Question 1

Laxman must cover 6 km in 45 minutes (0.75 h). He covers the first half of the distance (3 km) in two-thirds of the total time (30 minutes). What speed (in km/h) is required to cover the remaining 3 km in the remaining 15 minutes?

Accepted Answer

Introduction / Context: This question checks understanding of time partitioning and speed requirements when part of a journey uses a disproportionate share of the total time. The key is to compute remaining time and distance, then apply speed = distance / time. Given Data / Assumptions: Total distance = 6 km; total time = 45 minutes. First half (3 km) takes 2/3 of total time = 30 minutes. Remaining half = 3 km; remaining time = 15 minutes. Concept / Approach: For the leftover segment, convert minutes to hours and compute the constant speed needed to finish on time. Step-by-Step Solution: Remaining distance = 3 kmRemaining time = 15 min = 0.25 hRequired speed = 3 / 0.25 = 12 km/h Verification / Alternative check: Check total: First 3 km use 30 minutes; last 3 km at 12 km/h uses 15 minutes; total 45 minutes as required. Why Other Options Are Wrong: 16 km/h and 8 km/h: Over- or under-estimate the needed pace for 3 km in 15 minutes. 3 km/h and 10 km/h: Do not meet the 15-minute remainder constraint. Common Pitfalls: Forgetting to convert minutes to hours, or misallocating the remaining time. Final Answer: 12 km/h

Question 2

A train covers a certain distance in 50 minutes at an average speed of 48 km/h. To reduce the total journey time to 40 minutes for the same distance, at what constant speed must the train run?

Accepted Answer

Introduction / Context: Speed, distance, and time are related by distance = speed * time. For a fixed distance, if time decreases, speed must increase proportionally to keep distance constant. Given Data / Assumptions: Initial speed v1 = 48 km/h; time t1 = 50 minutes = 5/6 h. Target time t2 = 40 minutes = 2/3 h. Distance remains the same. Concept / Approach: Compute the distance from the first leg, then divide by the new time to get the required speed. Step-by-Step Solution: Distance D = 48 * (5/6) = 40 kmRequired speed v2 = D / (2/3) = 40 / (2/3) = 60 km/h Verification / Alternative check: Proportion method: v2 / v1 = t1 / t2 = (50) / (40) = 1.25, hence v2 = 48 * 1.25 = 60 km/h. Why Other Options Are Wrong: All other speeds fail to achieve 40 km in 40 minutes. Common Pitfalls: Not converting minutes to hours; mixing up proportionality (speed inversely proportional to time for fixed distance). Final Answer: 60 km/h

Question 3

A bus runs at 54 km/h while moving and averages 45 km/h overall when stoppages are included. How many minutes per hour does the bus stop?

Accepted Answer

Introduction / Context: Overall average speed with stoppages equals distance covered in one whole hour divided by one hour. If the running speed is higher, the bus must have stopped for some portion of the hour. Given Data / Assumptions: Running speed = 54 km/h (no stops). Overall speed including stops = 45 km/h. Assume constant running speed whenever moving. Concept / Approach: Let moving time per hour be t hours. Then distance in 1 clock-hour is 54*t. Since average is 45 km/h, we have 54*t = 45, giving t. Stoppage time = 60 minutes * (1 - t). Step-by-Step Solution: 54 * t = 45 → t = 45/54 = 5/6 hourStopped fraction = 1 - 5/6 = 1/6 hour = 10 minutes Verification / Alternative check: Distance in 50 minutes at 54 km/h is 45 km, matching the overall hourly average. Why Other Options Are Wrong: 9 or 12 minutes are close distractors but do not satisfy 54*t = 45; 20 and 8 minutes deviate further. Common Pitfalls: Confusing average of speeds with time-weighted average; the bus’s average depends on time moving vs. stopped. Final Answer: 10 minutes

Question 4

Two walkers start together for the same destination. One walks at 3.75 km/h and the other at 3 km/h. If the faster walker arrives 30 minutes earlier, what is the distance to the destination?

Accepted Answer

Introduction / Context: When two travelers cover the same distance at different constant speeds, the difference in their arrival times equals the distance times the difference of reciprocals of their speeds. Given Data / Assumptions: Speeds: v1 = 3.75 km/h, v2 = 3 km/h. Time difference = 0.5 hour. Both start simultaneously and go directly to the same point. Concept / Approach: Let distance be D. Then arrival time difference: D/3 - D/3.75 = 0.5. Solve for D. Step-by-Step Solution: D/3 - D/3.75 = 0.5D*(1/3 - 1/3.75) = 0.51/3.75 = 0.266666..., difference = 1/3 - 0.266666... = 0.066666...D = 0.5 / 0.066666... = 7.5 km Verification / Alternative check: Times: 7.5/3 = 2.5 h; 7.5/3.75 = 2 h; difference = 0.5 h as stated. Why Other Options Are Wrong: They do not satisfy the exact time-difference equation for the given pair of speeds. Common Pitfalls: Using average of speeds or average of times; the correct relation uses reciprocals of speeds for fixed distance. Final Answer: 7.5 km

Question 5

A car covers four successive stretches of 3 km each at speeds of 10 km/h, 20 km/h, 30 km/h, and 60 km/h, respectively. What is the average speed over the entire 12 km journey?

Accepted Answer

Introduction / Context: When equal distances are traveled at different speeds, the average speed is found from total distance divided by total time—effectively a harmonic-mean type across the segments. Given Data / Assumptions: Four segments, each 3 km (total 12 km). Speeds: 10, 20, 30, 60 km/h. Concept / Approach: Total time equals the sum of segment times d/vi. Average speed = total distance / total time. Step-by-Step Solution: Times: 3/10 + 3/20 + 3/30 + 3/60 hours= 0.3 + 0.15 + 0.1 + 0.05 = 0.6 hoursAverage speed = 12 km / 0.6 h = 20 km/h Verification / Alternative check: Harmonic-mean style: V = 4 / (1/10 + 1/20 + 1/30 + 1/60) = 20. Why Other Options Are Wrong: Arithmetic means (like 30 km/h) do not account for the longer time spent at slower speeds; 25 or 24 km/h are near-miss distractors. Common Pitfalls: Averaging the speeds directly instead of using time-weighted averaging via total time. Final Answer: 20 km/h

Question 6

A car completes a journey in 8 hours, covering half the distance at 40 km/h and the remaining half at 60 km/h. What is the total length of the journey (in km)?

Accepted Answer

Introduction / Context: When half the distance is covered at one speed and half at another, the overall average speed for the full trip is the harmonic mean of the two speeds (not the arithmetic mean). Given Data / Assumptions: First half at 40 km/h, second half at 60 km/h. Total time for the whole distance = 8 hours. Concept / Approach: For equal halves by distance, average speed V = 2 / (1/40 + 1/60) = 48 km/h. Distance D = V * total time. Step-by-Step Solution: Harmonic-mean style average: V = 2 / (1/40 + 1/60)Compute: 1/40 + 1/60 = (3 + 2) / 120 = 5/120 = 1/24V = 2 / (1/24) = 48 km/hTotal distance D = 48 * 8 = 384 km Verification / Alternative check: Let distance be 2x. Times: x/40 + x/60 = 8 → x(1/40 + 1/60) = 8 → x/24 = 8 → x = 192 → D = 384 km. Why Other Options Are Wrong: Values like 400 or 420 km correspond to using an incorrect average (e.g., arithmetic mean) or miscomputing time fractions. Common Pitfalls: Averaging speeds arithmetically for equal distance halves; always use total-distance over total-time logic. Final Answer: 384 km

Question 7

Rahim travels a certain distance in 14 hours 40 minutes. He covers half the distance by train at 60 km/h and the other half by road at 50 km/h. What is the total distance traveled (in km)?

Accepted Answer

Introduction / Context: Splitting a journey into equal halves by distance but different speeds requires summing the two leg times and equating to the given total time to solve for the half-distance. Given Data / Assumptions: Total time T = 14 h 40 min = 14.666... h. First half at 60 km/h, second half at 50 km/h. Let half-distance be x km. Concept / Approach: Total time = x/60 + x/50. Solve for x, then total distance = 2x. Step-by-Step Solution: x/60 + x/50 = 14.666... hCommon denominator 300: (5x + 6x)/300 = 14.666... → 11x/300 = 44/3x = (44/3) * (300/11) = 400 kmTotal distance = 2x = 800 km Verification / Alternative check: Time check: 400/60 = 6.666... h; 400/50 = 8 h; sum = 14.666... h as given. Why Other Options Are Wrong: They result from arithmetic slips or misinterpreting “half the distance” vs. “half the time.” Common Pitfalls: Confusing equal halves by distance with equal halves by time; forgetting to convert minutes to fractional hours. Final Answer: 800 km

Question 8

A person covers 2/15 of a journey by rail, 9/20 by tonga, and the remaining 10 km on foot. What is the total length of the journey?

Accepted Answer

Introduction / Context: Journeys expressed in fractional parts can be reconstructed by summing known fractions and tying the leftover fraction to a given absolute remainder. Given Data / Assumptions: By rail = 2/15 of total D. By tonga = 9/20 of D. On foot = 10 km = remaining fraction of D. Concept / Approach: Compute remaining fraction: 1 - (2/15 + 9/20). Set that fraction equal to 10/D and solve for D. Step-by-Step Solution: 2/15 + 9/20 = 8/60 + 27/60 = 35/60 = 7/12Remaining fraction = 1 - 7/12 = 5/125/12 of D = 10 km → D = 10 * (12/5) = 24 km Verification / Alternative check: Check parts: Rail = 2/15 * 24 = 3.2 km; Tonga = 9/20 * 24 = 10.8 km; Remaining = 10 km; total = 24 km. Why Other Options Are Wrong: They result from arithmetic errors in fraction addition or inversion. Common Pitfalls: Failing to use common denominators when adding fractions; forgetting to equate the leftover fraction to 10/D. Final Answer: 24 km

Question 9

Shard covers two-thirds of a journey at 4 km/h and the remaining one-third at 5 km/h. If his total time is 42 minutes, what is the total distance?

Accepted Answer

Introduction / Context: When segments of a trip are specified as fractions of the total distance with different speeds, compute total time as a sum of fractional-distance time contributions to solve for the total distance. Given Data / Assumptions: (2/3) of distance at 4 km/h. (1/3) of distance at 5 km/h. Total time = 42 minutes = 0.7 hour. Concept / Approach: Let total distance be D. Then time = (2D/3)/4 + (D/3)/5. Set equal to 0.7 hour and solve for D. Step-by-Step Solution: (2D/3)/4 + (D/3)/5 = 0.7D*(1/6 + 1/15) = 0.71/6 + 1/15 = (5 + 2)/30 = 7/30D = 0.7 * (30/7) = 3 km Verification / Alternative check: Time check: (2 km at 4) = 0.5 h; (1 km at 5) = 0.2 h; total = 0.7 h (42 minutes). Why Other Options Are Wrong: They correspond to misadding the fractional time terms or misreading minutes vs hours. Common Pitfalls: Forgetting to convert 42 minutes to hours; mixing up fractional distance weights. Final Answer: 3 km

Question 10

A boy goes from his village to school at 3 km/h and returns at 2 km/h. If the total time for the round trip is 5 hours, what is the one-way distance between the village and the school?

Accepted Answer

Introduction / Context: For out-and-back journeys with different speeds over the same distance, total time is the sum of the two leg times using the same one-way distance. Given Data / Assumptions: Forward speed = 3 km/h; return speed = 2 km/h. Total time (to and fro) = 5 h. Let one-way distance be D km. Concept / Approach: Total time = D/3 + D/2. Solve for D. Step-by-Step Solution: D/3 + D/2 = 5D*(1/3 + 1/2) = 5 → D*(5/6) = 5D = 5 * (6/5) = 6 km Verification / Alternative check: Times: 6/3 = 2 h out; 6/2 = 3 h back; total = 5 h as given. Why Other Options Are Wrong: They do not satisfy the exact time equation with the stated speeds. Common Pitfalls: Using average speed naively (e.g., arithmetic mean) instead of computing time per leg and summing. Final Answer: 6 km

Question 11

A certain distance is covered in 2 hours 45 minutes at 4 km/h. How much time (in minutes and seconds) will be needed to cover the same distance at 16.5 km/h?

Accepted Answer

Introduction / Context: The distance remains identical; thus time scales inversely with speed. First compute the distance using the initial speed and time, then divide by the new speed to find the new time. Given Data / Assumptions: Speed1 = 4 km/h; Time1 = 2 h 45 min = 2.75 h. Speed2 = 16.5 km/h. Concept / Approach: Distance D = v1 * t1. New time t2 = D / v2. Step-by-Step Solution: D = 4 * 2.75 = 11 kmt2 = 11 / 16.5 h = 2/3 hConvert 2/3 h = 40 minutes Verification / Alternative check: Proportion: t2 / t1 = v1 / v2 = 4 / 16.5 ≈ 0.2424; 0.2424 * 165 minutes ≈ 40 minutes. Why Other Options Are Wrong: 41 min 15 sec is a near-miss; 45 or 90 minutes contradict the inverse-speed relationship. Common Pitfalls: Not converting hours/minutes consistently; rounding before the final step. Final Answer: 40 min

Question 12

A person takes 4 hours to cover 15 km at constant speed. At the same speed, how much time will be required to cover 63 km (express your answer in hours and minutes)?

Accepted Answer

Introduction / Context: Uniform speed implies distance is proportional to time. You can compute the speed first, or scale time by the distance ratio. Given Data / Assumptions: 15 km in 4 h → speed = 3.75 km/h. Target distance = 63 km. Concept / Approach: Time = distance / speed; or scale 4 h by the factor 63/15. Step-by-Step Solution: Speed v = 15 / 4 = 3.75 km/hTime t = 63 / 3.75 = 16.8 hoursConvert 0.8 h to minutes: 0.8 * 60 = 48 min → 16 h 48 min Verification / Alternative check: Proportion method: 4 h * (63/15) = 4 * 4.2 = 16.8 h = 16 h 48 min. Why Other Options Are Wrong: They correspond to other multipliers or rounding errors. Common Pitfalls: Misconverting decimal hours to minutes; using 3.8 instead of 3.75 by mistake. Final Answer: 16 h 48 min

Question 13

A train moves at 92.4 km/h. How many metres will it cover in 10 minutes?

Accepted Answer

Introduction / Context: Convert the speed from km/h to km per minute, then multiply by 10 minutes and convert to meters. Given Data / Assumptions: Speed = 92.4 km/h. Time = 10 minutes = 1/6 hour. Concept / Approach: Distance = speed * time. Convert final result to meters (1 km = 1000 m). Step-by-Step Solution: Distance = 92.4 * (1/6) km = 15.4 kmConvert to meters: 15.4 * 1000 = 15400 m Verification / Alternative check: Per-minute rate: 92.4/60 = 1.54 km/min; in 10 minutes → 15.4 km → 15400 m. Why Other Options Are Wrong: 1540 m or 154 m are off by factors of 10 or 100; 15.4 m is absurdly small for a train in 10 minutes. Common Pitfalls: Unit mistakes (minutes vs hours, km vs m) leading to order-of-magnitude errors. Final Answer: 15400 m

Question 14

If a man covers 10.2 km in 3 hours at constant speed, how much distance will he cover in 5 hours at the same speed?

Accepted Answer

Introduction / Context: Uniform-speed scaling: distance is directly proportional to time when speed is fixed. Compute speed first, then scale to the new time. Given Data / Assumptions: Distance = 10.2 km in 3 h → speed = 10.2 / 3 = 3.4 km/h. Time required = 5 h. Concept / Approach: Distance in 5 h = speed * time = 3.4 * 5. Step-by-Step Solution: v = 10.2 / 3 = 3.4 km/hD = 3.4 * 5 = 17 km Verification / Alternative check: Proportional scaling: 10.2 * (5/3) = 17 km. Why Other Options Are Wrong: They reflect alternative (incorrect) multipliers or rounding slips. Common Pitfalls: Arithmetic errors when dividing decimals or scaling by 5/3. Final Answer: 17 km

Question 15

A man crosses a straight 600 m street in 5 minutes at constant pace. What is his speed in km/h?

Accepted Answer

Introduction / Context: Convert a meters-per-minute rate to km/h by appropriate unit conversions (m → km, min → h). Given Data / Assumptions: Distance = 600 m. Time = 5 minutes. Concept / Approach: First compute m/s or m/min, then convert to km/h. 1 m/s equals 3.6 km/h. Step-by-Step Solution: Speed = 600 m / 5 min = 120 m/minConvert to m/s: 120 / 60 = 2 m/sConvert to km/h: 2 * 3.6 = 7.2 km/h Verification / Alternative check: Direct conversion: 0.6 km in 1/12 h → 0.6 * 12 = 7.2 km/h. Why Other Options Are Wrong: They reflect wrong conversion factors between m/s and km/h, or minutes-to-hours errors. Common Pitfalls: Using 3.0 instead of 3.6 for the m/s to km/h factor; forgetting to divide by 60 to get m/s. Final Answer: 7.2 km/h

Question 16

Towns A and B are connected. Mr. Faruqui cycles from A to B at 16 km/h and then returns from B to A on foot at 9 km/h, covering the same distance each way. What is his average speed for the entire round trip?

Accepted Answer

Introduction / Context: For equal distances out and back at speeds v1 and v2, the average speed is the harmonic mean: V = 2*v1*v2 / (v1 + v2). This arises from total distance divided by total time. Given Data / Assumptions: v1 = 16 km/h (A to B), v2 = 9 km/h (B to A). Distances on both legs are equal. Concept / Approach: Apply the harmonic-mean formula directly for two equal-distance legs. Step-by-Step Solution: V = 2 * 16 * 9 / (16 + 9)= 288 / 25 = 11.52 km/h Verification / Alternative check: Let one-way distance be D: total time = D/16 + D/9 = (25D/144). Average = total distance (2D) / time = 2D / (25D/144) = 288/25. Why Other Options Are Wrong: They contradict the exact harmonic mean for v1 = 16 and v2 = 9. Common Pitfalls: Taking the arithmetic mean (12.5) instead of using total-time logic for equal distances. Final Answer: 11.52 km/h

Question 17

Convert a speed of 22.5 m/s to km/h.

Accepted Answer

Introduction / Context: Converting metres per second to kilometres per hour uses the constant factor 3.6 because 1 m/s = 3.6 km/h. Given Data / Assumptions: Speed = 22.5 m/s. Concept / Approach: Multiply by 3.6 to obtain km/h. Step-by-Step Solution: 22.5 * 3.6 = 81.0Therefore, 22.5 m/s = 81 km/h Verification / Alternative check: Breakdown: 22.5 * (36/10) = (22.5*36)/10 = 810/10 = 81. Why Other Options Are Wrong: They misuse the conversion factor or include random distractors. Common Pitfalls: Using 3.0 or 3.2 instead of the correct 3.6 factor. Final Answer: 81 km/h

Question 18

Convert a speed of 55 m/s to km/h.

Accepted Answer

Introduction / Context: Use the standard conversion 1 m/s = 3.6 km/h, multiplying metres-per-second by 3.6 to obtain km/h. Given Data / Assumptions: Speed = 55 m/s. Concept / Approach: Multiply 55 by 3.6. Step-by-Step Solution: 55 * 3.6 = 198Hence, 55 m/s = 198 km/h Verification / Alternative check: Compute as 55 * (18/5) = 55 * 3.6 = 198. Why Other Options Are Wrong: They ignore the correct conversion factor; 275 km/h would imply 5.0 factor, which is incorrect. Common Pitfalls: Using approximate or incorrect multipliers when converting units. Final Answer: 198 km/h

Question 19

Convert a speed of 30.6 km/h to m/s.

Accepted Answer

Introduction / Context: The conversion between km/h and m/s uses the factor 1 km/h = 1000/3600 m/s = 5/18 m/s. Equivalently, divide km/h by 3.6 to get m/s. Given Data / Assumptions: Speed = 30.6 km/h. Concept / Approach: m/s = (km/h) / 3.6. Step-by-Step Solution: 30.6 / 3.6 = 8.5Therefore, 30.6 km/h = 8.5 m/s Verification / Alternative check: Using 5/18: 30.6 * (5/18) = 153/18 = 8.5 m/s. Why Other Options Are Wrong: Large values like 110.16 m/s are off by a factor of 12.96; small values like 1.7 m/s ignore proper conversion. Common Pitfalls: Multiplying by 3.6 instead of dividing when converting from km/h to m/s. Final Answer: 8.5 m/s

Question 20

A speed of 36 km/hr. is the same as?

Accepted Answer

Speed in m/sec = speed in km/hr x (5 / 18 ) Speed in m/sec = ( 36 x 5/18 ) m/sec = 10 m/sec

Time and Distance Questions