Question 1

Uniform speed scaling A man cycles 12 km in 90 minutes at a uniform speed. How far will he travel in 3 hours at the same speed?

Accepted Answer

Introduction / Context: When speed is uniform, distance is directly proportional to time. Doubling the time doubles the distance; tripling the time triples the distance, and so on. Given Data / Assumptions: Covers 12 km in 90 minutes. Required time = 3 hours = 180 minutes. Speed remains constant throughout. Concept / Approach: Compute speed first, then multiply by the new time. Speed (km/h) = distance / time. Step-by-Step Solution: Time for 12 km = 90 min = 1.5 hSpeed = 12 / 1.5 = 8 km/hDistance in 3 h = 8 * 3 = 24 km Verification / Alternative check: Proportion method: 180 min is twice 90 min, so distance doubles from 12 km to 24 km. Why Other Options Are Wrong: 27 km (needs 3.375 h at 8 km/h), 30 km (needs 3.75 h), 36 km (needs 4.5 h). Common Pitfalls: Mixing minutes and hours without converting consistently. Final Answer: 24 km

Question 2

Speed–time inverse relation Two trains A and B travel from X to Y with speed ratio A:B = 2:7. What is the ratio of their times to cover the same distance?

Accepted Answer

Introduction / Context: For fixed distance, time is inversely proportional to speed. If one vehicle is faster by a factor, it takes proportionally less time. Given Data / Assumptions: Speed ratio A:B = 2:7. Distance X→Y is the same for both. Concept / Approach: If vA:vB = 2:7, then tA:tB = 1/2 : 1/7 = 7 : 2 after scaling to integers. Step-by-Step Solution: Time ratio = inverse of speed ratio = 7 : 2 Verification / Alternative check: Assume any convenient common distance, e.g., 14 units. A takes 14/2 = 7 units of time; B takes 14/7 = 2 units ⇒ 7:2. Why Other Options Are Wrong: Other ratios don’t match the strict inverse of 2:7. Common Pitfalls: Keeping the same 2:7 ratio for time; that’s only valid when distances differ, not here. Final Answer: 7 : 2

Question 3

Average speed over unequal segments A man covers the first 30 km at 6 km/h and the remaining 40 km in 5 hours. What is his average speed for the 70 km journey?

Accepted Answer

Introduction / Context: Average speed over multiple legs equals total distance divided by total time, not the average of individual speeds. When legs have different durations, using total distance / total time ensures correct weighting. Given Data / Assumptions: Leg 1: 30 km at 6 km/h. Leg 2: 40 km in 5 h (so the speed on leg 2 is 8 km/h). Total distance = 70 km. Concept / Approach: Compute time per leg, sum to total time, then divide total distance by total time. Step-by-Step Solution: Time for first 30 km = 30 / 6 = 5 hTime for last 40 km = 5 hTotal time = 5 + 5 = 10 hAverage speed = Total distance / Total time = 70 / 10 = 7 km/h Verification / Alternative check: Weighted-by-time average of speeds: (6 for 5 h and 8 for 5 h) ⇒ (6*5 + 8*5) / (5+5) = 70/10 = 7. Why Other Options Are Wrong: 8 km/h uses the last leg speed; 64/11 ≈ 5.82 and 7.5 km/h mis-handle weighting. Common Pitfalls: Taking the arithmetic mean (6+8)/2 = 7 (which coincidentally equals the correct value only because the times are equal here; in general you must use total distance / total time). Final Answer: 7 km/h

Question 4

Comparing speeds from distance–time change A distance D is covered at speed v1 in time T. If half the distance (D/2) is covered in double the time (2T), what is the ratio v1 : v2 of the two speeds?

Accepted Answer

Introduction / Context: Speed equals distance divided by time. Changing distance and time scales the speed accordingly. Here, the second scenario uses half the distance but takes double the time versus the original benchmark. Given Data / Assumptions: Original: distance D in time T ⇒ v1 = D / T. New: distance D/2 in time 2T ⇒ v2 = (D/2) / (2T). Concept / Approach: Compute v2 relative to v1 by simplifying the fraction. Then form the ratio v1 : v2 and reduce to simplest integers. Step-by-Step Solution: v1 = D / Tv2 = (D/2) / (2T) = D / (4T) = v1 / 4Therefore v1 : v2 = v1 : (v1/4) = 4 : 1 Verification / Alternative check: Choose D = 4 and T = 1. Then v1 = 4, v2 = (2)/(2) = 1 ⇒ 4:1. Why Other Options Are Wrong: 2:1 or 1:2 would occur under different scalings; here both distance and time changed. Common Pitfalls: Dividing 1/2 by 2 and incorrectly getting 1 instead of 1/4 for the speed factor. Final Answer: 4 : 1

Question 5

Late arrival due to reduced speed A train runs at 6/7 of its usual speed and arrives 10 minutes late. What is its usual time for the journey?

Accepted Answer

Introduction / Context: For a fixed distance, time is inversely proportional to speed. If the speed is scaled by a factor, the time is scaled by the reciprocal factor. The difference between new and usual time gives the delay. Given Data / Assumptions: New speed = (6/7) of usual speed. Delay = 10 minutes = 10/60 h. Let usual time be T hours. Concept / Approach: New time T' = T / (6/7) = (7/6)T. Delay = T' − T = (1/6)T equals 10 minutes. Step-by-Step Solution: (1/6)T = 10 minT = 60 min Verification / Alternative check: Usual time 60 min; at 6/7 speed time becomes (7/6)*60 = 70 min, exactly 10 min late. Why Other Options Are Wrong: 25, 15, 35 min cannot produce a 10-min increase when multiplied by 7/6. Common Pitfalls: Multiplying by 6/7 instead of dividing by it when converting time. Final Answer: 60 min

Question 6

Meeting point and speed ratio John starts from A towards B; Vinod from B towards A. Distance AB = 125 km. They meet at a point 75 km from A. What is John’s speed : Vinod’s speed?

Accepted Answer

Introduction / Context: When two travelers start at the same time toward each other and meet, the distances covered by them up to the meeting point are in the same ratio as their speeds. Given Data / Assumptions: AB = 125 km. Meeting point is 75 km from A and 50 km from B. They start simultaneously and travel uniformly. Concept / Approach: Speed ratio = distance ratio to meeting point. Step-by-Step Solution: Distance by John = 75 kmDistance by Vinod = 50 kmSpeed ratio (John : Vinod) = 75 : 50 = 3 : 2 Verification / Alternative check: If John’s speed is 30 and Vinod’s 20 (3:2), they meet after t such that 30t + 20t = 125 ⇒ t = 2.5 h; John covers 75, Vinod 50. Why Other Options Are Wrong: Other ratios don’t fit 75:50 once simplified. Common Pitfalls: Accidentally using remaining distances after meeting; here the meeting-point distances are the ones to compare. Final Answer: 3 : 2

Question 7

Average speed for a round trip (different speeds) A person goes from P to Q at 5 km/h and returns from Q to P at 3 km/h. What is the average speed for the whole journey?

Accepted Answer

Introduction / Context: For a round trip with different speeds in the two directions over the same distance, the average speed is the harmonic mean: 2uv / (u + v). It is not the arithmetic mean unless the times match, which they usually do not when speeds differ. Given Data / Assumptions: Speed P→Q = 5 km/h. Speed Q→P = 3 km/h. Distances each way are equal. Concept / Approach: Average speed over total journey = total distance / total time. For equal distances, this simplifies to 2uv/(u+v). Step-by-Step Solution: Average speed = 2 * 5 * 3 / (5 + 3) = 30 / 8 = 3.75 km/h Verification / Alternative check: Assume each leg is 24 km. Times: 24/5 = 4.8 h and 24/3 = 8 h. Total distance 48 km, total time 12.8 h ⇒ 48/12.8 = 3.75 km/h. Why Other Options Are Wrong: 4.5 and 4.25 are arithmetic-mean style answers; 4.0 is still too high. Common Pitfalls: Taking the simple average (5 + 3)/2 = 4, which is incorrect for average speed over equal distances. Final Answer: 3.75 km/h

Question 8

Distance fixed; required speed for shorter time A train runs at 96 km/h and covers a trip in 1 h 40 min (100 min). To reduce the journey time to 1 h 20 min (80 min), what average speed is required?

Accepted Answer

Introduction / Context: For the same route, the distance remains constant. If you shorten the time, the speed must increase in the ratio distance/time. Compute the original distance and then divide by the new time to get the required speed. Given Data / Assumptions: Original speed = 96 km/h. Original time = 100 min = 5/3 h. Target time = 80 min = 4/3 h. Concept / Approach: Distance D = v * t. Required speed v' = D / t'. Step-by-Step Solution: D = 96 * (5/3) = 160 kmv_required = 160 / (4/3) = 160 * 3 / 4 = 120 km/h Verification / Alternative check: At 120 km/h, time = 160 / 120 = 4/3 h = 80 min. Why Other Options Are Wrong: 90 is slower than before; 144 or 150 overshoot the need. Common Pitfalls: Mixing minutes and hours; always convert times to the same unit. Final Answer: 120 km/h

Question 9

Successive percent changes on average speed A car's average speed is 75 km/h. It is first decreased by 40%, then increased by 50%. What is the new average speed?

Accepted Answer

Introduction / Context: Successive percentage changes multiply, not add. After reducing a quantity by p% and then increasing by q%, the net factor equals (1 − p/100) * (1 + q/100) applied to the original value. Given Data / Assumptions: Initial speed = 75 km/h. Decrease by 40% ⇒ multiply by 0.60. Then increase by 50% ⇒ multiply by 1.50. Concept / Approach: Compute sequentially: v1 = 75 * 0.60; then v2 = v1 * 1.50. Step-by-Step Solution: After −40%: 75 * 0.60 = 45 km/hAfter +50%: 45 * 1.50 = 67.5 km/h Verification / Alternative check: Net factor = 0.60 * 1.50 = 0.90 ⇒ 75 * 0.90 = 67.5 km/h. Why Other Options Are Wrong: 60 km/h and 90 km/h incorrectly treat changes as additive or apply them in the wrong order; 60.5 is arbitrary. Common Pitfalls: Adding −40% and +50% to get +10% overall; percentage changes compound multiplicatively. Final Answer: 67.5 km/h

Question 10

Usual time from speed drop and delay A train runs at 5/6 of its usual speed and is late by 10 minutes. What is its usual time for the trip?

Accepted Answer

Introduction / Context: With distance fixed, time is inversely proportional to speed. Reducing speed by a factor 5/6 increases time by 6/5. The difference between new and usual times is the observed delay. Given Data / Assumptions: New speed = (5/6) of usual. Delay = 10 minutes. Let usual time be T minutes. Concept / Approach: New time = (6/5)T. Delay = (6/5)T − T = (1/5)T = 10. Step-by-Step Solution: (1/5)T = 10 ⇒ T = 50 minutes Verification / Alternative check: At usual time 50 min, reduced speed time = (6/5)*50 = 60 min, which is 10 minutes late. Why Other Options Are Wrong: 40, 45, 55 min won’t produce exactly a 10-minute increase under a 6/5 factor. Common Pitfalls: Multiplying the time by 5/6 instead of dividing by 5/6 when speed is reduced. Final Answer: 50 min

Question 11

Wheels and revolutions A car moves at 47.52 km/h. Its wheel radius is 21 cm. Approximately how many revolutions per minute does the wheel make?

Accepted Answer

Introduction / Context: Revolutions per minute (rpm) equals the linear distance traveled per minute divided by the wheel circumference. Convert speed units to meters per minute; compute circumference from radius. Given Data / Assumptions: Speed = 47.52 km/h. Radius r = 21 cm = 0.21 m. Use circumference C = 2 * pi * r; with pi ≈ 22/7 for neat arithmetic. Concept / Approach: rpm = (linear speed in m/min) / C. Step-by-Step Solution: Linear speed = 47.52 * 1000 / 60 = 792 m/minCircumference ≈ 2 * (22/7) * 0.21 = 1.32 mrpm ≈ 792 / 1.32 = 600 Verification / Alternative check: Using pi ≈ 3.1416 gives C ≈ 1.319, rpm ≈ 600.2, still rounds to 600 rpm. Why Other Options Are Wrong: 200, 250, 300 are far too small given the circumference and speed. Common Pitfalls: Forgetting to convert km/h to m/min or using diameter instead of radius in the circumference formula. Final Answer: 600 rpm

Question 12

Square field via diagonal crossing A man walking at 3 km/h crosses a square field along the diagonal in 2 minutes. What is the area of the field (approximately, in acres)?

Accepted Answer

Introduction / Context: When crossing a square field along its diagonal at uniform speed, the walking distance equals the diagonal length. From the diagonal, we can find the side and thus the area. Convert units consistently and, if asked in acres, convert from square meters (or hectares) at the end. Given Data / Assumptions: Speed = 3 km/h. Time = 2 minutes = 2/60 h. 1 acre = 4046.856 m^2. Concept / Approach: Distance = speed * time. For a square with side s and diagonal d, d = s * sqrt(2). Then area = s^2. Convert meters to acres. Step-by-Step Solution: Distance walked (diagonal) = 3 * (2/60) = 0.1 km = 100 mSide s = d / sqrt(2) = 100 / 1.4142 ≈ 70.71 mArea = s^2 ≈ (70.71)^2 ≈ 5000 m^2In acres = 5000 / 4046.856 ≈ 1.235 acres ≈ 1.24 acres Verification / Alternative check: In hectares: 5000 m^2 = 0.5 ha. 0.5 ha equals ≈ 1.2355 acres, confirming the conversion. Why Other Options Are Wrong: 0.50 and 0.75 acres are too small; 2.00 acres is too large for a 100 m diagonal. Common Pitfalls: Using the side instead of the diagonal as the 100 m distance, or failing to convert time to hours before multiplying by speed. Final Answer: ≈ 1.24 acres

Question 13

Comparing speeds when half distance takes double time A distance is covered at speed v1. If half that distance is covered in double the original time, what is the ratio v1 : v2 of the two speeds?

Accepted Answer

Introduction / Context: This revisits proportional reasoning with distance and time. When the new scenario covers half the distance in twice the time, the speed reduces by a factor of 4 relative to the original. Given Data / Assumptions: Original speed v1 = D / T. New speed v2 = (D/2) / (2T). Concept / Approach: Simplify v2 to express it as a multiple of v1 and form the ratio. Step-by-Step Solution: v2 = (D/2) / (2T) = D / (4T) = v1 / 4So v1 : v2 = 4 : 1 Verification / Alternative check: Plug D = 4, T = 1 ⇒ v1 = 4. New case: 2 units in 2 h ⇒ v2 = 1. Ratio 4:1. Why Other Options Are Wrong: 1:2 or 2:1 confuse which quantity changed and by what factor. Common Pitfalls: Thinking “half distance, double time ⇒ half/double = quarter” but then inverting incorrectly. Final Answer: 4 : 1

Question 14

Relative speeds chained (A twice B, B thrice C) A is twice as fast as B, and B is thrice as fast as C. If C covers a journey in 42 minutes, how long will A take for the same journey?

Accepted Answer

Introduction / Context: Composite speed relations can be chained by multiplication. If A is 2 times B and B is 3 times C, then A is 6 times C. Time for the same distance scales inversely with speed. Given Data / Assumptions: A = 2B; B = 3C ⇒ A = 6C (in speed). Time(C) = 42 min. Concept / Approach: Time(A) = Time(C) / (A/C) since time ∝ 1/speed. Step-by-Step Solution: Speed ratio A:C = 6:1Time ratio A:C = 1:6Time(A) = 42 / 6 = 7 min Verification / Alternative check: If C's speed is 1 unit, A's is 6 units; for a fixed distance, time scales reciprocally. Why Other Options Are Wrong: 14 and 28 are larger than the reciprocal factor implies; 63 min would be slower than C. Common Pitfalls: Adding factors (2 + 3) instead of multiplying (2 * 3) for chained relations. Final Answer: 7 min

Question 15

Early vs late arrival; find distance Ram travels a certain distance at 3 km/h and arrives 15 min late. At 4 km/h he arrives 15 min early. What is the distance?

Accepted Answer

Introduction / Context: When two different speeds produce equal and opposite schedule deviations, set up equations with the same scheduled time and eliminate it to solve for distance directly. Given Data / Assumptions: At 3 km/h: arrival is 15 min late. At 4 km/h: arrival is 15 min early. Let scheduled time be T hours and distance be D km. Concept / Approach: Form two equations: D/3 = T + 0.25 and D/4 = T − 0.25. Subtract to remove T and solve for D. Step-by-Step Solution: D/3 − D/4 = 0.25 + 0.25 = 0.5D * (1/12) = 0.5 ⇒ D = 6 km Verification / Alternative check: Compute T: From D/4 = T − 0.25 ⇒ 6/4 = T − 0.25 ⇒ T = 1.75 h. Check D/3 = 2 h = T + 0.25. Why Other Options Are Wrong: 7.2, 4.5, 12 contradict the equal 15-min deviations with the given speeds. Common Pitfalls: Using minutes in one equation and hours in the other; always convert consistently. Final Answer: 6 km

Question 16

Pursuit problem with start delay A thief steals a car at 1:30 p.m. driving 45 km/h. The theft is discovered at 2:00 p.m., and the owner starts at 50 km/h. At what time will the owner catch the thief?

Accepted Answer

Introduction / Context: Classic chase problems compare the initial head start to the relative speed. The pursuer closes the gap at the difference of speeds, provided the pursuer is faster. Given Data / Assumptions: Thief speed = 45 km/h, starts at 1:30 p.m. Owner speed = 50 km/h, starts at 2:00 p.m. Uniform speeds; straight route. Concept / Approach: Head start distance = thief speed * start gap (0.5 h). Catch-up time after the owner starts equals head start / relative speed. Step-by-Step Solution: Head start = 45 * 0.5 = 22.5 kmRelative speed = 50 − 45 = 5 km/hCatch-up time = 22.5 / 5 = 4.5 hCatch-up clock time = 2:00 p.m. + 4.5 h = 6:30 p.m. Verification / Alternative check: Distances at 6:30 p.m.: thief drives 5 h from 1:30 ⇒ 225 km; owner drives 4.5 h from 2:00 ⇒ 225 km. Why Other Options Are Wrong: Earlier times ignore the 22.5 km head start or use the wrong relative speed. Common Pitfalls: Adding speeds (for opposite directions) or forgetting the 30-minute start delay. Final Answer: 6:30 p.m.

Question 17

Opposite directions; distance difference at meeting Two trains start simultaneously from Aligarh and Delhi toward each other at 16 km/h and 21 km/h. When they meet, one has traveled 60 km more than the other. What is the distance between the station…

Accepted Answer

Introduction / Context: For two objects moving toward each other, the distances each has covered at meeting are proportional to their speeds. If the faster covers 60 km more, the time until meeting can be deduced from the speed difference. Given Data / Assumptions: Speeds: 16 km/h and 21 km/h. Difference in distances at meeting: 60 km (the faster train covers more). Concept / Approach: Let t be the time to meet. Distance difference = (21 − 16) * t = 5t = 60 ⇒ t = 12 h. Total distance = (16 + 21) * t. Step-by-Step Solution: t = 60 / 5 = 12 hTotal distance = (16 + 21) * 12 = 37 * 12 = 444 km Verification / Alternative check: Individual distances: 16*12 = 192 km and 21*12 = 252 km; difference 60 km; sum 444 km. Why Other Options Are Wrong: 440 and 445 are near misses; 450 ignores exact arithmetic. Common Pitfalls: Using the sum of speeds to compute the difference in distances; the difference uses (v2 − v1). Final Answer: 444 km

Question 18

Speed from time-change over fixed distance A car covers 840 km at uniform speed. If its speed were 10 km/h faster, the trip would take 2 hours less. What was the original speed?

Accepted Answer

Introduction / Context: When speed increases by a fixed amount for the same distance, the time decreases in a way that leads to a rational equation. Solving for the original speed yields a quadratic that typically factors neatly in standard test problems. Given Data / Assumptions: Distance D = 840 km. Original speed v; new speed v + 10. Time saved = 2 h. Concept / Approach: Equation: 840/v − 840/(v + 10) = 2. Clear denominators to solve for v and take the positive root. Step-by-Step Solution: 840/v − 840/(v + 10) = 2840[(v + 10) − v] / [v(v + 10)] = 2840 * 10 / [v(v + 10)] = 2 ⇒ v(v + 10) = 4200v^2 + 10v − 4200 = 0 ⇒ (v + 70)(v − 60) = 0 ⇒ v = 60 (valid) Verification / Alternative check: Times: 840/60 = 14 h; 840/70 = 12 h; difference 2 h. Why Other Options Are Wrong: 45, 50, 75 do not satisfy the time-difference equation. Common Pitfalls: Setting up 840/(v + 10) − 840/v = 2 (sign reversed) or canceling incorrectly. Final Answer: 60 km/h

Question 19

Chase with start lead; find cat’s speed A dog starts chasing a cat 2 hours after the cat starts. The dog runs at 30 km/h and catches the cat in 2 hours. What is the cat’s speed?

Accepted Answer

Introduction / Context: In chase problems with a start delay, the lead gained equals lead time multiplied by the evader’s speed. During the chase, the pursuer closes the lead at the relative speed (difference of speeds). Given Data / Assumptions: Dog speed = 30 km/h. Dog starts 2 h after the cat. Dog catches the cat in 2 h of chasing. Let cat speed be s km/h. Concept / Approach: Lead distance at the start of the chase = s * 2. Distance closed during chase = (30 − s) * 2. Set these equal (catch-up condition) and solve for s. Step-by-Step Solution: 2s = 2(30 − s)2s = 60 − 2s ⇒ 4s = 60 ⇒ s = 15 km/h Verification / Alternative check: Positions after 2 h of chase: cat travels 2 more hours at 15 → 30 km; dog travels 2 h at 30 → 60 km, exactly closing the 30 km head start. Why Other Options Are Wrong: 10 or 20 yield unequal lead and catch-up distances; “Can't be determined” is incorrect because the equations are solvable uniquely. Common Pitfalls: Using sum of speeds instead of difference; the sum applies when moving toward each other from different starting points without a chase. Final Answer: 15 km/h

Question 20

A 350 m long train is moving at 20 km/h. A man is walking toward the train from the opposite direction at 1 km/h (treat the man as a point). In how many seconds will the train completely cross the man?

Accepted Answer

Introduction / Context: When two bodies move toward each other, their relative speed is the sum of individual speeds. For a point-like man and a finite-length train, the crossing time equals (train length) / (relative speed). Given Data / Assumptions: Train length = 350 m. Train speed = 20 km/h. Man speed (opposite) = 1 km/h. 1 km/h = 1000/3600 m/s. Concept / Approach: Compute relative speed in m/s, then use time = distance / speed. Only the train’s length matters because the man is a point object. Step-by-Step Solution: Relative speed = (20 + 1) km/h = 21 km/h.Convert: 21 km/h = 21 * 1000 / 3600 = 5.833... m/s.Time = 350 / 5.833... = 60 s. Verification / Alternative check: Multiply 60 s by 5.833... m/s ≈ 350 m, matching the train length. Why Other Options Are Wrong: 27/35/45 s correspond to higher relative speeds that are not possible with the given velocities. Common Pitfalls: Forgetting to add speeds for opposite motion, or not converting km/h to m/s before dividing. Final Answer: 60 sec

Time and Distance Questions