance over total time. Common Pitfalls: Averaging speeds arithmetically without weighting by time intervals. Final Answer: 63 km/h

Question 20

Time and Distance – Train at reduced speed vs own speed: A train running at 8/11 of its normal speed reaches a destination in 5 h 30 min. If the train had run at its full normal speed, how much time would have been saved compared to 5 h 30 min?

Accepted Answer

Introduction / Context: This is a speed-time-distance comparison. When a vehicle runs at a fractional multiple of its normal speed, the time taken scales inversely by the reciprocal fraction. We compare the slower trip with the normal-speed trip to compute the time saved. Given Data / Assumptions: Reduced speed = 8/11 of normal speed. Time at reduced speed = 5 h 30 min = 5.5 h. Distance is the same on both trips. Concept / Approach: For fixed distance D and normal speed v, time at normal speed is T = D / v. At reduced speed v' = (8/11) v, time is T' = D / v' = (11/8) T. Hence T = (8/11) T'. Step-by-Step Solution: T = (8/11) * 5.5 h.Since 5.5 = 11/2, T = (8/11) * (11/2) = 8/2 = 4 h.Time saved = T' − T = 5.5 − 4 = 1.5 h = 3/2 h. Verification / Alternative check: If normal time is 4 h, then at 8/11 speed the required time should be (11/8)*4 = 5.5 h, which exactly matches the given slow trip time. Why Other Options Are Wrong: 2 h and 1 h are over or under the calculated saving; 5/4 h = 1.25 h and 1 h 20 min = 1.333 h do not equal 1.5 h. Common Pitfalls: Applying the 8/11 factor to time directly (should invert), or mixing minutes and hours during subtraction. Final Answer: 3/2 h

Time and Distance Questions