Question 1

To cover a distance of 333 km in 2 hours by car, what should be the average speed of the car in metres per second?

Accepted Answer

Introduction / Context: This question combines the average speed formula with a basic unit conversion from kilometres per hour to metres per second. It reflects real life travel planning, where one may know the total distance and desired time and must find the necessary speed. Such conversions are heavily tested in quantitative aptitude because they reinforce understanding of the relationships between distance, time and speed in different unit systems. Given Data / Assumptions: Required distance to cover = 333 km. Available time = 2 hours. We need the required average speed in metres per second (m/s). 1 kilometre = 1000 metres, 1 hour = 3600 seconds. The car is assumed to maintain constant average speed over the whole journey. Concept / Approach: First, we find the average speed in km/h using the formula speed = distance / time. Once we have the speed in km/h, we convert it into m/s using the standard factor 5 / 18, or equivalently multiply by 1000 / 3600. Performing the calculation in two clear stages helps reduce mistakes and makes the reasoning transparent. Finally, we match the numerical value with the given options. Step-by-Step Solution: Step 1: Compute the average speed in km/h. speed_kmph = 333 / 2 = 166.5 km/h. Step 2: Convert km/h to m/s. We use speed_mps = speed_kmph * 1000 / 3600. So, speed_mps = 166.5 * 1000 / 3600. Compute numerator and denominator carefully: 166.5 * 1000 = 166500. Now divide: 166500 / 3600 = 46.25. Therefore, required average speed = 46.25 m/s. Verification / Alternative check: We know that 1 km/h is approximately 0.2778 m/s. Multiplying 166.5 km/h by about 0.28 should give roughly 46.6 m/s, which is very close to 46.25 m/s. The slight difference is due to rounding 0.2778 to 0.28. This quick mental check confirms that our detailed computation is consistent and the order of magnitude is reasonable for a high but not impossible highway speed. Why Other Options Are Wrong: Option a (166.5) is the speed in km/h, not in m/s. Options c (83.25) and d (92.5) correspond to incorrect conversions or misapplied factors and would represent unrealistically high speeds in m/s. They stem from mistakes such as multiplying instead of dividing by 3.6. Only 46.25 correctly reflects the conversion from 166.5 km/h to m/s. Common Pitfalls: Frequent errors include forgetting to convert hours into seconds, using 60 instead of 3600, or stopping after finding the speed in km/h and directly marking it as the answer. Some students also mix up the conversion and multiply by 3600 / 1000 rather than 1000 / 3600. Always write the step as (km/h) * (1000 m / 1 km) * (1 h / 3600 s) and cancel units to confirm that the final unit is m/s. Final Answer: Thus, the car must travel at an average speed of 46.25 m/s to cover 333 km in 2 hours.

Question 2

A missile travels at a speed of 1260 km/h. How many metres does it travel in one second?

Accepted Answer

Introduction / Context: Once again we work with speed in km/h and convert it to metres per second to find how far an object moves in one second. Missile speed problems mirror real world high speed travel and are useful to test if a candidate can comfortably move between units and interpret the magnitude of a speed correctly. The central idea is to apply the unit conversion formula carefully and then apply distance = speed * time for one second. Given Data / Assumptions: Missile speed = 1260 km/h. We need the distance covered in one second, in metres. The missile is assumed to move at uniform speed over this short interval. Use 1 kilometre = 1000 metres and 1 hour = 3600 seconds. Concept / Approach: The conversion from km/h to m/s uses the same principle: multiply by 1000 to change kilometres to metres and divide by 3600 to convert hours to seconds. Once we convert 1260 km/h to m/s, the distance travelled in one second is numerically equal to that speed in m/s, because distance in one second equals speed times 1 second. We should also sanity check the value against a known reference, such as 360 km/h being 100 m/s, to see whether the final number is realistic. Step-by-Step Solution: Step 1: Convert speed from km/h to m/s. speed_mps = 1260 * 1000 / 3600. This gives speed_mps = 1260000 / 3600. Step 2: Simplify the fraction. 1260000 / 3600 = 350. So the missile moves at 350 m/s. Step 3: Distance in one second. distance = speed * time = 350 * 1 = 350 metres. Verification / Alternative check: We know 360 km/h corresponds to 100 m/s. The speed 1260 km/h is 3.5 times 360 km/h. Therefore, the speed in m/s should be 3.5 * 100 = 350 m/s. This quick proportional reasoning confirms the detailed numerical conversion. The value also looks reasonable for a fast projectile through air and matches the chosen option exactly. Why Other Options Are Wrong: Options a (322 metres), b (369 metres), and c (384 metres) are not equal to 1260000 / 3600 and likely arise from approximate or incorrect mental arithmetic. They may appear close to 350 but they do not respect the clear ratio between 1260 km/h and 360 km/h. Only 350 metres correctly matches the exact conversion and the proportional argument. Common Pitfalls: Typical mistakes include miscalculating 1260000 / 3600, forgetting to cancel common factors, or doing the conversion with rounded intermediate values that accumulate error. Another error is to divide by 60 instead of 3600, which would give a speed ten times too large. Writing the conversion as two distinct steps and simplifying fractions carefully helps avoid these issues. Final Answer: Therefore, the missile travels 350 metres in one second.

Question 3

To travel 612 km, an Express train takes 9 hours more than a Rajdhani train. If the speed of the Express train is doubled, it takes 3 hours less than the Rajdhani to cover the same distance. What is the speed of the Rajdhani train (in km/h)?

Accepted Answer

Introduction / Context: This is a typical time, speed and distance word problem with two moving objects and unknown speeds. The question describes how long an Express train and a Rajdhani train take for the same journey under different conditions. By comparing the times taken with original and modified speeds, we can set up simultaneous equations in two unknowns. Solving these equations gives the required speed. These types of problems test algebraic modelling from verbal descriptions. Given Data / Assumptions: Common distance for both trains = 612 km. Let the speed of the Express train be u km/h and that of the Rajdhani train be v km/h. The Express train takes 9 hours more than the Rajdhani, so 612 / u = 612 / v + 9. When the Express speed is doubled, it takes 3 hours less than the Rajdhani, so 612 / (2u) = 612 / v - 3. We assume both trains maintain constant speeds on the route. Concept / Approach: We use the fundamental relationship time = distance / speed to translate the statements into equations. Each condition produces one equation involving u and v. This gives a system of two simultaneous equations in two unknowns. We then solve using algebraic manipulation or substitution. Finally, we choose the value of v that matches one of the options. Checking the solution back in the original conditions ensures that no algebraic mistakes have been made. Step-by-Step Solution: Step 1: Write the time equations. 612 / u = 612 / v + 9. 612 / (2u) = 612 / v - 3. Step 2: Simplify equation one. 612 / u - 612 / v = 9. Step 3: Simplify equation two. 612 / (2u) - 612 / v = -3. Step 4: Let us solve the system. Subtract the second simplified equation from the first. (612 / u - 612 / v) - (612 / (2u) - 612 / v) = 9 - ( -3 ). This becomes 612 / u - 612 / (2u) = 12. Take 612 / u common: 612 / u * (1 - 1 / 2) = 12. 612 / u * (1 / 2) = 12, so 612 / (2u) = 12. Therefore, 612 = 24u, so u = 612 / 24 = 25.5 km/h. Step 5: Find v using equation 612 / u = 612 / v + 9. 612 / 25.5 = 612 / v + 9. 612 / 25.5 = 24, so 24 = 612 / v + 9. 612 / v = 15, hence v = 612 / 15 = 40.8 km/h. Verification / Alternative check: Check the original conditions with u = 25.5 and v = 40.8. Time for Rajdhani = 612 / 40.8 = 15 hours. Time for Express = 612 / 25.5 = 24 hours, which is indeed 9 hours more. With Express speed doubled to 51 km/h, its new time = 612 / 51 = 12 hours. This is 3 hours less than Rajdhani's 15 hours, exactly as described. Thus both conditions are satisfied, confirming the solution. Why Other Options Are Wrong: Options b (51), c (30.6) and d (61.2) do not satisfy both equations when substituted as v. They may correspond to one of the train speeds or arise from partial manipulation of the equations but do not meet all given time relations. Only v = 40.8 km/h satisfies the conditions consistently. Common Pitfalls: Students often confuse which train is faster or mix up the 9 hours more and 3 hours less conditions. Another trap is to attempt to solve directly with decimals and make arithmetic mistakes. It is safer to manipulate the equations symbolically and only plug in numbers at the end. Always verify the final speeds by substituting them back into the original word conditions, not just the algebraic form. Final Answer: The speed of the Rajdhani train is 40.8 km/h.

Question 4

Two cyclists P and Q cycle towards each other from some distance apart. P's speed is 20 km/h and Q's speed is 16 km/h. They meet after 40 minutes. What was the distance between them (in km) when they started?

Accepted Answer

Introduction / Context: This problem uses the idea of relative speed when two objects move towards each other. The combined effect of their motions makes them close the gap between them faster than either one alone. By knowing how long it takes for them to meet, we can find the original separation. This sort of question appears frequently in time and distance topics and is fundamental for understanding relative motion. Given Data / Assumptions: Speed of cyclist P = 20 km/h. Speed of cyclist Q = 16 km/h. They move towards each other and meet after 40 minutes. 40 minutes = 40 / 60 hours = 2 / 3 hours. We assume both ride at constant speeds along a straight path. Concept / Approach: When two objects move directly towards each other, their relative speed is the sum of their individual speeds. The distance between them shrinks at this combined speed. If we know the relative speed and the time until they meet, the initial distance is simply distance = relative speed * time. Working in hours and km/h keeps the units consistent and the calculation straightforward. Step-by-Step Solution: Step 1: Compute relative speed. relative speed = 20 + 16 = 36 km/h. Step 2: Convert meeting time into hours. Time = 40 minutes = 40 / 60 = 2 / 3 hours. Step 3: Use distance = speed * time. distance = 36 * (2 / 3) km. distance = 36 * 2 / 3 = 72 / 3 = 24 km. Therefore, the cyclists started 24 km apart. Verification / Alternative check: If the initial distance is 24 km and the relative speed is 36 km/h, then the time to meet should be distance / speed = 24 / 36 = 2 / 3 hours, which is 40 minutes. This matches the given condition exactly, so the computation is correct. Both a forward calculation and a reverse check agree on the value 24 km. Why Other Options Are Wrong: Options a (36), b (30), and c (25) would imply different meeting times if combined with a relative speed of 36 km/h. For example, 36 km at 36 km/h means 1 hour to meet, not 40 minutes. Similarly, 30 km at 36 km/h gives 50 minutes, and 25 km gives about 41.7 minutes. Only 24 km corresponds precisely to 40 minutes. Common Pitfalls: A typical mistake is to subtract the speeds instead of adding them, which would be appropriate only if the cyclists were moving in the same direction. Another error is to keep time in minutes while using speed in km/h without converting units, leading to incorrect distance values. Always ensure that speed and time are expressed in compatible units, usually km/h with hours or m/s with seconds. Final Answer: The starting distance between the two cyclists was 24 km.

Question 5

A man travels three fourths (3⁄4) of the total distance of his journey by bus, one sixth (1⁄6) by rickshaw, and the remaining 2 km on foot. What is the total distance travelled by the man (in km)?

Accepted Answer

Introduction / Context: This is a fractional distance problem that tests the ability to model a total quantity when parts of it are given as fractions plus a fixed remainder. Such problems are common in aptitude tests and help reinforce the idea that fractions represent parts of a whole and must sum appropriately. The task is to express all segments of the journey in terms of the unknown total distance and then use an equation to solve for that total. Given Data / Assumptions: Let the total distance be D km. Distance by bus = (3 / 4) * D km. Distance by rickshaw = (1 / 6) * D km. Distance on foot = 2 km. All three segments together equal the total distance D. Concept / Approach: We use the fact that the sum of the parts equals the whole. In algebraic form, this means bus distance plus rickshaw distance plus walking distance equals D. That gives an equation in D. We simplify the fractional coefficients, combine like terms, and solve for D. Once D is known, we simply match it with the correct option. Carefully adding the fractions (3 / 4 and 1 / 6) is the main arithmetic step. Step-by-Step Solution: Step 1: Write the total distance equation. (3 / 4) * D + (1 / 6) * D + 2 = D. Step 2: Combine fractional coefficients of D. (3 / 4) + (1 / 6) = (9 / 12) + (2 / 12) = 11 / 12. So the equation becomes (11 / 12) * D + 2 = D. Step 3: Move the fractional term to the other side. 2 = D - (11 / 12) * D = (1 / 12) * D. Step 4: Solve for D. D = 2 * 12 = 24 km. Verification / Alternative check: If D = 24 km, then bus distance is (3 / 4) * 24 = 18 km, rickshaw distance is (1 / 6) * 24 = 4 km, and walking distance is 2 km. Adding them gives 18 + 4 + 2 = 24 km, which matches the assumed total. Thus, the value D = 24 km is consistent with all parts of the journey and satisfies the equation exactly. Why Other Options Are Wrong: Option a (12 km) would give 9 km by bus and 2 km by rickshaw, leaving only 1 km for walking, which contradicts the given 2 km. Option b (18 km) and option c (20 km) similarly fail to satisfy the fractions and the leftover walking distance simultaneously. Only 24 km splits correctly into the described fractional parts plus 2 km on foot. Common Pitfalls: Common errors include adding the fractions (3 / 4 and 1 / 6) incorrectly or forgetting to express the walking distance separately instead of as another fraction. Some candidates also mistakenly set (3 / 4 + 1 / 6) * D equal to 2 instead of setting the sum plus 2 equal to D. Always start from the fundamental relationship that sum of all parts equals the total and then solve carefully. Final Answer: The total distance travelled by the man is 24 km.

Question 6

Excluding stoppages, the speed of a bus is 72 km/h. Including stoppages, its effective speed becomes 60 km/h. For how many minutes does the bus stop in each hour?

Accepted Answer

Introduction / Context: This question is about effective speed when a vehicle stops periodically. The difference between the running speed and the average speed including stoppages tells us how much time is spent stationary. Problems of this type are very common in time and distance and test whether a candidate can convert between speed and time using the same distance as a reference and correctly interpret a word description involving stoppages. Given Data / Assumptions: Running speed of the bus (when it is moving) = 72 km/h. Average speed including stoppages = 60 km/h. We need to find how many minutes per hour the bus remains stopped. The bus is assumed to repeat a similar pattern of motion and stoppage every hour. Concept / Approach: Consider a period of one hour of clock time. At the effective average speed of 60 km/h, the bus covers 60 km in that one hour. If it were to travel continuously at its running speed of 72 km/h without any stop, it would take less than one hour to cover 60 km. The difference between one hour and that smaller travel time is exactly the stoppage time within the hour. We use time = distance / speed with the same distance (60 km) for both running and effective conditions. Step-by-Step Solution: Step 1: Assume a reference distance. In one hour including stoppages, at 60 km/h, the bus covers 60 km. Step 2: Compute the moving time needed to cover 60 km at 72 km/h. time_moving = 60 / 72 hours. Simplify: 60 / 72 = 5 / 6 hours. Step 3: Convert moving time into minutes. (5 / 6) hours = (5 / 6) * 60 minutes = 50 minutes. Step 4: Find stoppage time per hour. Total clock time = 60 minutes, moving time = 50 minutes. Stoppage time = 60 - 50 = 10 minutes. Verification / Alternative check: If the bus runs 50 minutes at 72 km/h, the distance covered is (72 * 50 / 60) = 60 km. Over the full 60 minutes, including the 10 minutes of stoppage, the average speed is distance / time = 60 km / 1 hour = 60 km/h. This matches the given effective speed, confirming that 10 minutes per hour of stoppage is correct. Why Other Options Are Wrong: If stoppage time were 12 minutes, the bus would only run for 48 minutes, covering 72 * 48 / 60 = 57.6 km, giving an effective speed of 57.6 km/h, not 60 km/h. Similar mismatches occur with 8 or 15 minutes. Only a 10 minute stoppage yields an effective 60 km/h from a running speed of 72 km/h with the same one hour period. Common Pitfalls: Some students mistakenly compare 72 and 60 directly and try to work with a ratio of speeds without fixing a reference distance or time. Others compute how much distance is lost instead of time lost, which can be confusing. The safest approach is to pick a convenient distance based on effective speed in exactly one hour, then recompute the time needed at the running speed and subtract to find stoppage time. Final Answer: The bus stops for 10 minutes in every hour.

Question 7

A man travels a total distance of 60 km in 7 hours. He walks part of the way at 6 km/h and rides a bicycle for the remaining part at 12 km/h. What distance (in km) does he travel on foot?

Accepted Answer

Introduction / Context: This question mixes two different speeds over different portions of a journey and gives the total distance and total time. We need to determine how much of the journey was done at each speed. This is a standard type of time and distance problem where we translate the description into simultaneous equations in terms of time or distance. Solving these equations reveals how long or how far the person travelled by each mode of transport. Given Data / Assumptions: Total distance = 60 km. Total time = 7 hours. Walking speed = 6 km/h. Cycling speed = 12 km/h. Let the walking time be t1 hours and the cycling time be t2 hours. We assume speeds remain constant for each mode of travel. Concept / Approach: We build two equations: one from the total time and one from the total distance. The time equation is t1 + t2 = 7. The distance equation is 6 * t1 + 12 * t2 = 60. These two linear equations in t1 and t2 can be solved using substitution or elimination. Once we have t1, the walking time, we can compute walking distance as 6 * t1. This systematic algebraic approach is reliable and prevents guesswork. Step-by-Step Solution: Step 1: Write the time equation. t1 + t2 = 7. Step 2: Write the distance equation. 6 * t1 + 12 * t2 = 60. Step 3: Use substitution from t1 + t2 = 7, so t1 = 7 - t2. Step 4: Substitute into the distance equation. 6 * (7 - t2) + 12 * t2 = 60. 42 - 6 * t2 + 12 * t2 = 60. 42 + 6 * t2 = 60. 6 * t2 = 18, so t2 = 3 hours. Step 5: Find t1. t1 = 7 - 3 = 4 hours. Step 6: Compute walking distance. Walking distance = 6 * t1 = 6 * 4 = 24 km. Verification / Alternative check: Cycling distance is 12 * t2 = 12 * 3 = 36 km. Total distance = 24 + 36 = 60 km, which matches the given distance. Total time = 4 + 3 = 7 hours, also matching the given time. This confirms that the breakdown of 24 km walking and 36 km cycling is correct and that the algebraic solution is consistent with the story of the problem. Why Other Options Are Wrong: Option a (15 km) and option b (9 km) would require much longer cycling distances and would not sum up to 60 km with only 7 hours available at the given speeds. Option c (48 km) would leave only 12 km for cycling, which at 12 km/h takes 1 hour, making the total time 9 hours instead of 7. Only 24 km fits both the time and distance constraints simultaneously. Common Pitfalls: A common mistake is to assume equal time or equal distance for walking and cycling without justification. Another frequent error is to miswrite the distance equation, for example 6 + 12 equals 60, which is obviously incorrect. Always define variables clearly, write the equations step by step and double check that they accurately reflect the problem statement before solving them. Final Answer: The man travels 24 km on foot.

Question 8

Points A and B are 15 km apart. When travelling towards each other, two persons starting from A and B meet after half an hour. When they travel in the same direction, they meet 2.5 hours later. What is the speed of the faster person (in km/h)?

Accepted Answer

Introduction / Context: This question uses relative speed concepts both when two people move towards each other and when they move in the same direction. The same pair of speeds must satisfy both situations, which allows us to form two equations. Solving these equations gives the individual speeds. Identifying which person is faster and translating that into the final answer is the goal. Such problems are classic examples of how relative motion can simplify time and distance calculations. Given Data / Assumptions: Distance between A and B = 15 km. When travelling towards each other, they meet after 0.5 hours. When travelling in the same direction, they meet after 2.5 hours. Let the speeds be a km/h and b km/h, where a is the faster speed and b is the slower speed. They move along a straight line and maintain constant speeds. Concept / Approach: When two people move towards each other, their relative speed is the sum of their speeds (a + b). When they move in the same direction, the relative speed is the difference between their speeds (a - b), assuming a is greater than b. Using distance = speed * time, we write two equations: one with a + b and one with a - b, both with the same distance of 15 km. Solving these two equations simultaneously gives us a and b, and we then choose the larger one as the speed of the faster person. Step-by-Step Solution: Step 1: Use the case when they travel towards each other. Distance = 15 km, time = 0.5 hours. 15 = (a + b) * 0.5, so a + b = 30. Step 2: Use the case when they travel in the same direction. Distance = 15 km, time = 2.5 hours. 15 = (a - b) * 2.5, so a - b = 15 / 2.5 = 6. Step 3: Solve the system of equations: a + b = 30 and a - b = 6. Add the two equations: 2a = 36, so a = 18 km/h. Substitute back to find b: 18 + b = 30, so b = 12 km/h. Verification / Alternative check: Check the opposite direction scenario: relative speed = 18 + 12 = 30 km/h. Time = 15 / 30 = 0.5 hours, correct. Check the same direction scenario: relative speed = 18 - 12 = 6 km/h. Time = 15 / 6 = 2.5 hours, also correct. Both conditions are satisfied, confirming that the speeds are 18 km/h and 12 km/h and that 18 km/h is the faster speed. Why Other Options Are Wrong: Options a (15 km/hr), c (10 km/hr) and d (8 km/hr) do not produce the correct pair of relative speeds and times when paired with any other reasonable speed. Only 18 km/hr works smoothly with a second speed of 12 km/hr to satisfy both the head-on and same direction meeting conditions. Common Pitfalls: A frequent error is mixing up which equation belongs to which scenario, or incorrectly using a + b for both. Another mistake is to forget that the slower person's speed must be positive, which acts as a quick check on algebraic manipulations. Some students also compute 15 / 2.5 incorrectly. Taking care with basic arithmetic and clearly labelling equations helps avoid these problems. Final Answer: The faster person travels at a speed of 18 km/hr.

Question 9

A train has to cover a distance of 900 km in 25 hours. What should be its average speed in metres per second?

Accepted Answer

Introduction / Context: This problem is a straightforward application of the speed formula combined with unit conversion. Average speed is defined as total distance divided by total time. The twist is that the required answer must be in metres per second, while the distance and time are given in kilometres and hours. Converting units correctly is a critical skill for both exams and everyday use of measurements. Given Data / Assumptions: Distance to be covered = 900 km. Total time allowed = 25 hours. We want the average speed in metres per second (m/s). Use 1 kilometre = 1000 metres, 1 hour = 3600 seconds. We assume constant average speed over the entire journey. Concept / Approach: We first compute the average speed in km/h using speed = distance / time. Then we convert that speed to m/s by multiplying by 1000 / 3600, or equivalently by 5 / 18. Breaking the problem into these two clear steps avoids mixing units. Finally, we match the final numerical value with the correct option. A quick reasonableness check of the magnitude is also beneficial. Step-by-Step Solution: Step 1: Compute average speed in km/h. speed_kmph = 900 / 25 = 36 km/h. Step 2: Convert to m/s. speed_mps = 36 * 1000 / 3600. Calculate: 36 * 1000 = 36000. speed_mps = 36000 / 3600 = 10 m/s. Verification / Alternative check: We know 36 km/h is a classic reference speed that equals exactly 10 m/s because 36 * 5 / 18 equals 10. This known identity provides an immediate confirmation that the conversion is correct. Also, 900 km in 25 hours is a moderate train speed of 36 km/h, and 10 m/s is a realistic equivalent speed for a relatively slow passenger train or freight train. Why Other Options Are Wrong: Option c (18) and option d (36) would correspond to much higher speeds in km/h than 36 km/h if converted back, and thus do not match the original ratio of 900 km in 25 hours. Option a (20) is exactly double the correct value and may appear if someone mistakenly multiplies instead of dividing during conversion. Only 10 m/s exactly matches 36 km/h and fits all the given data. Common Pitfalls: Common errors include miscomputing 900 / 25, using 60 instead of 3600 when converting hours to seconds, or applying the 5 / 18 factor backwards (multiplying when one should divide, or vice versa). Carefully writing out each step with units and checking that the final units are m/s helps prevent such mistakes. Final Answer: The required average speed of the train is 10 m/s.

Question 10

Two cyclists A and B start cycling towards each other at speeds of 21 km/h and 24 km/h respectively. They meet after 1 hour 12 minutes. How far apart (in km) were they when they started?

Accepted Answer

Introduction / Context: This is another relative speed question involving two cyclists moving towards each other. Given their individual speeds and the time until they meet, we are asked to determine the initial distance between them. This type of question routinely appears in time and distance sections of aptitude tests and illustrates how relative motion simplifies the calculation of encounter times and distances. Given Data / Assumptions: Speed of cyclist A = 21 km/h. Speed of cyclist B = 24 km/h. Meeting time = 1 hour 12 minutes = 1.2 hours (since 12 minutes = 12 / 60 hours = 0.2 hours). They ride directly towards each other along a straight line. We assume both speeds are constant until they meet. Concept / Approach: When two bodies move towards each other, their relative speed is the sum of their speeds. The gap between them closes at this relative speed. The distance between them at the start is therefore equal to relative speed multiplied by the time taken to meet. So, we compute relative speed = 21 + 24 km/h and then multiply by 1.2 hours to obtain the initial separation in kilometres. Step-by-Step Solution: Step 1: Compute relative speed. relative speed = 21 + 24 = 45 km/h. Step 2: Convert time to hours. Time = 1 hour 12 minutes = 1 + 12 / 60 = 1.2 hours. Step 3: Use distance = speed * time. distance = 45 * 1.2 km. Compute: 45 * 1.2 = 54 km. Therefore, they were 54 km apart initially. Verification / Alternative check: With a starting distance of 54 km and relative speed 45 km/h, the time to meet is distance divided by relative speed = 54 / 45 hours. This equals 1.2 hours, which matches 1 hour 12 minutes. Thus, the initial distance of 54 km is consistent with the given speeds and meeting time. Why Other Options Are Wrong: Option a (48 km) would give a meeting time of 48 / 45 = 1.0667 hours (about 1 hour 4 minutes), not 1 hour 12 minutes. Option b (42 km) yields 42 / 45 = 0.9333 hours (about 56 minutes). Option d (36 km) gives 36 / 45 = 0.8 hours (48 minutes). None of these match the provided 1 hour 12 minutes. Only 54 km fits the data correctly. Common Pitfalls: Possible mistakes include misreading 1 hour 12 minutes as 1.12 hours (which is wrong), or forgetting to convert minutes to a fraction of an hour. Another error is subtracting speeds instead of adding them, which is appropriate only when moving in the same direction. Always check whether the objects are moving towards or away from each other to choose the correct relative speed expression. Final Answer: The cyclists were initially 54 km apart.

Question 11

Excluding stoppages, the speed of a bus is 60 km/h. Including stoppages, its effective speed is 45 km/h. For how many minutes does the bus stop in each hour?

Accepted Answer

Introduction / Context: This question is similar to an earlier stoppage problem and again examines how pauses affect the effective speed of a vehicle. The difference between the moving speed and the average speed tells us how much time is spent at rest. Understanding this relationship is valuable for interpreting timetables, planning trips and solving many standard aptitude problems where vehicles stop at intermediate stations. Given Data / Assumptions: Running speed of the bus (excluding stoppages) = 60 km/h. Average speed including stoppages = 45 km/h. We want to know how many minutes per hour the bus is not moving. The bus is assumed to repeat a similar pattern each hour, with constant running speed when moving. Concept / Approach: As before, consider one hour of total clock time. At an effective speed of 45 km/h, the bus covers 45 km in that hour. If it travelled the same 45 km at its running speed of 60 km/h continuously, it would require less than one hour. The difference between one hour and the required moving time to cover 45 km at 60 km/h gives the stoppage time per hour. We apply time = distance / speed under both conditions using the same distance of 45 km. Step-by-Step Solution: Step 1: Assume reference distance based on effective speed. In 1 hour including stoppages at 45 km/h, the bus travels 45 km. Step 2: Compute moving time at 60 km/h for 45 km. time_moving = 45 / 60 hours. This simplifies to 3 / 4 hours. Step 3: Convert this moving time to minutes. (3 / 4) hours = (3 / 4) * 60 minutes = 45 minutes. Step 4: Find stoppage time per hour. Total time = 60 minutes, moving time = 45 minutes. Stoppage time = 60 - 45 = 15 minutes. Verification / Alternative check: If the bus runs 45 minutes at 60 km/h, it covers 60 * 45 / 60 = 45 km. Over the full 60 minutes, including 15 minutes of rest, the average speed is distance / time = 45 km / 1 hour = 45 km/h. This matches the given effective speed including stoppages, confirming that 15 minutes per hour is correct. Why Other Options Are Wrong: If stoppage time were 12 minutes, the bus would run for 48 minutes and cover 60 * 48 / 60 = 48 km in an hour, making the average speed 48 km/h, not 45 km/h. Similarly, 9 minutes or 10 minutes would also lead to incorrect effective speeds. Only 15 minutes of stoppage yields the specified 45 km/h effective speed from a running speed of 60 km/h. Common Pitfalls: Many candidates incorrectly try to adjust speed by a direct ratio, such as thinking that since 45 is three quarters of 60, stoppage time must be one quarter of an hour but then forgetting to convert that to minutes. That reasoning is actually correct here, but if applied carelessly in other contexts without checking distance and time, it can mislead. In all such problems, explicitly computing the time for a chosen distance using both speeds is the safest method. Final Answer: The bus stops for 15 minutes in every hour.

Question 12

Two cars travel from city A to city B. One car moves at 30 km/h and the other at 36 km/h. The faster car takes 3 hours less than the slower car to cover the journey. What is the distance between city A and city B (in km)?

Accepted Answer

Introduction / Context: This problem involves two vehicles travelling the same route at different speeds with a specified time difference between their journeys. It tests whether a candidate can relate speed, time, and distance through the simple formula distance = speed * time and construct an equation from a verbal description. By expressing both travel times in terms of the same distance, we can solve for that distance using basic algebra. Given Data / Assumptions: Speed of the slower car = 30 km/h. Speed of the faster car = 36 km/h. The faster car takes 3 hours less than the slower car for the same journey. Let the distance between city A and city B be D km. The cars travel at constant speeds along the same route. Concept / Approach: If the slower car takes time t hours, then the faster car takes t - 3 hours. Using time = distance / speed, we write one equation for each car in terms of D and t. Since both distances are the same and equal to D, we set 30 * t equal to 36 * (t - 3). Solving this equation yields t, and then D is found by multiplying 30 * t or 36 * (t - 3). This approach directly aligns with the relationship linking distance, speed and time. Step-by-Step Solution: Step 1: Let the time taken by the slower car be t hours. Then, time taken by the faster car = t - 3 hours. Step 2: Write expressions for the distance. For the slower car: D = 30 * t. For the faster car: D = 36 * (t - 3). Step 3: Equate the two expressions for D. 30 * t = 36 * (t - 3). Step 4: Solve for t. 30t = 36t - 108. Bring like terms together: 108 = 36t - 30t = 6t. t = 108 / 6 = 18 hours. Step 5: Find D. D = 30 * t = 30 * 18 = 540 km. Verification / Alternative check: Check the faster car's travel time: t - 3 = 18 - 3 = 15 hours. Distance covered at 36 km/h in 15 hours is 36 * 15 = 540 km, which matches the distance computed with the slower car. Thus, both cars travel the same distance, confirming D = 540 km as consistent with all given conditions. Why Other Options Are Wrong: If the distance were 648 km, then the slower car at 30 km/h would take 21.6 hours, and the faster car at 36 km/h would take 18 hours, giving a difference of 3.6 hours, not 3. Similarly, 810 km and 432 km do not produce a 3 hour difference at these speeds. Only 540 km yields exactly 3 hours difference between 18 and 15 hours. Common Pitfalls: Some students mistakenly set up the equation as 30 * (t - 3) = 36 * t, reversing the time difference. Others misinterpret the phrase "takes 3 hours lesser time" and subtract 3 from the slower car's time instead of the faster car's time. Carefully reading and translating the word description into algebra is crucial. Always verify which car is faster and which time must be smaller. Final Answer: The distance between city A and city B is 540 km.

Question 13

What should be the average speed of a car (in metres per second) to cover a distance of 342 km in 1.9 hours?

Accepted Answer

Introduction / Context: This question again tests the basic formula for average speed along with a unit conversion, but with slightly less friendly numbers. The candidate must compute the average speed in km/h first and then convert it carefully to m/s. Problems like this ensure fluency in handling decimal hours and prevent careless errors when dividing by non round numbers. Given Data / Assumptions: Distance to be covered = 342 km. Time allowed = 1.9 hours. We need the average speed in metres per second (m/s). 1 kilometre = 1000 metres, 1 hour = 3600 seconds. The car is assumed to maintain the required average speed for the whole journey. Concept / Approach: Average speed is distance divided by time. So we first compute speed_kmph = 342 / 1.9 km/h. Then, to convert km/h to m/s, we multiply by 1000 / 3600. Doing the division 342 / 1.9 accurately is crucial. After we get the speed in m/s, we select the option that matches this value. A quick magnitude check ensures that the computed speed is realistic for a fast car on a highway. Step-by-Step Solution: Step 1: Compute average speed in km/h. speed_kmph = 342 / 1.9. 342 / 1.9 = 180 km/h. Step 2: Convert to m/s. speed_mps = 180 * 1000 / 3600. Compute numerator: 180 * 1000 = 180000. speed_mps = 180000 / 3600 = 50 m/s. Verification / Alternative check: We know that 36 km/h equals 10 m/s. Since 180 km/h is exactly 5 times 36 km/h, the corresponding speed in m/s should be 5 * 10 = 50 m/s. This proportional reasoning provides a very quick check of the conversion. Also, 342 km in 1.9 hours corresponds to a very fast car at highway speeds, and 50 m/s is consistent with 180 km/h, confirming that our computation is reasonable. Why Other Options Are Wrong: Option a (180 m/s) would correspond to an impossible road speed of 648 km/h. Option b (90 m/s) and option d (100 m/s) are even larger, suggesting unrealistically high speeds. They represent likely errors where someone directly uses the km/h number as m/s or multiplies incorrectly during conversion. Only 50 m/s correctly reflects the ratio 342 / 1.9 with proper unit conversion. Common Pitfalls: Common mistakes include miscalculating 342 / 1.9, forgetting to convert to m/s, or treating 1.9 hours as 1 hour 90 minutes which is incorrect. Also, some candidates may simply divide 342000 metres by 1.9 hours but forget to convert hours into seconds. Writing the formulas clearly and converting units step by step prevents such errors. Final Answer: The car must travel at an average speed of 50 m/s to cover 342 km in 1.9 hours.

Question 14

Abhay and Chintan start riding their motorcycles from the same point. Abhay rides 7 km north and then turns left and rides 4 km. Chintan rides 6 km west, then turns north and rides 7 km, and finally turns left and rides 8 km. Where is Abhay with res…

Accepted Answer

Introduction / Context: This question tests two dimensional movement and relative position using directions on a plane. It is essentially a coordinate geometry or vector movement problem disguised as a word puzzle. By choosing a convenient reference frame and tracking each rider's displacement, we can determine their final coordinates and hence the relative position of one with respect to the other. Such problems are common in reasoning sections related to directions and movement. Given Data / Assumptions: Both riders start from the same point, taken as the origin (0, 0). Positive x-axis represents east and positive y-axis represents north. Abhay rides 7 km north, then 4 km west (since turning left from north points west). Chintan rides 6 km west, then 7 km north, then 8 km west (left turn from north is also west). We assume straight line segments for all movements. Concept / Approach: We treat each movement as a vector displacement and add them component wise. Each north or south movement changes the y coordinate, while each east or west movement changes the x coordinate. After obtaining the final coordinates of both Abhay and Chintan relative to the origin, we subtract the coordinates to find the position of Abhay relative to Chintan. The sign and magnitude of the difference in x coordinates directly give the east or west offset between them. Step-by-Step Solution: Step 1: Track Abhay's position. Start at (0, 0). Move 7 km north: new position = (0, 7). Turn left from north (towards west) and move 4 km: new position = (-4, 7). So Abhay ends at (-4, 7). Step 2: Track Chintan's position. Start at (0, 0). Move 6 km west: new position = (-6, 0). Turn north and move 7 km: new position = (-6, 7). Turn left from north (towards west) and move 8 km: new position = (-14, 7). So Chintan ends at (-14, 7). Step 3: Find Abhay's position relative to Chintan. Abhay = (-4, 7), Chintan = (-14, 7). Difference in x coordinates = -4 - ( -14 ) = 10. Same y coordinate means they are horizontally aligned. A positive difference in x from Chintan to Abhay means Abhay is 10 km east of Chintan. Verification / Alternative check: Visualising a simple grid helps. From the origin, Chintan ends 14 units left (west) and 7 units up (north). Abhay ends 4 units left and 7 units up. That is, Abhay is 10 units to the right (east) of Chintan, while both share the same vertical level. This picture matches the coordinate calculation exactly and confirms the result. Why Other Options Are Wrong: Option a (4 km East) incorrectly uses the difference between 6 and 4, ignoring the additional 8 km Chintan travels west in the last leg. Option b (10 km West) and option c (4 km Wes) both place Abhay on the wrong side of Chintan. The coordinates clearly show Abhay is east, not west, of Chintan by 10 km, so only 10 km East is correct. Common Pitfalls: Many students lose track of left and right turns, especially when the current direction is not east. Another common mistake is to add distances without considering direction signs, effectively ignoring positive and negative coordinates. Using a simple sketch or writing coordinates after each move can greatly reduce errors and makes the final relative position much easier to determine. Final Answer: Abhay is 10 km East of Chintan.

Question 15

A person claps his hands near a cliff and hears the echo after 6 seconds. Taking the speed of sound v = 346 m/s, what is the distance of the cliff from the person (in metres)?

Accepted Answer

Introduction / Context: This problem is about sound waves and echoes. When a person claps near a cliff, sound waves travel to the cliff, reflect and travel back to the person. The total time recorded is for the journey to the cliff and back. Using the known speed of sound, we can determine the one way distance to the cliff. This is a common type of question in physics related aptitude sections and connects motion concepts with wave phenomena. Given Data / Assumptions: Time between clap and echo = 6 seconds. Speed of sound v = 346 m/s. The sound travels from the person to the cliff and back (round trip) during these 6 seconds. We assume air conditions are uniform so the speed of sound remains constant. We need the distance from the person to the cliff (one way distance). Concept / Approach: The total distance covered by the sound wave during the echo is twice the distance from the person to the cliff, because the sound travels that distance going forward and the same distance returning. We first calculate the total distance using distance_total = speed * time. Then we divide this total by 2 to get the one way distance to the cliff. This is similar to round trip distance problems in time and distance topics. Step-by-Step Solution: Step 1: Compute total distance travelled by sound. distance_total = v * t = 346 * 6 metres. 346 * 6 = 2076 metres. Step 2: Find the one way distance to the cliff. distance_one_way = distance_total / 2 = 2076 / 2 metres. distance_one_way = 1038 metres. Verification / Alternative check: If the cliff is 1038 metres away, then a sound wave must travel 1038 metres to reach the cliff and another 1038 metres to return, giving a total of 2076 metres. At 346 m/s, the time taken for this round trip is distance_total / speed = 2076 / 346 seconds, which equals 6 seconds. This matches the information given in the question, confirming that 1038 metres is correct. Why Other Options Are Wrong: Option c (2076 m) represents the total distance travelled by the sound, not the distance to the cliff. Option b (1083 m) and option d (2706 m) do not satisfy the relation distance_total = 346 * 6 when doubled or halved appropriately. Only 1038 m is exactly half of 2076 m and corresponds to the correct one way distance when the round trip time is 6 seconds. Common Pitfalls: A common mistake is to forget that the sound travels to the cliff and back, leading some to use distance = speed * time directly as the distance to the cliff instead of the round trip distance. This gives 2076 m, which is double the correct value. To avoid this, always visualise the path: from the person to the reflecting surface and back, making it clear that the total distance is twice the one way distance. Final Answer: The cliff is 1038 m away from the person.

Question 16

Tribeni has to travel 175 km in 3.5 hours. She actually travels at 58 km/h for 2.5 hours, then rests for 20 minutes, and then completes the remaining journey. If she still manages to reach exactly on time, what must have been her speed during the la…

Accepted Answer

Introduction / Context: This is a time and distance planning problem involving different phases of a journey: travelling at one speed, resting, and then travelling at another speed, with a fixed overall time limit. It tests the ability to break the journey into segments, compute distance and time for each part, and then deduce the required speed in the last segment so that the total schedule is met. Such questions are common in aptitude tests and mirror real life travel planning where delays must be compensated by faster travel later. Given Data / Assumptions: Total distance required = 175 km. Total time available = 3.5 hours = 3 hours 30 minutes. First segment: speed = 58 km/h for 2.5 hours. Rest period: 20 minutes = 1 / 3 hour, during which she covers zero distance. Last segment: unknown speed v km/h for the remaining distance and time. She reaches exactly on time, so the sum of all segment times equals 3.5 hours. Concept / Approach: We first calculate the distance covered during the first segment at 58 km/h and subtract this from the total required distance to get the remaining distance. Next, we determine how much time is left after the first segment and the rest period are accounted for. The remaining distance must be covered in this remaining time at a constant speed v. We use v = remaining_distance / remaining_time to find the required speed in the last stretch. Step-by-Step Solution: Step 1: Compute distance during the first segment. distance_first = 58 * 2.5 km. 58 * 2.5 = 145 km. Step 2: Compute remaining distance. remaining_distance = total_distance - distance_first. remaining_distance = 175 - 145 = 30 km. Step 3: Compute time used in first segment and rest. time_first = 2.5 hours. rest_time = 20 minutes = 20 / 60 = 1 / 3 hour. total_used_time = 2.5 + 1 / 3 hours. 2.5 = 7 / 2, 1 / 3 = 1 / 3, so total_used_time = 7 / 2 + 1 / 3 = (21 + 2) / 6 = 23 / 6 hours ≈ 3.833 / 6 * 6. Numerically, 2.5 + 0.333... = 2.833... hours. Step 4: Compute remaining time. total_time = 3.5 hours = 3.5. remaining_time = 3.5 - 2.833... = 0.666... hours. So remaining_time = 2 / 3 hour (which is 40 minutes). Step 5: Compute required speed for the last stretch. v = remaining_distance / remaining_time. v = 30 / (2 / 3) = 30 * (3 / 2) = 45 km/h. Verification / Alternative check: At 45 km/h, travelling for 2 / 3 hour covers 45 * 2 / 3 = 30 km, which is exactly the remaining distance. The total time spent is 2.5 hours in the first segment, 1 / 3 hour resting, and 2 / 3 hour in the last segment. Adding these gives 2.5 + 1 / 3 + 2 / 3 = 2.5 + 1 = 3.5 hours, which matches the total time allowed. Thus the journey is completed exactly on schedule at this speed. Why Other Options Are Wrong: At 36 km/h, in 2 / 3 hour Tribeni would cover only 24 km, falling short of the required 30 km. At 30 km/h, she would cover only 20 km, which is even further from the target. At 42.5 km/h, in 2 / 3 hour she would cover 42.5 * 2 / 3 ≈ 28.33 km, still not enough. Only 45 km/h allows her to cover exactly 30 km in the remaining 40 minutes and arrive on time. Common Pitfalls: One common mistake is to forget to convert 20 minutes into hours and attempt to add minutes directly to hours. Another is to subtract the rest time from the total time without first including the travel time, which mixes steps. It is also easy to compute the distance in the first part incorrectly by mishandling 2.5 hours. Writing all times in hours and handling them as fractions or decimals consistently helps avoid these issues. Final Answer: Tribeni's speed during the last stretch must have been 45 km/hr.

Question 17

Hema takes a total of 9 hours 55 minutes to walk a certain distance and then cycle back to the same place from where she started. She could walk both ways in 12 hours 30 minutes. How much time will she take to cycle the same distance both ways?

Accepted Answer

Introduction / Context: This is a classic time, speed, and distance problem involving two different modes of travel: walking and cycling. Hema travels along the same route, once in a mixed mode (walking one way and cycling back) and once hypothetically by walking both ways. From the total times given in these two scenarios, we need to determine how long she would take if she cycled both ways. This type of question tests your ability to translate word statements into equations and to use simple algebra to solve for unknown times. Given Data / Assumptions: - Let the one-way distance between the two points be D km. - Let the time taken to walk one way be T_w hours. - Let the time taken to cycle one way be T_c hours. - Walking both ways takes 12 hours 30 minutes, i.e. 12.5 hours. - Walking one way and cycling back takes 9 hours 55 minutes, i.e. 9 + 55/60 = 9.9167 hours (approximately). - Hema walks and cycles at constant speeds on a straight route, and there are no breaks other than the mode change at the turning point. Concept / Approach: The key idea is that time for a journey is equal to distance divided by speed, but here we work directly with times. We express the two given total times in terms of T_w and T_c. From the walking-both-ways scenario, we get a direct value for T_w. From the mixed scenario, we then find T_c. Once T_c (time to cycle one way) is known, cycling both ways will simply take 2 * T_c. The actual distance D does not need to be calculated because it cancels out when we work with times only. Step-by-Step Solution: Step 1: From “she could walk both ways in 12 hours 30 minutes”, we have 2 * T_w = 12.5, so T_w = 12.5 / 2 = 6.25 hours.Step 2: From “walking one way and cycling back takes 9 hours 55 minutes”, we have T_w + T_c = 9.9167 hours.Step 3: Substitute T_w = 6.25 into the second equation: 6.25 + T_c = 9.9167.Step 4: Solve for T_c: T_c = 9.9167 - 6.25 = 3.6667 hours, which is 3 hours 40 minutes.Step 5: Time to cycle both ways is 2 * T_c = 2 * 3.6667 ≈ 7.3333 hours, which is 7 hours 20 minutes. Verification / Alternative check: Check the numbers logically. One-way walking time is 6.25 hours, so walking both ways indeed gives 12.5 hours. If cycling one way takes 3.6667 hours, then walking out (6.25 hours) and cycling back (3.6667 hours) gives approximately 9.9167 hours, which is exactly 9 hours 55 minutes. Doubling 3 hours 40 minutes clearly gives 7 hours 20 minutes, so the calculation is consistent with all the given conditions. Why Other Options Are Wrong: - 7 hours 15 minutes corresponds to a one-way cycling time slightly less than 3 hours 40 minutes and would not add up correctly with the given walking time to produce 9 hours 55 minutes. - 7 hours 35 minutes and 7 hours 40 minutes correspond to longer cycling times that would make the mixed journey exceed 9 hours 55 minutes when combined with 6.25 hours of walking time. Common Pitfalls: Candidates sometimes try to find the actual walking and cycling speeds, which is unnecessary and can introduce rounding errors. Another typical mistake is to mishandle the time conversion between hours and minutes, especially for 9 hours 55 minutes and 12 hours 30 minutes. Always convert mixed times to decimal hours carefully and use them consistently in equations. Working symbolically with T_w and T_c before substituting numerical values reduces confusion. Final Answer: Hema will take 7 hours 20 minutes to cycle both ways.

Question 18

A man travels 50 km at a speed of 25 km/h, the next 40 km at 20 km/h and thereafter 90 km at 15 km/h. What is his average speed (in km/h) for the entire journey?

Accepted Answer

Introduction / Context: This problem is a straightforward application of the concept of average speed over multiple segments of a journey with different speeds. Many students mistakenly average the given speeds directly, but the correct approach is always to use total distance divided by total time. Here, the man travels three segments of known distances at three different constant speeds, and we must compute his overall average speed for the complete trip. Given Data / Assumptions: - Segment 1 distance = 50 km, speed = 25 km/h. - Segment 2 distance = 40 km, speed = 20 km/h. - Segment 3 distance = 90 km, speed = 15 km/h. - The man travels in a straight path and there are no breaks or waiting times between segments. - Each speed is constant over its respective segment. Concept / Approach: Average speed for a complete journey is defined as: average speed = total distance / total time. It is never the simple arithmetic mean of the speeds, unless the time spent at each speed is exactly the same. In this problem, distances are different, so we must calculate the time taken for each leg using time = distance / speed, then sum these times to find the total time, and then divide the total distance by this total time. Step-by-Step Solution: Step 1: Compute the total distance travelled.Total distance = 50 + 40 + 90 = 180 km.Step 2: Compute time for each segment.Time for first 50 km = 50 / 25 = 2 hours.Time for next 40 km = 40 / 20 = 2 hours.Time for last 90 km = 90 / 15 = 6 hours.Step 3: Total time = 2 + 2 + 6 = 10 hours.Step 4: Average speed = total distance / total time = 180 / 10 = 18 km/h. Verification / Alternative check: You can check the result by noticing that the man spends most of his time at the lowest speed of 15 km/h (6 hours), so the overall average must be closer to 15 km/h than to 25 km/h. The value 18 km/h fits this intuition: it is slightly above 15 but well below 25. This qualitative check supports the calculated answer, confirming that the arithmetic was done correctly. Why Other Options Are Wrong: - 25 km/h is simply the first segment speed and is higher than any realistic average, given that later segments are slower. - 20 km/h might look like a rough central value, but it ignores the longer time spent at 15 km/h and is not based on total distance and time. - 15 km/h is the slowest segment speed; to get an average of 15 km/h, the man would need to travel at 15 km/h for the entire journey, which is not the case here. Common Pitfalls: The most common mistake is to compute (25 + 20 + 15) / 3 = 20 km/h and assume it is the average speed. This is incorrect because average speed is weighted by time (or distance), not by the number of speed values. Another error is forgetting to add all distances and times carefully, leading to arithmetic mistakes. Always follow the total distance over total time formula rigorously for mixed-speed journeys. Final Answer: The man’s average speed for the entire journey is 18 km/h.

Question 19

Two cars travel from City A to City B at constant speeds of 30 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 (in km)?

Accepted Answer

Introduction / Context: This is a standard time and distance problem with two vehicles travelling the same route at different constant speeds. We are told that the faster car takes 3 hours less than the slower car to complete the journey. Using the relationship between distance, speed, and time, we can set up an equation and solve for the unknown distance between the two cities. Such questions are very common in aptitude exams and help in reinforcing algebraic modelling skills. Given Data / Assumptions: - Speed of slower car = 30 km/h. - Speed of faster car = 48 km/h. - Distance between City A and City B = D km (unknown). - The difference in time taken by the two cars for the same distance is 3 hours. - Both cars travel along the same straight route with constant speeds and no stops. Concept / Approach: The basic formula is time = distance / speed. Let the time taken by the slower car be T_s and by the faster car be T_f. Then T_s = D / 30 and T_f = D / 48. We are told that the slower car takes 3 hours more, so the time difference T_s - T_f equals 3 hours. This gives a single equation in terms of D, which can be solved easily using fractions. Once D is found, we compare it with the options provided. Step-by-Step Solution: Step 1: Write time expressions.Time taken by slower car = D / 30 hours.Time taken by faster car = D / 48 hours.Step 2: The slower car takes 3 hours more, so:D / 30 - D / 48 = 3.Step 3: Compute the difference in fractions.1/30 - 1/48 = (48 - 30) / (30 * 48) = 18 / 1440 = 1 / 80.Step 4: Substitute into the equation: D * (1 / 80) = 3, so D / 80 = 3.Step 5: Solve for D: D = 3 * 80 = 240 km. Verification / Alternative check: If the distance is 240 km, then time taken by the slower car is 240 / 30 = 8 hours, and time taken by the faster car is 240 / 48 = 5 hours. The difference in time is 8 - 5 = 3 hours, which exactly matches the condition in the problem. This confirms that the distance of 240 km is correct and consistent with all the given information. Why Other Options Are Wrong: - 288 km would give times of 9.6 hours and 6 hours, with a difference of 3.6 hours, not 3 hours. - 360 km would give times of 12 hours and 7.5 hours, with a difference of 4.5 hours. - 192 km would give times of 6.4 hours and 4 hours, with a difference of 2.4 hours. None of these match the required 3-hour difference. Common Pitfalls: Students often mistakenly set up the equation with the wrong order of subtraction or forget that the slower car has the larger time. Another mistake is to attempt to average speeds instead of using the time difference condition. Being systematic about which car is slower and writing the equation as (time of slower) minus (time of faster) helps avoid sign errors. Working carefully with fractions also prevents arithmetic mistakes. Final Answer: The distance between City A and City B is 240 km.

Question 20

Each question contains Quantity I and Quantity II. Read the information carefully and answer accordingly. Quantity I: Vipin can swim at 6 km/h in still water. The river flows at 3 km/h, and it takes him 8 hours more to swim upstream than downstream …

Accepted Answer

Introduction / Context: This problem is a comparative question involving two separate time and distance situations with boats or swimmers in running water. We are required to compute the one-way distance in each scenario and then compare the two quantities. It tests the concepts of upstream and downstream speeds, the use of time differences, and the average time for a round trip on a river. These ideas appear frequently in aptitude exams under “boats and streams” or “swimming in a river” topics. Given Data / Assumptions: - Quantity I: Swimming speed of Vipin in still water = 6 km/h. - River flow speed = 3 km/h. - It takes 8 hours more to swim upstream than downstream for the same distance. - Quantity II: Rowing speed in still water = 25 km/h. - River flow speed = 15 km/h. - Total time for rowing to a place and back = 2 hours. - All speeds are constant and the river flows uniformly in a straight path in both cases. Concept / Approach: For motion in a river, the effective upstream speed is (still water speed - stream speed), and the downstream speed is (still water speed + stream speed). Time is distance divided by speed. For Quantity I, we set up an equation using the given time difference between upstream and downstream journeys. For Quantity II, we use the total round-trip time to find the one-way distance. After compute both distances, we compare them to decide which quantity is greater. Step-by-Step Solution: Quantity I:Upstream speed = 6 - 3 = 3 km/h.Downstream speed = 6 + 3 = 9 km/h.Let the one-way distance be d km.Time upstream = d / 3 hours; time downstream = d / 9 hours.Given that upstream time is 8 hours more: d / 3 - d / 9 = 8.Compute: (1/3 - 1/9) * d = (2/9) * d = 8, so d = 8 * 9 / 2 = 36 km.Quantity II:Downstream speed = 25 + 15 = 40 km/h.Upstream speed = 25 - 15 = 10 km/h.Let one-way distance be x km.Total time: x / 40 + x / 10 = 2.Compute: (1/40 + 1/10) * x = (1/40 + 4/40) * x = (5/40) * x = x / 8 = 2, so x = 16 km. Verification / Alternative check: Recheck both distances. For Quantity I: upstream time = 36 / 3 = 12 hours; downstream time = 36 / 9 = 4 hours; difference = 8 hours, which matches the condition. For Quantity II: downstream time = 16 / 40 = 0.4 hours; upstream time = 16 / 10 = 1.6 hours; total = 2 hours, which again matches. Therefore, Quantity I is 36 km and Quantity II is 16 km, making Quantity I clearly larger. Why Other Options Are Wrong: - Quantity I = Quantity II would require both distances to be equal, but 36 km is not equal to 16 km. - Quantity I < Quantity II is incorrect because 36 is greater than 16, not smaller. - “Relationship cannot be determined” is wrong because the data in both quantities is sufficient to determine unique distances and compare them directly. Common Pitfalls: Some candidates mistakenly use still-water speeds instead of upstream or downstream speeds or mix up which is which. Others may mis-handle the time difference equation in Quantity I or the total time equation in Quantity II. It is also common to think that the comparison is tricky and that it cannot be determined, but as the calculations show, both distances are clearly defined. Always convert verbal statements about “more time” or “round-trip time” into explicit equations before comparing quantities. Final Answer: Quantity I is greater than Quantity II.

Time and Distance Questions