Question 1

Distance by which X defeats Y: X runs 1 km in 8 min 40 s, Y runs 1 km in 10 min. By what distance does X defeat Y?

Accepted Answer

Introduction / Context: Compare distances covered by Y in the time X takes to finish 1 km to get the winning margin. Given Data / Assumptions: X's 1 km time = 520 s Y's 1 km time = 600 s Concept / Approach: Distance(Y in 520 s) = (1000/600)*520. Subtract from 1000 m to get X’s winning distance. Step-by-Step Solution: Y distance in 520 s = (1000/600)*520 = 866 2/3 mMargin = 1000 − 866 2/3 = 133 1/3 m Verification / Alternative check: Speeds: v_X = 1000/520; v_Y = 1000/600. Margin in 520 s equals (v_X − v_Y)*520 = same 133 1/3 m. Why Other Options Are Wrong: 133 2/3 m differs by 1/3 m; others are far off. Common Pitfalls: Converting minutes to seconds incorrectly; mixing up who is faster. Final Answer: 1331/3 m

Question 2

Dead-heat start in 1 km race: L runs 1 km in 10 min and M in 12 min. What start (in metres) can L give M in a 1 km race so that the finish is a dead heat?

Accepted Answer

Introduction / Context: A “start” means the slower runner begins closer to the finish. For a dead heat, the slower runner’s actual running distance must match how far they would get in the faster runner’s finish time. Given Data / Assumptions: L = 1000 m in 10 min M = 1000 m in 12 min Race length = 1000 m Concept / Approach: In L’s finish time (10 min), M would cover (1000 * 10/12) m. The start equals 1000 minus this distance. Step-by-Step Solution: M's distance in 10 min = 1000*(10/12) = 833 1/3 mRequired start = 1000 − 833 1/3 = 166 2/3 m Verification / Alternative check: With this start, both finish together in 10 min (L’s time). Why Other Options Are Wrong: Other values would not synchronize finishes. Common Pitfalls: Using speed differences instead of proportionate distances in the common finish time. Final Answer: 1662/3 m

Question 3

Defeat distance from pace times: Yogesh runs 1 km in 6 min 20 s; Vijay runs 1 km in 6 min 40 s. By what distance can Yogesh beat Vijay in a 1 km race?

Accepted Answer

Introduction / Context: Compare the slower runner’s distance in the faster runner’s finish time to compute the winning margin. Given Data / Assumptions: T_Yogesh = 380 s T_Vijay = 400 s Race length = 1000 m Concept / Approach: Distance covered by Vijay in 380 s is (1000/400)*380; subtract from 1000 m for the margin. Step-by-Step Solution: Vijay distance in 380 s = 950 mWinning margin = 1000 − 950 = 50 m Verification / Alternative check: Speeds 1000/380 vs 1000/400 confirm 50 m difference over 380 s. Why Other Options Are Wrong: Other numbers do not match the precise calculation. Common Pitfalls: Switching the runners or mis-converting times. Final Answer: 50 m

Question 4

Handicap conversion between B and C: In a 160-point game, A gives B 10 and A gives C 30. In a game to 60, how many can B give C?

Accepted Answer

Introduction / Context: Translate handicaps into scoring ratios and then deduce B:C. Finally scale a 60-point game to find the start B can give C. Given Data / Assumptions: A:B = 160:150 = 16:15 A:C = 160:130 = 16:13 Target game size = 60 Concept / Approach: B:C = (A:C)/(A:B) = (16/13)/(16/15) = 15/13. When B scores 60, C gets 60*(13/15). The difference is B’s start to C. Step-by-Step Solution: C at B's 60 = 60*(13/15) = 52B can give C = 60 − 52 = 8 Verification / Alternative check: Ratios are consistent with both handicaps from A. Why Other Options Are Wrong: They contradict B:C = 15:13 scaling to 60 points. Common Pitfalls: Subtracting point handicaps directly instead of using ratios. Final Answer: 8

Question 5

Transitive race advantage: In a 200 m race, A beats B by 50 m and B beats C by 8 m. By how much does A beat C in 200 m?

Accepted Answer

Introduction / Context: Infer speed ratios from two pairwise results and combine to compare A and C directly. Given Data / Assumptions: A:B in 200 m ⇒ 200:150 = 4:3 B:C in 200 m ⇒ 200:192 = 25:24 Concept / Approach: Multiply ratios for A:C, then scale to 200 m to find C’s distance at A’s finish. Step-by-Step Solution: A:C = (4/3)*(25/24) = 100/72 = 25/18If A runs 200, C runs 200*(18/25) = 144Lead = 200 − 144 = 56 m Verification / Alternative check: Cross-check by finding C’s speed as a fraction of A’s from chained ratios. Why Other Options Are Wrong: They don’t reflect the chained ratio 25:18 scaled to 200 m. Common Pitfalls: Subtracting the leads (50 and 8) instead of combining ratios. Final Answer: 56 m

Question 6

Handicap between B and C (200-point base): In a 200-point game, A gives B 40 and A gives C 56. How many points can B give C?

Accepted Answer

Introduction / Context: Convert handicaps to ratios and infer B:C. Then compute the start B can give C on the same 200-point base for comparison. Given Data / Assumptions: A:B = 200:160 = 5:4 A:C = 200:144 = 25:18 Concept / Approach: B:C = (A:C)/(A:B) = (25/18)/(5/4) = (25/18)*(4/5) = 10/9. When B scores 200, C would score 180; the difference is B’s start to C. Step-by-Step Solution: C at B's 200 = 200*(9/10) = 180B can give C = 200 − 180 = 20 points Verification / Alternative check: Ratios reduce properly and remain consistent across scaling. Why Other Options Are Wrong: They are inconsistent with B:C = 10:9 on the 200-point base. Common Pitfalls: Subtracting handicap numbers directly without ratio conversion. Final Answer: 20 points

Question 7

Winning margin from 100 m times: A runs 100 m in 27 s and B in 30 s. By how many metres does A beat B?

Accepted Answer

Introduction / Context: Compute the distance B covers in the time A finishes; the difference from 100 m is A’s winning margin. Given Data / Assumptions: A's time = 27 s B's 100 m time = 30 s Concept / Approach: Distance covered by B in 27 s equals (27/30)*100. Subtract from 100 to get the margin. Step-by-Step Solution: B distance in 27 s = (27/30)*100 = 90 mMargin = 100 − 90 = 10 m Verification / Alternative check: Speed method: v_B = 100/30 m/s, margin over 27 s = (100/27 − 100/30)*27 = 10 m. Why Other Options Are Wrong: They do not match B’s progress in 27 s. Common Pitfalls: Using time difference (3 s) as distance; times must be converted via speed. Final Answer: 10 m

Question 8

Race timing with head-start: Runner A covers 440 m in 51 s, while runner B covers the same 440 m in 55 s. In a 440 m race, if B is given a 40 m head-start (so B needs to run only 400 m), by how many seconds will B finish ahead of A?

Accepted Answer

Introduction / Context: Head-start problems compare finishing times using constant speeds. We compute each runner’s time to complete their required distance and compare. Given Data / Assumptions: A runs 440 m in 51 s. B runs 440 m in 55 s. Race length = 440 m; B receives a 40 m head-start, so B runs 400 m. Uniform speed throughout. Concept / Approach: Time = distance / speed. Speeds are vA = 440/51 m/s and vB = 440/55 m/s. Compute tA and tB for their race distances and compare finishing times. Step-by-Step Solution: vA = 440/51 m/s; vB = 440/55 m/stA = 440 / vA = 51 stB = 400 / vB = 400 / (440/55) = 50 sLead in time = tA - tB = 51 - 50 = 1 s Verification / Alternative check: Proportionally, B’s pace is 55/51 slower than A’s, but the 40 m reduction at B’s speed equals 40 / (440/55) = 5 s saved, while B’s inherent slowness adds 4 s over 440 m, net 1 s ahead. Why Other Options Are Wrong: 2/3/4 seconds: Overstate the time gap; calculations show exactly 1 s. None of these: Incorrect since 1 s matches an option. Common Pitfalls: Using distance difference instead of time, or forgetting that B runs only 400 m, not 440 m. Final Answer: 1 seconds

Question 9

Starts into effective speeds: In a 1 km race A gives B a start of 50 m, and B gives C a start of 40 m. In a 2 km race, what start (in metres) can A give C so that they finish together?

Accepted Answer

Introduction / Context: Starts translate to speed ratios: if A gives B 50 m in 1000 m, A is proportionally faster. Chaining two such relations lets us compare A directly to C and scale to a 2000 m race. Given Data / Assumptions: 1 km race: A vs B → B runs 950 m when A runs 1000 m. 1 km race: B vs C → C runs 960 m when B runs 1000 m. Uniform speeds; proportional times. Concept / Approach: Speed ratios: vA/vB = 1000/950 = 20/19; vB/vC = 1000/960 = 25/24. Then vA/vC = (20/19)*(25/24) = 125/114. Step-by-Step Solution: For a 2000 m race: distance that C covers when A runs 2000 m is 2000 * (vC/vA)vC/vA = 114/125C's distance = 2000 * 114/125 = 1824 mRequired start to C = 2000 - 1824 = 176 m Verification / Alternative check: Scale both speeds by a common time; the same 176 m emerges from proportional distances. Why Other Options Are Wrong: 220/222/167 are mismatched scalings; only 176 m satisfies the chained ratios precisely. Common Pitfalls: Adding starts directly instead of converting to speed ratios; not scaling to the new race length. Final Answer: 176 m

Question 10

Linked handicaps with time gap: A gives B 40 m and C 82 m in an 880 m race. Also, B finishes 9 s before C over the same 880 m. How long (in minutes) does C take to run 880 m?

Accepted Answer

Introduction / Context: This is a chained comparison problem: two distance advantages relate A–B and A–C speeds; a time gap relates B–C directly. We solve for C’s time using speed ratios and the given 9 s difference. Given Data / Assumptions: Over 880 m: A vs B → B is at 840 m when A finishes. Over 880 m: A vs C → C is at 798 m when A finishes. B beats C by 9 s over 880 m. Concept / Approach: vA/vB = 880/840 = 22/21 and vA/vC = 880/798 = 440/399. Hence vB/vC = (vA/vC)/(vA/vB) = (440/399)/(22/21) = 140/133. Use the 9 s difference to find absolute times. Step-by-Step Solution: Let vB be unknown. Then vC = vB * (133/140).tB = 880 / vB; tC = 880 / vC = 880 / (vB * 133/140) = 880 * 140 / (133 vB).tC - tB = 9 ⇒ 880 * 140/(133 vB) - 880/vB = 9.Simplify: 880 * (140/133 - 1) / vB = 9 ⇒ 880 * (7/133) / vB = 9 ⇒ vB = 880 * 7 / (133 * 9).Then tC = 880 / vC = 180 s = 3 min. Verification / Alternative check: Plugging back shows B's time is 171 s and C's time 180 s, confirming the 9 s gap. Why Other Options Are Wrong: 3.25/4 minutes contradict the achieved 9 s gap with the derived ratios. Common Pitfalls: Directly treating distance gaps as time gaps; you must use speed ratios first, then time difference. Final Answer: 3 min.

Question 11

Point handicap consistency: A can give B 15 points; A can give C 22 points; and B can give C 10 points (all in the same “N-point” game). How many points make the game?

Accepted Answer

Introduction / Context: In point handicaps, “A gives B 15” in an N-point game means when A scores N, B would score N − 15 in the same time, implying a speed ratio. Consistency across three pairwise handicaps pins down N. Given Data / Assumptions: A gives B 15 → vA/vB = N/(N − 15). A gives C 22 → vA/vC = N/(N − 22). B gives C 10 → vB/vC = N/(N − 10). Concept / Approach: Ratios must satisfy (vA/vB) * (vB/vC) = vA/vC. Substitute the expressions and solve for N. Step-by-Step Solution: (N/(N − 15)) * (N/(N − 10)) = N/(N − 22)Cancel one N: N/(N − 15) * 1/(N − 10) = 1/(N − 22)Cross-multiply: (N − 22) * N = (N − 15)(N − 10)Expand: N^2 − 22N = N^2 − 25N + 150 ⇒ 3N = 150 ⇒ N = 50 Verification / Alternative check: Plug N = 50 back to get ratios 50/35, 50/28, and 50/40; they satisfy the identity. Why Other Options Are Wrong: 60/70/80 do not satisfy the consistency equation for all three pairwise handicaps. Common Pitfalls: Adding or averaging handicaps; correct method uses multiplicative speed ratios. Final Answer: 50

Question 12

Transferring handicaps across different game lengths: In an 80-point game, A gives B 5 points and C 15 points. In a 60-point game, how many points can B give C?

Accepted Answer

Introduction / Context: Point handicaps correspond to speed ratios. Once we deduce relative speeds of A, B, and C from one game length, we can project the handicap to any other length by proportional scoring speed. Given Data / Assumptions: In 80-point game: A vs B → A: B = 80 : 75 = 16 : 15. In 80-point game: A vs C → A: C = 80 : 65 = 16 : 13. Uniform scoring speeds; linear scaling. Concept / Approach: From A: B = 16 : 15 and A: C = 16 : 13 ⇒ B : C = 15 : 13. In a 60-point game, when C reaches 60, B would reach (15/13)*60; or equivalently, when B reaches 60, C would be at (13/15)*60. Step-by-Step Solution: B : C = 15 : 13At B = 60, C = 60 * (13/15) = 52So, B can give C = 60 − 52 = 8 points Verification / Alternative check: Alternatively, at C = 60, B = 60 * (15/13) ≈ 69.23, meaning B is 9.23 ahead at C's 60; rounded handicap in B's frame at total 60 for B is exactly 8 points given to C. Why Other Options Are Wrong: 6/10/12 don’t match the precise 15:13 ratio at 60 points. Common Pitfalls: Assuming differences transfer linearly; it’s the ratio that transfers, not absolute gap. Final Answer: 8

Question 13

Head-start with speed ratio: Two men A and B run a 500 m race. A has a 140 m head-start, and their speeds are in the ratio 3 : 4 (A : B). By how many metres does A win?

Accepted Answer

Introduction / Context: In head-start problems with speed ratios, determine where the opponent is when the leading runner finishes. That distance shortfall is the winning margin. Given Data / Assumptions: Race length = 500 m. A starts 140 m ahead, so A must run only 360 m to finish. Speeds A : B = 3 : 4 (constant). Concept / Approach: Equal times to the finish for A and B when A wins: if A runs 360 m, B runs (4/3)*360 = 480 m. The deficit from 500 m is the margin by which A wins. Step-by-Step Solution: A must cover 360 mB's distance in same time = (4/3) * 360 = 480 mWinning margin = 500 − 480 = 20 m Verification / Alternative check: At the instant B reaches 500 m, A would have covered (3/4)*500 = 375 m from his own start; since A only needed 360 m, he would have finished 15 m earlier. Both checks are consistent with a 20 m win margin definition (distance B is short when A finishes). Why Other Options Are Wrong: 10/40/60 do not match the direct proportional calculation. Common Pitfalls: Subtracting the head-start from the race length without using the speed ratio. Final Answer: 20 m

Question 14

Chaining two race advantages: In a 600 m race, A beats B by 60 m. In a 500 m race, B beats C by 50 m. By how many metres will A beat C in a 400 m race?

Accepted Answer

Introduction / Context: When advantages are given for different race lengths, convert each advantage to a speed ratio, multiply the ratios to compare non-adjacent contestants, and then scale to the desired length. Given Data / Assumptions: A vs B at 600 m: B at 540 m when A finishes → vA/vB = 600/540 = 10/9. B vs C at 500 m: C at 450 m when B finishes → vB/vC = 500/450 = 10/9. Uniform speeds. Concept / Approach: Combine: vA/vC = (10/9)*(10/9) = 100/81. For 400 m, compute the distance C covers when A finishes. Step-by-Step Solution: C's distance when A runs 400 = 400 * (vC/vA) = 400 * 81/100 = 324 mA's win margin = 400 − 324 = 76 m Verification / Alternative check: Repeating the derivation numerically yields the same 76 m difference for any proportional scaling. Why Other Options Are Wrong: 78/56/86 are off the exact 100/81 ratio result. Common Pitfalls: Adding raw metre advantages; only ratios transfer correctly across distances. Final Answer: 76 m

Question 15

Another chained advantage check: In a 600 m race, A beats B by 60 m. In a 500 m race, B beats C by 50 m. By how many metres will A beat C in a 400 m race?

Accepted Answer

Introduction / Context: Identical to the previous setup; we apply speed ratios derived from given metre advantages and scale to the new race length. Given Data / Assumptions: As above: vA/vB = 10/9 and vB/vC = 10/9, hence vA/vC = 100/81; uniform speeds. Concept / Approach: Compute C's distance when A completes 400 m, then subtract from 400 m to obtain the margin. Step-by-Step Solution: C at A's finish = 400 * 81/100 = 324 mWin margin = 400 − 324 = 76 m Verification / Alternative check: Proportional checks confirm 76 m. Why Other Options Are Wrong: Numbers not consistent with 100/81 ratio. Common Pitfalls: Using arithmetic differences instead of ratios across race lengths. Final Answer: 76 m

Question 16

Translating comparative strength across games: In a game, A can give B 25 points in 75 and C 18 points in 90. In a 120-point game, how many points can C give B?

Accepted Answer

Introduction / Context: We convert handicaps into scoring speed ratios, compare B and C, then project to the requested game length. Given Data / Assumptions: A vs B (75): A:B = 75:50 = 3:2. A vs C (90): A:C = 90:72 = 5:4. Uniform scoring rates. Concept / Approach: From A:B = 3:2 and A:C = 5:4 ⇒ B:C = (2/3)/(4/5) = (2/3)*(5/4) = 10/12 = 5/6. C is faster than B (denominator smaller time for same points). Step-by-Step Solution: At B = 120, C's equivalent = 120 * (6/5) = 144 (faster)Alternatively, when C = 120, B = 120 * (5/6) = 100Thus C can give B = 120 − 100 = 20 points Verification / Alternative check: Both frames (B finishing at 120 or C finishing at 120) are consistent with a 20-point handicap. Why Other Options Are Wrong: 30/40/50 arise from mis-scaling or difference-based thinking rather than ratio-based. Common Pitfalls: Using absolute point gaps instead of relative scoring speeds when changing game length. Final Answer: 20

Question 17

Duplicate head-start scenario (sanity check): In a 500 m race, speeds of A and B are in the ratio 3 : 4. A has a start of 140 m. By how many metres does A win?

Accepted Answer

Introduction / Context: This repeats the earlier head-start calculation to confirm consistency across similar items. Given Data / Assumptions: Race 500 m; A has 140 m start → A runs 360 m. Speeds A:B = 3:4. Concept / Approach: B covers (4/3) times A’s distance in the same time. Compute B’s position when A finishes 360 m. Step-by-Step Solution: B's distance at A's finish = (4/3)*360 = 480 mWin margin = 500 − 480 = 20 m Verification / Alternative check: Consistent with proportional reasoning across similar tasks. Why Other Options Are Wrong: Mismatch the exact ratio implication. Common Pitfalls: Ignoring the head-start in required distance for A. Final Answer: 20 m

Question 18

Start vs catch calculation: Two boys A and B run at 4.5 km/h and 6 km/h, respectively. Over a 1 km course, A has a 190 m start. By how many metres does B still win?

Accepted Answer

Introduction / Context: With different speeds and a head-start, compute where the faster runner is when the slower finishes. The difference to the finish line is the winning margin. Given Data / Assumptions: Course = 1000 m; A starts 190 m ahead → A needs 810 m. vA = 4.5 km/h; vB = 6 km/h; constant speeds. Concept / Approach: Time for A to finish tA = distance / speed. In that time, B covers vB * tA. Compare B’s covered distance with 1000 m. Step-by-Step Solution: Convert speeds to m/s: vA = 4.5 * 1000 / 3600 = 1.25 m/s; vB = 6 * 1000 / 3600 = 1.666… m/stA = 810 / 1.25 = 648 sB's distance in tA = 1.666… * 648 = 1080 mWinning margin = 1080 − 1000 = 80 m Verification / Alternative check: Ratio method: In A's finish time, B covers (vB/vA) * 810 = (6/4.5) * 810 = (4/3)*810 = 1080 → margin 80 m. Why Other Options Are Wrong: 60/65/45/75 do not match the computed 80 m margin. Common Pitfalls: Subtracting the 190 m head-start from 1000 m directly without applying speed difference. Final Answer: 80 m

Question 19

100 m race with start and time gap: A and B run a 100 m race. A runs at 5 km/h, gives B an 8 m start, and still beats B by 8 s. What is B’s speed?

Accepted Answer

Introduction / Context: We know A's speed and the time gap at finish. Using the 8 m start, we compute B's required time and hence B's speed. Given Data / Assumptions: A's speed = 5 km/h = 5000/3600 = 1.3889 m/s. Race length 100 m for A; B starts 8 m ahead → B needs 92 m. A finishes 8 s before B. Concept / Approach: Compute A's finish time; add 8 s to get B's finish time for 92 m; then speed = distance/time. Step-by-Step Solution: tA = 100 / 1.3889 ≈ 72 stB = tA + 8 = 80 svB = 92 / 80 = 1.15 m/s ⇒ 1.15 * 3.6 = 4.14 km/h Verification / Alternative check: Using exact fractions yields the same 4.14 km/h (to two decimals). Why Other Options Are Wrong: Other values correspond to incorrect conversion or ignoring the 8 m start. Common Pitfalls: Converting km/h to m/s incorrectly; misreading the start distance as time advantage. Final Answer: 4.14 km/h

Question 20

Meeting again at the start on a circular track: On a 5 km circular track, A, B, and C start together in the same direction at 2.5 km/h, 3 km/h, and 2 km/h, respectively. After how many hours will all three meet again at the starting point?

Accepted Answer

Introduction / Context: When multiple runners set off on a circular track, they are all back at the start together at the least common multiple (LCM) of their individual lap times. Given Data / Assumptions: Track length = 5 km. Speeds: 2.5, 3, and 2 km/h; constant speeds. Concept / Approach: Compute lap periods: T1 = 5/2.5 = 2 h; T2 = 5/3 = 5/3 h; T3 = 5/2 = 2.5 h. The required time is LCM(T1, T2, T3). Step-by-Step Solution: T1 = 2 h; T2 = 5/3 h; T3 = 5/2 hLCM(2, 5/3, 5/2) = 10 h (smallest t with t/Ti all integers)Check: 10/2 = 5; 10/(5/3) = 6; 10/(5/2) = 4 — all integers Verification / Alternative check: LCM via integerizing denominators also yields 10 h. Why Other Options Are Wrong: 6/15/30 are common distractors but fail integrality for at least one runner. Common Pitfalls: Using arithmetic mean or max instead of LCM of lap periods. Final Answer: 10 hours

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