Question 1

A thief drives away with a Maruti car at a constant speed of 40 km/h. The theft is discovered after 30 minutes, and the owner immediately starts chasing on a bike at 50 km/h. After how much time from the moment the thief left will the owner overtake…

Accepted Answer

Introduction / Context:This is a typical chase or pursuit problem in time and distance. One vehicle starts earlier at a lower speed, and another starts later at a higher speed. The key idea is to understand the concept of head start and use relative speed to determine when the faster chaser will catch the slower vehicle that started earlier. Given Data / Assumptions: Speed of the thief in the car = 40 km/h. The theft is discovered 30 minutes (0.5 hour) later. Speed of the owner on the bike = 50 km/h. The owner starts the chase exactly 0.5 hour after the thief starts. Both travel along the same straight route in the same direction. We are asked for the time from the thief's start until the owner catches him. Concept / Approach:When a faster vehicle is chasing a slower one, the catch up time depends on the initial lead and the difference in speeds. The lead is the distance already covered by the thief before the owner starts. After that, the relative speed between them is the difference of their speeds. Time to catch up is equal to lead divided by relative speed. Then we can convert that time back to the total time from the thief's start. Step-by-Step Solution:Let t be the time in hours from the thief's start until the catch up. Owner starts 0.5 hour later, so owner travel time = t - 0.5 hours. Distance traveled by thief = 40 * t km. Distance traveled by owner = 50 * (t - 0.5) km. At catch up, distances are equal: 40 * t = 50 * (t - 0.5). 40t = 50t - 25 10t = 25, so t = 25 / 10 = 2.5 hours. 2.5 hours = 2 hours 30 minutes. Verification / Alternative check:The thief has a 0.5 hour head start at 40 km/h, so the lead distance is 40 * 0.5 = 20 km. The relative speed between the owner and the thief is 50 - 40 = 10 km/h. Time to close a 20 km gap at 10 km/h is 20 / 10 = 2 hours. These 2 hours are measured from the moment the owner starts. Therefore, total time from the thief's start is 0.5 hour of head start plus 2 hours of chase, which again gives 2.5 hours or 2 hours 30 minutes. Why Other Options Are Wrong:2 hours and 2 hours 5 minutes are both too short because they ignore or underestimate the head start and the relative speed. 2 hours 10 minutes is still smaller than the correct 2.5 hours and does not satisfy the distance equality. The value 1 hour 50 minutes is much too small and clearly inconsistent with the head start and speed difference. Only 2 hours 30 minutes satisfies the correct equations. Common Pitfalls:Students often forget that the owner travels for less time than the thief and incorrectly set the travel times equal. Another frequent error is to add the speeds instead of subtracting them when calculating relative speed in a same-direction chase. Careful attention to who starts first, how long each travels, and whether the directions are the same or opposite is crucial for these problems. Final Answer:The owner will overtake the thief 2 hours 30 minutes after the thief starts driving away.

Question 2

Two cars, a Maruti and a Toyota, start at the same time from points A and B and move towards each other. The Maruti, starting from A, reaches B in 6 hours 20 minutes, and the Toyota, starting from B, reaches A in 4 hours 45 minutes. If the speed of …

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Introduction / Context:This question combines the ideas of distance, speed, and time for two vehicles traveling towards each other between two fixed points. It checks whether a student can find the distance between the points from one vehicle's data and then use that distance to compute the speed of the other vehicle. Such problems are very typical in aptitude and entrance examinations. Given Data / Assumptions: Maruti starts from point A and reaches point B in 6 hours 20 minutes. Toyota starts from point B and reaches point A in 4 hours 45 minutes. Speed of the Maruti = 60 km/h. Both cars travel along the same straight road between A and B. We assume constant speeds for both vehicles. We need to find the speed of the Toyota. Concept / Approach:The distance between A and B can be found by using the speed and time of the Maruti. Once the distance is known, the same distance can be used with the time taken by the Toyota to reach the other point to find its speed. We must convert mixed hours and minutes to pure hours for correct calculation. The formula used is distance = speed * time and then speed = distance / time. Step-by-Step Solution:Time taken by Maruti = 6 hours 20 minutes = 6 + 20 / 60 hours = 6 + 1 / 3 hours = 19 / 3 hours. Speed of Maruti = 60 km/h. Distance AB = speed * time = 60 * (19 / 3) = 20 * 19 = 380 km. Time taken by Toyota = 4 hours 45 minutes = 4 + 45 / 60 hours = 4.75 hours = 19 / 4 hours. Speed of Toyota = distance / time = 380 / (19 / 4) km/h. Speed of Toyota = 380 * 4 / 19 = 80 km/h. Verification / Alternative check:We can check the consistency by verifying that Toyota with speed 80 km/h indeed covers the 380 km distance in 4.75 hours. Compute time = 380 / 80 = 4.75 hours, which exactly matches the given time for Toyota. This confirms that the calculation of both distance and speed is correct and that 80 km/h is the correct speed. Why Other Options Are Wrong:A speed of 75 km/h or 72 km/h would require more than 4.75 hours to cover 380 km, conflicting with the given data. Speeds of 86 km/h and 90 km/h would result in shorter travel times than 4.75 hours, again contradicting the problem statement. Only 80 km/h produces the correct distance-time relationship for the Toyota car. Common Pitfalls:Common mistakes include not converting the time into fractional hours correctly or incorrectly adding or averaging the two times instead of using them separately. Another error is assuming that the two cars meet at the same time rather than using the complete journeys to compute the distance. Careful unit conversion and strict application of distance = speed * time avoid these issues. Final Answer:The speed of the Toyota is 80 km/h.

Question 3

In a 200 meter race, runner A can beat runner B by 31 meters and runner C by 18 meters. If there is a 350 meter race between C and B alone, by how many meters will C beat B?

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Introduction / Context:This is a standard athletics style aptitude question involving races and relative speeds. When one runner beats another by a certain distance in the same time, their speeds are in the ratio of the distances each covers. The problem then asks you to use this ratio to predict the outcome for a different race distance between two of the runners. Given Data / Assumptions: Race distance in the first case = 200 meters. When A finishes 200 meters, B has covered 200 - 31 = 169 meters. When A finishes 200 meters, C has covered 200 - 18 = 182 meters. The speeds of A, B, and C are proportional to the distances they cover in the same time. We need to compare C and B in a new race of 350 meters. Concept / Approach:Because A, B, and C run for the same time in the reference 200 meter race, their speeds are directly proportional to the distances they each cover in that time. Therefore, the speed ratio A:B:C is 200:169:182. For the second race, only C and B are running, and they keep these same speeds. Using the speed ratio for C and B, we calculate how far B has run when C finishes 350 meters, then subtract to find the lead of C over B. Step-by-Step Solution:Speed of B is proportional to 169 units, and speed of C is proportional to 182 units. Speed ratio B:C = 169:182. In the new race, C covers 350 meters. Let distance covered by B when C finishes 350 meters be d meters. Because speeds are proportional, d / 350 = 169 / 182. So d = 350 * 169 / 182. Compute d = (350 * 169) / 182 = 325 / ? we simplify numerically. 350 * 169 = 59150 and 59150 / 182 = 325 meters. Lead of C over B = 350 - d = 350 - 325 = 25 meters. Verification / Alternative check:We can cross check the arithmetic by noting that 182 is close to 200 and 169 is significantly smaller, making C clearly faster than B. For a longer race, C should pull further ahead. A quick approximate ratio 169 / 182 is slightly less than 0.93. Multiplying 350 by about 0.93 gives approximately 325 meters for B, leaving about 25 meters difference, which matches the exact calculation. Why Other Options Are Wrong:Differences of 21 meters, 22 meters, and 19 meters underestimate the lead that C will achieve over a 350 meter race given the significant speed difference implied by the original race. A difference of 28 meters is slightly larger than the correct value and does not match the precise proportional calculation. Only 25 meters accurately follows from the exact ratio of the speeds. Common Pitfalls:Students sometimes confuse the distances by which A beats B and C with the distances between B and C directly. Others try to scale up the given leads without using ratios, which leads to incorrect answers. The safe method is always to convert performance in one race into a speed ratio, then apply that ratio to the new race distance to find the finish positions. Final Answer:In the 350 meter race, runner C will beat runner B by 25 m.

Question 4

Two men A and B start from the same place X and walk in the same direction. A walks at 4.5 km/h and B walks at 5.75 km/h. How far apart will they be after 3.5 hours of walking?

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Introduction / Context:This question explores the concept of relative speed when two people move in the same direction. It checks if students can compute individual distances traveled, use the difference in speed to find how far apart the walkers are, and handle fractional speeds and times correctly. This type of problem is very common in time and distance sections of aptitude exams. Given Data / Assumptions: Speed of A = 4.5 km/h. Speed of B = 5.75 km/h. Both start from the same point X at the same time. They walk in the same direction for 3.5 hours. We assume constant speeds throughout. We need the distance between A and B after 3.5 hours. Concept / Approach:When two people move in the same direction, the distance between them after a certain time is based on their relative speed. The relative speed in this case is the difference between their individual speeds. The separation between them after a given time is equal to relative speed multiplied by time. It is often easier to work with decimal or fractional representations of mixed numbers. Step-by-Step Solution:Speed of A = 4.5 km/h. Speed of B = 5.75 km/h. Relative speed (B compared to A) = 5.75 - 4.5 = 1.25 km/h. Time of walking = 3.5 hours. Distance between them after 3.5 hours = relative speed * time. Distance between them = 1.25 * 3.5 km. 1.25 * 3.5 = (5 / 4) * (7 / 2) = 35 / 8 = 4.375 km. 4.375 km is equal to 4 3/8 km. Verification / Alternative check:We can also compute individual distances. In 3.5 hours, A covers 4.5 * 3.5 = 15.75 km, and B covers 5.75 * 3.5 = 20.125 km. The difference between 20.125 km and 15.75 km is 4.375 km, again equal to 4 3/8 km. This confirms that our relative speed method is correct and that the two approaches give the same result. Why Other Options Are Wrong:2 9/7 km, 3 7/5 km, 1 3/4 km, and 3 1/2 km are all distances that do not match the product of 1.25 km/h and 3.5 hours. Substituting these values backward into speed = distance / time does not give the correct relative speed of 1.25 km/h, so these options are inconsistent with the given data. Common Pitfalls:Students sometimes add the two speeds instead of subtracting them when the movement is in the same direction. Another common error is misreading 4.5 and 5.75 as whole numbers or mismanaging the fractional multiplication. Converting mixed numbers to improper fractions or decimals before performing operations helps to avoid calculation mistakes. Final Answer:After 3.5 hours, A and B are 4 3/8 km apart.

Question 5

Ramu rides his bike at an average speed of 45 km/h and reaches his destination in 4 hours. Somu covers the same distance in 6 hours. If Ramu rides at a speed that is 9 km/h less than 45 km/h and Somu rides at a speed that is 10 km/h more than 30 km/…

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Introduction / Context:This question tests understanding of average speed, total distance, and how changes in speed affect travel time. Two people travel the same distance under different conditions, and the question asks for the difference in their new travel times. Such problems help students practice using the fundamental formula time = distance / speed. Given Data / Assumptions: Ramu's original speed = 45 km/h. Ramu's travel time = 4 hours. Somu's travel time for the same distance = 6 hours. Somu's original speed = distance / time. New speed of Ramu = 45 - 9 = 36 km/h. New speed of Somu = his original speed + 10 km/h. We must find the difference between the new travel times in minutes. Concept / Approach:The distance between the two locations is the same for both riders, so we can calculate it once using any one rider's initial data. After we know the distance, we can compute new times at the changed speeds using time = distance / speed. The required difference in time is then converted from hours to minutes by multiplying by 60. Step-by-Step Solution:Distance D = Ramu's speed * time = 45 * 4 = 180 km. Somu's original speed = distance / time = 180 / 6 = 30 km/h. New speed of Ramu = 45 - 9 = 36 km/h. New speed of Somu = 30 + 10 = 40 km/h. New time for Ramu = 180 / 36 = 5 hours. New time for Somu = 180 / 40 = 4.5 hours. Difference in time = 5 - 4.5 = 0.5 hours. Convert to minutes: 0.5 * 60 = 30 minutes. Verification / Alternative check:We can quickly verify the logic by checking if the new speeds are reasonable. Ramu slows down from 45 km/h to 36 km/h, so his travel time should increase from 4 hours to 5 hours. Somu speeds up from 30 km/h to 40 km/h, so his travel time should reduce from 6 hours to 4.5 hours. The difference between 5 hours and 4.5 hours is clearly 0.5 hours or 30 minutes, confirming our calculation. Why Other Options Are Wrong:18 minutes, 27 minutes, 32 minutes, and 24 minutes do not correspond to the actual difference in travel times obtained from the correct distance and speed calculations. These values may result from arithmetic mistakes, such as incorrect computation of new speeds or using the original times instead of the new ones. Only 30 minutes matches the correct time difference. Common Pitfalls:Students sometimes forget to compute Somu's initial speed correctly and instead guess it. Another mistake is to modify times instead of speeds, or to subtract the original times rather than the new ones. A clear step-by-step calculation starting with the common distance and then finding each new time helps prevent these errors. Final Answer:The difference in their times to reach the destination under the new speeds is 30 minutes.

Question 6

A man can walk at a speed of 2 km/h and run at a speed of 6 km/h. He must cover a total distance of 20.5 km, walking exactly half the distance (10.25 km) and running the other half. How much total time will he take to complete the journey?

Accepted Answer

Introduction / Context:This question combines two different speeds over equal distances and asks for the total time taken. It is a typical time and distance problem where a person covers half the distance at one speed and the remaining half at another speed. The problem checks whether the student can correctly divide the journey, compute the time for each part using time = distance / speed, and then add those times accurately. Given Data / Assumptions: Total distance to be covered = 20.5 km. Half the distance is covered on foot (walking) and half by running. Half of 20.5 km is 10.25 km. Walking speed = 2 km/h. Running speed = 6 km/h. Speeds are constant throughout each part of the journey. Concept / Approach:When the total distance is split into two equal parts but with different speeds, we compute the time for each part separately. The formula used is time = distance / speed. The total time is the sum of the walking time and the running time. Since the question asks for the total time, we finally convert the answer from hours into hours and minutes for easier interpretation. Step-by-Step Solution:Half distance for walking = 20.5 / 2 = 10.25 km. Half distance for running = 10.25 km. Walking time = distance / speed = 10.25 / 2 hours. Walking time = 5.125 hours. Running time = distance / speed = 10.25 / 6 hours. Running time = 1.7083 hours approximately. Total time in hours = 5.125 + 1.7083 ≈ 6.8333 hours. Convert 0.8333 hours to minutes: 0.8333 * 60 ≈ 50 minutes. So total time ≈ 6 hours 50 minutes. Verification / Alternative check:We can calculate the exact value using fractions. 10.25 km is 41 / 4 km. Walking time = (41 / 4) / 2 = 41 / 8 hours. Running time = (41 / 4) / 6 = 41 / 24 hours. Total time = 41 / 8 + 41 / 24 = (123 + 41) / 24 = 164 / 24 = 41 / 6 hours. 41 / 6 hours equals 6 and 5 / 6 hours, and 5 / 6 of an hour is 50 minutes, again giving 6 hours 50 minutes. This confirms our approximate decimal calculation. Why Other Options Are Wrong:5 hours 40 minutes and 7 hours 20 minutes are too small or too large compared to the exact value of 6 hours 50 minutes. 8 hours 10 minutes and 9 hours are clearly overestimates and do not match the computed total time. Only 6 hours 50 minutes is consistent with the exact calculations based on the given speeds and distances. Common Pitfalls:A common mistake is to average the speeds (2 km/h and 6 km/h) and multiply by the total distance, which is not valid when the distances are equal but speeds differ. Another error is to split the total time instead of the total distance, which changes the structure of the problem. Carefully following time = distance / speed for each part of the journey avoids these issues. Final Answer:The man will require a total of approximately 6 hours 50 minutes to cover the 20.5 km journey.

Question 7

Two motorbikes cover the same unknown distance. One travels at 60 km/h and the other at 64 km/h. If the slower bike takes exactly 1 hour more than the faster bike, what is the distance traveled by each bike?

Accepted Answer

Introduction / Context:This question checks the understanding of how time varies with speed when the distance is fixed. Two vehicles cover the same distance at different speeds, and the relation between their travel times is given. From this relationship, we need to determine the common distance. This is a standard kind of problem in time, speed, and distance sections of aptitude exams. Given Data / Assumptions: Both motorbikes cover the same distance, say D km. Speed of slower bike = 60 km/h. Speed of faster bike = 64 km/h. Time taken by slower bike is 1 hour more than time taken by faster bike. Speeds are constant across the journey. Concept / Approach:We denote the time taken by each bike in hours and use the relation time = distance / speed. Since both cover the same distance, the difference in times comes only from their speeds. The problem gives the difference in times explicitly as 1 hour. Forming an equation based on this difference allows us to solve for the common distance D. This is an application of simple algebra with rational expressions. Step-by-Step Solution:Let D be the distance in km. Time taken by slower bike = D / 60 hours. Time taken by faster bike = D / 64 hours. Given: D / 60 = D / 64 + 1. Rearrange: D / 60 - D / 64 = 1. D * (1 / 60 - 1 / 64) = 1. 1 / 60 - 1 / 64 = (64 - 60) / (60 * 64) = 4 / 3840 = 1 / 960. So D * (1 / 960) = 1, hence D = 960 km. Verification / Alternative check:Check the travel times with D = 960 km. Time for slower bike = 960 / 60 = 16 hours. Time for faster bike = 960 / 64 = 15 hours. The difference is exactly 1 hour, which satisfies the condition in the problem. Therefore, 960 km is consistent with all given information. Why Other Options Are Wrong:For distances like 1102 km, 1060 km, 1250 km, and 840 km, the calculated times for each bike would not differ by exactly 1 hour. Substituting these values into D / 60 and D / 64 yields differences that are either less than or greater than 1 hour, so they do not satisfy the equation. Only 960 km produces the correct 1 hour difference in travel times. Common Pitfalls:Some students mistakenly add the speeds or try to average them instead of forming an equation using the time difference. Others might set the difference of speeds equal to the difference of times, which is incorrect. Always use time = distance / speed and set up the equation based on given time relations, then solve for distance or speed as required. Final Answer:The distance traveled by each motorbike is 960 km.

Question 8

Two persons start running simultaneously around a circular track of length 600 meters from the same point. Their speeds are 25 km/h and 35 km/h respectively, and they run in the same direction. After how many seconds will they meet side by side for …

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Introduction / Context:This question is about two runners moving on a circular track at different speeds, starting from the same point and running in the same direction. The problem tests the concept of relative speed and how to compute the time taken for the faster runner to gain exactly one full lap on the slower runner. Such circular track problems are standard in time and distance topics for aptitude exams. Given Data / Assumptions: Length of circular track = 600 meters. Speed of first runner = 25 km/h. Speed of second runner = 35 km/h. Both start together from the same point and move in the same direction. We need the time when they next come together side by side, anywhere on the track. Concept / Approach:When two runners start together in the same direction on a circular track, they will be side by side again when the faster runner has gained exactly one lap on the slower runner. The time to gain one lap is equal to the track length divided by the relative speed between them. We must convert speeds from km/h to m/s or track length from meters to kilometers so that the units are consistent. Step-by-Step Solution:Speed of first runner = 25 km/h. Speed of second runner = 35 km/h. Relative speed = 35 - 25 = 10 km/h. Convert relative speed to m/s: 10 km/h = (10 * 1000) / 3600 m/s. Relative speed = 10000 / 3600 m/s ≈ 2.7778 m/s. Track length = 600 meters. Time to gain one lap = distance / relative speed = 600 / 2.7778 seconds. Approximately, time ≈ 216 seconds. Verification / Alternative check:We can work directly in kilometers. Track length = 600 meters = 0.6 km. Relative speed = 10 km/h. Time in hours = distance / speed = 0.6 / 10 = 0.06 hours. Convert hours to seconds: 0.06 * 3600 = 216 seconds. This matches the previous calculation and confirms the result. Why Other Options Are Wrong:180 seconds, 144 seconds, 252 seconds, and 300 seconds do not satisfy the relationship between distance and relative speed. If we multiply the relative speed in m/s by any of these times, we do not get exactly 600 meters. Only 216 seconds leads to the correct lap distance, so the other options represent incorrect or approximate guesses. Common Pitfalls:One common mistake is to use the sum of the speeds instead of the difference when the runners are moving in the same direction. Another error is forgetting to convert units consistently, for example using km/h with meters, which gives wrong time values. Always ensure consistent units and remember that for same-direction movement, relative speed is the difference of speeds, not the sum. Final Answer:The two runners will meet side by side again for the first time after 216 seconds.

Question 9

Radha and Hema are neighbours who cycle to the same school along the same route. Radha's cycling speed is 8 km/h, and Hema's cycling speed is 10 km/h. If Hema takes exactly 9 minutes less than Radha to reach the school, how far is the school from th…

Accepted Answer

Introduction / Context:This is a straightforward time and distance problem comparing two people traveling the same route to school at different speeds. The given difference in travel times allows us to calculate the actual distance. This kind of question is very common in aptitude exams to test basic understanding of the relation time = distance / speed. Given Data / Assumptions: Radha's speed = 8 km/h. Hema's speed = 10 km/h. Both travel from the same locality to the same school. Hema takes 9 minutes less than Radha. 9 minutes = 9 / 60 hours = 0.15 hours. We assume constant speeds for both girls. Concept / Approach:Let the distance to school be D km. The time taken by each is distance divided by speed. The question gives the difference in these times. We set up an equation using the difference in times and solve for D. Then, we check which option matches this value. This process uses the fundamental formula time = distance / speed in an algebraic way. Step-by-Step Solution:Let D be the distance to the school in km. Time taken by Radha = D / 8 hours. Time taken by Hema = D / 10 hours. Hema takes 0.15 hours less, so D / 8 - D / 10 = 0.15. Compute the left side: D * (1 / 8 - 1 / 10) = D * (10 - 8) / 80 = D * (2 / 80) = D / 40. So D / 40 = 0.15. Therefore, D = 0.15 * 40 = 6 km. Verification / Alternative check:Check the times using D = 6 km. Radha's time = 6 / 8 = 0.75 hours = 45 minutes. Hema's time = 6 / 10 = 0.6 hours = 36 minutes. The difference is 45 - 36 = 9 minutes, which exactly matches the given condition. This confirms that the distance is correctly calculated. Why Other Options Are Wrong:Distances of 5 km, 5.5 km, 6.5 km, and 4.5 km do not produce a time difference of exactly 9 minutes when substituted into D / 8 and D / 10. They yield different time differences and therefore cannot satisfy the condition stated in the problem. Only 6 km produces the correct 9 minute difference in travel time. Common Pitfalls:Students sometimes mistakenly subtract speeds instead of computing the difference in times. Others may forget to convert minutes into hours properly before forming the equation. Carefully converting 9 minutes to 0.15 hours and applying the formula time = distance / speed helps avoid these errors and leads to the correct result. Final Answer:The distance from their locality to the school is 6 km.

Question 10

Tilak rides his cycle from his home to a certain place at a speed of 22 km/h and returns along the same route at a speed of 20 km/h. The time taken for the return journey is 36 minutes more than the time taken for the onward journey. What is the tot…

Accepted Answer

Introduction / Context:This question deals with a two way journey where the distances are equal but the speeds are different in each direction. The problem is designed to test a student's understanding of how changing speed affects travel time and how to work backwards from a time difference to find the actual distance. It is a common pattern in aptitude exams involving time, speed, and distance. Given Data / Assumptions: Speed from home to the place (onward journey) = 22 km/h. Speed on the return journey = 20 km/h. Time for return journey is 36 minutes more than time for onward journey. 36 minutes = 36 / 60 hours = 0.6 hours. The path is the same in both directions, so the one way distance is the same. We must find the total distance traveled, which is twice the one way distance. Concept / Approach:Let the one way distance be D km. Then, the time taken for the onward journey is D / 22 hours and the time for the return journey is D / 20 hours. The difference between these two times is given as 0.6 hours. We can set up an equation for the difference in times and solve for D. The total distance traveled is 2D = D going plus D coming back. Step-by-Step Solution:Let D be the one way distance in km. Time onward = D / 22 hours. Time return = D / 20 hours. Time difference = D / 20 - D / 22 = 0.6 hours. Factor D: D * (1 / 20 - 1 / 22) = 0.6. Compute 1 / 20 - 1 / 22 = (22 - 20) / (20 * 22) = 2 / 440 = 1 / 220. So D * (1 / 220) = 0.6, hence D = 0.6 * 220 = 132 km. Total distance traveled = 2 * D = 2 * 132 = 264 km. Verification / Alternative check:Using D = 132 km, time onward = 132 / 22 = 6 hours, and time return = 132 / 20 = 6.6 hours. The difference is 0.6 hours, which is 36 minutes. This matches the problem statement perfectly and confirms that the distances and times are consistent. Why Other Options Are Wrong:132 km is the one way distance, not the total distance, so option 132 km corresponds to half of the journey only. Values such as 134 km, 236 km, and 200 km do not satisfy the given 36 minute time difference when substituted into the time expressions. Only 264 km correctly corresponds to twice the one way distance of 132 km and respects the given condition. Common Pitfalls:Students sometimes treat 132 km as the final answer by forgetting that the question asks for the total distance. Another mistake is to incorrectly compute the difference of reciprocals of the speeds. Carefully setting up the equation for the time difference and then remembering to double the one way distance helps avoid these errors. Final Answer:The total distance traveled by Tilak (going and coming back) is 264 km.

Question 11

In an 800 meter race, Kaushik gives Piyush a head start of 200 meters. Even after this head start, Kaushik loses the race by 20 seconds. If the ratio of the speeds of Piyush and Kaushik is 3 : 2, what are the speeds of Piyush and Kaushik respectivel…

Accepted Answer

Introduction / Context:This is a more challenging race problem combining head start, time difference at the finish, and a given ratio of speeds. It tests whether students can model the situation with algebraic equations, convert word statements into mathematical relationships, and handle ratios and units correctly. These types of problems are common in higher level aptitude and quantitative exams. Given Data / Assumptions: Total race distance from the starting line = 800 meters for Kaushik. Piyush is given a head start of 200 meters, so Piyush runs only 600 meters. Kaushik loses the race by 20 seconds, meaning Kaushik finishes 20 seconds after Piyush. The ratio of speeds of Piyush to Kaushik is 3 : 2. We assume constant speeds for both runners throughout the race. We need the speeds of Piyush and Kaushik in meters per second. Concept / Approach:Let the speed of Piyush be 3k and the speed of Kaushik be 2k, in m/s, according to the given ratio. The time taken by each runner is distance divided by speed. Since Kaushik loses by 20 seconds, his time is 20 seconds more than that of Piyush. We set up an equation relating the two times and then solve for k. Once k is known, the speeds follow directly from the ratio. Step-by-Step Solution:Let speed of Piyush = 3k m/s. Let speed of Kaushik = 2k m/s. Distance run by Piyush = 600 m. Distance run by Kaushik = 800 m. Time taken by Piyush = 600 / (3k) seconds. Time taken by Kaushik = 800 / (2k) seconds. Kaushik loses by 20 seconds, so 800 / (2k) = 600 / (3k) + 20. Simplify: 800 / (2k) = 400 / k and 600 / (3k) = 200 / k. So 400 / k = 200 / k + 20. Subtract 200 / k from both sides: 200 / k = 20. Therefore k = 200 / 20 = 10. Speed of Piyush = 3k = 3 * 10 = 30 m/s. Speed of Kaushik = 2k = 2 * 10 = 20 m/s. Verification / Alternative check:Check the times with these speeds. Time of Piyush = 600 / 30 = 20 seconds. Time of Kaushik = 800 / 20 = 40 seconds. The difference is 40 - 20 = 20 seconds, which matches the condition that Kaushik loses by 20 seconds. The ratio of speeds is 30 : 20, which simplifies to 3 : 2, matching the given ratio. Therefore, the speeds are consistent with the entire problem statement. Why Other Options Are Wrong:The other speed pairs do not satisfy all conditions simultaneously. For example, 24 m/s and 16 m/s still have a ratio of 3 : 2 but do not produce the correct 20 second time difference when applied to the distances of 600 m and 800 m. Similarly, 27 m/s and 18 m/s, 36 m/s and 24 m/s, and 18 m/s and 12 m/s fail either the head start condition, the time difference, or both. Only 30 m/s and 20 m/s satisfy all given constraints. Common Pitfalls:Students may misinterpret who is faster or misread the ratio as Kaushik to Piyush instead of Piyush to Kaushik. Others forget that the race distances are different due to the head start and incorrectly use 800 meters for both runners. Careful attention to the wording of the problem and systematic use of the ratio in the equations prevents these mistakes. Final Answer:The speeds of Piyush and Kaushik are 30 m/s and 20 m/s respectively.

Question 12

Two trains start at the same time from two stations that are 200 km apart and travel towards each other on the same straight track. They cross each other at a point which is 110 km from one of the stations. What is the ratio of their speeds?

Accepted Answer

Introduction / Context:This question is a standard application of the idea that if two trains start at the same time and move towards each other, the distances they each cover before meeting are proportional to their speeds. By using the distances from the stations to the meeting point, we can directly compute the ratio of their speeds without needing the actual speeds or times. Given Data / Assumptions: Distance between the two stations = 200 km. The trains start at the same time from opposite stations and move towards each other. They cross at a point that is 110 km from one station. Therefore, the distance from the other station to the meeting point is 200 - 110 = 90 km. Both trains move at constant speeds. Concept / Approach:When two trains or vehicles start at the same time and travel towards each other, the time taken to meet is the same for both. Since speed = distance / time, and time is equal, their speeds are directly proportional to the distances they travel before they meet. Hence, the ratio of the speeds is equal to the ratio of the distances from the starting points to the meeting point. Step-by-Step Solution:Let Train 1 start from Station A and Train 2 start from Station B. Distance from Station A to meeting point = 110 km. Distance from Station B to meeting point = 200 - 110 = 90 km. Let speed of Train 1 = v1 and speed of Train 2 = v2. Time to meet is the same for both trains. So v1 / v2 = distance covered by Train 1 / distance covered by Train 2. Therefore, v1 / v2 = 110 / 90 = 11 / 9. Hence, the ratio of their speeds is 11 : 9. Verification / Alternative check:We can quickly test this logic by imagining a common meeting time t. Then v1 * t = 110 and v2 * t = 90. Dividing these gives (v1 * t) / (v2 * t) = 110 / 90, so v1 / v2 = 11 / 9. This confirms that the speed ratio is indeed equal to the ratio of distances traveled before meeting. Why Other Options Are Wrong:The ratio 9 : 20 and 11 : 20 involve 20 as a term, which does not reflect the distance ratio of 110 and 90. The answer 20 : 11 is the reverse of what comes directly from the distances and does not represent the correct train ordering based on how many kilometers each travels before the meeting. Therefore, these options are incorrect, and only 11 : 9 matches the distance ratio. Common Pitfalls:Students sometimes mistakenly invert the ratio or subtract distances instead of taking the ratio. Another common error is to misread the question and assume that 110 km is the distance from both stations, which is impossible. Remember that in such problems, the speeds are always proportional to the distances covered by each train in the same time before they meet. Final Answer:The ratio of the speeds of the two trains is 11 : 9.

Question 13

A student walks from home towards school. If the student walks at 4 meters per minute, arrival is 4 minutes late. If the student walks at 6 meters per minute, arrival is 30 minutes early. How far is the school from the student's home, in meters?

Accepted Answer

Introduction / Context:This is a classic punctuality problem involving two different walking speeds and early or late arrival times. The question tests the ability to set up equations using distance, speed, and time, while carefully interpreting what early and late mean in terms of time differences relative to a fixed schedule. Such problems appear frequently in quantitative aptitude tests. Given Data / Assumptions: Walking speed in the first case = 4 meters per minute. Walking speed in the second case = 6 meters per minute. At 4 m/min, the student is 4 minutes late for school. At 6 m/min, the student is 30 minutes early. There is a fixed scheduled arrival time, and we assume a fixed starting time from home in both cases. We assume straight line travel and constant speeds in both scenarios. Concept / Approach:Let D be the distance to school in meters and T be the ideal time (in minutes) that would result in exactly on time arrival at school. At 4 m/min, the actual travel time is T + 4 minutes. At 6 m/min, the travel time is T - 30 minutes. We can express both travel times as distance divided by speed and then solve the resulting system of two equations for D and T. Step-by-Step Solution:Let D be the distance in meters and T be the ideal on time travel duration in minutes. Case 1: At 4 m/min, student is 4 minutes late, so travel time = T + 4. Thus, D = 4 * (T + 4). Case 2: At 6 m/min, student is 30 minutes early, so travel time = T - 30. Thus, D = 6 * (T - 30). Equate distances: 4 * (T + 4) = 6 * (T - 30). Expand: 4T + 16 = 6T - 180. Rearrange: 16 + 180 = 6T - 4T, so 196 = 2T. Therefore, T = 98 minutes. Substitute back: D = 4 * (T + 4) = 4 * (98 + 4) = 4 * 102 = 408 meters. Verification / Alternative check:Check both scenarios with D = 408 meters. At 4 m/min, time = 408 / 4 = 102 minutes, which is 4 minutes more than T = 98 minutes, so the student is 4 minutes late. At 6 m/min, time = 408 / 6 = 68 minutes, which is 30 minutes less than 98 minutes, meaning the student arrives 30 minutes early. Both conditions are satisfied, confirming that the distance is correct. Why Other Options Are Wrong:Distances such as 360 m, 392 m, 420 m, and 450 m do not satisfy both the early and late conditions simultaneously. Substituting any of these values leads to times that are not separated by exactly 4 minutes late and 30 minutes early with respect to the same ideal time T. Only 408 m fits the equations for both speeds and timing conditions. Common Pitfalls:Students may confuse early and late conditions, treating both as positive time differences. Another frequent error is mixing minutes and seconds or forgetting that both journeys start at the same clock time. Properly defining an ideal on time duration and writing separate equations for each speed scenario is the most reliable way to solve these problems. Final Answer:The distance from the student's home to the school is 408 m.

Question 14

Person K is 50% faster than person L. L starts a journey at 9:00 a.m. from one point, and K starts at 10:00 a.m. from another point such that L and K are initially 300 km apart. L travels at 50 km/h, and both travel towards each other in opposite di…

Accepted Answer

Introduction / Context:This question deals with two travelers starting at different times from locations that are a fixed distance apart and moving towards each other. It examines the ability to handle different starting times, compute speeds from percentage relationships, and determine the meeting time using relative speed. Such problems are common in quantitative aptitude tests focusing on time, speed, and distance. Given Data / Assumptions: L starts at 9:00 a.m. K starts at 10:00 a.m. Initial distance between them = 300 km. L travels at 50 km/h. K is 50% faster than L, so K's speed = 1.5 * 50 km/h = 75 km/h. They travel towards each other in opposite directions along a straight route. Concept / Approach:First, we compute how far L travels alone in the one hour between 9:00 a.m. and 10:00 a.m. This reduces the distance between them. After 10:00 a.m., both K and L move towards each other, and we consider their combined (relative) speed. The time taken for them to meet from 10:00 a.m. onward is equal to remaining distance divided by the sum of their speeds. Finally, we add this time to 10:00 a.m. to get the actual meeting time. Step-by-Step Solution:Speed of L = 50 km/h. Speed of K = 1.5 * 50 = 75 km/h. Distance between them at 9:00 a.m. = 300 km. From 9:00 a.m. to 10:00 a.m., only L travels. Distance L covers in 1 hour = 50 * 1 = 50 km. Remaining distance at 10:00 a.m. = 300 - 50 = 250 km. From 10:00 a.m., both travel towards each other. Relative speed = 50 + 75 = 125 km/h. Time to meet from 10:00 a.m. = remaining distance / relative speed = 250 / 125 = 2 hours. Meeting time = 10:00 a.m. + 2 hours = 12:00 p.m. (noon). Verification / Alternative check:We can validate by computing their positions at the meeting time. L travels from 9:00 a.m. to 12:00 p.m., which is 3 hours, covering 3 * 50 = 150 km. K travels from 10:00 a.m. to 12:00 p.m., which is 2 hours, covering 2 * 75 = 150 km. Combined, they cover 150 + 150 = 300 km, which equals the initial separation. This confirms that they meet exactly at 12:00 p.m. Why Other Options Are Wrong:Meeting times of 11:00 a.m., 11:30 a.m., 1:00 p.m., or 2:00 p.m. do not match the distances covered given the stated speeds and initial separation. If they met earlier than 12:00 p.m., the joint distance would be less than 300 km; if they met later, the joint distance would exceed 300 km. Only 12:00 p.m. matches all conditions. Common Pitfalls:Some students forget to account for the one hour head start of L and apply relative speed from 9:00 a.m., which gives an incorrect meeting time. Others misinterpret 50% faster and add 50 km/h instead of multiplying by 1.5. Carefully computing the head start distance and then using the sum of speeds from the second start time is the correct approach. Final Answer:K and L will meet at 12 p.m. (noon).

Question 15

Jagan travels from his city to another town in three stages. He covers 240 km by car at 60 km/h, then 400 km by train at 100 km/h, and finally 200 km by bus at 50 km/h. What is his average speed for the entire 840 km journey?

Accepted Answer

Introduction / Context:This is a typical average speed problem over multiple segments of a journey, where each segment has a different speed and distance. The question tests whether students understand that average speed is total distance divided by total time, not the simple arithmetic mean of individual speeds. This concept is crucial in many quantitative reasoning problems involving journeys with varying speeds. Given Data / Assumptions: Segment 1: 240 km by car at 60 km/h. Segment 2: 400 km by train at 100 km/h. Segment 3: 200 km by bus at 50 km/h. Total distance = 240 + 400 + 200 km. Speeds are constant within each segment. We ignore stoppage times and assume continuous travel. Concept / Approach:Average speed for the entire trip is defined as total distance divided by total time. We first compute the time taken for each segment using time = distance / speed, then sum these times to get total time. Once total distance and total time are known, the average speed is calculated as total distance / total time. This approach correctly weights each speed by the distance or time spent at that speed. Step-by-Step Solution:Total distance = 240 + 400 + 200 = 840 km. Time for car segment = 240 / 60 = 4 hours. Time for train segment = 400 / 100 = 4 hours. Time for bus segment = 200 / 50 = 4 hours. Total time = 4 + 4 + 4 = 12 hours. Average speed = total distance / total time = 840 / 12 km/h. Average speed = 70 km/h. Verification / Alternative check:Because each segment takes exactly 4 hours, Jagan spends equal time at speeds of 60 km/h, 100 km/h, and 50 km/h. The distance based weighted average is therefore equal to the time based weighted average. Another quick check is to multiply the average speed 70 km/h by the total time 12 hours to get 840 km, which matches the total distance, confirming the correctness of the answer. Why Other Options Are Wrong:Values like 72 km/h, 64 km/h, 36 km/h, or 35 km/h arise if one incorrectly averages the speeds directly or mistakenly adds and divides by the number of segments without accounting for distances and times. These do not satisfy the relationship average speed = total distance / total time for the given journey. Only 70 km/h correctly reproduces the total distance when multiplied by total time. Common Pitfalls:Students often incorrectly compute the average speed by simply averaging 60, 100, and 50, which gives (60 + 100 + 50) / 3 = 70 km/h in this special case, but that works here only because each segment takes the same amount of time. In general, this approach can be wrong if time segments are not equal. Remember that the definition of average speed is always total distance divided by total time. Final Answer:The average speed for the entire journey is 70 km/h.

Question 16

An express train travels at an average speed of 100 km/h and stops for 4 minutes after every 75 km. How long does it take the train to reach a destination that is 600 km away from the starting point, including all stoppages?

Accepted Answer

Introduction / Context:This question involves a train that travels with a given average running speed but includes regular stops. The goal is to compute the total journey time including these stoppages. Problems like this require careful calculation of both running time and stopping time and are common in aptitude tests dealing with practical applications of speed, distance, and time. Given Data / Assumptions: Running speed of express train = 100 km/h. The train stops for 4 minutes after every 75 km traveled. Total distance to destination = 600 km. We assume no stop is counted after the final arrival at 600 km. Speeds between stops are constant. Concept / Approach:First, we compute the running time without considering stops by dividing the total distance by the running speed. Then we determine how many 75 km segments fit into 600 km and how many stops are made before the final destination. For each such segment (except after the final arrival), we add 4 minutes of stoppage time. The total time is the sum of running time and stoppage time, which is finally expressed in hours and minutes. Step-by-Step Solution:Running time without stops = total distance / speed = 600 / 100 = 6 hours. Number of 75 km segments in 600 km = 600 / 75 = 8. The train stops after each 75 km except after the final arrival. So number of stops = 8 - 1 = 7 stops. Stopping time per stop = 4 minutes. Total stopping time = 7 * 4 = 28 minutes. Total journey time = 6 hours running + 28 minutes stopping. Total time = 6 hours 28 minutes. Verification / Alternative check:We can check that the number of stops is correct by listing the points at which the train stops: after 75 km, 150 km, 225 km, 300 km, 375 km, 450 km, and 525 km. After reaching 600 km, the train has reached its destination and no further stop is needed for this calculation. There are indeed 7 intermediate stops, giving a total of 28 minutes stoppage time. Why Other Options Are Wrong:6 hours ignores the stoppage time completely. 6 hours 24 minutes and 6 hours 32 minutes correspond to fewer or more stops than actually occur. 7 hours 21 minutes is far too large and does not correspond to the given stopping pattern. Only 6 hours 28 minutes correctly accounts for 6 hours of running and 28 minutes of stoppage time. Common Pitfalls:Some students count a stop after the final 600 km segment, leading to 8 stops instead of 7. Others divide 600 by 75 incorrectly or forget to convert 4 minutes to a fraction of an hour when working entirely in hours. Carefully counting intermediate stops and handling units consistently avoids these mistakes. Final Answer:The express train takes 6 hours 28 minutes to reach the destination, including all stops.

Question 17

Robert travels by cycle towards a point A. He will reach A at 2:00 p.m. if he cycles at 10 km/h, and he will reach A at 12:00 noon if he cycles at 15 km/h. At what constant speed must he cycle in order to reach A exactly at 1:00 p.m.?

Accepted Answer

Introduction / Context:This question is a classic example of using two scenarios with different speeds and arrival times to deduce the total distance and starting time. Once the distance is known, we can find the speed needed to arrive at a new specified time. Problems of this type are common in aptitude tests and help reinforce the relation between distance, speed, and time in real life scheduling situations. Given Data / Assumptions: If Robert cycles at 10 km/h, he reaches A at 2:00 p.m. If he cycles at 15 km/h, he reaches A at 12:00 noon. The starting time from his home is the same in both cases. The distance from his home to point A is fixed and does not change. We assume straight line travel and constant speeds. Concept / Approach:Let D be the distance from Robert's home to A, and let t10 and t15 be the travel times at 10 km/h and 15 km/h respectively. We can express D as 10 * t10 and also as 15 * t15. The difference between t10 and t15 is 2 hours because 2:00 p.m. is 2 hours after 12:00 noon. Solving this system we get both D and the starting time. Once the distance is known, we compute the time available to arrive at 1:00 p.m. and then find the required speed using speed = distance / time. Step-by-Step Solution:Let t10 be the travel time at 10 km/h. Let t15 be the travel time at 15 km/h. Distance D = 10 * t10 = 15 * t15. Arrival at 2:00 p.m. and 12:00 noon from the same start implies t10 - t15 = 2 hours. From D = 10 * t10 and D = 15 * t15, we have 10 * t10 = 15 * t15. Thus, t10 = (3 / 2) * t15. Substitute into t10 - t15 = 2: (3 / 2) * t15 - t15 = 2. This simplifies to (1 / 2) * t15 = 2, so t15 = 4 hours. Then t10 = t15 + 2 = 6 hours. Distance D = 10 * t10 = 10 * 6 = 60 km. If Robert wants to arrive at 1:00 p.m., the travel time from his start must be 5 hours (since he starts 6 hours before 2:00 p.m., that is at 8:00 a.m.). Required speed = distance / time = 60 / 5 = 12 km/h. Verification / Alternative check:We can verify the starting time. At 10 km/h, Robert takes 6 hours and reaches at 2:00 p.m., so he must have started at 8:00 a.m. At 15 km/h, he takes 4 hours and reaches at 12:00 noon, also starting at 8:00 a.m. This is self consistent. To arrive at 1:00 p.m. starting from 8:00 a.m., he has 5 hours of travel time. Traveling 60 km in 5 hours clearly requires a speed of 12 km/h, which confirms the answer. Why Other Options Are Wrong:Speeds like 20 km/h, 18 km/h, 16 km/h, or 14 km/h would not allow Robert to both cover the 60 km distance and arrive exactly at 1:00 p.m. For example, at 20 km/h, he would arrive much earlier, and at 16 km/h he would not arrive on time. Only 12 km/h exactly matches the distance and required travel time. Common Pitfalls:Some students incorrectly average the two given speeds or the two arrival times instead of solving the equations properly. Others forget that the difference between arrival times reflects the difference between the two travel durations. Clearly writing the equations for distance and time in each scenario avoids these mistakes and leads to the correct required speed. Final Answer:Robert must travel at a constant speed of 12 km/h to reach point A at 1:00 p.m.

Question 18

Time, speed and distance: If Hema walks at 12 km/h instead of 8 km/h and would then cover 20 km more in the same time, what is the actual distance (in km) that she really walked at 8 km/h?

Accepted Answer

Introduction / Context:Questions of this type test the core relationship between speed, distance and time. When a person changes speed but walks for the same duration, the difference in distances covered directly reflects the difference in speeds. This is a standard time, speed and distance concept that appears frequently in aptitude exams. Given Data / Assumptions: Actual walking speed of Hema = 8 km/h. Hypothetical walking speed of Hema = 12 km/h. Time of walking in both situations is the same. At the higher speed she would walk 20 km more than at the lower speed. Concept / Approach:We use the basic formula distance = speed * time. If the time is the same, the difference in distance is equal to the difference in speed multiplied by that common time. Once we find the time, we can compute the actual distance covered at 8 km/h. Step-by-Step Solution: Let the common time of walking be t hours. Distance at 8 km/h = 8 * t km. Distance at 12 km/h = 12 * t km. Given that the faster option would cover 20 km more: 12 * t - 8 * t = 20. Simplify: 4 * t = 20. So t = 20 / 4 = 5 hours. Actual distance at 8 km/h = 8 * 5 = 40 km. Verification / Alternative check:If Hema walked 5 hours at 8 km/h, she covers 40 km. At 12 km/h for 5 hours, she would cover 60 km. The difference is 60 - 40 = 20 km, which matches the given condition, so the solution is consistent. Why Other Options Are Wrong: 32 km: Would imply a much shorter time and would not give a 20 km difference when speed is raised to 12 km/h. 30 km: The difference between distances at 8 km/h and 12 km/h would not be exactly 20 km. 46 km: Leads to a non-integer or inconsistent time value when back-calculated. Common Pitfalls:Candidates sometimes equate the ratio of speeds directly with the ratio of distances without considering the extra distance. Another common error is to treat 20 km as the total distance and not as the difference between the two scenarios. Forgetting that the time remains the same in both cases leads to wrong equations. Final Answer:The actual distance Hema walked at 8 km/h is 40 km.

Question 19

A train 250 m long crosses a bridge 150 m long in 20 seconds. What is the speed of the train in km/h?

Accepted Answer

Introduction / Context:This question tests understanding of how to compute speed when a train crosses a bridge. The key idea is that the train must cover its own length plus the length of the bridge. After obtaining the speed in metres per second, we convert it to kilometres per hour, which is a common conversion in aptitude questions on time and distance. Given Data / Assumptions: Length of the train = 250 m. Length of the bridge = 150 m. Total time to cross the bridge completely = 20 seconds. Speed is assumed to be uniform. Concept / Approach:When a train crosses a bridge, the distance covered before it has completely cleared the bridge is equal to the sum of the length of the train and the length of the bridge. Then we use the formula speed = distance / time. Finally, we convert the result from m/s to km/h using the standard factor 18/5. Step-by-Step Solution: Total distance to be covered = length of train + length of bridge. Total distance = 250 m + 150 m = 400 m. Time taken = 20 s. Speed in m/s = distance / time = 400 / 20 = 20 m/s. To convert m/s to km/h, multiply by 18/5. Speed in km/h = 20 * (18 / 5) = 20 * 3.6 = 72 km/h. Verification / Alternative check:We can reverse check: at 72 km/h, speed in m/s is 72 * 5 / 18 = 20 m/s. In 20 seconds, distance covered = 20 * 20 = 400 m, which matches the sum of the train and bridge lengths. Therefore, the calculation is correct. Why Other Options Are Wrong: 80 km/h: Gives speed in m/s as 80 * 5 / 18 which would cover more than 400 m in 20 s. 70 km/h: Corresponds to a slightly smaller speed and would lead to less than 400 m in 20 s. 92 km/h: Is far higher and clearly exceeds the required distance in the given time. Common Pitfalls:Many candidates forget to add both lengths and only use the train length, which underestimates the speed. Another frequent mistake is converting between m/s and km/h incorrectly, for example by multiplying with 5/18 instead of 18/5 when going from m/s to km/h. Final Answer:The speed of the train is 72 km/h.

Question 20

Two trains running in opposite directions cross a pole in 47 seconds and 31 seconds respectively. If they cross each other completely in 33 seconds, what is the ratio of their speeds?

Accepted Answer

Introduction / Context:This time and distance problem involves two trains with different speeds and lengths. You are given their times to cross a pole and the time taken to cross each other. The question tests your ability to form equations linking length, speed and time, and then derive the ratio of their speeds using algebraic manipulation. Given Data / Assumptions: Train 1 crosses a pole in 47 seconds. Train 2 crosses a pole in 31 seconds. They run in opposite directions. They cross each other completely in 33 seconds. Speeds of trains are constant. Concept / Approach:Let v1 and v2 be the speeds (in m/s) and L1 and L2 be the lengths (in metres) of trains 1 and 2. When a train crosses a pole, its own length is covered, so L1 = v1 * 47 and L2 = v2 * 31. When they cross each other in opposite directions, the distance covered is L1 + L2 and the relative speed is v1 + v2, so (L1 + L2) / (v1 + v2) = 33. Substituting the lengths in terms of speeds allows finding the ratio v1 : v2. Step-by-Step Solution: L1 = v1 * 47. L2 = v2 * 31. Time to cross each other: 33 s, so (L1 + L2) / (v1 + v2) = 33. Substitute: (47 * v1 + 31 * v2) / (v1 + v2) = 33. Cross multiply: 47 * v1 + 31 * v2 = 33 * v1 + 33 * v2. Rearrange: (47 - 33) * v1 = (33 - 31) * v2. 14 * v1 = 2 * v2, so v1 / v2 = 2 / 14 = 1 / 7. Therefore, the ratio of speeds of train 1 and train 2 is 1 : 7. Verification / Alternative check:You can take any convenient value satisfying v1 : v2 = 1 : 7, for example v1 = 1 m/s and v2 = 7 m/s, and compute all times again to see that the three conditions hold. This confirms that the derived ratio is consistent. Why Other Options Are Wrong: 7 : 1 implies the faster and slower trains are interchanged and contradicts the derived equation. 2 : 3 and 3 : 2 do not satisfy (47 * v1 + 31 * v2) / (v1 + v2) = 33 when substituted in proportional speeds. Common Pitfalls:A common mistake is to assume that time to cross each other is the average of 47 and 31 seconds, which is not correct. Another error is forgetting that when two bodies move in opposite directions, their relative speed is the sum of their individual speeds, not the difference. Final Answer:The required ratio of the speeds of the two trains is 1 : 7.

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