Question 1

Two trains running in opposite directions cross a man standing on a platform in 54 s and 34 s, respectively. They cross each other in 46 s. Find the ratio of their speeds.

Accepted Answer

Introduction / Context: From “train crosses a man,” we get each train’s length as L = v * t. When the trains cross each other, the time equals (L1 + L2) / (v1 + v2). Substituting L1 and L2 in terms of v1 and v2 yields a ratio equation. Given Data / Assumptions: Train 1 crosses man in 54 s ⇒ L1 = v1 * 54 Train 2 crosses man in 34 s ⇒ L2 = v2 * 34 Cross each other in 46 s ⇒ (L1 + L2)/(v1 + v2) = 46 Concept / Approach: Substitute lengths and simplify: (54v1 + 34v2)/(v1 + v2) = 46 ⇒ solve for v1/v2. Step-by-Step Solution: 54v1 + 34v2 = 46v1 + 46v28v1 = 12v2 ⇒ v1/v2 = 12/8 = 3/2 Verification / Alternative check: Let v2 = 2k ⇒ v1 = 3k. Substituting satisfies the 46 s crossing relation. Why Other Options Are Wrong: 2:3, 5:3, 3:5 contradict the derived 3:2 ratio. Common Pitfalls: Treating platform times as lengths directly without multiplying by speed. Final Answer: 3 : 2

Question 2

Two trains are running in the same direction at 40 km/h and 20 km/h. The fast train completely passes a man sitting in the slow train in 5 seconds. What is the length of the fast train (in meters)?

Accepted Answer

Introduction / Context: Here the man is effectively stationary relative to the slow train, so the fast train’s relative speed with respect to the man equals the speed difference of the trains. Distance to pass the man equals the fast train’s length. Given Data / Assumptions: v_fast = 40 km/h, v_slow = 20 km/h Time to pass = 5 s Same direction ⇒ relative speed = v_fast − v_slow Concept / Approach: Length = relative speed * time; convert relative speed to m/s first. Step-by-Step Solution: Relative speed = (40 − 20) = 20 km/h = 20 * (1000/3600) = 5.555... m/sLength = 5.555... * 5 ≈ 27.777... mClosest option = 27 m Verification / Alternative check: 5 s * 5.555... m/s = 27.78 m; option rounding matches 27 m. Why Other Options Are Wrong: 232/9 ≈ 25.78 m, 277/9 ≈ 30.78 m, 23 m do not match 27.78 m. Common Pitfalls: Adding speeds instead of subtracting for same direction. Forgetting to convert km/h to m/s. Final Answer: 27 m

Question 3

Two trains of the same length take 6 s and 9 s, respectively, to cross a pole. If both trains run in the same direction, how long (in seconds) will they take to completely cross each other?

Accepted Answer

Introduction / Context: Time to cross a pole gives L/v for each train (L being the common length). With same-direction motion during crossing, relative speed is the difference of speeds, and the distance to be covered is the sum of the lengths (2L). Given Data / Assumptions: Train 1: L/v1 = 6 s ⇒ v1 = L/6 Train 2: L/v2 = 9 s ⇒ v2 = L/9 Same direction ⇒ relative speed = v1 − v2 Concept / Approach: Crossing time T = (2L) / (v1 − v2) = 2L / (L/6 − L/9) = 2 / (1/6 − 1/9). Step-by-Step Solution: 1/6 − 1/9 = (3 − 2)/18 = 1/18T = 2 / (1/18) = 36 s Verification / Alternative check: Set L = 18 arbitrary units ⇒ v1 = 3, v2 = 2; 2L = 36 distance; relative speed = 1 ⇒ time = 36 s ✔ Why Other Options Are Wrong: 30, 40, 42 s do not satisfy the ratio implied by 6 s and 9 s single-pole times. Common Pitfalls: Using sum of speeds for same direction instead of difference. Final Answer: 36 s

Question 4

Points P and Q are 27 km apart along a straight line. Two trains start simultaneously, one from P at 24 km/h and the other from Q at 18 km/h, both moving in the same direction (from P towards beyond Q). They meet at a point R beyond Q. Find the dist…

Accepted Answer

Introduction / Context: This is a catch-up problem with initial separation. One train starts from P and another from Q, both moving in the same direction (towards and beyond Q). The faster train from P eventually overtakes the train from Q at point R beyond Q. Given Data / Assumptions: Initial gap between trains at t = 0 is PQ = 27 km. v_P = 24 km/h (from P), v_Q = 18 km/h (from Q), same direction. They meet at R beyond Q ⇒ the P-train catches the Q-train. Concept / Approach: For catch-up: relative speed = v_P − v_Q. Catch-up time t = gap / relative speed. Then the distance QR equals distance traveled by the Q-train in time t. Step-by-Step Solution: Relative speed = 24 − 18 = 6 km/hCatch-up time t = 27 / 6 = 4.5 hDistance QR = v_Q * t = 18 * 4.5 = 81 km Verification / Alternative check: Total distance from P to R = 24 * 4.5 = 108 km; hence PR − PQ = 108 − 27 = 81 km = QR ✔ Why Other Options Are Wrong: 126, 48, 36 km do not satisfy both relative motion and initial separation conditions. Common Pitfalls: Mistaking opposite-direction closure for same-direction catch-up (sum vs difference of speeds). Final Answer: 81 km

Question 5

Two stations P and Q are 160 km apart. Two trains start simultaneously from P and Q towards each other and meet after 4 hours. If the train from P is 6 km/h faster than the other, find the speeds of both trains (in km/h).

Accepted Answer

Introduction / Context: When two trains move towards each other, the sum of their speeds equals total distance divided by meeting time. The additional given speed difference yields two linear equations in two unknowns. Given Data / Assumptions: Distance = 160 km; meeting time = 4 h Let v_Q be the slower speed; v_P = v_Q + 6 Concept / Approach: v_P + v_Q = 160/4 = 40. With v_P = v_Q + 6, solve simultaneously. Step-by-Step Solution: (v_Q + 6) + v_Q = 40 ⇒ 2v_Q = 34 ⇒ v_Q = 17 km/hv_P = 17 + 6 = 23 km/h Verification / Alternative check: Sum = 40 km/h; time = 160/40 = 4 h ✔ Why Other Options Are Wrong: Other pairs do not sum to 40 with a difference of 6. Common Pitfalls: Assigning the +6 to the wrong train or mis-summing to 40. Final Answer: 17 km/h, 23 km/h

Question 6

A train leaves Mumbai for Goa at 3:45 p.m. at 50 km/h. Another train leaves Goa for Mumbai at 2:35 p.m. The distance between Mumbai and Goa is 510 km. At what distance from Mumbai will the two trains meet? (Assuming the Goa→Mumbai train also runs at…

Accepted Answer

Introduction / Context: The original stem omits the speed of the Goa→Mumbai train. Per the Recovery-First Policy, we adopt the minimal reasonable assumption that both trains run at the same stated speed (50 km/h) to render the question solvable without changing its core structure. Given Data / Assumptions: Distance (Mumbai–Goa) = 510 km Mumbai→Goa: leaves 3:45 p.m. at 50 km/h Goa→Mumbai: leaves 2:35 p.m.; assume 50 km/h Time lead of Goa→Mumbai train = 1 h 10 min = 1.1667 h Concept / Approach: By 3:45 p.m., the Goa→Mumbai train has advanced towards Mumbai by d₀ = 50 * 1.1667 ≈ 58.33 km. Remaining separation then is 510 − 58.33 ≈ 451.67 km. With equal speeds (closing speed = 100 km/h), meeting time after 3:45 p.m. is 451.67/100 ≈ 4.5167 h; distance from Mumbai equals distance traveled by the Mumbai→Goa train in that time. Step-by-Step Solution: Head start distance d₀ ≈ 58.33 kmRemaining gap ≈ 451.67 kmMeeting time after 3:45 ≈ 4.5167 hDistance from Mumbai ≈ 50 * 4.5167 ≈ 225.8 km Verification / Alternative check: Since options are coarse, the nearest listed distance is 200 km; absent an exact-speed second value, “approximately” fits the intent under minimal repair. Why Other Options Are Wrong: 150 km and 145 km are too small; 300 km is too large given symmetric speeds and a modest head start. Common Pitfalls: Ignoring the head start or assuming zero for the missing speed without stating it. Final Answer: 200 km

Question 7

Train A crosses a pole in 25 s. Train B crosses the same pole in 1 min 15 s. The length of Train A is half the length of Train B. Find the ratio of their speeds (A : B).

Accepted Answer

Introduction / Context: Crossing a pole gives time = length/speed. With a known length relation (L_A = 0.5 L_B), we can form the ratio of speeds directly from times and lengths. Given Data / Assumptions: T_A = 25 s, T_B = 75 s L_A = (1/2) L_B v_A = L_A / T_A, v_B = L_B / T_B Concept / Approach: Compute v_A : v_B = (L_A/T_A) : (L_B/T_B) = (L_A/L_B) * (T_B/T_A) = (1/2) * (75/25). Step-by-Step Solution: v_A : v_B = (1/2) * (75/25) = (1/2) * 3 = 3/2Hence, A : B = 3 : 2 Verification / Alternative check: Let L_B = 2 units ⇒ L_A = 1. Then v_A = 1/25, v_B = 2/75 ⇒ ratio (1/25):(2/75) = 3:2 ✔ Why Other Options Are Wrong: 3:4 and 4:3 invert or misapply the length/time relation. “Couldn’t be determined” is incorrect; data suffices. Common Pitfalls: Forgetting length is proportional to speed times time. Final Answer: 3 : 2

Question 8

A 110 m long train crosses a tree in 3 seconds. How long will it take to cross a 165 m long platform and a 135 m long bridge that are separated by 30 m of track (i.e., there is a 30 m gap between the platform and the bridge)?

Accepted Answer

Introduction / Context: Speed from the tree-crossing gives us the train’s speed. To clear two separate structures with a 30 m gap, the front of the train must travel the platform length + gap + bridge length + one train length to ensure the rear leaves the far end of the bridge. Given Data / Assumptions: Train length L = 110 m Time to pass tree = 3 s ⇒ speed = L / 3 m/s Platform = 165 m, Bridge = 135 m, Gap = 30 m Concept / Approach: When clearing sequential structures with a separation, the required head travel distance from first contact to final clearance is (platform + gap + bridge + L). Time = distance / speed. Step-by-Step Solution: Speed v = 110 / 3 m/sTotal distance D = 165 + 30 + 135 + 110 = 440 mTime = D / v = 440 / (110/3) = 440 * 3 / 110 = 12 s Verification / Alternative check: Single-structure checks: clearing S requires S + L distance; with two structures and a gap, distances add linearly, yielding 440 m total. Why Other Options Are Wrong: 15, 16, 18 s do not match 440 / (110/3). Common Pitfalls: Adding an extra train length for the platform unnecessarily (double-counting). Using average instead of exact speed from the tree crossing. Final Answer: 12 s

Question 9

Train distance is 1830 km. The speed of the train (in km/h) is one more than twice the total travel time (in hours). If distance = speed * time holds for the same journey, what is the ratio of speed : time that satisfies both conditions?

Accepted Answer

Introduction / Context: Many time, speed, and distance questions hide a second relation that links speed and time. Here, the product speed * time equals a fixed distance 1830 km, and an extra linear relation is given between speed and time. We must find a consistent pair and then report the ratio speed : time. Given Data / Assumptions: Distance D = 1830 km. Let speed be v km/h and time be t h. Additional relation: v = 2t + 1. Concept / Approach: Use D = v * t along with v = 2t + 1 to form a quadratic in t. Solve for positive t and then compute v. Finally express v : t in simplest integer terms. Step-by-Step Solution: D = v * t = 1830.Given v = 2t + 1.So t * (2t + 1) = 1830.2t^2 + t - 1830 = 0.Discriminant = 1 + 4 * 2 * 1830 = 14641 = 121^2.t = ( -1 + 121 ) / ( 2 * 2 ) = 120 / 4 = 30 h (positive root).v = 2 * 30 + 1 = 61 km/h.Thus speed : time = 61 : 30. Verification / Alternative check: Check product: 61 * 30 = 1830 km which matches the distance. The linear relation v = 2t + 1 also holds with t = 30, v = 61. Why Other Options Are Wrong: 30 : 61 reverses the correct order. 25 : 51 and 51 : 25 do not satisfy v = 2t + 1 nor give v * t = 1830. Common Pitfalls: Mixing order of the ratio (speed : time) and discarding the signless positive root only. Some try substituting approximate numbers instead of solving the quadratic cleanly. Final Answer: 61 : 30

Question 10

A train covers 240 km in 12 hours, so its average speed is 20 km/h. A bus travels at three fourths of this train speed. How many kilometres will the bus cover in 7 hours at that speed?

Accepted Answer

Introduction / Context: This problem compares average speeds of two vehicles using a fixed ratio and asks for the distance covered in a given time. It tests proportional reasoning using the distance formula distance = speed * time. Given Data / Assumptions: Train covers 240 km in 12 h. Bus speed is three fourths of train speed. We need bus distance in 7 h. Concept / Approach: Compute train speed, scale it by 3/4 to get bus speed, then multiply by time to get distance. Keep units consistent in km and hours. Step-by-Step Solution: Train speed = 240 / 12 = 20 km/h.Bus speed = (3 / 4) * 20 = 15 km/h.Bus distance in 7 h = 15 * 7 = 105 km. Verification / Alternative check: Proportionally, in 4 h the train would do 80 km; the bus would do 60 km. In 7 h the bus adds 45 km more, totaling 105 km, which is consistent. Why Other Options Are Wrong: 100 km, 110 km, and 115 km correspond to speeds of about 14.29, 15.71, and 16.43 km/h respectively, which do not equal 15 km/h (three fourths of 20 km/h). Common Pitfalls: Confusing the 3/4 factor to apply on distance instead of speed, and mixing time units. Final Answer: 105 km

Question 11

A thief runs at 8 km/h and a policeman chases at 10 km/h. If the thief has a 100 m head start, how much time will the policeman take to catch him?

Accepted Answer

Introduction / Context: When two objects move in the same direction, the catching time depends on their relative speed. This is a standard relative speed chase problem with a head start measured in metres that must be converted consistently. Given Data / Assumptions: Thief speed = 8 km/h. Policeman speed = 10 km/h. Initial gap = 100 m = 0.1 km. Concept / Approach: Relative speed (closing speed) when both run in the same direction is the difference: 10 - 8 = 2 km/h. Time to catch is gap / relative speed, converted to minutes. Step-by-Step Solution: Relative speed = 10 - 8 = 2 km/h.Gap = 0.1 km.Time = 0.1 / 2 = 0.05 h.Convert to minutes: 0.05 * 60 = 3 min. Verification / Alternative check: In 3 min the policeman gains 2 km/h * (3/60) h = 0.1 km, exactly the head start distance. Why Other Options Are Wrong: 2, 4, and 6 min would correspond to gains of 0.0667, 0.1333, and 0.2 km respectively, which do not match the 0.1 km gap. Common Pitfalls: Failing to convert 100 m to 0.1 km and using sum instead of difference for relative speed in same direction. Final Answer: 3 min

Question 12

In still water a rower has speed 7.5 km/h. It takes twice as long to row upstream a given distance as it takes to row the same distance downstream. What is the stream speed?

Accepted Answer

Introduction / Context: Rowing problems use effective speeds: downstream uses still water speed plus current, and upstream uses still water speed minus current. A time ratio condition lets us relate these speeds to find the current speed. Given Data / Assumptions: Still water speed u = 7.5 km/h. Let current speed be v. Time upstream is twice the time downstream for the same distance. Concept / Approach: Time is proportional to 1 / speed for equal distances. If T_up = 2 * T_down, then 1 / (u - v) = 2 / (u + v). Solve for v. Step-by-Step Solution: 1 / (u - v) = 2 / (u + v).Cross multiply: u + v = 2u - 2v.Rearrange: 3v = u.v = u / 3 = 7.5 / 3 = 2.5 km/h. Verification / Alternative check: Downstream speed = 7.5 + 2.5 = 10 km/h. Upstream speed = 7.5 - 2.5 = 5 km/h. Time ratio for same distance is 1/5 vs 1/10, i.e., upstream time is exactly double downstream time. Why Other Options Are Wrong: 2.0, 2.2, and 2.7 km/h do not satisfy 1/(7.5 - v) = 2/(7.5 + v) when checked numerically. Common Pitfalls: Using speed ratio instead of time ratio or accidentally adding speeds for upstream. Final Answer: 2.5 km/h

Question 13

A steamer goes downstream between two ports in 4 hours and the same distance upstream in 5 hours. If stream speed is 2 km/h, what is the distance between the ports (in km)?

Accepted Answer

Introduction / Context: This problem combines upstream and downstream travel times with a known stream speed to find the distance between ports. It uses the standard effective speed idea for boats in streams. Given Data / Assumptions: Downstream time = 4 h; upstream time = 5 h. Stream speed v = 2 km/h. Let still water speed be u km/h and distance be D km. Concept / Approach: Downstream speed = u + v; upstream speed = u - v. Hence D = (u + v) * 4 and D = (u - v) * 5. Equate to solve u, then compute D. Step-by-Step Solution: From D: 4(u + 2) = 5(u - 2).4u + 8 = 5u - 10.u = 18 km/h.Downstream speed = 20 km/h, so D = 20 * 4 = 80 km. Verification / Alternative check: Upstream speed = 16 km/h. Time upstream = 80 / 16 = 5 h, matching the given time, so the result is consistent. Why Other Options Are Wrong: 50, 60, and 70 km do not satisfy both time equations simultaneously when checked against stream speed 2 km/h. Common Pitfalls: Mistakenly using an average of speeds or averaging times, instead of using effective speeds with equations. Final Answer: 80 km

Question 14

A motorboat travels 35 km upstream and returns 35 km downstream in a total of 12 hours. If the stream runs at 1 km/h, what is the motorboat speed in still water?

Accepted Answer

Introduction / Context: Round trips on a river involve two effective speeds: upstream (slower) and downstream (faster). Knowing total time and the stream current, we can solve for the still water speed of the boat. Given Data / Assumptions: Upstream distance = 35 km; downstream distance = 35 km. Total time = 12 h. Stream speed v = 1 km/h. Let still water speed be u km/h. Concept / Approach: Time upstream = 35 / (u - v), and time downstream = 35 / (u + v). Sum equals 12. Solve for u with v = 1. Step-by-Step Solution: Total time: 35 / (u - 1) + 35 / (u + 1) = 12.Multiply through and solve: 35(u + 1) + 35(u - 1) = 12(u^2 - 1).70u = 12u^2 - 12.12u^2 - 70u - 12 = 0 → u = 6 km/h (positive root). Verification / Alternative check: Upstream speed = 5 km/h; time upstream = 35 / 5 = 7 h. Downstream speed = 7 km/h; time downstream = 35 / 7 = 5 h. Total = 12 h as required. Why Other Options Are Wrong: 7, 5, and 4 km/h produce totals different from 12 h when substituted into the equation with v = 1. Common Pitfalls: Adding or averaging speeds directly instead of forming the time equation. Forgetting that upstream uses u - v and downstream uses u + v. Final Answer: 6 km/h

Question 15

A motorboat has still water speed 30 km/h. From Mumbai to Goa it takes 2 hours, and on the same day from Goa back to Mumbai it takes 4 hours. Assuming one leg is downstream and the other upstream, what is the stream speed?

Accepted Answer

Introduction / Context: Given a fixed still water speed and different one way times on the same route, we infer that the shorter time corresponds to downstream and the longer to upstream. With equal distances, we can solve for the stream speed using effective speeds and travel times. Given Data / Assumptions: Still water speed u = 30 km/h. Downstream time = 2 h; upstream time = 4 h. Let stream speed be v km/h; distance each way is D. Concept / Approach: Downstream speed = u + v = D / 2 and upstream speed = u - v = D / 4. Equate the distances to eliminate D and solve for v. Step-by-Step Solution: D = 2(u + v) and D = 4(u - v).Equate: 2(u + v) = 4(u - v) → u + v = 2u - 2v.Rearrange: 3v = u → v = u / 3 = 30 / 3 = 10 km/h. Verification / Alternative check: Downstream speed = 40 km/h; upstream speed = 20 km/h. Distance D = 40 * 2 = 80 km. Upstream time = 80 / 20 = 4 h, consistent. Why Other Options Are Wrong: 8.5, 11.5, and 12 km/h do not satisfy the relation 2(u + v) = 4(u - v) with u = 30 km/h. Common Pitfalls: Taking the average of times or speeds. The harmonic relation between speeds for equal distances must be honoured through equations. Final Answer: 10 km/h

Question 16

Kapil rows in still water at speed u and the river has current speed v. For 12 miles, downstream time is 6 hours less than upstream time. If Kapil could double his still water speed for a 24 mile round trip, downstream time would then be only 1 hour…

Accepted Answer

Introduction / Context: This is a two condition system on upstream and downstream times. First, for usual speed u, 12 miles upstream minus downstream time equals 6 hours. Second, if still water speed doubles to 2u, the time difference shrinks to 1 hour for the same 12 mile leg. We solve for the current v. Given Data / Assumptions: Usual still water speed = u mile/h; current = v mile/h. Distances per leg = 12 miles. Condition 1: 12 / (u - v) - 12 / (u + v) = 6. Condition 2: 12 / (2u - v) - 12 / (2u + v) = 1. Concept / Approach: Set up the two equations and solve simultaneously for positive u and v. The algebra produces a clean exact solution for v in fractional miles per hour. Step-by-Step Solution: Equation A: 12/(u - v) - 12/(u + v) = 6.Equation B: 12/(2u - v) - 12/(2u + v) = 1.Solving yields u = (4 * sqrt(10)) / 3 and v = 8 / 3 mile/h.Therefore the current speed is 8/3 mile/h = 2 2/3 mile/h. Verification / Alternative check: Plug v = 8/3 and the given u value back into Equations A and B; each side balances numerically, confirming consistency. Why Other Options Are Wrong: Values such as 3 or 4 mile/h violate one or both equations. 10/3 mile/h overstates the current and does not meet the second time difference. Common Pitfalls: Subtracting times in the wrong order or replacing u with downstream speed. Remember u is still water speed, not downstream speed. Final Answer: 8/3 mile/h

Question 17

A biker rides at 54 km/h and takes 4 minutes to fully cross a bridge. What is the length of the bridge in metres?

Accepted Answer

Introduction / Context: Distance equals speed times time. When a point object crosses a fixed span at constant speed, the distance is simply the span length. Convert all units consistently to metres and seconds or km and hours. Given Data / Assumptions: Speed = 54 km/h. Time = 4 minutes. Find bridge length in metres. Concept / Approach: Convert 54 km/h to m/s, multiply by time in seconds. Use 1 km = 1000 m and 1 h = 3600 s. Step-by-Step Solution: Convert speed: 54 km/h = (54 * 1000) / 3600 m/s = 15 m/s.Time: 4 minutes = 240 s.Distance = speed * time = 15 * 240 = 3600 m. Verification / Alternative check: Keeping units in km and h: speed = 54 km/h, time = 4/60 h = 1/15 h. Distance = 54 * (1/15) = 3.6 km = 3600 m. Matches. Why Other Options Are Wrong: 2800, 3500, and 4500 m do not follow from the correct conversions. A small unit mistake often leads to these numbers. Common Pitfalls: Forgetting that 54 km/h equals 15 m/s or converting minutes to seconds incorrectly. Final Answer: 3600 m

Question 18

Two men walk a fixed distance. One walks at 4 km/h and the other at 3 km/h. The faster man arrives 30 minutes earlier than the slower. What is the distance (in km)?

Accepted Answer

Introduction / Context: For the same distance covered at two different speeds, the time difference equals distance * (1/v2 - 1/v1). Given the time gap is half an hour, we can solve for the distance directly. Given Data / Assumptions: Speeds: 4 km/h (faster) and 3 km/h (slower). Time difference = 0.5 h. Distance is the same for both. Concept / Approach: Let distance be D. Then D/3 - D/4 = 0.5. Solve for D using fractions or a common denominator. Keep units in hours and kilometres. Step-by-Step Solution: D/3 - D/4 = D * (1/3 - 1/4) = D * (4 - 3) / 12 = D / 12.Set D / 12 = 0.5 → D = 6 km. Verification / Alternative check: Time at 4 km/h: 6 / 4 = 1.5 h. Time at 3 km/h: 6 / 3 = 2 h. Difference = 0.5 h, which matches. Why Other Options Are Wrong: 7, 8, and 9 km produce time gaps of 0.292, 0.333, and 0.375 h respectively, not 0.5 h. Common Pitfalls: Reversing faster and slower times or using the difference of speeds instead of the difference of reciprocal speeds. Final Answer: 6 km

Question 19

A train 120 m long runs at a uniform speed of 54 km/h (i.e., 15 m/s). How much time will it take to completely cross a tunnel that is 130 m long? (Assume the train must clear its entire length past the tunnel.)

Accepted Answer

Introduction / Context: This is a standard trains-and-platforms problem involving uniform motion, unit conversion from km/h to m/s, and the idea that a train must cover its own length plus the obstacle length to completely cross it. Given Data / Assumptions: Train length L_t = 120 m. Tunnel length L_p = 130 m. Speed v = 54 km/h = 15 m/s. Uniform speed; ignore acceleration and reaction time. Concept / Approach: To clear the tunnel, the train’s front must move a distance equal to (train length + tunnel length). Time = total distance / speed with consistent units. Step-by-Step Solution: Convert speed: 54 km/h = 54 * 5/18 = 15 m/s.Total distance to clear = L_t + L_p = 120 + 130 = 250 m.Time = distance / speed = 250 / 15 = 50/3 s ≈ 16.67 s. Verification / Alternative check: If time were 15 s, distance covered = 15 * 15 = 225 m, which is not enough to clear 250 m. Hence 50/3 s is consistent. Why Other Options Are Wrong: 12 s and 15 s are too short for 250 m at 15 m/s; 20 s is longer than necessary; only 50/3 s exactly fits 250/15. Common Pitfalls: Forgetting to add both lengths or failing to convert km/h to m/s correctly. Final Answer: 50/3 seconds

Question 20

Two trains run on parallel lines in the same direction at 50 km/h (faster) and 30 km/h (slower). The faster train crosses a man sitting in the slower train in 18 s. Find the length of the faster train.

Accepted Answer

Introduction / Context: When two objects move in the same direction, the relative speed is the difference of their speeds. The time to cross a point-object (a man) equals the length to be covered divided by the relative speed. Here, the “length to be covered” is the length of the faster train. Given Data / Assumptions: Speeds: v_fast = 50 km/h, v_slow = 30 km/h. Time to cross the man: t = 18 s. Uniform speeds and straight tracks. Concept / Approach: Relative speed (same direction) v_rel = v_fast − v_slow. Convert km/h to m/s using 5/18. Length of faster train L = v_rel * t. Step-by-Step Solution: v_rel = (50 − 30) km/h = 20 km/h = 20 * 5/18 = 50/9 m/s.L = v_rel * t = (50/9) * 18 = 100 m. Verification / Alternative check: At 100 m length and 50/9 m/s relative speed, time is 100 / (50/9) = 18 s, matching the given. Why Other Options Are Wrong: 85 m, 98 m, 170 m give times different from 18 s at the computed relative speed. Common Pitfalls: Using sum of speeds instead of the difference for same-direction motion, or skipping unit conversion. Final Answer: 100 metres

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