Question 1

Kiran swims in a 90 m long pool. He covers a total distance of 360 m in 2 minutes by swimming from one end to the other and back along the same straight path twice. What is Kiran’s average speed (in m/s)?

Accepted Answer

Introduction / Context:This question is about average speed in a swimming pool. Kiran swims back and forth along a straight path in a 90 m pool, making multiple lengths, and we are given his total distance and total time. We must compute his average speed in metres per second. Such problems are designed to reinforce the basic definition of average speed and to ensure that students can handle unit conversions between minutes and seconds correctly. Given Data / Assumptions:- Length of the pool = 90 m.- Kiran swims from one end to the other and back along the same straight path twice, covering 360 m total.- Total time taken = 2 minutes.- Speeds are taken as uniform for the purpose of computing average speed, even though real swimmers may vary speed slightly. Concept / Approach:The key concept is the definition of average speed:average speed = total distance / total time.Here, the total distance is given directly as 360 m. However, the time is given in minutes, while the desired unit for speed is metres per second, so we need to convert 2 minutes into seconds first. Once both total distance and total time are in compatible units, we can compute a single ratio to get the average speed. Step-by-Step Solution:Step 1: Confirm the total distance.Each full length of the pool is 90 m. Going from one end to the other and back is 2 * 90 = 180 m.Doing this twice gives 2 * 180 = 360 m, which matches the statement.Step 2: Convert total time into seconds.Total time = 2 minutes = 2 * 60 = 120 seconds.Step 3: Apply the average speed formula.Average speed = total distance / total time = 360 m / 120 s = 3 m/s. Verification / Alternative check:We can think: if Kiran swims at 3 m/s, then in 120 seconds he would cover 3 * 120 = 360 m, exactly matching the total distance stated. If his speed were 4 m/s, he would cover 480 m in 2 minutes, which is too much, and if it were 2 m/s, he would cover only 240 m, which is too little. Therefore, 3 m/s is the only value consistent with the given numbers. Why Other Options Are Wrong:- 4 m/s would imply a total distance of 480 m in 2 minutes, greater than the 360 m given.- 6 m/s would correspond to 720 m in 2 minutes, which is clearly impossible for this problem setup.- 8 m/s is even higher and unrealistically large for both the numbers and a typical swimming speed. Common Pitfalls:One common mistake is to forget to convert minutes into seconds, leading to a speed computed as metres per minute instead of metres per second. Another pitfall is to misunderstand the description of the back-and-forth motions and miscalculate the total distance. Carefully counting the number of lengths and converting time units correctly ensures an accurate solution. Final Answer:Kiran’s average speed is 3 m/s.

Question 2

Sharmi starts from Pune and Ritesh starts from Mumbai to meet each other. The speed of Sharmi is 37 km/h and the speed of Ritesh is 41 km/h. If the distance between Pune and Mumbai is 234 km and both start their journeys at the same time, after how …

Accepted Answer

Introduction / Context:This is a classic “two people moving towards each other” time and distance problem. Sharmi and Ritesh start from two different cities, Pune and Mumbai, at the same time and travel towards each other along the same straight route. We are given their speeds and the distance between the cities, and we must determine the time after the start at which they will meet. This concept is widely used in aptitude tests to check understanding of relative speed. Given Data / Assumptions:- Distance between Pune and Mumbai = 234 km.- Speed of Sharmi = 37 km/h.- Speed of Ritesh = 41 km/h.- Both start at the same time and move directly towards each other along a straight path.- Their speeds remain constant throughout the motion. Concept / Approach:When two objects move directly towards each other on a straight line, their relative speed is the sum of their individual speeds. The total distance between them initially is 234 km, and they close this gap at the rate of (37 + 41) km/h. The time taken to meet is simply the initial distance divided by this combined speed. This method avoids tracking each traveller separately because their combined progress towards reducing the distance is what matters. Step-by-Step Solution:Step 1: Compute the relative speed.Relative speed = 37 km/h + 41 km/h = 78 km/h.Step 2: Use the formula time = distance / speed.Time to meet = 234 km / 78 km/h.Step 3: Simplify 234 / 78.Both numerator and denominator are divisible by 78, so 234 / 78 = 3 hours. Verification / Alternative check:In 3 hours, Sharmi travels 37 * 3 = 111 km and Ritesh travels 41 * 3 = 123 km. The sum of these distances is 111 + 123 = 234 km, exactly matching the distance between Pune and Mumbai. This means that after 3 hours they will meet somewhere in between the two cities. Since the numbers match perfectly, our result is consistent with the problem conditions. Why Other Options Are Wrong:- 2.5 hours would give a total closed distance of 78 * 2.5 = 195 km, which is less than 234 km, so they would not have met yet.- 4 hours would give 78 * 4 = 312 km, which exceeds the actual distance, so this cannot be the meeting time.- 3.5 hours would give 78 * 3.5 = 273 km, again exceeding the initial separation. Common Pitfalls:Some candidates mistakenly subtract speeds instead of adding them when the objects move towards each other. Others may calculate the distances for each person separately and then struggle to coordinate the time. Remember that when objects approach each other on a straight line, you always add their speeds to get the relative speed. Using this idea keeps the calculation short and reliable. Final Answer:They will meet 3 hours after they start their journeys.

Question 3

Walking at a speed of 10 km/h, Madhu reaches her university 15 minutes late. Next time she increases her walking speed to 12 km/h but is still late by 5 minutes. What is the distance of the university from her house (in km)?

Accepted Answer

Introduction / Context:This problem links time, speed, and punctuality. Madhu walks to her university at two different speeds on two different days, and we are told how late she is in each case. From this, we must deduce the distance between her home and the university. The question tests your ability to model lateness or earliness as a difference between actual travel time and scheduled (on-time) travel time, and then use algebra to solve for distance. Given Data / Assumptions:- Speed on the first day = 10 km/h; she arrives 15 minutes late.- Speed on the second day = 12 km/h; she arrives 5 minutes late.- Let the scheduled (on-time) travel time be T hours.- Let the distance between her home and the university be d km.- 15 minutes is 15/60 = 0.25 hours; 5 minutes is 5/60 = 1/12 hours.- Madhu walks at constant speed on a fixed, straight route. Concept / Approach:We express the actual travel times using time = distance / speed. On each day, this actual time is equal to the scheduled time plus a certain lateness. This gives us two equations linking d and T. Subtracting one equation from the other eliminates T and allows us to solve for d directly. Once we have the distance, we can check that both lateness conditions are satisfied. Step-by-Step Solution:Step 1: Let the exact on-time journey time be T hours.On the first day: d / 10 = T + 0.25.On the second day: d / 12 = T + 1/12.Step 2: Subtract the second equation from the first:d / 10 - d / 12 = (T + 0.25) - (T + 1/12).Left side: (1/10 - 1/12) * d = (6/60 - 5/60) * d = (1/60) * d.Right side: 0.25 - 1/12 = 1/4 - 1/12 = 3/12 - 1/12 = 2/12 = 1/6.So d / 60 = 1/6 ⇒ d = 60 * 1/6 = 10 km. Verification / Alternative check:With d = 10 km, first day time at 10 km/h is 10 / 10 = 1 hour. If she should have arrived in T hours, then T = 1 - 0.25 = 0.75 hours (45 minutes). Second day time at 12 km/h is 10 / 12 ≈ 0.8333 hours (50 minutes). Compared to the on-time 0.75 hours, she is now 0.0833 hours late, which is 5 minutes. Both lateness values match the problem statement perfectly. Why Other Options Are Wrong:- 15 km, 20 km, or 30 km, when substituted into the two speed conditions, do not yield lateness values of exactly 15 minutes and 5 minutes. Either one or both conditions fail, so these distances cannot be correct. Common Pitfalls:Some students mistakenly assume the scheduled time is the same as one of the actual times, or they attempt to work directly with differences between speeds without explicitly writing equations. Others misconvert minutes to hours. Expressing everything in hours and setting up two clear equations is the safest path to the correct answer. Always remember that “late” means the actual travel time is longer than the scheduled time by the given amount. Final Answer:The distance of the university from Madhu’s house is 10 km.

Question 4

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at a speed of 4 km/h and partly on a bicycle at a speed of 9 km/h. What distance (in km) did he travel on foot?

Accepted Answer

Introduction / Context:This is a standard mixed-speed travel problem where a person covers part of the distance at one speed and the remaining distance at another speed. We are given the total distance, the total time, and both speeds, and we must find how much of the journey was covered at the slower speed (on foot). This problem tests your ability to set up and solve a linear equation involving two unknown partial distances or times. Given Data / Assumptions:- Total distance travelled = 61 km.- Total time taken = 9 hours.- Walking speed = 4 km/h.- Cycling speed = 9 km/h.- Let the distance travelled on foot be x km.- Then the distance travelled by bicycle is (61 - x) km.- Motion along each segment is at constant speed with no additional breaks. Concept / Approach:The total time is the sum of the time spent on foot plus the time spent on the bicycle. Time is distance divided by speed, so we can express each part of the total time in terms of x and then equate their sum to the given 9 hours. This yields a linear equation in x, which can be solved easily using basic algebraic steps. Once x is found, it directly gives the distance travelled on foot. Step-by-Step Solution:Step 1: Let distance on foot be x km.Then distance by bicycle = 61 - x km.Step 2: Express the total time using time = distance / speed.Time on foot = x / 4 hours.Time on bicycle = (61 - x) / 9 hours.Step 3: Total time is given as 9 hours, so:x / 4 + (61 - x) / 9 = 9.Step 4: Multiply through by the common denominator, 36, to clear fractions:36 * (x / 4) + 36 * ((61 - x) / 9) = 36 * 9.This simplifies to 9x + 4(61 - x) = 324.Step 5: Expand and solve: 9x + 244 - 4x = 324 ⇒ 5x + 244 = 324 ⇒ 5x = 80 ⇒ x = 16 km. Verification / Alternative check:Check the times using x = 16 km. Time on foot = 16 / 4 = 4 hours. Distance by bicycle = 61 - 16 = 45 km. Time on bicycle = 45 / 9 = 5 hours. Total time = 4 + 5 = 9 hours, exactly matching the problem statement. This confirms that 16 km on foot and 45 km by bicycle satisfy both the time and distance conditions. Why Other Options Are Wrong:- 14 km, 18 km, and 20 km, when used in the time equation, lead to total times that are not exactly 9 hours; either the total time becomes more than 9 hours or less than 9 hours.- Only 16 km yields a total time of precisely 9 hours while still adding up to a total distance of 61 km. Common Pitfalls:Some candidates confuse time and distance and try to distribute 9 hours directly in proportion to 4 km/h and 9 km/h without setting up a proper equation. Another common issue is arithmetic mistakes when clearing fractions or solving the resulting linear equation. Writing the equation step by step and checking the solution by substituting it back is an effective way to avoid and catch such errors. Final Answer:The farmer travelled 16 km on foot.

Question 5

Aman and Kapil start from Delhi and Gwalior respectively towards each other at the same time. They meet at Mathura. After meeting, Aman takes 196 minutes to reach Gwalior, while Kapil takes 225 minutes to reach Delhi. If Aman’s speed is 30 km/h, wha…

Accepted Answer

Introduction / Context:This is a more advanced relative-speed problem involving two people travelling towards each other from different cities and meeting at an intermediate point, Mathura. After meeting, they continue to the opposite cities, and we are given the times they take to complete the remaining parts of their journeys. Using this post-meeting information along with the fact that they met at the same instant, we can deduce the speed of Kapil. This problem requires careful use of distances before and after meeting and a bit of algebra. Given Data / Assumptions:- Aman starts from Delhi and Kapil from Gwalior at the same time, moving towards each other.- They meet at Mathura, an intermediate point on the straight route between Delhi and Gwalior.- Aman’s speed = 30 km/h (constant).- After meeting, Aman takes 196 minutes to reach Gwalior.- After meeting, Kapil takes 225 minutes to reach Delhi.- 196 minutes = 196/60 hours; 225 minutes = 225/60 hours.- Let Kapil’s speed be v km/h.- Distances and speeds remain constant throughout their journeys. Concept / Approach:When two travellers meet, the point of meeting splits the total route in two parts. The segment from Mathura to Gwalior is covered by Aman after meeting and by Kapil before meeting. Similarly, the segment from Mathura to Delhi is covered by Kapil after meeting and by Aman before meeting. By using the times and speeds after meeting, we can compute the lengths of these segments. Then, equating the time taken to reach Mathura for each traveller before meeting gives an equation involving v. Solving this equation yields Kapil’s speed. Step-by-Step Solution:Step 1: Compute the distance from Mathura to Gwalior.Aman covers this distance after meeting at 30 km/h in 196/60 hours.Distance Mathura–Gwalior = 30 * (196/60) = 30 * (49/15) = 98 km.Step 2: Let the distance from Mathura to Delhi be x km.Kapil covers this distance after meeting at v km/h in 225/60 hours.So x = v * (225/60) = v * (15/4).Step 3: Use the fact that before meeting, Aman covers x km to reach Mathura from Delhi and Kapil covers 98 km to reach Mathura from Gwalior.Time taken by Aman to reach Mathura: x / 30 hours.Time taken by Kapil to reach Mathura: 98 / v hours.These times must be equal because they start together and meet at the same instant: x / 30 = 98 / v.Step 4: Substitute x = (15v) / 4 into x / 30 = 98 / v.( (15v) / 4 ) / 30 = 98 / v ⇒ (15v) / 120 = 98 / v.Simplify: 15v / 120 = v / 8.So v / 8 = 98 / v ⇒ v^2 = 784 ⇒ v = 28 km/h (taking the positive root). Verification / Alternative check:With v = 28 km/h, x = (15 * 28) / 4 = 15 * 7 = 105 km from Mathura to Delhi. Aman’s time to reach Mathura originally is 105 / 30 = 3.5 hours. Kapil’s time to reach Mathura originally is 98 / 28 = 3.5 hours as well, so they meet at the same time. After meeting, Aman needs 196/60 ≈ 3.27 hours to reach Gwalior, and Kapil needs 225/60 = 3.75 hours to reach Delhi, consistent with the given values. Everything fits perfectly. Why Other Options Are Wrong:- 30 km/h would imply Kapil’s speed equals Aman’s, which destroys the asymmetry created by the different post-meeting times.- 225/7 km/h and 392/15 km/h, when substituted, do not keep the pre-meeting times equal and violate at least one of the time constraints given for travelling from Mathura to the endpoints. Common Pitfalls:Several errors commonly occur: mixing up which segment is 98 km, forgetting to convert minutes into hours, or not equating the pre-meeting travel times correctly. Some students also try to involve the total distance unnecessarily, which complicates the algebra. The cleanest path is to use post-meeting distances first, then rely on equal pre-meeting times to solve for Kapil’s speed. Final Answer:Kapil’s speed is 28 km/h.

Question 6

A train leaves Delhi at 10:00 a.m. and reaches Jaipur at 4:00 p.m. on the same day. Another train leaves Jaipur at 12:00 noon and reaches Delhi at 5:00 p.m. the same day. Approximately at what time of day will the two trains meet?

Accepted Answer

Introduction / Context:This problem involves two trains travelling between the same two cities, Delhi and Jaipur, at different departure times and with different total journey durations. We need to determine the clock time at which they meet on the way. It is a typical relative speed and time-line question frequently found in aptitude tests and requires careful handling of departure times and travel durations. Given Data / Assumptions:- Train 1 (Delhi to Jaipur) leaves Delhi at 10:00 a.m. and reaches Jaipur at 4:00 p.m., so its journey time is 6 hours.- Train 2 (Jaipur to Delhi) leaves Jaipur at 12:00 noon and reaches Delhi at 5:00 p.m., so its journey time is 5 hours.- Let the distance between Delhi and Jaipur be D km.- Both trains travel along the same straight route and at constant speeds throughout their journeys. Concept / Approach:We first find each train’s speed in terms of the unknown distance D. Then we consider a variable time t (in hours) after 10:00 a.m. when the trains meet. By that time, Train 1 will have travelled for t hours, while Train 2, which departs later at 12:00 noon, will have travelled for (t - 2) hours, as long as t is greater than 2 hours. The sum of the distances they have covered by the meeting time must equal the total distance D. This leads to an equation in t, from which we can find the meeting time and convert it to a clock time. Step-by-Step Solution:Step 1: Compute speeds in terms of D.Train 1 speed = D / 6 km/h.Train 2 speed = D / 5 km/h.Step 2: Let t be the time in hours after 10:00 a.m. when the trains meet.Then Train 1 has travelled t hours; Train 2, which starts at 12:00 noon, has travelled t - 2 hours.Step 3: Distances covered by meeting time:Train 1: (D / 6) * t.Train 2: (D / 5) * (t - 2).Step 4: Their combined distance equals D:(D / 6) * t + (D / 5) * (t - 2) = D.Divide by D (assuming D > 0): t / 6 + (t - 2) / 5 = 1.Step 5: Solve t / 6 + (t - 2) / 5 = 1.Multiply by 30: 5t + 6(t - 2) = 30 ⇒ 5t + 6t - 12 = 30 ⇒ 11t = 42 ⇒ t = 42 / 11 ≈ 3.818 hours.Step 6: Convert 0.818 hours to minutes: 0.818 * 60 ≈ 49 minutes.Thus, meeting time ≈ 10:00 a.m. + 3 hours 49 minutes = 1:49 p.m. Verification / Alternative check:At 1:49 p.m., Train 1 has been travelling for approximately 3.818 hours and Train 2 for 1.818 hours. Train 1 covers (D / 6) * 3.818 ≈ 0.636D, while Train 2 covers (D / 5) * 1.818 ≈ 0.364D. The sum is very close to D (allowing for rounding), showing consistency. The approximate nature of the options also matches the approximate nature of the calculation. Why Other Options Are Wrong:- 1:42 p.m., 1:27 p.m., and 2:04 p.m. correspond to different values of t which, when plugged into the equation, do not satisfy t / 6 + (t - 2) / 5 = 1. They either make the combined travelled distance less than D (meeting earlier) or greater than D (meeting later than 2:00 p.m.). Common Pitfalls:Common mistakes include ignoring the different start times and treating both trains as if they started at 10:00 a.m., or miscomputing the speeds by assuming a specific distance instead of leaving it symbolic as D. Another issue is working with approximate decimals too early, which can introduce rounding errors before the final step. Keeping t as a fraction until the end helps maintain accuracy. Final Answer:The two trains will meet at approximately 1:49 p.m.

Question 7

A bus starts running with an initial speed of 21 km/h, and its speed increases every hour by 3 km/h. How many hours will it take to cover a distance of 252 km?

Accepted Answer

Introduction / Context:This is a time and distance problem involving a bus whose speed does not remain constant. Instead, the bus starts at a given initial speed and increases its speed by a fixed amount every hour. We must calculate how long it takes to cover a known distance. This is essentially a question about the sum of distances covered in consecutive hours with speeds forming an arithmetic progression. Given Data / Assumptions:- Initial speed of the bus = 21 km/h in the first hour.- Every hour the speed increases by 3 km/h.- Total distance to be covered = 252 km.- The change in speed is stepwise at the end of each hour, and within each hour the speed is constant. Concept / Approach:The speed in each hour forms an arithmetic sequence: 21 km/h in the first hour, 24 km/h in the second hour, 27 km/h in the third hour, and so on, with common difference 3 km/h. The distance covered in each hour is simply the speed for that hour, because time per hour is 1 hour. Thus, the total distance covered in n hours is the sum of the first n terms of that arithmetic sequence. We use the formula for the sum of an arithmetic series and then equate it to 252 km to solve for n. Step-by-Step Solution:Step 1: Write the sequence of speeds by hour:Hour 1: 21 km/h, Hour 2: 24 km/h, Hour 3: 27 km/h, etc.Step 2: Let n be the number of hours required to cover 252 km.Speed in the k-th hour = 21 + 3(k - 1).Step 3: Distance covered in each hour = speed (since time is 1 hour per term).Total distance in n hours = sum of first n terms of the arithmetic sequence:S_n = n/2 * [2 * 21 + (n - 1) * 3].Step 4: Set S_n = 252 and solve for n:n/2 * [42 + 3(n - 1)] = 252.Simplify: 42 + 3(n - 1) = 42 + 3n - 3 = 39 + 3n.So S_n = n/2 * (39 + 3n) = (3n(13 + n)) / 2 = 252.Step 5: Multiply both sides by 2/3:n(13 + n) = 168.This gives n^2 + 13n - 168 = 0.Step 6: Factor or use quadratic formula; the equation factors as (n - 8)(n + 21) = 0, so n = 8 (positive root). Verification / Alternative check:Compute the distance covered in 8 hours explicitly: speeds are 21, 24, 27, 30, 33, 36, 39, 42 km/h. Their sum is (21 + 42) * 8 / 2 = 63 * 4 = 252 km. This confirms that the bus covers exactly 252 km in 8 hours. Therefore, 8 is the required number of hours. Why Other Options Are Wrong:- 3, 5, and 10 hours, when plugged into the arithmetic series sum formula, yield total distances different from 252 km. For example, 5 hours gives a much smaller total distance, while 10 hours gives a larger distance than 252 km. Common Pitfalls:Some students treat the speed as increasing continuously instead of in hourly steps, or they mistakenly assume the speed remains fixed at the final value. Others forget to use the arithmetic progression formula and instead try to sum incorrectly. Writing out the pattern of speeds clearly and using the series sum formula helps avoid confusion and ensures accuracy. Final Answer:The bus will take 8 hours to cover 252 km.

Question 8

A bus travels 2/5 of a total journey at its usual speed. The remaining distance is covered at 6/7 of its usual speed. Because of the reduced speed on the remaining part, it reaches its destination 50 minutes late. If the total distance is 200 km, wh…

Accepted Answer

Introduction / Context:This problem combines average speed concepts with a change in speed over different segments of a fixed journey. The bus travels part of the distance at its usual speed and the remaining part at a reduced speed, causing it to arrive late. We must compute the usual speed using the information about the delay and the distances covered at each speed. This type of question is a staple in time and distance sections of aptitude exams. Given Data / Assumptions:- Total distance = 200 km.- Usual speed of the bus = v km/h (unknown).- First 2/5 of the journey is covered at usual speed v.- Remaining 3/5 of the journey is covered at reduced speed (6/7) * v.- The bus reaches 50 minutes late due to the slower speed on the last part.- 50 minutes = 50/60 = 5/6 hours.- If the bus had travelled the whole journey at usual speed v, it would arrive on time. Concept / Approach:We first calculate the time it would have taken to cover the 200 km at the usual speed v. Then we express the actual time with mixed speeds: one time for the first 2/5 of the distance at v and another time for the remaining 3/5 at (6/7)v. The difference between the actual time and the ideal time is given as 5/6 hours. Setting up this equation allows us to solve for v. The key formula throughout is time = distance / speed. Step-by-Step Solution:Step 1: Compute the “on-time” travel time if the entire 200 km is covered at usual speed v.Ideal time T_ideal = 200 / v hours.Step 2: Compute segment distances.First part (2/5 of 200) = 80 km.Second part (3/5 of 200) = 120 km.Step 3: Time for first part at speed v: 80 / v hours.Step 4: Time for second part at speed (6/7)v: 120 / [(6/7)v] = 120 * 7 / (6v) = 140 / v hours.Step 5: Actual time T_actual = 80 / v + 140 / v = 220 / v hours.Step 6: The delay is T_actual - T_ideal = 5/6 hours.So 220 / v - 200 / v = 5/6 ⇒ (20 / v) = 5/6.Step 7: Solve for v: v = 20 * 6 / 5 = 120 / 5 = 24 km/h. Verification / Alternative check:Check the times using v = 24 km/h. Ideal time = 200 / 24 ≈ 8.333 hours (8 hours 20 minutes). Actual time: first part 80 / 24 ≈ 3.333 hours, second part at reduced speed (6/7 * 24 = 144/7 ≈ 20.571 km/h) gives 120 / (144/7) = 120 * 7 / 144 = 5.833 hours. Total actual time ≈ 3.333 + 5.833 = 9.166 hours. Difference ≈ 9.166 - 8.333 = 0.833 hours, which is exactly 5/6 hours or 50 minutes, matching the problem statement. Why Other Options Are Wrong:- The speeds 20.57 km/h, 28 km/h, and 26.52 km/h, when used in the time calculations, do not produce a delay of 50 minutes. The delay is either more or less than 5/6 hours.- Only 24 km/h leads to the exact delay specified in the question while covering 200 km. Common Pitfalls:Typical mistakes include mixing up the fractional distances (2/5 and 3/5) or incorrectly computing the reduced speed segment. Some students also forget to convert 50 minutes into hours or mistakenly treat 50 minutes as 0.5 hours instead of 5/6. Accuracy with fractions and step-by-step calculation of times makes the solution straightforward. Final Answer:The usual speed of the bus is 24 km/h.

Question 9

A person goes from point N to point P and comes back. His average speed for the whole journey is 60 km/h. If his speed while going from N to P is 40 km/h, what is his speed (in km/h) while coming back from P to N?

Accepted Answer

Introduction / Context:This problem is a classic example of average speed over a round trip when the speeds in the two directions are different. Many students incorrectly assume that the average speed is the simple mean of the two speeds, but that is only true when the times are equal, not when the distances are equal. Here, the person travels the same distance from N to P and from P back to N, with different speeds in each direction, and we are told the average for the entire journey. We must determine the return speed. Given Data / Assumptions:- One-way distance between N and P = d km (same both ways).- Speed from N to P (going) = 40 km/h.- Speed from P to N (return) = v km/h (unknown).- Average speed for the entire to-and-fro trip = 60 km/h.- Motion is along a straight path with constant speeds and no additional stops. Concept / Approach:Average speed for a round trip with equal distances is defined as total distance divided by total time. The total distance is 2d. The total time is the sum of the time taken for the outward and return journeys: d / 40 + d / v. We are told that the average speed is 60 km/h, so 60 must equal 2d divided by this total time. This provides an equation that we can solve for v. The distance d cancels out, so the actual numerical value of d is not needed. Step-by-Step Solution:Step 1: Express total distance and total time.Total distance = 2d km.Time going = d / 40 hours; time returning = d / v hours.Total time = d / 40 + d / v.Step 2: Use the formula for average speed.Average speed = total distance / total time = 60 km/h.So 60 = 2d / (d / 40 + d / v).Step 3: Simplify by cancelling d (since d > 0):60 = 2 / (1 / 40 + 1 / v).Step 4: Invert both sides after rearranging:1 / 60 = (1 / 2) * (1 / 40 + 1 / v).So 1 / 40 + 1 / v = 2 / 60 = 1 / 30.Step 5: Solve for 1 / v:1 / v = 1 / 30 - 1 / 40 = (4 / 120 - 3 / 120) = 1 / 120.Therefore v = 120 km/h. Verification / Alternative check:Assume for convenience that d = 120 km. Time going at 40 km/h = 120 / 40 = 3 hours. Time returning at 120 km/h = 120 / 120 = 1 hour. Total distance = 240 km, total time = 4 hours. Average speed = 240 / 4 = 60 km/h, which matches the given average. This confirms that 120 km/h is correct for the return speed. Why Other Options Are Wrong:- 80 km/h or 100 km/h, when used as return speeds, give total times that produce average speeds different from 60 km/h.- 140 km/h is larger than necessary and would yield an average speed greater than 60 km/h for any fixed distance d. Common Pitfalls:A very common error is to compute the average speed as (40 + v) / 2 = 60, which leads to v = 80 km/h. This misapplies the arithmetic mean, which is not appropriate when distances, not times, are fixed. Always return to the fundamental definition of average speed as total distance divided by total time, especially for round-trip problems. Final Answer:The person’s speed while coming back from P to N is 120 km/h.

Question 10

A starts from a point with a speed of 30 metres/second. After 3 seconds, B starts chasing A from the same point with a speed of 50 metres/second. What will be the total distance (in metres) travelled by A and B together before A is caught by B?

Accepted Answer

Introduction / Context:This is a relative speed and catch-up problem in metres and seconds. One runner, A, starts first, and after a delay B starts from the same point with a higher speed. Because B is faster, he will eventually catch up with A. The question asks for the total distance travelled by both runners combined up to the moment of catching, not just the distance from the starting point to the meeting point. This twist requires careful thinking after solving for the catch-up time. Given Data / Assumptions:- Speed of A = 30 m/s.- Speed of B = 50 m/s.- A starts at t = 0 seconds from a point.- B starts 3 seconds later from the same point, at t = 3 seconds.- Both run along the same straight line in the same direction with constant speeds.- We need the sum of the distances A and B have individually travelled at the instant B catches A. Concept / Approach:First, we find the time at which B catches A. Let t be the time (in seconds) from A’s start until they meet. During this time, A runs for t seconds and B runs for (t - 3) seconds. They are at the same position at the catch-up moment, so their distances from the starting point are equal. This gives an equation in t. Once we solve for t, we can compute the distance each has travelled from the start and then add those two distances to get the total distance travelled by both runners together. Step-by-Step Solution:Step 1: Let t be the time in seconds from A’s start until B catches A.Distance covered by A in time t: d_A = 30 * t metres.Distance covered by B in time t - 3: d_B = 50 * (t - 3) metres.Step 2: At the catch-up moment, d_A = d_B.So 30 * t = 50 * (t - 3).Step 3: Solve for t: 30t = 50t - 150 ⇒ 20t = 150 ⇒ t = 150 / 20 = 7.5 seconds.Step 4: Compute the distance from the start to the catch point.d = 30 * 7.5 = 225 metres (A’s distance; B’s will match).Step 5: Total distance travelled by both runners together is d_A + d_B = 225 + 225 = 450 metres. Verification / Alternative check:Check B’s distance using t = 7.5 seconds. B’s running time is t - 3 = 4.5 seconds, so d_B = 50 * 4.5 = 225 metres. This matches A’s distance, confirming the catch-up point. The total combined distance of 450 metres is therefore just twice 225 metres, consistent with both having reached the same spot at the same time. Why Other Options Are Wrong:- 360 metres, 600 metres, and 720 metres do not correspond to 2 * 225 metres. To reach these totals, the runners would have to meet at distances different from 225 metres or at different times, which would contradict the equality of path lengths at the catch-up moment. Common Pitfalls:Many students mistakenly give 225 metres as the answer, thinking only about the distance from the origin to the meeting point. However, the question explicitly asks for the total distance travelled by both A and B, so we must add the two individual distances. Another common error is forgetting that B runs for (t - 3) seconds, not t seconds, or mixing up speeds and times when solving the equation. Final Answer:The total distance travelled by A and B before A is caught is 450 metres.

Question 11

After repairing, a scooter runs at a speed of 54 km/h, and before repairing it runs at a speed of 48 km/h. It covers a certain distance in 6 hours after repairing. How much time will it take to cover the same distance before repairing?

Accepted Answer

Introduction / Context:This is a speed and time comparison problem. The scooter’s speed increases after repair, and we know how long it takes to travel a certain fixed distance at the higher speed. We are asked how long it would take to travel the same distance at the lower pre-repair speed. This type of question reinforces the relationship between speed, distance, and time and the idea that for a fixed distance, time is inversely proportional to speed. Given Data / Assumptions:- Speed after repairing = 54 km/h.- Speed before repairing = 48 km/h.- Time taken after repairing to cover a certain distance = 6 hours.- Let the fixed distance be D km.- The road and driving conditions are assumed to be the same in both cases, with constant speeds over the journey. Concept / Approach:First, we determine the actual distance D covered at 54 km/h in 6 hours using D = speed * time. Then, using this same distance and the lower speed of 48 km/h, we find the required time using time = distance / speed. Finally, we convert the resulting decimal hours into hours and minutes to match the answer options. Because the distance is identical in both scenarios, any changes in time are purely due to the difference in speed. Step-by-Step Solution:Step 1: Find the distance D using the post-repair speed.D = 54 km/h * 6 h = 324 km.Step 2: Compute the time needed at the pre-repair speed (48 km/h).Time_before = D / 48 = 324 / 48 hours.Step 3: Simplify 324 / 48.Divide numerator and denominator by 12: 324 / 48 = 27 / 4 hours.Step 4: Convert 27 / 4 hours to hours and minutes.27 / 4 = 6.75 hours = 6 hours + 0.75 hours.0.75 hours = 0.75 * 60 = 45 minutes.So time_before = 6 hours 45 minutes. Verification / Alternative check:Check consistency by reversing the logic. If the scooter travels 324 km in 6 hours, the speed is 324 / 6 = 54 km/h, which matches the given after-repair speed. At 48 km/h, over 6 hours 45 minutes (which is 6.75 hours), the distance covered is 48 * 6.75 = 324 km, which again matches. Both directions confirm that the computations are correct. Why Other Options Are Wrong:- 6 hours 15 minutes corresponds to 6.25 hours, giving a distance of 48 * 6.25 = 300 km, too short.- 7 hours corresponds to 48 * 7 = 336 km, more than 324 km, so it does not represent the same distance.- 7 hours 30 minutes is 7.5 hours, giving 48 * 7.5 = 360 km, again larger than the distance covered after repair. Common Pitfalls:A common mistake is to try to scale the time directly in proportion to the ratio of speeds without checking the exact arithmetic. While time does scale inversely with speed for a fixed distance, working through the distance first and then recomputing time usually reduces errors. Another pitfall is mishandling decimal hours when converting to minutes; always multiply the fractional part by 60 to get minutes. Final Answer:The scooter will take 6 hours 45 minutes to cover the same distance before repairing.

Question 12

A runner starts from a point at 6:00 a.m. with a speed of 8 km/h. Another racer starts from the same point at 8:30 a.m. in the same direction with a speed of 10 km/h. At what time of the day (in p.m.) will the second racer overtake the first runner?

Accepted Answer

Introduction / Context:This question is a classic catch-up problem in which one runner starts earlier at a slower speed, and a second racer starts later but runs faster along the same line and in the same direction. The key requirement is to determine when the faster racer catches up with the slower runner. This situation is common in relative speed questions and tests your understanding of head start and relative speed concepts. Given Data / Assumptions:- First runner starts at 6:00 a.m. with speed 8 km/h.- Second racer starts at 8:30 a.m. with speed 10 km/h.- Both move along the same straight route in the same direction at constant speeds.- We need the clock time (in p.m.) when the second racer overtakes the first runner. Concept / Approach:Because the first runner starts earlier, he gains a head start distance by the time the second racer begins. After 8:30 a.m., the difference between their positions changes at the rate of the relative speed (difference of speeds) since both are moving in the same direction. The catch-up time from 8:30 a.m. can be found by dividing the head start distance by the relative speed. Finally, we add this catch-up time to 8:30 a.m. to obtain the overtaking time on the clock. Step-by-Step Solution:Step 1: Calculate the head start gained by the first runner before 8:30 a.m.Time from 6:00 a.m. to 8:30 a.m. = 2.5 hours.Distance covered by first runner in 2.5 hours = 8 * 2.5 = 20 km.Step 2: Compute the relative speed after 8:30 a.m.Speed of second racer = 10 km/h.Speed of first runner = 8 km/h.Relative speed (closing speed) = 10 - 8 = 2 km/h.Step 3: Time taken for the second racer to make up the 20 km head start at 2 km/h.Catch-up time = head start distance / relative speed = 20 / 2 = 10 hours.Step 4: Add this time to the second racer’s start time.Second racer starts at 8:30 a.m., so overtaking time = 8:30 a.m. + 10 hours = 6:30 p.m. Verification / Alternative check:Check distances at 6:30 p.m. From 6:00 a.m. to 6:30 p.m. is 12.5 hours. Distance of first runner = 8 * 12.5 = 100 km. From 8:30 a.m. to 6:30 p.m. is 10 hours. Distance of second racer = 10 * 10 = 100 km. Since both have covered 100 km by that time, they are at the same position, confirming that the overtaking happens at 6:30 p.m. Why Other Options Are Wrong:- 4:00 p.m. would give the second racer only 7.5 hours of running, covering 75 km, while the first runner would have covered 8 * 10 = 80 km, so they would not yet meet.- 5:30 p.m. similarly yields unequal distances that do not match at the same instant.- 8:00 p.m. is too late; by then, the faster racer would have passed the first runner earlier, not at that time. Common Pitfalls:Some students forget to convert the 2.5 hours head start correctly or mistakenly add speeds instead of subtracting them for same-direction motion. Another error is to compute the catch-up time from 6:00 a.m. instead of 8:30 a.m., which misaligns the time-line. Carefully distinguishing between the head start period and the catch-up period helps avoid these issues. Final Answer:The second racer will overtake the first runner at 6:30 p.m.

Question 13

A man makes four trips of equal distances. His speed on the first trip is 60 km/h, and on each subsequent trip his speed is half of the previous trip. What is his average speed (in km/h) over all four trips?

Accepted Answer

Introduction / Context:This question deals with average speed over several trips of equal distance but different speeds. The man makes four trips, and on each subsequent trip his speed is half of the previous one. We must determine the overall average speed across all four trips. Because the distances are equal for each leg, average speed is not the simple mean of the speeds; instead, it is the total distance divided by the total time across all legs. Given Data / Assumptions:- Distance of each trip = d km (same for all four trips).- Speed on trip 1 = 60 km/h.- Speed on trip 2 = 30 km/h (half of 60).- Speed on trip 3 = 15 km/h (half of 30).- Speed on trip 4 = 7.5 km/h (half of 15).- No extra waiting time between trips; only travel times are considered. Concept / Approach:Let the distance of each trip be d. The total distance travelled is 4d. The time taken for each trip is distance / speed, so the total time is the sum of all four individual times. The overall average speed is then computed as total distance / total time. Since d will appear in every term, it will cancel out, and we can find the average speed purely from speed values. This is a typical harmonic-mean style situation for equal distances. Step-by-Step Solution:Step 1: Express time for each trip.Trip 1 time = d / 60 hours.Trip 2 time = d / 30 hours.Trip 3 time = d / 15 hours.Trip 4 time = d / 7.5 hours.Step 2: Simplify the fourth time term.7.5 km/h = 15/2 km/h, so d / 7.5 = d / (15/2) = (2d) / 15.Step 3: Sum all times.Total time T = d / 60 + d / 30 + d / 15 + 2d / 15.Convert to a common denominator of 60:d / 60 = d/60.d / 30 = 2d / 60.d / 15 = 4d / 60.2d / 15 = 8d / 60.So T = (1 + 2 + 4 + 8)d / 60 = 15d / 60 = d / 4 hours.Step 4: Total distance = 4d.Average speed V_avg = total distance / total time = 4d / (d / 4) = 4d * 4 / d = 16 km/h. Verification / Alternative check:If we assume a specific distance, say d = 60 km for simplicity, then trip distances are 60 km each, total distance = 240 km. Times: 60/60 = 1 hour; 60/30 = 2 hours; 60/15 = 4 hours; 60/7.5 = 8 hours. Total time = 1 + 2 + 4 + 8 = 15 hours. Average speed = 240 / 15 = 16 km/h, confirming our result. Why Other Options Are Wrong:- 30 km/h is close to the first two speeds but ignores the very slow final trips which heavily increase total time and reduce average speed.- 28.125 km/h and 27.5 km/h are too high because they do not properly account for how the extremely slow speeds on later trips dominate the total time when distances are equal. Common Pitfalls:Many students incorrectly compute the arithmetic mean of the speeds: (60 + 30 + 15 + 7.5) / 4 = 28.125 km/h, which is not the correct overall average speed. This ignores the fact that equal distances at very different speeds do not contribute equally to the total time. Always use total distance divided by total time for problems involving varying speeds over equal distances. Final Answer:The man’s average speed over the four trips is 16 km/h.

Question 14

Three cars A, B, and C start from a point at 5:00 p.m., 6:00 p.m., and 7:00 p.m. respectively and travel in the same direction at uniform speeds of 60 km/h, 80 km/h, and x km/h respectively. If all three cars meet at another point at the same instan…

Accepted Answer

Introduction / Context:This problem involves three cars starting from the same point at different times and travelling in the same direction at constant speeds. Despite the staggered start times and different speeds, all three meet at some other point at the same moment. We are asked to find the speed of the third car, C. This is a relative time and distance question that uses the idea that the distances travelled by all three cars to the meeting point must be equal at the instant of meeting. Given Data / Assumptions:- Car A starts at 5:00 p.m. with speed 60 km/h.- Car B starts at 6:00 p.m. with speed 80 km/h.- Car C starts at 7:00 p.m. with speed x km/h (unknown).- All cars travel in the same direction along a straight road at constant speeds.- They meet at some point at the same instant during their journeys. Concept / Approach:Let the meeting happen at time T, measured in hours after 5:00 p.m. By that time, each car has been in motion for a different duration: A for T hours, B for T - 1 hours (since B starts an hour later), and C for T - 2 hours (since C starts two hours later). The distance travelled by each car to the meeting point must be equal, so we set their distance expressions equal and solve. First we use A and B to find T, then use A and C (or B and C) to find x. Step-by-Step Solution:Step 1: Let T be the time in hours after 5:00 p.m. when they all meet.Distance travelled by A: D = 60 * T.Distance travelled by B: D = 80 * (T - 1), since B starts at 6:00 p.m.Step 2: Set A’s and B’s distances equal:60T = 80(T - 1).Solve: 60T = 80T - 80 ⇒ 20T = 80 ⇒ T = 4 hours.Thus, the meeting occurs at 5:00 p.m. + 4 hours = 9:00 p.m.Step 3: Distance to the meeting point is D = 60 * 4 = 240 km.Step 4: Use Car C’s data to find x.C starts at 7:00 p.m., so its travel time by 9:00 p.m. is T - 2 = 2 hours.Distance of C to the meeting point is also D, so D = x * 2.Therefore, 240 = 2x ⇒ x = 120 km/h. Verification / Alternative check:Check all three cars: Car A runs from 5:00 p.m. to 9:00 p.m. (4 hours) at 60 km/h → 240 km. Car B runs from 6:00 p.m. to 9:00 p.m. (3 hours) at 80 km/h → 240 km. Car C runs from 7:00 p.m. to 9:00 p.m. (2 hours) at 120 km/h → 240 km. All three distances match, confirming that the computed x is correct and that they meet at the same point at the same time. Why Other Options Are Wrong:- Values like 110, 105, or 100 km/h would lead to distances for Car C that are not 240 km in 2 hours. For example, at 100 km/h for 2 hours, Car C would travel only 200 km, which is less than the 240 km travelled by A and B. Common Pitfalls:Students sometimes forget to adjust the time for each car correctly (for example, using T instead of T - 1 or T - 2), or they attempt to equate speeds instead of distances. The crucial point is that at the meeting instant, all cars have travelled the same distance from the starting point, even though they have been moving for different lengths of time. Carefully tracking start times and travel durations prevents such errors. Final Answer:The value of x, the speed of car C, is 120 km/h.

Question 15

A man travelled a distance of 50 km in 8 hours. He travelled partly on foot at 5 km/h and partly on a bicycle at 7 km/h. What is the distance (in km) that he travelled on foot?

Accepted Answer

Introduction / Context:This is a mixed-mode travel problem where a man covers a certain total distance using two different modes of transport with different speeds: walking and cycling. We are given the total distance and total time, along with both speeds, and we must determine how much of the distance was covered on foot. This type of question tests the ability to set up and solve a simple system involving distances and times. Given Data / Assumptions:- Total distance = 50 km.- Total time = 8 hours.- Walking speed = 5 km/h.- Bicycling speed = 7 km/h.- Let the distance travelled on foot be x km.- Then the distance travelled by bicycle is (50 - x) km.- The man travels at constant speeds on a straight route, with no extra waiting or resting time included in the 8 hours. Concept / Approach:The key idea is that the total time is the sum of the time spent walking and the time spent bicycling. Time is distance divided by speed, so we express the two components of the total time as x / 5 and (50 - x) / 7. Equating their sum to 8 hours yields a linear equation in x. Solving that equation gives the distance walked. We then verify that this value is reasonable and satisfies both total distance and total time constraints. Step-by-Step Solution:Step 1: Let distance on foot be x km.Then distance by bicycle is 50 - x km.Step 2: Express the time spent in each mode.Time on foot = x / 5 hours.Time on bicycle = (50 - x) / 7 hours.Step 3: Total time is given as 8 hours.So x / 5 + (50 - x) / 7 = 8.Step 4: Clear denominators by multiplying through by 35 (the least common multiple of 5 and 7):35 * (x / 5) + 35 * ((50 - x) / 7) = 35 * 8.This simplifies to 7x + 5(50 - x) = 280.Step 5: Expand and simplify:7x + 250 - 5x = 280 ⇒ 2x + 250 = 280 ⇒ 2x = 30 ⇒ x = 15 km. Verification / Alternative check:With x = 15 km walked, the bicycling distance is 50 - 15 = 35 km. Time on foot = 15 / 5 = 3 hours. Time on bicycle = 35 / 7 = 5 hours. Total time = 3 + 5 = 8 hours, exactly as given. The distance sum is 15 + 35 = 50 km, confirming that both total distance and total time constraints are satisfied. Why Other Options Are Wrong:- 20 km on foot would lead to total time 20/5 + 30/7 = 4 + 30/7 ≈ 8.286 hours, which is more than 8 hours.- 25 km on foot gives 25/5 + 25/7 = 5 + 25/7 ≈ 8.571 hours, again larger than 8 hours.- 30 km on foot gives 30/5 + 20/7 = 6 + 20/7 ≈ 8.857 hours, still not equal to 8 hours. Common Pitfalls:Some students attempt to directly proportion the distances based on speeds without using the time information correctly, or they treat 8 hours as an average instead of a total. Another common mistake is arithmetic error when clearing fractions or solving the linear equation. Writing each step clearly and verifying by substitution helps to avoid these issues. Final Answer:The man travelled 15 km on foot.

Question 16

To travel 720 km, an Express train takes 6 hours more than the Duronto train. If, however, the speed of the Express train is doubled, it then takes 2 hours less than the Duronto train to cover the same 720 km. What is the speed of the Duronto train …

Accepted Answer

Introduction / Context:This is a two-variable time and speed problem involving two trains, Express and Duronto, travelling the same fixed distance under different speed conditions. The Express train is slower initially, taking 6 hours more than Duronto, but when its speed is doubled, it becomes faster and arrives 2 hours earlier than Duronto. Using these comparative time statements, we can form equations and solve for the speed of the Duronto train. Such questions are very common in railway aptitude and competitive exams. Given Data / Assumptions:- Distance travelled by both trains = 720 km.- Let the speed of Duronto be v_D km/h.- Let the speed of Express initially be v_E km/h.- Express initially takes 6 hours more than Duronto: time_E1 = time_D + 6.- When Express speed is doubled to 2v_E, its time becomes 2 hours less than Duronto: time_E2 = time_D - 2.- Both trains run at constant speeds on the same route. Concept / Approach:We first express the travel times in terms of the speeds and the fixed distance using time = distance / speed. For Duronto, time_D = 720 / v_D. For the Express train, time_E1 = 720 / v_E. Doubling the Express speed leads to a new time time_E2 = 720 / (2v_E). Using the two given relationships between these times, we obtain two equations. Solving these, we find time_D first and then compute v_D as distance divided by time_D. Step-by-Step Solution:Step 1: Let time_D = 720 / v_D and time_E1 = 720 / v_E.From the first condition: time_E1 = time_D + 6.So 720 / v_E = 720 / v_D + 6. (Equation 1)Step 2: When Express speed is doubled, its time is time_E2 = 720 / (2v_E) = (1/2) * (720 / v_E) = (1/2) * time_E1.Given that time_E2 = time_D - 2, so (1/2) * time_E1 = time_D - 2. (Equation 2)Step 3: Use Equation 1 to rewrite time_E1 in terms of time_D: time_E1 = time_D + 6.Substitute into Equation 2: (1/2) * (time_D + 6) = time_D - 2.Step 4: Solve for time_D:(time_D / 2) + 3 = time_D - 2.Rearrange: 3 + 2 = time_D - time_D / 2 ⇒ 5 = time_D / 2 ⇒ time_D = 10 hours.Step 5: Now find v_D using time_D = 720 / v_D.So v_D = 720 / time_D = 720 / 10 = 72 km/h. Verification / Alternative check:With v_D = 72 km/h, time_D = 10 hours. From Equation 1, time_E1 = time_D + 6 = 16 hours, so v_E = 720 / 16 = 45 km/h. When Express speed is doubled to 90 km/h, its new time is time_E2 = 720 / 90 = 8 hours. Compare with Duronto: Duronto time = 10 hours. Indeed, 8 hours is 2 hours less than 10 hours, matching the second condition. Both conditions are satisfied, confirming v_D = 72 km/h is correct. Why Other Options Are Wrong:- A Duronto speed of 60 km/h would give time_D = 12 hours, which does not satisfy both relationships when you recompute Express times.- Speeds such as 66 km/h or 78 km/h similarly fail either the “6 hours more” condition or the “2 hours less after doubling” condition. Common Pitfalls:Some candidates confuse which train is faster in each scenario and may swap the “more” and “less” conditions. Others try to guess speeds that “look nice” without forming proper equations, leading to inconsistent results. Carefully defining time_D and time_E1, translating the word statements literally into equations, and solving step by step is the most reliable method. Final Answer:The speed of the Duronto train is 72 km/h.

Question 17

A train is moving at a constant speed of 72 km/h. How many seconds will it take to cover a distance of 1400 m?

Accepted Answer

Introduction / Context: In many aptitude and competitive exams, problems involving speed, distance and time are very common. This question asks you to determine how many seconds a train, moving with a uniform speed of 72 km/h, will take to cover a distance of 1400 m. To solve such questions correctly and quickly, you must be comfortable with unit conversions (especially km/h to m/s) and with the basic relationship between speed, distance and time. Given Data / Assumptions: The speed of the train is 72 km/h. The distance to be covered is 1400 m. The train moves at a constant (uniform) speed. We assume there are no stops or changes in speed during the motion. Concept / Approach: The fundamental formula relating speed, distance and time is: speed = distance / time From this, time can be written as: time = distance / speed However, the distance is given in metres (m) and the speed is given in kilometres per hour (km/h). To apply the formula consistently, we must convert the speed to metres per second (m/s). The standard relation is: 1 km/h = (1000 / 3600) m/s = 5 / 18 m/s Step-by-Step Solution: Step 1: Convert the speed from km/h to m/s. Speed in m/s = 72 * (5 / 18). 72 / 18 = 4, so speed = 4 * 5 = 20 m/s. Step 2: Use the time formula time = distance / speed. Distance = 1400 m, speed = 20 m/s. Time = 1400 / 20 seconds. 1400 / 20 = 70 seconds. Therefore, the train takes 70 seconds to cover 1400 m. Verification / Alternative check: We can perform a quick reasonableness check. At 20 m/s, the train covers 20 m every second. In 70 seconds, it will cover: distance = speed * time = 20 * 70 = 1400 m This matches the required distance exactly, so the calculation is consistent and correct. Why Other Options Are Wrong: 70 seconds is correct, so the other values are incorrect for the given data. 60 seconds would correspond to a speed of 1400 / 60 ≈ 23.33 m/s, which is more than 20 m/s. 72 seconds would mean a lower speed of about 19.44 m/s, not 20 m/s. 80 seconds or 84 seconds would correspond to even smaller speeds and do not match a constant speed of 72 km/h. Common Pitfalls: A common mistake is to directly substitute speed = 72 into the formula without converting to m/s, mixing kilometres, metres, hours and seconds incorrectly. Another frequent error is to misapply the km/h to m/s conversion factor, for example using 18 / 5 instead of 5 / 18. Students also sometimes invert the formula and compute speed = time / distance instead of time = distance / speed. Carefully keeping track of units and the correct formula avoids these issues. Final Answer: The train takes 70 seconds to cover a distance of 1400 m at a speed of 72 km/h.

Question 18

To cover a total distance of 90 km in 2.5 hours, what should be the average speed of a car in metres per second (m/s)?

Accepted Answer

Introduction / Context: Average speed questions are a staple of time, speed and distance topics in aptitude exams. Here, a car needs to cover 90 km in 2.5 hours, and we are asked to express the required average speed in metres per second (m/s). This demands both an understanding of how to compute average speed and how to convert from km/h to m/s accurately. Given Data / Assumptions: Total distance to be covered = 90 km. Total time allowed = 2.5 hours. The car is assumed to travel at a constant average speed. We need the answer in m/s, not in km/h. Concept / Approach: The basic formula for average speed is: average speed = total distance / total time First we will find the average speed in kilometres per hour (km/h). After that, we will convert the result to metres per second (m/s) using the standard conversion: 1 km/h = 5 / 18 m/s Step-by-Step Solution: Step 1: Compute average speed in km/h. Distance = 90 km, time = 2.5 hours. Average speed (km/h) = 90 / 2.5. 2.5 = 5 / 2, so 90 / 2.5 = 90 * (2 / 5) = 180 / 5 = 36 km/h. Step 2: Convert 36 km/h to m/s. Speed in m/s = 36 * (5 / 18). 36 / 18 = 2, so speed = 2 * 5 = 10 m/s. Therefore, the required average speed is 10 m/s. Verification / Alternative check: We can check quickly by reversing the process. Convert 10 m/s back to km/h: speed (km/h) = 10 * (18 / 5) = 10 * 3.6 = 36 km/h. Now confirm that a car travelling at 36 km/h for 2.5 hours covers 90 km: distance = speed * time = 36 * 2.5 = 36 * (5 / 2) = 90 km. This matches the original distance, so our answer is consistent. Why Other Options Are Wrong: 12 m/s corresponds to 43.2 km/h, which would cover more than 90 km in 2.5 hours. 15 m/s corresponds to 54 km/h and would also cover a greater distance than needed. 20 m/s is 72 km/h, which is much faster and would cover 180 km in 2.5 hours. 8 m/s is 28.8 km/h, which is too slow and would cover only 72 km in 2.5 hours. Only 10 m/s exactly matches a 90 km journey completed in 2.5 hours. Common Pitfalls: A common mistake is to convert 2.5 hours into minutes or seconds unnecessarily and then mis-handle the units. Another error is to use an incorrect conversion factor between km/h and m/s, such as 18 / 5 instead of 5 / 18, which reverses the conversion. Some learners also mistakenly average speeds or times directly without using the correct formula. Here the calculation is straightforward as long as you compute 90 / 2.5 correctly and then convert units carefully. Final Answer: The car must travel at an average speed of 10 m/s to cover 90 km in 2.5 hours.

Question 19

Two cars travel from city A to city B at speeds of 30 km/h and 44 km/h respectively. If one car takes 3.5 hours less than the other, what is the distance between city A and city B?

Accepted Answer

Introduction / Context: This time and distance problem compares travel times of two cars between the same two cities. One car travels at 30 km/h and the other at 44 km/h. We are told that one car takes 3.5 hours less than the other, and we must find the distance between city A and city B. Such questions test your ability to set up equations using speed, distance and time relationships and to handle fractional time differences correctly. Given Data / Assumptions: Speed of the slower car = 30 km/h. Speed of the faster car = 44 km/h. Both cars travel the same distance between city A and city B. Difference in their travel times = 3.5 hours. Motion is along the same route with constant speeds. Concept / Approach: Let the distance between the cities be D km. For each car, time taken is distance divided by speed. If the slower car takes more time, the difference in times is: D / 30 - D / 44 = 3.5 We solve this equation for D. The main idea is to use a single variable for the distance and relate the two times through the given time difference. Step-by-Step Solution: Step 1: Write the time expressions. Time taken by car 1 (30 km/h) = D / 30 hours. Time taken by car 2 (44 km/h) = D / 44 hours. Step 2: Use the time difference condition. D / 30 - D / 44 = 3.5. Step 3: Simplify the left-hand side using a common denominator. Common denominator of 30 and 44 is 660. D / 30 = (22D) / 660 and D / 44 = (15D) / 660. So (22D / 660) - (15D / 660) = 3.5. This gives (7D / 660) = 3.5. Step 4: Solve for D. 7D = 660 * 3.5. 3.5 = 7 / 2, so 660 * 3.5 = 660 * (7 / 2) = (660 * 7) / 2 = 4620 / 2 = 2310. Therefore 7D = 2310, so D = 2310 / 7 = 330 km. Verification / Alternative check: Check the times for D = 330 km: Time at 30 km/h = 330 / 30 = 11 hours. Time at 44 km/h = 330 / 44 = 7.5 hours. Difference in times = 11 - 7.5 = 3.5 hours, which matches the given condition. So D = 330 km is confirmed. Why Other Options Are Wrong: 396 km: Time difference would be 396 / 30 - 396 / 44, which is greater than 3.5 hours. 264 km: Time difference for this distance is less than 3.5 hours. 495 km and 360 km similarly do not yield exactly 3.5 hours difference when substituted into the equation. Only 330 km gives the correct time difference. Common Pitfalls: A frequent mistake is to write the time difference equation incorrectly, such as reversing which time is larger or adding times instead of subtracting them. Another issue is handling the fraction 3.5 incorrectly, for example treating it as 3.05. Some learners also approximate the fractions too early and end up with rounding errors. It is safer to work exactly with fractions, find a common denominator and only simplify at the end. Final Answer: The distance between city A and city B is 330 km.

Question 20

Two numbers are 10% and 20% less than a third number respectively. By what percentage must the second number be increased so that it becomes equal to the first number?

Accepted Answer

Introduction / Context: This question belongs to percentage comparison, a very important topic in quantitative aptitude. Instead of giving actual values, the problem describes two numbers in terms of their percentage reduction from a third number. We then need to find by what percentage the smaller number must be increased so that it becomes equal to the larger number. This tests your understanding of percentage decrease and percentage increase, and how they relate to each other with respect to a common base value. Given Data / Assumptions: Let the third (reference) number be T. The first number is 10% less than T. The second number is 20% less than T. We need the percentage increase required on the second number so that it equals the first number. We assume all numbers are positive real numbers. Concept / Approach: We use the idea of expressing all numbers in terms of T. Then we compare the first and second numbers directly. If we call the first number A and the second number B, we are essentially asked: by what percentage should B be increased so that B becomes equal to A? The formula for percentage increase from B to A is: percentage increase = [(A - B) / B] * 100% Step-by-Step Solution: Step 1: Express the first number A. 10% less than T means A = 90% of T = 0.9T. Step 2: Express the second number B. 20% less than T means B = 80% of T = 0.8T. Step 3: Find the difference A - B. A - B = 0.9T - 0.8T = 0.1T. Step 4: Compute the required percentage increase on B. percentage increase = [(A - B) / B] * 100%. = [0.1T / 0.8T] * 100%. The T cancels out, giving (0.1 / 0.8) * 100%. 0.1 / 0.8 = 1 / 8 = 0.125. So percentage increase = 0.125 * 100% = 12.5%. Therefore, the second number must be increased by 12.5% to become equal to the first number. Verification / Alternative check: Pick a convenient value for T, such as T = 100. Then: A = 10% less than 100 = 90. B = 20% less than 100 = 80. To go from 80 to 90, the increase is 10. The percentage increase based on 80 is: (10 / 80) * 100% = (1 / 8) * 100% = 12.5%. This numerical example confirms our algebraic solution. Why Other Options Are Wrong: 25% would mean increasing 80 by 25% to 100, not to 90, so it is too large. 10% gives 80 + 10% of 80 = 80 + 8 = 88, which is less than 90. 15% gives 80 + 12 = 92, which is more than 90. 20% gives 80 + 16 = 96, which is also more than 90. Only 12.5% makes the second number exactly equal to the first number. Common Pitfalls: One common mistake is to compute the percentage difference with respect to T instead of with respect to the second number B. Another frequent error is to subtract percentages directly (for example, taking 10% - 20% = -10%) without considering the base values. Students also sometimes confuse 12.5% with 12%, forgetting that 1 / 8 is exactly 12.5%. Writing everything symbolically in terms of T, then cancelling T, keeps the solution clear and avoids these mistakes. Final Answer: The second number must be increased by 12.5% to become equal to the first number.

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