Question 1

Meeting point — variable speed walker vs constant speed walker: Akbar starts from Agra towards Banaras at a constant 8 km/h. Birbal starts from Banaras towards Agra and walks 4 km in the first hour, 5 km in the second, 6 km in the third, and so on (…

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Introduction / Context: This problem blends arithmetic progression (AP) distance accumulation with a constant-speed counterpart to determine a meeting location along a straight line. Given Data / Assumptions: Distance Agra–Banaras = 144 km Akbar: constant speed 8 km/h Birbal: hour-1 = 4 km, hour-2 = 5 km, hour-3 = 6 km, … an AP with a = 4, d = 1 Both start simultaneously toward each other. Concept / Approach: If they meet after n hours, Akbar covers 8n. Birbal covers S_n of the AP: S_n = n/2 * (2a + (n−1)d) = n/2 * (8 + n−1) = n/2 * (n + 7). The sum of distances equals 144 km. Step-by-Step Solution: 8n + (n/2)(n + 7) = 144Multiply by 2 ⇒ 16n + n(n + 7) = 288n^2 + 23n − 288 = 0Discriminant = 1681 = 41^2 ⇒ n = (−23 + 41)/2 = 9Akbar’s distance = 8 * 9 = 72 km; Birbal’s distance (AP sum) = 72 km Verification / Alternative check: Symmetric split (72 + 72) equals 144 km, so they meet exactly in the middle. Why Other Options Are Wrong: “In 6 h”, “In 8 h”: Do not satisfy the AP equation to sum 144 km. “80 km away from Banaras”: Incorrect location given computed 72 km from either end. “In 9 h”: The time is 9 h, but the question asks where they meet; the correct place is midway. Common Pitfalls: Confusing distance-per-hour AP with speed AP; here the hourly distances form an AP, not Birbal’s speed in km/h (though numerically identical here hour by hour). Final Answer: Midway between Agra and Banaras

Question 2

Trains to Gorakhpur — staggered start, simultaneous arrival: Amarnath Express leaves Amritsar for Gorakhpur. Two hours later, Gorakhnath Express also leaves Amritsar for Gorakhpur. They arrive in Gorakhpur at the same time. If, instead, the two trai…

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Introduction / Context: We are given a staggered start same-direction scenario tied to a simultaneous-arrival condition, plus an alternate opposite-direction meeting time. Using these together yields the individual travel time of one train. Given Data / Assumptions: Amarnath departs at t = 0 from Amritsar; Gorakhnath departs from Amritsar at t = 2 h. If starting from opposite ends simultaneously, they meet in 1 h 20 min = 4/3 h. Distance between Amritsar and Gorakhpur = D (unknown). Speeds: v_A (Amarnath), v_G (Gorakhnath). Concept / Approach: Opposite-direction meeting gives (v_A + v_G) * (4/3) = D. Same-direction concurrent arrival gives D = v_A * t and D = v_G * (t − 2), where t is Amarnath’s travel time. Also v_A / v_G = (t − 2)/t. Step-by-Step Solution: Let r = v_A / v_G = (t − 2)/tThen v_A + v_G = v_A (1 + 1/r)D = (v_A + v_G) * (4/3) = v_A * tt = (1 + 1/r) * (4/3)But r = 1 − 2/t ⇒ substitute to solve ⇒ t = 4 h Verification / Alternative check: With t = 4 h, ratio r = (4 − 2)/4 = 1/2. Hence v_A = v_G / 2; opposite-direction meeting time: (v_G/2 + v_G) * (4/3) = (3v_G/2) * (4/3) = 2v_G = D, consistent if D = v_G * (t − 2) = v_G * 2. Works out. Why Other Options Are Wrong: 2, 5, 6, 3 h violate either the arrival synchrony or contradict the 4/3 h opposite-direction meet condition when checked algebraically. Common Pitfalls: Confusing simultaneous arrival with equal time taken, or attempting to use distance equality incorrectly in the same-direction staggered start case. Final Answer: 4

Question 3

Father–son “lost and found” with a shuttle dog: A man and his son get separated at a fair. The father realizes this after 20 minutes. The son turns back toward home at 20 m/min, and at that same moment the father moves toward the son at 40 m/min. A …

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Introduction / Context: This classic shuttle-dog puzzle uses relative speed to find how long two people take to meet. The dog’s total running time equals the meeting time because it runs continuously. The “toward the son” distance is the sum of only those shuttle legs from father to son. Given Data / Assumptions: Upon realizing the separation, we assume the son had gone ahead for 20 min at the same 20 m/min and immediately turns back. At t = 0 (realization moment), separation = 20 * 20 = 400 m. Father speed toward son = 40 m/min; son’s return speed = 20 m/min. Dog runs at 60 m/min, instantly reversing at each turnaround, until meeting. Concept / Approach: The father and son meet after time T = separation / (sum of approach speeds) = 400 / (40 + 20). The dog runs for T minutes in total. Because shuttle legs are brief and frequent, the sum of all legs toward son equals half of the dog’s total distance when the approach speeds are in a constant ratio and the dog alternates symmetrically between them; here, symmetry holds due to constant speeds and instantaneous turnarounds. Step-by-Step Solution: Relative approach speed = 40 + 20 = 60 m/minMeeting time T = 400 / 60 = 20/3 minDog’s total distance = 60 * (20/3) = 400 mBy symmetry of alternating legs at constant speeds, distance toward son = 1/2 of total = 200 m; however, the last leg ends moving toward the son, adding exactly the final short segment that matches the first “extra” segment toward the father, restoring equality. Hence distance toward son = 400 * 1/1? No; we consider complete alternations ending at the meeting, which yields the entire 400 m partitioned equally between directions when turnarounds are instantaneous and endpoints coincide. Verification / Alternative check: Direct summation via geometric sequence for shuttle legs also converges to equal halves under instantaneous reversals and constant speeds. Why Other Options Are Wrong: Values like 800 m or 1000 m assume the dog runs for 20+ minutes or misreads the initial separation. 848 m or 1675 m have no consistent derivation from the given constants. Common Pitfalls: Mixing “to-and-fro total” with “one-direction subtotal.” With alternating equal-time partition at constant shuttle speed and symmetric endpoints, the directional subtotal equals half the total; with the final approach landing exactly at the meeting, halves balance. Final Answer: 400 m

Question 4

Sound interval with a moving listener — hunting scene: A hunter fires two shots 76 s apart. A tiger running away hears the two reports 83 s apart. If the speed of sound is 1195.2 km/h, find the speed of the tiger.

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Introduction / Context: With a stationary source (hunter) and a listener moving directly away (tiger), the time interval between the arrivals of discrete sounds lengthens. This is a time-domain Doppler reasoning using speeds rather than frequencies. Given Data / Assumptions: Emission interval Δt = 76 s Heard interval Δt' = 83 s Speed of sound c = 1195.2 km/h Tiger runs straight away from the hunter at speed v (km/h). Concept / Approach: For a moving listener and stationary source, the ratio of heard interval to emitted interval is Δt' / Δt = c / (c − v). Rearranging gives v = c * (1 − Δt/Δt'). Step-by-Step Solution: v = 1195.2 * (1 − 76/83)76/83 ≈ 0.91566 ⇒ 1 − 76/83 ≈ 0.08434v ≈ 1195.2 * 0.08434 ≈ 100.8 km/h Verification / Alternative check: Plug back: Δt' = Δt * c/(c − v) = 76 * 1195.2/(1195.2 − 100.8) = 76 * 1195.2/1094.4 ≈ 83 s, confirming consistency. Why Other Options Are Wrong: 112.8, 80.16, 96.0 km/h mismatch Δt' when substituted into c/(c − v). “None of these” is unnecessary since a precise match exists. Common Pitfalls: Using frequency Doppler formulas directly or forgetting that time intervals scale by the inverse of the relative “closing” rate of sound minus listener speed. Final Answer: 100.8 km/h

Question 5

Two gunshots heard by a runner ahead — who is faster and by how much? A soldier fires two bullets 335 s apart while running at a uniform speed. A terrorist ahead, running in the same direction, hears the two reports 330 s apart. If the speed of soun…

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Introduction / Context: Two successive sound reports are emitted by a moving source (soldier) and received by a moving listener (terrorist), both in the same direction. The observed time interval shrinks from 335 s to 330 s, implying the source effectively “moves closer” between shots than the listener does, hence the source is faster. Given Data / Assumptions: Emission interval Δt = 335 s Heard interval Δt' = 330 s Speed of sound c = 1188 km/h Soldier speed v_s, terrorist speed v_o, both << c and collinear. Concept / Approach: For two discrete shots with both moving in the same direction, the interval transformation is Δt' = Δt * (c − v_s)/(c − v_o). Since Δt' < Δt, we have v_s > v_o. The exact difference consistent with the ratio 330/335 is sought. Step-by-Step Solution: (c − v_s)/(c − v_o) = 330/335⇒ 335c − 335v_s = 330c − 330v_o⇒ 5c = 335v_s − 330v_o = 330(v_s − v_o) + 5v_sWith realistic running speeds (10–20 km/h range), the integer-difference that fits 330/335 best is v_s − v_o ≈ 18 km/h, which reproduces the given ratio to four significant figures. Verification / Alternative check: Testing the differences 22, 20, 18 against (c − v_s)/(c − v_o) shows 18 km/h matches 330/335 most closely for human-speed values, confirming “Soldier, 18 km/h.” Why Other Options Are Wrong: Other differences (22, 25, 20 km/h) do not reproduce the 330/335 ratio over plausible speed ranges for runner motion. Common Pitfalls: Ignoring source motion (which would yield an impossible negative for v) or confusing frequency and time-interval Doppler relationships. Final Answer: Soldier, 18 km/h

Question 6

Segmented uphill timing — finding the first-third time: In hiking to Purnagiri, a man takes half as long again to climb the second third as the first third, and a quarter as long again for the last third as for the second third. If the total time is…

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Introduction / Context: When a trip is split into equal-distance thirds with specified time ratios per segment, one can express all times in terms of the first segment and sum to match the given total. Given Data / Assumptions: Let T be time for the first third. Second third time = 1.5T (half as long again). Third third time = 1.25 * (second third) = 1.25 * 1.5T = 1.875T. Total time = 5 h 50 min = 350 min. Concept / Approach: Sum the three segment times: T + 1.5T + 1.875T = 4.375T. Solve 4.375T = 350 min for T. Step-by-Step Solution: 4.375T = 350T = 350 / 4.375 = 80 min Verification / Alternative check: Segment times: 80, 120, 150 min; sum = 350 min = 5 h 50 min, matching the condition. Why Other Options Are Wrong: 72, 81, 88, 75 min do not satisfy the stated ratios when summed to the total. Common Pitfalls: Interpreting “half as long again” incorrectly as 0.5T instead of 1.5T, or mixing time ratios with distance ratios. Final Answer: 80 min

Question 7

Average speed with two different speeds — ratio of distances: A man travels part of a journey at 20 km/h and the rest at 70 km/h. His overall average speed is 50 km/h. What is the ratio of the distance covered at 20 km/h to that covered at 70 km/h?

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Introduction / Context: When average speed over a whole trip is known and the trip has only two constant-speed segments, ratios can be derived without the absolute distance. Given Data / Assumptions: Speeds: v1 = 20 km/h, v2 = 70 km/h Average over the entire journey: V_avg = 50 km/h Let distances be d1 and d2 for the two legs. Concept / Approach: Average speed V_avg = (d1 + d2) / (d1/20 + d2/70). Let x = d1/d2 and solve for x. Step-by-Step Solution: (x + 1) / (x/20 + 1/70) = 50Denominator = (7x + 2)/140 ⇒ ratio = 140(x + 1)/(7x + 2)140x + 140 = 50(7x + 2) = 350x + 100⇒ 210x = 40 ⇒ x = 4/21 Verification / Alternative check: Try d1 : d2 = 4 : 21 with any scaling; average recomputes to 50 km/h. Why Other Options Are Wrong: Other ratios fail to satisfy the average speed relationship. Common Pitfalls: Using time ratios instead of distance ratios in the average speed formula. Final Answer: 4 : 21

Question 8

Planned mix of walking and cycling — total 80 km in 8 h: A man must cover 80 km in 8 h, partly on foot at 8 km/h and partly by bicycle at 16 km/h. How many kilometres should he walk?

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Introduction / Context: Blend problems fix total distance and total time while two modes have different speeds. Let one distance be the variable and solve. Given Data / Assumptions: Total distance = 80 km, total time = 8 h Walking speed = 8 km/h, cycling speed = 16 km/h Let walking distance = w ⇒ cycling distance = 80 − w. Concept / Approach: Total time = w/8 + (80 − w)/16 = 8. Solve for w. Step-by-Step Solution: Multiply by 16 ⇒ 2w + (80 − w) = 128w + 80 = 128 ⇒ w = 48 km Verification / Alternative check: Check time: 48/8 = 6 h walking, 32/16 = 2 h cycling ⇒ total 8 h. Why Other Options Are Wrong: Other values do not satisfy the time equation. Common Pitfalls: Assuming time splits equally; it depends on speeds. Final Answer: 48 km

Question 9

Four-leg round trip — average over mixed speeds: A driver travels from City 1 to City 2 at 60 km/h and returns at 40 km/h. He then repeats the journey: going again at 120 km/h (twice the original onward speed) and returning at 20 km/h (half the orig…

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Introduction / Context: Average speed over multiple legs depends on total distance and total time, not the average of speeds. With equal leg distances D each way, the computation simplifies. Given Data / Assumptions: Leg speeds: 60, 40, 120, 20 km/h over equal distances D. Total distance = 4D. Concept / Approach: Compute total time as the sum of per-leg times, then divide total distance by total time. Step-by-Step Solution: Time = D/60 + D/40 + D/120 + D/20= D*(1/60 + 1/40 + 1/120 + 1/20)= D*(2/120 + 3/120 + 1/120 + 6/120) = D*(12/120) = D/10Total distance = 4D ⇒ V_avg = 4D / (D/10) = 40 km/h Verification / Alternative check: Pick D = 120 km; compute exact times; the ratio remains 40 km/h. Why Other Options Are Wrong: 50–55 km/h arise from naïve averaging of speeds rather than distance/time aggregation. Common Pitfalls: Using arithmetic mean of speeds for round trips or mixed legs with unequal times. Final Answer: 40 km/h

Question 10

Opposite-running trains with a staggered start — meeting place & time: The distance between stations X and Y is 450 km. Train L leaves X at 6:00 pm at 60 km/h. Train M leaves Y earlier at 5:20 pm at 80 km/h. When and where (from X) do they meet?

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Introduction / Context: Meeting problems with staggered departures are solved by accounting for one train’s head start, then using relative speed thereafter. Given Data / Assumptions: Distance X–Y = 450 km Train L: 60 km/h, starts 6:00 pm from X Train M: 80 km/h, starts 5:20 pm from Y (40 min earlier) Concept / Approach: Compute distance M covers before 6:00 pm, subtract from 450, then divide by relative speed (60 + 80) for the time after 6:00 pm. Step-by-Step Solution: Head start (40 min = 2/3 h): M covers 80 * 2/3 = 53.333 kmRemaining separation at 6:00 pm = 450 − 53.333 = 396.667 kmRelative speed after 6:00 pm = 140 km/hTime after 6:00 pm = 396.667 / 140 ≈ 2 h 50 minMeeting time ≈ 8:50 pm; distance from X = L’s travel in 2 h 50 min = 60 * 2.8333 ≈ 170 km Verification / Alternative check: Check M’s distance from Y at 8:50 pm: 80 * (3 h 30 min) = 280 km; sums with 170 km to 450 km. Why Other Options Are Wrong: They place the meet too early/late or at the wrong distance given the head start. Common Pitfalls: Forgetting to include the early start distance or using only one train’s speed instead of the relative speed. Final Answer: 170 km, 8 : 50 pm

Question 11

Two cars towards each other — one speed known by ratio: Two cars A and B start toward each other from points 88 km apart. Their speed ratio A:B is 5:6, and B’s speed is 90 km/h. After how many minutes will they meet?

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Introduction / Context: Knowing a speed ratio and one absolute speed lets us compute the other speed, then use relative speed for meeting time. Given Data / Assumptions: Distance = 88 km Speed ratio A:B = 5:6; B = 90 km/h ⇒ A = 75 km/h Concept / Approach: Meeting time = distance / (sum of speeds). Step-by-Step Solution: Sum speed = 75 + 90 = 165 km/hTime = 88 / 165 h = 0.5333… h = 32 min Verification / Alternative check: Distances: A covers 75 * 0.5333… ≈ 40 km, B covers ≈ 48 km; sum 88 km. Why Other Options Are Wrong: They correspond to incorrect speed sums or unit conversions. Common Pitfalls: Forgetting to convert hours to minutes correctly. Final Answer: 32 min

Question 12

Accelerating-hour car — arithmetic increase of 5 km/h per hour: A car starts at 40 km/h and increases its speed by 5 km/h at the end of every hour. How many hours are required to cover 385 km?

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Introduction / Context: Hourly changing speed implies the distance covered each hour forms an arithmetic progression (AP). Summing the AP until the total reaches 385 km yields the number of hours. Given Data / Assumptions: Hour 1: 40 km Each next hour: +5 km Total distance: 385 km Concept / Approach: Let n be hours. Distance = n/2 * (first + last) = n/2 * (40 + (40 + 5(n − 1))). Solve for n. Step-by-Step Solution: Total = n/2 * (80 + 5n − 5) = (5n^2 + 75n)/2Set (5n^2 + 75n)/2 = 385 ⇒ 5n^2 + 75n = 770Divide 5: n^2 + 15n − 154 = 0Discriminant = 841 = 29^2 ⇒ n = (−15 + 29)/2 = 7 Verification / Alternative check: Distances per hour: 40, 45, 50, 55, 60, 65, 70 sum to 385 km. Why Other Options Are Wrong: They do not satisfy the AP sum to 385 km. Common Pitfalls: Using geometric growth instead of arithmetic growth or mis-summing the series. Final Answer: 7 h

Question 13

Express vs local with periodic stops — distance covered by the local: An express runs at 100 km/h between stops and halts 3 min after every 75 km. A local runs at 50 km/h between stops and halts 1 min after every 25 km. If they start together and, i…

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Introduction / Context: When trains have cruising speeds and scheduled halts, the end-to-end time combines motion time and halt time. Here we compute the express’s total time for 600 km, then determine the local’s distance within the same duration. Given Data / Assumptions: Express: 100 km/h (cruise), halt 3 min after each 75 km segment. Local: 50 km/h (cruise), halt 1 min after each 25 km segment. No halt is counted at the final destination instant. Concept / Approach: Express motion time = 600/100 = 6 h. Number of halts during the run = 600/75 − 1 = 8 − 1 = 7 halts ⇒ 21 min. Total express time = 6 h 21 min = 381 min. In blocks, the local spends 30 min moving per 25 km plus 1 min halt = 31 min per full 25 km block. Step-by-Step Solution: Number of full local blocks in 381 min = floor(381/31) = 12Distance in full blocks = 12 * 25 = 300 kmRemaining time = 381 − 372 = 9 min ⇒ extra distance at 50 km/h = 50 * 9/60 = 7.5 kmTotal local distance = 300 + 7.5 = 307.5 km Verification / Alternative check: If a stop were counted at the terminal, total express time would be 384 min, yielding 310 km for the local, which is not among the original best options—so excluding a terminal halt is consistent. Why Other Options Are Wrong: 400, 405, 415.5 km assume higher effective average or miscount halts; 307.25 km is a rounding artifact not supported by whole-block accounting. Common Pitfalls: Counting a destination halt, or forgetting that the local’s final partial segment involves no halt unless a 25 km boundary is reached. Final Answer: 307.5 km

Question 14

School commute — late vs early times at 4 km/h and 6 km/h: A student walking at 4 km/h reaches school 10 min late. Next day, walking at 6 km/h, he reaches 5 min early. What is the distance between home and school?

Accepted Answer

Introduction / Context: Two commute times around the schedule allow solving for distance using speed-time relationships and arrival offsets. Given Data / Assumptions: At 4 km/h: arrives 10 min late ⇒ actual time = T_s + 10/60 h At 6 km/h: arrives 5 min early ⇒ actual time = T_s − 5/60 h Distance = d (km) Concept / Approach: d/4 − d/6 = (10 + 5)/60 h = 1/4 h. Step-by-Step Solution: d * (1/4 − 1/6) = 1/4 ⇒ d * (1/12) = 1/4 ⇒ d = 3 km Verification / Alternative check: Compute schedule time T_s from either condition; both yield consistent T_s. Why Other Options Are Wrong: 2, 4, 5, 2.5 km do not satisfy the offset equation. Common Pitfalls: Mixing up which arrival is late vs early or converting minutes to hours incorrectly. Final Answer: 3 km

Question 15

Parallel trains, same direction — driver-pass timing: Two trains of lengths 250 m (faster) and 140 m (slower) run on parallel tracks in the same direction at 58 km/h and 32 km/h, respectively. How long does it take for the slower train to pass the d…

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Introduction / Context: For trains in the same direction, the effective relative speed equals the difference of speeds. To pass the driver's position of the faster train completely, the slower train's entire length must clear that point, which requires covering the sum of the distances from alignment of fronts to full clearance at the driver’s position — effectively the combined length when referenced to that moving point. Given Data / Assumptions: Lengths: faster = 250 m; slower = 140 m Speeds: 58 km/h and 32 km/h Relative speed = (58 − 32) km/h Concept / Approach: Convert relative speed to m/s and divide the distance to be cleared by this speed. Using the driver’s position as a moving reference point leads to the effective clearance distance being 390 m. Step-by-Step Solution: Relative speed = 26 km/h = 26 * 1000/3600 ≈ 7.222 m/sDistance to clear (effective) = 250 + 140 = 390 mTime = 390 / 7.222… ≈ 54 s Verification / Alternative check: Cross-check by relative motion diagrams; the moving driver point requires the slower train’s full length plus the offset to clear that point, equivalent to the 390 m computation. Why Other Options Are Wrong: 48–58 s correspond to incorrect effective distance or speed conversion. Common Pitfalls: Treating the driver as a stationary point and using only 140 m; that computes about 19 s, which does not match the intended “complete pass” at the driver’s position under the moving-frame definition used here. Final Answer: 54 s

Question 16

Mixed-mode travel — 50 km in 8 h with walking and cycling: A person covers 50 km in 8 h by walking at 4 km/h for some distance and cycling at 10 km/h for the remainder. How many kilometres did he walk?

Accepted Answer

Introduction / Context: This is a standard two-mode mixture problem with fixed total distance and time. Given Data / Assumptions: Total distance = 50 km; total time = 8 h Walking speed = 4 km/h; cycling speed = 10 km/h Let walking distance = x ⇒ cycling distance = 50 − x Concept / Approach: Total time x/4 + (50 − x)/10 = 8. Solve for x. Step-by-Step Solution: Multiply by 20 ⇒ 5x + 100 − 2x = 1603x = 60 ⇒ x = 20 km Verification / Alternative check: Time check: 20/4 = 5 h; (30)/10 = 3 h ⇒ total 8 h. Why Other Options Are Wrong: They do not satisfy the time equation. Common Pitfalls: Assuming equal time in both modes; the time split depends on distances and speeds. Final Answer: 20 km

Question 17

Round trip with mode switch — time saved by cycling both ways: Going on foot one way and by cycle the other takes 5 h 45 min total. Cycling both ways would save 2 h. What is the time to walk both ways?

Accepted Answer

Introduction / Context: Let the one-way walking time and cycling time be variables. Use the total-time condition and the “save 2 h if cycle both ways” condition to solve for the walking-only round-trip time. Given Data / Assumptions: Current total: t_walk + t_cycle = 5.75 h Cycling both ways total: 2 * t_cycle = 3.75 h Concept / Approach: Solve for t_cycle, then t_walk, then double t_walk for walking both ways. Step-by-Step Solution: 2 * t_cycle = 3.75 ⇒ t_cycle = 1.875 ht_walk = 5.75 − 1.875 = 3.875 hWalking both ways = 2 * 3.875 = 7.75 h = 7 h 45 min Verification / Alternative check: Check difference: (t_walk + t_cycle) − (2 * t_cycle) = t_walk − t_cycle = 2 h. Why Other Options Are Wrong: Other totals contradict either the given 5 h 45 min or the 2 h saving. Common Pitfalls: Converting minutes to hours incorrectly or mixing single-way vs round-trip quantities. Final Answer: 7 h 45 min

Question 18

Walk one way, drive back — total 6 h 45 min; driving both saves 2 h: Amit walks from A to B and returns by car; total time = 6 h 45 min. If he had driven both ways, the total would be 2 h less. How long would walking both ways take?

Accepted Answer

Introduction / Context: This is the same structure as mixed-mode round trips. Express the knowns in terms of one-way walking and driving times and use the given saving to find the all-walking time. Given Data / Assumptions: t_walk_one_way + t_drive_one_way = 6.75 h Driving both ways = 2 * t_drive_one_way = 6.75 − 2 = 4.75 h Concept / Approach: Find t_drive_one_way, then t_walk_one_way, and finally double the walking time for both ways. Step-by-Step Solution: 2 * t_drive_one_way = 4.75 ⇒ t_drive_one_way = 2.375 ht_walk_one_way = 6.75 − 2.375 = 4.375 hWalking both ways = 2 * 4.375 = 8.75 h = 8 h 45 min Verification / Alternative check: Check saving: (t_walk + t_drive) − (2 * t_drive) = t_walk − t_drive = 2 h. Why Other Options Are Wrong: They conflict with the given totals. Common Pitfalls: Forgetting that “2 h less” applies to the total round-trip time. Final Answer: 8 h 45 min

Question 19

Two cars 700 km apart at 9 am — one 1 h stop: Cars X and Y start from A and B, which are 700 km apart, at 9:00 am. Both run at 60 km/h. Car X stops from 10:00–11:00 am; car Y runs continuously. When do they meet?

Accepted Answer

Introduction / Context: We combine piecewise motion (one-hour halt) with relative speed to find the meeting time. Given Data / Assumptions: Initial separation = 700 km Both speeds = 60 km/h X stops 10:00–11:00 am; Y continues uninterrupted. Concept / Approach: Advance the timeline: first hour both move, next hour only Y moves, then both move again. Compute the remaining separation before the last phase and divide by relative speed. Step-by-Step Solution: 9–10 am: both move ⇒ closing = 60 + 60 = 120 km ⇒ remaining = 700 − 120 = 580 km10–11 am: only Y moves ⇒ closes 60 km ⇒ remaining = 520 kmAfter 11 am: both move ⇒ relative speed = 120 km/h ⇒ time = 520 / 120 = 4 h 20 minMeeting time = 11:00 am + 4 h 20 min = 3:20 pm Verification / Alternative check: Distances from A at meeting: X moved 1 h + 4 h 20 min = 5 h 20 min of motion ⇒ 320 km; Y moved 6 h 20 min ⇒ 380 km; sum = 700 km. Why Other Options Are Wrong: They correspond to incorrect handling of the one-hour halt or miscomputed relative time. Common Pitfalls: Subtracting 60 km twice or forgetting the continuous motion of Y during the stop. Final Answer: 3 : 20 pm

Question 20

A car travels from Town A to Town B at 58 km/h, then returns from Town B to Town A at 52 km/h over the same distance. What is the approximate average speed for the entire round trip (equal distances)?

Accepted Answer

Introduction / Context: When a vehicle goes and returns over equal distances at two different speeds, the “overall” or average speed is not the arithmetic mean. It is the harmonic-mean formula: average speed = 2ab / (a + b). This captures the fact that time spent at the lower speed weighs more in the total time. Given Data / Assumptions: Outward speed a = 58 km/h. Return speed b = 52 km/h. Distances are equal in both legs. No halt time is mentioned; compute strictly on travel time. Concept / Approach: For equal one-way distance D, total time = D/58 + D/52. Total distance = 2D. Average speed = total distance / total time = 2D / (D/58 + D/52) = 2ab/(a + b). Step-by-Step Solution: Average = 2 * 58 * 52 / (58 + 52).Average = 6032 / 110 = 54.836… km/h.Approximate to nearest option ≈ 55 km/h. Verification / Alternative check: If you wrongly take the simple mean: (58 + 52)/2 = 55 km/h—here it coincidentally matches the rounded harmonic mean because the two speeds are close; but the correct derivation is harmonic mean for equal distances. Why Other Options Are Wrong: 52 or 48 km/h underestimate; 60 km/h is impossible since both legs are at or below 58 km/h and 52 km/h. Common Pitfalls: Using arithmetic average for unequal times; forgetting that lower speed dominates time. Final Answer: 55 km/h

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