Question 1

Time and Distance – Early or late arrival with two walking speeds: A student walks from home to school at 3 km/h and arrives 5 minutes late relative to the scheduled time. The next day, walking at 4 km/h, the student arrives 3 minutes early. What is…

Accepted Answer

Introduction / Context: Problems with early and late arrivals hinge on equating the same distance under different speeds and different arrival offsets. Translating minutes into hours keeps units consistent and allows a clean equation for distance. Given Data / Assumptions: Speed case 1: 3 km/h and arrival is 5 min late. Speed case 2: 4 km/h and arrival is 3 min early. Scheduled travel time is the same reference in both cases. Concept / Approach: Let D be distance and T be the scheduled travel time (in hours). Then D/3 = T + 5/60 and D/4 = T − 3/60. Subtracting eliminates T and solves D directly. Step-by-Step Solution: D/3 − D/4 = (5/60) + (3/60) = 8/60 = 2/15.Left side = D*(1/12) ⇒ D/12 = 2/15.Therefore D = 24/15 = 8/5 = 1.6 km. Verification / Alternative check: Check quickly: At 3 km/h, time = 1.6/3 ≈ 0.533 h = 32 min; at 4 km/h, time = 1.6/4 = 0.4 h = 24 min. The 8-minute gap matches the 5 min late vs 3 min early difference. Why Other Options Are Wrong: 1.24, 1.36, 1.5, and 1.8 km do not satisfy the pair of equations built from the arrival offsets. Common Pitfalls: Using 5 − 3 = 2 minutes rather than summing offsets to 8 minutes, or forgetting to convert minutes to hours. Final Answer: 1.6 km

Question 2

Time and Distance – Pursuit and relative speed (policeman chasing thief): A thief runs at 8 km/h and is 100 m ahead of a policeman who runs at 10 km/h. Assuming both maintain their speeds, how long will it take the policeman to catch the thief?

Accepted Answer

Introduction / Context: In linear pursuit with both runners on the same line, the catch-up time equals the head start distance divided by the relative speed. Converting the head start to the same units as speeds avoids mistakes. Given Data / Assumptions: Thief speed = 8 km/h. Policeman speed = 10 km/h. Head start = 100 m = 0.1 km. Concept / Approach: Relative speed (closing speed) = 10 − 8 = 2 km/h. Time = gap / relative speed, then convert hours to minutes. Step-by-Step Solution: Time (h) = 0.1 / 2 = 0.05 h.Time (min) = 0.05 * 60 = 3 min. Verification / Alternative check: In 3 minutes, policeman covers 0.5 km, thief covers 0.4 km, so the 0.1 km head start is exactly closed. Why Other Options Are Wrong: 2 or 4 minutes correspond to relative speeds of 3 or 1.5 km/h, not present here; 6 minutes doubles the correct value; 90 seconds halves it. Common Pitfalls: Failing to convert 100 m to 0.1 km or subtracting speeds in the wrong order. Final Answer: 3 min

Question 3

Boats and Streams – Upstream time is twice downstream time: A person rows at 7.5 km/h in still water. For a fixed distance, rowing upstream takes twice as long as rowing downstream. What is the speed of the stream?

Accepted Answer

Introduction / Context: In boats-and-streams, still-water speed and stream speed combine linearly: downstream = u + v and upstream = u − v. When times differ by a given factor for the same distance, we set up a ratio of speeds using time = distance / speed. Given Data / Assumptions: Still-water rowing speed u = 7.5 km/h. Upstream time is twice downstream time for the same distance. Concept / Approach: Let speeds be S_up = u − v and S_down = u + v. Since time is inversely proportional to speed for fixed distance, 1/(u − v) = 2 * 1/(u + v). Rearranging gives a linear equation in v. Step-by-Step Solution: u + v = 2(u − v).u + v = 2u − 2v ⇒ 3v = u ⇒ v = u / 3.Therefore v = 7.5 / 3 = 2.5 km/h. Verification / Alternative check: Speeds become S_down = 10 km/h and S_up = 5 km/h, so upstream time is indeed double downstream time for any fixed distance. Why Other Options Are Wrong: 2.0, 2.2, 2.7, 3.0 km/h do not satisfy the 2:1 time ratio when combined with u = 7.5 km/h. Common Pitfalls: Using time ratio directly on u without accounting for stream v, or mixing the direction for addition and subtraction. Final Answer: 2.5 km/h

Question 4

Boats and Streams – Find port-to-port distance from up/down times and stream speed: A steamer covers a downstream trip in 4 hours and the same route upstream in 5 hours. If the stream speed is 2 km/h, what is the distance between the two ports?

Accepted Answer

Introduction / Context: This uses the standard downstream and upstream speed relations with a known stream speed. With both travel times given, we can solve for still-water speed and then compute the distance directly from either leg. Given Data / Assumptions: Downstream time = 4 h; upstream time = 5 h. Stream speed v = 2 km/h; still-water speed u is unknown. Distance D is the same each way. Concept / Approach: Downstream speed = u + v; upstream speed = u − v. Thus D = (u + 2)*4 = (u − 2)*5. Solve for u, then compute D. Step-by-Step Solution: 4u + 8 = 5u − 10 ⇒ u = 18 km/h.Distance D = (u + 2)*4 = 20 * 4 = 80 km. Verification / Alternative check: Using upstream: (u − 2)*5 = 16 * 5 = 80 km, consistent with downstream calculation. Why Other Options Are Wrong: 60, 70, 72, and 50 km fail to satisfy both leg equations simultaneously with stream speed 2 km/h. Common Pitfalls: Confusing which leg is faster or failing to use the same distance for both directions. Final Answer: 80 km

Question 5

Boats and Streams – Round trip time with a known current: The stream current is 1 km/h. A motorboat goes 35 km upstream and returns 35 km downstream to the start in a total of 12 hours. What is the speed of the motorboat in still water?

Accepted Answer

Introduction / Context: This is a symmetric up-and-down trip with a fixed current. Total time is the sum of upstream and downstream legs. The unknown still-water speed u is found by solving a rational equation built from the two legs. Given Data / Assumptions: Current v = 1 km/h. Upstream distance = downstream distance = 35 km. Total time upstream + downstream = 12 h. Concept / Approach: Time upstream = 35/(u − 1); time downstream = 35/(u + 1). Set their sum to 12 and solve for u by clearing denominators. Step-by-Step Solution: 35/(u − 1) + 35/(u + 1) = 12.35[(u + 1) + (u − 1)]/(u^2 − 1) = 12 ⇒ 70u/(u^2 − 1) = 12.12u^2 − 12 − 70u = 0 ⇒ 12u^2 − 70u − 12 = 0.Discriminant = 5476 = 74^2 ⇒ u = (70 + 74)/24 = 144/24 = 6 km/h. Verification / Alternative check: Times: 35/5 = 7 h upstream and 35/7 = 5 h downstream sum to 12 h. Why Other Options Are Wrong: Values other than 6 km/h do not satisfy the time equation when substituted. Common Pitfalls: Forgetting to use the same distance for both legs or miscomputing the discriminant when solving the quadratic. Final Answer: 6 km/h

Question 6

Boats and Streams – Infer stream speed from unequal travel times: A motorboat has still-water speed 30 km/h. It travels from Mumbai to Goa in 2 hours, then later from Goa back to Mumbai in 4 hours. Assuming the stream speed is constant in both trips…

Accepted Answer

Introduction / Context: Different times for the same route in opposite directions reflect downstream and upstream effects. With known still-water speed u and unknown current v, the distance D is equal on both legs, allowing a simple equation in v. Given Data / Assumptions: Still-water speed u = 30 km/h. Downstream time = 2 h; upstream time = 4 h. Distance Mumbai–Goa is the same both ways. Concept / Approach: Let D be the common distance. Then D = (u + v)*2 and D = (u − v)*4. Equate and solve for v. Step-by-Step Solution: 2(30 + v) = 4(30 − v).60 + 2v = 120 − 4v ⇒ 6v = 60 ⇒ v = 10 km/h. Verification / Alternative check: Using v = 10, downstream speed 40 km/h for 2 h gives D = 80 km; upstream speed 20 km/h for 4 h also gives 80 km. Why Other Options Are Wrong: Any value not equal to 10 km/h breaks the equality of distances for the two legs. Common Pitfalls: Assuming different distances or averaging speeds arithmetically without considering direction and current. Final Answer: 10 km/h

Question 7

Boats and Streams – Two conditions on upstream vs downstream times: Kapil rows in a river where his usual still-water speed is u mile/h and the current is c mile/h. For 12 miles, he needs 6 hours less upstream than downstream. If he could double his…

Accepted Answer

Introduction / Context: This problem gives two time-gap conditions across upstream and downstream legs for the same distances under two different rowing capabilities. The unknowns are the usual still-water speed u and the current c. We translate the time gaps into equations in u and c and solve simultaneously. Given Data / Assumptions: Distance per leg = 12 miles. Usual speeds: downstream u + c and upstream u − c. Condition 1: 12/(u − c) − 12/(u + c) = 6 hours. Condition 2 (with 2u): 12/(2u − c) − 12/(2u + c) = 1 hour. Concept / Approach: For fixed distances, time differences convert to rational expressions in u and c. Multiply through to clear denominators to form polynomial equations, then eliminate u to solve for c. Step-by-Step Solution: From Condition 1: 12[(u + c) − (u − c)]/(u^2 − c^2) = 6 ⇒ 24c = 6(u^2 − c^2) ⇒ u^2 − c^2 − 4c = 0.From Condition 2: 12[(2u + c) − (2u − c)]/(4u^2 − c^2) = 1 ⇒ 24c = 4u^2 − c^2 ⇒ 4u^2 − c^2 − 24c = 0.Eliminate u by subtracting 4*(first) from second: 3c^2 − 8c = 0 ⇒ c(3c − 8) = 0 ⇒ c = 8/3 mile/h (positive solution). Verification / Alternative check: Back-substitute c = 8/3 into either equation to find u and verify both conditions hold numerically. Why Other Options Are Wrong: 2, 3, 5/2, and 10/3 mile/h do not satisfy both time-gap equations simultaneously. Common Pitfalls: Algebraic slips when clearing denominators, or incorrectly setting the direction of the time differences. Final Answer: 8/3 mile/h

Question 8

Time and Distance – Crossing a bridge at uniform speed: A rider travels over a bridge at 54 km/h and takes 4 minutes to completely cross it. What is the length of the bridge in metres?

Accepted Answer

Introduction / Context: Distance equals speed multiplied by time. With speed in km/h and time in minutes, convert to consistent units and then compute metres for the bridge length. Given Data / Assumptions: Speed = 54 km/h. Time = 4 minutes. Bridge length equals total distance traveled during the crossing interval. Concept / Approach: Convert time to hours or speed to m/s. Here we convert time: 4 min = 4/60 h, then compute distance in km and convert to metres. Step-by-Step Solution: Time in hours = 4/60 = 1/15 h.Distance in km = 54 * (1/15) = 3.6 km.Distance in metres = 3.6 * 1000 = 3600 m. Verification / Alternative check: Converting 54 km/h to 15 m/s, distance = 15 m/s * 240 s = 3600 m. Why Other Options Are Wrong: Any value other than 3600 m conflicts with the direct unit-consistent computation. Common Pitfalls: Forgetting to convert minutes to hours or misusing 54 km/h as 54 m/s. Final Answer: 3600 m

Question 9

Time and Distance – Two walkers, arrival-time difference: Two men walk the same distance; one at 4 km/h and the other at 3 km/h. The faster man arrives 30 minutes earlier than the slower. What is the distance?

Accepted Answer

Introduction / Context: The same distance covered at different speeds yields different times. The difference in times is given, so we can write a single equation in the unknown distance and solve directly. Given Data / Assumptions: Speeds = 4 km/h and 3 km/h. Time difference = 30 min = 0.5 h. Distance is common for both walkers. Concept / Approach: If the distance is D, then times are D/4 and D/3. The slower takes longer, so D/3 − D/4 = 0.5. Solve for D. Step-by-Step Solution: D/3 − D/4 = D*(1/12) = 0.5.Hence D = 0.5 * 12 = 6 km. Verification / Alternative check: Times: 6/4 = 1.5 h and 6/3 = 2 h; the difference is 0.5 h as required. Why Other Options Are Wrong: 7, 8, 9, and 5 km do not produce a 0.5 h time difference at the given speeds. Common Pitfalls: Reversing the subtraction order or failing to convert minutes to hours. Final Answer: 6 km

Question 10

Time and Distance – Mixed speeds over fixed fractions of a journey: Aashutosh covers 2/3 of a trip at 4 km/h and the remaining 1/3 at 5 km/h, taking a total of 84 minutes. What is the total distance?

Accepted Answer

Introduction / Context: When different segments of a journey are covered at different speeds, total time is the sum of segment times. Here the fractions are of distance, not time, so harmonic-mean shortcuts do not apply; we compute each segment time explicitly. Given Data / Assumptions: Total time = 84 min = 1.4 h. First segment = 2/3 of distance at 4 km/h. Second segment = 1/3 of distance at 5 km/h. Concept / Approach: Let total distance be D. Then total time T = ( (2/3)D / 4 ) + ( (1/3)D / 5 ). Solve T = 1.4 for D. Step-by-Step Solution: T = D*( (2/3)/4 + (1/3)/5 ) = D*( 1/6 + 1/15 ) = D*( 5/30 + 2/30 ) = D*(7/30).Set D*(7/30) = 1.4 = 7/5 ⇒ D = (7/5)*(30/7) = 6 km. Verification / Alternative check: Times: (2/3*6)/4 = 1 h and (1/3*6)/5 = 0.4 h; total 1.4 h, consistent. Why Other Options Are Wrong: 8, 9, 15, and 7 km do not satisfy the time equation with the given segment speeds. Common Pitfalls: Using the average of speeds rather than computing weighted segment times by distance fractions. Final Answer: 6 km

Question 11

Time and Distance – Finish a target journey with time already spent: A bullock cart must cover 80 km in 10 hours. It completes half the journey in 3/5 of the total time. What speed is required for the remaining half to finish on time?

Accepted Answer

Introduction / Context: This is a remaining-time problem. After a portion of the journey consumes a specified fraction of the total time, we compute the leftover time and required speed to complete the rest of the distance on schedule. Given Data / Assumptions: Total distance = 80 km; total time = 10 h. Half the distance (40 km) is done in 3/5 of total time = 6 h. Remaining distance = 40 km; remaining time = 4 h. Concept / Approach: Required speed for the remainder = remaining distance / remaining time. Step-by-Step Solution: Remaining speed = 40 km / 4 h = 10 km/h. Verification / Alternative check: First half average = 40/6 ≈ 6.67 km/h; second half at 10 km/h yields total average 80/10 = 8 km/h, which meets the schedule. Why Other Options Are Wrong: Speeds other than 10 km/h miss the 4-hour remainder constraint for the remaining 40 km. Common Pitfalls: Halving time instead of distance or confusing 3/5 of time with 3/5 of distance. Final Answer: 10 km/h

Question 12

Average Speed – Outward and return with different speeds: Rani travels from the origin to a point by car at 5 km/h and returns by scooter at 2 km/h. What is her average speed over the entire round trip?

Accepted Answer

Introduction / Context: Average speed for a round trip with equal distances but different speeds is the harmonic mean of the two speeds. Direct arithmetic averaging is incorrect because travel times per leg differ. Given Data / Assumptions: Outward speed v1 = 5 km/h. Return speed v2 = 2 km/h. Distances are equal for both legs. Concept / Approach: Average speed over equal distances is 2*v1*v2 / (v1 + v2). Derivation uses total distance divided by total time, with each leg's time equal to distance over speed. Step-by-Step Solution: Average speed = 2*5*2 / (5 + 2) = 20/7 km/h. Verification / Alternative check: Let distance each way be 1 km. Total distance = 2 km. Time = 1/5 + 1/2 = 0.2 + 0.5 = 0.7 h. 2 / 0.7 = 20/7 km/h. Why Other Options Are Wrong: Values like 26/7 and 36/7 arise from incorrect arithmetic means or misapplied formulas. Common Pitfalls: Averaging the speeds ( (5 + 2)/2 ) rather than using the harmonic mean for equal distances. Final Answer: 20/7 km/h

Question 13

Time and Distance – Compare walking and driving times: A person can go one way by walking and return by driving in a total of 6 hours. Walking both ways would take 10 hours. How long would it take if the person drives both ways?

Accepted Answer

Introduction / Context: Let the one-way distance be D, walking speed be w, and driving speed be v. Translate each scenario into time equations and solve for the unknown driving-only round-trip time 2D/v. Given Data / Assumptions: D/w + D/v = 6 h for walk one way and drive back. 2D/w = 10 h for walking both ways. All speeds are constant. Concept / Approach: From 2D/w = 10, we get D/w = 5. Substitute into the mixed trip to find D/v, which then yields the driving-only round-trip time 2D/v. Step-by-Step Solution: D/w = 5 ⇒ D/v = 6 − 5 = 1.Hence 2D/v = 2 * 1 = 2 h. Verification / Alternative check: Interpretation: One-way drive time is 1 h; round-trip drive is 2 h. With D/w = 5 h, the mixed trip is 5 + 1 = 6 h, consistent. Why Other Options Are Wrong: Any value other than 2 h fails to satisfy both original equations simultaneously. Common Pitfalls: Solving for speeds rather than times or misinterpreting which equation corresponds to which scenario. Final Answer: 2 h

Question 14

Relative Motion – Meet after turnaround on a straight route: A and B start from point P to Q (84 km) at 12 km/h and 16 km/h, respectively. B reaches Q, immediately turns back, and meets A at point R. What is the distance PR?

Accepted Answer

Introduction / Context: This is a classic chase-and-return problem. After B reaches Q and turns back, they meet when A's position from P equals B's position from P (computed as 84 km minus B's return distance). We compute the meeting time and then multiply by A's speed. Given Data / Assumptions: Distance P to Q = 84 km. Speeds: A = 12 km/h, B = 16 km/h. B turns back immediately on reaching Q. Concept / Approach: Let t be the time from the start to the meeting. For t ≥ 84/16, B has already turned. After turning, B's position from P is 84 − 16(t − 84/16). Set this equal to A's position 12t and solve for t. Step-by-Step Solution: B reaches Q at 84/16 = 5.25 h.Meeting condition: 12t = 84 − 16(t − 5.25) = 168 − 16t.28t = 168 ⇒ t = 6 h ⇒ PR = 12 * 6 = 72 km. Verification / Alternative check: At t = 6 h, B has been returning for 0.75 h and covered 12 km back from Q, so B's position is 72 km from P, matching A. Why Other Options Are Wrong: 76, 78, 70, and 68 km do not satisfy the meeting equality under the given speeds. Common Pitfalls: Assuming they meet before B turns or forgetting to subtract the return distance from 84 km. Final Answer: 72 km

Question 15

Pursuit – Overtaking distance with head start: A thief is spotted 200 m ahead. The thief runs at 16 km/h while a policeman chases at 20 km/h. Assuming constant speeds, how far will the thief run before being caught?

Accepted Answer

Introduction / Context: With a fixed lead and higher chaser speed, the catch occurs when the chaser closes the head start at the relative speed. The distance run by the thief is his speed multiplied by the catch-up time. Given Data / Assumptions: Lead = 200 m = 0.2 km. Thief speed = 16 km/h. Policeman speed = 20 km/h. Concept / Approach: Relative speed = 20 − 16 = 4 km/h. Time to catch = 0.2 / 4 hours. Distance thief runs = 16 * time. Step-by-Step Solution: Time = 0.2 / 4 = 0.05 h = 3 min.Distance thief runs = 16 * 0.05 = 0.8 km = 800 m. Verification / Alternative check: Chaser covers 1.0 km in 3 min more than thief's 0.8 km; the extra 0.2 km equals the head start. Why Other Options Are Wrong: 650, 700, 850, 900 m do not match the computed product speed × time. Common Pitfalls: Forgetting to convert 200 m to km or computing the relative speed in the wrong direction. Final Answer: 800 m

Question 16

Opposite Travel – Ratio of speeds from post-meeting completion times: Two cyclists start simultaneously from A to B and from B to A. After meeting, they take 16 h and 25 h, respectively, to complete the remaining distances. What is the ratio of the …

Accepted Answer

Introduction / Context: When two travelers start toward each other, the times they take to finish after meeting encode the ratio of their speeds. A standard identity says v1/v2 = sqrt(t2/t1), where t1 and t2 are the post-meeting finish times of travelers 1 and 2. Given Data / Assumptions: After meeting, first finishes in t1 = 16 h. After meeting, second finishes in t2 = 25 h. They started simultaneously and travel on the same straight route. Concept / Approach: Let initial distance be D and meeting time be T. Then t1 = (v2*T)/v1 and t2 = (v1*T)/v2, giving t1/t2 = (v2/v1)^2 and thus v1/v2 = sqrt(t2/t1). Step-by-Step Solution: v1/v2 = sqrt(25/16) = 5/4. Verification / Alternative check: Square both sides: (v1/v2)^2 = 25/16 = t2/t1, which matches the identity derived from meeting-time relations. Why Other Options Are Wrong: Ratios other than 5 : 4 conflict with the square-root relationship forced by the 16 h and 25 h data. Common Pitfalls: Reversing t1 and t2, which would invert the ratio, or directly dividing times rather than using the square-root relation. Final Answer: 5 : 4

Question 17

Speed Comparison – Half distance takes four times the time: A certain distance is covered at a speed v1. If half of this distance is covered in four times as long as before, what is the ratio of the two speeds (v1 : v2)?

Accepted Answer

Introduction / Context: This tests proportional reasoning. Time is inversely proportional to speed for the same distance. The problem states that for half the original distance, the new time is four times the old time for that half-distance at v1, which fixes the speed ratio directly. Given Data / Assumptions: Original total distance = D and original speed = v1. Half distance D/2 at v1 takes time t1 = (D/2)/v1. New time for the same half distance is t2 = 4 * t1 at speed v2. Concept / Approach: v2 = (D/2)/t2 = (D/2)/(4*(D/2)/v1) = v1/4. Ratio v1 : v2 = 4 : 1. Step-by-Step Solution: t1 = (D/2)/v1.t2 = 4t1 ⇒ v2 = (D/2)/t2 = v1/4.Therefore v1 : v2 = 4 : 1. Verification / Alternative check: Pick numbers: D = 8, v1 = 8 ⇒ half distance time t1 = 0.5 h. New time t2 = 2 h ⇒ v2 = 2 km/h. Ratio 8 : 2 = 4 : 1. Why Other Options Are Wrong: 1 : 8, 1 : 4, 8 : 1, and 2 : 1 contradict the direct proportionality established. Common Pitfalls: Applying the factor of 4 to distance instead of time or forgetting that the comparison is for the same half distance. Final Answer: 4 : 1

Question 18

Average Speed – Two equal halves at 6 km/h and 3 km/h: A man covers half his journey at 6 km/h and the remaining half at 3 km/h. What is his average speed for the whole journey?

Accepted Answer

Introduction / Context: For equal distances at two speeds, average speed is the harmonic mean. Direct averaging of speeds is invalid because the slower portion consumes more time. Given Data / Assumptions: First half speed = 6 km/h. Second half speed = 3 km/h. Distances of both halves are equal. Concept / Approach: Average speed over equal distances = 2ab / (a + b). Here a = 6 and b = 3. Step-by-Step Solution: Average speed = 2*6*3 / (6 + 3) = 36/9 = 4 km/h. Verification / Alternative check: Let each half be 9 km. Times are 9/6 = 1.5 h and 9/3 = 3 h; total time = 4.5 h for 18 km ⇒ 18/4.5 = 4 km/h. Why Other Options Are Wrong: 3 or 4.5 km/h are arithmetic means of parts or times misapplied; 9 km/h is the maximum leg speed not the round-trip average. Common Pitfalls: Using the arithmetic mean ( (6 + 3)/2 = 4.5 ) rather than the harmonic mean. Final Answer: 4 km/h

Question 19

Proportional Speeds – Time scaling across linked speeds: A is twice as fast as B, and B is three times as fast as C. If C covers a certain distance in 56 minutes, how long will A take to cover the same distance?

Accepted Answer

Introduction / Context: The times taken to cover the same distance are inversely proportional to speeds. Given chained speed relations among A, B, and C, we can express A as a multiple of C and scale the time accordingly. Given Data / Assumptions: A = 2B and B = 3C ⇒ A = 6C. Time for C to cover the distance = 56 min. Distance is the same for all individuals. Concept / Approach: With A = 6C in speed terms, time scales inversely, so T_A = T_C / 6. This uses the constant-distance relationship time = distance / speed. Step-by-Step Solution: T_C = 56 min.T_A = 56 / 6 = 28/3 min = 9 1/3 min. Verification / Alternative check: Because A is 6 times faster than C, A should take one-sixth of C's time. 56/6 equals 9 minutes 20 seconds, which is 28/3 minutes. Why Other Options Are Wrong: 7 min is too small (would imply A is 8 times C); other options do not reflect the 6x speed relationship. Common Pitfalls: Multiplying instead of dividing the time by the speed factor or misreading the chained speed multipliers. Final Answer: 28/3 min

Question 20

The speeds of A and B are in the ratio 3 : 4 (A is slower). A takes 20 minutes more than B to reach the same place. Find the actual times taken by A and by B, respectively, to reach that place (give answers in minutes).

Accepted Answer

Introduction / Context: This is a classic time–speed–distance ratio problem. If two people cover the same distance with speeds in a fixed ratio, then their times are in the inverse ratio. The given 20-minute gap anchors the absolute values. Given Data / Assumptions: Speed(A) : Speed(B) = 3 : 4. A is slower; A takes 20 minutes more than B. Both cover the same distance D. Concept / Approach: Times are inversely proportional to speeds: Time(A) : Time(B) = 1/3 : 1/4 = 4 : 3. Let times be 4x and 3x. Their difference equals 20 minutes, which gives x and hence the actual times. Step-by-Step Solution: Let Time(A) = 4x, Time(B) = 3x.4x − 3x = 20 ⇒ x = 20.Therefore Time(A) = 80 min; Time(B) = 60 min. Verification / Alternative check: Let speeds be 3k and 4k; times are D/(3k) and D/(4k). Difference D(1/3k − 1/4k) = D/(12k) must equal 20 min; choosing D/k = 240 gives times 80 and 60 min, consistent with the ratio 4:3. Why Other Options Are Wrong: 40 & 30, 90 & 45, 90 & 50, 72 & 54 do not keep a 4:3 time ratio while also differing by exactly 20 minutes. Common Pitfalls: Inverting the ratio incorrectly; applying the 20-minute difference to speeds instead of times. Final Answer: 80 min and 60 min

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