Question 1

In a 300 m race, A beats B by 22.5 m or 6 seconds. What is B's time over the course?

Accepted Answer

Problem restatement When A finishes 300 m, B is 22.5 m behind; equivalently, B takes 6 s longer to finish the race. Find B's total time for 300 m. Concept/Approach The 22.5 m deficit equals B's distance covered in 6 s, which gives B's speed. Then compute time for 300 m. Step-by-step calculation B's speed = 22.5 / 6 = 3.75 m/sB's time for 300 m = 300 / 3.75 = 80 s Verification/Alternative If B's time is 80 s, then A's time is 74 s (6 s ahead). In 74 s, B covers 3.75 × 74 = 277.5 m, which is 22.5 m short of 300 m (checks). Common pitfalls Assuming constant difference in distance translates directly to average speed without using the 6 s information. Final Answer 80 seconds (i.e., 1 min 20 s)

Question 2

In a 200 m race, A beats B by 35 m or 7 seconds. What is A's time to finish the 200 m course?

Accepted Answer

Problem restatement A finishes 200 m when B is at 165 m (beaten by 35 m). The same margin corresponds to 7 seconds. Find A's finishing time for 200 m. Given data Race length = 200 m. Margin: 35 m equals 7 seconds for B. Concept/Approach Use B's speed from the given margin-time, compute B's 200 m time, then subtract the 7 s lead to get A's time. Step-by-step calculation B's speed = 35 ÷ 7 = 5 m/s B's 200 m time = 200 ÷ 5 = 40 s A's time = B's time − lead = 40 − 7 = 33 s Verification/Alternative At 33 s, B covers 5 × 33 = 165 m, matching the 35 m deficit when A finishes. Common pitfalls Interpreting 7 s as A's advantage in speed rather than time. Using 165 m for A by mistake; A runs the full 200 m. Final Answer 33 seconds

Question 3

In a 100 m race, A can give B a start of 10 m and C a start of 28 m. In the same race, how many metres can B give C?

Accepted Answer

Problem restatement A's 100 m time equals B's 90 m time and equals C's 72 m time. Find B's head start over C in a 100 m race. Given data Speeds proportional to covered distance in the same time: vA : vB : vC = 100 : 90 : 72. Concept/Approach For the time B runs 100 m, C covers (vC/vB) × 100. Step-by-step calculation vC/vB = 72 ÷ 90 = 4 ÷ 5 C's distance when B runs 100 m = (4/5) × 100 = 80 m Hence B can give C = 100 − 80 = 20 m Verification/Alternative Ratio check: A:B:C speeds = 100:90:72 ⇒ for any equal time, distances scale the same way; result is consistent. Common pitfalls Using 28 − 10 = 18 m directly; head starts do not combine linearly across different racers. Final Answer 20 m

Question 4

A can run 22.5 m while B runs 25 m (same time). In a 1 km race, by how many metres does B beat A?

Accepted Answer

Problem restatement Given a constant-time distance ratio, find B's winning margin in a 1000 m race. Given data vA : vB = 22.5 : 25 = 9 : 10. Concept/Approach In the time B finishes 1000 m, A covers (9/10) of that distance. Step-by-step calculation A's distance when B finishes = (9/10) × 1000 = 900 m Winning margin = 1000 − 900 = 100 m Verification/Alternative Scale the 9:10 ratio to 900:1000; the difference is the margin. Final Answer 100 m

Question 5

In a 200 m race, A beats B by 31 m and C by 18 m. In a 350 m race, by how many metres will C beat B?

Accepted Answer

Problem restatement From the 200 m results, infer the speed ratios and predict the B–C gap over 350 m. Given data When A runs 200 m, B runs 169 m and C runs 182 m. Thus vA : vB = 200 : 169 and vA : vC = 200 : 182 = 100 : 91. Concept/Approach For the time B completes 350 m, C covers (vC/vB) × 350. Compute vC/vB from the two ratios. Step-by-step calculation vC/vB = (vC/vA) × (vA/vB) = (91/100) × (200/169) = 182/169 = 14/13 C's distance in B's 350 m time = (14/13) × 350 = 350 × 14 ÷ 13 = 376 12⁄13 m Winning margin (C over B) = 376 12⁄13 − 350 = 350 ÷ 13 ≈ 26.92 m Verification/Alternative Direct proportion: when B runs 169, C runs 182 ⇒ per 169 m of B, C is 13 m ahead; scale by 350/169 gives the same 350/13. Common pitfalls Assuming margins scale linearly (31 vs. 18) without accounting for speed ratios. Final Answer 350/13 m (≈ 26.92 m)

Question 6

In a 100 m race, A beats B by 25 m and B beats C by 4 m. In the same race, by how many metres does A beat C?

Accepted Answer

Problem restatement From two head-start results, deduce A vs. C gap over 100 m. Given data A : B speeds = 100 : 75 = 4 : 3. B : C speeds = 100 : 96 = 25 : 24. Concept/Approach Multiply ratios to get A : C speed ratio, then compute C's distance when A finishes 100 m. Step-by-step calculation A : C = (A : B) × (B : C) = (4/3) × (25/24) = 100/72 = 25/18 C's distance when A finishes = 100 × (18/25) = 72 m Margin (A over C) = 100 − 72 = 28 m Verification/Alternative Chain-speed method ensures consistency across racers; the 28 m result matches standard composition of ratios. Final Answer 28 m

Question 7

In a 500 m race, A and B's speed ratio is 3 : 4. A gets a start of 140 m. By how many metres does A win?

Accepted Answer

Problem restatement A starts 140 m ahead; A's speed : B's speed = 3 : 4. Find the distance by which A wins the 500 m race. Given data A's required run = 500 − 140 = 360 m. Speeds: vA : vB = 3 : 4. Concept/Approach Compare positions at the instant A finishes: compute B's covered distance by then. Step-by-step calculation Time for A = 360 ÷ (3k) = 120 ÷ k B's distance in this time = (4k) × (120/k) = 480 m Margin (A over B) = 500 − 480 = 20 m Verification/Alternative Alternatively, compute times to finish: A = 120/k, B = 500/(4k) = 125/k ⇒ A indeed finishes first by 5/k seconds. Final Answer 20 m

Question 8

A and B run a 100 m race. A runs at 5 km/h and gives B a start of 8 m, yet beats B by 8 seconds. What is B's speed?

Accepted Answer

Problem restatement B starts 8 m ahead (needs to run 92 m). A runs 100 m at 5 km/h and finishes 8 s before B. Find B's speed. Given data A's speed = 5 km/h = 5 × 1000 ÷ 3600 = 25/18 m/s ≈ 1.3889 m/s. A's distance = 100 m; B's distance = 92 m. Finish-time gap = 8 s (B is slower). Concept/Approach Compute A's time, then set B's time = A's time + 8 to solve for B's speed. Step-by-step calculation tA = 100 ÷ (25/18) = 100 × 18 ÷ 25 = 72 s tB = tA + 8 = 80 s vB = 92 ÷ 80 = 1.15 m/s Convert to km/h: 1.15 × 3.6 = 4.14 km/h Verification/Alternative Check: In 80 s at 1.15 m/s, B runs 92 m; A's 72 s time confirms the 8 s margin. Common pitfalls Letting B run 100 m despite the 8 m start. Incorrect unit conversion between m/s and km/h. Final Answer 4.14 km/h

Question 9

In billiards, A can give B 15 points in a game to 60, and A can give C 20 points in a game to 60. In a game to 90, how many points can B give C?

Accepted Answer

Problem restatement Handicaps: when A reaches 60, B has 45 and C has 40. Determine B's handicap over C in a game to 90. Given data Relative scoring in same time: A:B:C = 60:45:40 = 12:9:8. Concept/Approach Use proportional scoring rates: in the time B scores 90, C scores (rate ratio) × 90. Step-by-step calculation vB : vC = 9 : 8 C's score when B reaches 90 = (8/9) × 90 = 80 Therefore, B can give C = 90 − 80 = 10 points Verification/Alternative Consistency with 60-point data: scaling 45 and 40 by 2 gives 90 and 80 respectively. Final Answer 10 points

Question 10

In a 100 m race, A beats B by 10 m and C by 13 m. In a 180 m race, by how many metres will B beat C?

Accepted Answer

Problem restatement Use two margins against A to infer B vs. C speed ratio, then scale to 180 m. Given data vA : vB = 100 : 90 = 10 : 9. vA : vC = 100 : 87. Concept/Approach Compute vB : vC = (vB/vA) × (vA/vC) and then the distance C covers when B runs 180 m. Step-by-step calculation vB/vC = (90/100) × (100/87) = 90/87 = 30/29 C's distance when B runs 180 m = (29/30) × 180 = 174 m Margin (B over C) = 180 − 174 = 6 m Verification/Alternative Direct scaling from 100 m margins leads to same ratio; 6 m is the consistent result. Final Answer 6 m

Question 11

A runs 1 2/3 times as fast as B. If A gives B a start of 80 m, how far from A's start should the winning post be so that they finish together?

Accepted Answer

Problem restatement A's speed is 5/3 of B's. B starts 80 m ahead. Find the race length (from A's start) so that both finish simultaneously. Given data vA = (5/3) vB (i.e., A is 1 2/3 times as fast). B's head start = 80 m. Concept/Approach Equal finishing times ⇒ distance/speed equal for both, with B's distance shorter by the head start. Step-by-step calculation Let race length from A's start be D. Time equality: D ÷ vA = (D − 80) ÷ vB Substitute vA = (5/3)vB ⇒ D ÷ [(5/3)vB] = (D − 80) ÷ vB (3/5)D = D − 80 ⇒ (2/5)D = 80 ⇒ D = 200 m Verification/Alternative Times: A takes 200 ÷ (5/3 vB) = 120 ÷ vB; B takes (200 − 80) ÷ vB = 120 ÷ vB (same). Common pitfalls Adding the head start to the race length instead of subtracting from B's distance. Final Answer 200 m

Question 12

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

Accepted Answer

Given data On a 100-point game scale: when A scores 100, B scores 80 (A gives B 20 points). When A scores 100, C scores 72 (A gives C 28 points). Concept/Approach Use relative strengths: A : B : C = 100 : 80 : 72 ⇒ B : C = 10 : 9. If B plays to 100, C will score proportionally 90; hence B can spot C (100 − 90) points. Step-by-step calculation B : C = 80 : 72 = 10 : 9If B's target = 100, then C's corresponding score = 100 × (9/10) = 90Points B can give C = 100 − 90 = 10 points Verification/Alternative Check transitivity: A:B = 100:80 and A:C = 100:72 ⇒ B:C = (80:72) = 10:9, consistent. Common pitfalls Subtracting given advantages directly (20 − 28) instead of using ratios; always compare via common baseline. Final Answer 10 points

Question 13

In a 100 m race, A finishes in 36 s and B in 45 s. By how many metres does A beat B?

Accepted Answer

Given data A's time for 100 m = 36 s ⇒ speedA = 100/36 m/s. B's time for 100 m = 45 s ⇒ speedB = 100/45 m/s. Concept/Approach When A finishes (at 36 s), B has run distance = speedB × 36. The lead = 100 − that distance. Step-by-step calculation Distance by B in 36 s = (100/45) × 36 = 80 mLead of A = 100 − 80 = 20 m Verification/Alternative Time gap = 45 − 36 = 9 s; at 3.75 m/s (B's speed), 9 s corresponds to 33.75 m for B to finish; equivalently, he was 20 m behind when A finished. Common pitfalls Using average of times or subtracting speeds directly without aligning times. Final Answer 20 m

Question 14

Raman covers 1 km in 8 min while suman covers the same distance in 10 min. By what distance does Raman beat Suman?

Accepted Answer

Raman covers 1 km in 8 min and suman cover 1 km in 10 min. If they starts together, then distance covered by suman in 8 min = (1000/10) x 8 = 800 m ∴ Raman will beat suman by (1000 - 800 ) m = 200 m

Question 15

In a race of 150 m, A gives B a start of 20 m. What distance will be covered by B?

Accepted Answer

Required distance = (150 - 20) m = 130 m

Question 16

In a race, P gives Q a start of 25 m making length of race for Q a distance of 175 m. Find the total length of race.?

Accepted Answer

Required length = (175 + 25)m = 200 m

Question 17

In a 800 m race, A gives some start to B and this makes the length of race for B 725 m. What start does B get from A?

Accepted Answer

Start given by A to B = (800 - 725)m = 75 m

Question 18

In a game of 100 points, A scores 100 points while B scores only 65 points. In this game, how much points can A give to B?

Accepted Answer

Score of A = 100 points Score of B = 65 points ∴ A can give (100 - 65 ) = 35 points to B.

Question 19

In 1 km race, A beats B by 18 m or 9 s. Find the A' s time over the course.

Accepted Answer

Clearly, B covers 18 m in 9 s. ∴ B's time over the course = 9/18 x 1000 = 500 s ∴ A' s time over the course = (500 - 9) = 491 s

Question 20

In a 1000 m race, X beats Y by 140 m or 14 s. What will be the X's time over the course?

Accepted Answer

Clearly, Y covers 140 m in 14 s ∴ Y' s time over the course = (14/140) x 1000 = 100 s ∴ X' s time over the course = 100 - 14 = 86 s

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