Question 1

A car travels from one place to another at 24 km/h in one direction and returns at 36 km/h. What is the average speed of the car for the entire journey?

Accepted Answer

Introduction / Context:This question deals with average speed over a round trip where the distances in both directions are the same but the speeds are different. It tests your understanding that the average speed is not the simple arithmetic mean of the two speeds. Instead, a special harmonic mean formula must be used when the distance covered in each direction is equal. Given Data / Assumptions:Speed while going from the starting point to the destination = 24 km/h.Speed while returning from the destination back to the starting point = 36 km/h.The distance between the two places in each direction is the same.The car moves without any stoppages or delays between the two speeds.We are required to find the overall average speed for the complete journey, that is, going and coming back. Concept / Approach:When a vehicle travels the same distance at two different speeds, the average speed is computed using the harmonic mean: average speed = 2 * v1 * v2 / (v1 + v2). This arises because total time is distance over speed for each leg, and we must divide total distance by total time. Directly averaging the speeds gives the wrong answer and is a common trap in exams. Step-by-Step Solution:Step 1: Let the one way distance be D km.Step 2: Time taken to go from place 1 to place 2 = D / 24 hours.Step 3: Time taken to return = D / 36 hours.Step 4: Total distance for the round trip = D + D = 2D km.Step 5: Total time for the round trip = D / 24 + D / 36 hours.Step 6: Combine time: D * (1 / 24 + 1 / 36) = D * (3 / 72 + 2 / 72) = D * (5 / 72) hours.Step 7: Average speed = total distance / total time = 2D / (D * 5 / 72).Step 8: Simplify: 2D * 72 / (5D) = 144 / 5 = 28.8 km/h. Verification / Alternative check:The direct harmonic mean formula for equal distances is average speed = 2 * v1 * v2 / (v1 + v2). Substituting v1 = 24 and v2 = 36 gives 2 * 24 * 36 / (24 + 36) = 1728 / 60 = 28.8 km/h, which matches our detailed computation. This confirms that the answer is correct and the reasoning is consistent. Why Other Options Are Wrong:30 km/h is close to the arithmetic mean of the two speeds but not the correct harmonic mean, so it overestimates the true average speed.27.6 km/h and 26.4 km/h do not match the result of the correct formula and arise from incorrect mixing of distance and time values.31.2 km/h is higher than both the harmonic mean and the lower individual speed, which is impossible for a round trip where some part of the journey is done at only 24 km/h. Common Pitfalls:Many students mistakenly take the average of 24 and 36 as (24 + 36) / 2 = 30 km/h without considering that time spent at each speed is different. Remember that average speed is always total distance divided by total time, and for equal distances with two speeds, the harmonic mean formula should be applied. This understanding is critical for mastering time and distance problems. Final Answer:The average speed of the car for the entire journey is 28.8 km/h.

Question 2

A gun is fired at a distance of 1.7 km from Ram, and he hears the sound after 25 seconds. What is the speed of sound in metres per second based on this situation?

Accepted Answer

Introduction / Context:This question is a straightforward application of the basic speed, distance, and time relationship. It uses a real life scenario involving the sound of a gunshot to estimate the speed of sound in air. Such questions help reinforce the formula speed = distance divided by time and the importance of unit conversion. Given Data / Assumptions:The horizontal distance between Ram and the gun is 1.7 km.Ram hears the sound 25 seconds after the gun is fired.The speed of sound is assumed to be constant over this distance.We ignore any effects of wind, temperature variation, or obstacles.The answer is required in metres per second (m/s). Concept / Approach:The basic formula used is speed = distance / time. However, distance is given in kilometres while time is in seconds. To get speed in metres per second, we must first convert kilometres to metres. One kilometre equals 1000 metres. After the conversion, we divide the distance in metres by the time in seconds to get the required speed. This is a routine but very important conversion step in physics and aptitude problems. Step-by-Step Solution:Step 1: Convert the given distance from kilometres to metres.Distance = 1.7 km = 1.7 * 1000 m.Distance = 1700 m.Step 2: Note the time taken for the sound to reach Ram.Time = 25 seconds.Step 3: Use the formula speed = distance / time.Speed of sound = 1700 / 25 m/s.Step 4: Simplify the division: 1700 / 25 = 68.Step 5: Therefore, the speed of sound based on this observation is 68 m/s. Verification / Alternative check:You can check by multiplication. If the speed is 68 m/s and time is 25 seconds, distance covered by sound = 68 * 25 = 1700 m, which matches the converted distance from 1.7 km. This confirms that the calculation is consistent with the given data. Although actual physical speed of sound in air is usually higher, aptitude questions expect you to trust and use the provided numbers only. Why Other Options Are Wrong:60 m/s, 62 m/s, and 64 m/s are all lower than the correct computed value and do not satisfy distance = speed * time for the given data.72 m/s is greater than the correct result and would produce a larger distance than 1700 m for 25 seconds, which contradicts the problem statement. Common Pitfalls:Common mistakes include forgetting to convert kilometres to metres and directly dividing 1.7 by 25, which would give a speed in km per second and not in m/s. Another error is to use approximate physical values for speed of sound instead of strictly following the numbers given in the problem. Exam questions like this are designed to test consistent use of the formula with proper units. Final Answer:The speed of sound in this scenario is 68 m/s.

Question 3

Two cars travel from city A to city B, one at 42 km/h and the other at 60 km/h. If the faster car takes 2 hours less than the slower car to cover the journey, what is the distance between city A and city B?

Accepted Answer

Introduction / Context:This time and distance problem involves two vehicles traveling the same route at different constant speeds. The key information given is the difference in time taken by each vehicle. Such questions are standard in aptitude tests and help measure your ability to translate verbal conditions into algebraic equations. Given Data / Assumptions:Speed of the slower car = 42 km/h.Speed of the faster car = 60 km/h.Both cars travel the same distance from city A to city B.The faster car takes 2 hours less than the slower car to cover this distance.We need to find the distance between the two cities in kilometres. Concept / Approach:Let the common distance be D km. Time taken by the slower car is D / 42 hours. Time taken by the faster car is D / 60 hours. The problem states that the difference in these times is 2 hours. This leads to the equation D / 42 − D / 60 = 2. Solving this equation gives the value of D. The method relies purely on the standard relationship time = distance divided by speed. Step-by-Step Solution:Step 1: Assume the distance between city A and city B is D km.Step 2: Time taken by slower car = D / 42 hours.Step 3: Time taken by faster car = D / 60 hours.Step 4: According to the question, D / 42 − D / 60 = 2.Step 5: Factor out D: D * (1 / 42 − 1 / 60) = 2.Step 6: Compute 1 / 42 − 1 / 60: that is (60 − 42) / (42 * 60) = 18 / 2520.Step 7: Simplify 18 / 2520 to 3 / 420 or 1 / 140.Step 8: So the equation becomes D * (1 / 140) = 2.Step 9: Therefore D = 2 * 140 = 280 km. Verification / Alternative check:Check the times using D = 280 km. Time for slower car = 280 / 42 = 20 / 3 hours, which is 6 hours and 40 minutes. Time for faster car = 280 / 60 = 14 / 3 hours, which is 4 hours and 40 minutes. The difference is exactly 2 hours, as stated in the problem. This confirms that the distance is correct and that the algebraic steps were carried out properly. Why Other Options Are Wrong:336 km, 420 km, and 224 km do not satisfy the exact 2 hour time difference when divided by the given speeds. If you plug those distances into the expressions D / 42 and D / 60, the time difference is not 2 hours.300 km is a tempting round number, but substituting it yields times of 300 / 42 and 300 / 60, whose difference is not equal to 2 hours either. Common Pitfalls:One common mistake is to assume the distance is directly proportional to the difference in speeds without carefully using time. Another error is to subtract speeds instead of working with reciprocals of speeds when comparing times. Always remember that time equals distance divided by speed, so a condition on time must translate into an equation involving the reciprocals of speeds. Final Answer:The distance between city A and city B is 280 km.

Question 4

A man travels a total distance of 61 km in 9 hours, walking part of the way at 4 km/h and cycling the remaining distance at 9 km/h. How many kilometres does he cover by walking?

Accepted Answer

Introduction / Context:This problem is a mixture of two constant speeds over different segments of a journey: walking and cycling. It tests your ability to set up and solve simultaneous equations based on both total distance and total time. Such questions are very frequent in aptitude tests and strengthen algebraic reasoning connected to real life travel situations. Given Data / Assumptions:Total distance traveled = 61 km.Walking speed = 4 km/h.Cycling speed = 9 km/h.Total time taken for the full journey = 9 hours.The man walks some distance and cycles the remaining distance in a straight journey.We must find the distance covered while walking. Concept / Approach:Let the walking distance be W km and the cycling distance be C km. We know W + C = 61. Time taken for each part is distance divided by speed, so total time is W / 4 + C / 9 = 9 hours. This gives a system of two linear equations in two variables. Solving this system yields W, the walking distance. Substitution is usually the easiest method in such straightforward linear setups. Step-by-Step Solution:Step 1: Let W be the distance walked and C be the distance cycled.Step 2: Total distance equation: W + C = 61.Step 3: Total time equation: W / 4 + C / 9 = 9 hours.Step 4: From W + C = 61, express C as C = 61 − W.Step 5: Substitute C into the time equation: W / 4 + (61 − W) / 9 = 9.Step 6: Take the least common multiple of 4 and 9, which is 36, and multiply both sides by 36.Step 7: This gives 9W + 4(61 − W) = 9 * 36.Step 8: Expand: 9W + 244 − 4W = 324.Step 9: Combine like terms: 5W + 244 = 324.Step 10: Subtract 244 from both sides: 5W = 80.Step 11: So W = 80 / 5 = 16 km. Verification / Alternative check:Using W = 16 km, we get C = 61 − 16 = 45 km. Check the total time: time walking = 16 / 4 = 4 hours; time cycling = 45 / 9 = 5 hours. Total time = 4 + 5 = 9 hours, which matches the given information. The distances also add to 61 km, so both conditions are satisfied perfectly. Why Other Options Are Wrong:12 km or 15 km would make the cycling distance too large and the total time would no longer equal 9 hours when you compute with the given speeds.17 km and 20 km also fail when substituted back, because either the total distance or total time comes out different from the given values in the question. Common Pitfalls:Some learners mistakenly add the speeds or average them without using the given time condition. Others confuse which variable stands for which distance and mix up the equations. A systematic approach with clearly defined variables and careful substitution prevents these errors. Always verify your solution by checking both the distance and the time conditions at the end. Final Answer:The man covers 16 km of the journey by walking.

Question 5

Two people A and B start from the same point and head toward the same destination 3.5 miles away. B starts 4.5 minutes after A. A reaches the destination, walks back 1 mile, and meets B there. If A walks 1 mile in 6 minutes, how many minutes does B …

Accepted Answer

Introduction / Context:This question involves two walkers starting at different times and meeting after one of them turns back from the destination. It tests your understanding of relative motion with staggered start times and meeting points that are not at the end of the path. Careful tracking of distances and times is essential in this type of time and distance problem. Given Data / Assumptions:A and B start from the same point toward a destination 3.5 miles away.A starts first and walks at a constant rate of 1 mile every 6 minutes.B starts 4.5 minutes after A, from the same starting point.A reaches the destination, turns back immediately, and walks 1 mile back toward the starting point. Given Data / Assumptions:They meet at the point where A has walked 1 mile back from the destination, that is 2.5 miles from the start.We need to determine the time B takes to walk 1 mile, expressed in minutes per mile. Concept / Approach:The main idea is to compare distances from the starting point at the meeting time. A has travelled a total of 4.5 miles by the time they meet (3.5 miles going plus 1 mile returning). At his known speed, this gives the total time until the meeting. For B, the distance from the start to the meeting point is 2.5 miles, and time for B is slightly less because B started later. The ratio of B distance to B time provides B speed and thus the time per mile. Step-by-Step Solution:Step 1: Speed of A = 1 mile in 6 minutes, so A walks at a constant rate.Step 2: At the moment they meet, A has gone from the start to the destination (3.5 miles) and then 1 mile back, so total distance walked by A is 3.5 + 1 = 4.5 miles.Step 3: Time taken by A to walk 4.5 miles at 1 mile in 6 minutes = 4.5 * 6 = 27 minutes.Step 4: Therefore, the meeting happens 27 minutes after A starts.Step 5: B starts 4.5 minutes after A, so B walks for 27 − 4.5 = 22.5 minutes until the meeting.Step 6: The meeting point is 1 mile back from the destination, that is 3.5 − 1 = 2.5 miles from the start.Step 7: So B covers 2.5 miles in 22.5 minutes.Step 8: Time taken by B per mile = 22.5 / 2.5 minutes.Step 9: 22.5 / 2.5 = 9 minutes per mile. Verification / Alternative check:Using B speed of 1 mile in 9 minutes, in 22.5 minutes B covers 22.5 / 9 = 2.5 miles, exactly reaching the 2.5 mile meeting point. At the same time, A has walked 27 minutes at 1 mile every 6 minutes, which is 27 / 6 = 4.5 miles, equal to 3.5 miles forward and 1 mile back. The geometry of their positions and times perfectly matches the description in the problem, confirming our result. Why Other Options Are Wrong:8 minutes per mile would make B too fast; he would travel more than 2.5 miles in 22.5 minutes and pass the meeting point.10 or 12 minutes per mile would make B slower than required; he would not reach the meeting point in 22.5 minutes.6 minutes per mile would mean B has the same speed as A, and with the delayed start they would not meet exactly where A has walked back 1 mile from the destination. Common Pitfalls:Students may confuse the total distance travelled by A with the distance from the start to the meeting point. Another common mistake is to forget that B starts later, so B time is less than A time. Carefully accounting for offset start times and distances is essential in chase and meeting problems like this one. Final Answer:B takes 9 minutes to walk 1 mile.

Question 6

Two stations A and B are 200 km apart. A train starts from station A at 7:00 a.m. at 20 km/h towards B. Another train starts from station B at 8:00 a.m. at 25 km/h towards A. At what time will the two trains meet?

Accepted Answer

Introduction / Context:This problem is a classic time and distance question involving two trains starting at different times from opposite ends of a straight track. It checks your understanding of relative speed, staggered starting times, and the ability to express times in clock format after solving the equation. Given Data / Assumptions:Distance between station A and station B = 200 km.Train 1 starts from station A at 7:00 a.m. and travels towards B at 20 km/h.Train 2 starts from station B at 8:00 a.m. and travels towards A at 25 km/h.Both trains move along the same straight track without stopping.We need to find the clock time when the two trains meet. Concept / Approach:The idea is to fix a time reference and express how far each train has travelled from that reference moment. We can measure time from 7:00 a.m. onwards. The first train moves for a longer period, while the second starts after 1 hour. At the meeting time, the sum of the distances travelled by both trains must equal the total distance between the stations, which is 200 km. This leads to a simple linear equation in one variable representing the time from 7:00 a.m. Step-by-Step Solution:Step 1: Let t be the time in hours after 7:00 a.m. when the trains meet.Step 2: Train 1 travels the full duration t at 20 km/h.Step 3: Train 2 starts at 8:00 a.m., which is 1 hour after 7:00 a.m., so it travels for (t − 1) hours at 25 km/h, provided t is greater than 1.Step 4: Distance covered by Train 1 = 20 * t km.Step 5: Distance covered by Train 2 = 25 * (t − 1) km.Step 6: At the meeting point, the sum of the distances is equal to the total distance between the stations: 20 * t + 25 * (t − 1) = 200.Step 7: Expand and simplify: 20t + 25t − 25 = 200.Step 8: Combine like terms: 45t − 25 = 200.Step 9: Add 25 to both sides: 45t = 225.Step 10: Divide by 45: t = 225 / 45 = 5 hours.Step 11: The meeting time is 5 hours after 7:00 a.m., which is 12:00 p.m. (noon). Verification / Alternative check:At 12:00 p.m., Train 1 has been moving from 7:00 a.m. for 5 hours at 20 km/h, so its distance is 20 * 5 = 100 km. Train 2 has been moving from 8:00 a.m. for 4 hours at 25 km/h, so its distance is 25 * 4 = 100 km. Together they cover 100 + 100 = 200 km, exactly the distance between stations A and B. Hence they meet at noon. Why Other Options Are Wrong:11 a.m. would correspond to 4 hours after 7:00 a.m., which gives 20 * 4 + 25 * 3 = 80 + 75 = 155 km, less than 200 km.1 p.m. would be 6 hours after 7:00 a.m., giving 20 * 6 + 25 * 5 = 120 + 125 = 245 km, which exceeds the required 200 km separation.10 a.m. is only 3 hours after 7:00 a.m., and the second train has been running only 2 hours, so the total distance is far less than 200 km.12 a.m. is midnight and does not fit the time frame described in the problem. Common Pitfalls:Sometimes students forget that the second train starts one hour later and mistakenly use t for both travel times. Others may add or subtract the speeds incorrectly. Always define t clearly, keep track of which train starts when, and remember that the condition for meeting is that the sum of distances equals the total distance between the stations. Final Answer:The two trains meet at 12:00 p.m. (noon).

Question 7

A bus travels half of the total distance between two places at 40 km/h and the remaining half at 60 km/h. What is the average speed of the bus for the entire journey?

Accepted Answer

Introduction / Context:This question is another example of an average speed problem where the distance is split into two equal halves with different speeds. It tests whether you know how to handle average speed correctly when distance segments, rather than time segments, are equal. The harmonic mean formula for equal distances is again the central idea. Given Data / Assumptions:The bus travels the first half of the distance at 40 km/h.The bus travels the second half of the distance at 60 km/h.Both halves represent equal distances, not equal times.There are no intermediate stops or changes in speed apart from the two given speeds.We are asked to find the average speed for the whole journey in km/h. Concept / Approach:When a vehicle covers equal distances at two different speeds v1 and v2, the average speed is not the arithmetic mean but the harmonic mean: average speed = 2 * v1 * v2 / (v1 + v2). This formula comes from adding the times for each half distance and dividing the total distance by the total time. Applying it correctly yields the average speed over the complete trip. Step-by-Step Solution:Step 1: Let the total distance be 2D km, so each half is D km.Step 2: Time for first half at 40 km/h = D / 40 hours.Step 3: Time for second half at 60 km/h = D / 60 hours.Step 4: Total distance = 2D km.Step 5: Total time = D / 40 + D / 60 hours.Step 6: Combine time: D * (1 / 40 + 1 / 60) = D * (3 / 120 + 2 / 120) = D * (5 / 120) = D * (1 / 24).Step 7: Average speed = total distance / total time = 2D / (D / 24).Step 8: Simplify: 2D * 24 / D = 48 km/h. Verification / Alternative check:Using the harmonic mean formula directly with v1 = 40 and v2 = 60 gives average speed = 2 * 40 * 60 / (40 + 60) = 4800 / 100 = 48 km/h. This matches the detailed derivation. Hence both methods confirm that the average speed is 48 km/h. Why Other Options Are Wrong:50 km/h is the arithmetic mean of 40 and 60, which is not applicable for equal distances.46 km/h and 42 km/h are values that might arise from incorrect handling of the time formula or arithmetic mistakes.52 km/h is too high and cannot be the overall average when the bus travels part of the journey at only 40 km/h. Common Pitfalls:The most common error is taking the simple average (40 + 60) / 2 = 50 km/h. Another frequent mistake is to assume equal times were spent at each speed, which is not given here. Always check whether distances or times are equal, and pick the correct method accordingly. For equal distances with two speeds, the harmonic mean approach is the safest and most accurate. Final Answer:The average speed of the bus for the entire journey is 48 km/h.

Question 8

A bus running at 9/10 of its usual speed reaches a place in 22 hours. How much time could be saved on the same journey if the bus travelled at its full usual speed?

Accepted Answer

Introduction / Context:This question examines the inverse relationship between speed and time for a fixed distance. When speed is reduced to a fraction of its usual value, time increases. You are asked to work backward from the longer time to find the time that would have been taken at the full speed and then calculate the saving in time. Given Data / Assumptions:The bus travels at 9/10 of its usual speed and takes 22 hours for the journey.The distance between the two places remains the same in both situations.We need to find how many hours would be saved if the bus travelled at its full usual speed instead of 9/10 of it.Speeds and times are constant during each trip without stops or variations. Concept / Approach:Let the usual speed be v km/h and the distance be D km. At reduced speed 9v / 10, the time taken is distance divided by speed. The relationship D = speed * time holds. Using the fact that D is the same in both scenarios, we can equate the expressions and solve for the usual time. The time saved is simply the difference between the slower time and the faster time. Step-by-Step Solution:Step 1: Let the usual speed be v km/h and the distance be D km.Step 2: At reduced speed, the bus travels at 9v / 10 and takes 22 hours.Step 3: Distance D = (9v / 10) * 22.Step 4: At full usual speed v, let the time taken be T hours.Step 5: Then D = v * T.Step 6: Equate the two expressions for D: v * T = (9v / 10) * 22.Step 7: Cancel v (since v is not zero): T = (9 / 10) * 22.Step 8: Compute T: T = 9 * 22 / 10 = 198 / 10 = 19.8 hours.Step 9: Time saved = slower time − usual time = 22 − 19.8 = 2.2 hours. Verification / Alternative check:We can also compute the ratio of times. Since time is inversely proportional to speed for a fixed distance, time at reduced speed divided by time at usual speed = usual speed divided by reduced speed = v / (9v / 10) = 10 / 9. So 22 / T = 10 / 9, giving T = 22 * 9 / 10 = 19.8 hours. Subtracting from 22 gives 2.2 hours saved, which agrees with the earlier calculation. Why Other Options Are Wrong:1.5 hours and 1.7 hours underestimate the time saving and would correspond to incorrect ratios or arithmetic errors.3.5 hours overestimates the saving and would mean a very large difference in speed not supported by the factor of 9/10 given in the question.2.0 hours seems reasonable but does not match the exact proportional calculation. Common Pitfalls:Students sometimes assume that a 10 percent change in speed leads to a 10 percent change in time, which is incorrect because time is inversely proportional to speed. Others attempt to guess a time saving instead of solving it algebraically. Carefully setting up the equations using distance = speed * time and solving step by step avoids such mistakes. Final Answer:If the bus ran at its full usual speed, it would save 2.2 hours on the journey.

Question 9

A man runs at three fourths of his usual speed and as a result reaches his destination 15 minutes late. What is his usual travel time in hours for this journey?

Accepted Answer

Introduction / Context:This question highlights the relationship between speed and time when the distance remains fixed. Reducing speed increases travel time in a predictable way. You are given the amount of delay caused by running at three fourths of the usual speed and asked to determine the original, or usual, travel time. Given Data / Assumptions:The man usually travels a fixed distance in some time T hours.On a particular day, he runs at three fourths of his usual speed. Given Data / Assumptions:Running at this reduced speed makes him 15 minutes late compared to his usual time.Fifteen minutes is equal to 15 / 60 = 1 / 4 hours.We must find the usual travel time T in hours. Concept / Approach:For a given distance, time is inversely proportional to speed. If the usual speed is v and usual time is T, then distance D = v * T. At reduced speed 3v / 4, the new time becomes D divided by (3v / 4). This new time is given as T plus 1 / 4 hour. Setting up an equation using these relationships allows us to solve for T directly. Step-by-Step Solution:Step 1: Let the usual speed be v km/h and the usual time be T hours.Step 2: Distance D = v * T.Step 3: On the slower day, speed = 3v / 4.Step 4: New time taken = D / (3v / 4) hours.Step 5: Simplify D / (3v / 4) = D * (4 / 3v) = (v * T) * 4 / 3v = 4T / 3.Step 6: The new time is 15 minutes, or 1 / 4 hour, more than T, so 4T / 3 = T + 1 / 4.Step 7: Multiply both sides by 12 to clear denominators: 12 * (4T / 3) = 12 * (T + 1 / 4).Step 8: Left side becomes 16T, right side becomes 12T + 3.Step 9: So 16T = 12T + 3.Step 10: Subtract 12T from both sides: 4T = 3.Step 11: Therefore, T = 3 / 4 hours. Verification / Alternative check:If usual time T is 3 / 4 hours, then usual speed v satisfies D = v * 3 / 4. At reduced speed 3v / 4, time taken is D divided by (3v / 4) = 4D / 3v. Substituting D = v * 3 / 4 gives 4 * (v * 3 / 4) / 3v = 1 hour. So the new time is 1 hour, which is indeed 15 minutes more than 3 / 4 hour. This confirms that our calculation for T is correct. Why Other Options Are Wrong:2/3 hours and 1/3 hours do not yield exactly 15 minutes of delay when you apply the same reduction in speed.1/4 hour is far too short to be a realistic usual travel time given such a delay.1 hour as the usual time would produce a different relationship between reduced time and delay, not matching the 15 minutes given in the question. Common Pitfalls:Many students misinterpret three fourths of the speed as meaning the time becomes three fourths of the original, when in fact time increases by a factor of four thirds. Another mistake is to convert 15 minutes incorrectly or to mix hours and minutes in the same equation without converting units properly. Working consistently in hours and carefully applying inverse proportionality avoids these issues. Final Answer:The man usually takes 3/4 hours to cover the distance.

Question 10

An inspector is 228 metres behind a thief. The inspector runs 42 metres per minute and the thief runs 30 metres per minute. After how many minutes will the inspector catch the thief?

Accepted Answer

Introduction / Context:This is a classic chase or pursuit problem. It involves two people running in the same direction at different speeds, with one starting behind the other. The key idea is relative speed, which determines how quickly the distance between them closes. These problems are common in aptitude and reasoning tests. Given Data / Assumptions:The inspector is 228 metres behind the thief at the start. Given Data / Assumptions:The inspector runs at 42 metres per minute.The thief runs at 30 metres per minute in the same direction.Both maintain constant speeds and run on a straight path.We are asked to find the time in minutes after which the inspector catches the thief. Concept / Approach:In a chase situation where both move in the same direction, the effective speed at which the gap closes is the difference between their speeds. This is called the relative speed. Once the relative speed is known, time to catch up can be found using time = distance divided by relative speed. Here, the initial distance is the head start of the thief over the inspector. Step-by-Step Solution:Step 1: Find the relative speed of the inspector with respect to the thief.Step 2: Relative speed = speed of inspector − speed of thief.Step 3: Relative speed = 42 − 30 = 12 metres per minute.Step 4: Initial distance between them is 228 metres.Step 5: Time taken to catch up = distance / relative speed.Step 6: Time = 228 / 12 minutes.Step 7: Compute 228 / 12 = 19 minutes. Verification / Alternative check:You can verify by computing how far each runs in 19 minutes. Inspector distance = 42 * 19 = 798 metres. Thief distance = 30 * 19 = 570 metres. The difference in the distances is 798 − 570 = 228 metres, which exactly equals the initial separation. This shows that after 19 minutes, the inspector has covered exactly 228 metres more than the thief and therefore catches him. Why Other Options Are Wrong:20 minutes would produce a distance difference of 12 * 20 = 240 metres, which overshoots the initial gap, so it does not match the exact moment of catching.18 minutes would close only 12 * 18 = 216 metres, which leaves a remaining gap of 12 metres.21 minutes gives 12 * 21 = 252 metres, again overshooting the required difference.17 minutes closes just 12 * 17 = 204 metres, which is not enough for the inspector to catch the thief. Common Pitfalls:A typical mistake is to add the speeds instead of subtracting them, which would only be correct if they were running toward each other instead of in the same direction. Another error is mixing units, but here both speed and distance use metres and minutes consistently. Remember that for objects moving in the same direction, relative speed is the difference in their speeds. Final Answer:The inspector will catch the thief in 19 minutes.

Question 11

An object travels 23 metres in 3 seconds and then another 15 metres in 2 seconds. What is the average speed of the object in metres per second over the entire journey?

Accepted Answer

Introduction / Context:This question illustrates the concept of average speed when an object covers different distances in different time intervals. It emphasizes that average speed must always be calculated as total distance divided by total time, not by averaging individual speeds or distances separately. The units of the final answer are also part of the test. Given Data / Assumptions:The object travels 23 metres in the first 3 seconds.It travels an additional 15 metres in the next 2 seconds.The motion is along a straight line and we treat each segment as uniform speed over that interval.We are required to find the average speed for the full 5 second interval in metres per second. Concept / Approach:Average speed is defined as total distance divided by total time. We must sum the distances of all segments of motion and sum the times taken, then compute the ratio. We do not need to calculate individual speeds for each segment separately, although we could do that if desired. Units should be carefully maintained as metres and seconds to produce the final answer in m/s. Step-by-Step Solution:Step 1: Compute the total distance travelled.Total distance = 23 m + 15 m = 38 m.Step 2: Compute the total time taken.Total time = 3 s + 2 s = 5 s.Step 3: Use the formula average speed = total distance / total time.Average speed = 38 / 5 m/s.Step 4: Simplify 38 / 5 = 7.6.Step 5: Therefore, average speed = 7.6 m/s. Verification / Alternative check:We can compute the speed in each segment for understanding. In the first segment, speed1 = 23 / 3 m/s, which is approximately 7.67 m/s. In the second segment, speed2 = 15 / 2 = 7.5 m/s. The average of these numbers is about 7.585 m/s, which is close to but not exactly 7.6 m/s. The exact average speed must be 7.6 m/s from the total distance and total time calculation, so we rely on the correct formula, not the simple average of segment speeds. Why Other Options Are Wrong:8.0 m/s would require a total distance of 8 * 5 = 40 m, which is more than the actual 38 m travelled.7.6 m is incorrect because it uses distance units instead of speed units and ignores the time factor.8.0 s is only a time value, not a speed, and does not answer the question asked.6.5 m/s does not match the ratio 38 / 5 and appears from an incorrect division or guess. Common Pitfalls:Some students mistakenly average the speeds of the segments rather than using total distance over total time. Others forget to include all time intervals or misread units, answering with metres or seconds instead of metres per second. Always apply the definition of average speed carefully and verify that the final unit matches what is requested. Final Answer:The average speed of the object over the entire motion is 7.6 m/s.

Question 12

Vishnu travels three equal distances at speeds of 10 km/h, 30 km/h, and 8 km/h, taking a total of 15.5 minutes to complete the journey. What is the total distance he travels in kilometres?

Accepted Answer

Introduction / Context:This question is about motion in three stages where equal distances are travelled at different speeds. It tests your ability to set up an equation based on total time and to convert between minutes and hours correctly. Such problems are typical in competitive exams and require both algebraic thinking and unit management. Given Data / Assumptions:Vishnu travels an equal distance d in each of three segments.Speed in the first segment = 10 km/h.Speed in the second segment = 30 km/h.Speed in the third segment = 8 km/h.Total time for all three segments combined = 15.5 minutes.We must find the total distance travelled, that is 3d, in kilometres. Concept / Approach:Time taken for each segment is distance divided by speed. Since the distances are equal, we can express total time as the sum of three such fractions. The total time is given in minutes, so we first convert it to hours. Then we solve the equation for d. Finally, total distance becomes 3d. This approach is a direct application of the time formula and algebraic solving of a linear equation. Step-by-Step Solution:Step 1: Let each segment distance be d km.Step 2: Convert total time from minutes to hours. 15.5 minutes = 15.5 / 60 hours.Step 3: 15.5 minutes is 31 / 2 minutes, so in hours it is (31 / 2) / 60 = 31 / 120 hours.Step 4: Time for first segment = d / 10 hours.Step 5: Time for second segment = d / 30 hours.Step 6: Time for third segment = d / 8 hours.Step 7: Total time equation: d / 10 + d / 30 + d / 8 = 31 / 120.Step 8: Factor out d: d * (1 / 10 + 1 / 30 + 1 / 8) = 31 / 120.Step 9: Compute 1 / 10 + 1 / 30 + 1 / 8. The common pattern gives this sum as 31 / 120.Step 10: So d * (31 / 120) = 31 / 120.Step 11: Divide both sides by 31 / 120 to get d = 1 km.Step 12: Total distance travelled = 3d = 3 * 1 = 3 km. Verification / Alternative check:If each segment is 1 km, then times are: first segment 1 / 10 h = 6 minutes; second segment 1 / 30 h = 2 minutes; third segment 1 / 8 h = 7.5 minutes. Total time = 6 + 2 + 7.5 = 15.5 minutes, which matches the problem statement. This confirms that the distances and times are consistent, and the total distance is indeed 3 km. Why Other Options Are Wrong:1 km would correspond to the distance of only one segment, not the entire journey.2 km or 4 km do not satisfy the total time equation when you recompute the segment times using the given speeds.5 km is too large and leads to times that exceed 15.5 minutes when recalculated correctly. Common Pitfalls:Students often forget to convert minutes into hours, leading to wrong equations. Some also mistakenly assume the same time is spent at each speed instead of the same distance. Another error is to treat 15.5 minutes incorrectly as 15 minutes and 50 seconds. Always convert time into a fraction of an hour carefully and then form the correct equation. Final Answer:Vishnu travels a total distance of 3 km.

Question 13

Two buses travel the same route to a destination at speeds of 45 km/h and 60 km/h. If the faster bus takes 5.5 hours less than the slower bus, what is the total length of the journey?

Accepted Answer

Introduction / Context:This problem is similar to earlier time and distance questions with two vehicles moving at constant but different speeds over the same distance. The important information is the time difference between the two journeys. Converting this information into an algebraic equation is the key step toward finding the length of the journey. Given Data / Assumptions:Speed of the slower bus = 45 km/h.Speed of the faster bus = 60 km/h.Both buses cover the same distance from the starting city to the destination. Given Data / Assumptions:The faster bus takes 5.5 hours less time than the slower bus to cover the journey.We are required to find the length of the journey in kilometres. Concept / Approach:Let the distance be D km. Then time taken by the slower bus is D / 45 hours and by the faster bus is D / 60 hours. The condition given is that the difference between these two times equals 5.5 hours. This gives the equation D / 45 − D / 60 = 5.5. Solving this equation for D gives the required distance. Step-by-Step Solution:Step 1: Let the journey length be D km.Step 2: Time taken by the slower bus = D / 45 hours.Step 3: Time taken by the faster bus = D / 60 hours.Step 4: According to the problem, D / 45 − D / 60 = 5.5.Step 5: Factor out D: D * (1 / 45 − 1 / 60) = 5.5.Step 6: Compute 1 / 45 − 1 / 60 = (60 − 45) / (45 * 60) = 15 / 2700.Step 7: Simplify 15 / 2700 to 1 / 180.Step 8: So D * (1 / 180) = 5.5.Step 9: Therefore D = 5.5 * 180.Step 10: 5.5 * 180 = 5 * 180 + 0.5 * 180 = 900 + 90 = 990 km. Verification / Alternative check:For D = 990 km, time for slower bus = 990 / 45 = 22 hours. Time for faster bus = 990 / 60 = 16.5 hours. The difference in time is 22 − 16.5 = 5.5 hours, which is exactly the difference stated in the problem. This confirms that the distance 990 km is correct. Why Other Options Are Wrong:900 km gives times of 900 / 45 and 900 / 60, whose difference is not 5.5 hours.940 km and 780 km also do not yield the required 5.5 hour time difference when substituted into the time expressions.1300 km is too large and results in a time difference much greater than 5.5 hours. Common Pitfalls:One typical error is to subtract speeds instead of times, which leads to a wrong relationship between the quantities. Another mistake is incorrect handling of the mixed decimal 5.5 hours, such as treating it as 5 hours and 50 minutes rather than 5 and a half hours. Carefully expressing the given condition and solving the linear equation step by step prevents such issues. Final Answer:The length of the journey is 990 km.

Question 14

A car runs at 50 km/h when it has not been serviced and at 60 km/h after servicing. After servicing, it covers a certain distance in 6 hours. How much time will it take to cover the same distance when it has not been serviced?

Accepted Answer

Introduction / Context:This question compares the time taken to cover the same distance at two different speeds, before and after servicing a car. It is a direct application of the basic formula time = distance divided by speed, and it reinforces the idea that higher speed reduces travel time when distance remains constant. Given Data / Assumptions:Speed of the car after servicing = 60 km/h.Speed of the car when not serviced = 50 km/h.After servicing, the car covers a certain distance in 6 hours.The distance is the same in both cases.We must find the time taken by the car to cover the same distance when not serviced, at 50 km/h. Concept / Approach:First we calculate the distance covered after servicing using distance = speed * time. Then we use that distance with the lower speed (before servicing) to calculate the new time required. No complicated algebra is needed, only straightforward application of formulas with consistent units. Step-by-Step Solution:Step 1: Compute the distance travelled after servicing.Step 2: Distance D = speed * time = 60 km/h * 6 h.Step 3: D = 360 km.Step 4: For the unserviced car, speed = 50 km/h.Step 5: Time taken at 50 km/h to cover 360 km is time = distance / speed.Step 6: Time = 360 / 50 hours.Step 7: Simplify 360 / 50 = 36 / 5 = 7.2 hours. Verification / Alternative check:To check, multiply the unserviced speed by the computed time: 50 * 7.2 = 360 km, which matches the distance covered after servicing. This confirms that the distance is consistent in both situations and that the calculated time is correct. The small increase in time compared to 6 hours is logical because 50 km/h is slightly slower than 60 km/h. Why Other Options Are Wrong:8 hours and 8.2 hours represent more time than necessary. If the car took 8 hours at 50 km/h, it would travel 400 km, not 360 km.6.5 hours is too small. At 50 km/h, 6.5 hours would give only 325 km, which is less than the required distance.7 hours would still be short, giving 350 km, not the full 360 km. Common Pitfalls:Some learners incorrectly use proportions without first finding the exact distance, or they mix up which speed belongs to which time. Others may guess by rough reasoning rather than applying the formula precisely. Always compute the distance first using the known speed and time, then derive the second time from that distance and the second speed. Final Answer:The car will take 7.2 hours to cover the same distance when it has not been serviced.

Question 15

A boy walks from his house to school at 5 km/h and reaches 3 minutes late. On the next day, he starts at the same time but walks at 9 km/h and reaches 3 minutes early. What is the distance between his house and the school in kilometres?

Accepted Answer

Introduction / Context:This question explores the effect of changing walking speed on arrival time when the start time is fixed. The boy arrives late on one day and early on another day by changing his speed. This difference in arrival times leads to an equation that lets us compute the distance to school. It is a standard type of time and distance problem in aptitude examinations. Given Data / Assumptions:On the first day, walking speed = 5 km/h and the boy is 3 minutes late. Given Data / Assumptions:On the second day, walking speed = 9 km/h and he is 3 minutes early.The start time on both days is the same.The school is at a fixed distance from his house.Three minutes is equal to 3 / 60 = 1 / 20 hours.We need to find the distance between house and school in kilometres. Concept / Approach:Let the distance be D km and the scheduled correct time to reach be T hours. On the slower day, time taken is D / 5, which equals T plus 1 / 20 hour because he is late. On the faster day, time taken is D / 9, which equals T minus 1 / 20 hour because he is early. Equating these two expressions for T allows us to eliminate T and solve for D. Step-by-Step Solution:Step 1: Let D be the distance in km and T be the correct on time travel duration in hours.Step 2: On the first day, D / 5 = T + 1 / 20.Step 3: On the second day, D / 9 = T − 1 / 20.Step 4: From the first equation, T = D / 5 − 1 / 20.Step 5: From the second equation, T = D / 9 + 1 / 20.Step 6: Equate the two expressions for T: D / 5 − 1 / 20 = D / 9 + 1 / 20.Step 7: Bring terms involving D to one side and constants to the other: D / 5 − D / 9 = 1 / 20 + 1 / 20.Step 8: Right side becomes 2 * (1 / 20) = 1 / 10.Step 9: Left side: D * (1 / 5 − 1 / 9) = D * ((9 − 5) / 45) = D * (4 / 45).Step 10: So D * 4 / 45 = 1 / 10.Step 11: Solve for D: D = (1 / 10) * (45 / 4) = 45 / 40 = 9 / 8.Step 12: 9 / 8 km equals 1.125 km. Verification / Alternative check:Using D = 1.125 km: Time at 5 km/h is 1.125 / 5 = 0.225 hours, which is 0.225 * 60 = 13.5 minutes. Time at 9 km/h is 1.125 / 9 = 0.125 hours, which is 0.125 * 60 = 7.5 minutes. The average or correct on time is the midpoint between these two times: (13.5 + 7.5) / 2 = 10.5 minutes. Thus, with 5 km/h he is 3 minutes late (13.5 − 10.5) and with 9 km/h he is 3 minutes early (10.5 − 7.5). This matches the problem statement exactly. Why Other Options Are Wrong:1 km, 1.25 km, and 1.5 km do not produce the correct 3 minute early and 3 minute late pattern when substituted into the speed calculations.0.75 km is too short and would yield smaller travel times that do not match the described early and late differences. Common Pitfalls:Students sometimes forget to convert minutes into hours and write 3 instead of 1 / 20, which breaks the equation. Another mistake is to assume the difference of times equals 3 minutes instead of 6 minutes, forgetting that he is late by 3 minutes one day and early by 3 minutes another day. Setting up the equations correctly and handling units carefully ensures success in such problems. Final Answer:The distance between his house and the school is 1.125 km.

Question 16

Raghu travels from his home to college at 4 km/h and returns at 3 km/h. If the total time for the to and fro journey is 5.95 hours, what is the distance between his home and the college?

Accepted Answer

Introduction / Context:This final question is a standard round trip problem with different speeds for going and returning. The total time for the entire journey is known, and you must find the one way distance. This reinforces the importance of treating each leg of the trip separately and then combining them using the time relationship. Given Data / Assumptions:Speed from home to college = 4 km/h.Speed from college back to home = 3 km/h.Total time for the round trip = 5.95 hours.There are no extra delays or stops apart from the travel times at the given constant speeds.We must find the one way distance between home and college in kilometres. Concept / Approach:Let the one way distance be D km. Time taken to go from home to college is D / 4 hours. Time taken to return is D / 3 hours. The sum of these times is equal to the total time 5.95 hours. This gives a single linear equation in D. Solving this equation produces the required distance. Step-by-Step Solution:Step 1: Let the distance between home and college be D km.Step 2: Time taken to go to college = D / 4 hours.Step 3: Time taken to return = D / 3 hours.Step 4: Total time for both journeys = D / 4 + D / 3.Step 5: According to the question, D / 4 + D / 3 = 5.95 hours.Step 6: Convert 5.95 into a fraction for cleaner calculation. 5.95 = 5 + 0.95 = 5 + 19 / 20 = (5 * 20 + 19) / 20 = 119 / 20 hours.Step 7: Write the equation as D / 4 + D / 3 = 119 / 20.Step 8: Take a common denominator for the left side: 12 is the least common multiple of 4 and 3.Step 9: D / 4 = 3D / 12 and D / 3 = 4D / 12, so D / 4 + D / 3 = 7D / 12.Step 10: Therefore 7D / 12 = 119 / 20.Step 11: Solve for D: multiply both sides by 12 to get 7D = 119 * 12 / 20.Step 12: Simplify 119 * 12 / 20 = 119 * 3 / 5 = 357 / 5.Step 13: So 7D = 357 / 5.Step 14: Divide both sides by 7: D = (357 / 5) / 7 = 357 / 35 = 51 / 5.Step 15: 51 / 5 km equals 10.2 km. Verification / Alternative check:If D = 10.2 km, time to college at 4 km/h is 10.2 / 4 = 2.55 hours. Time to return at 3 km/h is 10.2 / 3 = 3.4 hours. Adding these gives 2.55 + 3.4 = 5.95 hours, exactly matching the total time given in the question. Therefore the distance is consistent with all the provided information. Why Other Options Are Wrong:10 km would give a total time of 10 / 4 + 10 / 3 = 2.5 + 3.33..., which is not 5.95 hours.10.5 km gives times that add to more than 6 hours, exceeding the allowed total time.20 km doubles the distance and leads to a much larger total time than 5.95 hours.9.5 km gives too small a total time when the speeds are applied, so it also fails the condition. Common Pitfalls:Some students mistakenly use average speed instead of working with separate legs of the journey. Others round 5.95 too early, which can introduce errors. Always keep the time value accurate, work symbolically with fractions when possible, and substitute back to confirm your result. Final Answer:The distance between Raghu home and his college is 10.2 km.

Question 17

A gun is fired at a distance of 6.64 km from Ram. He hears the sound 20 seconds later. What is the speed of sound in air (in metres per second)?

Accepted Answer

Introduction / Context: This question tests the basic concept of speed, which links distance and time. The situation models how we estimate the speed of sound in air when we see an event first and hear it later, such as a gun being fired far away. Understanding how to convert units and apply the speed formula is very important in aptitude tests and physics based word problems. Here, the numbers are straightforward, so once the units are handled correctly, the calculation becomes simple. Given Data / Assumptions: Distance between the gun and Ram = 6.64 km. Time delay between seeing the gun fired and hearing the sound = 20 seconds. Sound travels in a straight line from the gun to Ram and its speed is constant over this short interval. We want the speed of sound expressed in metres per second (m/s). Concept / Approach: Speed is defined as distance divided by time. The key formula is: speed = distance / time Here the distance is given in kilometres, while time is in seconds. Since the options use metres per second, we first convert kilometres into metres. One kilometre is equal to 1000 metres. After the conversion, we substitute the values into the formula and simplify. Checking the order of magnitude is also useful, because the speed of sound in air is known to be a bit more than 300 m/s at normal conditions. Step-by-Step Solution: Step 1: Convert distance from kilometres to metres. 6.64 km = 6.64 * 1000 m = 6640 m. Step 2: Use the formula speed = distance / time. speed = 6640 / 20 m/s. Step 3: Perform the division. 6640 / 20 = 332. Therefore, speed of sound = 332 m/s. Verification / Alternative check: The commonly quoted approximate speed of sound in air at room temperature is around 330 m/s. Our computed value 332 m/s is very close to this standard reference value, which confirms that the calculation is reasonable. The distance of a few kilometres and a time lag of some tens of seconds are also consistent with this speed range, so both the numerical calculation and real world intuition agree. Why Other Options Are Wrong: Option a (664 m/s) is almost exactly double the correct speed and does not match standard physical values for sound in air. Option b (664 km/s) and option d (332 km/s) use kilometres per second, which would be extremely large and unrealistic, even faster than the speed of light in some cases. These are unit traps meant to catch candidates who ignore the units. Only 332 m/s matches both the calculation and realistic physics. Common Pitfalls: Common mistakes include forgetting to convert kilometres to metres, using minutes instead of seconds, or dividing the time by the distance instead of distance by time. Another frequent error is to choose an answer with correct numbers but incorrect units. Always check that the final unit matches the requirement of the problem and the answer choices. Working systematically through the unit conversion and formula avoids these issues. Final Answer: Therefore, the speed of sound in air in this scenario is 332 m/s.

Question 18

A runner completes a three-part race. The speeds (in km/h) for Part I, Part II and Part III are 9, 8 and 7.5 respectively, and the corresponding times are 50, 80 and 100 minutes. What is the runner's overall average speed for the entire race (in km/…

Accepted Answer

Introduction / Context: This question checks understanding of average speed when different parts of a journey are covered at different speeds for different durations. Many students incorrectly average the speeds directly, which gives the wrong result. The correct method is to compute the total distance and total time, then divide distance by time. Such race and travel based problems are very common in competitive exams and require careful handling of units and basic arithmetic. Given Data / Assumptions: Speed in Part I = 9 km/h, time in Part I = 50 minutes. Speed in Part II = 8 km/h, time in Part II = 80 minutes. Speed in Part III = 7.5 km/h, time in Part III = 100 minutes. The runner moves at constant speed within each part, and there are no breaks between parts. We need the overall average speed in km/h for the full race. Concept / Approach: Average speed over a journey is defined as total distance divided by total time. It is not the arithmetic average of different speeds unless the time spent at each speed is the same. Here, each stage has a different time duration, so we must calculate each distance separately. We convert times from minutes to hours to match the speed units in km/h. Then we find the total distance, the total time in hours, and finally compute the average speed as distance over time. Step-by-Step Solution: Step 1: Convert all times into hours. Part I: 50 minutes = 50 / 60 hours = 5 / 6 hours. Part II: 80 minutes = 80 / 60 hours = 4 / 3 hours. Part III: 100 minutes = 100 / 60 hours = 5 / 3 hours. Step 2: Compute the distance in each part. Distance I = 9 * (5 / 6) = 7.5 km. Distance II = 8 * (4 / 3) = 32 / 3 km ≈ 10.67 km. Distance III = 7.5 * (5 / 3) = 12.5 km. Step 3: Add distances and times. Total distance = 7.5 + 32 / 3 + 12.5 = 92 / 3 km ≈ 30.67 km. Total time = 5 / 6 + 4 / 3 + 5 / 3 = 23 / 6 hours ≈ 3.83 hours. Step 4: Compute average speed. Average speed = (92 / 3) / (23 / 6) = (92 / 3) * (6 / 23) = 8 km/h. Verification / Alternative check: As a quick check, note that all speeds are around 8 km/h, and the longest time segment (100 minutes) uses 7.5 km/h, which would drag the average slightly below 8 if times were symmetric. However, the other durations and distances balance such that the exact computation gives 8 km/h. Performing the fraction simplification carefully confirms that the ratio of total distance to total time is exactly 8. Why Other Options Are Wrong: Option a (8.17 kmph), option c (7.80 kmph), and option d (7.77 kmph) come from various incorrect ways of averaging, such as weighted incorrectly by distance, or by directly averaging the speeds without considering time. None of these values exactly match the correct ratio of total distance to total time. Only 8.00 kmph correctly reflects the proper computation. Common Pitfalls: Many candidates take the simple mean of 9, 8 and 7.5, which gives 8.17, and then choose that as the answer. This is wrong because the time spent in each segment is not equal. Another frequent error is forgetting to convert minutes to hours, which leads to very small or very large speeds. Always ensure consistent units and remember that average speed is total distance divided by total time, not an average of the given speeds. Final Answer: Therefore, the runner's overall average speed is 8.00 kmph.

Question 19

A plane is moving in circles around an airport and takes 3 minutes to complete one full round. The angle of elevation of the plane from a point P on the ground at time t seconds is equal to the angle of elevation at time (t + 30) seconds. At time (t…

Accepted Answer

Introduction / Context: This is a classic time and distance problem combined with circular motion and symmetry. The plane is flying in a horizontal circle around an airport, and an observer on the ground measures the angle of elevation of the plane at different times. When the angle of elevation is the same at two different instants, the corresponding positions of the plane on the circular path must be symmetric with respect to a particular line. The problem then asks at what later time the plane will be exactly vertically above the observer, which is a position directly related to the symmetry of the circular motion. Given Data / Assumptions: The plane takes 3 minutes to complete one round, so the time period T = 3 minutes = 180 seconds. The angle of elevation from point P at time t seconds equals the angle of elevation at time t + 30 seconds. At time t + x seconds, the plane is vertically above the observer at P. The plane moves at constant speed along a circular horizontal path around the airport. Concept / Approach: For uniform circular motion, equal angles of elevation as seen from a fixed point correspond to symmetric positions on the circle. Let the positions at times t and t + 30 seconds be B and C respectively. These positions are symmetric about the point O that is diametrically opposite to P on the circle, because from B and C the line of sight from P makes the same angle. This symmetry implies that the plane reaches O at the mid time between t and t + 30, namely t + 15 seconds. Once the plane is at O, it still has to travel half a circle, or 180 degrees of angular distance, to reach directly above P at point A. Step-by-Step Solution: Step 1: Compute the time for one full revolution. T = 3 minutes = 180 seconds. Step 2: Use symmetry of equal angles of elevation. If the angle from P is the same at t and t + 30 seconds, the plane is at two symmetric points B and C. The midpoint in time between these events is t + 15 seconds. At time t + 15 seconds, the plane is at the point O, diametrically opposite to P on the circle. Step 3: Time from O to P. From O to P is half of the circle, which takes T / 2 = 180 / 2 = 90 seconds. Step 4: Combine times. Total time from t to the plane being vertically above P is 15 + 90 = 105 seconds. Therefore x = 105 seconds. Verification / Alternative check: We can also think in terms of angular distance. Let B and C be symmetric with respect to the diameter through O and P. The plane takes 30 seconds to move from B to C and 180 seconds for a full 360 degree revolution. Thus, the angular spacing between B and C is proportional to 30 seconds, while the path from O to P corresponds to a half revolution or 90 seconds. Since O is the midpoint in time between B and C, we again get 15 seconds from B to O and 90 seconds from O to P, making a total of 105 seconds from t to when the plane is above P. Why Other Options Are Wrong: Option a (75 seconds) and option b (90 seconds) ignore either the additional half revolution from O to P or the mid interval of 15 seconds. Option d (135 seconds) overcounts the required time and does not follow from any symmetric splitting of the circle given the 30 second gap. Only 105 seconds satisfies both the symmetry of the angles and the total time needed to cover the remaining half revolution. Common Pitfalls: A common mistake is to assume that the plane is directly above P exactly halfway between t and t + 30 seconds, which would give 15 seconds, clearly not in the options. Another error is to forget that the point of symmetry for equal angles of elevation is diametrically opposite P, not directly above P. Some candidates also divide the full time period 180 seconds incorrectly or fail to account for the half revolution between O and P. Carefully identifying the geometry of the circle and the role of symmetry avoids these traps. Final Answer: Thus, the plane is vertically above point P after 105 seconds, so x = 105 seconds.

Question 20

A missile travels at a speed of 1440 km/h. How many metres does it travel in one second?

Accepted Answer

Introduction / Context: This problem focuses on converting a speed given in kilometres per hour into the distance covered per second in metres. Such unit conversion questions are standard in aptitude tests and help reinforce the understanding of how different time and distance units relate to each other. They also train the habit of checking units carefully before performing numerical calculations, which is essential in physics and engineering problems. Given Data / Assumptions: Missile speed = 1440 km/h. We are asked for the distance travelled in one second. We assume the missile maintains a constant speed during this time. 1 kilometre = 1000 metres and 1 hour = 3600 seconds. Concept / Approach: To find the distance in one second, it is convenient to first convert the speed from kilometres per hour to metres per second using a standard conversion factor. The key relationship is: 1 km/h = (1000 / 3600) m/s Once we convert 1440 km/h to m/s, the distance travelled in one second is simply that speed value, because distance = speed * time and the time is exactly one second. This makes the final step immediate once the unit conversion is completed correctly. Step-by-Step Solution: Step 1: Use the conversion factor. 1440 km/h = 1440 * (1000 / 3600) m/s. Step 2: Simplify the fraction 1000 / 3600. 1000 / 3600 = 1 / 3.6. Step 3: Compute the speed in m/s. 1440 * (1000 / 3600) = 1440000 / 3600. 1440000 / 3600 = 400. So the missile's speed is 400 m/s. Step 4: Distance in one second. distance = speed * time = 400 * 1 = 400 metres. Verification / Alternative check: We can mentally check the magnitude. A speed of about 360 km/h corresponds to roughly 100 m/s, because 360 km/h is known to be 100 m/s. Since 1440 km/h is four times 360 km/h, its speed in m/s should be four times 100 m/s, which is 400 m/s. This quick ratio argument matches the detailed computation and confirms that 400 metres in one second is correct. Why Other Options Are Wrong: Options a (374 metres), b (385 metres), and c (394 metres) are all close to 400 but do not correspond to any accurate conversion of 1440 km/h to m/s. They are likely included to trap students who make arithmetic rounding errors or approximate incorrectly. Only option d, 400 metres, precisely matches the correct conversion and the logical ratio check with 360 km/h. Common Pitfalls: Typical mistakes include dividing by 60 instead of 3600, forgetting to multiply by 1000, or mixing up whether to multiply or divide by the conversion factor. Another error is to compute 1440 / 3600 and forget to multiply by 1000 later, which yields a value that is too small. Always write out the conversion clearly as (km/h) * (1000 m / 1 km) * (1 h / 3600 s) and cancel units step by step to avoid confusion. Final Answer: Hence, the missile travels 400 metres in one second.

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