Q: Paschim Express left Delhi for Mumbai at 14.30 hrs travelling at a speed of 60 kmph and August Kranti Express left Delhi for Mumbai on the same day at 16.30 hrs travelling at a speed of 80 kmph. How far away from Delhi will the two trains meet (excl…

Concept & Trains in Same Direction: When two trains travel in the same direction, the faster train starting later will eventually overtake the slower train. The time taken to overtake depends on the initial head start distance and their relative speed. Time\ to\ Overtake = (Head\ Start\ Distance/Relative\ Speed) Step-by-Step Solution: 1. Determine the time gap: Paschim Express leaves at 14:30. August Kranti leaves at 16:30. Time difference = 2 hours. 2. Calculate initial head start: In these 2 hours, Paschim Express covers a certain distance. Head start distance = Speed of Paschim × Time = 60 km/h × 2 h = 120 km. 3. Find relative speed: Since both travel from Delhi to Mumbai (same direction), relative speed = Speed of August Kranti - Speed of Paschim. Relative speed = 80 km/h - 60 km/h = 20 km/h. 4. Calculate time to catch up: Time taken for August Kranti to catch Paschim = Head start / Relative speed. Time = 120 km / 20 km/h = 6 hours. 5. Calculate meeting distance: The meeting distance from Delhi is the distance August Kranti Express travels in those 6 hours. Distance = Speed of August Kranti × Time = 80 km/h × 6 h = 480 km. Exam Strategy & Shortcut: Let t be the time taken by August Kranti to catch up. Paschim Express travels for t + 2 hours. Distance is same: 60(t + 2) = 80t \implies 60t + 120 = 80t \implies 20t = 120 \implies t = 6 hrs. Distance from Delhi = 80 × 6 = 480 km. Common Pitfall: Often students correctly find the time it takes the second train to catch the first (6 hours) but then incorrectly multiply this time by the slower train's speed or forget to add the initial 2 hours if they use the slower train for the final calculation (60 × (6+2) = 480). Final Answer: Therefore, the correct answer is 480 km.

Q: A train started from station $A$ and proceeded towards station $B$ at a speed of $48$ km/hr. Forty-five minutes later another train started from station $B$ and proceeded towards station a at $50$ km/hr. If the distance between the two stations is $…

Concept & Relative Speed with Head Start: When two trains start at different times, calculate the distance covered by the first train during its "head start". Subtract this from the total distance to find the remaining gap. Then, use relative speed to find the time it takes to close the remaining gap. Step-by-Step Solution: • Given: Total distance = 232 km. Speed of Train from A = 48 km/hr. Speed of Train from B = 50 km/hr. Train A starts 45 minutes (3/4 hour) early. • Calculation: Distance traveled by the train from A in 45 minutes = 48 × (3/4) = 36 km. • Remaining distance between the two trains when the train from B starts = 232 - 36 = 196 km. • Relative speed of the two trains (moving towards each other) = 48 + 50 = 98 km/hr. • Time taken for them to meet after the train from B starts = (196/98) = 2 hours. • The question asks for the meeting distance from station A. The train from A traveled for 45 minutes initially, plus 2 hours more. • Distance covered by train from A in the 2 hours = 48 × 2 = 96 km. • Total distance from station A = Head start distance + Distance traveled until meeting = 36 + 96 = 132 km. Exam Strategy & Shortcut: Find the distance remaining after the head start: 232 - (48 × 0.75) = 196. Find meeting time: 196 / (48 + 50) = 2 hours. Calculate distance from A: Initial 36 km + (2 × 48) = 132 km. Common Pitfall: A common error is to calculate the distance traveled by train A only during the simultaneous travel time (96 km), forgetting to add the initial head start distance (36 km). Final Answer: Therefore, the correct answer is 132 km.

Q: A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, then the length of the train is

Concept & Formula: First, find the speed of the train using the given distance and time. Once the uniform speed is known, use it along with the time taken to pass the pole to find the train's length. Speed = \frac{Distance}{Time} Length of Train = Speed (m/s) × Time to cross post (s) Step-by-Step Solution: • Given for Speed: Distance = 12 km = 12000 m. Time = 10 minutes = 10 × 60 = 600 seconds. • Calculate Speed: Speed = (12000/600) = 20 m/s. • Given for Crossing: Time to pass post = 6 seconds. • Calculate Length: Length = 20 m/s × 6 s = 120 m. Exam Strategy & Shortcut: 12 km in 10 minutes is equivalent to 72 km in 60 minutes (72 km/hr). We know 72 km/hr is exactly 20 m/s (4 × 18 -> 4 × 5). Multiplying 20 m/s by 6 seconds immediately yields 120 m. Common Pitfall: Mixing units. If you use 12 km and 10 minutes directly, you might end up with speed in km/min and apply it incorrectly to the 6 seconds. Always standardize to meters and seconds. Final Answer: Therefore, the correct answer is 120 m.

Question 1

A train of length 100 m crosses another train of length 150 m, running on a parallel track in the opposite direction, in 9 seconds. If the speed of the 150 m long train is 40 km/hr, what is the speed of the other 100 m long train in km/hr?

Accepted Answer

Introduction / Context:This question uses the concept of relative speed for two trains moving in opposite directions on parallel tracks. When they cross each other, the effective distance is the sum of their lengths, and the relative speed is the sum of their individual speeds. We are given the speed of one train and the total time to cross, and must find the speed of the other train.Given Data / Assumptions:- Length of first train (unknown speed) = 100 m.- Length of second train (known speed) = 150 m.- Time taken for the trains to completely cross each other = 9 seconds.- Speed of the 150 m train = 40 km/hr.- Trains are moving in opposite directions at constant speeds.Concept / Approach:When two trains move in opposite directions, their relative speed is the sum of their speeds. The distance covered during the crossing is the sum of their lengths. Using distance = speed * time, we can compute the relative speed in meters per second, then subtract the known train's speed (converted to m/s) to get the unknown speed. Finally, convert that unknown speed back to km/hr.Step-by-Step Solution:Step 1: Total distance to be covered while crossing = 100 m + 150 m = 250 m.Step 2: Time taken to cross = 9 seconds.Step 3: Relative speed in m/s = distance / time = 250 / 9 m/s.Step 4: Convert speed of 150 m train to m/s: 40 km/hr = 40 * (5 / 18) = 200 / 18 = 100 / 9 m/s.Step 5: Let the speed of the 100 m train be v m/s. Since directions are opposite, relative speed = v + 100 / 9.Step 6: Set v + 100 / 9 = 250 / 9. So v = (250 / 9) - (100 / 9) = 150 / 9 = 50 / 3 m/s.Step 7: Convert v to km/hr by multiplying by 18 / 5: v = (50 / 3) * (18 / 5) = (50 * 18) / 15 = 900 / 15 = 60 km/hr.Verification / Alternative check:Check the combined speed. 60 km/hr + 40 km/hr = 100 km/hr. In m/s, 100 km/hr = 100 * (5 / 18) = 500 / 18 = 250 / 9 m/s. Time to cross 250 m at 250 / 9 m/s is distance / speed = 250 / (250 / 9) = 9 seconds, which matches the given value. This confirms that 60 km/hr is correct.Why Other Options Are Wrong:30 km/hr or 42 km/hr would make the relative speed too small, leading to a larger crossing time than 9 seconds. 48 km/hr or 50 km/hr similarly fail to produce the required 250 / 9 m/s relative speed when added to 40 km/hr. Only 60 km/hr gives the correct combination.Common Pitfalls:Common mistakes include forgetting to convert from km/hr to m/s, subtracting the speeds instead of adding them for opposite directions, or using only one train's length instead of the sum of both lengths. Consistently applying the relative speed concept and unit conversions avoids these errors.Final Answer:The speed of the 100 m long train is 60 km/hr.

Question 2

In what time (in seconds) will a train that is 100 metres long, running at a speed of 50 km/hr, completely cross a stationary pillar standing beside the track?

Accepted Answer

Introduction / Context:This question is a straightforward application of the distance–speed–time relationship for trains. When a train crosses a pillar (or any stationary point), the distance covered for the purpose of the crossing is simply the length of the train itself. The problem checks your ability to convert units correctly and apply the basic formula time = distance / speed.Given Data / Assumptions:- Length of the train = 100 m.- Speed of the train = 50 km/hr.- The pillar is considered a point object with negligible width.- The train travels at a uniform speed without acceleration or braking.Concept / Approach:To find the time taken for the train to pass a pillar, we use the fact that the train must move a distance equal to its own length. Since the speed is given in km/hr and the length is in metres, we must convert the speed into metres per second. Then, applying time = distance / speed gives us the crossing time in seconds.Step-by-Step Solution:Step 1: Distance relevant for crossing a pillar = length of the train = 100 m.Step 2: Convert 50 km/hr into m/s using the factor 5 / 18.Step 3: Speed in m/s = 50 * (5 / 18) = 250 / 18 = 125 / 9 m/s.Step 4: Use the time formula: time = distance / speed = 100 / (125 / 9) seconds.Step 5: Simplify: 100 / (125 / 9) = 100 * 9 / 125 = 900 / 125.Step 6: Divide numerator and denominator by 25: 900 / 125 = 36 / 5 = 7.2 seconds.Step 7: Thus, the train takes 7.2 seconds to completely cross the pillar.Verification / Alternative check:We can approximate to feel if the answer is reasonable. 50 km/hr is about 13.89 m/s. If the train travels at roughly 14 m/s, then to cover 100 m it would take about 100 / 14 ≈ 7.14 seconds, which is very close to 7.2 seconds. This confirms that the computed value is realistic and accurate.Why Other Options Are Wrong:7.0 seconds and 6.8 seconds are close but slightly smaller than the correct 7.2 seconds and result from rounding or ignoring exact conversion factors. 72 seconds and 70 seconds are far too large and would correspond to extremely slow speeds (much less than walking speed). Only 7.2 seconds matches the correct calculation with proper unit conversion.Common Pitfalls:Many students forget to convert km/hr to m/s and directly divide metres by km/hr, which mixes units and produces incorrect answers. Others mistakenly use the length of some imagined platform instead of just the train length when the object is a pillar. Always remember: crossing a point object involves only the train's length.Final Answer:The 100 metre long train will cross the pillar completely in 7.2 seconds.

Question 3

A train 150 m long is running at a speed of 68 km/hr. In what time (in seconds) will it completely pass a man who is running at 8 km/hr in the same direction as the train?

Accepted Answer

Introduction / Context:This problem is a classic relative speed question in which a train overtakes a man running in the same direction. The key idea is that the effective speed at which the train approaches the man is the difference of their speeds, not the individual speeds. The time taken to overtake depends on this relative speed and the train's length.Given Data / Assumptions:- Length of the train = 150 m.- Speed of the train = 68 km/hr.- Speed of the man = 8 km/hr.- Both are moving in the same direction along a straight path.- We must find the time in seconds for the train to completely pass the man.Concept / Approach:When two objects move in the same direction, their relative speed equals the difference of their speeds. The train must travel a distance equal to its own length relative to the man in order to overtake him fully. Once we find the relative speed in m/s, we apply time = distance / relative speed using the train's length as the distance.Step-by-Step Solution:Step 1: Compute the relative speed: 68 km/hr (train) minus 8 km/hr (man) = 60 km/hr.Step 2: Convert 60 km/hr to m/s using the factor 5 / 18.Step 3: Relative speed in m/s = 60 * (5 / 18) = 300 / 18 = 50 / 3 m/s.Step 4: Distance for overtaking = length of the train = 150 m.Step 5: Use time = distance / speed: time = 150 / (50 / 3) seconds.Step 6: Simplify: 150 / (50 / 3) = 150 * 3 / 50 = 450 / 50 = 9 seconds.Step 7: Hence, the train takes 9 seconds to completely pass the man.Verification / Alternative check:We can do a quick sense check. Relative speed of 60 km/hr is about 16.67 m/s. A 150 m train at 16.67 m/s would need about 150 / 16.67 ≈ 9 seconds to overtake fully, which aligns perfectly with the detailed calculation. This confirms both the method and the answer.Why Other Options Are Wrong:8 or 8.5 seconds would require a slightly higher relative speed than 60 km/hr. 9.5 seconds implies a slower relative speed; if substituted back, it would not match the given absolute speeds. 10 seconds similarly leads to a relative speed smaller than 60 km/hr, contradicting the given values. Only 9 seconds fits the relative speed calculation exactly.Common Pitfalls:A common mistake is to add the speeds instead of subtracting them, which is only valid when objects move in opposite directions. Another frequent error is using the sum of lengths of both objects; here, only the train has length, so we use 150 m, not more. Forgetting to convert from km/hr to m/s also leads to incorrect time values.Final Answer:The 150 m long train will pass the man completely in 9 seconds.

Question 4

A man covers a certain one-way distance by train at a speed of 40 km/hr and returns over the same distance on foot at a speed of 8 km/hr. If the total time for the entire journey (going and coming back) is 12 hours, what is the one-way distance in k…

Accepted Answer

Introduction / Context:This is a classic average speed and total time question involving different speeds on the onward and return journeys. The man travels one way by train and returns on foot at a lower speed. By expressing the total time as the sum of the two individual travel times, we can form an equation in terms of the distance and solve it.Given Data / Assumptions:- One-way distance between the two points = D km (unknown).- Speed of the train for the onward trip = 40 km/hr.- Speed of walking for the return trip = 8 km/hr.- Total time for the round trip = 12 hours.- The route and distance are the same in both directions, and speeds are constant.Concept / Approach:The total journey time equals the sum of the times for the onward and return journeys. For any leg, time = distance / speed. Hence the onward time is D / 40 hours and the return time is D / 8 hours. Setting their sum equal to 12 hours gives a linear equation in D, which we can solve easily.Step-by-Step Solution:Step 1: Let the one-way distance be D km.Step 2: Time taken by train (onward) = D / 40 hours.Step 3: Time taken by walking (return) = D / 8 hours.Step 4: Total time for the round trip is given as 12 hours, so D / 40 + D / 8 = 12.Step 5: Find a common denominator. The least common multiple of 40 and 8 is 40, so rewrite as D / 40 + 5D / 40 = 12.Step 6: Combine terms: (D + 5D) / 40 = 6D / 40 = 12.Step 7: Simplify the fraction: 6D / 40 = 3D / 20, so 3D / 20 = 12.Step 8: Multiply both sides by 20: 3D = 12 * 20 = 240.Step 9: Divide by 3: D = 240 / 3 = 80 km.Verification / Alternative check:Check the individual times with D = 80 km. Onward by train: 80 / 40 = 2 hours. Return on foot: 80 / 8 = 10 hours. Total time = 2 + 10 = 12 hours, which matches the given total journey time. Therefore, the distance calculation is consistent and correct.Why Other Options Are Wrong:If D = 60 km, total time would be 60 / 40 + 60 / 8 = 1.5 + 7.5 = 9 hours, not 12. For 90 km, total time becomes 90 / 40 + 90 / 8 = 2.25 + 11.25 = 13.5 hours. For 100 km, time is 2.5 + 12.5 = 15 hours. 72 km gives 72 / 40 + 72 / 8 = 1.8 + 9 = 10.8 hours. None of these equal 12 hours.Common Pitfalls:Learners sometimes average the speeds (40 and 8) and directly multiply by total time, which is incorrect because time at each speed is not equal. Others may forget that both legs share the same distance and mistakenly use different D values for onward and return. Always express the total time as the sum of distance / speed for each leg.Final Answer:The one-way distance between the two points is 80 km.

Question 5

Seven hours after a goods train passes a station, another train travelling in the same direction at a speed of 54 km/hr passes that station. If this faster train overtakes the goods train 11 hours after passing the station, what is the speed of the …

Accepted Answer

Introduction / Context:This question examines relative motion in the same direction, focusing on how a faster train catches up with a slower goods train that has departed earlier. By translating the situation into distances travelled over time, we can set the positions equal at the catch-up moment and solve for the unknown speed of the goods train.Given Data / Assumptions:- A goods train passes a station at time t = 0 with speed v km/hr (unknown).- Seven hours later, a faster train passes the same station at 54 km/hr in the same direction.- The faster train overtakes the goods train 11 hours after it passes the station.- Both trains move at constant speeds on a straight track.Concept / Approach:We work with the positions of both trains relative to the station. The goods train has a 7 hour head start. When the faster train has been running for 11 hours, the goods train has been running for 7 + 11 = 18 hours. At the moment of overtaking, both trains are at the same distance from the station. Setting the distances equal gives a simple linear equation in v.Step-by-Step Solution:Step 1: Let v be the speed of the goods train in km/hr.Step 2: Time traveled by the goods train when it is overtaken = 7 hours (lead) + 11 hours (while the faster train moves) = 18 hours.Step 3: Distance covered by the goods train by then = v * 18 km.Step 4: Time traveled by the faster train = 11 hours, so distance covered by it = 54 * 11 = 594 km.Step 5: At the overtaking point, both distances are equal: v * 18 = 594.Step 6: Solve for v: v = 594 / 18.Step 7: Compute 594 / 18: since 18 * 30 = 540 and 18 * 33 = 594, v = 33 km/hr.Verification / Alternative check:Check relative speed and head start. The speed difference between the faster train and the goods train is 54 - 33 = 21 km/hr. The goods train's head start distance after 7 hours is 33 * 7 = 231 km. Time needed to close a 231 km gap at 21 km/hr is 231 / 21 = 11 hours, which matches the given 11 hours after the faster train passes the station. This confirms the correctness of v = 33 km/hr.Why Other Options Are Wrong:39.6 km/hr, 49.5 km/hr, 26.4 km/hr, and 42 km/hr all fail when substituted back. For example, if v = 39.6 km/hr, the head start distance is 39.6 * 7 = 277.2 km, and the speed difference is 54 - 39.6 = 14.4 km/hr, giving a catch-up time of 277.2 / 14.4 ≈ 19.25 hours, not 11 hours. Similar inconsistencies occur with the other values.Common Pitfalls:Some learners mistakenly add the distances instead of equating them, or think the 11 hours is the total time for the goods train instead of the faster train. Another common error is to forget the 7 hour head start and set 18v equal to 54 * 11 without understanding where 18 comes from. Carefully tracking how long each train has been moving is crucial.Final Answer:The speed of the goods train is 33 km/hr.

Question 6

Four hours after a goods train passes a station, another train travelling in the same direction at a speed of 72 km/hr passes through the same station. If this faster train overtakes the goods train 8 hours after passing the station, what is the spe…

Accepted Answer

Introduction / Context:This is another catch-up problem involving a goods train and a faster passenger train moving in the same direction. The goods train leaves first, giving it a head start, and the second train starts from the same station later but at a higher speed. We use distances travelled over different times to find the unknown speed of the goods train.Given Data / Assumptions:- Goods train passes a station at time t = 0 with speed v km/hr (unknown).- Four hours later, a second train passes the station at 72 km/hr in the same direction.- The second train overtakes the goods train 8 hours after it passes the station.- Both trains move on straight tracks at constant speeds.Concept / Approach:The logic is similar to the previous overtaking problem. The goods train has a 4 hour head start. At the catch-up moment, the distance of each train from the station is equal. We express these distances in terms of their speeds and times, then solve the resulting equation for v, the speed of the goods train.Step-by-Step Solution:Step 1: Let v be the goods train's speed in km/hr.Step 2: Time travelled by the goods train at the overtaking moment = 4 hours (head start) + 8 hours (while second train moves) = 12 hours.Step 3: Distance covered by the goods train = v * 12 km.Step 4: Time travelled by the faster train at that moment = 8 hours, so its distance from the station = 72 * 8 = 576 km.Step 5: At the overtaking moment, distances are equal: v * 12 = 576.Step 6: Solve for v: v = 576 / 12 = 48 km/hr.Verification / Alternative check:Check using relative speed and head start. The speed difference is 72 - 48 = 24 km/hr. The goods train's head start distance after 4 hours is 48 * 4 = 192 km. Time needed to close a 192 km gap at 24 km/hr is 192 / 24 = 8 hours, which matches the given time that the faster train takes to catch up after passing the station. This confirms v = 48 km/hr.Why Other Options Are Wrong:57.6 km/hr would give a head start of 57.6 * 4 = 230.4 km and a speed difference of 14.4 km/hr, so the catch-up time would not be 8 hours. 72 km/hr would mean the trains have the same speed and the faster train would never catch up. 38.4 km/hr yields a different head start and catch-up time inconsistent with 8 hours. 42 km/hr similarly fails the distance–time equality.Common Pitfalls:Some students incorrectly add the times instead of considering how long each train has been in motion at the catch-up moment. Others forget the 4 hour head start and simply set 8v equal to 72 * 8. Always track timelines carefully and ensure that distances from the station at the catch-up time match for both trains.Final Answer:The speed of the goods train is 48 km/hr.

Question 7

A train 500 m long, running at a uniform speed, passes completely over a station platform 221 m long in 35 seconds. What is the speed of the train in km/hr?

Accepted Answer

Introduction / Context:This trains and platforms question examines the idea that when a train passes completely over a platform, the effective distance covered during the crossing is the sum of the train's length and the platform's length. Using this total distance and the crossing time, we can calculate the speed of the train and then express it in km/hr.Given Data / Assumptions:- Length of the train = 500 m.- Length of the platform = 221 m.- Time taken to pass the station (platform) completely = 35 seconds.- Train runs at a constant speed and there are no stops during the crossing.Concept / Approach:When a train crosses a platform, the distance it effectively travels, with respect to the platform, is the sum of its own length and the platform length. The basic relationship is speed = distance / time. We first compute the speed in m/s using total distance and time, and then convert this speed to km/hr by multiplying by 18 / 5.Step-by-Step Solution:Step 1: Total distance to be covered while crossing the platform = length of train + length of platform.Step 2: Total distance = 500 m + 221 m = 721 m.Step 3: Time taken = 35 seconds.Step 4: Speed in m/s = distance / time = 721 / 35 m/s.Step 5: Convert this speed to km/hr using the factor 18 / 5.Step 6: Speed in km/hr = (721 / 35) * (18 / 5) = (721 * 18) / 175.Step 7: Compute (721 * 18) = 12978, so speed = 12978 / 175 km/hr.Step 8: Divide 12978 by 175 to get 74.16 km/hr.Verification / Alternative check:As a quick check, note that 721 / 35 ≈ 20.6 m/s. Multiplying by 3.6 to convert to km/hr (which is equivalent to 18 / 5) gives 20.6 * 3.6 ≈ 74.16 km/hr. This matches the more detailed fraction-based calculation, confirming the correctness of the result.Why Other Options Are Wrong:72.1 km/hr and 68.4 km/hr are close but correspond to slightly different crossing times than 35 seconds when you recompute distance = speed * time. 24.76 km/hr is far too small for the given distance and time. 78.54 km/hr is a bit too high and would imply a shorter crossing time than 35 seconds. Only 74.16 km/hr matches the exact arithmetic.Common Pitfalls:One frequent mistake is to forget to include the platform length and use only the train length as the distance. Another is to convert the speed incorrectly, for example by using 5 / 18 instead of 18 / 5 when moving from m/s to km/hr. Some learners also round intermediate values prematurely, leading to slightly off answers instead of the exact 74.16 km/hr.Final Answer:The speed of the train is 74.16 km/hr.

Question 8

A train is running at a uniform speed of 116 km/hr. What distance in metres will the train cover in 18 seconds at this speed?

Accepted Answer

Introduction / Context:This problem is a straightforward distance calculation given a constant speed and a short time interval. You must convert the train's speed from km/hr to m/s, then use the basic formula distance = speed * time to find how many metres are covered in 18 seconds.Given Data / Assumptions:- Speed of the train = 116 km/hr.- Time interval considered = 18 seconds.- The train's speed remains constant throughout these 18 seconds.- We must report the distance in metres.Concept / Approach:Since the speed is in km/hr and time is in seconds, we must convert speed into metres per second. The standard conversion is: 1 km/hr = 5 / 18 m/s. After converting, we simply multiply speed (in m/s) by time (in seconds) to get distance in metres.Step-by-Step Solution:Step 1: Convert 116 km/hr to m/s using the factor 5 / 18.Step 2: Speed in m/s = 116 * (5 / 18) = 580 / 18 = 290 / 9 m/s.Step 3: The given time is 18 seconds.Step 4: Use distance = speed * time: distance = (290 / 9) * 18 metres.Step 5: Simplify: (290 / 9) * 18 = 290 * 2 = 580 metres.Step 6: Therefore, the train covers 580 m in 18 seconds at 116 km/hr.Verification / Alternative check:We can check using approximate values. 116 km/hr is close to 120 km/hr, which is about 33.33 m/s. Our converted speed 290 / 9 is about 32.22 m/s, which is close. At about 32 m/s for 18 seconds, the train would cover around 576 m, which is very close to the exact 580 m. The exact arithmetic confirms 580 m as the correct distance.Why Other Options Are Wrong:900 m or 1160 m would require significantly higher speeds or longer times than given. 508 m and 720 m do not correspond to the product of (290 / 9) m/s and 18 seconds. Only 580 m results from a correct conversion and multiplication.Common Pitfalls:Students sometimes mistakenly convert the time from seconds to hours and then try to use km/hr directly, which can lead to unnecessary complexity and errors. Another common error is in the km/hr to m/s conversion, such as multiplying by 18 / 5 instead of 5 / 18. Keeping the conversion factor straight and using the simple distance = speed * time formula makes this question easy.Final Answer:The train will cover 580 m in 18 seconds at a speed of 116 km/hr.

Question 9

A train travels 20% faster than a car. Both start together from point A and move towards point B, which is 180 km away. They reach point B at the same time, but the train halts for a total of 30 minutes at intermediate stations. What is the speed of…

Accepted Answer

Introduction / Context:This question combines percentage-based speed comparison with equal arrival times and stoppage delays. The train is faster than the car by 20%, but it also wastes time in station halts. Since both vehicles start together and reach the destination at the same time, we can use the equality of their total journey times to determine the actual speeds.Given Data / Assumptions:- Distance from A to B = 180 km.- Train speed is 20% faster than car speed.- Car speed = c km/hr (unknown).- Train speed = 1.2c km/hr.- Train halts for a total of 30 minutes = 0.5 hours on the way.- Both start together from A and arrive at B at the same time.Concept / Approach:The car travels continuously, so its total time is simply distance / speed. The train travels with a higher running speed but also accumulates 0.5 hours of stoppage time. Hence, total time for the train equals running time + 0.5. Since arrival times are equal, we set the car's travel time equal to the train's total time and solve for c, then compute the train's speed 1.2c.Step-by-Step Solution:Step 1: Let the car speed be c km/hr. Then the train speed is 1.2c km/hr.Step 2: Time taken by the car to travel 180 km = 180 / c hours.Step 3: Running time taken by the train without stops = 180 / (1.2c) hours.Step 4: Simplify 180 / (1.2c): 1.2c = (6/5)c, so 180 / (1.2c) = 180 * (5 / 6c) = 150 / c hours.Step 5: The train also stops for 0.5 hours, so total train time = 150 / c + 0.5 hours.Step 6: Since both arrive together, set car time equal to train time: 180 / c = 150 / c + 0.5.Step 7: Subtract 150 / c from both sides: (180 / c) - (150 / c) = 0.5, giving 30 / c = 0.5.Step 8: Solve for c: c = 30 / 0.5 = 60 km/hr.Step 9: Train speed = 1.2c = 1.2 * 60 = 72 km/hr.Verification / Alternative check:Check times using the found speeds. Car speed = 60 km/hr, so car time = 180 / 60 = 3 hours. Train speed = 72 km/hr, so running time = 180 / 72 = 2.5 hours. Including 0.5 hours of stops, total train time = 2.5 + 0.5 = 3 hours. Both times are equal, so the speeds are consistent with the problem statement.Why Other Options Are Wrong:56, 66, 84, and 60 km/hr give either too large or too small running times for the train. When you add the 0.5 hour stoppage, the train time will not match the car's time for those speeds. Only 72 km/hr simultaneously satisfies the 20% faster condition and the equal arrival time condition.Common Pitfalls:Some learners misinterpret "20% faster" as adding a fixed 20 km/hr instead of 20% of the car's speed. Others forget to add the 0.5 hour stoppage time to the train's running time. It is also easy to invert the ratio, mistakenly using c = 1.2 * train speed instead of train speed = 1.2c.Final Answer:The speed of the train is 72 km/hr.

Question 10

Two trains are running in opposite directions at the same speed. The length of each train is 486 metres, and they completely cross each other in 27 seconds. What is the speed of each train in km/hr?

Accepted Answer

Introduction / Context:This question tests your understanding of relative speed when two trains travel in opposite directions. Both trains have equal length and equal speed. When they cross, the total distance covered relative to each other is the sum of their lengths, and the relative speed is the sum of their individual speeds. From the crossing time, we can find the common speed of each train.Given Data / Assumptions:- Length of each train = 486 m.- Speed of each train = v km/hr (same for both, unknown).- Trains are moving in opposite directions on parallel tracks.- Time taken to completely cross each other = 27 seconds.- Speeds are constant and there is no acceleration.Concept / Approach:For two trains moving in opposite directions, relative speed = v1 + v2. Since both trains have the same speed v, relative speed in km/hr is 2v. The distance covered during crossing is the sum of both lengths: 486 m + 486 m. Using distance = speed * time in m/s, we compute the relative speed in m/s, then divide it by 2 and convert to km/hr to get each train's speed.Step-by-Step Solution:Step 1: Total distance to be covered during crossing = 486 m + 486 m = 972 m.Step 2: Time taken to cross = 27 seconds.Step 3: Relative speed in m/s = distance / time = 972 / 27 m/s.Step 4: Compute 972 / 27: since 27 * 36 = 972, relative speed = 36 m/s.Step 5: Since both trains have the same speed and are moving in opposite directions, 2 * (speed of each train in m/s) = 36.Step 6: So speed of each train in m/s = 36 / 2 = 18 m/s.Step 7: Convert 18 m/s to km/hr by multiplying by 18 / 5 or equivalently by 3.6.Step 8: Speed in km/hr = 18 * 3.6 = 64.8 km/hr.Verification / Alternative check:Using the found speed, convert 64.8 km/hr back to m/s: 64.8 * (5 / 18) = 18 m/s. The relative speed of two trains in opposite directions is 18 + 18 = 36 m/s. At 36 m/s, time to cover 972 m is 972 / 36 = 27 seconds, which exactly matches the given crossing time. This confirms the correctness of 64.8 km/hr.Why Other Options Are Wrong:55, 56.4, 73.2, or 60 km/hr all yield different m/s speeds and thus different crossing times when used. For example, 60 km/hr corresponds to 16.67 m/s, so relative speed would be about 33.34 m/s and crossing time would be longer than 27 seconds. Only 64.8 km/hr yields the correct relative speed and crossing time.Common Pitfalls:One common error is to forget that both train lengths must be added to get the total crossing distance. Another is to convert km/hr to m/s incorrectly, or to forget to divide the relative speed by 2 when determining each train's speed. Carefully tracking units and using the relative speed concept correctly avoids these mistakes.Final Answer:The speed of each train is 64.8 km/hr.

Question 11

In what time (in seconds) will a train 60 metres long, running at a speed of 36 km/hr, completely pass a stationary telegraph post beside the track?

Accepted Answer

Introduction / Context:This is a basic train and distance question. When a train passes a telegraph post or any fixed point, the distance relevant for the passing is simply the length of the train. We are given the speed and length of the train and need to find the time taken to pass the post.Given Data / Assumptions:- Length of the train = 60 m.- Speed of the train = 36 km/hr.- The telegraph post is a fixed point with negligible width.- The train moves with constant speed throughout the crossing.Concept / Approach:Time is given by time = distance / speed. The distance is the train length, and the speed must be in metres per second to match the distance units. We convert the speed from km/hr to m/s, then apply the formula to find the time in seconds.Step-by-Step Solution:Step 1: Distance to be covered relative to the post = length of the train = 60 m.Step 2: Convert speed from km/hr to m/s: 36 km/hr = 36 * (5 / 18) m/s.Step 3: Compute the conversion: 36 * (5 / 18) = (36 / 18) * 5 = 2 * 5 = 10 m/s.Step 4: Use time = distance / speed: time = 60 / 10 seconds.Step 5: This gives time = 6 seconds.Step 6: Therefore, the train takes 6 seconds to pass the telegraph post completely.Verification / Alternative check:We can quickly sanity-check the result. At 10 m/s, a train covers 10 metres each second. To travel 60 metres, it needs 60 / 10 = 6 seconds. This simple mental check matches the detailed calculation exactly.Why Other Options Are Wrong:9, 8, 7, and 5 seconds would correspond to effective speeds of 60 / 9, 60 / 8, 60 / 7, and 60 / 5 m/s respectively, none of which equal the actual 10 m/s converted from 36 km/hr. Only 6 seconds is consistent with the given length and speed.Common Pitfalls:Typical errors include incorrectly converting km/hr to m/s (for example, multiplying by 18 / 5), or using a wrong distance such as assuming a platform instead of a post. Remember: for crossing a fixed point, only the train's own length matters.Final Answer:The 60 metre long train will pass the telegraph post completely in 6 seconds.

Question 12

Direction: In the following question, two quantities I and II are given. Compare quantity I and quantity II on its basis. (Only quantity is to be considered)  Two trains A and B are running in the same direction and they cross a 75 m long platform i…

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Concept & Relative Speed: When dealing with trains crossing platforms or other trains, use the fundamental formula linking distance, speed, and time. Time = (Total Distance/Relative Speed) For opposite directions, relative speed is the sum of their speeds; for the same direction, it's the difference. Convert all units to meters and seconds. Step-by-Step Solution: • Convert Units: The speed difference is 25.2 km/h. Convert to m/s: 25.2 × (5/18) = 7 m/s. • Let v_A be the speed of Train A. Then v_B = v_A + 7. • Let L_B be the length of Train B. Then L_A = L_B + 45. • Set up Equations: • Train A crosses a 75m platform in 25s: (L_A + 75/v_A) = 25 \Rightarrow L_A + 75 = 25v_A. • Train B crosses a 75m platform in 15s: (L_B + 75/v_B) = 15 \Rightarrow L_B + 75 = 15v_B. • Solve for Variables: Substitute L_A and v_B into the equations: • (L_B + 45) + 75 = 25v_A \Rightarrow L_B + 120 = 25v_A \Rightarrow L_B = 25v_A - 120. • L_B + 75 = 15(v_A + 7) \Rightarrow L_B = 15v_A + 105 - 75 \Rightarrow L_B = 15v_A + 30. • Equate both expressions for L_B: 25v_A - 120 = 15v_A + 30. • 10v_A = 150 \Rightarrow v_A = 15 m/s. • Then, v_B = 15 + 7 = 22 m/s. • L_B = 15(15) + 30 = 255 m. • L_A = 255 + 45 = 300 m. • Calculate Quantity I: Trains crossing each other in opposite directions. • Time = (L_A + L_B/v_A + v_B) = (300 + 255/15 + 22) = (555/37) = 15 seconds. • Calculate Quantity II: Train B with speed increased by 3 m/s (22 + 3 = 25 m/s) crossing a 95 m tunnel. • Time = (L_B + 95/25) = (255 + 95/25) = (350/25) = 14 seconds. • Compare: Quantity I (15) > Quantity II (14). Exam Strategy & Shortcut: Whenever a problem involves "speed difference in km/h", immediately convert it to m/s before reading further. This prevents unit mismatch errors deep into the calculation. Also, set up a linear system of equations quickly by isolating the shared variable (L_B in this case). Common Pitfall: A classic mistake is forgetting to add the platform's length to the train's length when calculating the total distance covered. Final Answer: Therefore, the correct answer is Quantity I > Quantity II.

Question 13

Direction: In the following question, two quantities I and II are given. Compare quantity I and quantity II on its basis. (Only quantity is to be considered)  Train A crosses train B in 13 seconds while running in the opposite direction and train C …

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Concept & Relative Speed Dynamics: This problem merges relative speed kinematics with logical sequence generation. Remember that when crossing a platform, Distance = Train Length + Platform Length. When trains cross each other, Distance = Sum of their Lengths. Relative speed is additive in opposite directions and subtractive in the same direction. Step-by-Step Solution: • Determine Speeds: • Average speed of A and B is 15 m/s \Rightarrow (v_A + v_B/2) = 15 \Rightarrow v_A + v_B = 30 m/s. • Given v_B = 14 m/s, so v_A = 30 - 14 = 16 m/s. • Speed of Train C is 4 m/s more than Train A \Rightarrow v_C = 16 + 4 = 20 m/s. • Determine Lengths: • Train A crosses a 140 m platform in 20 s: (L_A + 140/16) = 20 \Rightarrow L_A + 140 = 320 \Rightarrow L_A = 180 m. • Train A crosses Train B in 13 s (opposite direction): (L_A + L_B/v_A + v_B) = 13 \Rightarrow (180 + L_B/30) = 13 \Rightarrow 180 + L_B = 390 \Rightarrow L_B = 210 m. • Train C crosses Train A in 100 s (same direction): (L_C + L_A/v_C - v_A) = 100 \Rightarrow (L_C + 180/20 - 16) = 100 \Rightarrow L_C + 180 = 400 \Rightarrow L_C = 220 m. • Calculate Quantity I: • New speed of Train C = 95% of 20 = 19 m/s. • Time for C to cross B in same direction: Time = (L_C + L_B/v_C - v_B) = (220 + 210/19 - 14) = (430/5) = 86 seconds. • Calculate Quantity II: • 1^{st} term = L_A = 180. • 2^{nd} term = 180 - 2^2 = 176. • 3^{rd} term = 176 - 3^2 = 176 - 9 = 167. • 4^{th} term = 167 - 5^2 = 167 - 25 = 142. (Subtracting squares of consecutive prime numbers). • 5^{th} term = 142 - 7^2 = 142 - 49 = 93. • Compare: Quantity I (86) < Quantity II (93). Exam Strategy & Shortcut: Organize your scratchpad systematically: first dedicate a section to speeds (v_A, v_B, v_C), then to lengths (L_A, L_B, L_C). This prevents confusion when multiple interacting variables exist. For series problems embedded in algebra, don't compute until you are sure of the subtraction logic (here, consecutive prime squares). Common Pitfall: A major pitfall is missing the fact that the series subtracts the squares of prime numbers (2, 3, 5, 7), not odd numbers. If a student subtracts 1^2, 3^2, 5^2, they will derive an incorrect sequence. Final Answer: Therefore, the correct answer is Quantity I < Quantity II.

Question 14

Paschim Express left Delhi for Mumbai at 14.30 hrs travelling at a speed of 60 kmph and August Kranti Express left Delhi for Mumbai on the same day at 16.30 hrs travelling at a speed of 80 kmph. How far away from Delhi will the two trains meet (excl…

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Concept & Trains in Same Direction: When two trains travel in the same direction, the faster train starting later will eventually overtake the slower train. The time taken to overtake depends on the initial head start distance and their relative speed. Time\ to\ Overtake = (Head\ Start\ Distance/Relative\ Speed) Step-by-Step Solution: 1. Determine the time gap: Paschim Express leaves at 14:30. August Kranti leaves at 16:30. Time difference = 2 hours. 2. Calculate initial head start: In these 2 hours, Paschim Express covers a certain distance. Head start distance = Speed of Paschim × Time = 60 km/h × 2 h = 120 km. 3. Find relative speed: Since both travel from Delhi to Mumbai (same direction), relative speed = Speed of August Kranti - Speed of Paschim. Relative speed = 80 km/h - 60 km/h = 20 km/h. 4. Calculate time to catch up: Time taken for August Kranti to catch Paschim = Head start / Relative speed. Time = 120 km / 20 km/h = 6 hours. 5. Calculate meeting distance: The meeting distance from Delhi is the distance August Kranti Express travels in those 6 hours. Distance = Speed of August Kranti × Time = 80 km/h × 6 h = 480 km. Exam Strategy & Shortcut: Let t be the time taken by August Kranti to catch up. Paschim Express travels for t + 2 hours. Distance is same: 60(t + 2) = 80t \implies 60t + 120 = 80t \implies 20t = 120 \implies t = 6 hrs. Distance from Delhi = 80 × 6 = 480 km. Common Pitfall: Often students correctly find the time it takes the second train to catch the first (6 hours) but then incorrectly multiply this time by the slower train's speed or forget to add the initial 2 hours if they use the slower train for the final calculation (60 × (6+2) = 480). Final Answer: Therefore, the correct answer is 480 km.

Question 15

A train started from station $A$ and proceeded towards station $B$ at a speed of $48$ km/hr. Forty-five minutes later another train started from station $B$ and proceeded towards station a at $50$ km/hr. If the distance between the two stations is $…

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Concept & Relative Speed with Head Start: When two trains start at different times, calculate the distance covered by the first train during its "head start". Subtract this from the total distance to find the remaining gap. Then, use relative speed to find the time it takes to close the remaining gap. Step-by-Step Solution: • Given: Total distance = 232 km. Speed of Train from A = 48 km/hr. Speed of Train from B = 50 km/hr. Train A starts 45 minutes (3/4 hour) early. • Calculation: Distance traveled by the train from A in 45 minutes = 48 × (3/4) = 36 km. • Remaining distance between the two trains when the train from B starts = 232 - 36 = 196 km. • Relative speed of the two trains (moving towards each other) = 48 + 50 = 98 km/hr. • Time taken for them to meet after the train from B starts = (196/98) = 2 hours. • The question asks for the meeting distance from station A. The train from A traveled for 45 minutes initially, plus 2 hours more. • Distance covered by train from A in the 2 hours = 48 × 2 = 96 km. • Total distance from station A = Head start distance + Distance traveled until meeting = 36 + 96 = 132 km. Exam Strategy & Shortcut: Find the distance remaining after the head start: 232 - (48 × 0.75) = 196. Find meeting time: 196 / (48 + 50) = 2 hours. Calculate distance from A: Initial 36 km + (2 × 48) = 132 km. Common Pitfall: A common error is to calculate the distance traveled by train A only during the simultaneous travel time (96 km), forgetting to add the initial head start distance (36 km). Final Answer: Therefore, the correct answer is 132 km.

Question 16

A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, then the length of the train is

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Concept & Formula: First, find the speed of the train using the given distance and time. Once the uniform speed is known, use it along with the time taken to pass the pole to find the train's length. Speed = \frac{Distance}{Time} Length of Train = Speed (m/s) × Time to cross post (s) Step-by-Step Solution: • Given for Speed: Distance = 12 km = 12000 m. Time = 10 minutes = 10 × 60 = 600 seconds. • Calculate Speed: Speed = (12000/600) = 20 m/s. • Given for Crossing: Time to pass post = 6 seconds. • Calculate Length: Length = 20 m/s × 6 s = 120 m. Exam Strategy & Shortcut: 12 km in 10 minutes is equivalent to 72 km in 60 minutes (72 km/hr). We know 72 km/hr is exactly 20 m/s (4 × 18 -> 4 × 5). Multiplying 20 m/s by 6 seconds immediately yields 120 m. Common Pitfall: Mixing units. If you use 12 km and 10 minutes directly, you might end up with speed in km/min and apply it incorrectly to the 6 seconds. Always standardize to meters and seconds. Final Answer: Therefore, the correct answer is 120 m.

Question 17

A 280-metre long train crosses a platform thrice its length in 50 seconds. What is the speed of the train in km/hr?

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Concept & Formula: The total distance covered by the train to cross the platform is the sum of the train's length and the platform's length. Total Distance = L_{train} + L_{platform} Step-by-Step Solution: • Given: • Length of the train (L) = 280 m • Length of the platform = 3 × 280 = 840 m • Time taken (T) = 50 seconds • Calculation: • Total distance = 280 + 840 = 1120 m. • Speed in m/s = (1120/50) = (112/5) m/s. • Convert to km/hr = (112/5) × (18/5) = (2016/25). • (2016/25) = (2016 × 4/100) = (8064/100) = 80.64 km/hr. Exam Strategy & Shortcut: Instead of calculating 3 × 280 and adding 280, recognize that the total length is simply 4 times the train's length (4 × 280 = 1120). When dividing by 25, multiply the numerator by 4 and place the decimal two spots to the left. Common Pitfall: Students might calculate the platform's length as 3 × 280 but forget to add the train's original length when determining the total distance covered. Final Answer: Therefore, the correct answer is 80.64.

Question 18

A train travelling with a constant speed crosses a 96-metre long platform in 12 seconds and another 141-metre long platform in 15 seconds. The length of the train and its speed are

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Concept & Formula: When a train crosses two platforms of different lengths, the difference in the time taken is entirely due to the difference in the lengths of the platforms. Speed = \frac{Difference in Platform Lengths}{Difference in Time} Step-by-Step Solution: • Given: • Platform 1: 96 m, Time 1: 12 seconds. • Platform 2: 141 m, Time 2: 15 seconds. • Calculation: • Difference in lengths = 141 - 96 = 45 m. • Difference in times = 15 - 12 = 3 seconds. • Speed of train = (45/3) = 15 m/s. • Convert speed to km/hr = 15 × (18/5) = 3 × 18 = 54 km/hr. • To find the length of the train (L): In the first scenario, total distance = L + 96. • L + 96 = Speed × Time = 15 × 12 = 180 m. • L = 180 - 96 = 84 m. Exam Strategy & Shortcut: Find the speed immediately: (141 - 96/15 - 12) = (45/3) = 15 m/s. Eliminate options with incorrect speeds (15 m/s = 54 km/hr, so options (b) and (d) are out). Then, simply calculate length: (15 × 12) - 96 = 84 m. Common Pitfall: Setting up two separate complex equations (L + 96 = 12S and L + 141 = 15S) and trying to solve them simultaneously wastes precious time compared to using the difference method. Final Answer: Therefore, the correct answer is 84 metres and 54 km/hr.

Question 19

A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

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Concept & Relative Speed: When two bodies move in the same direction, their relative speed is the difference between their individual speeds. To completely pass the jogger, the train must cover the initial distance gap plus its own entire length. Relative Speed = V_{train} - V_{jogger} Total Distance = Initial Gap + Length of Train Step-by-Step Solution: • Given: • Speed of jogger = 9 kmph. • Speed of train = 45 kmph. • Initial gap = 240 m. • Length of train = 120 m. • Calculation: • Relative speed = 45 - 9 = 36 kmph. • Convert to m/s: 36 × (5/18) = 10 m/s. • Total distance the train needs to cover relative to the jogger = 240 + 120 = 360 m. • Time taken = \frac{Total Distance}{Relative Speed} = (360/10) = 36 seconds. Exam Strategy & Shortcut: Instantly convert the relative speed of 36 kmph to 10 m/s. Add the gap and the train's length (360 m) in your head. Divide 360 by 10 to get 36 seconds immediately. Common Pitfall: A very common mistake is only calculating the time required to cover the 240m gap, forgetting that the train must also advance its own 120m length past the jogger to completely "pass" him. Final Answer: Therefore, the correct answer is 36 sec.

Question 20

A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

Accepted Answer

Concept & Formula: When two bodies move in opposite directions, their relative speed is the sum of their individual speeds. Since a man is considered a point object, the distance to be covered is simply the length of the train. Relative Speed = V_{train} + V_{man} Step-by-Step Solution: • Given: • Length of train = 110 m. • Speed of train = 60 kmph. • Speed of man = 6 kmph. • Calculation: • Relative speed = 60 + 6 = 66 kmph. • Convert to m/s: 66 × (5/18) = (330/18) = (55/3) m/s. • Total distance = 110 m. • Time taken = \frac{Distance}{Relative Speed} = (110/\frac{55){3}} seconds. • Time = 110 × (3/55) = 2 × 3 = 6 seconds. Exam Strategy & Shortcut: Instead of fully simplifying (330/18), you can write the time equation as (110 × 18/330). The 110 goes into 330 exactly 3 times, leaving (18/3) = 6 seconds. Common Pitfall: Students often subtract the speeds by mistake out of habit. Always look for the keyword "opposite" to remind yourself to add the speeds together. Final Answer: Therefore, the correct answer is 6 sec.

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