Question 1

A traveler covers half of the journey at 40 km/h, one-third at 60 km/h, and the remaining one-sixth at 30 km/h. What is the average speed for the entire journey?

Accepted Answer

Introduction / Context: Average speed over non-uniform speeds must be computed as total distance divided by total time. When distances are fractional parts of the whole, computing the time for each segment and summing is the most reliable method. Given Data / Assumptions: Fractions of total distance: 1/2, 1/3, 1/6 (which sum to 1). Speeds: 40 km/h, 60 km/h, 30 km/h respectively. Constant speeds on each leg; no stoppage time. Concept / Approach: Let total distance = D. Then times per leg are (D/2)/40, (D/3)/60, (D/6)/30. The average speed v_avg = D / (sum of times). D cancels, leaving a pure numeric result. Step-by-Step Solution: t1 = (D/2) / 40 = D/80.t2 = (D/3) / 60 = D/180.t3 = (D/6) / 30 = D/180.Total time T = D * (1/80 + 1/180 + 1/180) = D * (17/720).Average speed v_avg = D / T = 720/17 ≈ 42.35 km/h. Verification / Alternative check: Using any concrete D (e.g., 720 km) yields T = 17 h, so v_avg = 720/17 km/h as above. Why Other Options Are Wrong: 45 or 50 km/h come from averaging speeds directly; 40 km/h ignores that the slowest segment consumes disproportionate time. Common Pitfalls: Taking a simple arithmetic mean of 40, 60, and 30 without distance weights, or incorrectly weighting by time rather than distance. Final Answer: 42.35 km/h

Question 2

Mohan reaches the office 1 hour late when walking at 20 km/h. When he increases his speed to 25 km/h, he arrives 30 minutes early. What is the distance from home to office?

Accepted Answer

Introduction / Context: Schedule-variance problems compare actual travel time against scheduled time at two different speeds to deduce distance. The difference-of-reciprocals technique is standard. Given Data / Assumptions: At 20 km/h: arrival is 1 hour late. At 25 km/h: arrival is 30 minutes early. Same route and scheduled arrival time. Concept / Approach: Let distance be d and scheduled time be T_s (hours). Then d/20 = T_s + 1 and d/25 = T_s − 0.5. Subtract the equations to eliminate T_s and solve for d. Step-by-Step Solution: d/20 − d/25 = 1.5d * (1/20 − 1/25) = 1.5d * (5 − 4)/100 = 1.5 → d/100 = 1.5d = 150 km. Verification / Alternative check: T_s from the second equation: T_s = d/25 + 0.5 = 150/25 + 0.5 = 6 + 0.5 = 6.5 h. First equation check: 150/20 = 7.5 h = 6.5 + 1, consistent. Why Other Options Are Wrong: 100 km and 125 km do not satisfy both timing relations simultaneously. Common Pitfalls: Mixing up the early/late signs; using minutes without converting to hours consistently. Final Answer: 150 km

Question 3

A boy reaches school 20 minutes late at 15 km/h. When he walks at 20 km/h, he is still 5 minutes late. What is the distance from home to school?

Accepted Answer

Introduction / Context: Another schedule-variance setup: with two speeds and known lateness at each, we deduce the fixed distance by eliminating the (unknown) scheduled time. Given Data / Assumptions: d/15 = T_s + 20/60 h. d/20 = T_s + 5/60 h. Uniform route and speeds. Concept / Approach: Subtract the equations to remove the scheduled time T_s, leaving one equation in d. Convert minutes to hours to keep units consistent. Step-by-Step Solution: d/15 − d/20 = (20 − 5)/60 = 15/60 = 1/4 h.d * (1/15 − 1/20) = 1/4.d * (4 − 3)/60 = 1/4 → d/60 = 1/4.d = 15 km. Verification / Alternative check: At 15 km/h: 15/15 = 1 h = T_s + 0.333.. → T_s = 0.666.. h. At 20 km/h: 15/20 = 0.75 h = T_s + 0.0833.. → T_s = 0.666.. h; consistent. Why Other Options Are Wrong: 5 km, 10 km, 20 km violate one or both timing equations. Common Pitfalls: Accidentally treating 5 minutes as 0.5 h; forgetting to convert both lateness values to hours. Final Answer: 15 km

Question 4

A dog makes 4 leaps in the same time a cat makes 5 leaps. The distance of 3 dog leaps equals the distance of 4 cat leaps. What is the ratio of their speeds (dog : cat)?

Accepted Answer

Introduction / Context: Speed equals step frequency multiplied by step length. We are given a ratio of leaps per equal time and a relation between leap lengths. Combine them to get the speed ratio. Given Data / Assumptions: Leaps per equal time: dog : cat = 4 : 5. 3 dog leaps = 4 cat leaps → L_d / L_c = 4/3. Uniform steps; no acceleration. Concept / Approach: If n is leaps per time and L is length per leap, then speed v = n * L. Hence v_d / v_c = (n_d * L_d) / (n_c * L_c). Step-by-Step Solution: n_d / n_c = 4/5.L_d / L_c = 4/3 (from 3L_d = 4L_c).v_d / v_c = (4/5) * (4/3) = 16/15 → 16 : 15. Verification / Alternative check: Scale leap counts over one minute and multiply by lengths to confirm the same ratio numerically. Why Other Options Are Wrong: Other ratios arise from inverting the frequency or length relations or mixing 3 and 4 incorrectly. Common Pitfalls: Using 3/4 instead of 4/3 for lengths; adding ratios rather than multiplying. Final Answer: 16 : 15

Question 5

Suresh traveled 1200 km by air, which was 2/5 of the entire trip. One-third was by car and the rest by train. What distance did he travel by train?

Accepted Answer

Introduction / Context: Fraction-of-whole travel problems require reconstructing total distance from a known fraction, then allocating the remaining portions accordingly. Given Data / Assumptions: Air: 1200 km = 2/5 of total. Car: 1/3 of total. Train: remainder. Concept / Approach: Let total distance be D. From 1200 = (2/5)D, compute D. Then compute the car portion and finally the train portion as D minus the other two. Step-by-Step Solution: D = 1200 * (5/2) = 3000 km.Car distance = (1/3) * 3000 = 1000 km.Train distance = 3000 − 1200 − 1000 = 800 km. Verification / Alternative check: Fractions check: 1200/3000 = 2/5 and 1000/3000 = 1/3, leaving 800/3000 = 4/15 for train; all parts sum to 1. Why Other Options Are Wrong: 1600, 1800, 400 km are misallocations of the remaining 3/5 distance after accounting for the air and car portions. Common Pitfalls: Applying 1/3 to the remaining distance instead of the total; forgetting to subtract both known portions. Final Answer: 800 km

Question 6

Two cyclists start from the same point at the same time: one north at 18 km/h and the other south at 20 km/h. After how long will they be 95 km apart?

Accepted Answer

Introduction / Context: When two objects move in opposite directions along a straight line, their separation grows at the sum of their speeds. This yields time directly from distance = rate * time. Given Data / Assumptions: Cyclist A speed = 18 km/h (north). Cyclist B speed = 20 km/h (south). Desired separation = 95 km. Concept / Approach: Relative separation speed = 18 + 20 = 38 km/h. Then time t = distance / rate. Step-by-Step Solution: t = 95 / 38 h = 2.5 h.Convert 0.5 h = 30 min → t = 2 h 30 min. Verification / Alternative check: In 2.5 h, A covers 45 km and B covers 50 km, totaling 95 km separation. Why Other Options Are Wrong: Other times correspond to separations different from 95 km when multiplied by 38 km/h. Common Pitfalls: Subtracting speeds (used for same-direction chasing) instead of adding for opposite directions. Final Answer: 2 h 30 min

Question 7

A cart must cover 80 km in 10 hours. It covers half the distance in 3/5 of the total time. What speed is needed to cover the remaining distance in the time left?

Accepted Answer

Introduction / Context: This is a two-phase motion problem with a fixed total time. Once part of the journey is completed, compute the remaining time and the speed required to finish on schedule. Given Data / Assumptions: Total distance = 80 km; total time = 10 h. First 40 km (half the distance) is covered in (3/5)*10 = 6 h. Remaining distance = 40 km; remaining time = 4 h. Concept / Approach: Required speed for the second part equals remaining distance divided by remaining time. Step-by-Step Solution: Remaining speed = 40 km / 4 h = 10 km/h. Verification / Alternative check: Average speed overall = 80/10 = 8 km/h; doing 40 km in 6 h implies average 6.67 km/h for the first half, making 10 km/h necessary for the second half to meet the average. Why Other Options Are Wrong: 8 km/h maintains the overall average but does not finish in time given the slow first half; 20 km/h and 6.4 km/h misapply the time split. Common Pitfalls: Using the target average speed (8 km/h) for the remainder despite the earlier 6 h already spent on half the distance. Final Answer: 10 km/h

Question 8

A and B have travel-rate ratio A:B = 2:3. A takes 10 minutes more than B to cover the same distance. If A had walked at double speed, how long would A take?

Accepted Answer

Introduction / Context: Rate–time inverse proportionality says time is proportional to 1/speed for fixed distance. With a speed ratio and time difference, we can find the individual times and then compute the effect of doubling A's speed. Given Data / Assumptions: Speed ratio A:B = 2:3. A takes 10 minutes more than B for the same distance. Constant distance; uniform speeds. Concept / Approach: If speeds are 2k and 3k, times are proportional to 1/2k and 1/3k. Therefore t_A : t_B = 3 : 2. The difference t_A − t_B equals 10 minutes. Step-by-Step Solution: Let t_B = 20 min and t_A = 30 min (maintains 3:2 with 10 min difference).Doubling A's speed halves A's time → 30/2 = 15 min. Verification / Alternative check: Let distance be d. With speeds 2k and 3k, t_B = d/(3k) = 20 and t_A = d/(2k) = 30 are consistent for some d,k; the ratio holds and the difference is 10. Why Other Options Are Wrong: 20, 25, 30 min do not reflect the halving correctly from A's 30-min original time. Common Pitfalls: Assuming time ratio equals speed ratio rather than the inverse. Final Answer: 15 min

Question 9

A man travels for 35 km at two constant speeds: 4 km/h for some time and 5 km/h for the rest. If, instead, he had swapped the speeds for those same time portions, he would cover 2 km more in the same total time. What is the total time at the origina…

Accepted Answer

Introduction / Context: The statement implies fixed time portions at two speeds. Swapping the speeds across the same time portions changes total distance while total time remains unchanged. This yields two linear equations in the two time portions whose sum is the desired total time. Given Data / Assumptions: Original distance = 35 km at speeds 4 km/h and 5 km/h over time portions t1 and t2. Swapped speeds for the same times produce distance 37 km (2 km more) in the same total time. No stoppages; uniform speeds. Concept / Approach: Original distance: 4t1 + 5t2 = 35. Swapped distance: 5t1 + 4t2 = 37. Add equations to eliminate alternately and obtain t1 + t2 (the total time). Step-by-Step Solution: (4t1 + 5t2) + (5t1 + 4t2) = 35 + 37 → 9(t1 + t2) = 72.Total time T = t1 + t2 = 72/9 = 8 hours. Verification / Alternative check: Subtract equations: (5t1 + 4t2) − (4t1 + 5t2) = 37 − 35 → t1 − t2 = 2. With T = 8, solve t1 = 5, t2 = 3; original distance = 4*5 + 5*3 = 35; swapped = 5*5 + 4*3 = 37, as stated. Why Other Options Are Wrong: 7, 9, or 4.5 hours do not satisfy both distance equations simultaneously. Common Pitfalls: Interpreting the statement as swapping distances rather than speeds for the same times; failing to note total time remains fixed. Final Answer: 8 hours

Question 10

A train running at 40 km/h arrives 11 minutes late, but at 50 km/h it is only 5 minutes late. What is the correct scheduled time for the journey (without lateness)?

Accepted Answer

Introduction / Context: We compare actual travel times at two speeds to deduce the scheduled time. The arrival-lateness values serve as offsets from the schedule. Given Data / Assumptions: d/40 = T_s + 11/60 h. d/50 = T_s + 5/60 h. Same distance d; constant speeds. Concept / Approach: Subtract the equations to eliminate T_s and solve for d. Then compute T_s from either equation. Step-by-Step Solution: d*(1/40 − 1/50) = (11 − 5)/60 = 6/60 = 1/10 h.(1/40 − 1/50) = 1/200 → d * (1/200) = 0.1 → d = 20 km.T_s = d/50 − 5/60 = 20/50 − 1/12 = 0.4 − 0.0833.. = 0.3166.. h = 19 min. Verification / Alternative check: At 40 km/h: 20/40 = 0.5 h = 30 min → 30 − 19 = 11 min late; at 50 km/h: 24 min → 24 − 19 = 5 min late. Why Other Options Are Wrong: 13, 15, or 21 min do not satisfy both lateness conditions. Common Pitfalls: Mixing minutes and hours or subtracting late times from each other without anchoring to the schedule. Final Answer: 19 min

Question 11

Train M leaves Meerut at 6:00 a.m. and reaches Delhi at 10:00 a.m. Train D leaves Delhi at 8:00 a.m. and reaches Meerut at 11:30 a.m. At what time do they cross?

Accepted Answer

Introduction / Context: Two trains depart from opposite ends at different start times and travel times. Let the meet time be t hours after the first train starts; write a distance-balance equation using their constant speeds. Given Data / Assumptions: Meerut→Delhi: 6:00 to 10:00 → 4 h. Delhi→Meerut: 8:00 to 11:30 → 3.5 h. Constant speeds; same distance D. Concept / Approach: Speeds: v1 = D/4, v2 = D/3.5. If they meet t hours after 6:00, train 2 has traveled (t − 2) hours (for t ≥ 2). Total distance covered at meet equals D. Step-by-Step Solution: D = (D/4)*t + (D/3.5)*(t − 2).Divide by D: 1 = t/4 + (t − 2)/3.5.Solve → t = 44/15 h ≈ 2.933.. h after 6:00 → 8:56 a.m. Verification / Alternative check: Check distances: v1*t = D*(t/4); v2*(t − 2) = D*((t − 2)/3.5). Their sum equals D by construction. Why Other Options Are Wrong: Other times do not satisfy the proportionality with the given end-to-end durations. Common Pitfalls: Using the average of departure times or travel times instead of solving the linear equation. Final Answer: 8:56 a.m.

Question 12

A train leaves Delhi at 5:00 a.m. and reaches Kanpur at 10:00 a.m. Another leaves Kanpur at 7:00 a.m. and reaches Delhi at 2:00 p.m. When do they meet?

Accepted Answer

Introduction / Context: Opposite-end departures with different start times again reduce to a single linear equation by expressing each distance at meet in terms of t hours after the first start. Given Data / Assumptions: Delhi→Kanpur: 5:00 to 10:00 → 5 h. Kanpur→Delhi: 7:00 to 14:00 → 7 h. Constant speeds; same route distance D. Concept / Approach: Speeds: v1 = D/5, v2 = D/7. If they meet t hours after 5:00, the second has traveled for (t − 2) hours. Sum of distances equals D. Step-by-Step Solution: 1 = t/5 + (t − 2)/7 (after dividing by D).Multiply by 35: 35 = 7t + 5(t − 2) = 12t − 10 → 12t = 45 → t = 15/4 h = 3 h 45 min.Meet time = 5:00 + 3:45 = 8:45 a.m. Verification / Alternative check: Check proportions of remaining times to endpoints; they should match the computed speeds. Why Other Options Are Wrong: 6:45 a.m. ignores the 7:00 a.m. start of the second train; 3:45 p.m. is after both arrive. Common Pitfalls: Using average start times or assuming equal speeds. Final Answer: 8:45 a.m.

Question 13

A starts from P to Q (51.75 km) at 3.75 km/h. One hour later, B starts from Q to P at 4.25 km/h. When and where do they meet?

Accepted Answer

Introduction / Context: This is a meet-with-head-start problem. One traveler leaves earlier from one end; the other starts later from the opposite end. We form a single distance equation with their speeds and the staggering of start times. Given Data / Assumptions: Distance P–Q = 51.75 km. A: 3.75 km/h from P at t = 0. B: 4.25 km/h from Q at t = 1 hour. Concept / Approach: Let T be hours after A starts when they meet (T ≥ 1). Then A covers 3.75T and B covers 4.25(T − 1). Their sum equals 51.75 km. Step-by-Step Solution: 3.75T + 4.25(T − 1) = 51.75.8T − 4.25 = 51.75 → 8T = 56 → T = 7 h.Distance from P = 3.75 * 7 = 26.25 km → from Q = 51.75 − 26.25 = 25.50 km. Verification / Alternative check: In 6 hours (T − 1), B covers 25.5 km; A covers 26.25 km in 7 hours; sums to 51.75 km. Why Other Options Are Wrong: 26.25 km from Q confuses ends; 25.30 km is a rounding/inexact arithmetic. Common Pitfalls: Ignoring the 1-hour delay for B or summing distances to a value different from the total. Final Answer: 25.50 km from Q

Question 14

Two boys write a booklet of 817 lines from opposite ends: Boy 1 from line 1 at 200 lines/h; Boy 2 from line 817 backwards at 150 lines/h. On which line do they meet?

Accepted Answer

Introduction / Context: When two agents work from opposite ends toward each other at constant rates, they meet when the cumulative work equals the total. The meeting index corresponds to the work completed by either worker at that instant. Given Data / Assumptions: Total lines = 817. Rates: 200 and 150 lines/h. They start simultaneously. Concept / Approach: Combined rate = 350 lines/h. Meeting time t = total work / combined rate. The line number reached by Boy 1 then is 200t (counting from line 1). Rounding rules imply the first line not yet written by both. Step-by-Step Solution: t = 817 / 350 h.Boy 1's line index = 200 * (817/350) = (4/7)*817 ≈ 466.857.They meet between lines 466 and 467, so the meeting line is the 467th. Verification / Alternative check: Boy 2 writes 150t ≈ 350.143 lines from the end; 466.857 + 350.143 ≈ 817, consistent. Why Other Options Are Wrong: 465th or 466th undercount; 468th overshoots the exact balance point. Common Pitfalls: Taking arithmetic mean of 1 and 817 rather than using rates; rounding down instead of to the crossing point. Final Answer: 467th

Question 15

Cars A and B start 88 km apart and drive towards each other. Speed ratio A:B = 5:6, and B's speed is 90 km/h. After how long do they meet?

Accepted Answer

Introduction / Context: Opposite-direction approach with a known ratio allows computation of the other speed and thus the relative approach rate. Given Data / Assumptions: B = 90 km/h, A:B = 5:6 → A = 75 km/h. Initial separation = 88 km. Concept / Approach: Relative speed toward each other = A + B = 75 + 90 = 165 km/h. Time = distance / relative speed. Step-by-Step Solution: t = 88 / 165 h = 0.5333.. h = 32 minutes. Verification / Alternative check: Distance covered by A in 32 min: 75*(32/60) = 40 km; by B: 90*(32/60) = 48 km; total 88 km. Why Other Options Are Wrong: 24, 26, 36 min correspond to relative speeds inconsistent with 165 km/h. Common Pitfalls: Using difference of speeds by mistake (used for same-direction problems). Final Answer: 32 min

Question 16

In a circus, a leopard and a tiger walk in separate rings of the same radius. The leopard makes 3 steps while the tiger makes 5 in the same time. The distance of 5 leopard steps equals 4 tiger steps. When the tiger completes 100 rounds, how many rou…

Accepted Answer

Introduction / Context: Number of rounds is proportional to linear speed for equal-ring circumference. Speed here equals steps per unit time times step length. Given Data / Assumptions: Steps per equal time: leopard : tiger = 3 : 5. 5 leopard steps = 4 tiger steps → L_leopard / L_tiger = 4/5. Equal ring radius (equal circumference). Concept / Approach: v_leopard / v_tiger = (3/5) * (4/5) = 12/25. Rounds completed are in the same ratio as speeds over the same time. Step-by-Step Solution: Speed ratio = 12 : 25.If tiger makes 100 rounds, leopard makes 100 * (12/25) = 48 rounds. Verification / Alternative check: Compare distances per minute: scaling confirms leopard covers 48% of tiger's distance. Why Other Options Are Wrong: 75 or 120 assume inverted or additive ratios; none match the combined frequency-length effect. Common Pitfalls: Using 5/4 instead of 4/5 for step lengths or adding the ratios rather than multiplying. Final Answer: 48

Question 17

A thief spots a jeep 250 m away coming at 36 km/h. After 5 s, the thief runs away at 54 km/h. Ten seconds later, the police speed up to 72 km/h and give chase. How long after the thief first saw the jeep is he caught, and how far does the jeep trave…

Accepted Answer

Introduction / Context: This problem involves piecewise-constant speeds with reaction delays. We track the separation over successive intervals, switching relative speeds as the chase dynamics change. Given Data / Assumptions: Initial separation = 250 m. Jeep: 36 km/h = 10 m/s; later 72 km/h = 20 m/s. Thief: starts after 5 s at 54 km/h = 15 m/s away from jeep. Police react 10 s after the thief starts running. Concept / Approach: Compute separation changes interval by interval: (0–5 s) thief stationary; (5–15 s) thief faster than jeep; (≥15 s) jeep faster than thief. Use relative speed to update separation until closure. Step-by-Step Solution: 0–5 s: separation reduces by 10*5 = 50 m → 200 m remains.5–15 s: thief at 15 m/s, jeep at 10 m/s → separation increases 5 m/s for 10 s → +50 m → 250 m.t ≥ 15 s: jeep at 20 m/s vs thief 15 m/s → closing rate 5 m/s.Time to close 250 m at 5 m/s = 50 s → catch at 15 + 50 = 65 s after first sight.Jeep distance = 10*15 + 20*50 = 150 + 1000 = 1150 m. Verification / Alternative check: At catch, positions match: thief distance after start (from t = 5 s) is 15*(65 − 5) = 900 m from his start point, consistent with the relative closure. Why Other Options Are Wrong: Other time–distance pairs conflict with the interval-by-interval separation updates. Common Pitfalls: Assuming the jeep accelerates immediately or forgetting that separation grows during 5–15 s. Final Answer: 65 s, 1150 m

Question 18

Den Bosch and Eastbourne are 300 km apart. Maradona leaves Den Bosch at 8:24 a.m.; Pele leaves an hour later from the same place. After 1 hour of travel, Pele reaches Nottingham, which Maradona passed 40 minutes earlier. Find their speeds (Maradona,…

Accepted Answer

Introduction / Context: Two travelers start from the same origin at different times. A landmark (Nottingham) is reached by the later traveler in 1 hour and was passed by the earlier traveler 40 minutes earlier. This fixes a speed ratio; combining with the 1-hour start gap gives absolute speeds that also make sense over the full 300 km. Given Data / Assumptions: Pele reaches Nottingham 1 hour after starting. Maradona passed Nottingham 40 minutes earlier than that. Both depart Den Bosch; route passes Nottingham en route to Eastbourne (300 km). Concept / Approach: Let V_m, V_p be speeds. Nottingham distance d_N satisfies V_p * 1 = V_m * (4/3) because Maradona took 4/3 h to reach Nottingham (from 8:24 to 9:44). Hence V_p / V_m = 4/3. Use the 1-hour start-gap arrival alignment to select the consistent pair from options. Step-by-Step Solution: Ratio from Nottingham condition: V_m : V_p = 3 : 4.Option (75, 100) and (60, 80) fit the 3:4 ratio.Check same-arrival feasibility over 300 km with a 1-hour delayed start: 300/75 − 300/100 = 4 − 3 = 1 h → works for (75, 100); not for (60, 80), which yields 1.25 h. Verification / Alternative check: Nottingham distance at (75, 100): 100*1 = 100 km; 75*(4/3) = 100 km, consistent. Why Other Options Are Wrong: They either break the 3:4 ratio or fail the 1-hour start-gap arrival consistency. Common Pitfalls: Ignoring the 40-minute offset or assuming equal arrival times without checking the start-gap consistency. Final Answer: 75 km/h, 100 km/h

Question 19

Amethi and Bellari are 645 km apart. Sonia and Priyanka start simultaneously from Amethi and Bellari toward each other and meet after 15 hours. After meeting, Sonia increases her speed by 3 km/h and Priyanka reduces hers by 3 km/h; both reach their …

Accepted Answer

Introduction / Context: Two-way travel with a meeting point gives the sum of initial speeds. The symmetric post-meeting arrival-on-same-time condition, after adjusting speeds by ±3 km/h, fixes the individual initial speeds uniquely. Given Data / Assumptions: Total distance D = 645 km; meeting time = 15 h. Initial speeds s (Sonia), p (Priyanka) so s + p = D/15 = 43 km/h. After meeting: Sonia's speed = s + 3; Priyanka's = p − 3. They arrive at endpoints at the same time after meeting. Concept / Approach: Remaining distances after meeting are D − 15s (for Sonia to Bellari) and D − 15p (for Priyanka to Amethi). Equal arrival times imply (D − 15s)/(s + 3) = (D − 15p)/(p − 3), together with s + p = 43. Step-by-Step Solution: s + p = 43.(645 − 15s)/(s + 3) = (645 − 15p)/(p − 3).Solving yields s = 20 km/h, p = 23 km/h. Verification / Alternative check: Check remaining distances: Sonia has 645 − 300 = 345 km; Priyanka has 645 − 345 = 300 km. Times: 345/(20 + 3) = 15 h; 300/(23 − 3) = 15 h; equal as required. Why Other Options Are Wrong: Other pairs do not satisfy both s + p = 43 and the equal post-meeting arrival condition. Common Pitfalls: Using s − p or averaging speeds instead of leveraging the equal-time condition with adjusted speeds. Final Answer: 20 km/h and 23 km/h

Question 20

Chase problem — tiger vs deer (uniform leaps): A tiger is behind a deer by a head start equal to 50 of the tiger’s own leaps. The tiger makes 5 leaps per minute and covers 8 m per leap. The deer makes 4 leaps per minute and covers 5 m per leap. How …

Accepted Answer

Introduction / Context: Relative speed chase problems compare how quickly a pursuer closes the gap on a target when both move simultaneously. Here, movement is in “leaps per minute,” but we convert each to metres per minute to compute the catch-up distance cleanly. Given Data / Assumptions: Tiger: 5 leaps/min, 8 m/leap ⇒ 40 m/min Deer: 4 leaps/min, 5 m/leap ⇒ 20 m/min Initial head start = 50 tiger leaps = 50 * 8 = 400 m Both run in the same straight line with constant speeds. Concept / Approach: The time to catch equals initial lead / relative speed. Relative speed (tiger w.r.t. deer) = tiger speed − deer speed. Step-by-Step Solution: Tiger speed = 5 * 8 = 40 m/minDeer speed = 4 * 5 = 20 m/minRelative speed = 40 − 20 = 20 m/minInitial gap = 50 * 8 = 400 mTime to catch = 400 / 20 = 20 minDistance run by tiger until catch = tiger speed * time = 40 * 20 = 800 m Verification / Alternative check: If both run 20 min, the deer covers 20 * 20 = 400 m; adding to the tiger’s initial deficit of 400 m confirms equality at the catch point (tiger total 800 m vs deer total 400 m after the head start). Why Other Options Are Wrong: 600 m, 700 m, 1000 m, 500 m: Each ignores either the correct relative speed or the 50-leap (400 m) head start. Common Pitfalls: Using leaps/min directly without converting to m/min, or misreading “50 leaps” as metres. Another mistake is adding speeds instead of taking the difference for same-direction chase problems. Final Answer: 800 m

Time and Distance Questions