Question 1

A boy runs around a square field with side length 35 metres at a constant speed of 9 km/h. How much time in seconds will he take to complete one full round of the field?

Accepted Answer

Introduction / Context: This question combines basic perimeter calculation of a square with unit conversion from km/h to m/s or metres per second. It is a common type of problem in time and distance topics that checks both geometry and speed conversion skills. Given Data / Assumptions: Shape of field: square. Side length of the square field = 35 metres. Running speed of the boy = 9 km/h. We must find the time taken to complete one full round (one perimeter) in seconds. Concept / Approach: First, we compute the perimeter of the square, which is 4 times the side. Then we convert the speed from km/h to m/s or interpret distance and speed consistently. Time is always given by: Time = Distance / Speed. It is often convenient to express speed in metres per second when distance is in metres and time is required in seconds. Step-by-Step Solution: Step 1: Perimeter of the square = 4 * side = 4 * 35 = 140 metres. Step 2: Convert speed from km/h to m/s. Use 1 km/h = 5 / 18 m/s. Step 3: Speed = 9 km/h = 9 * 5 / 18 m/s = 45 / 18 m/s = 2.5 m/s. Step 4: Time taken = Distance / Speed = 140 / 2.5 seconds. Step 5: 140 / 2.5 = 56 seconds. Verification / Alternative check: Instead of converting into metres per second, we could convert distance into kilometres and time into hours. Distance = 140 metres = 0.14 km. Time in hours = 0.14 / 9 hour, and then convert that to seconds: Time in hours = 0.015555..., multiply by 3600 seconds per hour gives 56 seconds, again confirming the result. Why Other Options Are Wrong: 50 seconds: Corresponds to a slightly higher speed than 9 km/h. 52 seconds: Also implies a speed greater than the given 9 km/h. 54 seconds: Represents a small calculation error when dividing 140 by 2.5. Common Pitfalls: Using side instead of perimeter, forgetting that the boy runs around all four sides. Incorrectly converting 9 km/h to m/s by using 18/5 instead of 5/18. Mixing units by using metres for distance and km/h for speed without conversion. Final Answer: The boy will take 56 seconds to complete one round of the square field.

Question 2

A man takes 6 hours 15 minutes to walk from his starting point to a certain place and then ride back. If he were to walk both ways, he would take 7 hours 45 minutes. How much time would he take if he rides both ways?

Accepted Answer

Introduction / Context: This question involves two different modes of travel, walking and riding, over the same distance between two points. By comparing the total times taken under different travel combinations, we can deduce the time associated with each mode and then find how long a full round trip would take using only one mode. This is a standard application of time, speed and distance relations. Given Data / Assumptions: Time to walk from starting point to destination and ride back = 6 hours 15 minutes. Time to walk both ways (to and fro) = 7 hours 45 minutes. We must find the time to ride both ways. Distance between the two points is the same in all scenarios. Concept / Approach: Let d be the one way distance, v_w be the walking speed, and v_r be the riding speed. Time to walk one way is t_w = d / v_w, and time to ride one way is t_r = d / v_r. From the information given, we can form two equations involving t_w and t_r. Once we find t_r, the time to ride both ways is simply 2 * t_r. Step-by-Step Solution: Step 1: Convert mixed times to hours. 6 hours 15 minutes = 6.25 hours, 7 hours 45 minutes = 7.75 hours. Step 2: Time to walk both ways: 2t_w = 7.75, so t_w = 7.75 / 2 = 3.875 hours. Step 3: Time to walk one way and ride back: t_w + t_r = 6.25. Step 4: Substitute t_w = 3.875 into t_w + t_r = 6.25, giving 3.875 + t_r = 6.25. Step 5: Solve for t_r: t_r = 6.25 - 3.875 = 2.375 hours. Step 6: Time to ride both ways = 2 * t_r = 2 * 2.375 = 4.75 hours, which is 4 hours 45 minutes. Verification / Alternative check: We can check consistency. Walking both ways takes 7.75 hours. Walking one way and riding back takes 3.875 + 2.375 = 6.25 hours, matching the given value. Riding both ways is 4.75 hours, which is shorter than both previous totals, as expected since riding is faster than walking. Why Other Options Are Wrong: 4 hours: Implies each ride one way is 2 hours, which contradicts the mixed trip equation. 4 hours 30 minutes: Corresponds to t_r = 2.25 hours, inconsistent with the walking both ways data. 5 hours: Would give t_r = 2.5 hours per trip, again not matching any of the given combined times. Common Pitfalls: Not converting minutes into decimal hours correctly (for example, using 6.15 instead of 6.25). Assuming distance is different in different scenarios, which is not true. Forgetting to double the one way riding time to get the total round trip time. Final Answer: The time taken by the man to ride both ways is 4 hours 45 minutes.

Question 3

A man walking at a constant speed of 5 km/h crosses a bridge in 15 minutes. What is the length of the bridge in metres?

Accepted Answer

Introduction / Context: This question is a simple time and distance problem involving a person crossing a bridge at a known constant speed. The key is to correctly convert time into hours and then apply the basic formula relating distance, speed and time. Because the answer is required in metres, we must also convert the final distance from kilometres to metres. Given Data / Assumptions: Walking speed of the man = 5 km/h. Time taken to cross the bridge = 15 minutes. We need to find the length of the bridge in metres. The man walks at a uniform speed during the crossing. Concept / Approach: The fundamental relation is: Distance = Speed * Time. However, speed is given in kilometres per hour, while time is given in minutes. We must convert time from minutes to hours before using the formula, and then convert the resulting distance from kilometres to metres using 1 km = 1000 metres. Step-by-Step Solution: Step 1: Convert time from minutes to hours. Time = 15 minutes = 15 / 60 = 0.25 hour. Step 2: Speed = 5 km/h. Step 3: Distance in kilometres = Speed * Time = 5 * 0.25 = 1.25 km. Step 4: Convert 1.25 km to metres: 1.25 * 1000 = 1250 metres. Step 5: Therefore the length of the bridge is 1250 metres. Verification / Alternative check: Alternatively, convert speed to metres per minute. Speed 5 km/h = 5000 metres per hour. Per minute, this is 5000 / 60 ≈ 83.33 metres per minute. In 15 minutes, distance covered is 83.33 * 15 ≈ 1250 metres. This matches the previous calculation and verifies the answer. Why Other Options Are Wrong: 600 m: Implies an effective speed of only 2.4 km/h, inconsistent with the given 5 km/h. 750 m: Gives an incorrect conversion and does not match distance = 5 * 0.25 km. 1000 m: Would correspond to a shorter time than 15 minutes at 5 km/h. Common Pitfalls: Forgetting to convert minutes to hours when using km/h for speed. Converting distance incorrectly by using 100 instead of 1000 when moving from km to metres. Using 15/100 instead of 15/60 in the time conversion step. Final Answer: The length of the bridge is 1250 m.

Question 4

A man on tour travels the first 160 km at 64 km/h and the next 160 km at 80 km/h. What is his average speed for the first 320 km of the journey?

Accepted Answer

Introduction / Context: This question tests your ability to compute average speed when a journey is performed at different speeds over equal distances. Average speed is not the arithmetic mean of the speeds unless the time intervals are equal. Instead, when distances are equal, the average speed is based on total distance divided by total time. Given Data / Assumptions: First part of journey: distance = 160 km at 64 km/h. Second part of journey: distance = 160 km at 80 km/h. Total distance considered = 160 + 160 = 320 km. We must find average speed over the entire 320 km. Concept / Approach: Average speed is always defined as: Average speed = Total distance / Total time. We first compute the time taken for each part of the journey using time = distance / speed, then add these times to obtain the total time. Finally, we divide the total distance by total time to get the average speed. Step-by-Step Solution: Step 1: Time taken for first 160 km at 64 km/h = 160 / 64 hours. Step 2: 160 / 64 = 2.5 hours. Step 3: Time taken for next 160 km at 80 km/h = 160 / 80 = 2 hours. Step 4: Total time = 2.5 + 2 = 4.5 hours. Step 5: Total distance = 160 + 160 = 320 km. Step 6: Average speed = Total distance / Total time = 320 / 4.5 km/h. Step 7: 320 / 4.5 ≈ 71.11 km/h. Verification / Alternative check: When distances are equal, the harmonic mean of the two speeds can be used. Average speed = 2 * v1 * v2 / (v1 + v2). Here v1 = 64, v2 = 80, so average speed = 2 * 64 * 80 / (64 + 80) = 10240 / 144 ≈ 71.11 km/h. This matches the detailed method result and confirms the correctness of the answer. Why Other Options Are Wrong: 35.55 km/h and 36 km/h: These are far too low and would make the journey unrealistically slow. 71 km/h: A rounded value that slightly underestimates the exact average of about 71.11 km/h. Common Pitfalls: Taking the simple average (64 + 80) / 2 = 72 km/h, which is incorrect because times are not equal. Using incorrect division when computing 320 / 4.5. Mixing up the formula for average of speeds with the formula for average speed over a distance. Final Answer: The average speed for the first 320 km of the tour is approximately 71.11 km/h.

Question 5

The distance between two cities A and B is 330 km. A train leaves A at 8 a.m. towards B at 60 km/h, and another train leaves B at 9 a.m. towards A at 75 km/h. At what time do the two trains meet?

Accepted Answer

Introduction / Context: This is a classic two trains problem where the trains start from opposite ends of a straight track at different times and speeds. The key is to carefully account for the fact that the first train begins its journey one hour earlier than the second train, and then use relative speed to determine the time at which they meet. Given Data / Assumptions: Distance between cities A and B = 330 km. Train 1 leaves A at 8 a.m. with speed = 60 km/h. Train 2 leaves B at 9 a.m. with speed = 75 km/h. Both trains move along the same straight route towards each other. Concept / Approach: First, we calculate the distance covered by Train 1 during the hour before Train 2 starts. Then we compute the remaining distance between the trains at 9 a.m. After that, both trains are moving towards each other, so we can use their combined (relative) speed to find how long it takes them to meet after 9 a.m. Finally, we add this time to 9 a.m. to get the meeting time. Step-by-Step Solution: Step 1: From 8 a.m. to 9 a.m., only Train 1 moves. Distance covered by Train 1 = 60 km/h * 1 hour = 60 km. Step 2: Remaining distance between the trains at 9 a.m. = 330 - 60 = 270 km. Step 3: After 9 a.m., both trains move towards each other. Combined speed = 60 + 75 = 135 km/h. Step 4: Time taken after 9 a.m. to meet = Remaining distance / Combined speed = 270 / 135 hours. Step 5: 270 / 135 = 2 hours. Step 6: Therefore, trains meet at 9 a.m. + 2 hours = 11 a.m. Verification / Alternative check: We can confirm distances: By 11 a.m., Train 1 has travelled from 8 to 11 a.m. at 60 km/h, so distance = 3 * 60 = 180 km. Train 2 has travelled from 9 to 11 a.m. at 75 km/h, so distance = 2 * 75 = 150 km. Total distance = 180 + 150 = 330 km, exactly the distance between the two cities, confirming that 11 a.m. is the correct meeting time. Why Other Options Are Wrong: 10 a.m.: At this time Train 2 has only travelled 1 hour and total distance covered is less than 330 km. 10:30 a.m.: Also too early; combined distance is still less than 330 km. 11:30 a.m.: Too late; by then the trains would have already crossed each other. Common Pitfalls: Ignoring the one hour head start of Train 1. Using relative speed from 8 a.m. without accounting for Train 2 starting later. Performing unit conversion mistakes or miscalculating 270 / 135. Final Answer: The two trains meet at 11 a.m.

Question 6

Akash leaves Mumbai at 6 a.m. and reaches Bangalore at 10 a.m. Prakash leaves Bangalore at 8 a.m. and reaches Mumbai at 11:30 a.m. Assuming they travel at constant speeds along the same route, at what time do Akash and Prakash meet each other?

Accepted Answer

Introduction / Context: This is another relative speed problem, but instead of knowing the distance between the two cities directly, you are given the times that each person takes to travel the entire route. Akash and Prakash start at different times and travel at different constant speeds. The question asks at what time they cross each other on the route. Given Data / Assumptions: Akash travels from Mumbai to Bangalore, leaving at 6 a.m. and reaching at 10 a.m., so his total travel time is 4 hours. Prakash travels from Bangalore to Mumbai, leaving at 8 a.m. and reaching at 11:30 a.m., so his total travel time is 3.5 hours. They travel along the same straight route with constant speeds. We must find the time at which they meet. Concept / Approach: Let the distance between Mumbai and Bangalore be D km. Then Akash’s speed is D / 4 km/h and Prakash’s speed is D / 3.5 km/h. Let t hours be the time after 6 a.m. when they meet. At that time, Akash has been travelling for t hours, while Prakash has been travelling for t - 2 hours because he starts at 8 a.m. Their total distance covered at the meeting moment equals D. Step-by-Step Solution: Step 1: Speed of Akash v_A = D / 4 km/h. Step 2: Speed of Prakash v_P = D / 3.5 km/h. Step 3: Let t be the time in hours after 6 a.m. when they meet. Then Akash has travelled t hours, Prakash has travelled t - 2 hours. Step 4: Distance covered by Akash = v_A * t = (D / 4) * t. Step 5: Distance covered by Prakash = v_P * (t - 2) = (D / 3.5) * (t - 2). Step 6: At meeting, sum of distances = D, so (D / 4) * t + (D / 3.5) * (t - 2) = D. Step 7: Divide entire equation by D to get t / 4 + (t - 2) / 3.5 = 1. Step 8: Solve: t / 4 + (t - 2) / 3.5 = 1 gives t = 44 / 15 hours. Step 9: 44 / 15 hours ≈ 2.933 hours, which is 2 hours 56 minutes after 6 a.m., so meeting time is 8:56 a.m. Verification / Alternative check: At 8:56 a.m., Akash has travelled 2 hours 56 minutes, and Prakash has travelled 56 minutes. Akash’s speed is D / 4; distance covered by him is (D / 4) * (44 / 15) = 11D / 15. Prakash’s speed is D / 3.5 = 2D / 7; distance covered by him in 44 / 15 - 2 = 14 / 15 hours is (2D / 7) * (14 / 15) = 4D / 15. Total distance equals 11D / 15 + 4D / 15 = D, confirming that they meet at this time. Why Other Options Are Wrong: 10 a.m.: Too late; by this time they would already have crossed each other. 8:32 a.m.: Too early; the algebraic equation does not support this time. 9:20 a.m.: Also inconsistent with the ratio of their speeds and travel times. Common Pitfalls: Forgetting that Prakash starts 2 hours after Akash, leading to incorrect time expressions. Cancelling distance D incorrectly and making algebraic mistakes when solving for t. Converting fractional hours to minutes inaccurately. Final Answer: Akash and Prakash meet each other at 8:56 a.m.

Question 7

In a 1 km race, runner A gives runner B a start of 100 metres, and in another 1 km race, runner B gives runner C a start of 80 metres. In a 1 km race among A, B and C, who will win and by what distance will the winner beat the slowest runner?

Accepted Answer

Introduction / Context: This is a classic race and relative speed problem involving three runners and different head starts. The head start tells you about the ratio of speeds between runners. Once you know those ratios, you can find how far each runner covers in the time taken by the winner to finish a race and hence calculate the difference in distances between them. Given Data / Assumptions: In a 1 km race between A and B, A gives B a start of 100 metres and they finish together. In a 1 km race between B and C, B gives C a start of 80 metres and they finish together. In a 1 km race among A, B and C, we must find how much the winner beats the slowest runner by in terms of distance. All runners maintain constant speeds in all races. Concept / Approach: From the first condition, when A covers 1000 metres, B covers 900 metres in the same time. Thus, speed ratio v_B / v_A = 900 / 1000 = 9 / 10. From the second condition, when B covers 1000 metres, C covers 920 metres, so v_C / v_B = 920 / 1000 = 23 / 25. By multiplying these ratios, we can find v_C relative to v_A. Then, in a common 1 km race, we measure how far B and C run in the time it takes A to cover 1000 metres. Step-by-Step Solution: Step 1: From A versus B race: v_B / v_A = 900 / 1000 = 9 / 10. Step 2: From B versus C race: v_C / v_B = 920 / 1000 = 23 / 25. Step 3: Combine ratios to express v_C in terms of v_A: v_C / v_A = (v_C / v_B) * (v_B / v_A) = (23 / 25) * (9 / 10) = 207 / 250. Step 4: Now in a common race, let v_A = 1 unit speed. Then v_B = 9 / 10, v_C = 207 / 250. Step 5: Time taken by A to complete 1000 m = 1000 / v_A = 1000 units of time. Step 6: Distance covered by B in this time = v_B * time = (9 / 10) * 1000 = 900 metres. Step 7: Distance covered by C in this time = v_C * time = (207 / 250) * 1000 = 828 metres. Step 8: C is the slowest runner, covering 828 m when A finishes 1000 m. Difference = 1000 - 828 = 172 metres. Verification / Alternative check: We know that A is the fastest because A beats B with a head start, and B beats C with another head start. It is consistent that C has the smallest speed. The computed distances 1000, 900 and 828 clearly reflect that A is first, B is next and C is last. The gap between A and C is 172 metres, which matches our calculation. Why Other Options Are Wrong: 182 m: Slightly larger than the correct value, likely due to arithmetic errors when multiplying fractions. 152 m: Underestimates the gap between A and C. 162 m: Also underestimates the difference and does not align with the precise ratios. Common Pitfalls: Confusing which runner is faster when interpreting the given head starts. Using 1000 - 920 or 1000 - 900 directly without combining the two races through speed ratios. Making mistakes in fraction multiplication when computing v_C relative to v_A. Final Answer: In the 1 km race, the fastest runner A beats the slowest runner C by 172 m.

Question 8

A train overtakes two girls who are walking towards it at speeds of 3 km/h and 6 km/h and passes them completely in 36 seconds and 30 seconds respectively. What is the length of the train (in metres)?

Accepted Answer

Introduction / Context: This time and distance question tests the concept of relative speed when a train overtakes people walking in the opposite direction, along with correct unit conversion between kilometres per hour and metres per second. The key idea is that the length of the train is the distance it travels relative to each girl while completely passing her. Given Data / Assumptions: - Speed of the train = V km/h (unknown). - Speed of first girl = 3 km/h, walking towards the train. - Speed of second girl = 6 km/h, also walking towards the train. - Time taken to overtake first girl completely = 36 seconds. - Time taken to overtake second girl completely = 30 seconds. - Length of the train = L metres, the same in both situations. - All motions are along the same straight track and speeds are constant. Concept / Approach: When two bodies move towards each other, their relative speed is the sum of their individual speeds. The distance covered while completely overtaking a walker is equal to the length of the train. We must convert km/h to m/s using 5/18 and then form two equations for L using the two given times. Equating these expressions for L lets us solve first for the speed of the train and then for its length in metres. Step-by-Step Solution: Let the speed of the train be V km/h and its length be L metres. Relative speed with the first girl = (V + 3) km/h. Convert to m/s: (V + 3) * 5 / 18. Distance to overtake = L = relative speed * time = (V + 3) * 5 / 18 * 36. So L = (V + 3) * 10. (Equation 1) Relative speed with the second girl = (V + 6) km/h. In m/s: (V + 6) * 5 / 18. Again L = (V + 6) * 5 / 18 * 30. So L = (V + 6) * 25 / 3. (Equation 2) Equate both expressions for L: (V + 3) * 10 = (V + 6) * 25 / 3. Multiply both sides by 3: 30(V + 3) = 25(V + 6). 30V + 90 = 25V + 150. 5V = 60, so V = 12 km/h. Substitute in Equation 1: L = (12 + 3) * 10 = 15 * 10 = 150 metres. Verification / Alternative check: If the train is 150 m long and moves at 12 km/h, its speed in m/s is 12 * 5 / 18 = 10/3 m/s. With the first girl, relative speed is (12 + 3) km/h = 15 km/h = 25/3 m/s. Time to cover 150 m is 150 / (25/3) = 18 seconds * 2 = 36 seconds, which matches the question. With the second girl, relative speed is (12 + 6) = 18 km/h = 5 m/s. Time = 150 / 5 = 30 seconds, which also matches. So the result is consistent with both sets of data. Why Other Options Are Wrong: 120 m: Using 120 m as the length does not satisfy both overtaking times when you solve for the speed, so it contradicts the given numbers. 125 m: This length also fails to give 36 seconds and 30 seconds simultaneously for any single train speed. None of these: There is a valid consistent length of 150 m, so this option is incorrect. Common Pitfalls: A common mistake is to subtract the walking speeds instead of adding them, even though the girls walk towards the train, not in the same direction. Another frequent error is to forget to convert km/h to m/s before using time in seconds, which produces inconsistent answers. Some learners also assume that the two given times must be averaged instead of building proper equations from the two situations. Final Answer: The length of the train is 150 m.

Question 9

Rajeev travels 80 km from his home to college. One day he leaves 1 hour later than usual, so he increases his speed by 4 km/h and still reaches the college at the normal time. What is his increased speed (in km/h)?

Accepted Answer

Introduction / Context: This problem tests your understanding of the relationship between distance, speed, and time, and how a change in departure time can be compensated for by increasing speed. Rajeev travels a fixed distance every day, and on one particular day he leaves late but still manages to arrive at his college at the usual time by walking faster. Given Data / Assumptions: - Distance between home and college = 80 km. - Usual speed of Rajeev = v km/h (unknown). - On the late day, his speed = v + 4 km/h. - He leaves 1 hour later than normal but still reaches at the normal arrival time. - Travel is along a straight route at constant speed. Concept / Approach: The fundamental formula is distance = speed * time. Since the distance remains the same on both days, we can express the usual and increased travel times in terms of v. Because he departs 1 hour late but arrives at the normal time, the travel time on the late day must be exactly 1 hour less than his usual travel time. This gives a simple equation in v that we can solve to get the usual and then the increased speed. Step-by-Step Solution: Let the usual speed of Rajeev be v km/h. Usual time to travel 80 km = 80 / v hours. On the late day, his speed is v + 4 km/h. So time on the late day = 80 / (v + 4) hours. He leaves 1 hour late but still arrives at the same time, so: 80 / (v + 4) = 80 / v - 1. Rearrange: 80 / v - 80 / (v + 4) = 1. Multiply through by v (v + 4): 80 (v + 4) - 80 v = v (v + 4). 80 v + 320 - 80 v = v^2 + 4 v. 320 = v^2 + 4 v. Rearrange: v^2 + 4 v - 320 = 0. Solve the quadratic: v = 16 km/h (reject the negative root). Thus the increased speed = v + 4 = 20 km/h. Verification / Alternative check: At the usual speed of 16 km/h, time taken = 80 / 16 = 5 hours. At the increased speed of 20 km/h, time taken = 80 / 20 = 4 hours. The difference in travel time is 5 - 4 = 1 hour, which matches the given condition about leaving 1 hour late and still arriving on time. So our calculation is consistent. Why Other Options Are Wrong: 28 km/h: This would not give a travel time 1 hour less than the usual time for a consistent distance of 80 km. 30 km/h and 40 km/h: These speeds would give even shorter travel times, and no usual speed would satisfy the given 1 hour difference with these as v + 4. Common Pitfalls: A common mistake is to mix up which time is greater or to add 1 hour instead of subtracting it in the equation. Another frequent error is to plug in options directly without forming a proper time equation, which can be confusing when numbers are larger. Always relate changes in departure time to changes in travel time using the basic distance formula. Final Answer: The increased speed of Rajeev on that day is 20 km/h.

Question 10

A train with 120 wagons overtakes Arun, who is riding a horse in the same direction, and completely passes him in 36 seconds. From the instant it starts overtaking Arun, the train runs for 30 minutes before it starts overtaking Sriram, who is riding…

Accepted Answer

Introduction / Context: This is a multi step time and distance puzzle involving a long train and two riders on horses. The train first overtakes Arun, then after running for half an hour it overtakes Sriram coming from the opposite direction. The question asks for the time between the instant when the train finishes overtaking Sriram and the moment when Arun and Sriram meet each other on the track. The key ideas are relative speed and careful use of a time line. Given Data / Assumptions: - Train speed = V m/s (unknown). - Length of the train = L metres (unknown). - Speed of Arun on his horse = vA m/s, same direction as the train. - Speed of Sriram on his horse = vS m/s, opposite direction to the train. - Train overtakes Arun completely in 36 seconds when moving in the same direction. - Train overtakes Sriram completely in 24 seconds when moving in the opposite direction. - From the instant it starts overtaking Arun to the instant it starts overtaking Sriram, 30 minutes = 1800 seconds elapse. - All bodies move at constant speeds along a straight track. Concept / Approach: When the train overtakes someone moving in the same direction, the relative speed is V - vA. When it overtakes someone moving towards it, the relative speed is V + vS. In both cases, the distance covered relative to the rider equals the length of the train L. Using the two overtaking times, we build equations for L. Then we relate the positions of Arun and Sriram at the start of each overtaking event and find when they will meet each other. An important observation is that the final answer turns out to be independent of the actual numerical values of V, vA, vS, and L. Step-by-Step Solution: From overtaking Arun (same direction): (V - vA) * 36 = L. (1) From overtaking Sriram (opposite direction): (V + vS) * 24 = L. (2) Equate (1) and (2): (V - vA) * 36 = (V + vS) * 24. Divide by 12: 3(V - vA) = 2(V + vS). So 3V - 3 vA = 2V + 2 vS, which gives V = 3 vA + 2 vS. Take t = 0 at the instant the train starts overtaking Arun. Position of train front: xT(t) = x0 + V t. Position of Arun: xA(t) = x0 + vA t. At t = 1800 seconds, the train starts overtaking Sriram: xT(1800) = xS(1800). Let Sriram's position be xS(t) = s0 - vS t (coming from opposite direction). Then x0 + V * 1800 = s0 - vS * 1800, so s0 - x0 = 1800 (V + vS). Arun and Sriram meet when xA(t) = xS(t). So x0 + vA t = s0 - vS t, which gives (vA + vS) t = s0 - x0. Substitute s0 - x0 from above: (vA + vS) t = 1800 (V + vS). But V + vS = 3 vA + 3 vS = 3 (vA + vS). Therefore t = 1800 * 3 (vA + vS) / (vA + vS) = 5400 seconds. The train finishes overtaking Sriram at t = 1800 + 24 = 1824 seconds. Time after crossing Sriram when Arun meets Sriram = 5400 - 1824 = 3576 seconds. Verification / Alternative check: Notice that the final time 5400 seconds (for Arun and Sriram to meet) depends only on the 1800 second gap and the relationship V = 3 vA + 2 vS. The extra 24 seconds for overtaking Sriram simply shifts the time of crossing, and subtracting it from 5400 seconds gives 3576 seconds. No contradictory information arises, so the numerical answer is consistent and unique. Any numerical assignment of vA and vS satisfying V = 3 vA + 2 vS will reproduce the same time difference. Why Other Options Are Wrong: 3560 sec and 3600 sec: These values do not match the exact algebraic difference 5400 - 1824 derived from the equations and would imply inconsistent speeds or distances. Cannot be determined: Although the problem looks complicated, the information provided is sufficient, and unknown speeds cancel out, so the answer can be determined exactly. Common Pitfalls: The most common mistake is to try to find explicit speeds for the horses and the train unnecessarily, which can lead to algebraic mistakes. Another error is to misinterpret the 30 minutes as the time between the ends of overtaking rather than between the starts. Learners also sometimes forget that Arun and Sriram continue to move after all overtaking events, which is essential for finding when they meet each other. Final Answer: After the train has crossed Sriram, Arun and Sriram meet each other after 3576 seconds.

Question 11

Between 12 noon and 2:30 in the afternoon of the same day, through how many degrees does the hour hand of a clock move?

Accepted Answer

Introduction / Context: This question checks your understanding of how the hour hand of an analog clock moves and how to convert time intervals into angular movement in degrees. It is a simple but important concept that often appears in aptitude tests under the topic of clocks and time. Given Data / Assumptions: - Start time = 12:00 noon. - End time = 2:30 pm on the same day. - The hour hand completes one full circle, which is 360 degrees, in 12 hours. - The hour hand moves at a constant rate. Concept / Approach: The hour hand moves 360 degrees in 12 hours. Therefore, its angular speed is 360 / 12 degrees per hour. Once we know this rate, we find the total elapsed time between 12:00 and 2:30, convert that to hours, and multiply by the angular speed. This directly gives the number of degrees through which the hour hand has moved. Step-by-Step Solution: Time interval from 12:00 to 2:30 = 2 hours 30 minutes. Convert 2 hours 30 minutes to hours: 2.5 hours. Hour hand speed = 360 degrees in 12 hours. So angular speed = 360 / 12 = 30 degrees per hour. Total angle turned = angular speed * time. Total angle = 30 * 2.5 = 75 degrees. Verification / Alternative check: We can also reason by half hours. In one hour, the hour hand moves 30 degrees. In half an hour, it moves 15 degrees. From 12:00 to 2:00 pm is 2 hours, so the hour hand moves 2 * 30 = 60 degrees. From 2:00 to 2:30 pm is another 15 degrees. Total = 60 + 15 = 75 degrees, which matches the earlier calculation. Why Other Options Are Wrong: 720°: This is the angle for two full rotations of the clock, which would take 24 hours, not 2.5 hours. 180°: This corresponds to 6 hours of movement (half a day on the clock), not 2.5 hours. 65°: This does not correspond to any simple multiple of the per hour movement of the hour hand and does not match the correct calculation. Common Pitfalls: A frequent mistake is to confuse the movements of the hour hand and the minute hand. Another error is to forget to convert 30 minutes into 0.5 hours and instead treat it as 30 hours or ignore it entirely. Always convert the entire time interval into hours before multiplying by the hourly angular speed. Final Answer: Between noon and 2:30 pm, the hour hand moves through 75°.

Question 12

The ratio of the speeds of two trains is 7 : 8. If the second train covers 400 km in 4 hours, what is the speed (in km/h) of the first train?

Accepted Answer

Introduction / Context: This question examines the use of ratios in speed problems. When the speeds of two moving objects are in a given ratio and one of the actual speeds is known, the other can be found quickly using proportional reasoning. Here, we work with two trains whose speed ratio is 7 : 8. Given Data / Assumptions: - Ratio of speeds of the first train to the second train = 7 : 8. - The second train covers 400 km in 4 hours. - Both trains move at constant speeds. Concept / Approach: If the speeds of two trains are in the ratio 7 : 8, we can express their speeds as 7k and 8k for some common factor k. Once we compute the actual speed of the second train, we can set 8k equal to that value and find k. Then we use the value of k to find the speed of the first train as 7k. Step-by-Step Solution: Let the first train speed = 7k km/h. Let the second train speed = 8k km/h. The second train covers 400 km in 4 hours. Speed of second train = distance / time = 400 / 4 = 100 km/h. So 8k = 100, which gives k = 100 / 8 = 12.5. Speed of first train = 7k = 7 * 12.5 = 87.5 km/h. Verification / Alternative check: Check that the ratio of speeds matches 7 : 8. First train = 87.5 km/h, second train = 100 km/h. The ratio 87.5 : 100 simplifies by dividing both by 12.5 to 7 : 8. This confirms that our use of ratios is correct and consistent with the problem statement. Why Other Options Are Wrong: 83.5 km/h, 84.5 km/h, and 86.5 km/h: None of these values, when compared with 100 km/h, gives a simplified ratio of exactly 7 : 8. Each would lead to a different ratio, so they are not consistent with the original condition. Common Pitfalls: Learners sometimes reverse the ratio or assign 7k to the second train and 8k to the first train by mistake. Another error is to miscalculate the speed of the second train by mixing units or misreading 400 km in 4 hours. Always compute the real known speed first, then apply the ratio carefully. Final Answer: The speed of the first train is 87.5 km/h.

Question 13

A train covers a certain distance in 50 minutes when it runs at an average speed of 48 km/h. At what speed (in km/h) must the train run so that the journey time is reduced to 40 minutes for the same distance?

Accepted Answer

Introduction / Context: This question is a standard application of the distance, speed, and time relationship. The train first travels a certain distance at a known speed, and then we are asked to find a new speed that allows the same distance to be covered in less time. Problems of this type are common in aptitude tests for competitive exams. Given Data / Assumptions: - Initial average speed of the train = 48 km/h. - Time taken at this speed = 50 minutes. - New required time for the same distance = 40 minutes. - Distance of the journey remains the same in both cases. - Motion is along a straight track with constant speeds in each case. Concept / Approach: We first compute the distance covered using the initial speed and time. Then we use that fixed distance and the new travel time to determine the required new speed. The fundamental formula remains distance = speed * time. The main care needed is to keep all times in hours when using speeds in km/h. Step-by-Step Solution: Convert 50 minutes to hours: 50 / 60 = 5 / 6 hours. Distance of the journey = speed * time. Distance = 48 * (5 / 6) = 48 * 5 / 6 = 40 km. Now the same 40 km must be covered in 40 minutes. Convert 40 minutes to hours: 40 / 60 = 2 / 3 hours. Required new speed = distance / time = 40 / (2 / 3). Compute this: 40 * 3 / 2 = 60 km/h. Verification / Alternative check: At the original speed of 48 km/h, a 40 km journey would take 40 / 48 hours, which is 5 / 6 hours, equal to 50 minutes. At the new speed of 60 km/h, the time is 40 / 60 hours, equal to 2 / 3 hours, which is 40 minutes. This matches exactly the times mentioned in the problem, so our answer is consistent. Why Other Options Are Wrong: 50 km/h: This would give a time of 40 / 50 hours = 0.8 hours, that is 48 minutes, which is still too long. 65 km/h and 70 km/h: These speeds would make the train arrive even earlier than 40 minutes, not exactly in 40 minutes, so they do not satisfy the given condition. Common Pitfalls: A common error is to attempt to scale the speed directly by the ratio of the times without carefully computing the distance first. Another frequent mistake is forgetting to convert minutes into hours before applying the formula, which leads to incorrect numbers. Always keep time units consistent with speed units during calculations. Final Answer: The train must run at an average speed of 60 km/h to reduce the journey time to 40 minutes.

Question 14

A man reaches his office 20 minutes late if he walks from his home at 3 km/h, and he reaches 30 minutes early if he walks at 4 km/h. What is the distance between his home and his office?

Accepted Answer

Introduction / Context: This time and distance question involves early and late arrival based on two different walking speeds. The key idea is that there is a fixed scheduled time to reach the office, and changing the walking speed changes the arrival time relative to that schedule. We use this information to find the actual distance between home and office. Given Data / Assumptions: - Distance between home and office = D km (unknown). - Usual scheduled travel time = T hours (unknown). - If the man walks at 3 km/h, he is 20 minutes late, so time taken = T + 20/60 hours. - If he walks at 4 km/h, he is 30 minutes early, so time taken = T - 30/60 hours. - Motion is straight and speeds are constant. Concept / Approach: Distance is equal to speed multiplied by time. Since the distance from home to office is fixed, we can write two expressions for that distance using the two different speeds and arrival conditions. These give two equations in the two unknowns D and T. Solving the system of equations yields the distance D directly. Step-by-Step Solution: Let the distance be D km and the scheduled time be T hours. At 3 km/h: D = 3 * (T + 20/60). At 4 km/h: D = 4 * (T - 30/60). So we have: 3 (T + 1/3) = D. (1) 4 (T - 1/2) = D. (2) Set right sides equal: 3 (T + 1/3) = 4 (T - 1/2). Expand: 3T + 1 = 4T - 2. Rearrange: 1 + 2 = 4T - 3T, so 3 = T. Now substitute T = 3 in (1): D = 3 (3 + 1/3). D = 3 * (10/3) = 10 km. Verification / Alternative check: If D = 10 km, then at 3 km/h the time taken is 10 / 3 hours, which is 3 hours 20 minutes. The scheduled time T is 3 hours, so he is 20 minutes late, which matches the statement. At 4 km/h, the time taken is 10 / 4 = 2.5 hours, which is 2 hours 30 minutes. This is 30 minutes earlier than the scheduled 3 hours. Both conditions are satisfied, so D = 10 km is correct. Why Other Options Are Wrong: 20 km, 16 km, 14 km: If you plug any of these distances into the time calculations with 3 km/h and 4 km/h, you will not get arrival times that differ from the schedule by exactly 20 minutes late and 30 minutes early. Each fails at least one of the conditions given in the problem. Common Pitfalls: Some learners mistakenly add or subtract both 20 minutes and 30 minutes from the same travel time, which is incorrect. Others forget to convert minutes to hours, leading to inconsistent equations. Always express all times in hours when speeds are given in km/h, and clearly write two separate equations for the two situations. Final Answer: The distance between his home and office is 10 km.

Question 15

A man travelling in a train counts 41 telephone posts passing by in one minute. The posts are known to be spaced 50 metres apart along the track. At what speed (in km/h) is the train travelling?

Accepted Answer

Introduction / Context: This is a classic speed estimation question where an observer counts equally spaced objects along the track, such as telephone posts. By relating the number of posts, the distance between them, and the time taken, we can work out the speed of the train. The important detail is how to convert the counted posts into actual distance travelled. Given Data / Assumptions: - The man counts 41 telephone posts in 1 minute. - Distance between consecutive telephone posts = 50 metres. - The train moves at a constant speed along a straight track. - Counting starts as the train is opposite one post and continues until the 41st post passes him. Concept / Approach: If there are n posts, there are n - 1 equal gaps between them. When the observer starts counting at one post and ends when the 41st post passes, the train has effectively travelled across 40 gaps between posts. We multiply the number of gaps by the spacing to get distance, then divide by time to get speed in m/s, and finally convert to km/h. Step-by-Step Solution: Number of posts counted = 41. Number of gaps between them = 41 - 1 = 40. Distance between posts = 50 m. Total distance travelled in 1 minute = 40 * 50 = 2000 m. Time taken = 1 minute = 60 seconds. Speed in m/s = distance / time = 2000 / 60 = 100 / 3 m/s. Convert to km/h: multiply by 18 / 5. Speed in km/h = (100 / 3) * 18 / 5 = (100 * 18) / (15) = 1800 / 15 = 120 km/h. Verification / Alternative check: We can do an approximate mental check: if the train travels about 2 km (2000 m) in one minute, then in 60 minutes it would travel about 120 km, giving a speed of around 120 km/h. This quick check confirms that our exact calculation of 120 km/h is reasonable and fits the values in the problem. Why Other Options Are Wrong: 60 km/h: This would correspond to roughly 1 km in 1 minute, not 2 km, and would involve crossing far fewer posts. 100 km/h and 110 km/h: These values are lower than the correct speed and would imply fewer than 40 gaps of 50 m each in one minute, so they do not match the data. Common Pitfalls: Many students incorrectly multiply the number of posts by the spacing, using 41 * 50, without realising that the distance is determined by the gaps. Others forget to convert metres per second to kilometres per hour by using the factor 18 / 5, which leads to an answer in the wrong units. Being precise about gaps and unit conversion avoids these errors. Final Answer: The speed of the train is 120 km/h.

Question 16

Deepesh walks a distance of 60 km at his usual speed. If he had walked 20 km/h faster, he would have saved 1 hour in travel time. What is his usual walking speed (in km/h)?

Accepted Answer

Introduction / Context: This time and distance question focuses on how a change in speed affects total travel time over a fixed distance. Deepesh covers 60 km, and we are told how much time he would save if his speed increased by a fixed amount. The key is to set up a clear equation relating the two travel times and then solve for his usual speed. Given Data / Assumptions: - Distance travelled = 60 km. - Usual speed of Deepesh = v km/h (unknown). - Increased speed = v + 20 km/h. - Time saved by walking faster = 1 hour. - Motion is along a straight path with constant speeds in each scenario. Concept / Approach: Travel time is equal to distance divided by speed. The usual travel time is 60 / v hours, and the faster travel time is 60 / (v + 20) hours. The problem statement says that the time difference between these two is exactly 1 hour. This allows us to form an equation in a single variable v and solve it using algebraic methods for quadratic equations. Step-by-Step Solution: Usual time = 60 / v hours. Time at faster speed = 60 / (v + 20) hours. Given that the time saved is 1 hour: 60 / v - 60 / (v + 20) = 1. Multiply both sides by v (v + 20): 60 (v + 20) - 60 v = v (v + 20). Simplify left side: 60 v + 1200 - 60 v = 1200. So 1200 = v^2 + 20 v. Rearrange: v^2 + 20 v - 1200 = 0. Solve this quadratic equation. The positive root is approximately v = -10 + 10 * sqrt(13), which is about 26.06 km/h. This value is not equal to 100, 120, or 150 km/h. Verification / Alternative check: Using v approximately 26.06 km/h, usual time = 60 / 26.06 hours which is about 2.30 hours. Faster speed = v + 20 is about 46.06 km/h, so faster time = 60 / 46.06 which is about 1.30 hours. The difference between the two times is approximately 1 hour, matching the condition in the question. This confirms that our computed speed is correct even though it does not appear explicitly among the listed numeric options. Why Other Options Are Wrong: 100, 120, and 150: If any of these were the usual speed, then adding 20 km/h would give an even larger speed and the time difference between the two speeds for a 60 km journey would be much smaller than 1 hour. Substituting them into the equation 60 / v - 60 / (v + 20) = 1 quickly shows they do not satisfy the condition. Common Pitfalls: Many students guess one of the large integer speeds or attempt to treat 20 km/h as a small correction without solving the quadratic carefully. Others forget that both speeds involve the same distance of 60 km. Always write the two time expressions clearly and form the exact equation rather than relying on rough estimation. Also remember that the correct choice in multiple choice questions can be “None of these” when the calculated value does not match any given number. Final Answer: Deepesh's usual walking speed is approximately 26.06 km/h, so among the given options the correct choice is None of these.

Question 17

A man covers half of his journey by train at 60 km/h, half of the remaining distance by bus at 30 km/h and the rest by cycle at 10 km/h. What is his average speed (in km/h) for the entire journey?

Accepted Answer

Introduction / Context: This question tests your understanding of average speed when a journey is divided into segments with different speeds and different fractions of the total distance. The key is to handle distance fractions carefully and compute the total time taken before finding the average speed over the entire journey. Given Data / Assumptions: - Total journey distance = D km (unknown). - First half of the journey (D / 2) is covered by train at 60 km/h. - Half of the remaining distance, that is D / 4, is covered by bus at 30 km/h. - The final D / 4 is covered by cycle at 10 km/h. - All speeds are constant for their respective segments. Concept / Approach: Average speed over an entire trip is defined as total distance divided by total time. Here the total distance is D, so we need to compute the time taken on each segment and add them. Once the total time is known, average speed is simply D divided by that total time. The absolute value of D will cancel out in the computation, so we can leave it symbolic. Step-by-Step Solution: First segment distance = D / 2 at 60 km/h. Time for first segment = (D / 2) / 60 = D / 120 hours. Remaining distance after first segment = D / 2. Second segment distance = half of the remaining = D / 4 at 30 km/h. Time for second segment = (D / 4) / 30 = D / 120 hours. Third segment distance = remaining D / 4 at 10 km/h. Time for third segment = (D / 4) / 10 = D / 40 hours. Total time T = D / 120 + D / 120 + D / 40. Combine: D / 120 + D / 120 = D / 60. So T = D / 60 + D / 40. Take common denominator 120: T = 2D / 120 + 3D / 120 = 5D / 120 = D / 24 hours. Average speed = total distance / total time = D / (D / 24) = 24 km/h. Verification / Alternative check: We can choose a convenient value for D, for example D = 24 km. Then time by train = 12 / 60 = 0.2 hours, time by bus = 6 / 30 = 0.2 hours, time by cycle = 6 / 10 = 0.6 hours. Total time = 1.0 hour for 24 km, giving an average speed of 24 km/h. This numeric check matches the symbolic solution exactly. Why Other Options Are Wrong: 32 km/h, 20 km/h, and 18 km/h: These values do not result from the actual distance fractions and speeds. They could come from incorrectly averaging the speeds directly or misunderstanding the phrasing “half of the remaining distance.” Average speed must be based on total distance divided by total time, not a simple arithmetic mean of the speeds. Common Pitfalls: Many learners incorrectly take the average of 60, 30, and 10, or they treat each third of the distance equally instead of using half and quarter segments. Another error is to forget that the total distance is split as D / 2, D / 4, and D / 4. Always write each segment's distance and time clearly before computing average speed. Final Answer: The average speed over the entire journey is 24 km/h.

Question 18

Runners A and B run around a circular track. In a four-lap race, A finishes one full lap ahead of B, and this one-lap lead is also equal to a time advantage of 10 minutes. If the race had been only one lap, what time (in minutes) would A take to com…

Accepted Answer

Introduction / Context: This problem is a race and relative speed question on a circular track. Two runners complete four laps, and A finishes exactly one lap ahead of B. The problem also states that this one lap lead corresponds to a time difference of 10 minutes. We are asked to find how long A would take to complete a single lap if the race were only one lap long. Given Data / Assumptions: - Total race distance = 4 laps of the circular track. - Runner A completes 4 laps in time TA hours (or minutes). - Runner B completes 4 laps in time TB hours, with TB - TA = 10 minutes. - At the moment A finishes 4 laps, B has completed only 3 laps, so A beats B by one lap. - Speeds of both runners are constant. Concept / Approach: If A finishes 4 laps while B finishes only 3 laps in the same time, the ratio of their speeds equals the ratio of distances covered, that is 4 : 3. Then we express their times for 4 laps in terms of a common lap length and these speeds. The given information that the difference between their total times for 4 laps is 10 minutes allows us to find A's time for one lap. Step-by-Step Solution: Let length of one lap be L units. Let A's speed be vA and B's speed be vB. When A finishes 4 laps, A covers 4L and B covers 3L in the same time. So vA : vB = 4 : 3. Time taken by A for 4 laps = TA = 4L / vA. Time taken by B for 4 laps = TB = 4L / vB. Since vB = 3 vA / 4 from the ratio, TB = 4L / (3 vA / 4) = 16L / (3 vA). Difference in their times for 4 laps is TB - TA = 10 minutes. Compute TA first: TA = 4L / vA. Now TB - TA = 16L / (3 vA) - 4L / vA. Take common denominator 3 vA: TB - TA = (16L - 12L) / (3 vA) = 4L / (3 vA). This 4L / (3 vA) is given as 10 minutes. So 4L / (3 vA) = 10 minutes. Time taken by A for one lap = tA1 = L / vA. Multiply both sides of 4L / (3 vA) = 10 by 3 / 4: L / vA = 10 * 3 / 4 = 7.5 minutes. Verification / Alternative check: If A takes 7.5 minutes per lap, then A needs 4 * 7.5 = 30 minutes for 4 laps. The time difference of 10 minutes implies B needs 40 minutes for 4 laps, so B's per lap time is 10 minutes. In 30 minutes (when A has run 4 laps), B would have run 3 laps (3 * 10 = 30 minutes), matching the statement that A beats B by one lap. This confirms that 7.5 minutes for one lap is consistent. Why Other Options Are Wrong: 8 min and 12 min: These values do not produce the correct combination of a one lap lead and a 10 minute time difference when tested over 4 laps. 12.5 min: This would make A significantly slower, which contradicts the condition that A beats B by one lap and by 10 minutes in a 4 lap race. Common Pitfalls: Learners sometimes use the 10 minute difference as the time for one lap directly without relating it to the speed ratio. Another error is to assume that the time difference for one lap is constant instead of being derived from the total 4 lap race. Always start from distance and speed ratios, then use the time difference over the full race to compute the per lap time. Final Answer: If the race were only one lap, runner A would take 7.5 minutes to complete the course.

Question 19

Person X starts from place P three hours earlier at 40 km/h. Person Y starts from the same place P at 3:00 pm at 60 km/h in the same direction. What is the difference in time (in hours) between the moment when X is 30 km ahead of Y and the moment wh…

Accepted Answer

Introduction / Context: This question is about relative motion in the same direction. One person starts earlier at a lower speed, and another starts later at a higher speed. We are asked to find how much time separates two events: first when the earlier person leads by 30 km, and later when the faster person leads by 30 km. The key tool is relative speed between the two travellers. Given Data / Assumptions: - Person X starts from P three hours before Y and travels at 40 km/h. - Person Y starts from P at 3:00 pm and travels at 60 km/h in the same direction. - At some time after 3:00 pm, X is 30 km ahead of Y. - At a later time, Y is 30 km ahead of X. - We need the difference between these two times, measured from the 3:00 pm start of Y. Concept / Approach: Take Y's starting time as the reference time t = 0. Express the positions of X and Y as functions of time t (in hours) after 3:00 pm. We then set up equations for the two situations: X leading by 30 km and Y leading by 30 km. Solving for the two times and taking the difference gives the required answer. Step-by-Step Solution: Let t be the time in hours after 3:00 pm (when Y starts). X has already been travelling for 3 hours at 40 km/h. So at t = 0, X is 40 * 3 = 120 km ahead of P. Position of X at time t: xX(t) = 120 + 40 t. Position of Y at time t: xY(t) = 60 t. First condition: X is 30 km ahead of Y. So xX(t1) - xY(t1) = 30. That is, 120 + 40 t1 - 60 t1 = 30. Simplify: 120 - 20 t1 = 30, so 20 t1 = 90, giving t1 = 4.5 hours. Second condition: Y is 30 km ahead of X. So xY(t2) - xX(t2) = 30. That is, 60 t2 - (120 + 40 t2) = 30. Simplify: 60 t2 - 120 - 40 t2 = 30, so 20 t2 - 120 = 30. Thus 20 t2 = 150 and t2 = 7.5 hours. Required difference in time = t2 - t1 = 7.5 - 4.5 = 3 hours. Verification / Alternative check: At t1 = 4.5 hours after 3:00 pm, Y has travelled 60 * 4.5 = 270 km, and X has travelled 120 + 40 * 4.5 = 300 km, a difference of 30 km. At t2 = 7.5 hours, Y has travelled 60 * 7.5 = 450 km, while X has travelled 120 + 40 * 7.5 = 420 km, so now Y leads by 30 km. The two conditions are satisfied perfectly, confirming the calculations. Why Other Options Are Wrong: 2 h, 3.5 h, and 4.25 h: These values do not match the actual difference t2 - t1 obtained from correctly solving the position equations. Any other value would imply incorrect positions for X and Y at one or both of the described events. Common Pitfalls: Some learners mistakenly ignore the 3 hour head start of X or choose the wrong reference time origin, which leads to incorrect equations. Another common mistake is to misinterpret the phrase “ahead of” and reverse the sign in the equations. Always define positions clearly and be consistent with the direction and the reference time. Final Answer: The difference in time between the two events is 3 hours.

Question 20

An employee can claim Rs 7 per km when travelling by taxi and Rs 6 per km when using his own car. In one week he claims a total of Rs 595 for travelling 90 km in all. How many kilometres did he travel by taxi?

Accepted Answer

Introduction / Context: This is a typical linear equations problem mixing money, distance, and two different rates of reimbursement. The employee sometimes travels by taxi and sometimes by his own car, and we know the total distance and total amount claimed. We must find how many kilometres were travelled by taxi. Given Data / Assumptions: - Rate for taxi travel = Rs 7 per km. - Rate for own car travel = Rs 6 per km. - Total distance travelled in the week = 90 km. - Total amount claimed = Rs 595. - Let x be the kilometres travelled by taxi and y be the kilometres travelled by own car. Concept / Approach: We form a system of two linear equations in two variables using the distance and money information. The first equation comes from total distance (x + y = 90). The second equation comes from total reimbursement (7 x + 6 y = 595). Solving the system gives the value of x, the distance travelled by taxi. Step-by-Step Solution: Let taxi distance be x km and own car distance be y km. Total distance equation: x + y = 90. (1) Total claim equation: 7 x + 6 y = 595. (2) From (1), y = 90 - x. Substitute into (2): 7 x + 6 (90 - x) = 595. Compute: 7 x + 540 - 6 x = 595. So x + 540 = 595. Thus x = 595 - 540 = 55. Therefore, he travelled 55 km by taxi. Verification / Alternative check: If x = 55 km by taxi, then y = 90 - 55 = 35 km by own car. Money from taxi travel = 55 * 7 = Rs 385. Money from own car travel = 35 * 6 = Rs 210. Total claim = 385 + 210 = Rs 595, which matches the given amount. This confirms that the numbers are consistent and correct. Why Other Options Are Wrong: 35 km and 25 km: These values, if used as taxi distances, will not satisfy both the total distance and total claim equations simultaneously. 65 km: This would make taxi distance too large and require a negative distance by own car to keep the total at 90 km, which is impossible. Common Pitfalls: Some learners mix up which variable refers to which mode of travel and may accidentally assign Rs 7 to the own car or vice versa. Others forget to enforce the total distance equation and try to work only with the money equation, which is not sufficient. Always define your variables clearly and write both equations before solving. Final Answer: The employee travelled 55 km by taxi.

Time and Distance Questions