Free Topic-Wise General Studies MCQs
Conquer Time, Speed, and Distance problems. Essential UPSC CSAT quantitative practice with comprehensive explanations to improve your analytical pacing.
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Explanation: (a) GENERAL PRINCIPLE: Average speed for a journey is always Total Distance divided by Total Time, never the arithmetic mean of individual speeds. (b) STEP-BY-STEP APPLICATION: Time for first segment = 45/15 = 3 hours. Time for second segment = 75/25 = 3 hours. Total distance = 120 km. Total time = 6 hours. Average speed = 120/6 = 20 km/h. (c) TRANSFERABLE TIP: Whenever a journey has multiple segments with different speeds, compute the time for each segment individually, sum the distances and times, then divide.
Explanation: (a) GENERAL PRINCIPLE: For 'late/early' problems, define the scheduled time as T, express actual travel times as T ± delay/early, equate the distance expressions, and solve for D. (b) STEP-BY-STEP APPLICATION: Let correct time be T hours. Then D/4 = T + 15/60 and D/6 = T - 10/60. Subtracting: D/4 - D/6 = 0.25 + 0.1667 = 5/12. So D × (1/12) = 5/12, giving D = 5 km. (c) TRANSFERABLE TIP: Convert all time differences to hours, set up two equations with the same scheduled time, and eliminate T by subtraction to isolate distance directly.
Explanation: (a) GENERAL PRINCIPLE: When speeds are in ratio a:b and time difference is given for the same distance, use D/s₁ - D/s₂ = Δt, substitute s₁=ak and s₂=bk, and solve for D in terms of k, then choose a convenient k that yields a clean integer. (b) STEP-BY-STEP APPLICATION: Let speeds be 3k and 4k. Then D/(3k) - D/(4k) = 2. This gives D/k × (1/12) = 2, so D = 24k. Taking k = 5 gives speeds 15 km/h and 20 km/h, and distance = 120 km. Verification: 120/15 = 8 hours, 120/20 = 6 hours, difference = 2 hours. (c) TRANSFERABLE TIP: In ratio-based speed problems, introduce a unit variable k, derive the distance formula, then select k to make all quantities integers and match the answer choices.
Explanation: (a) GENERAL PRINCIPLE: When a faster object chases a slower one that started earlier, the overtaking time equals the head-start distance divided by the relative speed (difference of speeds). (b) STEP-BY-STEP APPLICATION: By 9:00 AM, the first bus has traveled 60 km. Relative speed = 75 - 60 = 15 km/h. Overtaking time = 60/15 = 4 hours after 9:00 AM, which is 1:00 PM. (c) TRANSFERABLE TIP: Always compute the distance gap at the moment the second object starts; then divide by the speed difference to find the exact overtaking time.
Explanation: (a) GENERAL PRINCIPLE: When speed changes and the corresponding time change is given for the same distance, equate the two distance expressions (original speed × original time = new speed × new time) and solve for the unknown speed. (b) STEP-BY-STEP APPLICATION: Let original speed be s. Distance = 8s. Reduced speed = (s - 10). New distance expression = 10(s - 10). Equating: 8s = 10s - 100, so 2s = 100, giving s = 50 km/h. Verification: Distance = 400 km; at 40 km/h, time = 400/40 = 10 hours. (c) TRANSFERABLE TIP: For 'same distance, different speed and time' problems, write Distance = Speed₁ × Time₁ = Speed₂ × Time₂ and solve the resulting linear equation.
Explanation: (a) GENERAL PRINCIPLE: For equal distances traveled at two different speeds a and b, average speed is the harmonic mean: 2ab/(a+b). This is always less than the arithmetic mean. (b) STEP-BY-STEP APPLICATION: Let one-way distance be D. Total distance = 2D. Total time = D/4 + D/6 = D(5/12). Average speed = 2D / (5D/12) = 24/5 = 4.8 km/h. (c) TRANSFERABLE TIP: For to-and-fro journeys or equal distances at two speeds, memorize 2ab/(a+b) and remember the result is always closer to the slower speed.
Explanation: (a) GENERAL PRINCIPLE: When equal distances are covered at different speeds, express each segment's time as (D/2)/speed, sum to equal total time, and solve for total distance D. (b) STEP-BY-STEP APPLICATION: Let total distance be D. Time = D/(2×30) + D/(2×50) = D/60 + D/100 = (5D + 3D)/300 = 8D/300 = 2D/75. Setting equal to 8 hours: 2D/75 = 8, so D = 300 km. Verification: 150/30 + 150/50 = 5 + 3 = 8 hours. (c) TRANSFERABLE TIP: When 'half distance' appears with different speeds, use D/2 for each segment, combine fractions carefully, and solve the linear equation for D.
Explanation: (a) GENERAL PRINCIPLE: For n equal segments with speeds v₁, v₂, ..., vₙ, average speed equals n divided by the sum of reciprocals: n / (Σ 1/vᵢ). (b) STEP-BY-STEP APPLICATION: Let each segment be D. Total distance = 3D. Total time = D/60 + D/80 + D/120 = D(4+3+2)/240 = 9D/240 = 3D/80. Average speed = 3D ÷ (3D/80) = 80 km/h. (c) TRANSFERABLE TIP: For three or more equal segments, compute the harmonic mean by dividing the number of segments by the sum of individual time-per-unit-distance values.
Explanation: (a) GENERAL PRINCIPLE: When different speeds are maintained for different durations, average speed is the weighted harmonic mean by time: total distance divided by total time, with distances computed as speed × time for each phase. (b) STEP-BY-STEP APPLICATION: Let time at 12 km/h be t hours. Then time at 8 km/h is 2t. Distance at 8 km/h = 16t km. Distance at 12 km/h = 12t km. Total distance = 28t km. Total time = 3t hours. Average speed = 28t/3t = 28/3 = 9.33 km/h. (c) TRANSFERABLE TIP: When speeds are given with time ratios, assign a variable to the shorter time, express all distances in terms of that variable, and cancel it out in the final ratio.
Explanation: (a) GENERAL PRINCIPLE: For fractional distances at different speeds, express each segment's time as (fraction × total distance) / speed, sum the times, and divide total distance by total time. (b) STEP-BY-STEP APPLICATION: Let total distance be D. Time for first part = (3D/5)/30 = D/50. Time for second part = (2D/5)/60 = D/150. Total time = (3D+D)/150 = 4D/150 = 2D/75. Average speed = D ÷ (2D/75) = 75/2 = 37.5 km/h. (c) TRANSFERABLE TIP: For fraction-based distance splits, keep D symbolic until the final cancellation; this prevents arithmetic errors and reveals the answer independent of the actual distance.
Explanation: (a) GENERAL PRINCIPLE: When two objects move towards each other, their relative speed is the sum of individual speeds. The time to cross each other is the sum of their lengths divided by this relative speed. (b) STEP-BY-STEP APPLICATION: Relative speed = 72 + 54 = 126 km/h = 126 × (5/18) = 35 m/s. Total length to cross = 300 + 400 = 700 m. Time = 700/35 = 20 seconds. (c) TRANSFERABLE TIP: Always convert km/h to m/s by multiplying by 5/18 when lengths are in meters; the total crossing distance is always the sum of both objects' lengths.
Explanation: (a) GENERAL PRINCIPLE: When two objects move towards each other from a known separation, meeting time equals initial distance divided by the sum of their speeds. (b) STEP-BY-STEP APPLICATION: Combined speed = 5 + 8 = 13 km/h. Initial distance = 78 km. Meeting time = 78/13 = 6 hours. Verification: In 6 hours, one covers 30 km and the other 48 km, totaling 78 km. (c) TRANSFERABLE TIP: For 'when will they meet' problems, simply divide the initial gap by the speed sum; individual distances can be verified afterward as a check.
Explanation: (a) GENERAL PRINCIPLE: In a meeting problem, first find the meeting time using total distance divided by relative speed, then multiply each object's speed by that time to find its individual distance covered. (b) STEP-BY-STEP APPLICATION: Relative speed = 60 + 90 = 150 km/h. Meeting time = 450/150 = 3 hours. Distance from X = 60 × 3 = 180 km. Distance from Y = 90 × 3 = 270 km. Check: 180 + 270 = 450 km. (c) TRANSFERABLE TIP: After finding meeting time, the distance from any starting point is simply that point's speed multiplied by the meeting time; the ratio of distances equals the ratio of speeds.
Explanation: (a) GENERAL PRINCIPLE: When objects start at different times and move towards each other, compute the distance covered by the early starter before the second begins, then treat the remaining distance as a standard relative-speed meeting problem. (b) STEP-BY-STEP APPLICATION: By 7 AM, Car A has covered 50 km. Remaining distance = 250 km. Relative speed = 50 + 70 = 120 km/h. Meeting time after 7 AM = 250/120 = 25/12 hours = 2 hours 5 minutes. So meeting time = 9:05 AM. (c) TRANSFERABLE TIP: Always reset the clock to when the second object starts; calculate the reduced gap at that moment and proceed with standard relative-speed division.
Explanation: (a) GENERAL PRINCIPLE: When two objects move towards each other on a closed track from diametrically opposite points, the initial separation is half the track length; meeting time equals this separation divided by the sum of speeds. (b) STEP-BY-STEP APPLICATION: Diametrically opposite means half the circumference apart: 600/2 = 300 m. Relative speed = 8 + 7 = 15 m/s. Meeting time = 300/15 = 20 seconds. (c) TRANSFERABLE TIP: On circular tracks, 'diametrically opposite' always implies a 180-degree separation equal to half the perimeter; treat it as a linear meeting problem with that initial gap.
Explanation: (a) GENERAL PRINCIPLE: When a faster object overtakes a slower one moving in the same direction, relative speed is the difference of speeds, and the total distance to cover is the sum of both objects' lengths. (b) STEP-BY-STEP APPLICATION: Relative speed = 90 - 60 = 30 km/h = 30 × (5/18) = 25/3 m/s. Total distance to cross = 150 + 100 = 250 m. Time = 250 ÷ (25/3) = 250 × 3/25 = 30 seconds. (c) TRANSFERABLE TIP: For overtaking, always subtract speeds for relative speed and add lengths for crossing distance; convert units before dividing to avoid errors.
Explanation: (a) GENERAL PRINCIPLE: In catch-up problems, the distance traveled by the slower object until overtaking equals the distance traveled by the faster object in its shorter travel time; equate and solve for the unknown speed. (b) STEP-BY-STEP APPLICATION: When the car catches up, the bus has traveled for 2 + 4 = 6 hours, covering 40 × 6 = 240 km. The car covers the same 240 km in 4 hours. So car speed = 240/4 = 60 km/h. (c) TRANSFERABLE TIP: The key insight is that both objects have covered identical distances at the overtaking moment; use the slower object's total elapsed time to find that common distance.
Explanation: (a) GENERAL PRINCIPLE: When two objects move in the same direction, their separation after any time t is the relative speed multiplied by time: (faster speed - slower speed) × t. (b) STEP-BY-STEP APPLICATION: Relative speed = 24 - 18 = 6 km/h. Time = 45 minutes = 45/60 = 0.75 hours. Separation = 6 × 0.75 = 4.5 km. (c) TRANSFERABLE TIP: For same-direction travel, the gap grows linearly at the rate of the speed difference; always convert minutes to hours before multiplying by km/h.
Explanation: (a) GENERAL PRINCIPLE: In pursuit problems with a delayed start, compute the head-start distance at the moment pursuit begins, then divide by the relative speed to find the catch-up duration. (b) STEP-BY-STEP APPLICATION: By 3:00 PM, the thief has traveled 60 × 0.5 = 30 km. Relative speed = 90 - 60 = 30 km/h. Catch-up time = 30/30 = 1 hour. So the police catch the thief at 4:00 PM. (c) TRANSFERABLE TIP: The critical calculation is the distance gap at the exact moment the chase starts; everything before that is merely establishing the initial conditions.
Explanation: (a) GENERAL PRINCIPLE: For same-direction overtaking with different start times, find the distance gap when the second train starts, compute overtaking time using relative speed, then multiply the faster train's speed by that time to find the distance from start. (b) STEP-BY-STEP APPLICATION: By 10:30 AM, the first train has covered 80 × 0.5 = 40 km. Relative speed = 100 - 80 = 20 km/h. Overtaking time = 40/20 = 2 hours after 10:30 AM. Distance from station = 100 × 2 = 200 km. Check: First train travels 2.5 hours at 80 km/h = 200 km. (c) TRANSFERABLE TIP: You can verify your answer by calculating the distance covered by the slower train over its longer duration; both should match at the overtaking point.
Explanation: (a) GENERAL PRINCIPLE: When a train crosses a platform, the total distance covered is the sum of the train's length and the platform's length. Use Distance = Speed × Time, then subtract the train's length. (b) STEP-BY-STEP APPLICATION: Speed = 54 km/h = 54 × (5/18) = 15 m/s. Total distance covered = 15 × 36 = 540 m. Platform length = 540 - 240 = 300 m. (c) TRANSFERABLE TIP: Always remember that 'crossing a platform' means the train must travel its own length plus the platform's length; the formula is Platform = (Speed × Time) - Train Length.
Explanation: (a) GENERAL PRINCIPLE: The total distance a train travels while completely crossing a bridge equals the bridge length plus the train length. Compute total distance from speed and time, then isolate the bridge length. (b) STEP-BY-STEP APPLICATION: Speed = 72 km/h = 72 × (5/18) = 20 m/s. Total distance = 20 × 24 = 480 m. Bridge length = 480 - 180 = 300 m. (c) TRANSFERABLE TIP: For bridge crossing problems, the total distance is always the bridge length plus the train length; never forget to subtract the train's own length at the end.
Explanation: (a) GENERAL PRINCIPLE: When a train crosses a stationary point object like a pole or a person, the distance traveled equals exactly the train's own length. (b) STEP-BY-STEP APPLICATION: Speed = 108 km/h = 108 × (5/18) = 30 m/s. Length = Speed × Time = 30 × 8 = 240 m. (c) TRANSFERABLE TIP: A pole or a stationary person has negligible width, so the crossing distance is exactly the train's length; this is the simplest train-crossing scenario.
Explanation: (a) GENERAL PRINCIPLE: To find speed when a train crosses a platform, add the train length and platform length to get total distance, divide by time to get speed in m/s, then convert to km/h by multiplying by 18/5. (b) STEP-BY-STEP APPLICATION: Total distance = 320 + 480 = 800 m. Speed = 800/40 = 20 m/s. Converting to km/h: 20 × (18/5) = 72 km/h. (c) TRANSFERABLE TIP: The conversion factor from m/s to km/h is 18/5 (or 3.6); memorize this and apply it after computing speed in m/s.
Explanation: (a) GENERAL PRINCIPLE: For late/early scenarios, express actual travel times as scheduled time plus/minus the deviation, equate the distance expressions, and eliminate the scheduled time to solve for distance directly. (b) STEP-BY-STEP APPLICATION: Let scheduled time be T hours. Then D/5 = T + 0.1 and D/6 = T - 1/12. Subtracting: D/5 - D/6 = 1/10 + 1/12 = 1/4. So D × (1/30) = 1/4, giving D = 5 km. Verification: 5/5 = 1 hour (6 min late means scheduled = 54 min); 5/6 = 50 min (4 min early). 54 min scheduled fits both. (c) TRANSFERABLE TIP: Convert all time deviations to hours, set up two equations with the same scheduled time, and subtract them to eliminate T and solve for D in one step.
Explanation: (a) GENERAL PRINCIPLE: When two different speeds produce different time deviations from schedule, the distance can be found by dividing the total time gap between the two scenarios by the difference in the reciprocals of the speeds. (b) STEP-BY-STEP APPLICATION: Let scheduled time be T. Then D/40 = T + 0.2 and D/60 = T - 2/15. Subtracting: D/40 - D/60 = 1/5 + 2/15 = 1/3. So D × (1/120) = 1/3, giving D = 40 km. Verification: 40/40 = 1 hour (12 min late → 48 min scheduled); 40/60 = 40 min (8 min early → 48 min scheduled). (c) TRANSFERABLE TIP: The formula D = (t₁ + t₂) / (1/s₁ - 1/s₂) works directly when t₁ is late time and t₂ is early time, both in hours.
Explanation: (a) GENERAL PRINCIPLE: When speed reduction causes lateness, equate the normal travel time to the delayed travel time minus the delay, then solve for distance using the difference of reciprocals. (b) STEP-BY-STEP APPLICATION: Let normal time be T. Then D/50 = T and D/40 = T + 0.5. Subtracting: D/40 - D/50 = 0.5. So D × (1/200) = 0.5, giving D = 100 km. Verification: Normal time = 100/50 = 2 hours; delayed time = 100/40 = 2.5 hours; difference = 0.5 hour = 30 minutes. (c) TRANSFERABLE TIP: For 'speed reduction causes delay' problems, the time difference divided by the difference of reciprocals gives the distance directly.
Explanation: (a) GENERAL PRINCIPLE: For percentage speed increase problems, express the new speed as a multiplier of the original, set up the time equation with the early arrival, and solve for the original time or distance. (b) STEP-BY-STEP APPLICATION: Let usual speed be s. Distance = 4s. New speed = 1.25s. New time = 4s / 1.25s = 3.2 hours = 3 hours 12 minutes. Early arrival = 4 hours - 3.2 hours = 0.8 hours = 48 minutes. To get a numerical distance, if usual speed is 60 km/h, distance = 240 km. Verification: 240/60 = 4 hours; 240/75 = 3.2 hours; difference = 48 minutes. (c) TRANSFERABLE TIP: A 25% speed increase means new speed is 5/4 of original, so new time is 4/5 of original; the time saved is 1/5 of original time.
Explanation: (a) GENERAL PRINCIPLE: To find scheduled time in late/early problems, first solve for distance by eliminating T, then substitute back into either time equation to find T. (b) STEP-BY-STEP APPLICATION: Let scheduled time be T hours. D/4 = T + 1/6 and D/5 = T - 1/12. Subtracting: D/4 - D/5 = 1/6 + 1/12 = 1/4. So D × (1/20) = 1/4, giving D = 5 km. Then T = 5/4 - 1/6 = 15/12 - 2/12 = 13/12 hours = 1 hour 5 minutes. Verification: 5/5 + 1/12 = 1 + 1/12 = 13/12 hours. (c) TRANSFERABLE TIP: After finding distance, substitute into the simpler equation to find T; always verify by plugging T back into the other equation for consistency.
Explanation: (a) GENERAL PRINCIPLE: When speed changes mid-journey and average speed is given, set up the equation: Total Distance / Total Time = Average Speed, with time expressed as the sum of segment times. (b) STEP-BY-STEP APPLICATION: Let remaining distance be x km. Total time = 120/60 + x/40 = 2 + x/40 hours. Total distance = 120 + x. Equation: (120+x)/(2+x/40) = 48. Cross-multiply: 120 + x = 96 + 48x/40 = 96 + 6x/5. So 24 = x/5, giving x = 120 km. Total distance = 240 km. Verification: Time = 2 + 3 = 5 hours; 240/5 = 48 km/h. (c) TRANSFERABLE TIP: For mid-journey speed changes with known average speed, assign a variable to the unknown segment, express total time as a sum, and solve the rational equation by cross-multiplication.
Explanation: (a) GENERAL PRINCIPLE: For journeys with a speed change at a known distance marker, express each segment's time separately, sum to equal total time, and solve the resulting equation for the initial speed. (b) STEP-BY-STEP APPLICATION: Let initial speed be s. Time for first part = 150/s. Time for second part = 200/(s+20). Equation: 150/s + 200/(s+20) = 5. Testing s = 60: 150/60 + 200/80 = 2.5 + 2.5 = 5 hours. This satisfies the equation. (c) TRANSFERABLE TIP: When the equation is complex, test answer choices by substituting back into the time equation; this is often faster than solving the quadratic algebraically.
Explanation: (a) GENERAL PRINCIPLE: When a speed change saves time compared to uniform speed, compute the time at uniform speed minus the actual time, equate to the time saved, and solve for distance. (b) STEP-BY-STEP APPLICATION: Let distance be D. Uniform time = D/60. Actual time = (3D/4)/60 + (D/4)/80 = 3D/240 + D/320 = D/80 + D/320 = 5D/320 = D/64. Time saved = D/60 - D/64 = D(64-60)/3840 = D/960. Setting equal to 15 minutes = 1/4 hour: D/960 = 1/4, so D = 240 km. Verification: Uniform time = 4 hours; actual time = 240/64 = 3.75 hours; saved = 0.25 hour = 15 minutes. (c) TRANSFERABLE TIP: For time-saved problems, always compute both the hypothetical uniform time and the actual segmented time, then set their difference equal to the given time savings.
Explanation: (a) GENERAL PRINCIPLE: For multi-segment journeys with different distances and speeds, average speed is always total distance divided by total time, computed as the sum of individual segment times. (b) STEP-BY-STEP APPLICATION: Time for first segment = 200/80 = 2.5 hours. Time for second = 150/60 = 2.5 hours. Time for third = 100/100 = 1 hour. Total time = 6 hours. Total distance = 450 km. Average speed = 450/6 = 75 km/h. (c) TRANSFERABLE TIP: For three or more segments, create a time table for each segment, sum the times, and divide total distance by that sum; never use the arithmetic mean of speeds.
Explanation: (a) GENERAL PRINCIPLE: To find the required speed for a segment given the overall average speed, use the harmonic relationship: 1/v_avg = total time / total distance, expressed as the sum of segment times divided by total distance. (b) STEP-BY-STEP APPLICATION: Let total distance be D and second-half speed be s. Total time = D/(2×20) + D/(2s) = D/40 + D/(2s). Average speed = D / (D/40 + D/(2s)) = 24. So 1/40 + 1/(2s) = 1/24. Thus 1/(2s) = 1/24 - 1/40 = 1/60. So 2s = 60, giving s = 30 km/h. Verification: For D=120, time = 3 + 2 = 5 hours; 120/5 = 24 km/h. (c) TRANSFERABLE TIP: For equal-distance segments with a target average speed, set up the equation 1/v_avg = 1/(2v₁) + 1/(2v₂) and solve for the unknown speed; this avoids dealing with the actual distance.
Explanation: (a) GENERAL PRINCIPLE: When multiple objects start together and loop back, they meet again at the start when each has completed an integer number of laps; find the time each takes for one lap, then compute the LCM of these times. (b) STEP-BY-STEP APPLICATION: Lap times are 12/8 = 1.5 hours, 12/10 = 1.2 hours, and 12/12 = 1 hour. Converting to fractions: 3/2, 6/5, 1. The LCM of numerators (3,6,1) is 6; the GCF of denominators (2,5,1) is 1. So LCM = 6/1 = 6 hours. Verification: In 6 hours, they complete 4, 5, and 6 laps respectively—all integers. (c) TRANSFERABLE TIP: For circular track meeting problems, convert lap times to fractions and use LCM of numerators over GCF of denominators to find the reunion time.
Explanation: (a) GENERAL PRINCIPLE: Between successive meetings of two travelers moving back and forth between two points, they together cover exactly twice the inter-point distance. Use relative speed to find the time between meetings, then track individual distances. (b) STEP-BY-STEP APPLICATION: First meeting time = 80/(10+15) = 3.2 hours. Between first and second meeting, combined distance = 160 km. Time = 160/25 = 6.4 hours. Total elapsed time = 9.6 hours. Distance covered by P-traveler = 10 × 9.6 = 96 km. Since 96 = 80 + 16, he is 16 km back from Q, which is 64 km from P. (c) TRANSFERABLE TIP: For second meetings on a linear back-and-forth route, remember the combined distance between meetings is 2D, and track each traveler's total distance modulo 2D to find their position.
Explanation: (a) GENERAL PRINCIPLE: For delayed-start overtaking, calculate the distance gap at the moment the second traveler starts, then divide by the relative speed to find the catch-up time after that moment. (b) STEP-BY-STEP APPLICATION: By 7:00 AM, A has covered 8 × 1 = 8 km. Relative speed = 12 - 8 = 4 km/h. Catch-up time = 8/4 = 2 hours after 7:00 AM. So B overtakes A at 9:00 AM. Verification: By 9 AM, A has walked 8 × 3 = 24 km; B has walked 12 × 2 = 24 km. (c) TRANSFERABLE TIP: The head-start distance is always the slower traveler's speed multiplied by the time difference; divide this by the speed difference to find overtaking time.
Explanation: (a) GENERAL PRINCIPLE: When a faster train starts later and must catch a slower train that has a scheduled stop, equate the total elapsed times from the slow train's departure, including the stop duration, to the fast train's elapsed time plus its delayed start. (b) STEP-BY-STEP APPLICATION: Let overtaking distance be x km (where x > 60). Slow train time = 60/60 + 0.5 + (x-60)/60 = 1.5 + (x-60)/60 hours from 8 AM. Fast train time = 1 + x/90 hours from 8 AM. Equating: 1.5 + (x-60)/60 = 1 + x/90. Simplify: 1.5 + x/60 - 1 = 1 + x/90. So 0.5 + x/60 = x/90. Thus x/60 - x/90 = 0.5, giving x × (1/180) = 0.5, so x = 90 km. Verification: Slow train reaches 90 km at 8:00 + 1 hour + 0.5 hour + 0.5 hour = 10:00 AM. Fast train reaches 90 km at 9:00 + 1 hour = 10:00 AM. (c) TRANSFERABLE TIP: For delayed-start catch-up with intermediate stops, write the slow train's total elapsed time as a piecewise function and equate it to the fast train's delayed elapsed time.
Explanation: (a) GENERAL PRINCIPLE: When one traveler reaches the destination and turns back while the other continues, their meeting occurs when the sum of distances traveled equals twice the inter-town distance. Divide by the speed sum to find meeting time. (b) STEP-BY-STEP APPLICATION: Combined distance to cover = 2 × 30 = 60 km. Relative speed = 10 + 15 = 25 km/h. Meeting time = 60/25 = 2.4 hours. Distance from A = 10 × 2.4 = 24 km. Verification: Faster friend travels 15 × 2.4 = 36 km = 30 km to B + 6 km back. 24 + 6 = 30 km. (c) TRANSFERABLE TIP: For 'reach and return' meeting problems, the total distance covered by both travelers until meeting is always 2D; use this to find the meeting time directly.
Explanation: (a) GENERAL PRINCIPLE: In race problems with head starts, the winner's time equals the loser's time minus the margin of victory (or plus, depending on framing). Compute the winner's time from distance and speed, then derive the loser's speed from the distance they actually covered. (b) STEP-BY-STEP APPLICATION: A runs 1000 m at 10 m/s, taking 100 seconds. B runs 900 m (1000 - 100 head start) in 100 + 20 = 120 seconds. So B's speed = 900/120 = 7.5 m/s. Verification: 7.5 × 120 = 900 m. (c) TRANSFERABLE TIP: Always identify the actual distance each runner covers, not just the race distance; the head-start reduces the effective distance for the beneficiary.
Explanation: (a) GENERAL PRINCIPLE: In race problems, the ratio of speeds equals the ratio of distances covered in the same time. Use this ratio to project how far the slower runner covers when the middle runner finishes the full distance. (b) STEP-BY-STEP APPLICATION: When A runs 100 m, B runs 80 m and C runs 60 m. So the speed ratio B:C = 80:60 = 4:3. When B runs 100 m, C runs 100 × (3/4) = 75 m. Thus B beats C by 100 - 75 = 25 m. (c) TRANSFERABLE TIP: First establish the speed ratio from the common race, then apply it to the new race distance by multiplying the full distance by the slower-to-faster speed ratio.
Explanation: (a) GENERAL PRINCIPLE: In races with both head starts and time margins, compute the winner's time from their speed and full distance, add the time margin to get the loser's time, then divide the loser's effective distance by this time. (b) STEP-BY-STEP APPLICATION: A's time = 2/8 = 0.25 hours = 15 minutes. B's time = 15 + 5 = 20 minutes = 1/3 hour. B's effective distance = 2 - 0.2 = 1.8 km. B's speed = 1.8 ÷ (1/3) = 5.4 km/h. Verification: 5.4 × (1/3) = 1.8 km. (c) TRANSFERABLE TIP: The loser's time is always the winner's time plus the margin of victory; the loser's distance is the race distance minus any head start.
Explanation: (a) GENERAL PRINCIPLE: When a head start is given in time rather than distance, compute the winner's running time, add the head-start time to get the loser's total running time, and divide the loser's covered distance by this total time. (b) STEP-BY-STEP APPLICATION: A's running time = 1000/5 = 200 seconds. B's total running time = 200 + 120 = 320 seconds. B's covered distance = 1000 - 200 = 800 m. B's speed = 800/320 = 2.5 m/s. Verification: 2.5 × 320 = 800 m. (c) TRANSFERABLE TIP: For time-based head starts, the loser's total elapsed time is the winner's race time plus the head-start duration; never confuse this with distance-based head starts.
Explanation: (a) GENERAL PRINCIPLE: When a head start is combined with a speed ratio, the effective race distance for the head-start recipient is reduced, and the opponent's distance covered in that time is found by applying the inverse speed ratio. (b) STEP-BY-STEP APPLICATION: A runs 1000 - 100 = 900 m. With speed ratio 5:4, when A covers 900 m, B covers 900 × (4/5) = 720 m. So B has 1000 - 720 = 280 m remaining when A finishes. A wins by 280 m. (c) TRANSFERABLE TIP: The head start reduces the distance the faster runner must cover; multiply this reduced distance by the slower-to-faster speed ratio to find how far the opponent has traveled.
Explanation: (a) GENERAL PRINCIPLE: The effective speed with stoppages reflects the actual distance covered per hour of real time. The stoppage time per hour equals the difference between the speed without stoppages and the effective speed, divided by the speed without stoppages, converted to minutes. (b) STEP-BY-STEP APPLICATION: In 1 hour of real time, the bus covers 45 km. Without stoppages, 45 km would take 45/60 = 0.75 hours = 45 minutes. So the bus moves for 45 minutes and stops for 60 - 45 = 15 minutes each hour. (c) TRANSFERABLE TIP: Stoppage time per hour = (Speed without stoppages - Effective speed) / Speed without stoppages × 60 minutes.
Explanation: (a) GENERAL PRINCIPLE: To find average stoppage time per hour, compare the actual travel time without stoppages to the total elapsed time with stoppages; the difference is total stoppage time, which is then divided by total hours. (b) STEP-BY-STEP APPLICATION: Without stoppages, speed = 120/2 = 60 km/h. Time without stoppages for 120 km = 2 hours. Total time with stoppages = 2.5 hours. Total stoppage time = 0.5 hours = 30 minutes. Average per hour = 30/2.5 = 12 minutes. (c) TRANSFERABLE TIP: Total stoppage time = Total time with stoppages - (Distance / Speed without stoppages); divide by total hours for the hourly rate.
Explanation: (a) GENERAL PRINCIPLE: The actual moving speed is higher than the average speed because it must cover the same distance in less moving time. Divide the distance covered in one hour by the actual moving time within that hour. (b) STEP-BY-STEP APPLICATION: In 1 hour of real time, the car covers 72 km but moves for only 50 minutes = 5/6 hour. Actual moving speed = 72 ÷ (5/6) = 72 × 6/5 = 86.4 km/h. Verification: In 50 minutes at 86.4 km/h, distance = 86.4 × (5/6) = 72 km. (c) TRANSFERABLE TIP: Actual moving speed = Average speed / (1 - stoppage fraction); for 10 minutes per hour, the moving fraction is 50/60 = 5/6.
Explanation: (a) GENERAL PRINCIPLE: Total stoppage time equals the difference between the actual journey time with stoppages and the pure travel time computed from the speed without stoppages. (b) STEP-BY-STEP APPLICATION: Pure travel time = 180/60 = 3 hours. Actual journey time = 180/45 = 4 hours. Total stoppage time = 4 - 3 = 1 hour = 60 minutes. (c) TRANSFERABLE TIP: For total stoppage time on a known route, simply subtract the no-stoppage time from the actual journey time; no need to compute per-hour rates.
Explanation: (a) GENERAL PRINCIPLE: To find speed without stoppages, divide the distance covered in a given real-time period by the actual moving time within that period, excluding all stoppages. (b) STEP-BY-STEP APPLICATION: In 1 hour of real time, the bus covers 40 km but moves for only 45 minutes = 0.75 hours. Speed without stoppages = 40/0.75 = 160/3 = 53.33 km/h. Verification: In 45 minutes at 53.33 km/h, distance = 53.33 × 0.75 = 40 km. (c) TRANSFERABLE TIP: Speed without stoppages = Average speed / (Moving time fraction); for 15 minutes per hour, the moving fraction is 45/60 = 3/4.
Explanation: (a) GENERAL PRINCIPLE: Average speed for a journey with stoppages is Total Distance divided by Total Elapsed Time, where total elapsed time includes both moving time and all stoppage durations. (b) STEP-BY-STEP APPLICATION: Moving time = 300/75 + 200/100 = 4 + 2 = 6 hours. Total elapsed time = 6 + 20/60 = 6 + 1/3 = 19/3 hours. Total distance = 500 km. Average speed = 500 ÷ (19/3) = 1500/19 = 78.95 km/h. (c) TRANSFERABLE TIP: Never forget to add stoppage time to the denominator when computing average speed; the numerator remains the total distance covered.
Explanation: (a) GENERAL PRINCIPLE: When the faster traveler stops, the gap increases at the slower traveler's speed during the stoppage duration, added to the gap already created before the stop. (b) STEP-BY-STEP APPLICATION: After 2 hours, the faster cyclist is 18 × 2 - 12 × 2 = 36 - 24 = 12 km ahead. During the 30-minute stop, the slower cyclist covers 12 × 0.5 = 6 km more. Total gap = 12 + 6 = 18 km. (c) TRANSFERABLE TIP: Compute the gap at the moment the stop begins, then add the slower traveler's speed multiplied by the stoppage duration to find the final separation.
Explanation: (a) GENERAL PRINCIPLE: For two travelers moving back and forth between two points, the total distance covered by both together until the second meeting is exactly three times the inter-point distance. Use this to find the meeting time, then compute individual distance. (b) STEP-BY-STEP APPLICATION: Combined speed = 60 + 90 = 150 km/h. For the second meeting, total combined distance = 3 × 300 = 900 km. Time = 900/150 = 6 hours. Distance covered by P-train = 60 × 6 = 360 km. Since 360 = 300 + 60, the P-train has reached Q and returned 60 km. So the second meeting point is 300 - 60 = 240 km from P. Verification: Q-train covers 90 × 6 = 540 km = 300 + 240, so it has reached P and returned 240 km, also at 240 km from P. (c) TRANSFERABLE TIP: For second meetings on a linear back-and-forth route, the combined distance is always 3D; for the nth meeting, it is (2n-1)D.
Explanation: (a) GENERAL PRINCIPLE: When two objects move in opposite directions from the same starting point, their separation after time t is the sum of their speeds multiplied by t. Set up an equation with the speed difference constraint. (b) STEP-BY-STEP APPLICATION: Let slower speed be s km/h. Then faster speed = s + 20. Combined speed = 2s + 20. Separation after 3 hours = 3(2s + 20) = 450. So 2s + 20 = 150, giving 2s = 130 and s = 65 km/h. Verification: Speeds are 65 and 85 km/h; in 3 hours, separation = 3 × 150 = 450 km. (c) TRANSFERABLE TIP: For opposite-direction separation problems, the combined speed equals the rate of separation; use the given time and distance to form a single equation with the speed variable.
Explanation: (a) GENERAL PRINCIPLE: For percentage speed changes with time deviations, express the new speed as a multiplier of the usual speed, set the distance equal for both scenarios, and solve for the usual time using the early/late constraints. (b) STEP-BY-STEP APPLICATION: Let usual time be T hours and usual speed be s. Distance = sT. At 20% faster: 1.2s × (T - 0.1) = sT, so 1.2(T - 0.1) = T, giving 0.2T = 0.12, so T = 0.6 hours = 36 minutes. Verification with slower speed: 0.8s × (0.6 + 0.15) = 0.8s × 0.75 = 0.6s = sT. (c) TRANSFERABLE TIP: For percentage speed changes, the new speed is (1 ± p/100)s; eliminate s by dividing both sides, leaving a linear equation in T alone.
Explanation: (a) GENERAL PRINCIPLE: When a train overtakes a moving person in the same direction, the relative speed is the train's speed minus the person's speed, and the crossing distance equals the train's length. (b) STEP-BY-STEP APPLICATION: Let train speed be s km/h. Relative speed = (s - 6) km/h = (s - 6) × (5/18) m/s. Distance = 180 m. Time = 18 seconds. So (s - 6) × (5/18) = 180/18 = 10. Thus (s - 6) × 5/18 = 10, giving s - 6 = 36, so s = 42 km/h. Verification: Relative speed = 36 km/h = 10 m/s. 180/10 = 18 seconds. (c) TRANSFERABLE TIP: For train-person overtaking, the relative speed in m/s equals train length divided by crossing time; convert back to km/h and add the person's speed.
Explanation: (a) GENERAL PRINCIPLE: For time-constrained journeys with an initial segment at known speed, subtract the initial distance and time from the totals, then divide the remaining distance by the remaining time to find the required speed. (b) STEP-BY-STEP APPLICATION: Distance covered in first 45 minutes = 80 × 0.75 = 60 km. Remaining distance = 60 km. Remaining time = 2 - 0.75 = 1.25 hours. Required speed = 60/1.25 = 48 km/h. Verification: Total time = 0.75 + 60/48 = 0.75 + 1.25 = 2 hours. Total distance = 60 + 60 = 120 km. (c) TRANSFERABLE TIP: In deadline problems, always compute the 'remaining distance' and 'remaining time' separately; the required speed is simply their ratio.
Explanation: (a) GENERAL PRINCIPLE: When speeds are in ratio and time difference is given for a fixed distance, express times as D/(ax) and D/(bx), subtract, equate to the given time difference, and solve for x. (b) STEP-BY-STEP APPLICATION: Let speeds be 4x and 5x. Time difference = 300/4x - 300/5x = 300/x × (1/4 - 1/5) = 300/(20x) = 15/x hours. Setting equal to 45 minutes = 3/4 hour: 15/x = 3/4, so x = 20. Faster speed = 5 × 20 = 100 km/h. Verification: 300/80 = 3.75 hours; 300/100 = 3 hours; difference = 0.75 hour = 45 minutes. (c) TRANSFERABLE TIP: For ratio-based speeds with time differences, the unit variable x cancels out the distance; solve for x first, then multiply by the ratio term for the required speed.
Explanation: (a) GENERAL PRINCIPLE: For multi-leg journeys with known total time, compute the time for each known leg, subtract from the total to find the remaining time, then multiply by the speed of the final leg to find its distance. (b) STEP-BY-STEP APPLICATION: Time A→B = 120/60 = 2 hours. Time B→A = 120/40 = 3 hours. Combined time = 5 hours. Remaining time for A→C = 8 - 5 = 3 hours. Distance A→C = 50 × 3 = 150 km. Verification: Total time = 2 + 3 + 3 = 8 hours. Total distance = 120 + 120 + 150 = 390 km. (c) TRANSFERABLE TIP: In multi-leg journey problems, create a time ledger for each known segment; the remaining time is the key to unlocking the unknown distance.
Explanation: (a) GENERAL PRINCIPLE: To find the required speed for a reduced travel time, first compute the distance from the usual speed and time, then divide that distance by the new desired time. (b) STEP-BY-STEP APPLICATION: Distance = 30 × (40/60) = 20 km. New time = 30 minutes = 0.5 hours. Required speed = 20/0.5 = 40 km/h. Verification: 40 × 0.5 = 20 km. (c) TRANSFERABLE TIP: Distance is the invariant in such problems; compute it from the original scenario and reuse it as the numerator for the new speed calculation.
Explanation: (a) GENERAL PRINCIPLE: When travelers start at different times and move towards each other, compute the distance covered by the early starter before the second begins, then treat the remaining distance as a standard relative-speed meeting problem. (b) STEP-BY-STEP APPLICATION: By 8:00 AM, the first cyclist has covered 20 × 1 = 20 km. Remaining distance = 60 km. Relative speed = 20 + 30 = 50 km/h. Meeting time after 8:00 AM = 60/50 = 1.2 hours = 1 hour 12 minutes. So they meet at 9:12 AM. Verification: First cyclist travels 2.2 hours × 20 = 44 km. Second travels 1.2 hours × 30 = 36 km. 44 + 36 = 80 km. (c) TRANSFERABLE TIP: Reset the problem to the later start time, compute the reduced gap, and divide by the speed sum to find the exact meeting time.