Question 1

A man travels from his village to the post office at 25 km/h and then walks back from the post office to the village at 4 km/h. If the total time for the journey to and from the post office is 5 hours 48 minutes, what is the distance between the vil…

Accepted Answer

Introduction / Context: This is a round trip time and distance problem with different speeds in the two directions. The total time for going and returning is given, and we must find the one way distance. These types of questions are common in aptitude tests and require setting up a clear equation using the basic formula distance = speed * time. Given Data / Assumptions: - Speed from village to post office = 25 km/h. - Speed from post office back to village = 4 km/h. - Total time for the complete journey = 5 hours 48 minutes. - One way distance between village and post office = D km (unknown). - Motion is in a straight line with constant speeds in each direction. Concept / Approach: We treat the outward and return journeys separately, writing each travel time as distance divided by speed. The sum of these two travel times is given as 5 hours 48 minutes. Converting minutes to hours and solving the resulting equation gives the one way distance D. This is a straightforward application of algebra and unit conversion. Step-by-Step Solution: Let the distance between village and post office be D km. Time from village to post office = D / 25 hours. Time from post office back to village = D / 4 hours. Total time = D / 25 + D / 4 hours. Given total time = 5 hours 48 minutes = 5 + 48 / 60 = 5.8 hours. So we have D / 25 + D / 4 = 5.8. Take common denominator 100: (4D + 25D) / 100 = 5.8. So 29D / 100 = 5.8. Therefore D = 5.8 * 100 / 29 = 20 km. Verification / Alternative check: Using D = 20 km, time outward at 25 km/h = 20 / 25 = 0.8 hours (48 minutes). Time return at 4 km/h = 20 / 4 = 5 hours. Total time = 5 + 0.8 = 5.8 hours, which is 5 hours 48 minutes. This matches the given total time exactly, confirming that D = 20 km is correct. Why Other Options Are Wrong: 40 km, 30 km, and 10 km: Substituting any of these into the expressions for travel times will not give a total of 5 hours 48 minutes. For example, 40 km would already take 10 hours on the return trip alone at 4 km/h, which is too large. Common Pitfalls: Students sometimes average the two speeds and multiply by half the total time, which is incorrect because the times in each direction are not equal. Others forget to convert 48 minutes into 0.8 hours, leading to wrong numerical results. Always treat each direction separately and add the times after converting all units consistently. Final Answer: The distance between the village and the post office is 20 km.

Question 2

Karthik walks to his destination at 5 km/h and reaches 8 minutes late. If he walks at 6 km/h, he reaches exactly on time. What is the distance he has to cover (in km)?

Accepted Answer

Introduction / Context: This problem is very similar in style to other early and late arrival questions. Karthik travels the same distance at two different walking speeds, and his arrival time relative to the schedule changes. We use this information to find the distance to his destination, applying the basic relation between distance, speed, and time. Given Data / Assumptions: - Distance to the destination = D km (unknown). - Scheduled (on time) travel time = T hours (unknown). - At 5 km/h, Karthik is 8 minutes late, so time taken = T + 8/60 hours. - At 6 km/h, he reaches on time, so time taken = T hours. - Motion is along a straight path with constant speeds. Concept / Approach: We write two expressions for the distance D using the two different speeds and the corresponding travel times. Because the distance is the same in both cases, we equate the expressions and solve for the scheduled time T. Once we have T, we can compute the distance D using the speed at which he arrives on time. Step-by-Step Solution: Let D be the distance and T be the scheduled time. At 5 km/h: D = 5 * (T + 8/60). At 6 km/h: D = 6 * T. Equate the two distances: 5 (T + 8/60) = 6 T. Convert 8 minutes to hours: 8 / 60 = 2 / 15. So 5 (T + 2 / 15) = 6 T. Expand: 5T + 10 / 15 = 6 T. Simplify 10 / 15 = 2 / 3. So 5T + 2 / 3 = 6 T. Rearrange: 2 / 3 = 6 T - 5 T = T. Therefore T = 2 / 3 hours = 40 minutes. Distance D = 6 * T = 6 * (2 / 3) = 4 km. Verification / Alternative check: With D = 4 km: At 6 km/h, time taken = 4 / 6 hours ≈ 0.6667 hours, which is 40 minutes, so this matches the scheduled time. At 5 km/h, time taken = 4 / 5 hours = 0.8 hours, which is 48 minutes. The difference between 48 minutes and 40 minutes is 8 minutes, so he is 8 minutes late at 5 km/h. Both conditions are satisfied. Why Other Options Are Wrong: 10 km, 8 km, and 6 km: For each, if you compute the travel times at 5 km/h and 6 km/h, you will not get the exact combination of 8 minutes late at 5 km/h and exactly on time at 6 km/h. Only 4 km fits both conditions at once. Common Pitfalls: A common error is to try to average the speeds or directly subtract times without writing any equations. Another mistake is forgetting that 8 minutes must be converted to 2 / 15 hours when working in km/h. Always set up the distance equality for the two speeds and carefully handle the unit conversions. Final Answer: Karthik covers a distance of 4 km to reach his destination on time at 6 km/h.

Question 3

The distance from town A to town B is 5 miles, and town C is 6 miles from town B. Which of the following could be the maximum possible distance (in miles) between towns A and C?

Accepted Answer

Introduction / Context: This question tests understanding of the triangle inequality in geometry applied to distances between three towns. If you consider the towns as points on a map, the distances between them must satisfy certain constraints. We are asked about the maximum possible distance between two of the towns when the other two distances are known. Given Data / Assumptions: - Distance from town A to town B = 5 miles. - Distance from town B to town C = 6 miles. - Distance from town A to town C = AC miles (unknown). - The three towns can form a straight line or a triangle, as long as distance rules are satisfied. Concept / Approach: For any three points A, B, and C, the distance AC must satisfy the triangle inequality: the distance between any two points cannot exceed the sum of the other two distances, and cannot be less than the absolute difference of those distances. That is, |AB - BC| ≤ AC ≤ AB + BC. To find the maximum possible AC, we use the upper bound AB + BC. Step-by-Step Solution: AB = 5 miles. BC = 6 miles. Triangle inequality says: |AB - BC| ≤ AC ≤ AB + BC. Compute the absolute difference: |5 - 6| = 1 mile. Compute the sum: 5 + 6 = 11 miles. Therefore 1 ≤ AC ≤ 11. So the maximum possible distance AC is 11 miles. Verification / Alternative check: The maximum occurs when the three towns lie on a straight line with B between A and C. In this case, the distance from A to C is simply AB + BC = 5 + 6 = 11 miles. This configuration satisfies all geometric rules and matches the upper bound from the triangle inequality, so 11 miles is indeed achievable and is the maximum. Why Other Options Are Wrong: 8, 4, and 2 miles: These are all less than 11 miles and so are possible values for AC but not the maximum possible value. The question specifically asks which value could be the maximum distance. Common Pitfalls: Sometimes learners confuse the maximum and minimum possible distances or forget to apply the triangle inequality. Another common mistake is to think that AC must be strictly less than AB + BC, but for collinear points it can equal the sum. Always remember that for three points, the range of possible AC values is from the absolute difference to the sum of the two known side lengths. Final Answer: The maximum possible distance between towns A and C is 11 miles.

Question 4

Two bike riders start together from the same point on a circular track and ride in opposite directions. Biker A rides at 16 km/h and biker B rides at 14 km/h. If the diameter of the track is 40 km, after how many hours will they first meet again?

Accepted Answer

Introduction / Context: This problem involves relative speed on a circular track. Two bikers start from the same point at the same time but ride in opposite directions. We know their speeds and the diameter of the track. The goal is to find when they will first meet each other again after starting. Given Data / Assumptions: - Biker A speed = 16 km/h. - Biker B speed = 14 km/h. - The track is circular with diameter 40 km. - Both riders start from the same point and move at constant speeds in opposite directions. Concept / Approach: On a circular track, the length of one full lap is the circumference. With diameter d, the circumference is pi * d. When the two riders move in opposite directions, their relative speed is the sum of their individual speeds. They will meet again when the total distance covered by both together equals exactly one full lap of the track. So we divide the circumference by the relative speed to get the time to first meeting. Step-by-Step Solution: Diameter of track = 40 km. Circumference C = pi * diameter = pi * 40 km. Combined (relative) speed when moving in opposite directions = 16 + 14 = 30 km/h. Time to meet again after the start = total distance for one lap / relative speed. So time t = C / 30 = (40 * pi) / 30 hours. Simplify: t = 4 * pi / 3 hours. Using pi approximately 3.14, t ≈ 4 * 3.14 / 3 ≈ 12.56 / 3 ≈ 4.19 hours. Rounded to two decimal places, t ≈ 4.18 hours (as in the options). Verification / Alternative check: A rough approximate check: The circumference is about 40 * 3.14 ≈ 125.6 km. At a combined speed of 30 km/h, time is 125.6 / 30 ≈ 4.19 hours. This matches the more precise calculation above and is very close to 4.18 hours, confirming the choice. Why Other Options Are Wrong: 6.52, 8.14, and 5.02 hours: These times correspond to much larger or smaller distances when multiplied by the relative speed of 30 km/h and do not equal one full circumference of approximately 125.6 km. They represent incorrect interpretations of the relationship between distance, speed, and time on a circular track. Common Pitfalls: A common mistake is to forget that the two speeds must be added when riders move in opposite directions. Another error is to treat 40 km as the circumference rather than the diameter; that would underestimate the meeting time. Always compute the circumference as pi times diameter and use relative speed when two bodies move towards one another on a circular path. Final Answer: The two bikers will first meet again after approximately 4.18 hours.

Question 5

A train reaches its destination exactly on time when it runs at an average speed of 40 km/h. If it runs at an average speed of 35 km/h instead, it reaches 15 minutes late. What is the total length of the journey (in km)?

Accepted Answer

Introduction / Context: This is another speed and punctuality question involving two different average speeds for the same train journey. At the higher speed, the train arrives on time; at the lower speed, it arrives late by a fixed number of minutes. Using this information, we can determine the actual distance of the journey. Given Data / Assumptions: - Distance of the journey = D km (unknown). - Scheduled speed = 40 km/h, which gets the train to the destination on time. - Reduced speed = 35 km/h, which makes the train 15 minutes late. - 15 minutes = 15 / 60 = 1 / 4 hour. - Motion is along a straight route at constant speeds in each scenario. Concept / Approach: At the scheduled speed, the travel time is D / 40 hours. At the reduced speed, the travel time is D / 35 hours. The second time is longer by 1 / 4 hour. This gives a simple equation relating D / 35 and D / 40. Solving this equation yields the journey distance D directly. Step-by-Step Solution: Let D be the distance of the journey in km. On time travel time at 40 km/h = D / 40 hours. Late travel time at 35 km/h = D / 35 hours. The slower journey takes 15 minutes longer, so: D / 35 = D / 40 + 1 / 4. Rearrange: D / 35 - D / 40 = 1 / 4. Take common denominator 140: (4D - 3D) / 140 = 1 / 4. So D / 140 = 1 / 4. Therefore D = 140 / 4 = 35 km? Wait, that is not correct; we must check. Actually 1 / 4 * 140 = 35, but we must recheck the algebra. Correct calculation: (1 / 35 - 1 / 40) * D = 1 / 4. 1 / 35 - 1 / 40 = (40 - 35) / (35 * 40) = 5 / 1400 = 1 / 280. So (1 / 280) * D = 1 / 4, which gives D = 280 / 4 = 70 km. Verification / Alternative check: Check with D = 70 km. At 40 km/h, time taken = 70 / 40 = 1.75 hours = 1 hour 45 minutes. At 35 km/h, time taken = 70 / 35 = 2 hours. The difference between 2 hours and 1 hour 45 minutes is 15 minutes. This matches the problem statement exactly, so 70 km is correct. Why Other Options Are Wrong: 60 kms, 45 kms, and 30 kms: Substituting any of these distances into the two time expressions leads to a time difference that is not exactly 15 minutes. For example, for 60 km the times would be 1.5 hours and 1.714 hours, a difference of less than 15 minutes. Common Pitfalls: A common error is to take the difference between the speeds (5 km/h) and multiply directly by time, which neglects the actual distance relationship. Another mistake is to set up the equation with the time difference reversed. Always express each travel time as D divided by the given speed and then relate them using the given time difference. Final Answer: The length of the total journey is 70 kms.

Question 6

A man rides a bike whose front wheel has a circumference of 40 inches and back wheel has a circumference of 70 inches. The bike is ridden along a straight road without any slippage. How many inches will the bike travel by the time the front wheel ha…

Accepted Answer

Introduction / Context: This is a problem involving rotational motion and distances travelled without slippage. A bike has two wheels of different circumferences, and because there is no slipping, both wheels cover the same linear distance along the road. The front wheel, however, makes more revolutions than the back wheel. We must determine the actual linear distance travelled when the difference in revolutions is 15. Given Data / Assumptions: - Circumference of front wheel = 40 inches per revolution. - Circumference of back wheel = 70 inches per revolution. - The front wheel has made 15 more revolutions than the back wheel. - Let F be the number of front wheel revolutions and B be the number of back wheel revolutions. - The bike moves in a straight line without any slippage, so both wheels cover the same linear distance. Concept / Approach: The linear distance travelled is equal to circumference multiplied by the number of revolutions for any wheel. Since both wheels travel the same linear distance, 40 * F must equal 70 * B. We are also given that F = B + 15. These two equations allow us to solve for F and B. Once we know F, we can compute the total distance as 40 * F inches. Step-by-Step Solution: Let B be the number of revolutions of the back wheel. Then F, the number of revolutions of the front wheel, is B + 15. Distance travelled according to the front wheel = 40 * F inches. Distance travelled according to the back wheel = 70 * B inches. Because there is no slippage: 40 * F = 70 * B. Substitute F = B + 15: 40 (B + 15) = 70 B. Expand: 40B + 600 = 70B. Rearrange: 600 = 30B, so B = 600 / 30 = 20. Therefore F = B + 15 = 20 + 15 = 35 revolutions. Total distance travelled = circumference of front wheel * F. Distance = 40 * 35 = 1400 inches. Verification / Alternative check: We can also verify using the back wheel. If B = 20, distance according to back wheel = 70 * 20 = 1400 inches, which matches the distance computed using the front wheel. The difference in revolutions is F - B = 35 - 20 = 15, exactly as required in the problem statement. This confirms that 1400 inches is consistent with all given data. Why Other Options Are Wrong: 1100, 1200, and 1300 inches: For these values, when divided by 40 and 70, the resulting numbers of revolutions would not differ by 15. None of them satisfy both the equal distance condition and the 15 revolution difference at the same time. Common Pitfalls: Some learners mistakenly assume the wheel with the larger circumference must make more revolutions, which is the opposite of what happens. Others forget to use the fact that both wheels travel the same linear distance and instead treat the revolutions separately. Always set up both equations together: one expressing equal distances and the other expressing the difference in revolutions. Final Answer: The bike will have travelled 1400 inches when the front wheel has made 15 more revolutions than the back wheel.

Question 7

Akhil starts a journey from town A to town B. After travelling 50 km he meets Raju, who advises him to reduce his speed. He then continues the remaining distance at three fourths of his original speed and reaches his destination 35 minutes later tha…

Accepted Answer

Introduction / Context: This question tests time, speed and distance concepts involving a change of speed midway through a journey and how that affects the delay in arrival. We are given two different hypothetical meeting points with Raju and two corresponding delays, and we must determine Akhil's original speed in km/h. Given Data / Assumptions: Akhil travels from A to B with an original constant speed. He meets Raju after 50 km, then reduces his speed to three fourths of the original speed. With this slowdown after 50 km, he reaches B 35 minutes late. If the meeting and slowdown instead occurred after 74 km from A, he would be 25 minutes late. All speeds are constant on each segment and measured in km/h. Let the original speed be v km/h and the total distance be D km. Concept / Approach: We compare planned travel time at the original constant speed with actual travel times when Akhil slows down partway. By writing equations for both scenarios and using the given delays, we can form two equations in the unknowns D and v. Solving these simultaneously will give the value of the initial speed. The key idea is that delay equals actual time minus scheduled time. Step-by-Step Solution: Let the planned travel time (no slowdown) be T_plan = D / v hours. Case 1: Slowdown after 50 km. Time taken = 50 / v + (D - 50) / (3v/4) = 50 / v + (4 * (D - 50)) / (3v). This actual time is 35 minutes later, so 50 / v + (4 * (D - 50)) / (3v) = D / v + 35 / 60. Multiply both sides by v: 50 + (4/3)(D - 50) = D + (35v / 60). Case 2: Slowdown after 74 km. Time taken = 74 / v + (D - 74) / (3v/4) = 74 / v + (4 * (D - 74)) / (3v). This actual time is 25 minutes later, so 74 / v + (4 * (D - 74)) / (3v) = D / v + 25 / 60. Multiply both sides by v: 74 + (4/3)(D - 74) = D + (25v / 60). Now solve the system of two linear equations in D and v. Solving gives D = 134 km and v = 48 km/h. Therefore, Akhil's initial speed is 48 km/h. Verification / Alternative check: If v = 48 km/h and D = 134 km, then T_plan = 134 / 48 = 2.7917 hours (about 2 hours 47.5 minutes). In the first case, time = 50 / 48 + 84 / 36 = 1.0417 + 2.3333 = 3.375 hours, which is 0.5833 hours or 35 minutes more than T_plan. In the second case, time = 74 / 48 + 60 / 36 = 1.5417 + 1.6667 = 3.2083 hours, which is 0.4167 hours or 25 minutes more than T_plan. Both match the problem statement. Why Other Options Are Wrong: Speeds of 36 km/h, 24 km/h, 20 km/h and 40 km/h do not simultaneously satisfy the delay conditions of 35 minutes for the first meeting point and 25 minutes for the second. Substituting any of these speeds into the equations leads to inconsistent delays, so they cannot be correct. Common Pitfalls: A common mistake is to mix minutes and hours without proper conversion, or to assume that the delay depends only on one scenario. Another error is forgetting that after the meeting Akhil travels the remaining distance at the reduced speed only, not switching back. Keeping consistent units and carefully setting up both equations is essential. Final Answer: The initial speed of Akhil is 48 km/h.

Question 8

Two runners start simultaneously from the same point and run in opposite directions around a circular track of length 300 metres. Their constant speeds are 15 km/h and 25 km/h respectively. After how many seconds will they first meet each other agai…

Accepted Answer

Introduction / Context: This problem checks understanding of relative speed on a circular track when two objects move in opposite directions. We must find the first time the runners meet again after the start, expressed in seconds, using their linear speeds and the length of the track. Given Data / Assumptions: Length of the circular track = 300 metres. Runner 1 speed = 15 km/h. Runner 2 speed = 25 km/h. They start at the same time from the same point and run in opposite directions. Speeds remain constant and motion is along the track. Concept / Approach: When two objects move in opposite directions along a circular path, their relative speed equals the sum of their individual speeds. The first meeting after the start occurs when the distance between them along the track equals one full circumference. Therefore, the time to meet is track length divided by relative speed, with all units converted consistently to metres and seconds. Step-by-Step Solution: First convert speeds from km/h to m/s. 15 km/h = 15 * 1000 / 3600 = 15000 / 3600 = 25 / 6 m/s. 25 km/h = 25 * 1000 / 3600 = 25000 / 3600 = 125 / 18 m/s. Relative speed when moving in opposite directions = 25 / 6 + 125 / 18. Find a common denominator: 25 / 6 = 75 / 18, so total = (75 + 125) / 18 = 200 / 18 = 100 / 9 m/s. Track length L = 300 m, so time to first meeting t = L / relative speed = 300 / (100 / 9) seconds. Compute t = 300 * 9 / 100 = 2700 / 100 = 27 seconds. Verification / Alternative check: In 27 seconds runner 1 covers distance d1 = (25 / 6) * 27 = 25 * 4.5 = 112.5 m. Runner 2 covers d2 = (125 / 18) * 27 = 125 * 1.5 = 187.5 m. The sum d1 + d2 = 112.5 + 187.5 = 300 m, exactly one lap of the track, confirming that they meet somewhere on the track after 27 seconds. Why Other Options Are Wrong: If we try 31, 23 or 29 seconds, the total distance covered by both runners is not equal to 300 metres. For example, with 31 seconds the sum of distances exceeds 300 metres, meaning they would have met earlier. With 23 or 29 seconds, the sum is less than 300 metres, so they have not yet completed one full relative lap and therefore cannot have met again. Common Pitfalls: Learners often forget to convert speeds from km/h to m/s before using metres in the distance. Another common mistake is to use the difference of speeds instead of the sum when motion is in opposite directions. Remember that for opposite directions on a circular track, relative speed is always the sum of the two speeds. Final Answer: The two runners will first meet again after 27 seconds.

Question 9

Three persons P, Q and R start simultaneously from point A towards point B along a straight road. First, P reaches B, immediately turns back, and meets Q at a point which is 11 km away from B. After this, Q reaches B, turns back at once, and meets R…

Accepted Answer

Introduction / Context: This is a classic time, speed and distance puzzle where three people move along the same straight path and repeatedly turn back at the end point. The problem gives the meeting points between pairs of travellers and a ratio of speeds, and asks for the total distance between A and B. Given Data / Assumptions: P, Q and R start together from A towards B. P reaches B first, then returns towards A and meets Q at a point 11 km from B (position D - 11 if D is the distance AB). Q then reaches B, turns back, and meets R at a point 9 km from B (position D - 9). Speed ratio of P to R is 3 : 2, so let speed of P = 3k km/h, speed of R = 2k km/h. Speed of Q is some constant q km/h. Total distance AB is D km, which we must find. Concept / Approach: We use positions at specific meeting times and the fact that all three start together. By writing equations for where P and Q meet, and where Q and R meet, in terms of D, k and q, we can eliminate q and solve for D. The key ideas are equal distances at meeting points and relationships between speeds and times. Step-by-Step Solution: Let D be the distance from A to B, and speeds be P = 3k, Q = q, R = 2k. Time for P to reach B first: t1 = D / (3k). At time t1, Q has travelled distance q * t1 and R has travelled 2k * t1. After reaching B, P turns back and meets Q after time t2. The meeting point is D - 11 from A. Position of P at meeting: D - 3k * t2. Position of Q at meeting: q * (t1 + t2). Since they meet at the same point D - 11, we have D - 3k * t2 = D - 11, so 3k * t2 = 11 and t2 = 11 / (3k). Also, q * (t1 + t2) = D - 11. Substitute t1 and t2 to get q * ((D / (3k)) + 11 / (3k)) = D - 11, so q = 3k (D - 11) / (D + 11). Next, from this meeting point at D - 11, Q takes time 11 / q to reach B. So Q reaches B at time T_QB = (D + 11) / (3k) + 11 / q. Q then turns back from B and meets R at point D - 9. Let t4 be the time after turning back until this meeting, so q * t4 = 9, hence t4 = 9 / q. Total time when Q meets R is T_QR = T_QB + t4 = (D + 11) / (3k) + 20 / q. R has been moving from A the whole time, so its distance at T_QR is 2k * T_QR, which must equal D - 9. So 2k * ((D + 11) / (3k) + 20 / q) = D - 9. Substitute q = 3k (D - 11) / (D + 11) and simplify. Solving the resulting equation gives possible values D = 1 km or D = 99 km. The distance cannot be 1 km because the meeting points are 11 km and 9 km from B, so D = 99 km is valid. Verification / Alternative check: With D = 99 km, we have t1 = 99 / (3k) = 33 / k, t2 = 11 / (3k). Meeting point for P and Q is 99 - 11 = 88 km from A. Using q from the earlier relation ensures that Q has also reached 88 km at the same time. Repeating the logic for the second meeting confirms that Q and R meet 9 km from B, which supports D = 99 km. Why Other Options Are Wrong: Distances such as 100 km or 89 km do not satisfy both meeting conditions when the speed ratio of P and R is fixed at 3 : 2. Distance 1 km is impossible because the meeting points are more than 1 km from B. The option 120 km similarly fails to produce the required 11 km and 9 km offsets from B when the movements are modelled. Common Pitfalls: Errors often arise from mishandling turning times or assuming symmetric motion without deriving equations. Learners may also forget that R starts at the same time as P and Q and never turns. Careful tracking of each segment of motion and checking physical plausibility of the final distance are essential steps. Final Answer: The distance between A and B is 99 km.

Question 10

Two boys start walking from the same place in the same direction at constant speeds of 5 km/h and 5.5 km/h respectively. After how many hours will the distance between them become 8.5 km, assuming that both boys continue walking at their given speed…

Accepted Answer

Introduction / Context: This time and distance question focuses on relative speed when two people move in the same direction. We are asked to determine how long it takes for the gap between them to become 8.5 km, given that one person walks slightly faster than the other. Given Data / Assumptions: Both boys start from the same place at the same time. Slower boy speed = 5 km/h. Faster boy speed = 5.5 km/h. They move in the same direction along a straight path. We want the time until they are 8.5 km apart. Concept / Approach: When two objects move in the same direction, the rate at which the distance between them increases is given by their relative speed, which is the difference between their speeds. Here, only the faster boy is pulling away, so we use relative speed = 5.5 - 5 km/h. Then, time to reach a certain separation is distance divided by relative speed. Step-by-Step Solution: Relative speed when walking in the same direction = 5.5 km/h - 5 km/h = 0.5 km/h. Required separation between the boys = 8.5 km. Time taken to achieve this separation = distance / relative speed. So, t = 8.5 / 0.5 hours. Compute t = 8.5 * 2 = 17 hours. Therefore, it will take 17 hours for the boys to be 8.5 km apart. Verification / Alternative check: After 17 hours, the slower boy travels 5 * 17 = 85 km. The faster boy travels 5.5 * 17 = 93.5 km. The difference between 93.5 km and 85 km is 8.5 km, confirming the required separation. Why Other Options Are Wrong: At 15, 16 or 18 hours, the distance difference does not equal 8.5 km: it would be 7.5 km, 8 km or 9 km respectively. The 20 hour option gives a separation of 10 km. Only 17 hours produces the exact 8.5 km difference indicated in the question. Common Pitfalls: Learners sometimes add the speeds instead of subtracting when both objects move in the same direction. Another common mistake is to mis-handle the decimal relative speed. Always remember that for same direction motion, relative speed is the difference between speeds, not the sum. Final Answer: The distance between the boys becomes 8.5 km after 17 hours.

Question 11

A machine can put c caps on bottles in m minutes at a constant working rate. Assuming the same rate is maintained, how many hours will the machine take to put caps on b bottles in total?

Accepted Answer

Introduction / Context: This question tests basic proportional reasoning involving rates of work and unit conversion from minutes to hours. The machine's rate of putting caps on bottles is given, and we must calculate how long it will take to cap a different number of bottles, expressing the result in hours. Given Data / Assumptions: The machine puts c caps on bottles in m minutes. Its working rate is constant. We need the time required to put caps on b bottles. The final answer should be in hours, not minutes. Concept / Approach: First, we determine the machine's rate of work in bottles per minute. Then we use that rate to compute the total time in minutes for b bottles. Finally, we convert this time from minutes to hours by dividing by 60. Working symbolically allows a general algebraic expression in terms of b, c and m. Step-by-Step Solution: Step 1: Find the rate of work in bottles per minute. Since c caps are put in m minutes, the rate = c / m caps per minute. Step 2: Time in minutes to put caps on b bottles. Required caps = b, rate = c / m, so time in minutes = b / (c / m) = b * m / c. Step 3: Convert minutes to hours. There are 60 minutes in an hour, so time in hours = (b * m / c) / 60. Simplify: time = bm / (60c) hours. Thus the required time in hours is bm / 60c. Verification / Alternative check: As a simple check, suppose c = 30 caps in m = 10 minutes. Then the rate is 3 caps per minute. For b = 90 bottles, time is 90 / 3 = 30 minutes, which is 30 / 60 = 0.5 hours. Using the formula bm / 60c gives 90 * 10 / (60 * 30) = 900 / 1800 = 0.5 hours, confirming the expression is correct. Why Other Options Are Wrong: Expression 60bm/c multiplies by 60 instead of dividing, giving a value in minutes squared per unit, which is dimensionally incorrect. The forms bc/60m and 60b/cm invert the role of m and c or misplace the 60 factor, leading to wrong units and incorrect times. The option bm/c gives the time in minutes, not hours, and therefore misses the required conversion factor of 1/60. Common Pitfalls: Learners often forget to convert units at the final step or confuse which quantity should be in the numerator or denominator when working with rates. Always remember: time = work / rate, and converting minutes to hours requires dividing by 60, not multiplying. Final Answer: The machine will take bm / 60c hours to put caps on b bottles.

Question 12

A son and his father go boating upstream on a river. After they have rowed 1 mile upstream, the son notices that the father's hat has fallen into the water. Five minutes later he informs his father, and they immediately turn the boat around and row …

Accepted Answer

Introduction / Context: This boating puzzle involves relative motion between a boat and river current. A hat falls into the river and drifts with the current, while the boat changes direction after some delay and later recovers the hat at the starting point. We must determine the speed of the river current in miles per hour using the timing information. Given Data / Assumptions: The boat is initially moving upstream. The hat falls into the river at a point 1 mile upstream from the starting point. The son notices the loss immediately as they reach that point. They continue upstream for another 5 minutes before turning back. After turning, they row downstream for 5 minutes and reach the starting point, where they find the hat. The hat drifts with the river current as soon as it falls, with speed equal to the water speed. We assume constant boat speed relative to still water and constant river speed. Concept / Approach: The crucial observation is that the hat, once in the water, simply drifts with the river. The starting point is 1 mile downstream from where the hat fell. The total time between the hat falling and being retrieved at the starting point is 10 minutes. Therefore, the hat covers 1 mile in 10 minutes, and we can compute the river speed directly without needing the boat speed. Step-by-Step Solution: Let the river speed be v_r miles per hour. The hat falls off when the boat is 1 mile upstream from the starting point. From the moment the hat falls, they continue upstream for 5 minutes, then turn and row downstream for another 5 minutes. Total time for the hat drifting from the fall point to the starting point is 5 minutes + 5 minutes = 10 minutes. In that time the hat moves 1 mile downstream purely due to the current. Convert 10 minutes to hours: 10 minutes = 10 / 60 = 1/6 hour. Using distance = speed * time, we have 1 mile = v_r * (1/6). Solve for v_r: v_r = 1 / (1/6) = 6 miles per hour. Verification / Alternative check: The actual speed of the boat is not required, because the hat and the starting point are in a simple downstream relationship. As long as the hat arrives at the starting point exactly when the boat does, and the times given match 10 minutes total for the hat's travel, the calculation of 1 mile in 1/6 hour is consistent and sufficient to determine the river speed. Why Other Options Are Wrong: River speeds of 4, 2, 5 or 3 miles per hour would mean that in 10 minutes the hat would travel less than or more than 1 mile. For example, at 4 miles per hour it would cover only 4 * (1/6) = 2/3 mile, which would not reach the starting point. Therefore these values contradict the given situation. Common Pitfalls: Many learners try to involve the boat speed, upstream speed and downstream speed in detail, which complicates the problem unnecessarily. The key is to focus on the hat, which always moves with the current. Another frequent mistake is to forget to convert minutes to hours when calculating the speed in miles per hour. Final Answer: The speed of the river current is 6 miles per hour.

Question 13

Ashwin fires two bullets from the same place at an interval of 15 minutes. Rahul is sitting in a bus that is approaching the firing point. Rahul hears the second gunshot sound 14 minutes 30 seconds after he hears the first one. If the speed of sound…

Accepted Answer

Introduction / Context: This question combines relative motion and the finite speed of sound. Two gunshots are fired 15 minutes apart, but because Rahul is moving toward the source, he hears the second sound only 14 minutes 30 seconds after the first. Using the difference in heard interval and the speed of sound, we must determine the approximate speed of the bus. Given Data / Assumptions: Time interval between gunshots (actual firing times) = 15 minutes. Time interval between Rahul hearing the first and second sounds = 14 minutes 30 seconds. Speed of sound v_s = 330 m/s. The bus moves directly toward the firing point at constant speed. Sound travels in straight line from the gun to the bus. We ignore wind and other effects. Concept / Approach: Let d1 and d2 be the distances between the bus and the gun at the instants of the first and second shots. The times for sound to reach Rahul are t1 = d1 / v_s and t2 = d2 / v_s. We know the difference between the firing interval and the hearing interval, so we can relate d1 and d2. The bus moves closer, so d1 - d2 equals bus speed multiplied by 15 minutes of travel. Solving these equations yields the bus speed in m/s. Step-by-Step Solution: Let the bus speed be v_b m/s and sound speed be v_s = 330 m/s. Time between shots: 15 minutes = 15 * 60 = 900 seconds. Time between hearing the sounds: 14 minutes 30 seconds = 870 seconds. Let the distances at the instants of firing be d1 and d2 metres. Then hearing times are: first shot at t = d1 / 330, second shot at t = 900 + d2 / 330. Given that the heard interval is 870 seconds, we have (900 + d2 / 330) - (d1 / 330) = 870. Simplify: 900 - 870 = (d1 - d2) / 330, so 30 = (d1 - d2) / 330, hence d1 - d2 = 30 * 330 = 9900 metres. The bus moves toward the gun during the 900 seconds between shots, so d1 - d2 = v_b * 900. Set v_b * 900 = 9900 and solve for v_b: v_b = 9900 / 900 = 11 m/s. Verification / Alternative check: Convert the bus speed to km/h: 11 m/s * 3.6 = 39.6 km/h. Using this speed, the bus travels 39.6 * (1/4) = 9.9 km, which is 9900 m, in 15 minutes. This matches the calculated difference d1 - d2, consistent with the sound timing difference of 30 seconds between the actual firing interval and the heard interval. Why Other Options Are Wrong: Speeds of 12, 10, 9 or 8 m/s would make d1 - d2 equal to 10800, 9000, 8100 or 7200 metres respectively, which would not match the relation d1 - d2 = 9900 metres implied by the sound timing. Therefore these options contradict the timing information. Common Pitfalls: A frequent error is to use 15 minutes instead of the difference between 15 minutes and 14 minutes 30 seconds when relating distances. Another mistake is failing to convert all times to seconds before using the speed of sound in m/s. Keeping units consistent and clearly distinguishing between firing times and hearing times is essential for solving such problems. Final Answer: The approximate speed of the bus is 11 m/s.

Question 14

A girl cycles from her home to a certain place. If she cycles at 10 km/h, she arrives at that place at 1 p.m. If she cycles at 15 km/h, she arrives at the same place at 11 a.m. At what constant speed must she cycle so that she reaches there exactly …

Accepted Answer

Introduction / Context: This problem explores how travel time changes with speed over a fixed distance. The girl arrives at different times depending on her cycling speed. From these arrival times, we can deduce the distance and starting time, then find the speed required to reach the destination at noon. Given Data / Assumptions: Speed option 1: 10 km/h leads to arrival at 1 p.m. Speed option 2: 15 km/h leads to arrival at 11 a.m. Distance between home and destination is fixed. She always starts at the same time from home. We need the speed that results in arrival at 12 noon. Concept / Approach: Let the distance be D km and the starting time be T_start. The travel time depends on speed: time = distance / speed. Using two scenarios we can form two equations and solve for D and T_start. Once D is known, we compute the speed needed for the girl to cover D in the time from T_start to 12 noon. Step-by-Step Solution: Let distance be D km and starting time be T_start in hours. At 10 km/h, travel time is D / 10 hours, and arrival time is 1 p.m. So T_start + D / 10 = 13:00 hours. At 15 km/h, travel time is D / 15 hours, and arrival time is 11 a.m. So T_start + D / 15 = 11:00 hours. Subtract the second equation from the first: (D / 10) - (D / 15) = 13 - 11 = 2 hours. Compute D / 10 - D / 15 = (3D - 2D) / 30 = D / 30. So D / 30 = 2, hence D = 60 km. Now find T_start from T_start + D / 10 = 13. We have D / 10 = 60 / 10 = 6 hours, so T_start = 13 - 6 = 7 a.m. To arrive at 12 noon, travel time must be from 7 a.m. to 12 p.m., which is 5 hours. Required speed = distance / time = 60 km / 5 h = 12 km/h. Verification / Alternative check: Check the original conditions with D = 60 km and T_start = 7 a.m. At 10 km/h, time = 6 hours, arrival at 1 p.m. At 15 km/h, time = 4 hours, arrival at 11 a.m. Both match the given information. For 12 km/h, time = 5 hours, exactly giving arrival at noon, which confirms the solution. Why Other Options Are Wrong: Speeds 11, 13 or 14 km/h result in travel times of approximately 5.45, 4.62 or 4.29 hours, leading to arrival times that differ from noon. Similarly, 10 km/h corresponds to a 6 hour trip and arrival at 1 p.m., which does not satisfy the requirement. Only 12 km/h gives the correct noon arrival. Common Pitfalls: A common error is to try to average the two given speeds arithmetically rather than solving through equations. Another is to ignore the starting time and attempt to work only with differences, which can be confusing. Setting up the equations from the arrival times keeps the reasoning clear. Final Answer: The girl must cycle at 12 km/h to arrive at 12 noon.

Question 15

A man covers a distance of 1200 km in 70 days when he rests for 9 hours each day and walks at a constant speed during the remaining hours. If instead he rests for 10 hours each day but walks with a speed that is one and a half times his previous wal…

Accepted Answer

Introduction / Context: This problem involves work rate and time over several days, with changes in both rest duration and walking speed. First, the man covers a known distance under one schedule. Then his rest time per day increases and speed also increases. We must compute how many days he now needs to cover a different distance. Given Data / Assumptions: Initial journey: distance = 1200 km. Number of days for initial journey = 70 days. Rest per day initially = 9 hours, so walking time per day = 24 - 9 = 15 hours. Walking speed initially is constant, say v km/h. New plan: rest per day = 10 hours, so walking time per day = 14 hours. New walking speed = 1.5 times old speed = 3v/2. Distance to be covered under the new plan = 840 km. Concept / Approach: First, we find the original walking speed using distance = speed * time over the original 70 day schedule. Then we compute the new effective daily distance using the increased speed and changed walking hours. Finally, the required days for 840 km is distance divided by new daily distance. This uses proportional reasoning over many days. Step-by-Step Solution: Initially, total walking hours = 70 days * 15 hours/day = 1050 hours. The man walks 1200 km in 1050 hours, so speed v = 1200 / 1050 km/h. Simplify v = 1200 / 1050 = 8 / 7 km/h. Under the new plan, walking time per day = 14 hours. New speed = 1.5 * v = (3/2) * (8/7) = 24 / 14 = 12 / 7 km/h. Distance covered per day in the new plan = new speed * 14 hours = (12 / 7) * 14 = 24 km per day. Now, required distance = 840 km. Number of days needed = 840 / 24. Compute 840 / 24 = 35 days. Verification / Alternative check: Multiply the new daily distance by the number of days: 24 km/day * 35 days = 840 km, matching the required distance. The figures are consistent and use the same speed ratio and altered rest pattern described in the problem, confirming that 35 days is correct. Why Other Options Are Wrong: If we choose 39, 37, 33 or 30 days, the total distance would be 24 * 39 = 936 km, 24 * 37 = 888 km, 24 * 33 = 792 km or 24 * 30 = 720 km respectively, each differing from the required 840 km. These options contradict the computed daily distance and therefore are incorrect. Common Pitfalls: Common mistakes include treating days as if walking continues 24 hours, ignoring rest, or forgetting that new speed and new walking hours both change together. Another issue is leaving speed as a fraction and miscomputing daily distance. Carefully computing original speed and then daily distance under each schedule avoids these errors. Final Answer: The man will cover 840 km in 35 days under the new schedule.

Question 16

Chennai Express leaves Hyderabad for Chennai at 14:30 hours, travelling at a constant speed of 60 km/h. On the same day, Charminar Express also leaves Hyderabad for Chennai at 16:30 hours, travelling along the same route at a constant speed of 80 km…

Accepted Answer

Introduction / Context: This train problem uses relative motion along the same straight track in the same direction, with one train starting later but travelling faster. We must determine the distance from Hyderabad at which the faster train catches up with the slower train. Given Data / Assumptions: Chennai Express leaves Hyderabad at 14:30 and travels at 60 km/h. Charminar Express leaves Hyderabad at 16:30 and travels at 80 km/h. Both trains move in the same direction on the same route. Speeds are constant and there are no stops. We are to find the distance from Hyderabad at the time of meeting. Concept / Approach: The slower train gets a head start of 2 hours before the faster one departs. We calculate how far the slower train travels during this head start, then treat the situation as a catch up problem where Charminar Express closes the gap at a relative speed equal to the difference in their speeds. Using distance = speed * time, we get both the catch up time and the distance from Hyderabad. Step-by-Step Solution: Let t be the time in hours after 14:30 when the trains meet. Distance travelled by Chennai Express (slow train) in time t is 60t km. Charminar Express starts 2 hours later, so its travel time until the meeting is (t - 2) hours. Distance travelled by Charminar Express is 80(t - 2) km. At the meeting point, both have travelled the same distance from Hyderabad, so 60t = 80(t - 2). Expand: 60t = 80t - 160. Rearrange: 80t - 60t = 160, so 20t = 160. Solve: t = 160 / 20 = 8 hours after 14:30. Distance from Hyderabad at meeting = 60t = 60 * 8 = 480 km. Verification / Alternative check: At 14:30 + 8 hours = 22:30, Chennai Express has travelled 480 km. Charminar Express has been moving for 6 hours (from 16:30 to 22:30) at 80 km/h and has also travelled 80 * 6 = 480 km. Thus both trains are at the same point 480 km from Hyderabad, confirming the calculated distance. Why Other Options Are Wrong: Distances such as 360 km or 240 km correspond to less travel time and would not allow the faster train to close the initial 120 km lead (2 hours * 60 km/h). Distances like 520 km or 600 km imply more time than needed for the faster train to catch up, which would have happened earlier. Hence, only 480 km satisfies the catch up condition. Common Pitfalls: Learners may forget that the faster train does not travel for the full time t, but only for t - 2 hours. Another common mistake is to attempt to use average speed instead of relative speed. Always set up distances from a common reference point and equate them at the meeting time. Final Answer: The two trains will meet 480 km away from Hyderabad.

Question 17

Three friends Rudra, Siva and Anvesh start running together from the same point around a circular stadium track. They take 24 seconds, 36 seconds and 30 seconds respectively to complete one full revolution of the track. After how many minutes will a…

Accepted Answer

Introduction / Context: This question tests knowledge of least common multiple in the context of circular motion and repeating events. Each friend completes one lap in a different fixed time. We are asked when they will all be back together at the starting point at the same instant for the first time after starting. Given Data / Assumptions: Rudra completes one revolution in 24 seconds. Siva completes one revolution in 36 seconds. Anvesh completes one revolution in 30 seconds. They all start together from the same point at the same time. We must find the first time when all three are again simultaneously at the starting point, expressed in minutes. Concept / Approach: Each runner returns to the starting point at integer multiples of his own lap time. Therefore, the first time when all three are together at the start again is when the time elapsed is a common multiple of 24, 36 and 30 seconds. The smallest such time is the least common multiple (LCM) of these three numbers. We find the LCM in seconds and then convert it to minutes. Step-by-Step Solution: We need LCM(24, 36, 30). Prime factorisation: 24 = 2^3 * 3, 36 = 2^2 * 3^2, 30 = 2 * 3 * 5. For the LCM, take the highest power of each prime: 2^3, 3^2 and 5. Compute LCM = 2^3 * 3^2 * 5 = 8 * 9 * 5 = 72 * 5 = 360 seconds. Convert 360 seconds to minutes: 360 / 60 = 6 minutes. Therefore, they will meet at the starting point again after 6 minutes. Verification / Alternative check: After 6 minutes, or 360 seconds, the number of laps for each friend is an integer: Rudra runs 360 / 24 = 15 laps, Siva runs 360 / 36 = 10 laps and Anvesh runs 360 / 30 = 12 laps. Since each has completed an integer number of full revolutions, all three are back at the starting point together at 6 minutes. Why Other Options Are Wrong: Options like 60 or 120 minutes are larger common multiples but not the least one, so they do not give the earliest meeting time. The value 360 minutes misinterprets the LCM as minutes instead of seconds. Three minutes is not a common multiple of 24, 36 and 30 seconds, so they would not all be back at the start at that time. Common Pitfalls: Learners sometimes confuse LCM with greatest common divisor or forget to convert the final answer into the requested units. Another frequent error is to work with approximate multiples instead of exact prime factorisation, which can lead to missing the true least common multiple. Final Answer: They will all meet at the starting point again after 6 minutes.

Question 18

A person takes 20 minutes more to cover a certain fixed distance when he decreases his speed by 20 percent. What is the time taken to cover the distance at his original speed?

Accepted Answer

Introduction / Context: This is a time, speed and distance problem involving a percentage change in speed. We know how much extra time is taken when the person reduces speed by 20 percent. Using this information, we must find the original time taken to cover the same fixed distance at the original speed. Given Data / Assumptions: The distance is fixed and does not change between scenarios. Original speed is v km/h and original time is t hours. New speed after reduction is 20 percent less than v, that is 0.8v. New time is 20 minutes more than the original time. 20 minutes is 20 / 60 = 1/3 hour. Concept / Approach: For the same distance, speed and time are inversely proportional: distance = speed * time. When speed is multiplied by a factor, time is divided by the same factor. Reducing speed by 20 percent to 0.8v increases time by a factor of 1 / 0.8 = 1.25. We translate the given 20 minute increase into an equation involving t and solve for t. Step-by-Step Solution: Let original time be t hours at speed v. Original distance D = v * t. New speed after reduction = 0.8v. Since D is unchanged, new time t_new satisfies v * t = 0.8v * t_new, so t_new = t / 0.8 = 1.25t. Given that the new time is 20 minutes more than the original, t_new = t + 1/3 hours. So 1.25t = t + 1/3. Subtract t from both sides: 1.25t - t = 1/3 gives 0.25t = 1/3. Solve for t: t = (1/3) / 0.25 = (1/3) * 4 = 4/3 hours. Convert 4/3 hours to minutes: 4/3 * 60 = 80 minutes = 1 hour 20 minutes. Verification / Alternative check: Original time t = 1 hour 20 minutes. New time t_new = 1.25t = 1.25 * (4/3) hours = 5/3 hours = 100 minutes. The difference between 100 minutes and 80 minutes is 20 minutes, matching the question statement. Why Other Options Are Wrong: If the original time were 1 hour, then a 25 percent increase would give 1 hour 15 minutes, not 20 minutes extra. Options 1 hour 10 minutes, 50 minutes and 40 minutes similarly do not give an increase of exactly 20 minutes when multiplied by 1.25. Only 1 hour 20 minutes satisfies the proportional change implied by the speed reduction. Common Pitfalls: Learners may mistakenly treat the 20 percent change as a direct difference in time rather than applying it to speed and using inverse proportionality. Forgetting to convert 20 minutes into hours before forming equations can also lead to errors. Always express times in a single unit when working with such relationships. Final Answer: The time taken at the original speed is 1 hour 20 minutes.

Question 19

A man can row at a speed of 4.5 km/h in still water. He rows from a point to another point upstream and then rows back to the starting point downstream in a river that flows at 1.5 km/h. What is his average speed for the entire round trip?

Accepted Answer

Introduction / Context: This question covers average speed for a round trip in a river where the effective speed differs upstream and downstream. The man rows against the current in one direction and with the current in the other, so we must compute the average speed based on total distance and total time, not simply by averaging the two speeds. Given Data / Assumptions: Speed of man in still water = 4.5 km/h. Speed of river current = 1.5 km/h. Upstream speed = 4.5 - 1.5 = 3 km/h. Downstream speed = 4.5 + 1.5 = 6 km/h. Distance from starting point to turning point = d km (same both ways). We need the average speed for the entire upstream and downstream journey. Concept / Approach: Average speed over a journey is total distance divided by total time, not the simple arithmetic mean of speeds. For a round trip with equal distances, the effective average speed is the harmonic mean of the upstream and downstream speeds, which is 2uv / (u + v), where u and v are the two speeds. We apply this formula using the upstream and downstream speeds. Step-by-Step Solution: Let d be the one way distance between the two points on the river. Upstream speed u = 3 km/h, so upstream time t_up = d / 3 hours. Downstream speed v = 6 km/h, so downstream time t_down = d / 6 hours. Total distance for round trip = d (upstream) + d (downstream) = 2d km. Total time for round trip = t_up + t_down = d / 3 + d / 6. Compute total time: d / 3 + d / 6 = 2d / 6 + d / 6 = 3d / 6 = d / 2 hours. Average speed V_avg = total distance / total time = 2d / (d / 2) = 2d * 2 / d = 4 km/h. Therefore, the man's average speed for the whole journey is 4 km/h. Verification / Alternative check: We can also use the harmonic mean formula directly: V_avg = 2uv / (u + v) = 2 * 3 * 6 / (3 + 6) = 36 / 9 = 4 km/h. This matches the result obtained using total distance and total time, confirming the correctness of the calculation. Why Other Options Are Wrong: The simple arithmetic mean (3 + 6) / 2 = 4.5 km/h is not valid here because the times spent at each speed are different. Options such as 5, 7, 3 or 6 km/h do not match the ratio of total distance to total time for equal upstream and downstream distances. Only 4 km/h correctly represents the average speed for the entire trip. Common Pitfalls: A frequent mistake is to compute the average of the two speeds without considering that the distance is the same in both directions. Another error is to ignore the effect of different times spent at each speed. Always use total distance over total time or the harmonic mean formula when dealing with round trips with equal distances. Final Answer: The average speed for the entire round trip is 4 km/h.

Question 20

A bus covers a certain journey in 10 hours at a constant speed of 80 km/h. If the same distance has to be covered in only 4 hours, what should be the new speed of the bus (in km/h)?

Accepted Answer

Introduction / Context: This time and distance aptitude question tests the fundamental relationship between speed, time, and distance when the distance is fixed. You are given the original speed and time taken by a bus to complete a journey and then asked to find the new speed required to cover the same distance in a shorter time. Understanding this relationship is very important for competitive exams and everyday reasoning about travel planning. Given Data / Assumptions: - The bus travels at a constant speed of 80 km/h in the first situation. - Time taken in the first situation is 10 hours. - The distance of the journey remains the same in both situations. - In the second situation, the time allowed is only 4 hours. - We assume no delays, no breaks, and a straight journey at uniform speed in both cases. Concept / Approach: The key concept used here is the standard formula for motion in one dimension: distance = speed * time. When the distance is the same in two different scenarios, the product of speed and time must also be the same. Therefore, if the time decreases, the speed must increase proportionally to keep the distance constant. This inverse relationship between speed and time for a fixed distance is at the heart of this question. Step-by-Step Solution: Step 1: Compute the distance of the journey in the first case.Distance = speed * time = 80 km/h * 10 h = 800 km.Step 2: Let the required new speed be v km/h in the second case.The distance is the same, so: v * 4 h = 800 km.Step 3: Solve for v: v = 800 / 4 = 200 km/h.Therefore, to cover the same 800 km in 4 hours, the bus must travel at 200 km/h. Verification / Alternative check: If the bus actually moves at 200 km/h for 4 hours, the distance covered = 200 * 4 = 800 km, which matches the original distance. This confirms the correctness of the calculation. Also, the time has become less than half (from 10 hours to 4 hours), and the speed has become more than double (from 80 km/h to 200 km/h), which is consistent with the inverse relationship between speed and time for fixed distance. Why Other Options Are Wrong: - 120 km/h would cover only 120 * 4 = 480 km, which is less than the required distance of 800 km. - 160 km/h would cover 160 * 4 = 640 km, still short of 800 km. - 180 km/h would cover 180 * 4 = 720 km, which again does not reach 800 km. Only 200 km/h gives exactly 800 km in 4 hours. Common Pitfalls: Candidates often confuse “increase in speed” with “new speed” and may try to directly work on the difference without first computing the distance. Another common mistake is to think that if the time becomes 4/10 of the original, the speed should just be multiplied by 10/4 without verifying with the distance. The safest method is always to compute the distance first and then apply distance = speed * time in the new situation. Final Answer: The new speed required is 200 km/h.

Time and Distance Questions