Question 1

What is the distance between the points (3, 6) and (-2, -6) in the Cartesian plane?

Accepted Answer

Introduction / Context: This question tests knowledge of the distance formula in coordinate geometry. You are asked to find the straight-line distance between two points in the Cartesian plane, namely (3, 6) and (-2, -6). Such questions appear frequently in aptitude tests, engineering entrance exams and school mathematics, and they build intuition for how distances are measured using coordinates. Given Data / Assumptions: First point: P(3, 6). Second point: Q(-2, -6). We are asked for the distance between P and Q. The distance is to be interpreted as the Euclidean distance (straight-line distance). Concept / Approach: The standard distance formula between two points (x1, y1) and (x2, y2) in the Cartesian plane is: distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2] This formula comes directly from the Pythagorean theorem, considering the line segment joining the two points as the hypotenuse of a right triangle whose legs represent the horizontal and vertical differences in coordinates. Step-by-Step Solution: Step 1: Identify x1, y1, x2 and y2. Let P(3, 6) be (x1, y1) and Q(-2, -6) be (x2, y2). So x1 = 3, y1 = 6, x2 = -2, y2 = -6. Step 2: Compute the differences in x and y. x2 - x1 = -2 - 3 = -5. y2 - y1 = -6 - 6 = -12. Step 3: Square the differences. (x2 - x1)^2 = (-5)^2 = 25. (y2 - y1)^2 = (-12)^2 = 144. Step 4: Add and take the square root. distance = sqrt[25 + 144] = sqrt[169]. sqrt[169] = 13. Therefore, the distance between the two points is 13 units. Verification / Alternative check: We can interpret the movement from P(3, 6) to Q(-2, -6) as moving 5 units left (from x = 3 to x = -2) and 12 units down (from y = 6 to y = -6). This forms a right triangle with legs 5 and 12. A well-known Pythagorean triple is (5, 12, 13), meaning a right triangle with legs 5 and 12 has a hypotenuse of 13. This matches our calculated distance, confirming the result. Why Other Options Are Wrong: 11 units and 12 units are both less than the true distance and do not satisfy the Pythagorean relation 5^2 + 12^2 = 169. 10 units would correspond to a much smaller hypotenuse and is clearly inconsistent with legs of 5 and 12. 15 units is larger than required and would correspond to a different set of leg lengths. Only 13 units exactly satisfies the distance formula for the given coordinates. Common Pitfalls: Students sometimes forget to square the differences, or they mistakenly add the coordinates instead of subtracting them. Another error is to take the absolute values of differences and then add without squaring, which is incorrect. Additionally, some learners forget to take the square root after summing the squares, leaving the squared distance instead of the actual distance. Carefully following the distance formula step by step avoids all of these mistakes. Final Answer: The distance between the points (3, 6) and (-2, -6) is 13 units.

Question 2

A car moves at a speed of 72 km/h after repairing and had been moving at 54 km/h before repairing. It covers a distance of X km in 6 hours after repairing. How much time (in hours) will it take to cover 3X km before repairing?

Accepted Answer

Introduction / Context: This problem combines speed, distance and proportional reasoning. A car travels faster after being repaired compared to before repairing. We know how far it goes in a certain time after repair, and we must find the time it would need to cover three times that distance at its old, slower speed. Problems like this are common in aptitude tests and help build comfort with manipulating ratios and the basic speed formula. Given Data / Assumptions: Speed after repair = 72 km/h. Speed before repair = 54 km/h. After repair, the car travels X km in 6 hours. We need the time to travel 3X km at 54 km/h. Speeds are constant in each phase and the same distance unit (km) is used throughout. Concept / Approach: We use the formula: speed = distance / time From this we get: distance = speed * time and time = distance / speed First we determine X using the faster speed and then compute 3X and the time needed at the slower speed. Step-by-Step Solution: Step 1: Find X using the speed after repair. Speed after repair = 72 km/h, time = 6 hours. X = speed * time = 72 * 6 km. 72 * 6 = 432, so X = 432 km. Step 2: Compute the new distance 3X. 3X = 3 * 432 = 1296 km. Step 3: Calculate the time at the old speed (54 km/h). time = distance / speed = 1296 / 54 hours. Compute 1296 / 54: 54 * 20 = 1080, 1296 - 1080 = 216, and 54 * 4 = 216. So 1296 / 54 = 24 hours. Therefore, at 54 km/h the car takes 24 hours to cover 3X km. Verification / Alternative check: We can check using proportional reasoning. At 72 km/h, the car covers 72 km each hour. In 6 hours, it covers 72 * 6 = 432 km, matching X. At 54 km/h, the time required to cover 432 km would be 432 / 54 = 8 hours. For 3X = 3 * 432 = 1296 km, the time is 3 * 8 = 24 hours. This confirms our calculation is consistent. Why Other Options Are Wrong: 18 hours corresponds to a distance of 54 * 18 = 972 km, which is less than 3X. 20 hours corresponds to 54 * 20 = 1080 km, still less than 1296 km. 21 hours gives 54 * 21 = 1134 km, also too small. 16 hours gives 864 km, which is even smaller. Only 24 hours gives exactly the required 1296 km. Common Pitfalls: A common mistake is to assume that since the speed reduced from 72 to 54, the time will change in the same simple ratio without carefully tracking the change in distance from X to 3X. Another error is miscalculating products such as 72 * 6 or 1296 / 54. Learners may also mix up which speed applies to which distance. Writing each step clearly and checking intermediate computations reduces these errors. Final Answer: The car will take 24 hours to cover 3X km at the old speed of 54 km/h.

Question 3

A spaceship travels at a speed of 1260 km/h. How many metres does it travel in one-tenth (1/10) of a second?

Accepted Answer

Introduction / Context: This question tests the ability to convert speeds given in kilometres per hour into distances covered in very short time intervals measured in seconds. The object here is a spaceship traveling at 1260 km/h, and we are asked how many metres it travels in one tenth (1/10) of a second. This is a classic unit conversion problem in the time, speed and distance chapter. Given Data / Assumptions: Speed of the spaceship = 1260 km/h. Time interval considered = 1/10 of a second. We need the distance covered in metres during this small time interval. The speed is constant throughout the time interval. Concept / Approach: We use the distance formula: distance = speed * time However, speed is given in km/h and time in seconds. To use the formula correctly, we must convert the speed to metres per second (m/s) and time to seconds. The standard conversion is: 1 km/h = 5 / 18 m/s Step-by-Step Solution: Step 1: Convert 1260 km/h to m/s. Speed (m/s) = 1260 * (5 / 18). First compute 1260 / 18 = 70. Then multiply by 5: 70 * 5 = 350 m/s. So the spaceship travels 350 metres every second. Step 2: Convert the time interval to seconds. One tenth of a second = 1 / 10 second = 0.1 second. Step 3: Compute the distance covered in 0.1 second. distance = speed * time = 350 * 0.1 metres. 350 * 0.1 = 35 metres. Therefore, the spaceship travels 35 m in one tenth of a second. Verification / Alternative check: We can cross-check by first finding the distance traveled in one hour and then scaling down. In one hour, the spaceship covers 1260 km, which is 1260 * 1000 = 1,260,000 metres. One hour is 3600 seconds. So in one second, distance = 1,260,000 / 3600 metres, which equals 350 metres per second. In 0.1 second, it will cover 350 / 10 = 35 metres. This matches our earlier calculation. Why Other Options Are Wrong: 126 m would require the spaceship to travel at 1260 m/s or a much higher km/h value, which does not match the given speed. 36 m is very close but comes from rounding 35 to the nearest whole number incorrectly when an exact calculation is possible. 42 m or 25 m correspond to other arbitrary speeds and do not result from the correct unit conversion of 1260 km/h to m/s. Only 35 m is consistent with 1260 km/h over 0.1 second. Common Pitfalls: Learners often forget to convert km/h to m/s and directly multiply 1260 by 0.1, which mixes units incorrectly. Another common mistake is to use the wrong conversion factor, such as 18 / 5 rather than 5 / 18, leading to an incorrect speed in m/s. Careless arithmetic with decimals can also create small but significant errors. Keeping track of units at every step is the best way to avoid these pitfalls. Final Answer: The spaceship travels 35 m in one-tenth of a second at a speed of 1260 km/h.

Question 4

A bus moves at a speed of 84 km/h after repairing and at 56 km/h before repairing. It covers a distance of 2X km in 8 hours after repairing. How much time (in hours) will it take to cover X km before repairing?

Accepted Answer

Introduction / Context: This problem is a variant of speed and distance questions where a vehicle moves with one speed after repair and a different speed before repair. We know how far it travels after repair and in how much time. From this we must find how long it would take to travel a related distance at the old speed. The question checks your understanding of distance = speed * time and how to relate distances in different scenarios using proportional reasoning. Given Data / Assumptions: Speed after repair = 84 km/h. Speed before repair = 56 km/h. After repair, the bus covers 2X km in 8 hours. We need the time required to cover X km before repair. Speeds are constant and distances are measured in kilometres. Concept / Approach: We first determine the value of 2X using the known speed and time after repair. From that we find X. Once we know X, we can calculate the time taken at the older, slower speed using: time = distance / speed Step-by-Step Solution: Step 1: Calculate 2X using the after-repair speed. Speed = 84 km/h, time = 8 hours. distance covered = speed * time = 84 * 8 km. 84 * 8 = 672 km. So 2X = 672 km. Step 2: Find X. X = 672 / 2 = 336 km. Step 3: Compute time to cover X km at 56 km/h. Speed before repair = 56 km/h. time = distance / speed = 336 / 56 hours. 56 * 6 = 336, so time = 6 hours. Therefore, it takes 6 hours to cover X km at the old speed. Verification / Alternative check: We can check using proportional reasoning. At 84 km/h, the bus covers 84 km every hour. Over 8 hours, that is 672 km, which is 2X. At the slower speed of 56 km/h, the time to cover 336 km is 336 / 56 hours. Dividing gives 6. This double-check confirms our arithmetic is correct. Why Other Options Are Wrong: 5 hours would cover 56 * 5 = 280 km, which is less than 336 km. 7 hours would cover 56 * 7 = 392 km, which is more than X. 8 hours would cover 56 * 8 = 448 km, again too much distance. 9 hours would cover 56 * 9 = 504 km, even farther from X. Only 6 hours corresponds exactly to travelling X km at 56 km/h. Common Pitfalls: One common mistake is to treat 2X as if it were just X and divide by the speed prematurely. Another error is miscalculating the product 84 * 8 or mis-dividing 336 by 56. Learners may also confuse which speed applies to which distance, especially when both before and after repair conditions are present. Writing each stage clearly, first solving for the unknown distance and then for time, helps avoid these difficulties. Final Answer: The bus will take 6 hours to cover X km at its older speed of 56 km/h.

Question 5

A man travels a total distance of 400 km partly by rail and partly by steamer. He spends 9 hours more time on the steamer than on the rail. If the speed of the steamer is 30 km/h and the speed of the train is 70 km/h, how many kilometres does he cov…

Accepted Answer

Introduction / Context: This is a classic combined journey problem. A traveller covers a fixed total distance partly by one mode of transport (rail) and partly by another (steamer). Each mode has a different constant speed. Additionally, we know the difference in time spent on each mode. From this information, we must determine the part of the journey completed by steamer. Such questions test the ability to translate word statements into algebraic equations involving speed, distance and time. Given Data / Assumptions: Total distance travelled = 400 km. Speed by rail = 70 km/h. Speed by steamer = 30 km/h. Time on steamer is 9 hours more than time on rail. We must find the distance travelled by steamer. Concept / Approach: Let the distance travelled by steamer be S km. Then the distance by rail is (400 - S) km. Time is distance divided by speed, so: Time on steamer = S / 30 hours. Time on rail = (400 - S) / 70 hours. We are told that time on steamer is 9 hours more than time on rail: S / 30 - (400 - S) / 70 = 9 We solve this equation for S. Step-by-Step Solution: Step 1: Write the time difference equation. S / 30 - (400 - S) / 70 = 9. Step 2: Use a common denominator for 30 and 70. The least common multiple of 30 and 70 is 210. S / 30 = (7S) / 210. (400 - S) / 70 = (3 * (400 - S)) / 210 = (1200 - 3S) / 210. Step 3: Substitute back into the equation. (7S / 210) - (1200 - 3S) / 210 = 9. Combine numerators: (7S - 1200 + 3S) / 210 = 9. (10S - 1200) / 210 = 9. Step 4: Clear the denominator. 10S - 1200 = 9 * 210. 9 * 210 = 1890. So 10S - 1200 = 1890. 10S = 1890 + 1200 = 3090. S = 3090 / 10 = 309 km. Therefore, the man travels 309 km by steamer. Verification / Alternative check: Time on steamer = 309 / 30 = 10.3 hours. Time on rail = (400 - 309) / 70 = 91 / 70 ≈ 1.3 hours. Difference in times = 10.3 - 1.3 = 9 hours, which matches the given condition. Also, 309 km + 91 km = 400 km, so the distances add up correctly. This confirms that the calculation is consistent. Why Other Options Are Wrong: 371 km: Then rail distance is only 29 km, leading to a very small time on rail and a time difference different from 9 hours. 280 km: Rail distance becomes 120 km; the resulting time difference is not 9 hours. 464 km or 556 km exceed or distort the total journey of 400 km when combined with the implied rail distances. Only 309 km satisfies both the total distance of 400 km and the 9-hour time difference condition. Common Pitfalls: A typical mistake is to mix up which time is greater and write the equation in reverse, leading to a negative or incorrect solution. Another issue is mishandling the algebraic manipulation when dealing with fractions, especially forgetting to use a common denominator or making arithmetic errors in multiplication. Some learners may also assume equal distances instead of using the given time difference. Carefully setting up the equation and solving step by step is crucial. Final Answer: The man covers 309 km of the journey by steamer.

Question 6

Rakesh goes on a trip on his motorcycle and covers a total distance of 368 km. If he rides for 5 hours at a speed of 49 km/h, at what speed (in km/h) must he travel for the remaining 3 hours of the journey?

Accepted Answer

Introduction / Context: This question is about average speed over different segments of a journey. Rakesh travels a fixed total distance, but his speed changes midway. We know his speed and time for the first part, and the total distance and total time for the entire trip. From this, we must determine the speed needed during the remaining time. This tests your ability to handle multi-stage journeys and to apply distance = speed * time correctly in each stage. Given Data / Assumptions: Total distance of the journey = 368 km. First part: time = 5 hours, speed = 49 km/h. Second part: time = 3 hours, speed = unknown (let it be v km/h). Total time = 5 + 3 = 8 hours. Speeds are constant within each segment. Concept / Approach: We split the total distance into two parts: distance covered in the first 5 hours and distance covered in the remaining 3 hours. For each part: distance = speed * time The sum of these distances must equal the total distance of 368 km. We solve for the unknown speed in the second segment. Step-by-Step Solution: Step 1: Compute distance covered in the first 5 hours. Speed = 49 km/h, time = 5 hours. distance_1 = 49 * 5 = 245 km. Step 2: Find remaining distance. Total distance = 368 km. distance_2 = 368 - 245 = 123 km. Step 3: Use the time for the second part to find the speed. Time for second part = 3 hours. Let speed for second part be v km/h. Then v = distance_2 / time = 123 / 3 = 41 km/h. Therefore, Rakesh must ride at 41 km/h during the last 3 hours. Verification / Alternative check: We can verify by recomputing the total distance using the found speed. First part: 49 km/h * 5 h = 245 km. Second part: 41 km/h * 3 h = 123 km. Total distance = 245 + 123 = 368 km, exactly as given. The total time is still 8 hours, so the information is consistent. Why Other Options Are Wrong: 54 km/h: Distance in last 3 hours would be 54 * 3 = 162 km, giving a total of 245 + 162 = 407 km, too large. 46 km/h: Distance in last 3 hours is 46 * 3 = 138 km, total 245 + 138 = 383 km, still too large. 58 km/h: Distance in last 3 hours is 174 km, total 245 + 174 = 419 km, much larger than 368 km. 50 km/h: Distance in last 3 hours is 150 km, total 245 + 150 = 395 km, also incorrect. Only 41 km/h yields the exact total of 368 km. Common Pitfalls: Some learners mistakenly average the two speeds directly or average the times instead of dealing with distances separately. Others miscompute the distance covered in the first segment due to multiplication errors. A few may forget that the total distance must be the sum of distances from both segments, not the sum of speeds or times. Keeping each journey segment separate and applying distance = speed * time to each is the safest strategy. Final Answer: Rakesh must travel at 41 km/h for the remaining 3 hours of his journey.

Question 7

A thief is 150 m ahead of a policeman when the policeman notices him. As soon as the policeman starts chasing, the thief also starts running. If the thief's speed is 7 km/h and the policeman's speed is 9 km/h, how far (in metres) will the thief have…

Accepted Answer

Introduction / Context: Chase problems are a standard application of relative speed. In this situation, a thief has a head start, and a policeman runs faster from behind. We know the initial gap and both speeds. The question asks how far the thief will have run by the time he is caught. This requires computing the relative speed between them and then using that to find the time to close the gap, and finally the distance the thief covers during that time. Given Data / Assumptions: Initial distance between thief and policeman = 150 m. Speed of thief = 7 km/h. Speed of policeman = 9 km/h. Both start moving at the same instant and in the same direction. Speeds remain constant until the thief is caught. Concept / Approach: When two objects move in the same direction, the closing speed (relative speed) is the difference between their speeds. Once we know the relative speed, we can compute the time needed to cover the initial gap and then the total distance run by the thief. The steps are: relative speed = speed of policeman - speed of thief time to catch = initial gap / relative speed distance run by thief = speed of thief * time to catch Step-by-Step Solution: Step 1: Compute the relative speed. Speeds are given in km/h. Relative speed = 9 - 7 = 2 km/h. Step 2: Convert the gap to kilometres. Initial gap = 150 m = 150 / 1000 km = 0.15 km. Step 3: Compute the time to catch up. time (hours) = gap / relative speed = 0.15 / 2 = 0.075 hours. Step 4: Compute the distance run by the thief. Thief's speed = 7 km/h. distance run by thief = 7 * 0.075 km. 7 * 0.075 = 0.525 km. Convert to metres: 0.525 km = 0.525 * 1000 = 525 m. Therefore, the thief will have run 525 m by the time he is caught. Verification / Alternative check: We can also compute the distance covered by the policeman in this time to confirm. In 0.075 hours, at 9 km/h, the policeman travels: distance = 9 * 0.075 = 0.675 km = 675 m. The difference between their distances should be the initial gap of 150 m: 675 m - 525 m = 150 m. This matches the initial separation, confirming our computations. Why Other Options Are Wrong: 420 m corresponds to a smaller time, implying a higher relative speed than is actually the case. 630 m would give a larger difference between their distances than 150 m at the moment of capture. 315 m is too small and does not satisfy the closing gap condition with the given speeds. 450 m also fails to maintain the required 150 m difference at the start and zero gap at the capture moment. Only 525 m fits all the conditions of the chase. Common Pitfalls: Some learners mistakenly add the speeds when the objects move in the same direction, instead of using the difference. Another common error is to forget to convert metres to kilometres or vice versa, leading to inconsistent units. Additionally, a few students directly compute the distance covered by the policeman and misinterpret that as the thief's distance. Always remember that we are asked for the distance run by the thief, not the policeman. Final Answer: The thief will have run 525 m before he is caught by the policeman.

Question 8

A car travels a certain distance at 32 km/h and returns over the same distance at 68 km/h. What is the average speed of the car for the entire journey?

Accepted Answer

Introduction / Context: Average speed for a round trip with different speeds in each direction is a very common type of aptitude question. Many candidates mistakenly average the two speeds arithmetically, which is incorrect. Instead, we must use the harmonic mean, derived from the formula for total distance divided by total time. Here, a car travels one way at 32 km/h and returns at 68 km/h. We must find the average speed for the whole journey. Given Data / Assumptions: One-way speed going from starting point to destination = 32 km/h. Return speed on the same route = 68 km/h. Distances in both directions are equal. We must find the average speed over the entire journey. Concept / Approach: Let the one-way distance be D km. Then total distance for the round trip is 2D. Corresponding times are: time_1 = D / 32 hours time_2 = D / 68 hours Total time = time_1 + time_2. Average speed is: average speed = total distance / total time = 2D / (D / 32 + D / 68) The D cancels out, leaving a pure expression in terms of the two speeds. Step-by-Step Solution: Step 1: Express the total distance. Total distance = D (going) + D (return) = 2D. Step 2: Express the total time. time_1 = D / 32, time_2 = D / 68. Total time = D / 32 + D / 68. Factor out D: total time = D * (1 / 32 + 1 / 68). Step 3: Compute the average speed. average speed = 2D / [D * (1 / 32 + 1 / 68)]. Cancel D: average speed = 2 / (1 / 32 + 1 / 68). Step 4: Simplify the denominator. 1 / 32 + 1 / 68 = (68 + 32) / (32 * 68) = 100 / (2176). So average speed = 2 / (100 / 2176) = 2 * (2176 / 100). 2 * 2176 = 4352, so average speed = 4352 / 100 = 43.52 km/h. Therefore, the average speed for the entire journey is 43.52 km/h. Verification / Alternative check: As an alternative, assume a convenient one-way distance, such as D = 2176 km. Then: time_1 = 2176 / 32 = 68 hours. time_2 = 2176 / 68 = 32 hours. Total distance = 4352 km, total time = 100 hours. Average speed = 4352 / 100 = 43.52 km/h. This matches the previous computation, confirming the result. Why Other Options Are Wrong: 50 km/h is the simple arithmetic mean of 32 and 68, which is not correct for average speed over equal distances with different speeds. 37.04 km/h, 40 km/h and 56.48 km/h result from incorrect manipulations of the formula or from mixing up distances and times. Only 43.52 km/h comes from the correct calculation using total distance divided by total time. Common Pitfalls: The biggest pitfall is averaging the speeds directly using (32 + 68) / 2. Average speed problems with equal distances require the harmonic mean: 2ab / (a + b), not the simple average. Another common error is to assume different distances for each leg of the trip, which is not given in the problem. Also, computational mistakes with fractions can lead to wrong decimals. Carefully applying the formula and checking with a concrete distance helps avoid these errors. Final Answer: The average speed of the car for the entire journey is 43.52 km/h.

Question 9

Two cars travel from city A to city B at speeds of 36 km/h and 48 km/h respectively. If one car takes 3 hours less than the other for the journey, what is the distance between city A and city B?

Accepted Answer

Introduction / Context: This question is another example of comparing travel times for the same journey at different speeds. Two cars go from city A to city B, one at 36 km/h and another at 48 km/h. The faster car takes 3 hours less than the slower car. We must use this information to find the distance between the two cities. Such questions reinforce the idea that time is distance divided by speed and that differences in times can be turned into equations for the unknown distance. Given Data / Assumptions: Speed of first car = 36 km/h. Speed of second car = 48 km/h. Both travel the same distance between city A and city B. The difference in their travel times is 3 hours. We assume constant speeds and a direct route. Concept / Approach: Let the distance between the cities be D km. Then: time_1 = D / 36 hours. time_2 = D / 48 hours. If the slower car takes more time, the difference is: D / 36 - D / 48 = 3. We solve this equation for D. Step-by-Step Solution: Step 1: Write the time difference equation. D / 36 - D / 48 = 3. Step 2: Use a common denominator for 36 and 48. The least common multiple of 36 and 48 is 144. D / 36 = (4D) / 144. D / 48 = (3D) / 144. So the equation becomes (4D / 144) - (3D / 144) = 3. This simplifies to (D / 144) = 3. Step 3: Solve for D. D = 3 * 144 = 432 km. Therefore, the distance between the two cities is 432 km. Verification / Alternative check: Check the times for D = 432 km: time_1 = 432 / 36 = 12 hours. time_2 = 432 / 48 = 9 hours. Difference in times = 12 - 9 = 3 hours, which matches the given information. Thus the solution is consistent. Why Other Options Are Wrong: 518 km, 648 km, 384 km and 346 km do not produce a time difference of exactly 3 hours when divided by 36 and 48. For example, 648 km gives 18 hours and 13.5 hours, a difference of 4.5 hours, not 3. Only 432 km produces 12 hours and 9 hours, giving the required 3-hour difference. Common Pitfalls: Some candidates may mistakenly add the times instead of subtracting them or may assume a wrong time difference sign. Others may choose an incorrect common denominator or make arithmetic errors when simplifying the fractions. A few may even attempt to use an average speed, which is not directly helpful here. Writing the equation carefully and checking each algebraic step prevents such mistakes. Final Answer: The distance between city A and city B is 432 km.

Question 10

A thief is 350 m ahead of a policeman when the policeman starts chasing. The thief also starts running at that moment. If the thief's speed is 5 km/h and the policeman's speed is 7 km/h, how far (in metres) will the thief have run before he is overt…

Accepted Answer

Introduction / Context: This is another chase problem involving relative speed. A thief has a 350 m head start when the policeman begins to chase him. We know both speeds and must determine how far the thief runs before he is caught. Handle such situations by focusing on the difference in speeds when the motion is in the same direction, and treat the initial head start as the gap that must be closed. Given Data / Assumptions: Initial lead of the thief = 350 m. Thief's speed = 5 km/h. Policeman's speed = 7 km/h. Both run in the same direction with constant speeds. The policeman starts chasing at the instant we start counting time. Concept / Approach: Relative speed when two bodies move in the same direction is the difference of their speeds. Once we know the relative speed, we can compute the time needed for the policeman to close the initial gap. That time, when multiplied by the thief's speed, gives the distance run by the thief. relative speed = 7 - 5 = 2 km/h Step-by-Step Solution: Step 1: Convert the initial gap to kilometres. Initial gap = 350 m = 350 / 1000 km = 0.35 km. Step 2: Compute the time to close the gap. Relative speed = 2 km/h. time (hours) = gap / relative speed = 0.35 / 2 = 0.175 hours. Step 3: Compute the distance run by the thief. Thief's speed = 5 km/h. distance_thief = speed * time = 5 * 0.175 km. 5 * 0.175 = 0.875 km. Convert to metres: 0.875 * 1000 = 875 m. Therefore, the thief runs 875 m before he is overtaken. Verification / Alternative check: We can also compute the distance covered by the policeman in the same time. In 0.175 hours at 7 km/h, the policeman covers: distance_policeman = 7 * 0.175 km = 1.225 km = 1225 m. The difference in their distances at the capture moment should equal the initial lead of 350 m: 1225 m - 875 m = 350 m. This matches the original lead, so our solution is consistent. Why Other Options Are Wrong: 700 m implies less time than required to close the 350 m gap at a relative speed of 2 km/h. 1050 m implies more time than necessary, which would give a greater initial lead than 350 m when combined with the policeman's distance. 525 m and 650 m similarly fail to maintain the required 350 m initial difference when we reconstruct the situation. Only 875 m fits both the head start and the speed conditions correctly. Common Pitfalls: Typical errors include adding the speeds instead of subtracting them when both are running in the same direction. Learners sometimes forget to convert metres to kilometres, causing mismatched units in the formula. Others may compute the distance covered by the policeman and mistakenly present it as the thief's distance. Focusing on relative speed and carefully tracking whose distance is being asked helps avoid these mistakes. Final Answer: The thief will have run 875 m before the policeman overtakes him.

Question 11

To cover a distance of 216 km in 3.2 hours, what should be the average speed of the car in metres per second (m/s)?

Accepted Answer

Introduction / Context: This question is another straightforward average speed calculation with a twist: the final answer must be in metres per second (m/s). A car must cover 216 km in 3.2 hours, and we are asked what average speed is required. Such problems are common in aptitude tests and require both the correct use of speed = distance / time and the conversion from km/h to m/s. Given Data / Assumptions: Total distance = 216 km. Total time = 3.2 hours. The car travels with constant average speed. We must express this speed in m/s. Concept / Approach: First compute the average speed in km/h using: average speed = distance / time Then convert the result from km/h to m/s using: 1 km/h = 5 / 18 m/s This two-step method keeps the calculation organised and clear. Step-by-Step Solution: Step 1: Compute average speed in km/h. Distance = 216 km, time = 3.2 hours. average speed (km/h) = 216 / 3.2. Write 3.2 as 32 / 10. So 216 / 3.2 = 216 / (32 / 10) = 216 * (10 / 32). Simplify 216 / 32: 216 / 32 = 6.75. Then 6.75 * 10 = 67.5 km/h. Step 2: Convert 67.5 km/h to m/s. Speed in m/s = 67.5 * (5 / 18). First compute 67.5 / 18. 67.5 / 18 = 3.75. Now multiply by 5: 3.75 * 5 = 18.75 m/s. Therefore, the required average speed is 18.75 m/s. Verification / Alternative check: We can reverse the conversion. Convert 18.75 m/s back to km/h: speed (km/h) = 18.75 * (18 / 5) = 18.75 * 3.6. 18.75 * 3.6 = 67.5 km/h. Now check if 67.5 km/h for 3.2 hours produces 216 km: distance = 67.5 * 3.2 = 67.5 * (32 / 10) = (67.5 * 32) / 10. 67.5 * 32 = 2160. 2160 / 10 = 216 km. The distance matches, so our calculations are correct. Why Other Options Are Wrong: 33.75 m/s corresponds to a much higher km/h value, so the car would cover more than 216 km in 3.2 hours. 37.5 m/s is even faster and clearly too big. 27 m/s and 16.5 m/s correspond to other average speeds that, when converted and multiplied by 3.2 hours, do not give 216 km. Only 18.75 m/s corresponds exactly to 216 km in 3.2 hours. Common Pitfalls: Errors often occur while dividing 216 by 3.2, particularly when handling decimals. Some learners forget to convert km/h to m/s or use the wrong conversion factor. Another mistake is rounding intermediate values too early, which can change the final answer. Keeping fractional calculations exact until the final step and remembering the factor 5 / 18 for km/h to m/s prevents most errors. Final Answer: The car must travel at an average speed of 18.75 m/s to cover 216 km in 3.2 hours.

Question 12

A missile travels at a speed of 1422 km/h. How many metres does it travel in one second?

Accepted Answer

Introduction / Context: Here we have a speed given in kilometres per hour for a missile, and we must find the distance it travels in just one second, expressed in metres. This involves a unit conversion from km/h to m/s and then an application of the distance formula. Such questions help strengthen your ability to move between different units of speed and time, which is vital for both exams and practical problem solving. Given Data / Assumptions: Speed of the missile = 1422 km/h. Time interval = 1 second. We must find the distance covered in metres during this one second. The missile's speed is constant over this short time. Concept / Approach: We use two key ideas: 1) Convert speed from km/h to m/s. 2) distance = speed * time. The standard conversion factor is: 1 km/h = 5 / 18 m/s. Step-by-Step Solution: Step 1: Convert 1422 km/h to m/s. Speed (m/s) = 1422 * (5 / 18). First compute 1422 / 18. 18 * 79 = 1422, so 1422 / 18 = 79. Then multiply by 5: 79 * 5 = 395 m/s. So the missile travels 395 metres every second. Step 2: Find the distance travelled in 1 second. time = 1 second. distance = speed * time = 395 * 1 = 395 m. Therefore, the missile travels 395 m in one second. Verification / Alternative check: We can confirm by converting back to km/h. If the missile covers 395 m in 1 second, then per hour (3600 seconds) it will cover: 395 * 3600 = 1,422,000 m = 1422 km. This is exactly the given speed in km/h, so our conversion is correct. That confirms the answer. Why Other Options Are Wrong: 400 m would correspond to 1440 km/h when converted back, which is slightly higher than the given speed. 364 m or 360 m correspond to slower speeds than 1422 km/h. 319 m is also too small and gives a significantly lower speed than 1422 km/h. Only 395 m matches a speed of 1422 km/h exactly. Common Pitfalls: Some students accidentally use 18 / 5 instead of 5 / 18 when converting km/h to m/s, which inverts the speed. Others mix up metres and kilometres, dividing by 3600 and 1000 in the wrong order. Rounding the conversion too early can also introduce small differences. The safest approach is to do the division 1422 / 18 exactly and then multiply by 5, making sure units are consistently handled. Final Answer: The missile travels 395 m in one second at a speed of 1422 km/h.

Question 13

If a person walks at 15 km/h instead of 9 km/h, he would cover 3 km more in the same time. What is the actual distance (in km) that he covers at 9 km/h?

Accepted Answer

Introduction / Context: This question deals with comparing distances covered at two different speeds over the same time interval. The person could walk at 9 km/h but instead walks at 15 km/h and covers 3 km more in the same time. From this information we must find the actual distance he covered at the slower speed of 9 km/h. This is a typical algebraic application of the formula distance = speed * time. Given Data / Assumptions: Actual intended speed = 9 km/h. Alternative speed = 15 km/h. Time of travel is the same in both cases. Distance at 15 km/h exceeds distance at 9 km/h by 3 km. We must find the distance at 9 km/h. Concept / Approach: Let the common time of travel be t hours. Then: Distance at 9 km/h = 9t. Distance at 15 km/h = 15t. We know that: 15t - 9t = 3. From this we can solve for t and then compute the actual distance 9t. Step-by-Step Solution: Step 1: Set up the equation using the distance difference. 15t - 9t = 3. Simplify: 6t = 3. Step 2: Solve for t. t = 3 / 6 = 0.5 hours. So the person travels for half an hour. Step 3: Compute the actual distance at 9 km/h. Distance = speed * time = 9 * 0.5 = 4.5 km. Therefore, the actual distance travelled at 9 km/h is 4.5 km. Verification / Alternative check: At 9 km/h for 0.5 hours, the distance is 4.5 km. At 15 km/h for the same 0.5 hours, the distance is 15 * 0.5 = 7.5 km. The difference between the two is: 7.5 - 4.5 = 3 km. This matches the problem statement, so the solution is consistent and correct. Why Other Options Are Wrong: 5.5 km or 6.5 km would require a different time t that does not make the distance difference exactly 3 km when combined with 15 km/h. 7.5 km is actually the distance at 15 km/h, not at 9 km/h. 3.5 km is too small and would not result in a 3 km difference when compared with the distance at 15 km/h over the same time. Only 4.5 km satisfies the given condition of a 3 km extra distance at the higher speed. Common Pitfalls: Some learners incorrectly assume the 3 km difference applies directly to speed rather than distance. Others forget to introduce the time variable t and attempt to solve with distances alone. Misinterpreting the statement “in the same time” is also common; it is crucial because it allows us to set up the difference equation. Carefully translating the words into algebraic expressions avoids these errors. Final Answer: The person actually travels 4.5 km at the speed of 9 km/h.

Question 14

A man goes to a place by bicycle at a speed of 16 km/h and returns at a lower speed. If his average speed for the entire journey is 6.4 km/h, what is his speed (in km/h) on the return journey?

Accepted Answer

Introduction / Context: This problem involves average speed over a round trip with different speeds in the outbound and return journeys. The man cycles to a place at 16 km/h and returns at some lower speed. The overall average speed for the entire journey is 6.4 km/h. We must find the return speed. This is a classic harmonic mean style question in the time, speed and distance chapter, but here we solve directly using total distance and total time. Given Data / Assumptions: Outbound speed = 16 km/h. Return speed = v km/h (unknown, but lower than 16 km/h). Average speed over the whole journey = 6.4 km/h. Distances outbound and return are equal. Concept / Approach: Let the one-way distance be D km. Then total distance = 2D km. Times for each leg are: time_out = D / 16 hours. time_back = D / v hours. Average speed is defined as: average speed = total distance / total time = 2D / (D / 16 + D / v). We know the average speed is 6.4 km/h, so: 6.4 = 2D / (D / 16 + D / v). Step-by-Step Solution: Step 1: Simplify the expression for average speed. 6.4 = 2D / [D * (1 / 16 + 1 / v)]. Cancel D from numerator and denominator. 6.4 = 2 / (1 / 16 + 1 / v). Step 2: Rearrange to solve for (1 / 16 + 1 / v). 1 / 16 + 1 / v = 2 / 6.4. 2 / 6.4 = 20 / 64 = 5 / 16. So 1 / 16 + 1 / v = 5 / 16. Step 3: Solve for 1 / v. 1 / v = 5 / 16 - 1 / 16 = 4 / 16 = 1 / 4. Step 4: Invert to find v. v = 4 km/h. Therefore, his return speed is 4 km/h. Verification / Alternative check: Assume D = 16 km for simplicity. Then outbound time = 16 / 16 = 1 hour. Return speed = 4 km/h, so return time = 16 / 4 = 4 hours. Total distance = 32 km, total time = 5 hours. Average speed = 32 / 5 = 6.4 km/h, which matches the given value. Hence, v = 4 km/h is consistent. Why Other Options Are Wrong: 6 km/h or 8 km/h would give different total times and hence different average speeds, not 6.4 km/h. 10 km/h would produce a relatively high average speed since both legs would be fairly fast. 5 km/h also fails to make the combined average exactly 6.4 km/h when checked with a sample distance. Only 4 km/h yields the required 6.4 km/h average speed. Common Pitfalls: A common mistake is to average the speeds directly, for example taking (16 + v) / 2 = 6.4, which is incorrect because average speed is distance-based, not speed-based. Another error is to forget that distances are equal for a round trip and to assign different distances for the two legs. Algebra mistakes when handling fractions in 1 / 16 + 1 / v are also frequent. Handling the problem symbolically and checking with an assumed distance helps avoid these issues. Final Answer: The man's speed on the return journey is 4 km/h.

Question 15

A car moving in the morning fog overtakes a man walking at 4 km/h in the same direction. The man can see the car clearly for 3 minutes, and the maximum distance at which the car is visible ahead of him is 130 m. What is the speed of the car (in km/h…

Accepted Answer

Introduction / Context: This is a relative speed and visibility problem set in morning fog. A man walks along a road at 4 km/h. A car moving in the same direction overtakes him. Because of fog, the man can only see objects up to 130 m ahead. He can see the car clearly for 3 minutes. We must find the car's speed. The key idea is to relate visibility distance and the time for which the car remains within that visible range while using relative speed between the car and the man. Given Data / Assumptions: Man's walking speed = 4 km/h. Car moves in the same direction as the man. Maximum distance at which the car is visible ahead of the man = 130 m. The man can see the car clearly for 3 minutes. We assume that the car becomes visible when it is next to the man and disappears when it is 130 m ahead. Concept / Approach: In this interpretation, the car remains visible from the moment it is level with the man until it is 130 m ahead. The relative distance covered between car and man during this time is therefore 130 m. The relative speed is the difference between their speeds because they are moving in the same direction. We find the relative speed from distance and time, and then add the man's speed to get the car's speed. Step-by-Step Solution: Step 1: Express visibility distance in kilometres. Visibility range ahead = 130 m = 130 / 1000 km = 0.13 km. Step 2: Express the visible time in hours. Time for which the car is visible = 3 minutes. 3 minutes = 3 / 60 hours = 0.05 hours. Step 3: Compute the relative speed. relative speed = distance / time = 0.13 / 0.05 km/h. 0.13 / 0.05 = 2.6 km/h. So the car is moving 2.6 km/h faster than the man. Step 4: Find the car's speed. Man's speed = 4 km/h. Car's speed = man's speed + relative speed = 4 + 2.6 = 6.6 km/h. Therefore, the car's speed is 6.6 km/h. Verification / Alternative check: Check by recomputing the relative distance using the discovered car speed. If the car moves at 6.6 km/h and the man walks at 4 km/h, their relative speed is 2.6 km/h. Over 3 minutes (0.05 hours), the relative distance covered is: relative distance = 2.6 * 0.05 = 0.13 km = 130 m. That matches the visibility distance ahead at which the car disappears from view. Hence, 6.6 km/h is consistent with the problem data. Why Other Options Are Wrong: 10 km/h or 8.9 km/h would give a higher relative speed and thus a larger visible distance or a shorter visible time than 130 m in 3 minutes. 7.5 km/h also produces a relative speed greater than 2.6 km/h and does not match the 130 m figure when recomputed. 5.2 km/h would give a relative speed of only 1.2 km/h, which is too small and would not carry the car far enough ahead in the given time. Only 6.6 km/h gives exactly 130 m of relative advance in 3 minutes. Common Pitfalls: A key misunderstanding arises if one assumes the car is visible from 130 m behind the man to 130 m ahead, making the relative distance 260 m. That interpretation yields a different speed that does not match any of the given options. Also, students sometimes add the speeds instead of subtracting them, which is incorrect for same-direction motion. Another frequent mistake is mishandling the time conversion from minutes to hours. Carefully choosing a consistent interpretation of visibility and checking with the numbers avoids these issues. Final Answer: The speed of the car is 6.6 km/h.

Question 16

A man travels a distance of 99 km in 9 hours, walking part of the way at 9 km/h and cycling the rest at 18 km/h. What distance (in km) does he travel on foot?

Accepted Answer

Introduction / Context: This problem involves mixed-mode travel with two different speeds: walking and cycling. The total distance and total time are given, and we know the speeds for each mode. We must determine how much of the journey is done on foot. These questions test your ability to set up and solve a system of equations derived from distance = speed * time relationships for different segments of a trip. Given Data / Assumptions: Total distance = 99 km. Total time = 9 hours. Walking speed = 9 km/h. Cycling speed = 18 km/h. Let time spent walking be t1 hours and time spent cycling be t2 hours. We must find the distance walked, which is 9 * t1 km. Concept / Approach: We form two equations. One equation uses the total time, and the other uses the total distance. Specifically: t1 + t2 = 9 9 * t1 + 18 * t2 = 99 Solving this system gives us t1 and t2. Then distance walked is speed multiplied by t1. Step-by-Step Solution: Step 1: Write the time equation. t1 + t2 = 9. Step 2: Write the distance equation. Distance walked = 9 * t1. Distance cycled = 18 * t2. Total distance: 9 * t1 + 18 * t2 = 99. Step 3: Use substitution from the time equation. t2 = 9 - t1. Substitute into the distance equation: 9 * t1 + 18 * (9 - t1) = 99. Step 4: Expand and simplify. 9 * t1 + 18 * 9 - 18 * t1 = 99. 9 * t1 - 18 * t1 = -9 * t1. So -9 * t1 + 162 = 99. Step 5: Solve for t1. -9 * t1 = 99 - 162 = -63. t1 = -63 / -9 = 7 hours. Step 6: Compute distance walked. Distance walked = 9 * t1 = 9 * 7 = 63 km. Therefore, the man walks 63 km and cycles the remaining part. Verification / Alternative check: If t1 = 7 hours, then t2 = 9 - 7 = 2 hours. Distance walked = 9 * 7 = 63 km. Distance cycled = 18 * 2 = 36 km. Total distance = 63 + 36 = 99 km, which matches the given total. Total time = 7 + 2 = 9 hours, so all conditions are satisfied. Why Other Options Are Wrong: 36 km, 45 km, 54 km and 27 km do not allow both the time and distance equations to be satisfied simultaneously when converted back to walking and cycling times at the given speeds. For example, if 54 km were walked, the remaining 45 km at 18 km/h would require 2.5 hours, giving a total time of more than 9 hours. Only 63 km gives times and distances consistent with all the data. Common Pitfalls: Some learners confuse time and distance, equating 9 * t1 + 18 * t2 with total time instead of total distance. Others miscalculate when substituting t2 = 9 - t1, especially during distribution and simplification. Another common error is to assume equal times or equal distances for walking and cycling, which is not supported by the problem statement. Carefully forming the two equations from first principles and solving methodically avoids these errors. Final Answer: The man travels 63 km on foot and the remainder by bicycle.

Question 17

A taxi travels at 90 km/h and takes 35 minutes to cover a certain distance. By how many kilometres per hour should it increase its speed to cover the same distance in only 21 minutes?

Accepted Answer

Introduction / Context: This question tests the core time, speed and distance relationship that is heavily used in quantitative aptitude and competitive exams. The situation describes a taxi that initially travels at a given constant speed and takes a known time to cover an unknown distance. Then we are asked how much the speed must be increased so that the same distance is covered in a shorter time. Understanding how speed changes when time changes for a fixed distance is a very important concept for solving many real life travel and motion problems in exams. Given Data / Assumptions: Initial speed of the taxi = 90 km/h. Time taken at this speed = 35 minutes. The distance remains the same in both journeys. Required time for the same distance = 21 minutes. We need the increase in speed in km/h, not just the new speed. Concept / Approach: For any uniform motion, the basic relation is distance = speed * time. When the distance is fixed, speed is inversely proportional to time. That means if time decreases, speed must increase in a specific ratio so that distance remains unchanged. We first compute the original distance using the initial speed and time. Then we compute the new speed required to cover that same distance in the shorter time. Finally, we subtract the old speed from the new speed to find the increase in speed in km/h. Step-by-Step Solution: Convert 35 minutes into hours: 35 minutes = 35/60 hour. Original distance = 90 * (35/60) km = 90 * 35 / 60 = 52.5 km. Convert 21 minutes into hours: 21 minutes = 21/60 hour. Required new speed = same distance / new time = 52.5 / (21/60) km/h. Compute new speed: 52.5 * 60 / 21 = 150 km/h. Increase in speed = 150 - 90 = 60 km/h. Verification / Alternative check: We can verify by checking that at 150 km/h the taxi really needs only 21 minutes. Time = distance / speed = 52.5 / 150 hours. This is equal to 0.35 hours. Converting 0.35 hours into minutes gives 0.35 * 60 = 21 minutes, which matches the required time. The calculations are therefore consistent and confirm that the increase in speed must be 60 km/h. Why Other Options Are Wrong: Option 150 represents the new speed, not the increase in speed, so it does not answer the exact question. Option 120 km/h would still not reduce the time enough, because 52.5 / 120 hours is more than 21 minutes. Option 90 km/h is simply the original speed and would not reduce the travel time at all. Therefore all other options fail to satisfy the condition of covering the same distance in only 21 minutes. Common Pitfalls: Many learners forget to convert minutes into hours before applying the speed formula, which leads to wrong numerical values. Another common mistake is to compute only the new speed and select that as the answer, even though the question specifically asks for the increase in speed. It is also easy to mix up the proportional reasoning and think that the speed increases by the same number of minutes reduced, which is not correct. Always compute the distance first or use ratios carefully. Final Answer: The taxi must increase its speed by 60 km/h to cover the same distance in 21 minutes.

Question 18

A bus travels at a constant speed of 36 km/h. What distance in metres does it cover in exactly one second?

Accepted Answer

Introduction / Context: This problem checks your understanding of basic unit conversion between kilometres per hour and metres per second. It is a standard type of speed conversion question that appears in many aptitude and entrance exams. Being comfortable with these conversions saves time and reduces errors in many time and distance problems. Here we are asked to find the distance travelled in a single second by a bus that moves at a known speed in km/h. Given Data / Assumptions: Speed of the bus = 36 km/h. We need the distance travelled in one second. One kilometre = 1000 metres. One hour = 3600 seconds. Motion is assumed to be uniform at the given speed. Concept / Approach: To find the distance covered in one second, we convert the given speed from kilometres per hour into metres per second. The standard factor is 5/18 or 18/5 depending on the direction of conversion. Once we know the speed in metres per second, the distance travelled in one second is simply equal to that speed, because distance = speed * time and time is one second. Step-by-Step Solution: Write the speed as 36 km/h. Use the conversion 1 km/h = 5/18 m/s. Speed in m/s = 36 * 5/18. Simplify: 36 * 5/18 = 2 * 5 = 10 m/s. Distance in one second = speed in m/s * 1 second = 10 metres. Verification / Alternative check: We can also convert directly using the definition. In one hour at 36 km/h, the bus covers 36 km which is 36 000 metres. One hour has 3600 seconds. So distance per second is 36 000 / 3600 = 10 metres. This matches the previous calculation and confirms that the answer is correct. Why Other Options Are Wrong: Option 15 m corresponds to a higher speed of 54 km/h, not 36 km/h. Option 12.5 m would correspond to 45 km/h. Option 13.5 m corresponds to approximately 48.6 km/h. None of these match the given speed of 36 km/h when converted back to km/h, so they are not correct for this specific problem. Common Pitfalls: Candidates often confuse the conversion factor and multiply by 18/5 instead of 5/18 when moving from km/h to m/s. Another common mistake is to forget that 1 km equals 1000 m or that 1 hour equals 3600 seconds. Some learners also mistakenly think that 36 km/h is 36 m/s, which is not true. Always write the conversion clearly to avoid such errors. Final Answer: The bus covers a distance of 10 metres in one second.

Question 19

Ram walks from his house to school. At a speed of 3 km/h he reaches 6 minutes late, but at a speed of 4 km/h he reaches 4 minutes early. What is the distance in kilometres between his house and the school?

Accepted Answer

Introduction / Context: This is a classic time and distance problem where early and late arrival times are given for two different speeds. Such questions test your ability to set up equations based on the relationship between time, speed and distance. Instead of giving the distance directly, the problem describes how the arrival time changes when Ram changes his walking speed. From this information, we must deduce the distance between his home and school. Given Data / Assumptions: Speed of Ram in the first case = 3 km/h and he is 6 minutes late. Speed of Ram in the second case = 4 km/h and he is 4 minutes early. Let the scheduled or correct time to reach school be T hours. Let the distance between home and school be D kilometres. The walking route is the same in both cases and speed remains constant during each walk. Concept / Approach: Time, speed and distance are connected by distance = speed * time, or time = distance / speed. In this problem there are two forms of the time taken: one when Ram walks slower and arrives late, and another when he walks faster and arrives early. The difference between these times equals the total difference between late and early arrival. We form two equations for time using D and the two speeds, relate them through the scheduled time T, and then solve for D. Step-by-Step Solution: Let D be the distance in kilometres and T be the scheduled time in hours. When Ram walks at 3 km/h, time taken = D / 3 hours = T + 6/60 hours. When Ram walks at 4 km/h, time taken = D / 4 hours = T - 4/60 hours. Subtract the second equation from the first: D/3 - D/4 = (T + 6/60) - (T - 4/60). Left side: D/3 - D/4 = (4D - 3D) / 12 = D / 12. Right side: T cancels, giving 6/60 + 4/60 = 10/60 = 1/6 hour. So D / 12 = 1/6 which gives D = 12 * 1/6 = 2 kilometres. Verification / Alternative check: Now we can calculate the actual times and see if the early and late conditions are satisfied. At 3 km/h, time = 2 / 3 hour which is 40 minutes. At 4 km/h, time = 2 / 4 hour which is 30 minutes. The difference between these times is 10 minutes. Since Ram is 6 minutes late in the first case and 4 minutes early in the second case, the total difference is 6 + 4 = 10 minutes. This matches our calculation, confirming that D = 2 kilometres is correct. Why Other Options Are Wrong: If the distance were 1.8 km, the times would not differ by exactly 10 minutes. A distance of 3.5 km or 4 km would produce even larger time differences and would not satisfy both the late and early arrival conditions together. Only 2 km leads to 40 and 30 minutes, which perfectly match the description in the problem. Common Pitfalls: Learners often forget to convert minutes into hours before forming equations. Another common mistake is to think that 6 minutes late minus 4 minutes early equals only 2 minutes, but in reality the total difference between the two arrival times is 10 minutes. Some students also attempt to guess distances without forming equations, which can easily go wrong. Always use the relationship time = distance / speed carefully. Final Answer: The distance between Ram’s house and his school is 2 kilometres.

Question 20

The Shatabdi Express runs from Kolkata to Ranchi at an average speed of 80 km/h and takes 7 hours for the journey. What is the distance between Kolkata and Ranchi in kilometres?

Accepted Answer

Introduction / Context: This question is a direct application of the basic time and distance formula. It provides the average speed of a train and the time taken for a journey between two cities and asks for the distance. Such problems are standard in aptitude tests and help in building confidence with straight forward calculations involving speed, time and distance. Given Data / Assumptions: Average speed of the Shatabdi Express = 80 km/h. Time taken for the journey = 7 hours. Motion is assumed to be uniform at the given average speed. We need the distance in kilometres between Kolkata and Ranchi. Concept / Approach: The foundational formula is distance = speed * time. When both speed and time are known and expressed in compatible units, distance can be found simply by multiplying. Here speed is in kilometres per hour and time is in hours, so the resulting distance will naturally be in kilometres. There is no need for unit conversions because the units already match. Step-by-Step Solution: Write the basic formula: distance = speed * time. Substitute the given values: distance = 80 km/h * 7 h. Perform the multiplication: 80 * 7 = 560. Therefore, distance = 560 kilometres. Verification / Alternative check: As a quick check, if the train travels at 80 km in each hour, in 5 hours it would cover 400 km and in 2 more hours it would cover 160 km more. Together this becomes 400 + 160 = 560 km. This mental verification confirms that the multiplication 80 * 7 has been done correctly and the obtained distance is logical for such a journey. Why Other Options Are Wrong: Option 506 km does not come from any simple product involving 80 and an integer number of hours, so it is not correct. Options 560 m and 650 m are in metres, which are far too small for a long distance train journey between cities like Kolkata and Ranchi. The units in those options do not match the realistic scale of the trip. Therefore only 560 km is consistent with the given speed and time. Common Pitfalls: Students sometimes confuse hours and minutes or think they need to convert units when it is not necessary. Here both speed and time are already in compatible units. Another mistake is to miscalculate 80 * 7 as 5600 or 480 by misplacing a zero or adding incorrectly. It is always a good idea to quickly break the multiplication into 8 * 7 = 56 and then append the zero. Final Answer: The distance between Kolkata and Ranchi is 560 kilometres.

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