Question 1

A bus has a running speed of 80 km/h when it is moving without any stops. Including regular stoppages, its average speed over the whole hour reduces to 60 km/h. For how many minutes in each hour does the bus remain stopped?

Accepted Answer

Introduction / Context: This problem examines the difference between running speed and average speed when stoppages are involved. It is a common type of question in time and distance where one must determine how long a vehicle is stationary based on its effective speed across a full hour. The concept closely relates to practical travel planning and timetable analysis. Given Data / Assumptions: Running speed of the bus when it is moving = 80 km/h. Average speed including stoppages = 60 km/h. The time interval considered is 1 hour, that is 60 minutes. When the bus is stopped, it covers no distance. Speeds are assumed constant while the bus is in motion. Concept / Approach: Average speed is defined as total distance divided by total time taken, including both running time and stoppage time. In one hour, at an average speed of 60 km/h, the bus covers a distance of 60 km. However, whenever it is moving, it does so at 80 km/h. Hence, we first compute how much running time is needed to cover 60 km at 80 km/h. The remaining fraction of the hour is the stoppage time. Converting that fraction into minutes gives the required answer. Step-by-Step Solution: Step 1: In one hour, at an average speed of 60 km/h, the bus covers 60 km. Step 2: Let the time in hours for which the bus is actually moving be t hours. While running, its speed is 80 km/h. Step 3: Distance covered during running is 80 * t kilometres. This must be equal to the total distance of 60 km, so 80 * t = 60. Step 4: Solve for t: t = 60 / 80 = 3 / 4 hour = 0.75 hour. Step 5: Total time in an hour is 1 hour, so stoppage time = 1 − 0.75 = 0.25 hour. Step 6: Convert 0.25 hour into minutes: 0.25 * 60 = 15 minutes. Verification / Alternative check: If the bus runs for 45 minutes and stops for 15 minutes, in 45 minutes (which is 0.75 hour) at 80 km/h it travels 80 * 0.75 = 60 km. Over the full hour, distance is 60 km and time is 1 hour, so average speed is 60 km/h, which matches the given value. This confirms that the stoppage time of 15 minutes per hour is correct. Why Other Options Are Wrong: A stoppage time of 12 minutes or 10 minutes would give more running time and therefore a higher effective average speed than 60 km/h. Stoppages of 18 or 20 minutes would reduce the distance covered in an hour below 60 km at a running speed of 80 km/h, giving a lower average speed. Only 15 minutes of stoppage yields exactly the required average speed of 60 km/h. Common Pitfalls: Learners sometimes confuse running speed and average speed and try to subtract speeds directly, which is incorrect. Another error is assuming that the average speed is simply the mean of two speeds, which does not hold when there are periods of rest. Correctly defining distance covered during the running time and dividing by the total time, including stops, is essential for solving such problems accurately. Final Answer: The bus stops for 15 minutes per hour.

Question 2

A bus covers a distance of 720 kilometres in 20 hours while travelling at a constant speed. Using the basic speed formula and appropriate unit conversion, what is its average speed expressed in metres per second?

Accepted Answer

Introduction / Context: This question assesses the understanding of average speed and unit conversion from kilometres per hour to metres per second. It is a standard numerical problem that strengthens the relationship between distance, time, and speed, and is particularly relevant for exams that require quick mental calculations with unit conversions. Given Data / Assumptions: Total distance travelled by the bus = 720 kilometres. Total time taken = 20 hours. The speed is assumed constant during the entire journey. We need to express the final answer in metres per second. Use 1 kilometre = 1000 metres and 1 hour = 3600 seconds as standard conversion factors. Concept / Approach: The formula for average speed is speed = distance / time. We first calculate the speed in kilometres per hour by dividing 720 by 20. After that, we convert the speed to metres per second using the standard relation between km/h and m/s. Specifically, to convert km/h to m/s, we multiply by 1000 and then divide by 3600, which is equivalent to dividing by 3.6. This results in the speed expressed in the desired unit. Step-by-Step Solution: Step 1: Compute the speed in km/h: speed_kmph = 720 / 20 = 36 km/h. Step 2: Use the relation 1 km/h = 1000 metres / 3600 seconds = 1 / 3.6 m/s. Step 3: Convert 36 km/h into m/s: speed_mps = 36 / 3.6. Step 4: Evaluate 36 / 3.6 = 10, so the bus moves at 10 metres per second on average. Step 5: Therefore, the average speed of the bus is 10 m/s. Verification / Alternative check: We can cross check by converting the total distance and time directly into metres and seconds. Distance = 720 kilometres = 720 * 1000 = 720000 metres. Time = 20 hours = 20 * 3600 = 72000 seconds. Hence, speed = 720000 / 72000 = 10 metres per second. This independent computation confirms that the calculated value is correct. Why Other Options Are Wrong: Values such as 12, 15, 18, or 20 m/s correspond to significantly different combinations of distance and time and are inconsistent with 720 kilometres in 20 hours. For example, 12 m/s would give a distance of 12 * 72000 = 864000 metres, which is 864 kilometres, not 720 kilometres. Only 10 m/s reproduces the exact given journey details. Common Pitfalls: Common mistakes include forgetting to divide by 3.6 when converting from km/h to m/s or incorrectly multiplying instead. Some learners also miscalculate 36 / 3.6 or treat 720 / 20 as 20 instead of 36. Carefully applying the formula speed = distance / time and paying attention to unit conversion steps prevents such errors. Final Answer: The average speed of the bus is 10 metres per second.

Question 3

A boat goes upstream at a speed of 21 km/h against the current and returns downstream over the same distance at a speed of 28 km/h with the current. Considering the complete to and fro journey, what is the average speed of the boat (in km/h) for the…

Accepted Answer

Introduction / Context: This question focuses on average speed when an object travels equal distances at different speeds in two directions. This is a classic concept in time and distance where the correct method is to use the harmonic mean rather than the simple arithmetic mean. It is particularly important in river boat and stream related problems in aptitude tests. Given Data / Assumptions: Upstream speed of the boat = 21 km/h. Downstream speed of the boat = 28 km/h. Distance upstream is equal to the distance downstream. The current speed is constant and the boat speed relative to water is constant. We need the average speed over the entire round trip. Concept / Approach: When the distances in both directions are equal and the speeds are different, average speed is not the arithmetic mean. Instead, average speed = (2 * u * v) / (u + v), where u and v are speeds in the two directions. This formula is derived from distance = speed * time for both legs of the journey. We apply this formula directly to the upstream and downstream speeds to find the overall average speed. Step-by-Step Solution: Step 1: Let the one-way distance be d kilometres. Upstream time = d / 21 hours and downstream time = d / 28 hours. Step 2: Total distance for the round trip = 2d kilometres. Step 3: Total time for the trip = d / 21 + d / 28 hours. Step 4: Average speed = total distance / total time = 2d / (d / 21 + d / 28). Step 5: Factor out d from the denominator to get 2d / [d * (1 / 21 + 1 / 28)] = 2 / (1 / 21 + 1 / 28). Step 6: Compute 1 / 21 + 1 / 28 = (4 + 3) / 84 = 7 / 84 = 1 / 12. Step 7: Hence average speed = 2 / (1 / 12) = 2 * 12 = 24 km/h. Verification / Alternative check: Choose a convenient value, for example, d = 84 kilometres, which is a common multiple of 21 and 28. Upstream time = 84 / 21 = 4 hours. Downstream time = 84 / 28 = 3 hours. Total distance = 168 kilometres. Total time = 7 hours. Average speed = 168 / 7 = 24 km/h. This direct calculation confirms that the formula based result is correct. Why Other Options Are Wrong: 24.5 km/h and 25 km/h are close to the arithmetic mean of 21 and 28 but do not match the correct harmonic mean calculation. 25.4 km/h and 23.5 km/h also do not satisfy the detailed time and distance relationship. Only 24 km/h is consistent with equal distances travelled at 21 km/h and 28 km/h. Common Pitfalls: A frequent error is to compute the simple average (21 + 28) / 2 = 24.5 and assume it is the answer. This ignores that time spent at the lower speed is more, which pulls the effective average speed down. Another mistake is to forget that total distance is double the one-way distance when computing the final average speed. Using the harmonic mean formula avoids these errors. Final Answer: The average speed of the boat for the entire journey is 24 km/h.

Question 4

Two runners A and B start from two different points and run towards each other along a straight track. Runner A runs at 12 km/h and runner B runs at 16 km/h. If they meet after 1 hour and 30 minutes, what was the distance (in kilometres) between the…

Accepted Answer

Introduction / Context: This problem is a straightforward application of the relative speed concept in time and distance. Two runners are moving directly towards each other, and we are given their individual speeds and the time taken to meet. The question asks for the initial separation distance between them, which is a classical and frequently asked type in quantitative exams. Given Data / Assumptions: Speed of runner A = 12 km/h. Speed of runner B = 16 km/h. They are running towards each other in a straight line. Time taken to meet = 1 hour 30 minutes = 1.5 hours. Speeds are assumed to be constant and there are no breaks in their motion. Concept / Approach: When two objects move towards each other along the same straight line, their relative speed is equal to the sum of their individual speeds. The total distance between them initially is equal to relative speed multiplied by the time until they meet. Because speeds are in kilometres per hour, we convert the time into hours and then apply distance = speed * time to find the initial separation between the two runners. Step-by-Step Solution: Step 1: Convert the meeting time into hours: 1 hour 30 minutes = 1.5 hours. Step 2: Compute the relative speed: relative speed = 12 + 16 = 28 km/h. Step 3: Use distance = speed * time. Here, distance between them initially = relative speed * time taken to meet. Step 4: Distance = 28 * 1.5 = 42 kilometres. Step 5: Therefore, the initial distance between runner A and runner B was 42 kilometres. Verification / Alternative check: We can check by calculating how far each runner travels in 1.5 hours. Runner A covers 12 * 1.5 = 18 kilometres. Runner B covers 16 * 1.5 = 24 kilometres. Adding these distances gives 18 + 24 = 42 kilometres, which must equal the original separation between them. This confirms that 42 kilometres is correct. Why Other Options Are Wrong: If the distance were 36 kilometres or 40 kilometres, the time required at a combined speed of 28 km/h would be 36 / 28 hours or 40 / 28 hours, which are not equal to 1.5 hours. A distance of 45 kilometres would require 45 / 28 hours, which is more than 1.5 hours. Similarly, 30 kilometres would result in a smaller time. Only 42 kilometres corresponds exactly to 1.5 hours at a relative speed of 28 km/h. Common Pitfalls: Common mistakes include subtracting the speeds instead of adding them for motion towards each other, or failing to convert 1 hour 30 minutes into 1.5 hours. Another typical error is mixing up which distance to attribute to each runner instead of summing them. A clear understanding of relative motion and careful unit handling avoids these issues. Final Answer: The runners were initially separated by 42 kilometres.

Question 5

Flight A and Flight B both travel the same long route of 7200 kilometres. Under normal conditions, Flight A takes 1 hour more than Flight B to cover this distance. Due to engine trouble, the speed of Flight B decreases by a factor of 1/6 (so it beco…

Accepted Answer

Introduction / Context: This question explores comparative speed and time for two flights covering the same long distance. It involves setting up equations based on the relationship between distance, speed, and time and then adjusting one flight speed due to an engine problem. The problem requires forming and solving linear equations rather than just direct substitution, so it tests algebraic manipulation in the context of time and distance. Given Data / Assumptions: Distance for both flights A and B = 7200 kilometres. Under normal conditions, time taken by A is 1 hour more than time taken by B. Speed of B under normal conditions is v_B km/h and speed of A is v_A km/h. Due to engine trouble, new speed of B becomes (5 / 6) * v_B. With this reduced speed, time taken by B is 36 minutes (0.6 hour) more than time taken by A. We need to find v_A, the speed of Flight A in km/h. Concept / Approach: If distance is D and speed is v, then time = D / v. Using this relation, we translate the two time comparisons into equations. First, we relate the original times of A and B. Second, we relate the time of B after the speed reduction to the time of A. This produces a system of two equations in two unknowns (the times or the speeds). Solving these equations yields the required speed of Flight A. Step-by-Step Solution: Step 1: Let the normal speeds be v_A for Flight A and v_B for Flight B. Then time taken by A is 7200 / v_A hours and time taken by B is 7200 / v_B hours. Step 2: From the first condition, A takes 1 hour more than B: 7200 / v_A = 7200 / v_B + 1. Step 3: When B's speed reduces to (5 / 6) * v_B, its new time is 7200 / [(5 / 6) * v_B] = (6 / 5) * 7200 / v_B. Step 4: Second condition states new time of B is 0.6 hour more than time of A: (6 / 5) * 7200 / v_B = 7200 / v_A + 0.6. Step 5: Solve these equations. From the first equation, express 7200 / v_A in terms of 7200 / v_B. Substitute into the second equation and solve to get time of A = 9 hours and time of B = 8 hours. Step 6: Speed of Flight A = distance / time = 7200 / 9 = 800 km/h. Verification / Alternative check: At 800 km/h, Flight A takes 9 hours. Since Flight B takes 1 hour less normally, its time is 8 hours, giving speed v_B = 7200 / 8 = 900 km/h. After the engine trouble, speed of B becomes (5 / 6) * 900 = 750 km/h. Now time of B becomes 7200 / 750 = 9.6 hours. Time of A is 9 hours, so B now takes 0.6 hour (36 minutes) more than A. Both conditions are satisfied, confirming that 800 km/h for Flight A is correct. Why Other Options Are Wrong: If Flight A had a speed of 900 km/h or 750 km/h or 720 km/h, the derived times and adjusted times would not satisfy both given time differences of 1 hour and 36 minutes. The numerical checks would fail one or both of the conditions. Only 800 km/h leads to consistent times before and after the speed change of Flight B. Common Pitfalls: Some learners attempt to compare speeds directly without forming proper equations or forget to convert 36 minutes into 0.6 hour. Others may mistakenly think the speed of Flight A also changes when only Flight B is affected. A systematic approach, carefully setting up equations from the language of the problem, avoids these mistakes and leads to the correct result. Final Answer: The speed of Flight A is 800 km/h.

Question 6

A car travelling at an average speed of 72 km/h takes 9 minutes to cover a certain distance on a straight road. If the car wants to cover the same distance in only 8 minutes, by how many km/h must it increase its average speed?

Accepted Answer

Introduction / Context: This question examines the relationship between speed and time when the distance is fixed. It is a very common type in aptitude tests where the candidate must determine the new speed required to cover the same distance in a shorter time. It requires careful unit conversion from minutes to hours and accurate use of the formula distance = speed * time. Given Data / Assumptions: Initial average speed of the car = 72 km/h. Time taken at this speed = 9 minutes. The distance remains the same for the second case. Desired time for the second case = 8 minutes. Speeds are assumed to be constant in each case. We need the increase in speed in km/h, that is new speed minus old speed. Concept / Approach: For a fixed distance, speed and time are inversely proportional. First, we compute the actual distance using the initial speed and time. Next, we divide this same distance by the new, shorter time to get the required new speed. The difference between this new speed and the original speed is the increase in speed. Time values must be converted from minutes to hours to maintain consistency with km/h units. Step-by-Step Solution: Step 1: Convert 9 minutes into hours: 9 minutes = 9 / 60 hours = 0.15 hours. Step 2: Compute the distance covered at 72 km/h in 0.15 hours: distance = 72 * 0.15. Step 3: Calculate 72 * 0.15 = 10.8 kilometres. Step 4: Convert the desired time of 8 minutes to hours: 8 minutes = 8 / 60 hours = 2 / 15 hours. Step 5: Required new speed = same distance / new time = 10.8 / (2 / 15) km/h. Step 6: Simplify 10.8 / (2 / 15) = 10.8 * (15 / 2) = 10.8 * 7.5 = 81 km/h. Step 7: Increase in speed = 81 − 72 = 9 km/h. Verification / Alternative check: Check that at 81 km/h for 8 minutes the car travels the same distance. Time in hours = 8 / 60 = 2 / 15 hours. Distance = 81 * (2 / 15) = 162 / 15 = 10.8 kilometres, which matches the original distance. This confirms that 81 km/h is the correct new speed and that the required increase is indeed 9 km/h. Why Other Options Are Wrong: An increase of 8 km/h would give a new speed of 80 km/h and a distance of 80 * (2 / 15) = 160 / 15 = 10.67 km, which is too small. Similarly, increases of 6 or 7 km/h produce distances smaller than 10.8 km in 8 minutes. An increase of 10 km/h results in a speed of 82 km/h, giving 82 * (2 / 15) ≠ 10.8. Only an increase of 9 km/h yields the exact required distance in the shorter time. Common Pitfalls: A common mistake is to directly scale the speeds by the ratio of times without computing the distance properly, which can lead to rounding or conceptual errors. Another frequent error is forgetting to convert minutes to hours or miscalculating 9 / 60 and 8 / 60. Careful attention to units and consistent use of the distance formula avoids such problems. Final Answer: The car must increase its speed by 9 km/h to cover the distance in 8 minutes instead of 9 minutes.

Question 7

Two cars A and B travel from one city to another along the same route. Car A travels at 60 km/h and car B travels at 108 km/h. If car B takes exactly 2 hours less than car A to complete the journey, what is the distance (in kilometres) between the t…

Accepted Answer

Introduction / Context: This problem tests relative travel times when two vehicles move over the same distance but at different constant speeds. It requires setting up an equation equating the difference in their travel times to a given value. The question is a classic example of solving for distance using the time difference between a slower and a faster vehicle. Given Data / Assumptions: Speed of car A = 60 km/h. Speed of car B = 108 km/h. Both cars travel the same distance between the two cities. Car B takes 2 hours less than car A to cover the journey. Speeds remain constant for the entire trip. We need to find the common distance D in kilometres. Concept / Approach: If distance is D kilometres and speed is v km/h, then time taken is D / v hours. Let time taken by car A be D / 60 and time taken by car B be D / 108. We are told that the difference in their travel times is 2 hours. This gives a linear equation in D that we can solve. Once D is found, we check that the time difference matches the given value. Step-by-Step Solution: Step 1: Let the distance between the two cities be D kilometres. Step 2: Time taken by car A = D / 60 hours. Step 3: Time taken by car B = D / 108 hours. Step 4: According to the question, the faster car B takes 2 hours less than car A, so D / 60 − D / 108 = 2. Step 5: Compute the difference: D * (1 / 60 − 1 / 108) = 2. Step 6: Find 1 / 60 − 1 / 108 = (108 − 60) / (60 * 108) = 48 / 6480 = 1 / 135. Step 7: So D * (1 / 135) = 2, hence D = 2 * 135 = 270 kilometres. Verification / Alternative check: Check the times for each car. For D = 270 km, time for car A = 270 / 60 = 4.5 hours. Time for car B = 270 / 108 = 2.5 hours. The difference in time is 4.5 − 2.5 = 2 hours, which matches the given condition. Therefore, the distance between the two cities must be 270 km. Why Other Options Are Wrong: If D were 240 km, times would be 240 / 60 = 4 hours and 240 / 108 ≈ 2.22 hours, with a difference not equal to 2 hours. Similarly, 300 km would give times 5 hours and approximately 2.78 hours, again not differing by 2 hours. Values of 210 km or 330 km also fail to produce the correct time difference. Only 270 km satisfies the time difference of 2 hours exactly. Common Pitfalls: Learners sometimes mistakenly equate the ratio of speeds to the ratio of times directly or attempt to use average speeds instead of forming the proper time difference equation. Another common error is in computing the fraction 1 / 60 − 1 / 108 incorrectly. Careful algebraic manipulation and accurate fraction work prevent these mistakes. Final Answer: The distance between the two cities is 270 kilometres.

Question 8

Two cars travel from city A to city B along the same route. One car moves at 24 km/h and the other moves at 32 km/h. If the faster car takes exactly 2.5 hours less time than the slower one to complete the journey, what is the distance (in kilometres…

Accepted Answer

Introduction / Context: This question is very similar in structure to other time and distance comparison problems involving two vehicles travelling the same route at different speeds. The relationship between speed, time, and distance is used, along with the given difference in time, to determine the unknown distance between two cities. Given Data / Assumptions: Speed of the slower car = 24 km/h. Speed of the faster car = 32 km/h. Both cars cover the same distance from city A to city B. The faster car takes 2.5 hours less than the slower car. Speeds are constant and there are no breaks or changes in speed. We need to find the distance in kilometres between the two cities. Concept / Approach: Let the common distance be D kilometres. The time taken by the slower car is D / 24 hours, and the time taken by the faster car is D / 32 hours. The difference between these times is given as 2.5 hours. Setting up the equation D / 24 − D / 32 = 2.5 allows us to solve for D. This type of equation uses basic algebra and fraction manipulation to find the unknown distance. Step-by-Step Solution: Step 1: Let the distance between city A and city B be D kilometres. Step 2: Time taken by the slower car = D / 24 hours. Step 3: Time taken by the faster car = D / 32 hours. Step 4: According to the question, D / 24 − D / 32 = 2.5. Step 5: Compute the difference in fractions: 1 / 24 − 1 / 32 = (32 − 24) / (24 * 32) = 8 / 768 = 1 / 96. Step 6: So D * (1 / 96) = 2.5, which implies D = 2.5 * 96. Step 7: Multiply 2.5 * 96 = (5 / 2) * 96 = 5 * 48 = 240. Step 8: Hence, D = 240 kilometres. Verification / Alternative check: Check the times for D = 240 km. Time for the slower car: 240 / 24 = 10 hours. Time for the faster car: 240 / 32 = 7.5 hours. The difference in times is 10 − 7.5 = 2.5 hours, which exactly matches the information in the question. Therefore, the distance of 240 km is correct. Why Other Options Are Wrong: For 288 km, times would be 288 / 24 = 12 hours and 288 / 32 = 9 hours, giving a 3 hour difference. For 360 km, the difference would be even larger. A distance of 192 km would give times of 8 hours and 6 hours, which differ by 2 hours. A distance of 320 km yields a difference that is not 2.5 hours. Only 240 km satisfies the exact 2.5 hour difference condition. Common Pitfalls: A common mistake is to attempt to use average speed or proportional reasoning without properly forming the time difference equation. Another error is miscalculating the fraction 1 / 24 − 1 / 32 or handling 2.5 incorrectly as a fraction. Writing 2.5 as 5 / 2 and proceeding step by step with fractional arithmetic helps avoid such mistakes. Final Answer: The distance between city A and city B is 240 kilometres.

Question 9

A man runs at a constant speed of 20 km/h around a circular garden that has a radius of 350 metres. How much time (in seconds) will he take to complete exactly one full round of the circular garden?

Accepted Answer

Introduction / Context: This problem combines circular geometry with time and speed concepts. It requires computing the circumference of a circle and then using the basic speed formula to compute the time needed to cover that circumference. It also tests the ability to handle unit conversions between kilometres per hour and metres per second, which is a common theme in time and distance problems. Given Data / Assumptions: Speed of the man = 20 km/h. Radius of the circular garden = 350 metres. The man runs along the boundary, so he covers the circumference of the circle. He runs at a constant speed without stopping. We need the time taken to complete one full round, expressed in seconds. Use π ≈ 22 / 7 for practical calculation of circumference. Concept / Approach: The distance around a circular path is its circumference, given by circumference = 2 * π * r. After computing the circumference in metres, we convert the runner's speed into metres per second to keep units consistent. Then we use time = distance / speed to find the required time. This structured approach ensures correct handling of both geometry and unit conversions. Step-by-Step Solution: Step 1: Compute the circumference of the garden. Radius r = 350 m. Use π = 22 / 7. Step 2: Circumference = 2 * π * r = 2 * (22 / 7) * 350 = 44 * 50 = 2200 metres. Step 3: Convert the man's speed from km/h to m/s. Speed = 20 km/h = 20 * 1000 / 3600 m/s = 20000 / 3600 m/s = 5.555... m/s. Step 4: Use the distance and speed to find time: time = distance / speed = 2200 / (20000 / 3600) seconds. Step 5: Simplify: 2200 * 3600 / 20000 = 2200 * 0.18 = 396 seconds. Step 6: Therefore, the man takes 396 seconds to complete one round of the garden. Verification / Alternative check: We can approximate using 5.56 m/s for the speed. Then time ≈ 2200 / 5.56 ≈ 395.7 seconds, which rounds to 396 seconds, matching our exact calculation using fractions. This confirms that the numerical result is consistent with both precise fractional and approximate decimal methods. Why Other Options Are Wrong: Times such as 336, 376, 360, or 412 seconds correspond to significantly different speeds for the same distance. For example, 2200 / 336 ≈ 6.55 m/s and 2200 / 412 ≈ 5.34 m/s, which are not equal to the given speed of 20 km/h (approximately 5.56 m/s). Only 396 seconds aligns with the given speed after proper unit conversion and circumference calculation. Common Pitfalls: A frequent mistake is using diameter instead of radius in the circumference formula or neglecting to multiply by 2. Another common error is forgetting to convert 20 km/h into m/s and instead using inconsistent units, which leads to incorrect time values. Approximating π too coarsely can also create slight discrepancies. Using π = 22 / 7 and carefully converting units ensures an accurate answer. Final Answer: The man will take 396 seconds to complete one round of the garden.

Question 10

A man covers a certain distance from his home to a destination by train at a speed of 50 km/h and returns over the same distance on foot, walking at 5 km/h. If the total time taken for the two-way journey is 5 hours 30 minutes, what is the distance …

Accepted Answer

Introduction / Context: This question is a classic example of travelling the same distance by two different modes of transport at different speeds, one for the onward journey and one for the return journey. The total time for the round trip is known, and we must find the one way distance. It uses the relationship between distance, speed, and time along with basic algebra to solve for the unknown distance. Given Data / Assumptions: Speed of the train on the onward journey = 50 km/h. Speed of walking on the return journey = 5 km/h. Distance from home to destination (one side) = d kilometres (unknown). Total two way time = 5 hours 30 minutes = 5.5 hours. Speeds are constant and there are no breaks during each leg of the journey. Concept / Approach: For a given distance, time = distance / speed. There are two legs: going by train and returning on foot. Thus, total time = d / 50 + d / 5. This total time is given as 5.5 hours. Setting up this equation and solving for d gives the required one way distance. Careful conversion of minutes to hours and correct algebraic manipulation are essential. Step-by-Step Solution: Step 1: Let the one way distance be d kilometres. Step 2: Time taken by train = d / 50 hours. Step 3: Time taken walking back = d / 5 hours. Step 4: Total time is given as 5.5 hours, so d / 50 + d / 5 = 5.5. Step 5: Express 1 / 5 as 10 / 50 to combine fractions: d / 50 + 10d / 50 = 5.5, so 11d / 50 = 5.5. Step 6: Multiply both sides by 50: 11d = 5.5 * 50 = 275. Step 7: Solve for d: d = 275 / 11 = 25 kilometres. Verification / Alternative check: Check the times for d = 25 km. Time by train = 25 / 50 = 0.5 hours (30 minutes). Time walking back = 25 / 5 = 5 hours. Total time = 0.5 + 5 = 5.5 hours, which is exactly 5 hours 30 minutes. This confirms that the one way distance is 25 km. Why Other Options Are Wrong: If the distance were 30 km, times would be 0.6 hours by train and 6 hours walking, totaling 6.6 hours, which is too long. For 20 km, total time would be 0.4 + 4 = 4.4 hours, which is too short. Similarly, 18 km or 22 km do not produce a total time of 5.5 hours. Only 25 km satisfies the exact time condition. Common Pitfalls: Learners sometimes forget to convert 5 hours 30 minutes into 5.5 hours or mishandle the fraction addition when combining d / 50 and d / 5. Another error is to attempt to use an average speed formula incorrectly instead of summing the individual times. Carefully forming the time equation and solving it systematically leads to the correct distance. Final Answer: The distance of one side is 25 kilometres.

Question 11

After repairing, a truck moves at a speed of 90 km/h, while before repairing it used to move at a speed of 80 km/h. After repairing, it covers a distance x in 8 hours. If the same truck were to cover four times that distance (4x) at its older speed …

Accepted Answer

Introduction / Context: This question examines the relationship between distance, speed, and time when a vehicle has two different speeds before and after repair. It tests the ability to compute distance from speed and time, and then reuse this distance to find another time value at a different speed. The structure is straightforward but reinforces accurate use of basic formulas in time and distance problems. Given Data / Assumptions: Speed of the truck after repairing = 90 km/h. Speed of the truck before repairing = 80 km/h. After repair, the truck covers a distance x in 8 hours. We need the time taken to cover 4x distance at the old speed of 80 km/h. Speeds are constant during each scenario and there are no breaks. Concept / Approach: First, we compute the distance x using distance = speed * time for the truck after repairing. Then we multiply this distance by 4 to obtain 4x. Next, we calculate the time needed to cover 4x at the older speed before repairing using time = distance / speed. This two step approach shows how distance acts as a bridge between different speed and time combinations. Step-by-Step Solution: Step 1: Compute distance x covered after repairing. Speed = 90 km/h and time = 8 hours. Step 2: Distance x = 90 * 8 = 720 kilometres. Step 3: We are asked for the time taken to cover 4x at the old speed. So 4x = 4 * 720 = 2880 kilometres. Step 4: Before repairing, the speed was 80 km/h. Step 5: Time to cover 2880 kilometres at 80 km/h = distance / speed = 2880 / 80 hours. Step 6: Evaluate 2880 / 80 = 36 hours. Step 7: Therefore, the truck would take 36 hours to cover 4x distance at the older speed. Verification / Alternative check: We can reduce the fraction differently. Time = 4x / 80 = (4 * 720) / 80 hours. Simplify: 720 / 80 = 9. So time = 4 * 9 = 36 hours. This alternative calculation arrives at the same result, confirming the correctness of the answer. Why Other Options Are Wrong: If the time were 40 hours or 48 hours, the implied distances would be 40 * 80 = 3200 km or 48 * 80 = 3840 km, which are not equal to 2880 km. A value of 28 hours would give only 2240 km, and 32 hours would give 2560 km, both inconsistent with four times the original distance of 720 km. Only 36 hours is consistent with 4x = 2880 km at 80 km/h. Common Pitfalls: A common error is to forget to multiply the distance x by 4 before calculating the time, or to confuse which speed corresponds to which part of the journey. Another mistake is to miscalculate 90 * 8 or 2880 / 80. Keeping track of the two scenarios separately and writing down each step avoids these issues. Final Answer: The truck would take 36 hours to cover four times the original distance at its older speed of 80 km/h.

Question 12

An Express train and a Rajdhani train both travel the same distance of 708 kilometres. The Express train takes 6 hours more than the Rajdhani train to cover this distance. If the speed of the Express train is doubled, it then takes 3 hours less than…

Accepted Answer

Introduction / Context: This problem involves two trains travelling the same distance at different speeds and compares their travel times under two scenarios. First, the Express train is slower and takes longer, and later its speed is doubled so that it becomes effectively faster in terms of time. The question requires setting up and solving a system of equations based on the distance speed time relationship, making it a more advanced time and distance problem. Given Data / Assumptions: Common distance for both trains = 708 kilometres. Let the speed of the Express train be v_E km/h and the speed of the Rajdhani train be v_R km/h. Initially, the Express train takes 6 hours more than Rajdhani: 708 / v_E = 708 / v_R + 6. After doubling the speed of the Express train, its speed becomes 2 * v_E km/h. With this new speed, the Express train takes 3 hours less than Rajdhani: 708 / (2 * v_E) = 708 / v_R − 3. We need to find v_R, the speed of the Rajdhani train in km/h. Concept / Approach: We use time = distance / speed for each train and interpret the given time differences as equations. This yields a system of two equations in two unknowns (v_E and v_R). Solving this system requires algebraic manipulation and substitution. Once v_R is found, we verify that both time difference conditions are satisfied. The key idea is to carefully translate the verbal statements into precise mathematical equations. Step-by-Step Solution: Step 1: Let v_E be the speed of the Express and v_R be the speed of the Rajdhani. Step 2: From the first condition, time of Express − time of Rajdhani = 6 hours: 708 / v_E − 708 / v_R = 6. Step 3: From the second condition after doubling Express speed: time of Express (new) = time of Rajdhani − 3 hours: 708 / (2 * v_E) = 708 / v_R − 3. Step 4: Multiply the first equation by appropriate factors or use algebraic tools to solve the system. Solving yields v_E = 118 / 3 km/h and v_R = 59 km/h. Step 5: So the speed of the Rajdhani train v_R is 59 km/h. Verification / Alternative check: Compute times with these speeds. Distance = 708 km. For Rajdhani: time_R = 708 / 59 = 12 hours. For Express at original speed v_E = 118 / 3 ≈ 39.33 km/h: time_E = 708 / (118 / 3) = 708 * 3 / 118 = 18 hours. The difference time_E − time_R = 18 − 12 = 6 hours, which matches the first condition. After doubling the Express speed, new Express speed = 2 * 118 / 3 = 236 / 3 ≈ 78.67 km/h, so new Express time = 708 / (236 / 3) = 708 * 3 / 236 = 9 hours. Time difference time_R − new time_E = 12 − 9 = 3 hours, matching the second condition. Both conditions are satisfied. Why Other Options Are Wrong: Speeds like 78.7 km/h or 39.3 km/h represent variations of the Express train speed, not the Rajdhani speed. Choosing them for v_R would give inconsistent times that fail one or both of the time difference conditions. A speed of 98.3 km/h or 65 km/h for Rajdhani also fails to satisfy both the 6 hour and 3 hour differences simultaneously. Only 59 km/h fits both conditions exactly. Common Pitfalls: Common mistakes include mixing up which train is faster in each scenario, misinterpreting the time differences, or incorrectly doubling the wrong speed. Another issue is trying to guess values instead of forming proper equations, which often leads to errors in complex problems. A careful translation of the problem into symbolic equations and their systematic solution is the most reliable approach. Final Answer: The speed of the Rajdhani train is 59 km/h.

Question 13

A man travels a total distance of 430 kilometres, partly by rail at a speed of 65 km/h and partly by steamer at a speed of 25 km/h. He spends 10 hours more time on the steamer than on the train. How many kilometres does he cover by steamer?

Accepted Answer

Introduction / Context: This question involves travelling a fixed total distance using two different modes of transport with different speeds and a known difference in travel times. It combines the basic distance speed time formula with an additional condition comparing time spent on each mode. The objective is to determine how much of the total distance is covered by steamer, which is a standard type of algebraic time and distance problem. Given Data / Assumptions: Total distance travelled = 430 kilometres. Speed by rail = 65 km/h. Speed by steamer = 25 km/h. Let distance covered by rail be x kilometres and by steamer be 430 − x kilometres. Time spent on steamer is 10 hours more than time spent on rail. Speeds are constant and there are no breaks during each part of the journey. Concept / Approach: Time taken on each mode is distance / speed. Time by rail is x / 65 hours and time by steamer is (430 − x) / 25 hours. The given condition is (430 − x) / 25 = x / 65 + 10. This equation can be solved for x, the distance by rail, and then 430 − x gives the distance by steamer. Careful algebraic manipulation and solving of linear equations are required. Step-by-Step Solution: Step 1: Let the distance covered by rail be x kilometres. Step 2: Then the distance covered by steamer is 430 − x kilometres. Step 3: Time by rail = x / 65 hours. Time by steamer = (430 − x) / 25 hours. Step 4: Given that time on steamer is 10 hours more than time on rail: (430 − x) / 25 = x / 65 + 10. Step 5: Multiply through by the common denominator 325 to clear fractions: 13 * (430 − x) = 5x + 3250. Step 6: Expand: 13 * 430 = 5590, so 5590 − 13x = 5x + 3250. Step 7: Combine like terms: 5590 − 3250 = 5x + 13x, so 2340 = 18x, hence x = 2340 / 18 = 130 kilometres. Step 8: Distance by steamer = 430 − 130 = 300 kilometres. Verification / Alternative check: Check the times with these distances. Time by rail = 130 / 65 = 2 hours. Distance by steamer = 300 km, so time by steamer = 300 / 25 = 12 hours. The difference in times is 12 − 2 = 10 hours, which matches the condition given in the question. Therefore the distance travelled by steamer is correctly found as 300 km. Why Other Options Are Wrong: If the man travelled 360 km by steamer, then the rail distance would be 70 km and the times would not differ by 10 hours. Values like 450 km or 540 km exceed the total distance of 430 km and are impossible. A value of 280 km would give a different time difference, not equal to 10 hours. Only 300 km satisfies both the total distance and the time difference constraint. Common Pitfalls: A common mistake is to misplace x and 430 − x with rail and steamer, or to forget that the time difference condition is steamer time minus rail time equals 10, not the other way around. Errors in clearing fractions and simplifying the equation also occur. Writing each step clearly and verifying that the final distances add up to 430 km helps to avoid these issues. Final Answer: The man covers 300 kilometres by steamer.

Question 14

A policeman is 400 metres behind a thief when he starts chasing. At that instant, the thief starts running at 5 km/h and the policeman runs at 9 km/h, both along the same straight road in the same direction. Assuming both maintain their speeds, how …

Accepted Answer

Introduction / Context: This problem is a classic chase or pursuit question in time and distance. A faster person (the policeman) chases a slower person (the thief) who has a head start. The goal is to find how far the thief runs before being caught. It tests the understanding of relative speed when two bodies move in the same direction along a straight line. Given Data / Assumptions: Initial distance between the policeman and the thief = 400 metres. Speed of the thief = 5 km/h. Speed of the policeman = 9 km/h. They run in the same direction along a straight road without changing speed. We must find the distance run by the thief by the time he is caught. 1 kilometre = 1000 metres for conversion if needed. Concept / Approach: When two objects move in the same direction, the relative speed between the faster and the slower is the difference of their speeds. The time taken by the faster object to catch the slower is equal to the initial gap divided by the relative speed. After finding this time, we multiply it by the speed of the thief to get the distance he covers before being caught. All speeds should be in the same units when computing time and distance. Step-by-Step Solution: Step 1: Convert the head start distance into kilometres for consistency. Initial gap = 400 metres = 0.4 kilometres. Step 2: Compute the relative speed when both move in the same direction: relative speed = speed of policeman − speed of thief = 9 − 5 = 4 km/h. Step 3: Time taken for the policeman to catch the thief = initial gap / relative speed = 0.4 / 4 hours = 0.1 hour. Step 4: Distance run by the thief in that time = speed * time = 5 km/h * 0.1 hour = 0.5 kilometres. Step 5: Convert 0.5 kilometres back into metres: 0.5 * 1000 = 500 metres. Step 6: Hence, the thief runs 500 metres before being caught. Verification / Alternative check: We can check by computing how far the policeman runs in the same 0.1 hour. Distance covered by the policeman = 9 km/h * 0.1 hour = 0.9 kilometres = 900 metres. Since the thief has run 500 metres, the distance between their starting positions is 400 metres plus 500 metres for the thief, which is 900 metres, matching the policeman's distance. This confirms that they meet after the thief has run 500 metres. Why Other Options Are Wrong: If the thief ran only 400 metres, the policeman would still be behind because he would have to cover more than the initial gap. Distances like 300 or 450 metres would not satisfy the equality of distances travelled when checked with their respective speeds and time. A distance of 600 metres would require a longer time, which would give the policeman more than enough time to close the gap. Only 500 metres leads to consistent positions for both at the meeting point. Common Pitfalls: Learners sometimes add the speeds instead of subtracting them when both move in the same direction, which is incorrect. Another common mistake is to use metres for the head start and kilometres per hour for speed without conversion, leading to unit inconsistency. Carefully using relative speed as a difference and maintaining consistent units avoids these errors. Final Answer: The thief will have run 500 metres before being caught by the policeman.

Question 15

Two cars travel from city A to city B along the same route at speeds of 60 km/h and 108 km/h respectively. One of the cars takes exactly 2 hours less time than the other to complete the journey. What is the distance (in kilometres) between city A an…

Accepted Answer

Introduction / Context: This problem is another example of comparing travel times for two vehicles moving over the same distance but at different constant speeds. The key is to interpret the statement about one car taking 2 hours less than the other as a precise relationship between their times and to translate that into an equation involving the unknown distance. Given Data / Assumptions: Speed of the slower car = 60 km/h. Speed of the faster car = 108 km/h. Both cars cover the same distance between city A and city B. The faster car takes 2 hours less time than the slower car. Speeds remain constant during the journey and there are no breaks. We need to find the distance between the two cities. Concept / Approach: If D is the distance, time taken by the slower car is D / 60 hours and time taken by the faster car is D / 108 hours. The time difference between them is D / 60 − D / 108, which is given to be 2 hours. By setting up and solving the equation D / 60 − D / 108 = 2, we can determine the value of D. This approach uses basic algebra and the fundamental distance speed time relationship. Step-by-Step Solution: Step 1: Let the common distance between city A and city B be D kilometres. Step 2: Time taken by the slower car (60 km/h) = D / 60 hours. Step 3: Time taken by the faster car (108 km/h) = D / 108 hours. Step 4: The faster car takes 2 hours less, so D / 60 − D / 108 = 2. Step 5: Compute the difference in fractions: 1 / 60 − 1 / 108 = (108 − 60) / (60 * 108) = 48 / 6480 = 1 / 135. Step 6: So D * (1 / 135) = 2, hence D = 2 * 135 = 270 kilometres. Step 7: Therefore, the distance between the two cities is 270 km. Verification / Alternative check: For D = 270 km, time for the slower car is 270 / 60 = 4.5 hours and time for the faster car is 270 / 108 = 2.5 hours. The difference is 4.5 − 2.5 = 2 hours, which matches the given information. Thus, the computed distance is correct. Why Other Options Are Wrong: If the distance were 324 km, times would be 5.4 and 3 hours, giving a difference of 2.4 hours. For 405 km, the difference would be even larger, not 2 hours. Distances like 216 km or 300 km also lead to time differences that are not equal to 2 hours. Only 270 km yields the exact 2 hour difference required by the problem statement. Common Pitfalls: Some learners attempt to guess the distance or directly manipulate speeds without forming the time difference equation. Others miscompute the fraction 1 / 60 − 1 / 108 or forget whether the faster car time should be subtracted from the slower car time or the reverse. Writing the relationship carefully and systematically solving the equation avoids these errors. Final Answer: The distance between city A and city B is 270 kilometres.

Question 16

Raghu travels from his home to his college on a motorbike, covering a total distance of 410 kilometres during a trip. For the first 5 hours he rides at a constant speed of 50 km/h. For the remaining 4 hours he rides at a different constant speed. Wh…

Accepted Answer

Introduction / Context: This question is a straightforward application of the distance speed time relationship for a journey divided into two segments with different speeds. The total distance and the time for each segment are given, and the task is to compute the unknown speed in the second segment. Such problems help reinforce the idea that total distance is the sum of segment distances and that each segment distance is speed multiplied by time. Given Data / Assumptions: Total distance of the trip = 410 kilometres. Time for the first part of the journey = 5 hours. Speed during the first part = 50 km/h. Time for the second part of the journey = 4 hours. Speed during the second part is constant and unknown. There are no breaks between segments and distances simply add up. Concept / Approach: For each segment, distance = speed * time. The distance for the first segment can be computed directly. Let the unknown speed for the second segment be v km/h. Then its distance is v * 4. Since the sum of these two distances equals the total distance of 410 km, we set up the equation first distance + second distance = 410 and solve for v. This simple algebraic equation provides the required speed. Step-by-Step Solution: Step 1: Compute the distance covered in the first 5 hours. Distance_1 = speed * time = 50 * 5 = 250 kilometres. Step 2: Let the speed during the remaining 4 hours be v km/h. Step 3: Distance covered in the remaining 4 hours is Distance_2 = v * 4 kilometres. Step 4: Total distance is the sum of the two segment distances: 250 + 4v = 410. Step 5: Solve this equation for v: 4v = 410 − 250 = 160. Step 6: v = 160 / 4 = 40 km/h. Step 7: Therefore, Raghu travels at 40 km/h in the last 4 hours of his journey. Verification / Alternative check: Check by recomputing the total distance. First segment: 50 km/h * 5 hours = 250 km. Second segment: 40 km/h * 4 hours = 160 km. Total distance = 250 + 160 = 410 km, which matches the given total. This confirms that the computed speed of 40 km/h for the second segment is correct. Why Other Options Are Wrong: If Raghu rode at 47 km/h for the second part, he would cover 188 km, giving a total of 438 km. At 56 km/h, he would cover 224 km, giving 474 km total. At 48 km/h, the second part distance would be 192 km, totalling 442 km. At 52 km/h, the distance would be 208 km, totalling 458 km. None of these match the required total of 410 km. Only 40 km/h gives the correct total distance. Common Pitfalls: A common mistake is to average the two speeds instead of using distance and time relations or to confuse total time with time for one segment. Some learners also miscalculate the first segment distance or forget that the unknown speed must be multiplied by 4 hours, not 5. Systematically writing the equation for total distance keeps the logic clear and leads to the correct result. Final Answer: During the remaining 4 hours of the journey, Raghu rides at 40 km/h.

Question 17

Two cars travel from city A to city B at constant speeds of 30 km/h and 45 km/h respectively. If one car takes 2.5 hours less time than the other car to cover the whole journey, then what is the distance between city A and city B?

Accepted Answer

Introduction / Context: This time and distance question tests your understanding of the basic relation distance = speed * time and how to set up equations when two different speeds produce a known time difference. Such problems are very common in aptitude exams as they check algebraic manipulation and interpretation of word statements. Given Data / Assumptions: Car 1 speed = 30 km/h. Car 2 speed = 45 km/h. Both travel the same distance between city A and city B. The faster car takes 2.5 hours less than the slower car. We assume both move at constant speeds without breaks. Concept / Approach: We use the formula distance = speed * time. Since the distance is the same for both cars, we can equate the two distance expressions. The key is to represent both travel times in terms of the same unknown distance and then use the given time difference to form an equation. Step-by-Step Solution: Let D be the distance between city A and city B in km. Time taken by slower car (30 km/h) = D / 30 hours. Time taken by faster car (45 km/h) = D / 45 hours. Given that the faster car saves 2.5 hours: D / 30 - D / 45 = 2.5. Compute left side: 1/30 - 1/45 = (3 - 2) / 90 = 1 / 90. So (1 / 90) * D = 2.5. Therefore D = 2.5 * 90 = 225 km. Verification / Alternative check: Check times with D = 225 km. Time at 30 km/h = 225 / 30 = 7.5 hours. Time at 45 km/h = 225 / 45 = 5 hours. Difference = 7.5 - 5 = 2.5 hours, which matches the given condition. Why Other Options Are Wrong: 270 km gives times 9 hours and 6 hours, difference 3 hours, not 2.5 hours. 338 km gives non simple decimal times and their difference is not 2.5 hours. 180 km gives times 6 hours and 4 hours, difference 2 hours, not 2.5 hours. Common Pitfalls: Students sometimes write D / 45 - D / 30 = 2.5, reversing the order of times and getting a negative value. Another mistake is to treat 2.5 as 2 hours 5 minutes instead of 2.5 hours, but here 2.5 clearly means two and a half hours. Also, using an approximate fraction for 2.5 too early can lead to rounding errors, so keep it exact as 5 / 2 if you prefer fractions. Final Answer: The distance between city A and city B is 225 km.

Question 18

Three hours after a goods train passes a station, an express train travelling at 88 km/h in the same direction passes the station and overtakes the goods train 8 hours after passing the station. What is the speed of the goods train in km/h?

Accepted Answer

Introduction / Context: This is a classic relative speed problem where a faster vehicle starts later, then catches up with a slower one. It checks understanding of how to equate distances travelled by two objects when one is trying to overtake the other, using basic time and distance formulas. Given Data / Assumptions: Goods train speed = v km/h (unknown). Goods train passes the station first. Express train passes the same station 3 hours later at 88 km/h. Express train overtakes the goods train 8 hours after it passes the station. Both trains move in the same direction on straight track with constant speeds. Concept / Approach: At the instant of overtaking, both trains have travelled the same distance from the station. The goods train has been moving for longer time, while the express train has a higher speed. We set up an equation based on equal distances and solve for the unknown speed v of the goods train. Step-by-Step Solution: Let v be the speed of the goods train in km/h. When overtaking happens, the express train has been travelling for 8 hours. During the same period from the station, the goods train has been travelling for 3 + 8 = 11 hours. Distance travelled by express train = 88 * 8 km. Distance travelled by goods train = v * 11 km. At overtaking, distances are equal: 88 * 8 = 11 * v. Compute: 88 * 8 = 704, so 704 = 11v. Therefore v = 704 / 11 = 64 km/h. Verification / Alternative check: Goods train distance from station at overtaking: 64 * 11 = 704 km. Express train distance from station: 88 * 8 = 704 km. Both match, so the calculation is consistent with the scenario. Why Other Options Are Wrong: 76.8 km/h would give distance 76.8 * 11 = 844.8 km, which does not match express distance 704 km. 96 km/h gives goods distance 96 * 11 = 1056 km, far larger than the express train distance. 51.2 km/h gives 51.2 * 11 = 563.2 km, less than the express train distance. Common Pitfalls: A common error is to subtract times incorrectly and set 88 * 5 equal to 11v by confusing the time difference. Another mistake is to work with relative speed directly without carefully tracking the starting time difference, which leads to wrong equations. Always align both motions with a common reference point, here the station, and equalize distances at the point of overtaking. Final Answer: The speed of the goods train is 64 km/h.

Question 19

A racing car runs around a racing track at an average speed of 108 km/h and currently takes 15 minutes to complete one full lap. By how much should the driver increase the speed (in km/h) so that the car completes the same lap in only 12 minutes?

Accepted Answer

Introduction / Context: This question checks your ability to convert between time and speed for a fixed distance. When the distance remains constant, average speed is inversely proportional to time taken. The problem asks for the increase in speed required to reduce the lap time from 15 minutes to 12 minutes for a racing car. Given Data / Assumptions: Initial average speed of the car = 108 km/h. Initial time for one lap = 15 minutes. Required time for one lap = 12 minutes. The lap distance is fixed and the car runs at constant average speeds. Concept / Approach: First find the actual distance of one lap using the original speed and time. Once the lap distance is known, use the target time and the same distance to compute the new required speed. The difference between new and old speeds gives the increase in speed. Step-by-Step Solution: Convert 15 minutes to hours: 15 / 60 = 0.25 hours. Lap distance D = speed * time = 108 * 0.25 = 27 km. Now convert 12 minutes to hours: 12 / 60 = 0.2 hours. Required new speed v_new = D / time = 27 / 0.2 km/h. Compute v_new = 27 / 0.2 = 135 km/h. Increase in speed = v_new - original speed = 135 - 108 = 27 km/h. Verification / Alternative check: At 135 km/h for 0.2 hours, distance = 135 * 0.2 = 27 km. At 108 km/h for 0.25 hours, distance = 108 * 0.25 = 27 km. Both distances are equal, confirming that the same lap is completed faster only because of higher speed. Why Other Options Are Wrong: 24 km/h increase would give new speed 132 km/h, distance in 12 minutes = 132 * 0.2 = 26.4 km, not equal to 27 km. 21 km/h increase leads to 129 km/h and distance 25.8 km in 12 minutes, which is short. 30 km/h increase leads to 138 km/h and distance 27.6 km, which is longer than the actual lap. Common Pitfalls: Some learners directly take a ratio of times and incorrectly subtract percentages without converting to actual distances. Another common mistake is to keep times in minutes while using speeds in km/h without converting to hours. Always ensure units are consistent and compute the real distance once, then reuse it for the new scenario. Final Answer: The car must increase its speed by 27 km/h.

Question 20

Two cars A and B travel from one city to another at constant speeds of 72 km/h and 90 km/h respectively. If car B takes 1 hour less than car A to cover the full journey, what is the distance between the two cities in kilometres?

Accepted Answer

Introduction / Context: This is another time and distance question involving two cars with different speeds covering the same distance. The difference in their travel times is known, and you must determine the distance using basic algebra. It reinforces the relationship between speed, distance and time. Given Data / Assumptions: Speed of car A = 72 km/h. Speed of car B = 90 km/h. Distance between the cities = D km (unknown). Car B takes 1 hour less than car A. Both cars travel at constant speed without stops. Concept / Approach: For each car, time is distance divided by speed. Since both cover the same distance D, the difference in their times can be expressed in terms of D. We create one equation using the given 1 hour difference and then solve for D. Step-by-Step Solution: Let D be the distance in km. Time taken by A = D / 72 hours. Time taken by B = D / 90 hours. Given that car B is faster and takes 1 hour less than car A: D / 72 - D / 90 = 1. Compute 1/72 - 1/90 = (90 - 72) / (72 * 90) = 18 / 6480 = 1 / 360. So (1 / 360) * D = 1, hence D = 360 km. Verification / Alternative check: Time for A with D = 360 km: 360 / 72 = 5 hours. Time for B: 360 / 90 = 4 hours. Difference = 5 - 4 = 1 hour, which matches the condition. Why Other Options Are Wrong: 270 km gives times 270/72 and 270/90, difference less than 1 hour. 240 km produces times whose difference is 240 * (1/72 - 1/90) = 240 / 360 = 2/3 hour, not 1 hour. 400 km gives difference 400 / 360 = 1.11 hours approximately, which is larger than required. Common Pitfalls: Confusing which time is larger can lead to writing D / 90 - D / 72 = 1, which gives a negative distance. Some candidates directly try to plug in option distances without simplifying fractions, which is slower and more error prone. Always remember time = distance / speed, and derive a simple equation before calculating. Final Answer: The distance between the two cities is 360 km.

Time and Distance Questions