Question 1

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

Accepted Answer

Introduction / Context: This is the same logical pattern as an earlier race problem about three runners. It checks whether you can combine two leads, A over B and B over C, correctly by working with speed ratios rather than just adding distances. The numbers are kept simple to highlight understanding rather than heavy computation. Given Data / Assumptions: - When A finishes 100 m, B has covered 75 m. - When B finishes 100 m, C has covered 96 m. - All runners maintain constant speeds during the race. - We must find the distance by which A beats C in the same 100 m race. Concept / Approach: We convert the given information into ratios of speeds vA, vB, and vC. From the first statement we get vB relative to vA, and from the second we get vC relative to vB. Multiplying the ratios gives vC relative to vA. Then we find the distance C covers while A runs 100 m and subtract to find the lead of A. Step-by-Step Solution: Step 1: From A vs B, vB / vA = 75 / 100 = 3 / 4. Step 2: From B vs C, vC / vB = 96 / 100 = 24 / 25. Step 3: Therefore vC / vA = (vC / vB) * (vB / vA) = (24 / 25) * (3 / 4) = 72 / 100 = 18 / 25. Step 4: When A runs 100 m, time taken is T = 100 / vA. Step 5: In the same time, C runs distance = vC * T = 100 * (vC / vA) = 100 * 18 / 25 = 72 m. Step 6: Lead of A over C = 100 - 72 = 28 m. Verification / Alternative check: Assume vA = 25 units. Then vB = 3/4 of 25 = 18.75 units, and vC = 24/25 of vB = 18 units. Time for A to complete 100 m is 100 / 25 = 4 units. In 4 units, C covers 18 * 4 = 72 m. The difference 28 m matches the earlier calculation, confirming the result. Why Other Options Are Wrong: - 21 m and 26 m: These are produced when ratios are handled incorrectly or distances are combined in a linear way without using speed ratios. - 29 m: This is the simple sum of 25 m and 4 m, but adding leads like this is not valid. Common Pitfalls: The main error is to believe that if A beats B by 25 m and B beats C by 4 m, then A beats C by 25 + 4 = 29 m. This ignores that all runners take different times to reach those distances. Always derive the ratio of speeds, then reapply that ratio to the race distance of interest. That is the only reliable approach for such questions. Final Answer: In the 100 m race, A beats C by 28 m.

Question 2

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

Accepted Answer

Introduction / Context: This race question uses the idea of giving a start, which is another way of expressing relative speeds. If one runner can give fixed starts to two different runners over the same race distance, then their speeds are related. We must find how much start the intermediate runner, B, can give to the slowest runner, C, in the same 100 m race. Given Data / Assumptions: - Race distance is 100 m. - A can give B a start of 10 m, meaning when A runs 100 m, B runs only 90 m in the same time. - A can give C a start of 28 m, meaning when A runs 100 m, C runs only 72 m in the same time. - Speeds are constant for all runners throughout the race. - We must find how many metres start B can give C so that both finish 100 m together. Concept / Approach: Again we use relative speeds. From the two given starts we find vB and vC in terms of vA, then get vC / vB. Once we know the speed ratio of B and C, we imagine B running the full 100 m and find the distance C would cover in the same time. The difference between 100 m and that distance is the start B can give C. Step-by-Step Solution: Step 1: From A versus B, when A runs 100 m, B runs 90 m. So vB / vA = 90 / 100 = 9 / 10. Step 2: From A versus C, when A runs 100 m, C runs 72 m. So vC / vA = 72 / 100 = 18 / 25. Step 3: Find vC / vB = (vC / vA) / (vB / vA) = (18 / 25) / (9 / 10) = (18 / 25) * (10 / 9) = 4 / 5. Step 4: Therefore vB / vC = 5 / 4. Step 5: Let B run the full 100 m. Time taken by B is T = 100 / vB. Step 6: In the same time, C covers distance = vC * T = vC * 100 / vB = 100 * (vC / vB) = 100 * 4 / 5 = 80 m. Step 7: B can therefore give C a start of 100 - 80 = 20 m. Verification / Alternative check: Assume vB = 5 units and vC = 4 units from the 5 : 4 ratio. Time for B to cover 100 m is 100 / 5 = 20 units. In 20 units, C covers 4 * 20 = 80 m. If C starts 20 m ahead, C needs to cover only 80 m to finish 100 m, and both will finish together. This confirms the logic and the computed start of 20 m. Why Other Options Are Wrong: - 18 m and 27 m: These values come from incorrect manipulation of ratios or from approximations that ignore exact fractional values. - 9 m: This is far too small and does not match the proportional relationship between the speeds. Common Pitfalls: One common error is to subtract 10 m and 28 m directly to obtain 18 m, assuming that B can give C the difference between the two starts. Another is to confuse which runner is getting the head start in the second stage of the problem. Always convert starts to speed ratios first and then apply those ratios to the full race distance to find the correct start between B and C. Final Answer: In the 100 m race, B can give C a start of 20 m.

Question 3

Runner A can give B a start of 100 m and can give C a start of 200 m in a 1 km race. In a 1 km race between B and C, how many metres start can B give C?

Accepted Answer

Introduction / Context: This question uses the idea of head starts in a 1 km race to compare three runners A, B, and C. A can give different starts to B and C, and we must deduce the start that B alone can give C. This problem is a good test of relative speed, ratio handling, and careful interpretation of what a head start means in distance races. Given Data / Assumptions: - Race length is 1 kilometre, that is 1000 m. - A can give B a start of 100 m in a kilometre race. - A can give C a start of 200 m in a kilometre race. - Head start means that when A runs the full 1000 m, the other runner has covered only 900 m or 800 m respectively. - Speeds of A, B, and C are constant in all races. Concept / Approach: If A can give B 100 m start in 1000 m, then in the time A runs 1000 m, B runs 900 m. This directly gives vB relative to vA. Similarly, C runs 800 m while A runs 1000 m, giving vC. From these, we obtain vB : vC. To find the start B can give C in a kilometre race, we allow B to run the full 1000 m, compute how far C runs in the same time, and take the difference as the start. Step-by-Step Solution: Step 1: From A versus B, vB / vA = 900 / 1000 = 9 / 10. Step 2: From A versus C, vC / vA = 800 / 1000 = 4 / 5. Step 3: Therefore vB : vC = (9 / 10) : (4 / 5) = (9 / 10) * (5 / 4) = 45 : 40 = 9 : 8. Step 4: Let vB = 9k and vC = 8k. Step 5: Time for B to run 1000 m is TB = 1000 / (9k). Step 6: In time TB, C runs distance = vC * TB = 8k * 1000 / (9k) = 8000 / 9 m. Step 7: Numerical value of 8000 / 9 is about 888.89 m. Step 8: For both to finish together in a 1000 m race, C must start ahead by d = 1000 - 8000 / 9 = 1000 / 9 ≈ 111.11 m. Verification / Alternative check: We can think in terms of total time. When B starts from 0 and C starts from d, C needs to run only 1000 - d metres. If d = 1000 / 9, then C runs 8000 / 9 metres, exactly what we computed as C's distance in B's finishing time. Thus both finish together. The approximate decimal value is about 111.11 m, which corresponds to the listed option of 111.12 m when rounded to two decimal places. Why Other Options Are Wrong: - 110.12 m, 112.12 m, 113.12 m: These are nearby decimal values but do not match the exact fraction 1000 / 9. Only 111.12 m correctly reflects the rounding of 111.11 repeated to two decimal places. Common Pitfalls: A frequent mistake is to subtract the starts 200 m minus 100 m and claim that B can give C a start of 100 m. This ignores the fact that starts represent performance over the same total distance and must be translated into speed ratios first. Another error is to think that A's lead over C is constant in all race setups, which is not true if race lengths change and different runners are being compared directly. Final Answer: In a kilometre race, B can give C a start of approximately 111.12 metres.

Question 4

In a 300 m race, runner A beats B by 22.5 m or by 6 seconds. What is the total time taken by B to complete the 300 m course?

Accepted Answer

Introduction / Context: This question combines a distance lead and a time lead to find the actual speed and total time for one runner. It is a typical aptitude question that links distance, speed, and time in a race where one runner finishes earlier and also covers a longer distance. Understanding how to convert the given 22.5 m and 6 seconds into the speed of the slower runner is the key. Given Data / Assumptions: - Race length is 300 m. - A beats B by 22.5 m. - A also beats B by 6 seconds. - Speeds of both runners are constant over the entire race. - We must find B's time to run the whole 300 m. Concept / Approach: At the moment when A finishes the race, B is still 22.5 m behind the finish line. The 6 second time difference tells us that B needs 6 more seconds to run those remaining 22.5 m. This directly gives B's speed as distance divided by time. Once we know B's speed, we can compute the time B takes to cover the full 300 m. We do not need A's speed explicitly. Step-by-Step Solution: Step 1: When A finishes 300 m, B has covered 300 - 22.5 = 277.5 m. Step 2: B still needs to run 22.5 m to finish. Step 3: The time B takes to run these 22.5 m is given as 6 seconds. Step 4: Therefore speed of B = distance / time = 22.5 / 6 m per second. Step 5: Simplify that value: 22.5 / 6 = 3.75 m per second. Step 6: Now time taken by B to run the total 300 m = 300 / 3.75 seconds. Step 7: Calculate this quotient: 300 / 3.75 = 80 seconds. Verification / Alternative check: If B runs at 3.75 m per second, then in 80 seconds B covers 3.75 * 80 = 300 m, which matches the race length. In 74 seconds B would cover 277.5 m, leaving 22.5 m to be run in the remaining 6 seconds. Thus, B is exactly 22.5 m behind after A finishes, and the time gap of 6 seconds is maintained, so the numbers are internally consistent. Why Other Options Are Wrong: - 86 sec and 76 sec: These times do not yield both a 22.5 m distance gap and a 6 second time gap when combined with any consistent speed. - None of these: This would be correct only if no option matched the consistent calculation, but 80 seconds clearly does. Common Pitfalls: Students sometimes assume that the 6 second difference is the total time of B to run 22.5 m plus some unknown time, and they overcomplicate the equation. Others try to derive both speeds first, which is not necessary. The clean approach is to use the final 22.5 m and 6 seconds directly to find B's speed, and then scale that speed up to the full 300 m distance. Final Answer: B takes 80 seconds to complete the 300 m race.

Question 5

In a game played to 100 points, player A can give B a start of 20 points and can give C a start of 28 points. How many points can B give C?

Accepted Answer

Introduction / Context: This problem repeats the same game setup as an earlier question but asks again for the handicap B can give C. It reinforces the idea that scoring abilities can be compared through ratios derived from how many points each player scores when another reaches a fixed target score. Given Data / Assumptions: - Target score in the game is 100 points. - When A scores 100, B scores 80 points. - When A scores 100, C scores 72 points. - Scoring rates are constant and proportional to these scores. - We must find how many points B can give C in a game to 100 points. Concept / Approach: We interpret these statements as ratios of scoring rates. A to B has ratio 100 : 80, and A to C has 100 : 72. Using these, we find the ratio B : C and then deduce how many points C will score while B scores 100. The difference between 100 and that score is the handicap B can give C. Step-by-Step Solution: Step 1: From A and B, vA : vB = 100 : 80 = 5 : 4. Step 2: From A and C, vA : vC = 100 : 72 = 25 : 18. Step 3: Express vB = (4/5) * vA and vC = (18/25) * vA. Step 4: Ratio vB : vC = (4/5) : (18/25) = (4/5) * (25/18) = 100 / 90 = 10 / 9. Step 5: If B scores 100 points, C scores (9/10) * 100 = 90 points in the same time. Step 6: Therefore B can give C a start of 10 points in a 100 point game. Verification / Alternative check: Take vA = 50 units. Then vB = 40 units and vC = 36 units. Time for B to reach 100 points is 100 / 40 = 2.5 units. In that time C scores 36 * 2.5 = 90 points. So if B starts from 0 and C starts from 10, both reach 100 together, which confirms that 10 points is the correct handicap. Why Other Options Are Wrong: - 8 points and 14 points: These do not match the exact 10 : 9 ratio and lead to mismatched finishing scores. - 40 points: This implies C is extremely weak, which contradicts the given ratios that show C is only slightly weaker than B. Common Pitfalls: Many learners skip the ratio step and manipulate raw point differences directly. Another mistake is reversing who gives whom points. Always start from the meaning of a handicap, convert it into a rate ratio, and then compute the handicap between the other pair using these derived ratios. Final Answer: B can give C a start of 10 points in a 100 point game.

Question 6

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

Accepted Answer

Introduction / Context: This question is a repeat of the earlier race setup with A, B, and C, emphasising again how head starts and relative speeds are related. Understanding this structure is useful because very similar problems appear in many exams with only slight number changes. Given Data / Assumptions: - Standard race length is 100 m. - A can give B 10 m start, so A runs 100 m while B runs 90 m in the same time. - A can give C 28 m start, so A runs 100 m while C runs 72 m in the same time. - Speeds remain constant for each runner. - We must find B's possible start to C in a 100 m race. Concept / Approach: A giving different starts to B and C lets us calculate the speed ratio of B and C. Once we know vB : vC, we can imagine B running the whole 100 m and calculate how far C travels in that time. The difference between 100 m and that distance is the head start that B can safely give to C without losing. Step-by-Step Solution: Step 1: From A and B, vB / vA = 90 / 100 = 9 / 10. Step 2: From A and C, vC / vA = 72 / 100 = 18 / 25. Step 3: Therefore vC / vB = (18 / 25) / (9 / 10) = (18 / 25) * (10 / 9) = 4 / 5. Step 4: So vB / vC = 5 / 4. Step 5: Let B run the full 100 m. Time taken is TB = 100 / vB. Step 6: In time TB, C runs distance = vC * TB = 100 * (vC / vB) = 100 * 4 / 5 = 80 m. Step 7: Hence B can give C a start of 100 - 80 = 20 m. Verification / Alternative check: Assume vB = 5 units and vC = 4 units. Then time for B to run 100 m is 100 / 5 = 20 time units. In 20 units, C covers 4 * 20 = 80 m. If C begins 20 m ahead, C only needs 80 m to reach 100 m, so both runners finish together. This confirms that 20 m is the correct start from B to C. Why Other Options Are Wrong: - 18 m and 27 m: These do not correspond to any correct manipulation of the speed ratios and would not make the finishing times equal. - 9 m: Much too small a start, and C would finish behind B, contradicting the equal finishing time requirement with that start. Common Pitfalls: Students often subtract the starts given by A to B and C, obtaining 18 m, without working through the speed ratio. Another confusion comes from mixing which distance corresponds to which runner in the ratio. Always structure the solution around vB / vA and vC / vA, then divide to get vB / vC and only then apply it to the actual race distance. Final Answer: In the 100 m race, B can give C a start of 20 m.

Question 7

In a 200 m race, runner A beats B by 35 m or by 7 seconds. What is the time taken by A to complete the 200 m course?

Accepted Answer

Introduction / Context: This is a standard distance and time comparison question. We know that A finishes a 200 m race 35 m ahead of B and 7 seconds earlier. From the time difference and the distance B covers in that extra time, we can find B's speed, then B's total time, and finally A's time over the course. The problem checks comfort with speed = distance / time and basic algebra. Given Data / Assumptions: - Race distance is 200 m. - A beats B by 35 m. - A finishes 7 seconds before B. - All motion is at constant speed for each runner. - We must find A's time to cover 200 m. Concept / Approach: When A finishes 200 m, B has covered only 165 m. B needs to run the remaining 35 m in 7 seconds. That directly gives B's speed. With B's speed known, we can find how long B takes to run 200 m. Since A finishes 7 seconds earlier than B, subtracting 7 from B's time gives A's finishing time. Step-by-Step Solution: Step 1: At the moment A finishes 200 m, B has covered 200 - 35 = 165 m. Step 2: B still needs to run 35 m and takes 7 seconds more to finish. Step 3: Speed of B = distance / time = 35 / 7 = 5 m per second. Step 4: Time taken by B to run the full 200 m = 200 / 5 = 40 seconds. Step 5: A is 7 seconds faster, so time taken by A = 40 - 7 = 33 seconds. Verification / Alternative check: If A takes 33 seconds for 200 m, speed of A is 200 / 33 m per second. B takes 40 seconds, so B's speed is 5 m per second. In 33 seconds B runs 5 * 33 = 165 m, which is 35 m behind the finish line, and to run the extra 35 m at 5 m per second, B needs exactly 7 more seconds. This matches both the given distance gap and time gap, confirming that A's time is 33 seconds. Why Other Options Are Wrong: - 40 sec: This is B's time, not A's. - 47 sec: Larger than B's time and inconsistent with the statement that A finishes earlier. - None of these: Incorrect because 33 seconds is a consistent and valid answer. Common Pitfalls: Students sometimes apply the 7 second difference the wrong way round and add it to B's time instead of subtracting it to find A's time. Another common error is using 35 m divided by 7 seconds to get a speed but then mistakenly treating that speed as A's rather than B's. Carefully tracking which runner each piece of information refers to avoids these issues. Final Answer: A takes 33 seconds to complete the 200 m race.

Question 8

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

Accepted Answer

Introduction / Context: Here we compare three runners using A as a reference. We know how much A beats B and C by in a 100 m race and must then predict the result of a longer race between B and C only. This is another application of relative speeds, where we first find B and C in terms of A and then obtain a direct comparison of B and C. Given Data / Assumptions: - A beats B by 10 m in a 100 m race, so at A's finish, B has run 90 m. - A beats C by 13 m in a 100 m race, so at A's finish, C has run 87 m. - Speeds of A, B, and C are constant and do not change across races. - We must find B's lead over C in a 180 m race. Concept / Approach: From the 100 m race, we find the speed ratios: vB / vA and vC / vA. Dividing these gives vB / vC directly. Knowing the ratio vB : vC lets us compute how far C runs in the time that B covers 180 m. The difference between 180 m and that distance is B's lead over C. Step-by-Step Solution: Step 1: From A vs B, vB / vA = 90 / 100 = 9 / 10. Step 2: From A vs C, vC / vA = 87 / 100. Step 3: Therefore vB / vC = (vB / vA) / (vC / vA) = (9 / 10) / (87 / 100) = (9 / 10) * (100 / 87) = 90 / 87 = 30 / 29. Step 4: So vC / vB = 29 / 30. Step 5: Let B run 180 m in time TB = 180 / vB. Step 6: In time TB, C covers distance = vC * TB = 180 * (vC / vB) = 180 * 29 / 30 = 174 m. Step 7: Lead of B over C = 180 - 174 = 6 m. Verification / Alternative check: Assume vB = 30 units and vC = 29 units. Time for B to cover 180 m is 180 / 30 = 6 units. In that time C covers 29 * 6 = 174 m. The difference is again 6 m, matching our earlier computation. Also, if we scale these speeds to a 100 m race with A, the original leads can be reproduced, which supports the internal consistency of the ratios used. Why Other Options Are Wrong: - 5.4 m and 4.5 m: These numbers usually arise from approximate or incorrect use of decimal ratios instead of exact fractions. - 5 m: A rounded guess close to the correct answer but not supported by the exact ratio 30 : 29. Common Pitfalls: The main trap is to treat the 10 m and 13 m as independent gaps and try to scale them directly to 180 m, instead of working through speed ratios. Another mistake is to subtract 10 and 13 in the wrong direction, leading to negative or meaningless values. Always convert the leads into ratios before scaling to a different race distance. Final Answer: In a 180 m race, B will beat C by 6 m.

Question 9

Runner A can run 22.5 m in the same time that runner B runs 25 m. In a 1 km race, by how many metres will B beat A?

Accepted Answer

Introduction / Context: This question directly compares the speeds of two runners A and B by giving the distances each covers in the same time. From this we can form a speed ratio and then predict the outcome of a longer race. Such questions are simple but powerful tests of ratio handling and basic proportional reasoning in aptitude exams. Given Data / Assumptions: - In some equal time interval, A runs 22.5 m while B runs 25 m. - Race length is 1 kilometre, that is 1000 m. - Both runners maintain constant speeds during the race. - We must find B's lead over A at the finish line. Concept / Approach: If distances covered in the same time are 22.5 m and 25 m, then the ratio of speeds vA : vB is 22.5 : 25. From that, we can find how far A runs in the time B takes to run 1000 m. The difference in distances at that time gives B's winning margin in metres. Step-by-Step Solution: Step 1: vA : vB = 22.5 : 25. Step 2: Divide numerator and denominator by 2.5 to simplify: 22.5 / 2.5 = 9 and 25 / 2.5 = 10. So vA : vB = 9 : 10. Step 3: Express vA = 9k and vB = 10k for some constant k. Step 4: Time for B to run 1000 m is TB = 1000 / (10k). Step 5: In time TB, A covers distance = vA * TB = 9k * 1000 / (10k) = 900 m. Step 6: Therefore, when B finishes 1000 m, A is at 900 m, so B beats A by 1000 - 900 = 100 m. Verification / Alternative check: We can normalise time so that the common interval is 1 unit. Then A's speed is 22.5 m per unit time and B's speed is 25 m per unit time. To find A's distance when B has run 1000 m, scale by the ratio 22.5 / 25 = 0.9. Then A's distance is 0.9 * 1000 = 900 m, leaving a gap of 100 m, which agrees with the previous calculation and confirms the result. Why Other Options Are Wrong: - 111 m and 50 m: These margins do not correspond to the 9 : 10 speed ratio and would imply different relative speeds. - 25 m: Much too small, and it would only match a ratio closer to 25 : 26 instead of 9 : 10. Common Pitfalls: An error seen often is to subtract 22.5 from 25 and treat the 2.5 m difference as a constant gap over the whole race, then scale it incorrectly. Another issue is forgetting to convert the 1 km race into 1000 m. The safe approach is always to use ratios of speeds and then apply them to the exact race distance mentioned. Final Answer: In the 1 km race, B will beat A by 100 m.

Question 10

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

Accepted Answer

Introduction / Context: This question is similar to the earlier point game problems but now includes a different target score. A can give handicaps to two other players in a 60 point game, and we must find the handicap one of those players can give to the other in a 90 point game. It checks the ability to use ratios of scoring rates and to scale them to a different game length. Given Data / Assumptions: - In a game to 60 points, A can give B 15 points. - In the same game type, A can give C 20 points. - This means that when A scores 60 points, B has scored only 45 and C has scored only 40. - Scores are proportional to scoring rates, which are constant for each player. - We must find how many points B can give C in a 90 point game. Concept / Approach: The ratio of scoring rates is derived from the scores at the moment A reaches 60 points. A to B is 60 : 45, and A to C is 60 : 40. We then compute B : C and use that ratio in a 90 point game where B plays to 90 points. By seeing how many points C scores in that time, we can work out the handicap B can offer to C so that both finish together at 90 points. Step-by-Step Solution: Step 1: From A and B, vA : vB = 60 : 45 = 4 : 3. Step 2: From A and C, vA : vC = 60 : 40 = 3 : 2. Step 3: Write vB = (3/4) * vA and vC = (2/3) * vA. Step 4: Ratio vB : vC = (3/4) : (2/3) = (3/4) * (3/2) = 9 : 8. Step 5: Thus vB = 9k and vC = 8k for some constant k. Step 6: In a 90 point game, time for B to score 90 points is TB = 90 / (9k) = 10 / k. Step 7: In this time, C scores points = vC * TB = 8k * (10 / k) = 80 points. Step 8: Therefore B can give C a start of 90 - 80 = 10 points in a 90 point game. Verification / Alternative check: Assume vB = 9 and vC = 8. When B scores 90, time taken is 90 / 9 = 10 units. In 10 units C scores 8 * 10 = 80. If C starts at 10 points, C also reaches 90 at the same time. This shows that a 10 point handicap from B to C in a 90 point game is exactly correct. Why Other Options Are Wrong: - 30 points and 20 points: These are too large and would give C an unrealistic advantage, causing C to finish well before B. - 12 points: This does not match the 9 : 8 scoring rate ratio and would lead to mismatched final scores. Common Pitfalls: A common error is to scale the handicaps directly from 60 to 90 by simple proportion, for example taking 15 or 20 and scaling them by 1.5. This approach ignores the one step where we must first convert the handicaps into scoring rate ratios. Another mistake is to forget that the target score has changed and still work with 60 instead of 90 when computing final points. Final Answer: In a 90 point game, B can give C a start of 10 points.

Question 11

In a 100 m race, runner A covers the distance in 36 seconds and runner B covers it in 45 seconds. In this race, by how many metres does A beat B?

Accepted Answer

Introduction / Context: This is a straightforward race question where we know the times taken by two runners to cover the same distance. We must find by how much distance the faster runner leads when he finishes the race. Such questions reinforce the idea that when distance is fixed, faster speed means less time, and we can work back to distance covered by the slower runner in the same time. Given Data / Assumptions: - Race distance is 100 m. - A takes 36 seconds to run 100 m. - B takes 45 seconds to run 100 m. - Each runner runs at constant speed. - We must find A's lead over B at the moment A finishes. Concept / Approach: We find the speed of B by using distance over time, then compute how far B runs in the 36 seconds that A needs to complete his 100 m. The difference between 100 m and this distance is A's winning margin. There is no need to compute A's speed directly, though we could do so as a check. Step-by-Step Solution: Step 1: Speed of B = distance / time = 100 / 45 m per second. Step 2: Time taken by A to finish the race is 36 seconds. Step 3: In 36 seconds, distance covered by B = (100 / 45) * 36. Step 4: Simplify: 36 / 45 = 4 / 5, so distance by B = 100 * 4 / 5 = 80 m. Step 5: Lead of A over B = 100 - 80 = 20 m. Verification / Alternative check: Compute A's speed also for confirmation. A's speed is 100 / 36 m per second. Time taken to cover 80 m at B's speed 100 / 45 is 80 * 45 / 100 = 36 seconds. That means that when A finishes in 36 seconds, B has indeed run 80 m, so the 20 m lead is consistent from both perspectives. Why Other Options Are Wrong: - 25 m and 22.5 m: These correspond to incorrect manipulation of the ratio 36 : 45, and do not match the exact 4 : 5 simplification. - 9 m: This is far too small and could arise from confusing time difference with distance difference. Common Pitfalls: Some learners directly subtract the times 45 and 36 and then try to convert that into a distance for A, not B, which creates confusion. Another mistake is to assume that the difference in times translates linearly into difference in distances at the same speed. Always compute the slower runner's distance in the faster runner's finishing time by using the slower runner's speed explicitly. Final Answer: In the 100 m race, A beats B by 20 m.

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