Question 1

A jeep travels from one place to another in 6 hours on the forward journey. On the return journey its speed is increased by 12 km/h and it takes only 4 hours. What is the distance (in km) travelled by the jeep each way?

Accepted Answer

Introduction / Context: This question examines understanding of how distance, speed and time relate when a vehicle travels the same route at two different speeds. The jeep covers the same distance going forward and coming back, but with different speeds and times. Using these relationships, we can form an equation and compute the distance travelled each way. Given Data / Assumptions: Time taken in the forward journey = 6 hours. Time taken in the return journey = 4 hours. Return speed is 12 km/h more than the forward speed. The distance from the starting point to the destination is the same for both journeys. Concept / Approach: Let the one way distance be D km. Then forward speed is D / 6 km/h and return speed is D / 4 km/h. The problem states that the difference between these speeds is 12 km/h. Setting up the equation (D / 4) - (D / 6) = 12 allows us to solve for D directly. Step-by-Step Solution: Let the distance between the two points be D km. Forward speed = D / 6 km/h. Return speed = D / 4 km/h. Given: return speed - forward speed = 12 km/h. So, D / 4 - D / 6 = 12. Take LCM of 4 and 6 which is 12: (3D - 2D) / 12 = 12. Therefore, D / 12 = 12. So D = 12 * 12 = 144 km. Verification / Alternative check: Forward speed = 144 / 6 = 24 km/h. Return speed = 144 / 4 = 36 km/h. The increase in speed is 36 - 24 = 12 km/h, which matches the given condition, and times 6 hours and 4 hours are consistent with the same distance of 144 km. So the solution is verified. Why Other Options Are Wrong: 244 km, 344 km, and 444 km all give different speed differences when divided by 6 and 4 hours. None of them produce a difference of exactly 12 km/h. Common Pitfalls: Some learners mistakenly treat 6 hours and 4 hours as separate distances rather than the same distance. Others add the speeds instead of subtracting them to form the equation. Mismanaging the LCM or arithmetic when solving D / 4 - D / 6 = 12 can also lead to incorrect results. Final Answer: The distance travelled by the jeep each way is 144 km.

Question 2

Person K cycles from point A to point B at 10 km/h and returns from B to A at 9 km/h. Person L cycles both ways at 12 km/h. Over the complete trip, L takes 10 minutes less than K. What is the distance (in km) between A and B?

Accepted Answer

Introduction / Context: This is a classic average speed and time comparison problem where two people travel the same route with different speeds. By comparing total times for round trips, we can find the distance between two points. The problem focuses on forming time equations from distance and speed, then using the given time difference to solve for the distance. Given Data / Assumptions: K goes from A to B at 10 km/h and from B to A at 9 km/h. L goes from A to B and back at 12 km/h in both directions. L takes 10 minutes (that is 10 / 60 = 1 / 6 hour) less than K for the round trip. Both travel the same distance between A and B. Concept / Approach: Let the one way distance be d km. Then the total time for K is d / 10 + d / 9 hours. The total time for L is 2d / 12 hours. The difference between these total times is 1 / 6 hour. Setting up this equation and solving for d gives the required distance. Step-by-Step Solution: Let distance A to B be d km. Time taken by K = d / 10 + d / 9 = (9d + 10d) / 90 = 19d / 90 hours. Time taken by L = 2d / 12 = d / 6 hours. Given that K takes 1 / 6 hour more: 19d / 90 - d / 6 = 1 / 6. Convert d / 6 to /90: d / 6 = 15d / 90. So 19d / 90 - 15d / 90 = 1 / 6. This gives 4d / 90 = 1 / 6. Simplify: 2d / 45 = 1 / 6 so 2d = 45 / 6. Therefore, d = 45 / 12 = 15 / 4 = 3.75 km. Verification / Alternative check: Time taken by K = 19 * 3.75 / 90 hours = 0.7917 hours (approximately 47.5 minutes). Time taken by L = 3.75 / 6 = 0.625 hours (37.5 minutes). The difference is about 10 minutes, which matches the condition. So the distance is correct. Why Other Options Are Wrong: 3.25 km, 4.00 km and 4.50 km all give time differences that are not equal to 10 minutes when substituted into the time equations. Common Pitfalls: Candidates sometimes average speeds instead of using time formulas, which is incorrect for round trips. Another common error is forgetting to convert 10 minutes to hours or mishandling fractional arithmetic, leading to an incorrect distance. Final Answer: The distance between points A and B is 3.75 km.

Question 3

Laxmi walks northwards at 20 km/h and Prasanna walks southwards at 30 km/h, starting from the same point at the same time. How far apart (in km) will they be after 60 minutes?

Accepted Answer

Introduction / Context: This problem is a straightforward application of relative speed when two people move in opposite directions. It checks if you correctly understand that when two objects move away from each other, the rate at which the distance between them increases is the sum of their individual speeds. Given Data / Assumptions: Laxmi walks northwards at 20 km/h. Prasanna walks southwards at 30 km/h. They start from the same point at the same time. Time interval considered = 60 minutes = 1 hour. Concept / Approach: When two people move in exactly opposite directions from the same starting point, their separation grows at the sum of their speeds. Hence, the relative speed between them is 20 km/h + 30 km/h. Multiplying this relative speed by the time gives the distance between them after the given time. Step-by-Step Solution: Speed of Laxmi = 20 km/h. Speed of Prasanna = 30 km/h. Relative speed (moving in opposite directions) = 20 + 30 = 50 km/h. Time of travel = 60 minutes = 1 hour. Distance between them after 1 hour = relative speed * time. Distance = 50 * 1 = 50 km. Verification / Alternative check: You can also compute separate distances: in 1 hour, Laxmi walks 20 km north, and Prasanna walks 30 km south. The total separation is 20 km + 30 km = 50 km, matching the previous calculation. Why Other Options Are Wrong: 24 km and 30 km ignore that both are moving and only consider one person's distance. 42 km is just a random value and does not correspond to the sum of the two distances after 1 hour. Common Pitfalls: A common mistake is to subtract the speeds instead of adding them when objects move in opposite directions. Another is to forget converting time correctly if it is not already in hours. Here, time is already 1 hour, so the calculation is simple. Final Answer: After 60 minutes, Laxmi and Prasanna are 50 km apart.

Question 4

Train A passes a stationary train B in 24 seconds and passes a pole in 9 seconds. If the speed of train A is 48 km/h, what is the length (in metres) of train B?

Accepted Answer

Introduction / Context: This question involves finding the length of a stationary train using information about how a moving train passes a pole and passes the stationary train. It uses the basic time, speed and distance relationship and the idea that when a train passes an object, the distance covered depends on what is being passed (a pole, a platform or another train). Given Data / Assumptions: Speed of train A = 48 km/h. Time taken by train A to pass a pole = 9 seconds. Time taken by train A to pass stationary train B completely = 24 seconds. Train B is stationary on the track. Concept / Approach: First convert the speed of train A from km/h to m/s. Using the time to cross a pole, we can find the length of train A. Then, using the time taken to pass train B, we can find the combined length of both trains. Subtracting the length of train A from the combined length gives the length of train B. Step-by-Step Solution: Speed of train A in m/s = 48 * 1000 / 3600 = 48 * 5 / 18 = 40 / 3 m/s. When passing a pole, distance covered = length of train A. Length of train A = speed * time = (40 / 3) * 9 = 120 m. When passing train B, distance covered = length of A + length of B. Total distance in 24 s = (40 / 3) * 24 = 320 m. So length of B = 320 - 120 = 200 m. Verification / Alternative check: Check by reversing: length of A = 120 m, length of B = 200 m. When A passes B, total distance is 320 m. At speed 40 / 3 m/s, time = 320 / (40 / 3) = 24 s, matching the given time. So the computed length of train B is correct. Why Other Options Are Wrong: 160 m and 180 m are both smaller than the correct value and do not satisfy the combined length calculation. 145 m also does not produce the given 24 seconds when substituted back. Common Pitfalls: Some learners forget to convert km/h to m/s, which leads to incorrect lengths. Others mistakenly consider only one train when computing the distance for the 24 second interval, rather than the sum of both trains' lengths. Careful unit conversion and understanding of what distance is covered in each scenario are crucial. Final Answer: The length of the stationary train B is 200 m.

Question 5

Two cyclists P and Q cycle towards each other along a straight road at 20 km/h and 25 km/h respectively. If they meet after 56 minutes, what was the distance (in km) between them when they started?

Accepted Answer

Introduction / Context: This is a direct application of relative speed when two objects move towards each other. The distance between them decreases at the sum of their speeds. By combining their speeds and using the time until they meet, we can easily compute the initial separation between the two cyclists. Given Data / Assumptions: Speed of cyclist P = 20 km/h. Speed of cyclist Q = 25 km/h. They ride towards each other. They meet after 56 minutes. Speeds are constant and along a straight line. Concept / Approach: When two bodies move towards each other, their relative speed is the sum of their individual speeds. The distance between them at the start equals relative speed multiplied by the time until they meet. Time must be converted into hours to match the speed units of km/h. Step-by-Step Solution: Speed of P = 20 km/h. Speed of Q = 25 km/h. Relative speed = 20 + 25 = 45 km/h. Time until they meet = 56 minutes = 56 / 60 hour = 14 / 15 hour. Distance between them = relative speed * time. Distance = 45 * (14 / 15) = 3 * 14 = 42 km. Verification / Alternative check: You can compute how far each cyclist travels: P travels 20 * (14 / 15) = 280 / 15 km, and Q travels 25 * (14 / 15) = 350 / 15 km. The sum of these is (280 + 350) / 15 = 630 / 15 = 42 km, matching the computed distance. Why Other Options Are Wrong: 36 km, 39 km and 45 km do not match the product of 45 km/h and 14 / 15 hour. If used, they produce a meeting time differing from 56 minutes. Common Pitfalls: Some students mistakenly subtract speeds when objects move towards each other, which is incorrect. Others forget to convert 56 minutes into hours, leading to unit mismatch. Keeping units consistent and remembering that for opposite directions we add speeds is key. Final Answer: The initial distance between the two cyclists was 42 km.

Question 6

A train covers 325 km in 3.5 hours on one journey and 470 km in 4 hours on another journey. What is the average speed (in km/h) of the train for the combined 795 km?

Accepted Answer

Introduction / Context: This question checks your understanding of average speed over multiple segments where distances and times differ. Average speed is not the simple average of two speeds but rather the total distance divided by the total time taken. This is a very common trap in aptitude tests, so carefully aggregating distances and times is important. Given Data / Assumptions: Distance in first journey = 325 km, time = 3.5 hours. Distance in second journey = 470 km, time = 4 hours. Total distance is the sum of both segments. Train speed on each segment is constant but may differ between segments. Concept / Approach: Average speed for a complete trip with varying speeds is defined as total distance divided by total time. Therefore, we add the distances and add the times, then divide. We do not average individual segment speeds directly, because each speed applies over a different distance or time. Step-by-Step Solution: First distance = 325 km. Second distance = 470 km. Total distance = 325 + 470 = 795 km. First time = 3.5 hours. Second time = 4 hours. Total time = 3.5 + 4 = 7.5 hours. Average speed = total distance / total time. Average speed = 795 / 7.5 = 106 km/h. Verification / Alternative check: We can observe that 7.5 * 100 = 750, so 7.5 * 106 = 795, confirming that dividing 795 by 7.5 correctly gives 106. This confirms that the average speed value is consistent. Why Other Options Are Wrong: 107 km/h is slightly higher than the correct value and does not satisfy distance = average speed * total time. 72 km/h and 76 km/h are much lower than what is consistent with covering 795 km in only 7.5 hours. Common Pitfalls: Many candidates average the individual speeds of each leg instead of using total distance and total time. Others may incorrectly handle the 3.5 hour representation, forgetting that 3.5 means three and a half hours, not three hours and fifty minutes. Being precise with decimal hours and the definition of average speed prevents such mistakes. Final Answer: The average speed of the train for the combined journey is 106 km/h.

Question 7

You drive from your home to a store at 20 km/h and return by the same route at 30 km/h, ignoring the time spent in the store. What is your average speed (in km/h) for the entire round trip?

Accepted Answer

Introduction / Context: This problem focuses on average speed for a round trip with different speeds in each direction. Even though you might be tempted to take the simple average of 20 km/h and 30 km/h, the correct method is to use the harmonic mean when distances are equal. The question evaluates whether you properly apply the total distance over total time concept. Given Data / Assumptions: Speed from home to store = 20 km/h. Speed from store back to home = 30 km/h. Distance from home to store is the same as distance from store to home. Time spent at the store is ignored. Concept / Approach: Let the one way distance be d km. Time taken going is d / 20 hours and time taken coming back is d / 30 hours. The total distance is 2d and total time is d / 20 + d / 30. The average speed is then (total distance) / (total time). For equal distances, this simplifies to the harmonic mean: (2 * v1 * v2) / (v1 + v2). Step-by-Step Solution: Let distance one way be d km. Time going = d / 20 hours. Time returning = d / 30 hours. Total distance = 2d km. Total time = d / 20 + d / 30 = (3d + 2d) / 60 = 5d / 60 = d / 12 hours. Average speed = total distance / total time = 2d / (d / 12) = 2d * 12 / d = 24 km/h. Verification / Alternative check: Using harmonic mean directly: 2 * 20 * 30 / (20 + 30) = 1200 / 50 = 24 km/h. This matches the previous calculation and confirms the correctness of the result. Why Other Options Are Wrong: 25 km/h and 26 km/h are close to the arithmetic mean but do not use the correct formula for average speed over equal distances. 22 km/h is simply an arbitrary value and does not satisfy 2d = average speed * total time when computed. Common Pitfalls: The most common error is to calculate (20 + 30) / 2 = 25 km/h and assume that is the correct average speed. This ignores the fact that the time spent at each speed is different. Remember that average speed depends on the total distance and total time, not just a simple mean of the speeds. Final Answer: The average speed for the round trip is 24 km/h.

Question 8

Ali and Faizer are 27 m apart. When they walk in the same direction, they meet after 9 hours. When they walk in opposite directions, they meet after 3 hours. What is the walking speed of Ali (in m/h)?

Accepted Answer

Introduction / Context: This problem tests understanding of relative speed in both same-direction and opposite-direction motion, with a twist that the speeds are in metres per hour and the initial separation is small compared to the time. The key idea is to form equations for the relative motion in both scenarios and solve for the individual speeds. Given Data / Assumptions: Initial distance between Ali and Faizer = 27 m. They meet after 9 hours when walking in the same direction. They meet after 3 hours when walking in opposite directions. Speeds of Ali and Faizer are constant and in m/h. Concept / Approach: Let Ali's speed be a m/h and Faizer's speed be f m/h. When they move in the same direction, the relative speed is |a - f| and they close the 27 m gap in 9 hours, so |a - f| = 27 / 9 = 3 m/h. When they move in opposite directions, the relative speed is a + f and they meet in 3 hours, so a + f = 27 / 3 = 9 m/h. Solving these two equations gives Ali's speed. Step-by-Step Solution: From opposite directions: a + f = 9. From same direction: |a - f| = 3. Assume Ali is faster, so a - f = 3. Add the two equations: (a + f) + (a - f) = 9 + 3. This gives 2a = 12, so a = 6 m/h. Substitute back: 6 + f = 9, so f = 3 m/h. Verification / Alternative check: Same direction: relative speed = a - f = 6 - 3 = 3 m/h. Time to meet = distance / relative speed = 27 / 3 = 9 hours, matches given. Opposite directions: relative speed = a + f = 6 + 3 = 9 m/h. Time to meet = 27 / 9 = 3 hours, also matches. So the values are consistent. Why Other Options Are Wrong: 9 m/h or 12 m/h as Ali's speed would not satisfy both meeting times when solving the equations. 3 m/h is actually Faizer's speed, not Ali's, under the assumption that Ali walks faster. Common Pitfalls: Some candidates mistakenly add speeds in the same direction instead of taking the difference. Others confuse which person is faster, but the question asks for Ali's speed, and the equations clearly show that a must be the larger one for consistent solutions. Keeping track of relative speeds in different directions is crucial. Final Answer: The walking speed of Ali is 6 m/h.

Question 9

Excluding stoppages, the average speed of a bus is 60 km/h. Including stoppages, the average speed drops to 40 km/h. For how many minutes per hour does the bus stop?

Accepted Answer

Introduction / Context: This question is about effective average speed when a vehicle stops intermittently. It checks understanding that average speed including stoppages is lower than the running speed and that the difference arises from time spent not moving. The problem asks you to interpret average speed values to compute total stoppage time in one hour of the schedule. Given Data / Assumptions: Running average speed without stoppages = 60 km/h. Overall average speed including stoppages = 40 km/h. We consider a 1 hour schedule (60 minutes). The bus either moves at 60 km/h or is stopped; there are no other speeds. Concept / Approach: In one hour of clock time, the bus covers 40 km on average. When it is moving at 60 km/h, the distance covered in the moving time is 60 * T hours, where T is the time in motion. Since this must equal 40 km, we solve 60 * T = 40 to find the moving time. The stoppage time is the remainder of the hour. Step-by-Step Solution: Let T be the time (in hours) for which the bus is moving in one hour of schedule. Distance covered while moving: 60 * T km. Overall average is 40 km/h, so in one hour the bus covers 40 km. Equate the distances: 60 * T = 40. So T = 40 / 60 = 2 / 3 hour. Total time in one hour = 1 hour. Stoppage time = 1 - 2 / 3 = 1 / 3 hour. Convert 1 / 3 hour to minutes: (1 / 3) * 60 = 20 minutes. Verification / Alternative check: If the bus moves 40 minutes at 60 km/h, distance covered is 60 * (40 / 60) = 40 km. For the remaining 20 minutes it stands still. Total time is 60 minutes, total distance is 40 km, so the overall average speed is 40 km/h, exactly as given. Why Other Options Are Wrong: 2 minutes, 30 minutes and 40 minutes do not satisfy the equation 60 * T = 40 when converted back to hours and tested. Common Pitfalls: Students sometimes try to use ratios of speeds directly without interpreting them as distances over time. Another common mistake is to assume linear proportionality between speed and stoppage time without forming an equation. Carefully equating the distance from the running speed with the distance from overall average speed avoids these errors. Final Answer: The bus stops for 20 minutes in every hour of its schedule.

Question 10

An 800 m long train is moving at 63 km/h and a man is walking at 3 km/h in the same direction along the track. How many seconds will the train take to completely cross the man?

Accepted Answer

Introduction / Context: This question tests understanding of relative speed in the same direction between a fast-moving train and a walking man. The train must cover its entire length relative to the man in order to completely cross him. The problem requires correct use of relative speed, unit conversion from km/h to m/s and then finding the time using distance and speed. Given Data / Assumptions: Length of train = 800 m. Speed of train = 63 km/h. Speed of man = 3 km/h. Both are moving in the same direction. Speeds are constant. Concept / Approach: When two objects move in the same direction, their relative speed is the difference between their speeds. The effective speed at which the train draws away from the man is (63 - 3) km/h. We convert this relative speed to m/s and then use time = distance / speed, where the distance is the length of the train, because the train must cover its own length relative to the man to cross him completely. Step-by-Step Solution: Relative speed = 63 - 3 = 60 km/h. Convert 60 km/h to m/s: 60 * 1000 / 3600 = 60 * 5 / 18 = 50 / 3 m/s. Distance to be covered relative to the man = length of train = 800 m. Time taken = distance / relative speed. Time = 800 / (50 / 3) = 800 * 3 / 50 = 2400 / 50 = 48 seconds. Verification / Alternative check: At a relative speed of 50 / 3 m/s, in 48 seconds the distance covered = (50 / 3) * 48 = 50 * 16 = 800 m, exactly the length of the train. This confirms that 48 seconds is the correct time for complete crossing. Why Other Options Are Wrong: 52, 38 and 42 seconds all give distances different from 800 m when multiplied by the relative speed of 50 / 3 m/s. Common Pitfalls: Common mistakes include adding the speeds instead of subtracting them when motion is in the same direction, and incorrect conversion between km/h and m/s. Some candidates also forget that the whole length of the train has to pass the man, not just its front end, leading to incorrect use of distance. Final Answer: The train will take 48 seconds to completely cross the man.

Question 11

A person travels from town K to town L at 50 km/h and returns from L to K by increasing his speed by 50 percent. What is his average speed (in km/h) for the entire journey?

Accepted Answer

Introduction / Context: This is a standard round trip average speed problem where the speeds in the two directions are different. The forward speed is given, and the return speed is a percentage increase over the forward speed. The question tests whether you can compute the average speed over equal distances correctly using the harmonic mean idea or by explicitly using total distance and total time. Given Data / Assumptions: Speed from K to L = 50 km/h. Return speed from L to K = 50 percent more than 50 km/h. Distances from K to L and L to K are equal. Speeds are constant on each leg of the journey. Concept / Approach: Find the return speed first. A 50 percent increase means multiplying 50 km/h by 1.5, giving 75 km/h. Let the one way distance be d km. Time going is d / 50 and time returning is d / 75. The total distance is 2d and total time is d / 50 + d / 75. Average speed is then (total distance) / (total time). For equal distances, this is equivalent to the harmonic mean: 2 * v1 * v2 / (v1 + v2). Step-by-Step Solution: Forward speed v1 = 50 km/h. Return speed v2 = 50 * 1.5 = 75 km/h. Using harmonic mean for equal distances: Average speed = 2 * v1 * v2 / (v1 + v2). Average speed = 2 * 50 * 75 / (50 + 75). Average speed = 7500 / 125 = 60 km/h. Verification / Alternative check: Let the one way distance be 150 km. Time going = 150 / 50 = 3 hours. Time returning = 150 / 75 = 2 hours. Total distance = 300 km, total time = 5 hours. Average speed = 300 / 5 = 60 km/h, which matches the harmonic mean result. Why Other Options Are Wrong: 55 km/h, 58 km/h and 66 km/h do not equal 2 * 50 * 75 / 125 and do not match total distance divided by total time when any reasonable distance is assumed. Common Pitfalls: Some students mistakenly average the two speeds arithmetically, (50 + 75) / 2 = 62.5 km/h, which is incorrect for equal distances. Others compute the percentage increase incorrectly or confuse it with a percentage of the final speed. Always remember that average speed over equal distances must be calculated from the total distance and total time, not by simply averaging the two speeds. Final Answer: The average speed for the entire journey is 60 km/h.

Question 12

Joel travels the first 3 hours of his journey at 60 miles per hour and the remaining 5 hours at 24 miles per hour. What is his average speed for the entire 8 hour journey, expressed in km/h (take 1 mile ≈ 1.6 km)?

Accepted Answer

Introduction / Context: This problem combines average speed over different time segments with a unit conversion from miles per hour to kilometres per hour. It checks if you can compute total distance using speeds and times in miles, then convert the overall average speed into km/h using the given conversion factor. Given Data / Assumptions: First part: speed = 60 mph for 3 hours. Second part: speed = 24 mph for 5 hours. Total journey time = 3 + 5 = 8 hours. Conversion: 1 mile ≈ 1.6 km. Concept / Approach: First compute the total distance travelled in miles using distance = speed * time for each segment. Then sum up the distances. Average speed in mph is total distance divided by total time. Finally, convert this average speed from mph to km/h by multiplying with 1.6. Step-by-Step Solution: Distance in first 3 hours = 60 * 3 = 180 miles. Distance in next 5 hours = 24 * 5 = 120 miles. Total distance = 180 + 120 = 300 miles. Total time = 8 hours. Average speed in mph = total distance / total time = 300 / 8 = 37.5 mph. Convert to km/h using 1 mile ≈ 1.6 km. Average speed in km/h = 37.5 * 1.6 = 60 km/h. Verification / Alternative check: Compute total distance directly in km: 180 miles ≈ 288 km, and 120 miles ≈ 192 km, so total ≈ 480 km. Average speed = 480 km / 8 hours = 60 km/h, which agrees with the previous result. This cross-check confirms the correctness. Why Other Options Are Wrong: 38 km/h and 40 km/h underestimate the total distance based on the actual speeds. 54 km/h also does not equate to 480 km over 8 hours and arises from incorrect arithmetic or conversion. Common Pitfalls: Common mistakes include averaging the two speeds without considering the different times spent at each speed, or converting speeds incorrectly between mph and km/h. Another error is computing distance in km for one segment only and then mixing units. Always keep units consistent and compute total distance and time before finding average speed. Final Answer: The average speed of Joel's travel is 60 km/h.

Question 13

Two trains are running in the same direction at 30 km/h and 58 km/h respectively. A man sitting in the slower train takes 18 seconds to completely pass the faster train. What is the length (in metres) of the faster train?

Accepted Answer

Introduction / Context: This question again involves relative speed, but now both trains are moving in the same direction and a person is sitting in the slower train. The man sees the faster train pass by, which effectively measures how long it takes to cross the length of the faster train at the relative speed between the two trains. Given Data / Assumptions: Speed of slower train = 30 km/h. Speed of faster train = 58 km/h. Direction of motion: same direction. Time for a man in slower train to pass the faster train = 18 seconds. The man's speed equals the speed of the slower train. Concept / Approach: In the frame of the man, the faster train appears to move past him at the relative speed between the two trains, which is their speed difference when they move in the same direction. The distance that needs to be covered relative to the man is the length of the faster train. Thus, length of faster train = relative speed * time, using m/s for the speed. Step-by-Step Solution: Relative speed = 58 - 30 = 28 km/h. Convert 28 km/h to m/s: 28 * 1000 / 3600 = 28 * 5 / 18 = 140 / 18 = 7.777... m/s. Time taken to pass faster train = 18 seconds. Length of faster train = relative speed * time. Length ≈ 7.777... * 18 = 140 m. Verification / Alternative check: Compute exactly using fraction: relative speed = 28 * 1000 / 3600 = 700 / 90 = 70 / 9 m/s. Multiply by 18 s: (70 / 9) * 18 = 70 * 2 = 140 m. This confirms the length of the faster train is exactly 140 m. Why Other Options Are Wrong: 105 m, 115 m and 120 m are all smaller and do not satisfy length = (70 / 9) * 18 when recalculated. Common Pitfalls: Some candidates add the speeds instead of subtracting them, which is correct only when trains move in opposite directions. Others forget the unit conversion from km/h to m/s, leading to serious calculation errors. It is also important to realise that the entire length of the faster train must pass the observer for the crossing to be complete. Final Answer: The length of the faster train is 140 m.

Question 14

Sabrina travels part of a journey on foot at 8 km/h and the rest by bicycle at 13 km/h. She covers a total distance of 84 km in 8 hours. For how many hours does she travel by bicycle?

Accepted Answer

Introduction / Context: This is a typical mixture of speeds problem where a person travels different parts of a journey at different constant speeds but the total time and total distance are known. The question checks your ability to form simultaneous equations from distance and time relationships and then solve for the time spent at each speed. Given Data / Assumptions: Speed on foot = 8 km/h. Speed on bicycle = 13 km/h. Total distance traveled = 84 km. Total time taken = 8 hours. Speeds are constant in each mode of travel. Concept / Approach: Let t1 be the time in hours spent walking, and t2 be the time spent cycling. Then t1 + t2 = 8 hours. Distance while walking is 8 * t1 and distance while cycling is 13 * t2. The total distance is 84 km, so 8 * t1 + 13 * t2 = 84. Solving these two equations gives the values of t1 and t2, and the question specifically asks for t2, the cycling time. Step-by-Step Solution: Let t1 = time on foot (hours), t2 = time on bicycle (hours). Equation 1: t1 + t2 = 8. Equation 2: 8 * t1 + 13 * t2 = 84. From equation 1, t1 = 8 - t2. Substitute into equation 2: 8 * (8 - t2) + 13 * t2 = 84. Compute: 64 - 8 * t2 + 13 * t2 = 84. Simplify: 64 + 5 * t2 = 84. So 5 * t2 = 20, giving t2 = 4 hours. Verification / Alternative check: Then t1 = 8 - 4 = 4 hours. Distance walking = 8 * 4 = 32 km. Distance cycling = 13 * 4 = 52 km. Total distance = 32 + 52 = 84 km and total time = 4 + 4 = 8 hours, so all given conditions are satisfied. Why Other Options Are Wrong: 5, 3 or 2 hours do not satisfy both the total time and total distance conditions when substituted into the equations. Common Pitfalls: Some students try to average speeds without considering the different times spent at each speed. Others may incorrectly assume equal distances at each speed and use a simpler formula, which does not fit the data here. Keeping track of both distance and time equations is essential. Final Answer: Sabrina travels by bicycle for 4 hours.

Question 15

Two cars leave a house 10 minutes apart and travel along the same road at 20 km/h in the same direction. A woman is walking towards the house along the same road. If she meets the two cars at an interval of 8 minutes, with what speed (in km/h) is sh…

Accepted Answer

Introduction / Context: This is a relative speed problem involving two cars leaving the same house at different times and a woman walking towards the house from the opposite direction. The key idea is that the time gap between the woman meeting the two cars is compressed compared to the starting time gap of the cars, because all are moving. By comparing these intervals, we can deduce the woman's walking speed. Given Data / Assumptions: Two cars leave the house 10 minutes (1 / 6 hour) apart. Both cars travel at 20 km/h in the same direction away from the house. A woman walks towards the house from the opposite direction at constant speed v km/h. The woman meets the two cars at an interval of 8 minutes (2 / 15 hour). Concept / Approach: Let t1 be the time when the woman meets the first car and t2 when she meets the second car, measured from the time the first car leaves the house. The cars are separated in time by 10 minutes when leaving, but they are seen by the woman at an 8 minute interval. Using relative motion equations for positions of car and woman at t1 and t2 and the given time difference, we can eliminate the unknown distance and solve directly for v, the woman's speed. Step-by-Step Solution: Let the woman's speed be v km/h. Let distance of the woman from the house at t = 0 be x km. Position of woman at time t: x - v * t (towards house). Position of first car at time t: 20 * t (away from house). They meet when x - v * t1 = 20 * t1. Second car starts at t = 1 / 6 hour, position at time t: 20 * (t - 1 / 6). They meet when x - v * t2 = 20 * (t2 - 1 / 6). Also, t2 - t1 = 2 / 15 hour (8 minutes). From the two position equations, eliminating x gives (2 / 15) * (v + 20) = 10 / 3. Solving gives v + 20 = 25, so v = 5 km/h. Verification / Alternative check: Assuming v = 5 km/h, the relative speed between each car and the woman when they move towards each other is 20 + 5 = 25 km/h. The change from 10 minutes departure gap to 8 minutes meeting gap is consistent with the derived equation. A more detailed algebraic verification reproduces the 5 km/h value exactly. Why Other Options Are Wrong: 4, 6 or 7 km/h do not satisfy the relationship between the 10 minute departure interval and the 8 minute meeting interval when used in the equations of relative motion. Common Pitfalls: Because there are three moving objects, many students attempt to guess based on proportions rather than writing equations. Another common error is to treat the 8 minutes as the same as the 10 minutes, ignoring the effect of the woman's motion. Carefully forming and simplifying the equations avoids these traps. Final Answer: The woman must walk at a speed of 5 km/h.

Question 16

Two cars travel from city A to city B at speeds of 36 km/h and 54 km/h respectively. One car takes 3.5 hours less than the other to cover the journey. What is the distance (in km) between city A and city B?

Accepted Answer

Introduction / Context: This time and distance problem involves two cars traveling the same route at different constant speeds. The time difference between their arrivals is known, and the speeds are known, but the distance is not. The question tests your ability to use the relationship between time, speed and distance and set up an equation using the difference in times. Given Data / Assumptions: Speed of slower car = 36 km/h. Speed of faster car = 54 km/h. Both travel from city A to city B over the same distance. The faster car takes 3.5 hours less than the slower car. Time difference = 3.5 hours = 7 / 2 hours. Concept / Approach: Let the distance between the two cities be D km. Time taken by the slower car is D / 36 hours, and time taken by the faster car is D / 54 hours. The slower car takes 7 / 2 hours more than the faster car. This gives the equation D / 36 - D / 54 = 7 / 2. Solving this equation yields the value of D. Step-by-Step Solution: Let distance between A and B be D km. Time taken by slower car = D / 36 hours. Time taken by faster car = D / 54 hours. Given: D / 36 - D / 54 = 7 / 2. Take LCM of 36 and 54 which is 108. Rewrite: (3D - 2D) / 108 = 7 / 2. So D / 108 = 7 / 2. Therefore, D = 108 * 7 / 2 = 54 * 7 = 378 km. Verification / Alternative check: If distance is 378 km, time for slower car = 378 / 36 = 10.5 hours. Time for faster car = 378 / 54 = 7 hours. Time difference = 10.5 - 7 = 3.5 hours, which matches the given condition exactly. Thus the distance is consistent. Why Other Options Are Wrong: 454 km, 567 km and 302 km produce time differences different from 3.5 hours when divided by 36 and 54 respectively. Common Pitfalls: Some students mistakenly add the times or set up the equation with reversed signs. Others mis-handle the fraction 3.5 hours or its conversion into 7 / 2. Accurate algebraic manipulation and careful use of LCM avoid these issues. Final Answer: The distance between city A and city B is 378 km.

Question 17

Ruchir walks at 20 km/h and Rukma cycles at 25 km/h directly towards each other. If they start at the same time and meet after 48 minutes, what was the distance between them at the start of the journey?

Accepted Answer

Introduction / Context: This question tests the fundamental concept of relative speed in time and distance problems. When two people move directly towards each other, their speeds add up, and this combined speed helps us find the initial distance between them. Such problems are very common in aptitude exams and are excellent for revising the link between speed, distance, and time. Given Data / Assumptions: Ruchir walks at a constant speed of 20 km/h. Rukma cycles at a constant speed of 25 km/h towards Ruchir. They start at the same time and meet after 48 minutes. Both are assumed to move in a straight line without stopping or changing speed. We are asked to find the distance between them at the moment they started moving. Concept / Approach: The key idea is the formula: distance = speed * time. When two objects move towards each other, the effective or relative speed is the sum of their individual speeds. Once we know the relative speed and the time taken to meet, we can directly compute the starting distance. We must also convert the time from minutes to hours so that the units match the speed in km/h. Step-by-Step Solution: Step 1: Relative speed = speed of Ruchir + speed of Rukma.Relative speed = 20 km/h + 25 km/h = 45 km/h.Step 2: Convert 48 minutes to hours: 48 minutes = 48 / 60 hours.48 / 60 = 4 / 5 hours.Step 3: Use distance = speed * time with the relative speed.Distance = 45 * (4 / 5) km.Distance = 45 * 4 / 5 = 180 / 5 = 36 km.So the initial distance between Ruchir and Rukma was 36 km. Verification / Alternative check: We can check by working backward. If the distance is 36 km, in 4 / 5 hours, Ruchir covers 20 * 4 / 5 = 16 km and Rukma covers 25 * 4 / 5 = 20 km. The total covered is 16 + 20 = 36 km, which matches the starting separation. Hence the calculation is consistent and confirms the answer. Why Other Options Are Wrong: 54 km would require a much larger time or higher speeds. At a combined speed of 45 km/h, 54 km would need 54 / 45 hours, which is 1.2 hours or 72 minutes, not 48 minutes. 48 km would require 48 / 45 hours, which is more than 1 hour, again not equal to 48 minutes. 30 km would mean 30 / 45 hours, that is 40 minutes, which is less than 48 minutes. 42 km would need 42 / 45 hours, which is 56 minutes, not 48 minutes. Common Pitfalls: Students sometimes subtract the speeds instead of adding them when two people move towards each other. Another common mistake is to forget converting minutes to hours before using the speed in km/h, which leads to incorrect results. Also, some learners directly pick an option that feels roughly close without checking with the exact formula, which is risky in competitive exams. Final Answer: The initial distance between Ruchir and Rukma was 36 km.

Question 18

Two men start from the same point P, where the first man runs from 11:00 a.m. with rest breaks, and a second man starts at 2:00 p.m. and runs non stop at 15 km/h towards him. At what time in the afternoon does the second man catch the first man?

Accepted Answer

Introduction / Context: This problem combines uniform speed motion with a repeated pattern of running and resting. It tests your understanding of how to track positions over time when one person moves in cycles and another moves continuously. It is a typical time and distance catch up scenario seen in competitive examinations. Given Data / Assumptions: The first man starts running from point P at 11:00 a.m. His running speed is 10 km/h. He runs for 2 hours, then rests for 1 hour, and repeats this run rest cycle. The second man starts from the same point P at 2:00 p.m. and runs non stop at 15 km/h. Both men run in the same direction along a straight path. We need to find the clock time in p.m. at which the second man catches the first man. Concept / Approach: We treat the motion piecewise. For the first man, we compute how far he has reached at different times according to the 2 hour running and 1 hour resting pattern. For the second man, the speed is constant, so his distance from P is simply speed multiplied by time after 2:00 p.m. They meet when both have covered the same distance from P. Setting these distances equal allows us to solve for the meeting time. Step-by-Step Solution: Step 1: From 11:00 a.m. to 1:00 p.m., the first man runs for 2 hours at 10 km/h, so distance = 10 * 2 = 20 km.Step 2: From 1:00 p.m. to 2:00 p.m., he rests, so his distance stays 20 km from P.Step 3: At 2:00 p.m., the second man starts at P. At that moment the first man is 20 km ahead.Step 4: From 2:00 p.m. to 4:00 p.m., the first man is again in a 2 hour running phase, so his speed remains 10 km/h.Step 5: Let t be the time in hours after 2:00 p.m. when the second man catches the first man, during this 2:00 to 4:00 running period.Step 6: Distance of first man from P at time t after 2:00 p.m. = 20 + 10 * t.Step 7: Distance of second man from P at the same moment = 15 * t.Step 8: At meeting, distances are equal, so 20 + 10 * t = 15 * t.Step 9: Rearranging: 20 = 5 * t, so t = 4 hours.Step 10: Add 4 hours to 2:00 p.m., we get 6:00 p.m. This seems to contradict the earlier computed pattern, so we must carefully place the meeting inside the correct segment.Step 11: Actually, we need to check the activity pattern around 4:00 p.m. The first man runs 11 to 1, rests 1 to 2, runs 2 to 4, and then rests 4 to 5.Step 12: If t = 4 hours after 2:00 p.m., that is 6:00 p.m., which lies outside the 2:00 to 4:00 running period, so our assumption that he is running during the full catch up interval is wrong.Step 13: Instead, calculate the positions at 4:00 p.m. First man has run 11 to 1, rested 1 to 2, and run 2 to 4, so total running time is 4 hours, giving distance 40 km from P.Step 14: At 4:00 p.m., second man has been running from 2:00 to 4:00, that is 2 hours at 15 km/h, so distance = 30 km.Step 15: From 4:00 to 5:00 p.m., the first man rests and stays at 40 km. The second man continues at 15 km/h.Step 16: Let u be the time in hours after 4:00 p.m. when the second man catches the first man. At that time, his distance from P is 30 + 15 * u and the first man remains at 40 km.Step 17: Set 30 + 15 * u = 40. This gives 15 * u = 10, so u = 10 / 15 = 2 / 3 hours.Step 18: 2 / 3 hours is 40 minutes. So the meeting time is 4:00 p.m. plus 40 minutes, that is 4:40 p.m. Verification / Alternative check: At 4:40 p.m., the time from 11:00 a.m. is 5 hours and 40 minutes, or 17 / 3 hours. The first man has run for 4 hours and rested for 1 hour and 40 minutes in total. His distance from P is 40 km. The second man has run from 2:00 p.m. to 4:40 p.m., which is 2 hours and 40 minutes or 8 / 3 hours. His distance at 15 km/h is 15 * 8 / 3 = 40 km. The positions match, confirming the solution. Why Other Options Are Wrong: 6:20 p.m. and 6:00 p.m. would place the meeting too late, when the second man has already gone far beyond 40 km while the first man is still near his earlier position. The equations for distances no longer balance there. 5:30 p.m. is also incorrect, because at that time the second man distance minus the first man distance does not equal zero when computed using their exact running and resting pattern. 5:00 p.m. is simply a midpoint guess and does not satisfy the distance equality either. Common Pitfalls: Many learners try to average speeds or ignore the rest periods, which leads to incorrect times. Another common mistake is to assume a single linear equation without checking which phase of the run rest cycle the first man is in. Carefully tracking time segments and checking whether the person is running or resting at each stage is essential for solving such problems correctly. Final Answer: The second man catches the first man at 4:40 p.m.

Question 19

A man walks in an increasing pattern, covering 1 km on the first day, 2 km on the second day, 3 km on the third day, and so on. What is the total distance he covers in 10 days?

Accepted Answer

Introduction / Context: This question checks your understanding of arithmetic progressions and summation of consecutive natural numbers. The daily walking distances form a simple increasing sequence where each term increases by a fixed amount. Recognizing this pattern helps you use a direct formula instead of adding each day manually. Given Data / Assumptions: Day 1 distance = 1 km. Day 2 distance = 2 km. Day 3 distance = 3 km, and the pattern continues with an increase of 1 km per day. The man follows this pattern for 10 consecutive days. We assume there are no breaks and no changes in the pattern during these 10 days. Concept / Approach: The distances walked each day form an arithmetic progression with first term a = 1, common difference d = 1, and number of terms n = 10. The sum of the first n natural numbers is a well known result: n * (n + 1) / 2. Using this formula makes the calculation fast and accurate. Applying it correctly is a common skill expected in aptitude exams. Step-by-Step Solution: Step 1: List the daily distances: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 km.Step 2: Recognize that this is the sequence of the first 10 natural numbers.Step 3: Use the formula for the sum of the first n natural numbers: Sum = n * (n + 1) / 2.Step 4: Substitute n = 10: Sum = 10 * (10 + 1) / 2.Step 5: Sum = 10 * 11 / 2 = 110 / 2.Step 6: Therefore, Sum = 55 km.Step 7: This is the total distance walked in 10 days. Verification / Alternative check: You can check by pairing terms from the beginning and end of the sequence: (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6). Each pair sums to 11, and there are 5 such pairs, so total distance = 5 * 11 = 55 km. This alternative method agrees with the formula based approach and confirms that the total distance is correct. Why Other Options Are Wrong: 40 km and 45 km underestimate the sum because they correspond to stopping the pattern earlier than 10 days or miscalculating the series. 50 km arises if someone mistakenly uses 10 * 10 / 2 or forgets to add the last term correctly, which is a common error. 58 km is an overestimate and does not match the sum of the exact daily distances from 1 to 10. Common Pitfalls: A common mistake is to assume that since the last term is 10 km, the average is 10 / 2 = 5 and then incorrectly multiply by 10 without following the proper formula. Another issue is adding the numbers quickly in the mind and missing one or more terms. Remembering and correctly applying the arithmetic progression sum formula is a reliable way to avoid these errors. Final Answer: The man covers a total distance of 55 km in 10 days.

Question 20

In a straight line, 37 trees are planted at equal spacing. A car takes 20 seconds to travel from the first tree to the 13th tree. How many additional seconds will the car take to reach the last (37th) tree from the first tree?

Accepted Answer

Introduction / Context: This problem focuses on constant speed motion over equally spaced points. The key idea is that if distance between consecutive points is equal and the speed of the car is constant, the travel time is directly proportional to the number of gaps between those points. Counting gaps instead of points is a crucial skill for such questions. Given Data / Assumptions: There are 37 trees in a straight line. The distance between any two consecutive trees is the same. The car takes 20 seconds to go from the first tree to the 13th tree. The car continues with the same constant speed and direction. We need the additional time required to go from the 13th tree position to the last 37th tree, expressed as extra seconds beyond the initial 20 seconds. Concept / Approach: When consecutive points are equally spaced, the number of intervals (gaps) between them is one less than the number of points. Time taken is proportional to distance, and distance is proportional to the number of gaps. Therefore, we can compare the ratios of gaps to find the ratio of times. This removes the need to know the exact physical distance between trees or the actual speed of the car. Step-by-Step Solution: Step 1: From the first tree to the 13th tree, the car passes through 13 − 1 = 12 equal gaps.Step 2: Time for these 12 gaps is given as 20 seconds.Step 3: From the first tree to the 37th tree, the car passes through 37 − 1 = 36 equal gaps.Step 4: The ratio of total gaps is 36 / 12 = 3.Step 5: Since speed is constant, time ratio is the same as the gap ratio, so total time from first to 37th tree = 3 * 20 seconds.Step 6: Total time = 60 seconds.Step 7: The question asks how much more time is needed beyond the 20 seconds already taken to reach the 13th tree.Step 8: Extra time = 60 − 20 = 40 seconds. Verification / Alternative check: We can calculate the time per gap. Time for 12 gaps is 20 seconds, so time per gap = 20 / 12 seconds. For 36 gaps, time = 36 * (20 / 12) = 3 * 20 = 60 seconds. Subtracting the initial 20 seconds still gives 40 seconds. This confirms that the extra time calculation is consistent with the proportional reasoning method. Why Other Options Are Wrong: 36 seconds corresponds to incorrectly assuming there are only 18 additional gaps or misusing the ratio of points instead of gaps. 57 seconds and 44 seconds come from arithmetic slips or from mixing up the total time with the extra time. 60 seconds would be the total time for the journey, not the additional time after the first 20 seconds, so it does not answer the question as stated. Common Pitfalls: A frequent mistake is to count the number of trees instead of the number of spaces between them. Another common error is to give the total time instead of the requested extra time. Carefully reading phrases like "how much more time" or "additional time" is essential. Also, some students try to assume a particular distance between trees, which is not needed and can cause confusion. Final Answer: The car takes an additional 40 seconds to reach the last tree from the 13th tree.

Time and Distance Questions