Question 1

A 360 m long train is moving at a speed of 44 km/h. In how much time (in seconds) will it completely cross a 140 m long bridge?

Accepted Answer

Introduction / Context:This is a classic time and distance problem involving a train crossing a bridge. When a train crosses a bridge, it must cover not only its own length but also the full length of the bridge. The question checks your understanding of effective distance, speed conversion from km/h to m/s, and the use of the basic distance formula. Such questions are very common in railway and aptitude examinations. Given Data / Assumptions:- Length of the train = 360 m.- Length of the bridge = 140 m.- Speed of the train = 44 km/h, assumed constant.- The train moves in a straight line with no acceleration or deceleration.- We need the total time taken for the train to completely clear the bridge, expressed in seconds. Concept / Approach:Two main ideas are used here:- Effective distance to be covered while crossing a bridge = length of train + length of bridge.- Relationship between distance, speed, and time: distance = speed * time.Because the lengths are given in metres and the answer is required in seconds, we must convert the speed from km/h to m/s using the standard factor: 1 km/h = (5/18) m/s. Step-by-Step Solution:Step 1: Compute the effective distance.Effective distance = 360 m + 140 m = 500 m.Step 2: Convert the speed from km/h to m/s.Speed in m/s = 44 * (5/18) = 220/18 m/s ≈ 12.22 m/s.Step 3: Use the formula time = distance / speed.Time = 500 / (220/18) = 500 * 18 / 220 = (500/220) * 18.Simplify: 500/220 = 50/22 = 25/11. So time = (25/11) * 18 = 450/11 seconds ≈ 40.91 seconds.Step 4: Round to the nearest whole second, which is 41 seconds. Verification / Alternative check:An approximate mental check can be done. If the train moves at about 12 m/s, then in 40 seconds it would cover roughly 480 m, and in 41 seconds it would cover about 492 m. Because the exact calculation gave slightly more than 40.9 seconds and the required distance is 500 m, rounding to 41 seconds is reasonable and consistent. This matches option “41 seconds”. Why Other Options Are Wrong:- 36 seconds and 39 seconds are too small; at roughly 12 m/s, they would cover only about 432 m or 468 m, which is less than 500 m.- 43 seconds is slightly larger than required and would correspond to covering more distance than necessary at the given speed.Only 41 seconds matches the computed and rounded value correctly. Common Pitfalls:Candidates sometimes forget to add the length of the bridge to the length of the train and use only the train length as distance. Another frequent mistake is to mix kilometres and metres or hours and seconds without converting units, leading to incorrect answers. Always make sure to convert speed to m/s when distances are in metres and time is required in seconds. Final Answer:The train will cross the bridge completely in 41 seconds.

Question 2

Two persons K and L start walking towards each other at 4:00 p.m. from points 17.5 km apart. K walks at 3 km/h and L walks at 4 km/h. At what time will they meet?

Accepted Answer

Introduction / Context:This question involves two people walking towards each other from two different points that are a known distance apart. It tests your understanding of relative speed when objects move towards each other, as well as your ability to convert time in hours to a clock format. Problems like this are very common in the time and distance section of aptitude exams. Given Data / Assumptions:- Distance between K and L at the start = 17.5 km.- K walks at a constant speed of 3 km/h.- L walks at a constant speed of 4 km/h.- Both start walking towards each other at exactly 4:00 p.m.- They continue to walk without stopping until they meet. Concept / Approach:When two people move towards each other along a straight line, their effective or relative speed is the sum of their individual speeds. The time needed to meet is then found by dividing the initial distance between them by this relative speed. Finally, we add this time interval to the starting time to get the exact meeting time on the clock. Step-by-Step Solution:Step 1: Calculate the relative speed.Relative speed = speed of K + speed of L = 3 km/h + 4 km/h = 7 km/h.Step 2: Use the formula time = distance / speed to find the time taken to meet.Time taken to meet = 17.5 km / 7 km/h = 2.5 hours.Step 3: Convert 2.5 hours to hours and minutes.2.5 hours = 2 hours 30 minutes.Step 4: Add this duration to the starting time of 4:00 p.m.4:00 p.m. + 2 hours 30 minutes = 6:30 p.m. Verification / Alternative check:In 2.5 hours, K will walk 3 * 2.5 = 7.5 km, and L will walk 4 * 2.5 = 10 km. The total distance covered by both together is 7.5 km + 10 km = 17.5 km, which is exactly the original separation between them. This confirms that they meet after 2.5 hours, that is, at 6:30 p.m. Why Other Options Are Wrong:- 6:00 p.m. would correspond to only 2 hours of walking, in which they would cover 3 * 2 + 4 * 2 = 14 km, less than 17.5 km.- 5:45 p.m. corresponds to 1 hour 45 minutes, which is too early; they would not have covered the full 17.5 km together.- 5:52 p.m. is an arbitrary time that does not match a precise division of the distance by their relative speed.Thus, 6:30 p.m. is the only correct meeting time. Common Pitfalls:A common mistake is to subtract the speeds instead of adding them when objects move towards each other. Another error is to mis-handle the conversion of decimal hours to minutes, for example confusing 0.5 hours with 50 minutes instead of 30 minutes. Always convert decimal fractions of an hour using the fact that 1 hour = 60 minutes, so 0.5 hour = 30 minutes and 0.25 hour = 15 minutes. Final Answer:K and L will meet at 6:30 p.m.

Question 3

Mr. Karthik drives to his office at a constant average speed of 48 km/h. The time taken to cover the first 60% of the distance is 20 minutes more than the time taken to cover the remaining 40% of the distance. What is the total distance from his hom…

Accepted Answer

Introduction / Context:This question tests understanding of average speed and the relationship between distance, speed, and time when different portions of a journey are considered. Although only one speed is mentioned, the question focuses on comparing the time taken for different fractions of the same total distance. This type of problem frequently appears in time and distance sections of aptitude tests. Given Data / Assumptions:- The overall average speed of Mr. Karthik is 48 km/h.- The journey from home to office is a single trip with constant speed 48 km/h.- The first 60% of the distance takes 20 minutes more than the last 40% of the distance.- We assume straight travel at uniform speed with no breaks or delays.- We need to find the entire distance between his home and office in kilometres. Concept / Approach:We use the basic formula distance = speed * time. If the speed is constant throughout the journey, then time is directly proportional to distance. That means the time taken to cover 60% of the distance and the time taken to cover 40% of the distance are proportional to 0.6 and 0.4 of the total distance, and their difference is given. By using the speed and the time difference, we can find the full distance. Step-by-Step Solution:Step 1: Let the total distance be D km.Step 2: Distance for the first part is 0.6D, and for the second part is 0.4D.Step 3: Since the speed is 48 km/h throughout, time for first 60% is t1 = 0.6D / 48 hours.Step 4: Time for last 40% is t2 = 0.4D / 48 hours.Step 5: The difference in time t1 - t2 equals 20 minutes, that is 1/3 hour.So, 0.6D / 48 - 0.4D / 48 = 1/3.Step 6: Simplify the left-hand side: (0.6D - 0.4D) / 48 = 0.2D / 48 = 1/3.Step 7: Therefore, 0.2D = 48 * (1/3) = 16, so D = 16 / 0.2 = 80 km. Verification / Alternative check:If the total distance is 80 km, then 60% of it is 48 km and 40% is 32 km. Time for the first 48 km at 48 km/h is 48 / 48 = 1 hour. Time for the last 32 km is 32 / 48 = 2/3 hour, which is 40 minutes. The difference between 1 hour and 40 minutes is 20 minutes, which matches the given condition, confirming that 80 km is correct. Why Other Options Are Wrong:- 40 km would make the time difference between the two parts smaller than 20 minutes.- 50 km and 70 km similarly do not satisfy the condition when you compute the times for 60% and 40% segments at 48 km/h.Only 80 km gives a time difference of exactly 20 minutes between the two parts of the journey. Common Pitfalls:One common error is to think that “average speed” means two different speeds were used, which can overcomplicate the problem. Here the average speed is the constant speed. Another pitfall is to convert 20 minutes incorrectly or to forget that 60% and 40% refer to fractions of the same total distance. Always work in consistent units and carefully translate percentage information into fractional distances. Final Answer:The distance from Karthik's home to his office is 80 km.

Question 4

A man covered a certain distance at a constant speed. If he had moved 3 km/h faster, he would have taken 40 minutes less. If he had moved 2 km/h slower, he would have taken 40 minutes more. What is the distance covered (in km)?

Accepted Answer

Introduction / Context:This is a classic time, speed, and distance problem where the same distance is travelled at three different hypothetical speeds. The question asks you to use the information about changes in speed and corresponding changes in time to determine the unknown distance. Such questions test algebraic formulation of time and distance relations and are very common in aptitude exams. Given Data / Assumptions:- Let the original speed of the man be v km/h and the distance be d km.- If he moves at (v + 3) km/h, his time decreases by 40 minutes (which is 2/3 hour).- If he moves at (v - 2) km/h, his time increases by 40 minutes (2/3 hour).- The distance d remains constant in all cases.- Motion is along a straight path with uniform speed, and there are no stops. Concept / Approach:We use time = distance / speed. The original time is d / v. When the speed is changed, the time also changes. The given statements about 40 minutes less or more allow us to set up two algebraic equations involving d and v. Solving these simultaneous equations gives the values of v and d. This approach is a standard technique for handling problems with changed speed and changed time for a fixed distance. Step-by-Step Solution:Step 1: Let the original time be T = d / v hours.Step 2: At speed (v + 3) km/h, time taken is d / (v + 3). This is 2/3 hour less than T.So, d / v - d / (v + 3) = 2/3.Step 3: At speed (v - 2) km/h, time taken is d / (v - 2). This is 2/3 hour more than T.So, d / (v - 2) - d / v = 2/3.Step 4: Now we solve the system of equations:(i) d / v - d / (v + 3) = 2/3.(ii) d / (v - 2) - d / v = 2/3.Step 5: Solving these equations (for example, by algebraic manipulation) gives v = 12 km/h and d = 40 km. Verification / Alternative check:With v = 12 km/h and d = 40 km, original time T = 40 / 12 = 10/3 hours (3 hours 20 minutes). At speed 15 km/h (12 + 3), time = 40 / 15 = 8/3 hours (2 hours 40 minutes), which is 40 minutes less than 3 hours 20 minutes. At speed 10 km/h (12 - 2), time = 40 / 10 = 4 hours, which is 40 minutes more than 3 hours 20 minutes. Thus both conditions are satisfied, confirming that the distance is 40 km. Why Other Options Are Wrong:- 30 km, 36 km, and 42 km do not simultaneously satisfy the two conditions when you set up and solve the equations with the given speed changes.- Only 40 km gives a consistent set of times that differ by exactly 40 minutes for the given increases and decreases in speed. Common Pitfalls:Many candidates incorrectly try to average speeds or directly subtract distances instead of using algebraic equations. Another frequent mistake is to convert 40 minutes incorrectly or to forget that the distance is the same in all three scenarios. Always translate the word statements into precise time equations using time = distance / speed and then solve the system step by step. Final Answer:The distance covered by the man is 40 km.

Question 5

Buses leave a bus terminal every 10 minutes, each travelling at a constant speed of 20 km/h along the same route. A man walks towards the terminal from the opposite direction and meets the buses at intervals of 8 minutes. What is the walking speed o…

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Introduction / Context:This problem tests the concept of relative speed and the idea of vehicles leaving a terminal at regular time intervals. The man is moving towards the terminal while the buses move away from it, and he meets them more frequently than they would pass a fixed point. Understanding the relationship between the spacing of buses and the relative speed between the man and the buses is the key to solving this question. Given Data / Assumptions:- Buses leave the terminal every 10 minutes, which is 10/60 = 1/6 hour.- Speed of each bus = 20 km/h.- The man walks towards the terminal from the opposite direction.- The man meets successive buses every 8 minutes, which is 8/60 = 2/15 hour.- All buses and the man move at constant speeds along the same straight route, with no stops. Concept / Approach:First, we compute the distance between two consecutive buses along the route in the ground frame. This distance depends on the speed of the buses and the time interval between their departures. Next, we consider the relative motion between the man and the buses. The time between successive meetings from the man's point of view equals the distance between buses divided by the sum of the speeds (because they are moving towards each other). We then solve for the man's speed using this relationship. Step-by-Step Solution:Step 1: Compute the distance between two consecutive buses.Time between departures = 1/6 hour. Bus speed = 20 km/h.Distance between buses = 20 * (1/6) = 20/6 = 10/3 km.Step 2: Let the man's speed be v km/h.Because they move towards each other, the relative speed = 20 + v km/h.Step 3: The man meets successive buses every 2/15 hour.So, time between meetings = distance between buses / relative speed.Therefore, 2/15 = (10/3) / (20 + v).Step 4: Solve for v: (10/3) / (20 + v) = 2/15.Cross multiply: (10/3) * 15 / 2 = 20 + v.This gives 10 * 5 / 2 = 20 + v, so 25 = 20 + v, hence v = 5 km/h. Verification / Alternative check:If the man walks at 5 km/h while the buses move at 20 km/h, their relative speed is 25 km/h. The separation between buses is 10/3 km. The time between meetings is then (10/3) / 25 hours = 10 / 75 hours = 2 / 15 hours, which is exactly 8 minutes. This matches the condition given in the question, confirming that the solution is correct. Why Other Options Are Wrong:- 6 km/h, 7.5 km/h, and 8 km/h give different relative speeds and hence different meeting intervals that do not match 8 minutes when calculated using the formula time = distance / relative speed.- Only 5 km/h gives a meeting interval of exactly 8 minutes. Common Pitfalls:A typical mistake is to treat the problem as if the man were standing still and to assume that the time between buses remains 10 minutes. Another error is forgetting that the relative speed is the sum of speeds when objects move towards each other. Make sure to convert all time units to hours or minutes consistently and carefully apply the relationship between distance, speed, and time. Final Answer:The walking speed of the man is 5 km/h.

Question 6

The speeds of a car, a jeep, and a tractor are in the ratio 3 : 5 : 2. The speed of the jeep is 250 percent of the speed of the tractor, and the tractor covers 360 km in 12 hours. What is the average of the speeds of the car and the jeep together (i…

Accepted Answer

Introduction / Context:This question checks your understanding of speed ratios and how to use them to find actual speeds when one real speed is known. You are also asked to find the simple average of two speeds. Such questions are frequently asked in aptitude exams under the topic of time, speed, and distance, and they help in strengthening ratio and proportion concepts. Given Data / Assumptions:- The speeds of the car, jeep, and tractor are in the ratio 3 : 5 : 2.- The speed of the jeep is 250 percent of the speed of the tractor, which is consistent with the ratio 5 : 2 (since 5/2 = 2.5 = 250 percent).- The tractor covers 360 km in 12 hours.- All speeds are assumed to be constant.- We need the average of the speeds of the car and the jeep in km/h. Concept / Approach:First, we find the actual speed of the tractor using distance and time. Then, using the ratio 3 : 5 : 2, we determine the speeds of the car and the jeep. Finally, we compute the simple arithmetic mean of these two speeds. The problem mainly uses proportional reasoning and the basic speed formula speed = distance / time. Step-by-Step Solution:Step 1: Compute the speed of the tractor.Speed of tractor = distance / time = 360 km / 12 h = 30 km/h.Step 2: Let the common ratio factor be k.Then speeds are: car = 3k, jeep = 5k, tractor = 2k.Step 3: Since the tractor's speed is 30 km/h, we have 2k = 30, so k = 15.Step 4: Compute actual speeds.Car speed = 3k = 3 * 15 = 45 km/h.Jeep speed = 5k = 5 * 15 = 75 km/h.Step 5: Find the average speed of car and jeep.Average speed = (45 + 75) / 2 = 120 / 2 = 60 km/h. Verification / Alternative check:The given statement that the jeep speed is 250 percent of the tractor speed is consistent: tractor speed = 30 km/h, jeep speed = 75 km/h, and 75 / 30 = 2.5, which is 250 percent. This confirms that the ratio and the calculated values are correct. Therefore, the average of 45 km/h and 75 km/h must be 60 km/h, matching our answer. Why Other Options Are Wrong:- 50 km/h, 70 km/h, and 80 km/h are simple averages of other wrongly assumed speed pairs and do not correspond to the correct values obtained from the ratio 3 : 5 : 2 with tractor speed 30 km/h.- Only 60 km/h is consistent with the ratio and the actual speeds derived from the given data. Common Pitfalls:Some candidates confuse the simple average speed of two different vehicles with weighted averages or harmonic means. Here, the question clearly asks for the average of two speeds, not the average speed over some combined journey. Another common error is to misinterpret percentage statements or to ignore the ratio information when real data is given. Always use both the ratio and the actual numerical data to find the scaling factor correctly. Final Answer:The average of the speeds of the car and the jeep is 60 km/h.

Question 7

A man travels a total distance of 480 km in 4 hours, partly by air and partly by train. If he had travelled the entire 480 km by air, he would have saved four fifths of the time he would have taken by train and would have arrived 2 hours earlier tha…

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Introduction / Context:This is a multi-step word problem involving two different modes of travel (air and train) and comparisons of hypothetical scenarios. It tests your ability to translate verbal conditions into mathematical equations involving distance, speed, and time. You also have to interpret a comparative statement about time saved as a fraction of another time. Such problems are considered higher level in aptitude tests because they require careful reading and algebraic reasoning. Given Data / Assumptions:- Total distance = 480 km.- Actual total time taken for the mixed journey = 4 hours.- If he had travelled all 480 km by air, he would have reached 2 hours earlier than he actually did, so the all-air time would have been 2 hours.- The time saved by travelling entirely by air is four fifths of the time he would have taken if he had travelled the whole distance by train.- Speeds of air travel and train travel are constant and positive. Concept / Approach:First, use the information about arriving 2 hours earlier to find the hypothetical all-air travel time. Then, interpret the statement about saving four fifths of the train time to relate the all-air time and the all-train time. From this, we find the speeds for air and train when they cover the full 480 km. Once we know these speeds, we set up an equation for the actual mixed journey: some distance x by air and the remaining 480 - x by train in a total of 4 hours. Solving this equation gives the distance travelled by train in the actual situation. Step-by-Step Solution:Step 1: Let T_train be the time to cover 480 km by train only and T_air be the time to cover 480 km by air only.We are told the man would arrive 2 hours earlier by air than in the mixed 4 hour journey, so T_air = 4 - 2 = 2 hours.Step 2: The time saved by choosing air over train is four fifths of T_train.Time saved = T_train - T_air = (4/5) * T_train.So T_train - T_air = (4/5) * T_train, which gives T_air = (1/5) * T_train.Step 3: Substitute T_air = 2 into T_air = (1/5) * T_train.2 = (1/5) * T_train, so T_train = 10 hours.Step 4: Compute speeds for air and train for the 480 km distance.Train speed v_train = 480 / 10 = 48 km/h.Air speed v_air = 480 / 2 = 240 km/h.Step 5: Let the actual distance travelled by air be x km, so the distance by train is 480 - x km.Total actual time is 4 hours, so:x / 240 + (480 - x) / 48 = 4.Step 6: Solve the equation: x / 240 + (480 - x) / 48 = 4.Multiply throughout by 240 to clear denominators: x + 5 * (480 - x) = 960.This gives x + 2400 - 5x = 960 ⇒ -4x = 960 - 2400 = -1440 ⇒ x = 360 km.Step 7: Distance by train = 480 - 360 = 120 km. Verification / Alternative check:Check the times with x = 360 km by air and 120 km by train. Time by air = 360 / 240 = 1.5 hours. Time by train = 120 / 48 = 2.5 hours. Total actual time = 1.5 + 2.5 = 4 hours, which matches the problem statement. Also, all-air travel takes 2 hours, which is indeed 2 hours less than 4 hours. For the all-train journey, time is 10 hours, and time saved by air is 10 - 2 = 8 hours, which is four fifths of 10 hours. Everything is consistent, so the calculation is correct. Why Other Options Are Wrong:- 80 km, 90 km, and 110 km do not satisfy the time equation x / 240 + (480 - x) / 48 = 4 when substituted as the train distance (480 - x). They lead to total times different from 4 hours.- Only 120 km gives a consistent solution that satisfies all conditions of the problem. Common Pitfalls:Many candidates misread the statement about saving four fifths of the train time and set up an incorrect equation. Others incorrectly assume that 4 hours is the time for either all-air or all-train travel, which is not the case. Always define the times clearly for each scenario and translate proportional statements carefully. Working step by step with clear algebraic equations helps avoid confusion in such multi-condition word problems. Final Answer:The man travelled 120 km by train in the actual journey.

Question 8

A person goes to his office at one third of the speed at which he returns from his office. If the average speed for the entire to-and-fro trip is 12 km/h, what is his speed while going to the office (in km/h)?

Accepted Answer

Introduction / Context:This question deals with average speed over a round trip where the speeds in the two directions are different. It tests your understanding of how to compute average speed when the distance is the same in each direction but the speeds are unequal. The relationship between the speeds and the given average speed must be used to determine the speed while going to the office. Given Data / Assumptions:- Let the speed while going to the office be v km/h.- The speed while returning is three times this, that is 3v km/h, because going speed is one third of returning speed.- The distance from home to office is the same in both directions and is positive, say d km one way.- The average speed for the whole trip (going and returning) is 12 km/h.- Motion is at constant speeds with no intermediate stops considered for calculation. Concept / Approach:Average speed for a round trip with unequal speeds is not the simple arithmetic mean of the speeds. Instead, average speed is defined as total distance divided by total time. For a round trip over the same distance d each way, total distance is 2d, and total time is d/v + d/(3v). Using the definition of average speed, we set up an equation with the given average speed (12 km/h) and then solve for v. Step-by-Step Solution:Step 1: Let one-way distance be d km.Speed while going = v km/h, speed while returning = 3v km/h.Step 2: Total distance for the round trip = 2d km.Step 3: Time taken to go = d / v hours. Time taken to return = d / (3v) hours.Total time for the round trip = d / v + d / (3v) = d * (1 / v + 1 / (3v)).Step 4: Compute the expression inside the brackets.1 / v + 1 / (3v) = (3 + 1) / (3v) = 4 / (3v).So total time = d * 4 / (3v) = 4d / (3v).Step 5: Average speed is total distance divided by total time.Average speed = 2d / (4d / (3v)) = 2d * (3v / 4d) = (6v / 4) = 3v / 2.Step 6: We are given that average speed is 12 km/h, so 3v / 2 = 12.Solve for v: 3v = 24 ⇒ v = 8 km/h. Verification / Alternative check:If v = 8 km/h, then the return speed is 3 * 8 = 24 km/h. Let d = 24 km as an example. Time to go = 24 / 8 = 3 hours. Time to return = 24 / 24 = 1 hour. Total distance = 48 km, total time = 4 hours, so average speed = 48 / 4 = 12 km/h, which matches the given data. Thus, v = 8 km/h is correct. Why Other Options Are Wrong:- 10 km/h and 6 km/h do not satisfy the equation 3v / 2 = 12 when substituted as v, and the resulting average speeds do not match 12 km/h.- “Cannot be determined” is incorrect because the given information is sufficient to form a unique equation and determine v exactly. Common Pitfalls:A frequent error is to take the simple average of the two speeds v and 3v, which would give (v + 3v) / 2 = 2v. Using the condition that 2v = 12 would wrongly lead to v = 6. This is incorrect because average speed over equal distances depends on the harmonic mean, not the arithmetic mean. Always use the definition of average speed as total distance divided by total time when dealing with unequal speeds in each direction. Final Answer:The person's speed while going to the office is 8 km/h.

Question 9

Den Bosch and Eastbourne are two cities 300 km apart. Maradona starts from Den Bosch towards Eastbourne at 8:24 a.m. One hour later, Pele also starts from Den Bosch towards Eastbourne. After travelling for 1 hour, Pele reaches Nottingham, a town on …

Accepted Answer

Introduction / Context:This is a challenging relative speed and time coordination problem involving two people starting from the same place at different times but arriving at a common destination together. The extra information about Nottingham, a point on the route, provides a relationship between their speeds. The question requires careful time-line reasoning and systematic use of distance = speed * time to deduce the speeds of Maradona and Pele. Given Data / Assumptions:- Distance between Den Bosch and Eastbourne = 300 km.- Maradona starts at 8:24 a.m. from Den Bosch towards Eastbourne.- Pele starts at 9:24 a.m. from Den Bosch towards Eastbourne, one hour after Maradona.- After travelling 1 hour, Pele reaches Nottingham at 10:24 a.m.- Maradona passed Nottingham 40 minutes earlier, at 9:44 a.m.- Nottingham lies on the straight line route between Den Bosch and Eastbourne.- Both travel at constant but different speeds and reach Eastbourne at the same time. Concept / Approach:The first key step is to use Nottingham to relate the speeds. The time taken by Maradona to reach Nottingham and the time taken by Pele to reach the same point are different but correspond to the same distance, so we can express their speeds in terms of a common ratio. The second key step is to use the fact that they both reach Eastbourne at the same time. Since Pele starts one hour later, Maradona's travel time to Eastbourne is one hour more than Pele's. Combining these two relationships lets us solve for both speeds and confirm the correct option. Step-by-Step Solution:Step 1: Let the speed of Maradona be v_m km/h and the speed of Pele be v_p km/h.Step 2: Time for Maradona to reach Nottingham is from 8:24 a.m. to 9:44 a.m., which is 1 hour 20 minutes = 4/3 hours.Step 3: Time for Pele to reach Nottingham is from 9:24 a.m. to 10:24 a.m., which is 1 hour.Step 4: Distances to Nottingham are the same for both, so v_m * (4/3) = v_p * 1.Therefore, v_p = (4/3) * v_m.Step 5: Let T_p be the total travel time for Pele to reach Eastbourne. Then Maradona travels for T_p + 1 hours (because he started one hour earlier and they arrive together).Step 6: Distance to Eastbourne for each is 300 km.So 300 = v_p * T_p and 300 = v_m * (T_p + 1).Step 7: Substitute v_p = (4/3) * v_m into 300 = v_p * T_p to get 300 = (4/3) * v_m * T_p.Also, from 300 = v_m * (T_p + 1), we have T_p + 1 = 300 / v_m.Step 8: From 300 = (4/3) * v_m * T_p, get T_p = 300 * 3 / (4 * v_m) = 225 / v_m.Step 9: Substitute into T_p + 1 = 300 / v_m:225 / v_m + 1 = 300 / v_m ⇒ 225 + v_m = 300 ⇒ v_m = 75 km/h.Step 10: Then v_p = (4/3) * 75 = 100 km/h. Verification / Alternative check:With v_m = 75 km/h and v_p = 100 km/h, distance to Nottingham covered by Maradona in 4/3 hours is 75 * 4/3 = 100 km. Pele covers the same 100 km in 1 hour at 100 km/h, which is consistent. For Eastbourne: Maradona takes T_m = 300 / 75 = 4 hours. Pele takes T_p = 300 / 100 = 3 hours. Because Pele starts 1 hour after Maradona, their arrival times both occur at 8:24 a.m. + 4 hours = 12:24 p.m., confirming the logic. Why Other Options Are Wrong:- 100 km/h, 125 km/h and the other pairs do not satisfy both the Nottingham timing condition and the equal arrival time condition when tested similarly.- Only the pair 75 km/h and 100 km/h satisfies all the constraints exactly. Common Pitfalls:It is very easy to confuse the starting times and travel durations, or to mis-handle the 40 minute difference at Nottingham. Another common error is to assume that Nottingham is at the midpoint of the route, which is never stated. Instead, always treat Nottingham as an unknown intermediate point and use the equality of distances with different times to relate speeds. Drawing a time-line can greatly reduce mistakes in such multi-step problems. Final Answer:The speeds are 75 km/h for Maradona and 100 km/h for Pele.

Question 10

Three persons go for a morning walk together and start stepping off at the same time. Their steps measure 75 cm, 80 cm, and 85 cm respectively. What is the minimum distance each should walk so that they all cover the distance in a whole number of st…

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Introduction / Context:This question tests your understanding of the least common multiple (LCM) in a practical context involving step lengths. Each person has a different step length, and we are asked to find the smallest distance that all three can cover in an exact number of steps. Such questions link number theory concepts like multiples and the LCM to real life scenarios, which is common in competitive exams. Given Data / Assumptions:- Step length of person 1 = 75 cm.- Step length of person 2 = 80 cm.- Step length of person 3 = 85 cm.- All start walking together and we want a minimum distance that is an exact multiple of each step length.- Distance is to be reported in metres and centimetres or just metres where appropriate. Concept / Approach:The distance each person covers is equal to number of steps multiplied by his or her step length. For each person to complete the same distance in a whole number of steps, the total distance must be a common multiple of all three step lengths. The smallest such distance is their least common multiple (LCM). After computing the LCM in centimetres, we convert that distance into metres (and centimetres if needed) and then match it with the given options. Step-by-Step Solution:Step 1: Write down the step lengths: 75 cm, 80 cm, and 85 cm.Step 2: Factorise the numbers if needed.75 = 3 * 5^2, 80 = 2^4 * 5, and 85 = 5 * 17.Step 3: The LCM is found by taking each prime factor at the highest power that appears.LCM = 2^4 * 3 * 5^2 * 17.Compute step by step: 2^4 = 16, 5^2 = 25.So LCM = 16 * 3 * 25 * 17.16 * 3 = 48, 25 * 17 = 425.Now 48 * 425 = (50 - 2) * 425 = 50 * 425 - 2 * 425 = 21250 - 850 = 20400 cm.Step 4: Convert 20400 cm to metres.Since 100 cm = 1 m, 20400 cm = 20400 / 100 = 204 m. Verification / Alternative check:Check if 204 m is divisible by each step length. Convert 204 m to centimetres: 20400 cm. Now 20400 / 75 = 272 steps, 20400 / 80 = 255 steps, and 20400 / 85 = 240 steps. All these quotients are whole numbers, so each person will indeed take an integer number of steps to cover 204 m. This confirms that 204 m is the smallest such distance because it is the LCM of the three step lengths. Why Other Options Are Wrong:- 222 m 44 cm, 201 m 21 cm, and 208 m do not correspond to the LCM of 75, 80, and 85 when converted to centimetres and checked for divisibility.- They either leave remainders when divided by one or more of the step lengths or are multiples of a larger common multiple, not the least common multiple. Common Pitfalls:Many students confuse greatest common divisor (GCD) with least common multiple (LCM). For this type of problem, always remember that when you want the smallest common size that can accommodate all step lengths without remainder, you should use the LCM. Errors in prime factorisation and arithmetic multiplications can also lead to wrong answers, so it is helpful to redo the multiplication step carefully or break it into smaller parts as shown. Final Answer:The minimum distance each should walk is 204 m.

Question 11

A bike runs at a speed of 35 km/h when it has not been serviced and at 55 km/h after servicing. After servicing, the bike covers a certain distance in 5 hours. Approximately how much time will it take to cover the same distance when it has not been …

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Introduction / Context:This question compares the time taken to cover the same distance at two different speeds. It tests the basic relationship between speed, distance, and time and also asks for an approximate value. Such problems are common in speed and distance sections and help you build intuition about how travel time changes with speed. Given Data / Assumptions:- Speed of the bike after servicing = 55 km/h.- Time taken after servicing to cover the distance = 5 hours.- Speed of the bike before servicing = 35 km/h.- The distance covered in both cases is the same.- We are asked for the approximate time taken at 35 km/h to cover that distance. Concept / Approach:First, we use the post-service speed and time to find the actual distance. Then, we use the pre-service speed and the same distance to find the required time. The core formula is distance = speed * time. The final answer is expressed as an approximate time, so a small rounding is acceptable if needed. Step-by-Step Solution:Step 1: Compute the distance using the post-service data.Distance = speed * time = 55 km/h * 5 h = 275 km.Step 2: Use the pre-service speed to find the time for the same distance.Pre-service speed = 35 km/h.Time at 35 km/h = distance / speed = 275 km / 35 km/h.Step 3: Simplify 275 / 35.275 / 35 = 55 / 7 ≈ 7.857 hours.Step 4: Round off 7.857 hours to the nearest option. The closest given option is 8 hours. Verification / Alternative check:We can roughly estimate: at 35 km/h, in 8 hours the bike would cover 35 * 8 = 280 km, which is very close to 275 km. At 7.5 hours, the distance would be 35 * 7.5 = 262.5 km, which is farther from 275 km. This comparison confirms that 8 hours is the best approximate choice among the options. Why Other Options Are Wrong:- 7.5 hours corresponds to about 262.5 km, which is too short compared to 275 km.- 7 hours gives 245 km, and 6.5 hours gives 227.5 km, both significantly smaller than 275 km.- Therefore, these options do not represent the correct time for covering the same distance at 35 km/h. Common Pitfalls:Some candidates forget that the distance remains constant and mistakenly try to average speeds instead of using distance and time properly. Others miscalculate 275 / 35 due to fraction handling errors. A good habit is to first find the distance using one scenario and then reuse that distance in the second scenario, rather than trying to jump directly to a ratio without clear steps. Final Answer:The bike will take approximately 8 hours to cover the same distance when not serviced.

Question 12

The respective ratio between the speeds of a bike, a van, and a lorry is 3 : 5 : 2. The speed of the van is 250 percent of the speed of the lorry, and the lorry covers 360 km in 12 hours. What is the average of the speeds of the bike and the van tog…

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Introduction / Context:This question again uses speed ratios to connect the speeds of three vehicles, and then asks for the simple average of the speeds of two of them. It reinforces the same style of reasoning as other ratio-based speed problems and is very common in aptitude exams under the time and distance topic. Given Data / Assumptions:- Speeds of bike, van, and lorry are in the ratio 3 : 5 : 2.- The van's speed is 250 percent of the lorry's speed, consistent with the ratio (5 : 2 = 2.5 times).- The lorry covers 360 km in 12 hours.- All vehicles travel at constant speeds.- We need the average (arithmetic mean) of the speeds of the bike and the van in km/h. Concept / Approach:First, find the actual speed of the lorry using distance and time. Then, treat the ratio 3 : 5 : 2 as 3k, 5k, and 2k for bike, van, and lorry respectively. Use the lorry's actual speed to find the value of k, and then compute the speeds of the bike and the van. Finally, calculate their average speed as (bike speed + van speed) / 2. Step-by-Step Solution:Step 1: Compute the speed of the lorry.Speed of lorry = distance / time = 360 km / 12 h = 30 km/h.Step 2: Let the common ratio factor be k.Bike speed = 3k, van speed = 5k, lorry speed = 2k.Step 3: Since 2k = 30, we have k = 15.Step 4: Compute the actual speeds.Bike speed = 3k = 3 * 15 = 45 km/h.Van speed = 5k = 5 * 15 = 75 km/h.Step 5: Compute the average speed of bike and van.Average speed = (45 + 75) / 2 = 120 / 2 = 60 km/h. Verification / Alternative check:The van speed being 75 km/h and the lorry speed being 30 km/h is consistent with the statement that van speed is 250 percent of lorry speed, since 75 / 30 = 2.5, or 250 percent. This confirms that the ratio scaling has been applied correctly. The average of 45 km/h and 75 km/h is clearly 60 km/h, matching the chosen answer. Why Other Options Are Wrong:- 62 km/h, 64 km/h, and 63 km/h do not equal the true average of 45 km/h and 75 km/h.- Each of these values would require different base speeds or ratios and would therefore contradict the given ratio and lorry speed data. Common Pitfalls:Some test takers mistakenly try to use weighted averages or combine distances, but the question only asks for the numerical average of two speeds. Another pitfall is misreading the ratio or miscalculating the value of k. Always carefully connect the ratio with the actual speed of one of the vehicles and then scale correctly. Final Answer:The average speed of the bike and the van together is 60 km/h.

Question 13

Two trains 85 m and 75 m long are running in the same direction on parallel tracks with speeds of 62 km/h and 44 km/h respectively. In how many seconds will the faster train completely cross the slower train?

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Introduction / Context:This problem deals with relative speed of two trains moving in the same direction on parallel tracks. When one train overtakes another, it must cover the combined length of both trains relative to the other train. The question checks your understanding of effective relative speed, conversion of units, and time calculation when two moving objects are involved. Given Data / Assumptions:- Length of the first (faster) train = 85 m.- Length of the second (slower) train = 75 m.- Speed of the first train = 62 km/h.- Speed of the second train = 44 km/h.- Both trains move in the same direction on parallel tracks at constant speeds.- We need the time for the faster train to completely cross the slower one, in seconds. Concept / Approach:When two trains move in the same direction, the relative speed is the difference of their speeds. For the faster train to fully overtake the slower train, it must cover a distance equal to the sum of both train lengths relative to the slower train. We convert the relative speed from km/h to m/s and then apply time = distance / speed to get the result in seconds. Step-by-Step Solution:Step 1: Compute the relative speed.Relative speed = 62 km/h - 44 km/h = 18 km/h.Step 2: Convert 18 km/h to m/s.Speed in m/s = 18 * (5/18) = 5 m/s.Step 3: Compute the total effective distance to be covered for overtaking.Total length = 85 m + 75 m = 160 m.Step 4: Use time = distance / speed.Time = 160 m / 5 m/s = 32 seconds. Verification / Alternative check:An approximate check confirms the result. At 5 m/s, in 10 seconds the relative distance covered is 50 m, in 20 seconds it is 100 m, and in 30 seconds it is 150 m. To cover 160 m, it should take just a little more than 30 seconds, which matches exactly with 32 seconds from the accurate calculation. Therefore, 32 seconds is consistent and correct. Why Other Options Are Wrong:- 22 seconds and 42 seconds would correspond to distances of 110 m and 210 m at 5 m/s, which do not match the 160 m needed.- 52 seconds would correspond to 260 m at 5 m/s, which is longer than required.Only 32 seconds is consistent with the correct relative speed and combined length of the trains. Common Pitfalls:It is easy to forget to add both train lengths when calculating the overtaking distance. Another frequent error is using the sum of the speeds instead of their difference when the trains move in the same direction. Also, using km/h directly without converting to m/s leads to incorrect time values in seconds. Always handle units and relative speed carefully. Final Answer:The faster train will completely cross the slower train in 32 seconds.

Question 14

Two trains start simultaneously from the same place but move along straight tracks at right angles to each other. Their speeds are 36 km/h and 48 km/h respectively. What will be the distance between them (in metres) after 30 seconds?

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Introduction / Context:This problem combines time and distance concepts with basic geometry. Two trains move away from the same starting point in perpendicular directions. The question asks for the distance between them after a certain time interval. This requires converting speeds, computing distances travelled along each track, and then using the Pythagorean theorem to find the straight line distance between the two trains. Given Data / Assumptions:- Speed of the first train = 36 km/h.- Speed of the second train = 48 km/h.- They start at the same time from the same point and move along tracks at right angles (forming a 90 degree angle between their paths).- Time elapsed = 30 seconds.- We need the distance between the trains after 30 seconds, in metres. Concept / Approach:We first convert the speeds from km/h to m/s, because the time is in seconds and the answer is required in metres. Then we calculate how far each train travels in 30 seconds. Since the trains move along perpendicular directions, the distance between them is the hypotenuse of a right-angled triangle whose legs are the distances travelled by each train. We then apply the Pythagorean theorem: hypotenuse^2 = leg1^2 + leg2^2. Step-by-Step Solution:Step 1: Convert speeds to m/s.First train: 36 km/h = 36 * (5/18) = 10 m/s.Second train: 48 km/h = 48 * (5/18) = 40/3 m/s ≈ 13.33 m/s.Step 2: Compute distances in 30 seconds.First train distance = 10 m/s * 30 s = 300 m.Second train distance = (40/3) m/s * 30 s = 40 * 10 = 400 m.Step 3: Use the Pythagorean theorem for the right-angled triangle.Let d be the distance between the trains.Then d^2 = 300^2 + 400^2 = 90000 + 160000 = 250000.So d = square root of 250000 = 500 m. Verification / Alternative check:The 3-4-5 triangle pattern appears here: the legs are in the ratio 3 : 4 (300 m and 400 m), and the hypotenuse is 5 parts, which gives 500 m. This is a well-known Pythagorean triple, so the calculation is clearly consistent and provides a quick mental check that the answer is correct. Why Other Options Are Wrong:- 300 m and 400 m are the individual distances travelled by each train but not the distance between them.- 900 m is much larger and would require both trains to travel much farther than they actually do in 30 seconds at the given speeds.- 250 m is too small, incorrect under the Pythagorean relation for 300 m and 400 m legs. Common Pitfalls:A frequent mistake is to add or subtract the distances directly instead of using the Pythagorean theorem when the paths are at right angles. Another common error is forgetting to convert speeds to m/s, which leads to unit inconsistencies. Always be careful with units and geometry when dealing with motion in two perpendicular directions. Final Answer:The distance between the two trains after 30 seconds is 500 m.

Question 15

P is faster than Q. P and Q each walk 24 km. The sum of their speeds is 7 km/h and the sum of the times taken by them is 14 hours. What is the speed of P (in km/h)?

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Introduction / Context:This problem involves two people walking the same distance at different speeds. You are given the sum of their speeds and the sum of the times taken by them and asked to determine the speed of the faster walker. The question tests algebraic setup using time = distance / speed and solving simultaneous equations, which is a very common pattern in aptitude exams. Given Data / Assumptions:- Distance walked by P = 24 km.- Distance walked by Q = 24 km.- Let speeds of P and Q be v_p and v_q km/h respectively.- v_p + v_q = 7 km/h.- Sum of times taken: (24 / v_p) + (24 / v_q) = 14 hours.- P is faster than Q, so v_p > v_q. Concept / Approach:We use the relationship time = distance / speed for each person and combine the information into two equations: one for the sum of speeds and another for the sum of times. Solving this system of equations will give both speeds. Then we use the condition that P is faster to identify P's speed. This is a straightforward application of simultaneous equations in algebra connected with time and distance. Step-by-Step Solution:Step 1: Let v_p be the speed of P and v_q be the speed of Q.We are given v_p + v_q = 7. (Equation 1)Step 2: Times taken are 24 / v_p and 24 / v_q hours.Sum of times is 14 hours:24 / v_p + 24 / v_q = 14. (Equation 2)Step 3: Divide Equation 2 by 24 to simplify:1 / v_p + 1 / v_q = 14 / 24 = 7 / 12.Step 4: Use the identity 1 / v_p + 1 / v_q = (v_p + v_q) / (v_p * v_q).So (v_p + v_q) / (v_p * v_q) = 7 / 12.Step 5: Substitute v_p + v_q = 7 into this:7 / (v_p * v_q) = 7 / 12 ⇒ v_p * v_q = 12.Step 6: Now we have a system: v_p + v_q = 7 and v_p * v_q = 12.Step 7: Solve the quadratic t^2 - 7t + 12 = 0 for t, whose roots are v_p and v_q.This factors as (t - 3)(t - 4) = 0, so t = 3 or t = 4.Step 8: Since P is faster, P's speed v_p = 4 km/h and Q's speed v_q = 3 km/h. Verification / Alternative check:Check the data with v_p = 4 km/h and v_q = 3 km/h. Sum of speeds = 4 + 3 = 7 km/h, which matches the given condition. Time for P = 24 / 4 = 6 hours. Time for Q = 24 / 3 = 8 hours. Sum of times = 6 + 8 = 14 hours, which also matches the given condition. Therefore the values are correct. Why Other Options Are Wrong:- 3 km/h would assign the faster speed incorrectly to P, contradicting the statement that P is faster than Q.- 5 km/h and 6 km/h do not satisfy both the sum of speeds and the sum of times simultaneously when you attempt to find a partner speed v_q that fits all conditions. Common Pitfalls:Some candidates misinterpret the average concepts or attempt to compute an average speed instead of using the sum of speeds and times correctly. Others may mistakenly treat 14 hours as the time taken by one person rather than the sum. Carefully translating the verbal description into precise equations and then using algebraic techniques avoids such errors. Final Answer:The speed of P is 4 km/h.

Question 16

Karthik can cover a distance of 200 km in 22 days while resting for 2 hours each day. How many days (approximately) will he take to cover a distance of 250 km if he still rests for 2 hours each day but moves at two thirds of his previous speed?

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Introduction / Context:This question explores the relationship between speed, distance, and time in a multi-day travel context. The person travels a certain distance in a given number of days, and then we consider a new situation where his speed is reduced but his daily rest pattern remains the same. The problem asks for the approximate number of days needed to cover a larger distance under the slower speed condition. It focuses on proportional reasoning rather than exact daily scheduling. Given Data / Assumptions:- Karthik covers 200 km in 22 days with 2 hours of rest per day.- In the new situation, he still rests for 2 hours per day, so his effective daily travel time pattern is similar. - His new speed is two thirds of his previous effective travel speed. - We want to know how many days he will take to cover 250 km at the reduced speed with the same daily rest habit. - We treat his effective travel hours per day as the same in both scenarios, so daily distance is directly proportional to speed. Concept / Approach:Because Karthik rests for the same amount of time each day in both scenarios, his actual travel hours per day are the same. Thus, the distance he covers per day is proportional to his speed. We first calculate his original average distance per day from the 200 km in 22 days. Then, we reduce that by the factor two thirds to reflect the slower speed and compute how many days would be required to cover 250 km at the new daily distance rate. Step-by-Step Solution:Step 1: Compute Karthik's original average distance per day.Average distance per day originally = 200 km / 22 days ≈ 9.09 km per day.Step 2: His new speed is two thirds of the old speed, so his new daily distance is also two thirds of the old daily distance (same travel hours per day).New daily distance ≈ (2/3) * 9.09 ≈ 6.06 km per day.Step 3: We now need 250 km to be covered at approximately 6.06 km per day.Number of days ≈ 250 / 6.06.Step 4: Compute the division: 250 / 6.06 is approximately 41.25 days. Verification / Alternative check:Another way to think about it is to use the ratio of speeds and distances. Original speed is some value S, new speed is (2/3)S. To travel 200 km at speed S takes 22 days. To travel 250 km at speed (2/3)S would multiply the time by both 250/200 (for increased distance) and 3/2 (for the slower speed). So new days = 22 * (250/200) * (3/2) = 22 * (5/4) * (3/2) = 22 * (15/8) = (22 * 15) / 8 = 330 / 8 = 41.25 days, which matches our previous calculation exactly. Why Other Options Are Wrong:- 37.5 days and 39.75 days are too small and would imply a higher effective speed than two thirds of the original when checked via proportional reasoning.- 40 days is also smaller than the correctly computed 41.25 days and does not satisfy the precise ratio calculations. Common Pitfalls:Some candidates try to handle the daily rest hours explicitly and overcomplicate the problem. However, because the rest period is unchanged in both scenarios, it cancels out when comparing daily effective distances. Another mistake is to apply the two thirds factor to the number of days instead of to the speed or daily distance. Always apply the speed change factor to the distance per day, not directly to time, unless you systematically invert it. Final Answer:Karthik will take approximately 41.25 days to cover 250 km at two thirds of his original speed.

Question 17

A 200 m long train passes a motorcyclist running in the same direction at 12 km/h in 15 seconds and also passes a jeep travelling in the same direction in 20 seconds. Assuming the lengths of both the motorcyclist and the jeep are negligible, at what…

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Introduction / Context:This question involves a train overtaking two different vehicles (a motorcyclist and a jeep) in the same direction. From the time taken to pass each, we infer relative speeds. First, we use the time taken to pass the motorcyclist to find the speed of the train. Then, we use the passing time with the jeep to determine the jeep's speed. This type of problem tests your understanding of relative motion and proper handling of units. Given Data / Assumptions:- Length of the train = 200 m.- The train and both vehicles move in the same direction in a straight line.- Speed of the motorcyclist = 12 km/h.- The train completely passes the motorcyclist in 15 seconds.- The train completely passes the jeep in 20 seconds.- Lengths of the motorcyclist and the jeep are negligible compared to the train, so only the train length is used as the passing distance. Concept / Approach:When the train overtakes another moving object in the same direction, the relevant speed is the relative speed, which is the difference between the train's speed and the other object's speed. Time taken to pass is distance divided by relative speed. We express speeds in m/s when working with distance in metres and time in seconds. First, we use the motorcyclist data to find the train speed. Then, using the jeep data and the known train speed, we find the jeep speed. Finally, we convert the jeep speed back into km/h and compare with the options. Step-by-Step Solution:Step 1: Find the relative speed between the train and the motorcyclist.Distance to be covered for passing = 200 m.Time taken = 15 seconds.Relative speed (train minus motorcyclist) in m/s = 200 / 15 ≈ 13.33 m/s.Step 2: Convert relative speed to km/h.Speed in km/h = 13.33 * (18/5) ≈ 48 km/h.So train speed - motorcyclist speed = 48 km/h.Given motorcyclist speed = 12 km/h, so train speed = 48 + 12 = 60 km/h.Step 3: Use the jeep data to find jeep speed.Again, passing distance = 200 m.Time taken to pass jeep = 20 seconds.Relative speed between train and jeep in m/s = 200 / 20 = 10 m/s.Convert 10 m/s to km/h: 10 * (18/5) = 36 km/h.Thus, train speed - jeep speed = 36 km/h.We already know train speed = 60 km/h, so jeep speed = 60 - 36 = 24 km/h. Verification / Alternative check:Check both passing times with the derived speeds. With train speed 60 km/h and motorcyclist speed 12 km/h, relative speed = 48 km/h. In m/s, that is 48 * (5/18) ≈ 13.33 m/s. Time to cover 200 m at 13.33 m/s is 200 / 13.33 ≈ 15 seconds, which matches the first condition. With jeep speed 24 km/h, relative speed with train is 60 - 24 = 36 km/h = 36 * (5/18) = 10 m/s, so time to cover 200 m is 200 / 10 = 20 seconds, matching the second condition. Why Other Options Are Wrong:- 36 km/h, 32 km/h, and 29 km/h do not fit both passing-time conditions when you recompute the relative speeds and times.- Only 24 km/h makes the 15 second and 20 second overtaking times both consistent with a single train speed of 60 km/h. Common Pitfalls:One common mistake is to forget that overtaking distance is just the length of the train when the other object's length is negligible. Another is to add speeds instead of subtracting them when vehicles move in the same direction. Also, mixing km/h and m/s without proper conversion often leads to wrong answers. Always convert to m/s when working with metres and seconds and back to km/h at the end if needed. Final Answer:The jeep is travelling at 24 km/h.

Question 18

Kamal runs 240 meters every day and increases his run to 400 meters on each Saturday. Over a continuous period of four weeks, how many kilometers does he run in total?

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Introduction / Context:This question is a basic time and distance problem that checks whether a student can convert meters into kilometers and correctly account for different distances run on different days of the week. Such questions are very common in aptitude and competitive exams to test comfort with unit conversion and simple arithmetic over multiple days. Given Data / Assumptions: Kamal runs 240 meters every day from Sunday to Friday. On every Saturday, he runs 400 meters instead of 240 meters. We consider a continuous period of four weeks, which is 4 * 7 = 28 days. There are exactly four Saturdays in these four weeks. We need the total distance in kilometers. Concept / Approach:The key idea is to separate the days into two groups: ordinary days with 240 meters and Saturdays with 400 meters. We multiply the daily distance by the number of corresponding days, add the results to get the total distance in meters, and finally convert meters into kilometers using the relation 1000 meters = 1 kilometer. This direct approach avoids any unnecessary complexity and is ideal for quick calculations in an exam setting. Step-by-Step Solution:Number of weeks = 4 Total days = 4 * 7 = 28 days Number of Saturdays in four weeks = 4 Number of non-Saturday days = 28 - 4 = 24 days Distance on each non-Saturday day = 240 meters Distance on each Saturday = 400 meters Total distance on non-Saturdays = 24 * 240 = 5760 meters Total distance on Saturdays = 4 * 400 = 1600 meters Total distance in meters = 5760 + 1600 = 7360 meters Convert to kilometers: distance in km = 7360 / 1000 = 7.36 km Verification / Alternative check:A quick mental check is useful. If Kamal ran 240 meters every day for 28 days, that would be 28 * 240 = 6720 meters. Because he actually runs an extra (400 - 240) = 160 meters on each of the four Saturdays, the additional distance is 4 * 160 = 640 meters. So the true total is 6720 + 640 = 7360 meters, which again converts to 7.36 kilometers, confirming our earlier calculation. Why Other Options Are Wrong:5.75 km underestimates the total distance and ignores either some days or the extra Saturday distance. 8.20 km and 6.98 km are overestimates or underestimates that do not match the exact multiplication and addition described above. The value 7.04 km is close but still incorrect because it does not correspond to any correct combination of 240 meter and 400 meter runs over four weeks. Common Pitfalls:Students often make two common mistakes: forgetting that four weeks contain four Saturdays, and mixing up meters and kilometers by not dividing by 1000 correctly. Another typical error is assuming that Kamal runs 240 meters plus 400 meters on Saturdays, instead of only 400 meters. Carefully reading the wording and systematically separating ordinary days and Saturdays helps avoid these issues. Final Answer:The total distance Kamal runs over four weeks is 7.36 km.

Question 19

Two different trains each cover the same distance of 180 km. The first train takes 5 hours, while the second train takes 1 hour less. If both trains are moving in the same direction, what is the difference between the distances they each cover in on…

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Introduction / Context:This question tests the basic relation between distance, speed, and time, and specifically focuses on interpreting speed as the distance covered in one hour. By comparing two trains that cover the same total distance in different times, we can directly compare their speeds and hence the distance each covers per hour. This type of problem is common in aptitude tests because it checks both conceptual understanding and quick calculation skills. Given Data / Assumptions: Distance covered by each train = 180 km. Train 1 takes 5 hours to cover 180 km. Train 2 takes 1 hour less, so it takes 4 hours to cover 180 km. We need the difference in distance covered by the two trains in one hour. Both trains are assumed to move at constant speeds. Concept / Approach:Speed is defined as distance divided by time. When the total distance is fixed and time changes, the speed changes accordingly. Since the question asks about the distances covered in one hour, we first calculate the speed of each train in km/h and then subtract the smaller speed from the larger speed. The result directly gives the required difference in distance covered in one hour because distance in one hour equals speed in km/h. Step-by-Step Solution:Speed of Train 1 = distance / time = 180 / 5 km/h Speed of Train 1 = 36 km/h Speed of Train 2 = distance / time = 180 / 4 km/h Speed of Train 2 = 45 km/h Difference in speeds = 45 - 36 km/h Difference in speeds = 9 km/h Since speed in km/h equals distance traveled in one hour, the difference in distance covered in one hour is 9 km. Verification / Alternative check:As a quick check, imagine observing both trains for exactly 1 hour. In that time, Train 1 will go 36 km, and Train 2 will go 45 km. The gap between them would then be 45 - 36 = 9 km. This matches our previous calculation and confirms that the answer is correct. Using simple mental arithmetic or checking by ratio of times also leads to the same result. Why Other Options Are Wrong:15 km and 18 km correspond to much larger speed differences than those produced by changing the travel time from 5 hours to 4 hours for a 180 km journey. The value 6 km is too small and does not match the correct difference in speeds. The option 12 km also arises from incorrect manipulation of the ratio or distance and time values. Only 9 km matches the correct calculation. Common Pitfalls:Many students mistakenly subtract the times (5 hours and 4 hours) and misinterpret that difference directly as a distance. Others compute average speeds incorrectly by mixing up the formulas or doing mental math too quickly. Carefully applying the formula speed = distance / time and then interpreting speed as distance in one hour helps avoid such errors. Final Answer:The difference in distance covered by the two trains in one hour is 9 km.

Question 20

A goods truck leaves a town and travels at a constant unknown speed. Six hours later, a second truck leaves the same town in the same direction at 90 km/h. If the second truck catches up with the goods truck after 4 hours of its own journey, what is…

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Introduction / Context:This problem is a classic example of a relative speed and catch up situation. One vehicle starts first at an unknown speed, and a faster vehicle starts later and eventually catches up. Such questions are frequently used in aptitude tests to check the understanding of distance equals speed multiplied by time and how to handle different starting times. Given Data / Assumptions: The goods truck starts from a town at time zero and travels at a constant speed v km/h. The second truck starts 6 hours later from the same point. The speed of the second truck is 90 km/h. The second truck catches up with the goods truck after 4 hours of its own travel. Both trucks move in the same direction along the same route. Concept / Approach:When one vehicle catches another, the distances traveled by both vehicles from the starting point are the same at the moment of catching up. Therefore, we can equate the distance expressions of both trucks. The goods truck has more travel time, while the chasing truck has a higher speed. The equation distance = speed * time is used for both, and we solve for the unknown speed v of the goods truck. Step-by-Step Solution:Let v be the speed of the goods truck in km/h. Time traveled by goods truck when caught = 6 + 4 = 10 hours. Distance traveled by goods truck = v * 10 km. Time traveled by faster truck = 4 hours. Distance traveled by faster truck = 90 * 4 = 360 km. At catch up, distances are equal, so v * 10 = 360. Solve for v: v = 360 / 10 = 36 km/h. Verification / Alternative check:We can verify the answer by simulating the motion. The goods truck travels for 10 hours at 36 km/h, so distance = 36 * 10 = 360 km. The faster truck travels 4 hours at 90 km/h, also covering 360 km. Since both distances are equal, the catch up condition is satisfied, which confirms that the speed of the goods truck is indeed 36 km/h. Why Other Options Are Wrong:Speeds like 32 km/h, 28 km/h, 34 km/h, and 40 km/h do not satisfy the equality of distances at the catch up time. If we substitute any of these values into distance = speed * time for the goods truck, the resulting distance would not be 360 km after 10 hours, and the trucks would not meet at the same position after the given times. Therefore, these options are incorrect. Common Pitfalls:A common mistake is using the relative speed directly (90 - v) without accounting properly for the 6 hour head start, or forgetting to add the initial 6 hours to the goods truck travel time. Another frequent error is to assume equal times instead of equal distances at the moment of meeting. Always remember that at the catch up point, distances from the starting location are equal for both vehicles. Final Answer:The speed of the goods truck is 36 km/h.

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