Question 1

In a 400 m race, A runs at 16 m/s. A gives B a 16 m start and still beats B by 40 s. What is B’s speed (m/s)?

Accepted Answer

Introduction / Context: Starts reduce the distance for one runner. If the faster runner still wins by a time margin, we can deduce the slower runner's average speed from course lengths and finish times. Given Data / Assumptions: A's speed = 16 m/s; A runs full 400 m. B gets a 16 m start ⇒ B's course = 400 − 16 = 384 m. A beats B by 40 s (B finishes 40 s after A). Concept / Approach: Compute A's finish time, then add 40 s to get B's finish time over 384 m, and finally obtain B's speed as distance/time. Step-by-Step Solution: A's time = 400 / 16 = 25 s B's time = 25 + 40 = 65 s B's speed = 384 / 65 ≈ 5.9077 m/s ≈ 5.9 m/s Verification / Alternative check: With 5.9 m/s, B's time ≈ 65 s; A's is 25 s, confirming the 40 s gap. The 16 m start is correctly accounted for by shortening B's course to 384 m. Why Other Options Are Wrong: 6 or 8 m/s yield different time gaps; 15 m/s would make B faster than A, contradicting the result that A wins by 40 s. Common Pitfalls: Forgetting to reduce B's course by the start; misreading the phrase “beats by 40 s” (it means B's time is 40 s longer). Final Answer: 5.9 m/s

Question 2

A can run 45 m while B runs 50 m. In a 1000 m race, by how many metres does B beat A?

Accepted Answer

Introduction / Context: Comparative speed statements like “A runs 45 m while B runs 50 m” define a speed ratio. Converting that ratio into a race margin requires scaling distances to the race length for the faster runner and reading off how far the slower runner gets in the same time. Given Data / Assumptions: Speed ratio A:B = 45:50 = 9:10. B completes 1000 m; find A's distance at that instant. Concept / Approach: For equal time, distances are proportional to speeds. When B runs 10 “parts,” A runs 9 “parts.” Scale to 1000 m for B and compute A's distance. Step-by-Step Solution: When B = 1000 m (10 parts), A = (9/10) * 1000 = 900 m Lead of B over A = 1000 − 900 = 100 m Verification / Alternative check: Equivalently, speed ratio 10:9 implies B is 11.11% faster; over 1000 m, that translates to a 100 m margin. Why Other Options Are Wrong: 150, 200, 125 m do not correspond to the exact 9:10 proportional distance in the same time frame. Common Pitfalls: Using time ratios instead of distance for equal-time comparison; mixing up which athlete is faster (here B is faster). Final Answer: 100 m

Question 3

A runs 200 m in 35 s while B runs 200 m in 38 s. By what distance does A beat B in a 200 m race?

Accepted Answer

Introduction / Context: When finish times over the same course are known, the distance lead at the winner's finish time equals the slower runner's distance covered by then. Compute that distance using the slower runner's speed and the winner's finishing time. Given Data / Assumptions: A's 200 m time = 35 s ⇒ A finishes at t = 35 s. B's 200 m time = 38 s ⇒ B's speed = 200/38 m/s. Concept / Approach: At t = 35 s, B has covered d = speed_B * 35. The lead is 200 − d. Express the exact value as a simplified fraction. Step-by-Step Solution: speed_B = 200/38 = 100/19 m/s B's distance at 35 s = (100/19)*35 = 3500/19 m Lead = 200 − 3500/19 = (3800 − 3500)/19 = 300/19 m 300/19 m = 15 + 15/19 m = 15 15/19 m Verification / Alternative check: As a decimal, 300/19 ≈ 15.789 m, consistent with A being slightly faster over 200 m. Why Other Options Are Wrong: 15 m and 152⁄3 m (≈ 15.667 m) are close but not exact; 154⁄19 m ≈ 8.105 m is far too small. Common Pitfalls: Using (38 − 35) seconds directly as meters; mixing up average vs instantaneous speeds; rounding too early and losing precision. Final Answer: 1515⁄19 m

Question 4

In a 1000 m race, A beats B by 100 m and C by 200 m. By how much does B beat C in a 1350 m race?

Accepted Answer

Introduction / Context: When margins versus a common competitor are known, convert them to speed ratios, then compare the two remaining competitors over a new distance by scaling with those ratios. Given Data / Assumptions: When A runs 1000 m, B runs 900 m, and C runs 800 m. Speeds are proportional to these distances over equal time: A:B:C = 1000:900:800 = 10:9:8. We compare B and C over 1350 m. Concept / Approach: For equal times, distance ∝ speed. When B finishes 1350 m, C covers (speed_C/speed_B) times that distance. Step-by-Step Solution: speed_B : speed_C = 9 : 8 C's distance when B finishes = 1350 * (8/9) = 1200 m Lead of B over C = 1350 − 1200 = 150 m Verification / Alternative check: Consistent with the 10:9:8 baseline; margins scale linearly with course length for uniform speeds. Why Other Options Are Wrong: 125, 130, 140 m are not the exact scaled difference implied by the 9:8 speed ratio over 1350 m. Common Pitfalls: Using 100 m vs 200 m differences directly without forming ratios; forgetting to scale the margin for the new race length. Final Answer: 150 m

Question 5

In a game of 100 points, A can give B 20 points and C 28 points. How many points can B give C in a 100-point game?

Accepted Answer

Introduction / Context: Point handicaps reflect scoring rate ratios. If A gives B and C certain points to 100, we can compute A:B and A:C rate ratios and then deduce B:C and the corresponding head start in a 100-point game. Given Data / Assumptions: When A scores 100, B scores 80 ⇒ A:B = 100:80 = 5:4. When A scores 100, C scores 72 ⇒ A:C = 100:72 = 25:18. Concept / Approach: Rates are proportional to scores in equal time. Use B/A = 4/5 and C/A = 18/25 to find B/C = (B/A)/(C/A). Translate the ratio into a points head start out of 100. Step-by-Step Solution: B/A = 4/5, C/A = 18/25 B/C = (4/5) / (18/25) = (4/5) * (25/18) = 100/90 = 10/9 If B scores 100, C scores 90 ⇒ B can give C 10 points in 100. Verification / Alternative check: Check consistency: A:B:C speeds ∝ 5:4 and 25:18 ⇒ B:C = 10:9 as derived. Why Other Options Are Wrong: 12, 15, 20 points imply different B:C ratios not supported by the given head starts versus A. Common Pitfalls: Adding/subtracting head starts directly instead of forming rate ratios; mis-scaling to a 100-point target. Final Answer: 10 points

Question 6

In a 1000 m race, A can give B a 50 m start and C an 80 m start. In a 1000 m race between B and C, how many metres start should B give C?

Accepted Answer

Introduction / Context: Head starts relative to a common runner (A) let us infer speed ratios. From A:B and A:C, derive B:C and convert that to the start B should give C over the same official distance. Given Data / Assumptions: In 1000 m for A, B covers 950 m ⇒ A:B = 1000:950 = 20:19. In 1000 m for A, C covers 920 m ⇒ A:C = 1000:920 = 25:23. Uniform speeds. Concept / Approach: Use B/A = 19/20 and C/A = 23/25 to get B/C = (19/20)/(23/25) = (19*25)/(20*23) = 95/92. When B runs 1000 m, C runs 1000*(92/95), so the start is the difference to 1000. Step-by-Step Solution: C's distance when B finishes = 1000 * (92/95) = 968 8/19 m Required start = 1000 − 1000*(92/95) = 1000*(3/95) = 31 11/19 m Verification / Alternative check: As a decimal, 31 11/19 m ≈ 31.5789 m; B:C ≈ 1.0326, consistent with B being slightly faster. Why Other Options Are Wrong: 31 m and 312⁄19 m (≈ 16.42 m) do not match the exact fractional start; 3011⁄19 m is far too large. Common Pitfalls: Subtracting 50 and 80 directly; forgetting to convert starts into speed ratios via a common reference. Final Answer: 3111⁄19 m

Question 7

In a 250-point game, A can give B 50 points and C 70 points. In the same game, how many points can B give C?

Accepted Answer

Introduction / Context: Head starts over a common target convert into scoring rate ratios. From A's advantages over B and C to 250 points, deduce B's advantage over C for the same target. Given Data / Assumptions: When A scores 250, B scores 200 ⇒ A:B = 250:200 = 5:4. When A scores 250, C scores 180 ⇒ A:C = 250:180 = 25:18. Concept / Approach: Compute B/C = (B/A)/(C/A) = (4/5)/(18/25) = 10/9. If B scores 250, C scores 250*(9/10) = 225, so B can give C a 25-point start to 250. Step-by-Step Solution: B/A = 4/5; C/A = 18/25 B/C = (4/5) / (18/25) = 10/9 C's score when B reaches 250 = 250*(9/10) = 225 Start B gives C to 250 = 250 − 225 = 25 points Verification / Alternative check: Scaling down to 100 points would yield 10 points; scaling to 250 multiplies by 2.5, giving 25, consistent with proportional rates. Why Other Options Are Wrong: 18, 20, 22 do not correspond to the 10:9 ratio at a 250-point target. Common Pitfalls: Adding/subtracting the given head starts directly; not converting starts to rate ratios first. Final Answer: 25 points

Question 8

Race timing and lead: In a 300 m race, runner A defeats runner B by 15 m (i.e., when A finishes, B is 15 m behind) and this deficit corresponds to 5 s for B. Find the total time A takes to complete the 300 m race.

Accepted Answer

Introduction / Context: We are given a 300 m race where A finishes first. At A’s finish, B is 15 m short of the finish line, and that 15 m takes B exactly 5 s. The task is to compute A’s finish time over 300 m. This tests relative speed and race timing interpretation. Given Data / Assumptions: Total distance for A = 300 m B is 15 m behind when A finishes For B, 15 m takes 5 s → B's speed = 3 m/s Concept / Approach: At the instant A finishes, B has covered 285 m. If we know B's speed, then the shared race time equals the time B took to cover 285 m. That common time equals A's finish time. Step-by-Step Solution: B speed = 15 / 5 = 3 m/sDistance B covered when A finishes = 300 − 15 = 285 mCommon time = 285 / 3 = 95 s Verification / Alternative check: In 95 s, B covers 95 * 3 = 285 m, exactly 15 m short, matching the statement. Why Other Options Are Wrong: 90 s, 100 s, 105 s would imply different B deficits than 15 m, contradicting the given 5 s for 15 m. Common Pitfalls: Using 300 m with B's speed directly; the common time corresponds to B’s 285 m, not 300 m. Final Answer: 95 sec.

Question 9

1 km race timing from lead: In a 1 km race, A beats B by 25 m or 5 s. Determine the time A takes to finish the 1 km course.

Accepted Answer

Introduction / Context: This is a standard race problem where a distance lead is also given as a time lead. We use B’s speed deduced from “25 m in 5 s” and then compute A’s finish time. Given Data / Assumptions: Total distance = 1000 m Lead at finish = 25 m = 5 s for B Concept / Approach: When A finishes, B is at 975 m. The shared race time equals the time B takes to cover 975 m at B’s computed speed. Step-by-Step Solution: B speed = 25 / 5 = 5 m/sCommon time = 975 / 5 = 195 s195 s = 3 min 15 s Verification / Alternative check: In 195 s, B would need 25 m more (5 s) to finish; consistent with the statement. Why Other Options Are Wrong: Other times do not produce a 25 m (5 s) shortfall for B at A’s finish. Common Pitfalls: Using 1000 m for B instead of 975 m for the common time reference. Final Answer: 3 minutes 15 seconds

Question 10

Billiards handicap consistency: A can give B 20 points, A can give C 32 points, and B can give C 15 points. Assuming consistent skill ratios, how many points make the game?

Accepted Answer

Introduction / Context: In handicap problems, “A gives B k in N” means when A scores N points, B scores N − k points in the same time. Consistency across pairs lets us solve for N (the game size). Given Data / Assumptions: A gives B 20 in N ⇒ A:B = N : (N − 20) A gives C 32 in N ⇒ A:C = N : (N − 32) B gives C 15 in N ⇒ B:C = N : (N − 15) Concept / Approach: When A scores N, B scores N − 20. In that time, C should score (N − 20)*(N − 15)/N by the B:C ratio. But also, by A:C, C must be N − 32. Equate to solve N. Step-by-Step Solution: (N − 20)(N − 15)/N = N − 32N^2 − 35N + 300 = N^2 − 32N−35N + 300 = −32N ⇒ −3N = −300 ⇒ N = 100 Verification / Alternative check: Plugging N = 100 satisfies all three handicaps. Why Other Options Are Wrong: They violate at least one of the three given pairwise handicaps. Common Pitfalls: Adding point differences linearly; the correct method uses proportional scoring in equal time. Final Answer: 100

Question 11

Fraction of race remaining: A runner completes 1 and 1/4 laps of a 5-lap race. What fractional part of the race remains to be run?

Accepted Answer

Introduction / Context: This is a simple fraction-of-total problem expressed in laps. We translate completed laps to a fraction of the whole and subtract from 1. Given Data / Assumptions: Total laps = 5 Completed laps = 1.25 Concept / Approach: Remaining laps = total − completed. The remaining fraction of the race equals remaining laps divided by total laps. Step-by-Step Solution: Remaining laps = 5 − 1.25 = 3.75Fraction remaining = 3.75 / 5 = 0.75 = 3/4 Verification / Alternative check: Completed fraction = 1.25/5 = 0.25; remaining = 1 − 0.25 = 0.75 = 3/4. Why Other Options Are Wrong: 4/5, 5/6, 2/3 represent 0.8, 0.833…, 0.666…, none equals 0.75. Common Pitfalls: Confusing “laps left” (15/4 laps) with “fraction of the race left” (3/4 of the race). Final Answer: 3⁄4

Question 12

Common meeting time around a track: A, B, and C complete one round in 252 s, 308 s, and 198 s respectively. Starting together from the same point, after how long will all three next be together at the start point?

Accepted Answer

Introduction / Context: When multiple persons move in loops and we want their next simultaneous meeting at the start point, we compute the least common multiple (LCM) of their individual round times. Given Data / Assumptions: T_A = 252 s T_B = 308 s T_C = 198 s Concept / Approach: Next simultaneous return to start occurs at LCM(T_A, T_B, T_C). Convert the result to minutes and seconds. Step-by-Step Solution: LCM(252, 308, 198) = 2772 s2772 s = 46 min 12 s Verification / Alternative check: Prime-factorize each time and take the maximum powers to confirm LCM = 2772 s. Why Other Options Are Wrong: Other durations are not multiples of all three given times. Common Pitfalls: Using GCD or pairwise meetings instead of LCM for all three together at the start point. Final Answer: 46 minutes 12 seconds

Question 13

Weekly running schedule: Raju runs 1250 m on Monday and Friday, runs 1500 m on other weekdays and Saturday, and rests on Sunday. Starting Monday, how many kilometres does he run in 3 weeks?

Accepted Answer

Introduction / Context: This is a schedule-based arithmetic problem summing weekly distances over multiple weeks, with varying daily amounts and weekly rest days. Given Data / Assumptions: Mon & Fri = 1250 m each Tue, Wed, Thu, Sat = 1500 m each Sun = 0 3 consecutive weeks starting Monday Concept / Approach: Compute one-week total, then multiply by 3. Convert metres to kilometres at the end. Step-by-Step Solution: Per week: 1250 + 1250 = 2500 m (Mon & Fri)Per week: 4 * 1500 = 6000 m (Tue, Wed, Thu, Sat)Weekly total = 2500 + 6000 = 8500 m = 8.5 kmThree weeks total = 3 * 8.5 = 25.5 km Verification / Alternative check: Counting days explicitly confirms 18 running days and 3 rest Sundays over 3 weeks. Why Other Options Are Wrong: They reflect incorrect weekly totals or omission of rest days. Common Pitfalls: Forgetting to include Saturday or excluding both Monday and Friday special distances correctly. Final Answer: 25.5 Km

Question 14

Race timing from lead (repeat form): In a 300 m race, A defeats B by 15 m or 5 s. Determine A’s time over the course.

Accepted Answer

Introduction / Context: Same structure as a classic race-time conversion problem: translate B's shortfall (distance and seconds) into B's speed and compute the shared finish time. Given Data / Assumptions: Total race = 300 m B is 15 m behind at A's finish 15 m = 5 s for B Concept / Approach: The finish time for A equals the time B took to run 285 m at 3 m/s. Step-by-Step Solution: B speed = 15 / 5 = 3 m/sA's finish time = 285 / 3 = 95 s Verification / Alternative check: In 95 s, B covers 285 m, 15 m behind the finish. Why Other Options Are Wrong: They contradict the given 15 m in 5 s relationship for B. Common Pitfalls: Using 300 m for B in the shared time instead of 285 m. Final Answer: 95 seconds

Question 15

Relative speed with head start: A runs 1 3⁄8 times as fast as B. If A gives B a start of 120 m and they finish together, find the race distance (the winning post distance from A’s start).

Accepted Answer

Introduction / Context: This problem is about equalizing finish times with a head start when one runner is faster by a known factor. Given Data / Assumptions: Speed ratio v_A : v_B = 11 : 8 Head start for B = 120 m They finish together Concept / Approach: Set equal times: distance/v_A = (distance − headstart)/v_B. Substitute the known ratio and solve for the distance. Step-by-Step Solution: Let D be race distance from A's start.D / (11k) = (D − 120) / (8k) ⇒ (8/11)D = D − 120D − (8/11)D = 120 ⇒ (3/11)D = 120 ⇒ D = 440 m Verification / Alternative check: Plug D = 440 to confirm both times equal: 440/(11k) vs. 320/(8k) = 40/k each. Why Other Options Are Wrong: Other distances do not satisfy the time equality for the given ratio. Common Pitfalls: Adding/subtracting speeds incorrectly or forgetting to subtract the head start from B’s running distance. Final Answer: 440 m

Question 16

Billiards handicaps transitivity: A gives B 6 in 50 and A gives C 13 in 65. In a game of 55, how many points can B give C?

Accepted Answer

Introduction / Context: We use proportional scoring in equal time. From A:B and A:C ratios we infer B:C and then scale to a 55-point game. Given Data / Assumptions: A gives B 6 in 50 ⇒ (A:B) = 50:44 = 25:22 A gives C 13 in 65 ⇒ (A:C) = 65:52 = 5:4 Game size in query = 55 points Concept / Approach: Compute B:C = (A:C)/(A:B). Then, when B scores 55, compute C’s score and find the points B can give C. Step-by-Step Solution: B:C = (5/4) / (25/22) = (5/4)*(22/25) = 11/10If B scores 55 ⇒ C = 55*(10/11) = 50So B can give C = 55 − 50 = 5 points Verification / Alternative check: Scaling either ratio to a common A-score yields the same B:C relation of 11:10. Why Other Options Are Wrong: They contradict the deduced B:C = 11:10. Common Pitfalls: Subtracting handicaps instead of converting them into proportional scoring rates. Final Answer: 5

Question 17

Linked race comparisons: In a 600 m race A beats B by 30 m; in a 500 m race B beats C by 25 m. By how many metres will A beat C in a 400 m race?

Accepted Answer

Introduction / Context: We infer speed ratios from separate races and combine them to compare A and C in a new race length. Given Data / Assumptions: A vs B in 600 m: A:B = 600:570 = 20:19 B vs C in 500 m: B:C = 500:475 = 20:19 Target race length for A = 400 m Concept / Approach: Multiply ratios to get A:C and then scale to 400 m to find C’s distance at A’s finish. Step-by-Step Solution: A:C = (20:19)*(20:19) = 400:361If A runs 400 m, C runs 361 mLead = 400 − 361 = 39 m Verification / Alternative check: Ratios reduce consistently; no units needed as they cancel in ratios. Why Other Options Are Wrong: Other leads mismatch the squared ratio 400:361 implied by chaining 20:19 twice. Common Pitfalls: Adding distance leads directly; proper method is via speeds/ratios. Final Answer: 39 m

Question 18

Head start with speed factor (corrected statement): A runs 1 2⁄3 times as fast as B. If A gives B a start of 30 m and they finish together, how far from the start must the winning post be?

Accepted Answer

Introduction / Context: The original text had a likely typo (“1% times”). By the Recovery-First Policy we minimally repair to a standard form: A runs 1 2/3 times as fast as B. With a 30 m head start to B, find the race distance so they tie. Given Data / Assumptions (after minimal repair): v_A : v_B = 5 : 3 Head start to B = 30 m Finish together Concept / Approach: Equalize times: D / v_A = (D − 30) / v_B. Solve for D with v_A/v_B = 5/3. Step-by-Step Solution: D / (5k) = (D − 30) / (3k) ⇒ (3/5)D = D − 30D − (3/5)D = 30 ⇒ (2/5)D = 30 ⇒ D = 75 m Verification / Alternative check: Times equal: 75/(5k) = 45/(3k) = 15/k. Why Other Options Are Wrong: They do not satisfy the time equality for ratio 5:3 with a 30 m head start. Common Pitfalls: Using speed difference instead of ratio, or forgetting to subtract the head start from B’s distance. Final Answer: 75 m

Question 19

Start advantage in short race: In a 250 m race, A:B speed ratio is 1:2. If A starts 140 m ahead, by how many metres does A win?

Accepted Answer

Introduction / Context: Given a significant head start and a speed disadvantage, we compute where A is when B finishes to get A’s winning margin. Given Data / Assumptions: Race length = 250 m v_A : v_B = 1 : 2 A starts at 140 m mark Concept / Approach: Time for B to finish = 250/(2u). In that time, A covers additional distance (time * u) beyond the 140 m head start. Step-by-Step Solution: t = 250/(2u) = 125/uA distance at B’s finish = 140 + u * (125/u) = 265 mMargin = 265 − 250 = 15 m Verification / Alternative check: Proportional scaling confirms A’s progress in half the time B needs (since v_A is half of v_B). Why Other Options Are Wrong: They mismatch the precise computation using ratio 1:2 and the given head start. Common Pitfalls: Adding/subtracting ratios directly instead of using time comparisons. Final Answer: 15 m

Question 20

1 km race lead with time info: In a 1 km race, P beats Q by 72 m in 12 s. Find P’s time for the whole course.

Accepted Answer

Introduction / Context: We infer Q’s speed from the fact that 72 m corresponds to 12 s. Then we evaluate the shared finish time when P ends the race and compute P’s course time. Given Data / Assumptions: Total = 1000 m Q is 72 m short when P finishes Q needs 12 s for 72 m ⇒ v_Q = 6 m/s Concept / Approach: At P’s finish, Q has covered 928 m at 6 m/s. That time equals P’s course time. Step-by-Step Solution: t_common = 928 / 6 = 154.666… s = 154 2/3 s Verification / Alternative check: Check that in an additional 12 s, Q would finish; 154 2/3 + 12 = 166 2/3 s for Q’s 1000 m, matching v_Q = 6 m/s (1000 / 166 2/3 = 6). Why Other Options Are Wrong: They do not align with v_Q = 6 m/s and the 72 m/12 s relationship. Common Pitfalls: Using 1000 m for Q in the shared time instead of 928 m. Final Answer: 1542/3 s

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