Question 1

A and B run a 5 km race on a 400 m circular course. Their speeds are in the ratio 5 : 4. How many times does the winner pass the other during the race?

Accepted Answer

Introduction / Context: Laps gained on a circular track equal the relative distance advantage divided by the track length. The number of passes equals the count of complete 400 m gains; the fractional part indicates a partial, not complete, pass near the end. Given Data / Assumptions: Total winner distance = 5,000 m. Speed ratio 5:4 ⇒ loser’s distance at winner’s finish = 5,000 * (4/5) = 4,000 m. Track length = 400 m. Concept / Approach: Relative lead at finish = 5,000 − 4,000 = 1,000 m. Full passes = 1,000 / 400 = 2.5 times (two complete passes and half a lap extra). Step-by-Step Solution: Passes = 2.5 ⇒ interpret as “2 1/2 times”. Verification / Alternative check: Laps by winner = 5,000/400 = 12.5; loser laps = 4,000/400 = 10; difference = 2.5 passes. Why Other Options Are Wrong: Other fractional counts do not match the 1,000 m relative lead. Common Pitfalls: Rounding 2.5 down to 2; the question allows fractional counts of passes. Final Answer: 2 1/2 times

Question 2

A and B run a 1 km race. Case 1: A gives B a start of 50 m and still wins by 14 s. Case 2: A gives B a start of 22 s and then B wins by 20 m. Find the time taken by A to run 1 km.

Accepted Answer

Introduction / Context: This blends two handicap conditions (distance head-start and time head-start) to solve the underlying speeds. With vA and vB extracted from the pair of equations, compute A’s 1 km time as 1,000/vA. Given Data / Assumptions: Scenario 1: A runs 1,000 m; B runs 950 m; B’s time − A’s time = 14 s. Scenario 2: A starts 22 s late; when B finishes 1,000 m, A is 20 m short (at 980 m). Concept / Approach: Let vA = a, vB = b (m/s). From Case 1: 950/b − 1000/a = 14. From Case 2: a*(1000/b − 22) = 980. Solve simultaneously to obtain a, then tA = 1000/a. Step-by-Step Solution: From Case 2: (1000a)/b − 22a = 980.From Case 1: 950/b = 14 + 1000/a.Eliminate 1/b to get a = 10 m/s ⇒ tA = 1000/10 = 100 s. Verification / Alternative check: Check Case 1: tA = 100 s; tB = 950/b. Using a = 10 in the relations reproduces the 14 s gap. Check Case 2: A’s distance at B’s finish equals 980 m with a 22 s delay. Why Other Options Are Wrong: Other times mismatch one or both scenarios when substituted back. Common Pitfalls: Treating both handicaps as distances or both as times; keep them distinct and set equations carefully. Final Answer: 100 sec

Question 3

Two runners start from the same point on a 4,200 m circular stadium and run in opposite directions at 500 m/min and 700 m/min. After how many minutes will they meet each other?

Accepted Answer

Introduction / Context: When two runners move towards each other on a circular track from the same point in opposite directions, they meet when their combined distance equals one full circumference. Use relative speed (sum) to find the meeting time. Given Data / Assumptions: Circumference = 4,200 m. Speeds: 500 m/min and 700 m/min. Concept / Approach: Relative speed (opposite directions) = 500 + 700 = 1,200 m/min. Meeting time = distance / relative speed = 4,200 / 1,200. Step-by-Step Solution: t = 4,200 / 1,200 = 3.5 minutes. Verification / Alternative check: Positions after 3.5 min: 500×3.5 = 1,750 m; 700×3.5 = 2,450 m; total = 4,200 m ⇒ meet. Why Other Options Are Wrong: Other times misapply relative speed or circumference. Common Pitfalls: Using difference of speeds (appropriate for same direction), not the sum. Final Answer: 3.5 minutes

Question 4

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

Introduction / Context: Point handicaps translate to proportional scoring strengths. “A gives B 20 in 100” means when A scores 100, B scores 80. We can find relative strengths and then deduce a fair handicap between B and C for the same target score. Given Data / Assumptions: A:B = 100:80 = 5:4. A:C = 100:72 = 25:18 (or 5:3.6). Concept / Approach: Normalize with A as common reference. From A:B = 5:4 and A:C = 25:18 ⇒ set A = 25 units ⇒ B = 20 units; C = 18 units. Then B:C = 20:18 = 10:9. Step-by-Step Solution: When B scores 100, C would score 100*(9/10) = 90.Thus, B can give C = 100 − 90 = 10 points. Verification / Alternative check: Scaling by any common factor preserves the 10:9 ratio. Why Other Options Are Wrong: 15 or 85/90 misrepresent the proportional difference. Common Pitfalls: Adding/subtracting 20 and 28 directly rather than converting to ratios first. Final Answer: 10

Question 5

In a 90-point game, A can give B 15 points and C 30 points. How many points can B give C in a 100-point game?

Accepted Answer

Introduction / Context: Again we convert handicaps to relative strengths and then translate to a new target total. “A gives B 15 in 90” ⇒ when A scores 90, B scores 75. “A gives C 30 in 90” ⇒ when A scores 90, C scores 60. Given Data / Assumptions: A:B:C = 90:75:60 = 6:5:4. Concept / Approach: Thus B:C = 5:4. In a game to 100 for B, C’s equivalent score would be 100*(4/5) = 80, so B can give C 20 points. Step-by-Step Solution: Fair mapping to 100: C = 80 when B = 100.Handicap B gives C = 20. Verification / Alternative check: Proportional scores (5:4) hold consistently. Why Other Options Are Wrong: 30 or 50 violate the 5:4 proportionality; 140 is nonsensical for a 100-point target. Common Pitfalls: Mixing the 90-point base with the 100-point question without re-scaling. Final Answer: 20

Question 6

The speed ratio of a car, a train, and a bus is 5 : 9 : 4. If their average speed taken together is 72 km/h, what is the average speed of the car and the train together?

Accepted Answer

Introduction / Context: Convert the ratio to concrete speeds using the given combined average. Then compute the simple average for the subset (car and train). Given Data / Assumptions: Car:Train:Bus = 5:9:4 ⇒ let speeds be 5k, 9k, 4k. Average of all three = (5k + 9k + 4k)/3 = 72 ⇒ 6k = 72. Concept / Approach: Solve k, then compute car and train speeds and take their mean. Step-by-Step Solution: 6k = 72 ⇒ k = 12.Car = 60 km/h, Train = 108 km/h.Average (car & train) = (60 + 108)/2 = 84 km/h. Verification / Alternative check: Bus would be 48 km/h; the mean of all three is (60 + 108 + 48)/3 = 72—consistent. Why Other Options Are Wrong: 78 or 82 km/h do not follow from k = 12; “Cannot be determined” is false because the ratio plus average is sufficient. Common Pitfalls: Averaging the ratio numbers directly without solving for k. Final Answer: 84 km/h

Question 7

In a 100 metre race, runner A beats runner B by 10 metres and beats runner C by 13 metres. If they now run a 180 metre race at the same constant speeds, by how many metres will B beat C?

Accepted Answer

Introduction / Context: This is a classic races and games question involving relative speeds of different runners. You know how much A is ahead of B and C in a 100 metre race. From this, you can find the ratio of their speeds. The question then asks you to use these speed ratios to determine by how much B will beat C in a longer race of 180 metres. It tests your understanding of proportional reasoning and constant speed motion. Given Data / Assumptions: In a 100 m race, A finishes first. When A has completed 100 m, B has completed 90 m (A beats B by 10 m). When A has completed 100 m, C has completed 87 m (A beats C by 13 m). All runners maintain constant speeds across races. We consider a new race of 180 m and must find by how many metres B beats C. Concept / Approach: From the first race: Speeds are proportional to distances covered in the same time. So v_A : v_B : v_C = 100 : 90 : 87. To find how much B beats C in a 180 m race, we: Consider the time when B has just finished 180 m. Compute how far C runs in that same time, using the speed ratio between B and C. The difference between 180 m and C's distance is the margin by which B beats C. Step-by-Step Solution: Step 1: From the first race, v_A : v_B : v_C = 100 : 90 : 87. Step 2: Focus on B and C. Their speed ratio v_B : v_C = 90 : 87 = 30 : 29. Step 3: In the new 180 m race, we consider the moment B finishes 180 m. Step 4: Let the time taken by B to run 180 m be T seconds. Step 5: Since distance = speed * time, we have 180 = v_B * T. Step 6: In time T, C covers distance = v_C * T. Step 7: Using ratio v_B / v_C = 30 / 29, we get v_C = (29 / 30) * v_B. Step 8: So distance run by C in time T = v_C * T = (29 / 30) * v_B * T = (29 / 30) * 180 = 174 m. Step 9: Therefore, when B has finished 180 m, C has run 174 m. Step 10: B beats C by 180 - 174 = 6 m. Verification / Alternative check: You can also think directly in terms of scaling: When A runs 100 m, B runs 90 m and C runs 87 m. Ratio of B's speed to C's speed = 90 : 87 = 30 : 29. If B runs 30 units of distance, C runs 29 units in the same time. Scale this up so that B runs 180 m: 30 units correspond to 180 m, so 1 unit = 6 m, and 29 units = 174 m. Again, B beats C by 6 m. Both methods agree perfectly and show the margin is 6 metres. Why Other Options Are Wrong: 5.4m, 4.5m, and 5m all come from incorrect ratios or from mixing up who beats whom. None of them arises when you correctly use the speed ratio 30 : 29 and scale distances to 180 m. Only 6m exactly matches the difference between 180 m and 174 m in the new race. Common Pitfalls: Learners often try to directly proportion the given 10 m and 13 m advantages to the new distance without first finding speed ratios, which leads to incorrect results. Another common error is to think the ratio is 100 : 90 : 87 and then scale everything incorrectly without isolating B and C. Always convert the scenario to speed ratios and then compute distances for the new race carefully. Final Answer: In the 180 metre race, runner B will beat runner C by 6 metres.

Question 8

In a 1 kilometre race, runner A beats runner B by 28 metres, and B takes 7 seconds to cover this remaining 28 metres. What is A's total time to complete the 1 km race?

Accepted Answer

Introduction / Context: This problem involves a race where one runner finishes ahead of another by a certain distance, and you know how long the slower runner takes to cover that remaining distance. From this information, you can determine the slower runner's speed and then find the faster runner's time for the entire race. The question tests your understanding of speed, distance, and time relationships and how to interpret race conditions carefully. Given Data / Assumptions: Total race distance = 1 km = 1000 metres. A beats B by 28 metres. This means when A finishes 1000 m, B has covered 972 m. B takes 7 seconds to cover the remaining 28 m. Speeds are assumed constant throughout the race. We are asked for A's time to complete 1000 m. Concept / Approach: Key ideas: First find B's speed from the information about the last 28 metres. Use B's speed to find the time B would take to run the full 1000 metres. Use the ratio of distances at the instant A finishes to relate A's and B's times. At the moment A finishes, B has covered only 972 m, so B's time to run 972 m equals A's time to run 1000 m. Step-by-Step Solution: Step 1: B covers 28 m in 7 seconds. Step 2: So B's speed v_B = distance / time = 28 / 7 = 4 m/s. Step 3: Time taken by B to run the entire 1000 m is T_B_total = 1000 / 4 = 250 seconds. Step 4: At the instant A finishes the race, B has covered only 972 m. Step 5: Time taken by B to cover 972 m is T_B_972 = 972 / 4 = 243 seconds. Step 6: This time T_B_972 is equal to A's time to cover 1000 m, since they started together and at that instant A finishes. Step 7: Therefore, A's time T_A = 243 seconds. Step 8: Convert 243 seconds to minutes and seconds: 243 / 60 = 4 minutes with a remainder of 3 seconds. Step 9: So A's time is 4 minutes 3 seconds. Verification / Alternative check: We can confirm consistency: A covers 1000 m in 243 seconds, so A's speed v_A = 1000 / 243 ≈ 4.115 m/s. B's speed is 4 m/s. In 243 seconds, B runs distance = v_B * time = 4 * 243 = 972 m. Difference between A and B at that instant = 1000 - 972 = 28 m. This matches the given condition that A beats B by 28 m, confirming that the computed time is correct. Why Other Options Are Wrong: 5min,4sec, 2min,3sec, and 3min,4sec do not satisfy the relationship between speeds and distances. If you plug these times into speed calculations, B's position at A's finish does not differ by exactly 28 m, so these times cannot be correct. Only 4min,3sec yields the correct 28 m lead at the finish line. Common Pitfalls: Some students misinterpret "A beats B by 28 metres in 7 seconds" as meaning A takes 7 seconds to run 28 metres more, which is not what the question says. The phrase actually means B needs 7 more seconds to cover the 28 metre gap after A has already finished. Another common error is to mis-handle the conversion between metres and kilometres or between seconds and minutes. Always convert units properly and carefully interpret who reaches which point when. Final Answer: Runner A completes the 1 kilometre race in 4 minutes 3 seconds.

Question 9

Runner A can cover 22.5 metres in the same time that runner B covers 25 metres. In a 1 kilometre race, by how many metres will B beat A?

Accepted Answer

Introduction / Context: This question is a straightforward application of speed ratios in a race. You are told that in the same time, A runs 22.5 m while B runs 25 m. From this, you can find the ratio of their speeds. The race distance is 1 kilometre, and you must determine by how much B beats A at the finish line. Such questions are important for building intuition about relative speed and proportional reasoning. Given Data / Assumptions: In equal time intervals, A covers 22.5 m and B covers 25 m. Race distance = 1 km = 1000 m. Both runners maintain constant speeds. We need the distance by which B beats A at the end of the 1000 m race. Concept / Approach: Speeds are proportional to distances covered in the same time: v_A : v_B = 22.5 : 25. Simplify this to a simple integer ratio. When B runs the full 1000 m, compute how much distance A covers in the same time using the speed ratio. The difference between 1000 m and A's distance is the winning margin of B. Step-by-Step Solution: Step 1: v_A : v_B = 22.5 : 25. Step 2: Write 22.5 as 45/2 and 25 as 25. Step 3: Ratio v_A : v_B = (45 / 2) : 25 = 45 : 50. Step 4: Simplify 45 : 50 by dividing both by 5 to get 9 : 10. Step 5: So v_A : v_B = 9 : 10. Step 6: Let B cover 10 speed units while A covers 9 units in the same time. Step 7: In the actual race, B runs 1000 m. This corresponds to 10 units of distance. Step 8: So 1 unit = 1000 / 10 = 100 m. Step 9: A's distance in that time = 9 units = 9 * 100 = 900 m. Step 10: Therefore, B beats A by 1000 - 900 = 100 m. Verification / Alternative check: We can verify with explicit speeds: Assume a unit time t where A covers 22.5 m and B covers 25 m. Speed v_A = 22.5 / t, v_B = 25 / t. To run 1000 m, B's time = 1000 / v_B = 1000t / 25 = 40t. In 40t, A covers distance = v_A * 40t = (22.5 / t) * 40t = 900 m. Difference = 1000 - 900 = 100 m. This consistency check confirms that B is indeed 100 metres ahead when B finishes the 1 km race. Why Other Options Are Wrong: 75 m, 25 m, and 50 m occur if one mis-scales the ratio 9 : 10 or forgets to map 10 units to 1000 m correctly. Any smaller difference would contradict the ratio v_B / v_A = 10 / 9, which predicts that A covers only 900 m when B completes 1000 m. Thus, only 100 m matches the correct calculation. Common Pitfalls: The most common error is failing to simplify the ratio correctly and making arithmetic mistakes when scaling it up to the actual distance. Some students also mistakenly average the distances 22.5 and 25 or think the difference 2.5 m directly translates to 1000 m without using ratios. Always express the relationship in terms of speed ratio and then scale it carefully to the given race distance. Final Answer: In the 1 kilometre race, runner B will beat runner A by 100 metres.

Question 10

A and B run a 100 metre race. A runs at a constant speed of 5 km/h and gives B a head start of 8 metres, yet still finishes 8 seconds before B. What is B's speed in km/h?

Accepted Answer

Introduction / Context: This is a relative speed problem in a race where one runner gives the other a head start and still wins. A has a known constant speed and gives B an 8 metre start in a 100 metre race. A still finishes 8 seconds before B. From this, you must determine B's speed. The question tests your ability to convert units, calculate time from speed and distance, and interpret race time differences correctly. Given Data / Assumptions: Race distance for A = 100 m. B gets a start of 8 m, so B runs only 92 m. A's speed = 5 km/h. A finishes 8 seconds earlier than B. Speeds are constant for both runners. We need B's speed in km/h. Concept / Approach: We use: Convert A's speed from km/h to m/s. Compute A's time to run 100 m. Since A finishes 8 seconds before B, B's time is A's time plus 8 seconds. B's speed = distance / time = 92 m / B's time in seconds. Convert B's speed back to km/h. Careful unit conversion is critical for accuracy. Step-by-Step Solution: Step 1: Convert A's speed from km/h to m/s. Step 2: 5 km/h = 5 * 1000 / 3600 m/s = 5000 / 3600 m/s ≈ 1.3889 m/s. Step 3: Time taken by A to run 100 m: T_A = distance / speed = 100 / (5000 / 3600) = 100 * 3600 / 5000 = 72 seconds. Step 4: A finishes 8 seconds earlier than B, so B's time T_B = T_A + 8 = 72 + 8 = 80 seconds. Step 5: B's distance = 92 m (because of 8 m head start). Step 6: B's speed in m/s: v_B = 92 / 80 = 1.15 m/s. Step 7: Convert v_B to km/h: v_B_kmh = 1.15 * 3600 / 1000 = 1.15 * 3.6 = 4.14 km/h. Step 8: Therefore, B's speed is approximately 4.14 km/h. Verification / Alternative check: Check times with computed speeds: A's speed ≈ 1.3889 m/s, so time for 100 m is 100 / 1.3889 ≈ 72 s (as computed). B's speed ≈ 1.15 m/s, so time for 92 m is 92 / 1.15 ≈ 80 s. Difference in finishing times = 80 - 72 = 8 seconds, which matches the given information. This confirms that our conversion and calculations are consistent with the conditions of the race. Why Other Options Are Wrong: 5.15 kmph is greater than A's speed, which would imply B could not finish after A if given a head start. 4.25 kmph and 4.4 kmph do not yield the exact 8 second difference when you compute the times. They produce finishing time gaps different from 8 seconds and therefore do not satisfy the race conditions exactly. Only about 4.14 kmph results in B finishing exactly 8 seconds after A when B runs 92 m and A runs 100 m. Common Pitfalls: Common errors include forgetting to convert km/h to m/s correctly, misinterpreting the head start so that both runners are assumed to run 100 m, or incorrectly adding the 8 seconds. Some students also mistakenly assume that B's distance is 100 - 8 = 92 without linking it to B's total time carefully. Always verify that the time difference derived from speeds and distances matches the given condition. Final Answer: Runner B's speed in the race is approximately 4.14 km/h.

Question 11

In a 100 metre race, A finishes the race in 36 seconds while B finishes it in 45 seconds. By how many metres does A beat B?

Accepted Answer

Introduction / Context: This is a basic races and games problem where the finishing times of two runners for the same distance are known. Because they run at constant speeds, the distances they cover in the same time are proportional to their speeds. The question asks how many metres A is ahead of B when A finishes the race. This tests your ability to use proportional reasoning with speed, distance, and time. Given Data / Assumptions: Race distance = 100 m. A's finishing time = 36 seconds. B's finishing time = 45 seconds. Both run at constant speeds. We need the distance between A and B when A finishes the race. Concept / Approach: We know: B takes 45 seconds to cover 100 m. Therefore, B's speed = 100 / 45 m/s. When A finishes in 36 seconds, we compute how far B has run in those 36 seconds using B's constant speed. The difference between 100 m and B's distance at that time is the lead by which A beats B. Step-by-Step Solution: Step 1: Compute B's speed. Step 2: v_B = 100 m / 45 s = 100 / 45 m/s. Step 3: In 36 seconds, distance covered by B = v_B * 36. Step 4: Distance_B_36 = (100 / 45) * 36 = 100 * 36 / 45. Step 5: Simplify 36 / 45 = 4 / 5, so Distance_B_36 = 100 * 4 / 5 = 80 m. Step 6: When A finishes 100 m, B has covered only 80 m. Step 7: Therefore, A beats B by 100 - 80 = 20 m. Verification / Alternative check: You can also compare speeds: A's speed v_A = 100 / 36 m/s. B's speed v_B = 100 / 45 m/s. The ratio of their speeds v_A : v_B = (100 / 36) : (100 / 45) = 45 : 36 = 5 : 4. This means when A runs 5 units of distance, B runs 4 units in the same time. At the moment A has completed 100 m (5 units), B has completed (4 / 5) * 100 = 80 m, so the gap is 20 m. Both methods give consistent results. Why Other Options Are Wrong: 25m, 22.5m, and 9m arise from incorrect handling of ratios or incorrect simplification. None of these values match the derived relation that B covers 80 m in the time A covers 100 m. Therefore, only 20m correctly captures the distance by which A is ahead. Common Pitfalls: A common mistake is to confuse the time difference (45 - 36 = 9 seconds) with a distance difference, and therefore answer 9m, which is incorrect. Another error is to try to average times or distances without using proper proportional reasoning. Always work either with speeds or with ratios of speeds for a precise solution. Final Answer: Runner A beats runner B by 20 metres in the 100 metre race.

Question 12

In a 500 metre race, the speeds of contestants A and B are in the ratio 3 : 4. A is given a head start of 140 metres. By how many metres does A win the race?

Accepted Answer

Introduction / Context: This problem combines speed ratios with a head start in a race. Two contestants, A and B, run a 500 metre race. Their speed ratio is known, and A is given a 140 metre head start. You need to find by how many metres A wins the race. Such questions assess understanding of relative speeds, distances, and head starts in competitive scenarios. Given Data / Assumptions: Total race distance = 500 m. Speed ratio v_A : v_B = 3 : 4. A is given a start of 140 m. Hence, A needs to run 500 - 140 = 360 m. B runs the full 500 m. Both run at constant speeds. Concept / Approach: We work with speed ratio and times: Let v_A = 3k and v_B = 4k for some k. Time taken by A to complete 360 m: T_A = 360 / (3k). Time taken by B to complete 500 m: T_B = 500 / (4k). Compare T_A and T_B to see who finishes first. Then determine how far the trailing runner has gone when the winner finishes. Step-by-Step Solution: Step 1: Let v_A = 3k and v_B = 4k. Step 2: Distance A runs = 360 m, so T_A = 360 / (3k) = 120 / k seconds. Step 3: Distance B runs = 500 m, so T_B = 500 / (4k) = 125 / k seconds. Step 4: Since 120 / k < 125 / k, A finishes earlier and wins the race. Step 5: At time T_A = 120 / k, distance covered by B = v_B * T_A = 4k * (120 / k) = 480 m. Step 6: At that moment, B is at 480 m while the race distance is 500 m. Step 7: So A wins by 500 - 480 = 20 m. Verification / Alternative check: You can use a scaling method: Let k = 1 for simplicity, so v_A = 3 m/s and v_B = 4 m/s. A's time: T_A = 360 / 3 = 120 s. B's time if he runs 500 m: T_B = 500 / 4 = 125 s. In 120 s, B travels 4 * 120 = 480 m. Difference between race finish line and B's position = 500 - 480 = 20 m. This numerical check yields the same margin of victory and confirms our solution. Why Other Options Are Wrong: 60m and 40m represent larger margins than what the speed ratio and distances allow. If A were this far ahead, the times would not match the given speeds. A margin of 10m is too small; recalculations show B would not be that close when A finishes. Only 20m is consistent with the ratio v_A : v_B = 3 : 4 and the 140 m head start. Common Pitfalls: Some students incorrectly adjust the starting distance rather than the distance A actually runs, or they assume both runners still cover 500 m. Others may average the speeds or distances erroneously. The correct method is to compute time for each runner based on the true distance they cover, then compare how far the slower runner has gone when the faster one finishes. Final Answer: Runner A wins the 500 metre race by 20 metres.

Question 13

In a 200 metre race, A beats B by 35 metres or 7 seconds. What is A's time to complete the 200 metre race?

Accepted Answer

Introduction / Context: This question describes a race result both in terms of distance (35 metres) and time (7 seconds). A beats B by 35 metres or 7 seconds in a 200 metre race. This means that when A finishes, B is 35 metres behind and needs 7 more seconds to finish the race. From this, you can find B's speed and then use it to calculate A's time for the race. Understanding such equivalences is essential in races and games problems. Given Data / Assumptions: Race distance = 200 m. A beats B by 35 m or 7 seconds. When A finishes 200 m, B has covered 200 - 35 = 165 m. B takes 7 seconds to cover the remaining 35 m. Speeds are constant for both runners. We must compute A's time to complete 200 m. Concept / Approach: Key steps: First find B's speed from the information about the last 35 m. Compute B's time to run the entire 200 m. The time A takes is equal to the time B needs to run 165 m, since at that instant A has finished 200 m and B is at 165 m. Using B's speed, find that time and hence A's race time. Step-by-Step Solution: Step 1: B covers 35 m in 7 seconds. Step 2: B's speed v_B = distance / time = 35 / 7 = 5 m/s. Step 3: B's time to cover the full 200 m = 200 / 5 = 40 seconds. Step 4: At the moment A finishes the race, B is only at 165 m (35 m behind). Step 5: Time taken by B to reach 165 m = distance / speed = 165 / 5 = 33 seconds. Step 6: Since both started together, A's time to run 200 m equals B's time to reach 165 m. Step 7: Therefore, A's race time is 33 seconds. Verification / Alternative check: Check with speeds: A's time = 33 s => A's speed v_A = 200 / 33 ≈ 6.06 m/s. B's speed v_B = 5 m/s. In 33 s, B covers distance = 5 * 33 = 165 m. Difference when A finishes = 200 - 165 = 35 m. B then takes additional time to cover 35 m = 35 / 5 = 7 seconds. This matches the statement that A beats B by 35 m or 7 seconds, confirming that the computed time is correct. Why Other Options Are Wrong: 40 sec is B's time, not A's, and gives no lead for A if misapplied. 47sec and "none of these" do not satisfy the condition of 35 m or 7 seconds gap. If you plug them into the speed equations, the resulting differences between A and B do not match the given race outcome. Only 33sec yields the correct combination of distance and time advantage. Common Pitfalls: Many students misinterpret "35 metres or 7 seconds" and incorrectly think the two pieces of information are separate rather than equivalent descriptions of the same race outcome. Another common mistake is to assume that 7 seconds is the time difference between A and B's start, which is not true. It actually represents the additional time B needs to finish after A. Always translate the wording into precise positions and times before doing any calculations. Final Answer: Runner A completes the 200 metre race in 33 seconds.

Question 14

At a game of billiards, A can give B 15 points in a game of 60, and A can give C 20 points in a game of 60. In a game of 90 points, how many points can B give C as a start so that they finish together?

Accepted Answer

Introduction / Context: This is a classic billiards handicap problem involving relative scoring rates. When we say that A can give B 15 points in 60, it means that when A reaches 60 points, B has only 45. Similarly, A can give C 20 points in 60, meaning that when A reaches 60, C has only 40. Using these relationships, you are asked to find how many points B can give C in a longer game of 90 so that they still finish at the same time. The question tests understanding of proportional reasoning and relative rates. Given Data / Assumptions: When A scores 60 points in a game, B has 45 points. When A scores 60 points, C has 40 points. This implies A, B, and C have different scoring rates. We consider a new game where the target is 90 points. We must find how many points B can give C as a start in this 90 point game so that they finish together. Concept / Approach: We interpret points given as a measure of relative scoring speed: Rate of scoring is directly proportional to points scored in a given time. When A scores 60, B scores 45, C scores 40 in the same time. This gives the ratio of their scoring rates: A : B : C = 60 : 45 : 40. Simplify to get base speed ratio, then compare B and C. Use these rates to find the handicap B can give C in a 90 point game. Step-by-Step Solution: Step 1: From given information, when A scores 60, B scores 45, C scores 40 in the same duration. Step 2: So scoring rate ratio A : B : C = 60 : 45 : 40. Step 3: Simplify, dividing by 5: A : B : C = 12 : 9 : 8. Step 4: Focus only on B and C. Their rate ratio B : C = 9 : 8. Step 5: In a game to 90 points, suppose B plays without a start and needs to reach 90 points from zero. Step 6: Let the time for B to reach 90 points be T. Step 7: Then B's rate r_B = 90 / T, and C's rate r_C = (8 / 9) * r_B since B : C = 9 : 8. Step 8: In the same time T, C will score points = r_C * T = (8 / 9) * 90 = 80 points. Step 9: So when B reaches 90 points, C would naturally reach 80 points if both started from zero at the same time. Step 10: To make them finish together in a 90 point game, B must give C a start equal to this difference: 90 - 80 = 10 points. Verification / Alternative check: You can verify by simulating the game: Let B start from 0 and C start from 10 points. B needs to score 90 points; C needs to go from 10 to 90, so C must score 80 points. Time for B to score 90 at rate r_B is T. At rate r_C = (8 / 9) * r_B, in time T, C scores (8 / 9) * 90 = 80 points. So C goes from 10 to 10 + 80 = 90 points exactly when B reaches 90. Therefore, with a 10 point start, both B and C finish the 90 point game together. Why Other Options Are Wrong: 30 points or 20 points would give C too large a head start, making C finish earlier than B. 12 points is slightly larger than needed and would again let C win early. The detailed rate calculations show that the exact difference in natural scores by time T is 10 points, so any larger or smaller handicap fails to produce a simultaneous finish. Common Pitfalls: Some learners misread "give 15 points in 60" as 15% or confuse points with percentages. Others try to subtract 15 and 20 directly or average them to find a start, which does not capture the underlying rate relationship. Always interpret such statements as "when A scores 60, B (or C) scores fewer points in the same time," then use those numbers to derive a proper rate ratio before analyzing the new game. Final Answer: In a 90 point game, B can give C a start of 10 points so that they finish together.

Question 15

In an athletics race of 1000 m, runner A can beat runner B by 100 m, and in a race of 800 m, runner B can beat runner C by 100 m. By how many metres will A beat C in a race of 600 m?

Accepted Answer

Introduction / Context: This question tests relative speed in running races and how results from different race lengths can be combined. Instead of working with time directly, we compare how far each runner covers when another runner finishes the race. This type of aptitude problem is very common in competitive exams because it checks comfort with proportional reasoning and careful handling of ratios. Given Data / Assumptions: - In a 1000 m race, A finishes when B has covered only 900 m. - In an 800 m race, B finishes when C has covered only 700 m. - All runners keep constant speeds in every race. - We must find how many metres A will be ahead of C in a 600 m race. Concept / Approach: The key concept is that speed is proportional to distance covered in the same time. From race results we can find ratios of speeds of A, B, and C. Once we know the ratio of speeds, we can calculate how far C runs while A completes 600 m, and the difference between 600 m and that distance gives the lead of A over C. Step-by-Step Solution: Step 1: From the first race, A runs 1000 m while B runs 900 m in the same time, so vA : vB = 1000 : 900 = 10 : 9. Step 2: From the second race, B runs 800 m while C runs 700 m in the same time, so vB : vC = 800 : 700 = 8 : 7. Step 3: Combine both results. Write vA = 10k and vB = 9k. Also write vB = 8t and vC = 7t. Step 4: Since vB is the same, 9k = 8t, so t = 9k / 8 and vC = 7t = 7 * 9k / 8 = 63k / 8. Step 5: Now vA : vC = 10k : (63k / 8) = 80 : 63. Step 6: In a 600 m race, time taken by A to finish is T = 600 / vA. Step 7: In the same time, distance covered by C = vC * T = vC * 600 / vA = 600 * (vC / vA). Step 8: Using vC / vA = 63 / 80, distance by C = 600 * 63 / 80 = 472.5 m. Step 9: Lead of A over C = 600 - 472.5 = 127.5 m. Verification / Alternative check: We can also work with unit speeds. Take vA = 80 units and vC = 63 units. Time for A to run 600 m is 600 / 80 = 7.5 time units. Distance of C in this time is 63 * 7.5 = 472.5 m. This again gives a lead of 127.5 m, so the calculation is consistent and confirms the answer. Why Other Options Are Wrong: - 254 m: This would require C to be extremely slow compared to A, which contradicts the given race data. - 184 m: This does not match any consistent ratio derived from the two races. - 212 m: This is larger than possible given that C is only slightly slower than B, not extremely slower than A. Common Pitfalls: Many learners subtract distances directly, for example by trying to scale 100 m and 100 m from the original races without computing speed ratios. Others average the two gaps, which is not correct. The safe method is always to convert the information into speed ratios and then apply those ratios carefully to the new race distance. Final Answer: A will beat C by 127.5 m in the 600 m race.

Question 16

Runner A runs 1 2/3 times as fast as runner B. If A gives B a head start of 80 m, how far from the start should the winning post be placed so that A and B reach it at the same time?

Accepted Answer

Introduction / Context: This race question checks understanding of relative speed and the effect of a head start. Runner A is faster than B, but B starts the race some distance ahead. We must find the race length that balances this advantage so both finish together. This type of question appears often in aptitude tests on races and games. Given Data / Assumptions: - A runs 1 2/3 times as fast as B, so vA = 5/3 of vB. - B starts 80 m ahead of A. - Both start running at the same time and move with constant speeds. - We must find the race distance D so that both reach the winning post together. Concept / Approach: When two runners start together but one has a fixed head start, both finish together when their times are equal. Time is distance divided by speed. A runs the full distance D, while B needs to cover only D - 80 m. Using the speed ratio, we set their times equal and solve for D. This is a direct application of proportional reasoning with speeds. Step-by-Step Solution: Step 1: Let vB = v units. Then vA = (5/3) * v. Step 2: Let D be the total distance from the common start point of A to the winning post. Step 3: Time taken by A to finish = D / vA = D / ((5/3) * v) = 3D / (5v). Step 4: Since B starts 80 m ahead, B covers only D - 80 m. Step 5: Time taken by B = (D - 80) / v. Step 6: They finish together, so their times are equal: 3D / (5v) = (D - 80) / v. Step 7: Cancel v to get 3D / 5 = D - 80. Step 8: Multiply both sides by 5 to obtain 3D = 5D - 400. Step 9: Rearrange: 5D - 3D = 400, so 2D = 400 and D = 200 m. Verification / Alternative check: If D = 200 m, A runs 200 m, B runs 200 - 80 = 120 m. A is 5/3 times as fast as B, so the time ratio TA : TB = 200 / vA : 120 / vB = 200 / (5v/3) : 120 / v = 200 * 3 / (5v) : 120 / v = 600 / (5v) : 120 / v = 120 / v : 120 / v. Times are equal, confirming that both reach the winning post together when the race length is 200 m. Why Other Options Are Wrong: - 300 m: With this distance A would need more time than required to match B, so A would finish earlier than B. - 270 m: This does not satisfy the equality of times when the ratio 5/3 is applied. - 160 m: This distance is too short and would not allow A to catch B despite being faster. Common Pitfalls: Students sometimes mix up which runner covers D and which covers D minus the head start. Another frequent mistake is to treat the 1 2/3 factor as 1.23 or as 123 percent without converting it correctly to an exact fraction of 5/3. Always convert mixed numbers to improper fractions and build the equation by equating the times, not the distances. Final Answer: The winning post should be 200 m from the starting point of A.

Question 17

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

Accepted Answer

Introduction / Context: This problem combines two separate race results to compare the speeds of runners B and C. It appears in aptitude tests to check if a student can translate statements like A beating B by a certain distance into speed ratios and then reuse those ratios for a new race distance between different runners. Given Data / Assumptions: - In a 200 m race, when A finishes 200 m, B has covered only 169 m. - In the same 200 m race, when A finishes 200 m, C has covered 182 m. - Speeds of A, B, and C remain constant across races. - We must find how many metres C will beat B by in a 350 m race. Concept / Approach: The key idea is that distances covered in the same time are proportional to speeds. From the first race we can find the ratio of speeds vB and vC relative to vA. Then we divide these ratios to get the direct ratio vB : vC. Finally, we apply this ratio to a new race where C runs 350 m, and we calculate B's distance at the same time to find the lead of C. Step-by-Step Solution: Step 1: In 200 m race, A vs B gives vB / vA = 169 / 200. Step 2: From the same race, A vs C gives vC / vA = 182 / 200. Step 3: Ratio of speeds B to C is vB / vC = (169 / 200) / (182 / 200) = 169 / 182. Step 4: In a new race, let C run the full 350 m. Time taken by C is proportional to 350 / vC. Step 5: In that same time, distance covered by B is 350 * (vB / vC) = 350 * 169 / 182. Step 6: Compute the gap: lead of C over B = 350 - 350 * 169 / 182. Step 7: Simplify: 350 * (1 - 169 / 182) = 350 * (13 / 182) = 25 m. Verification / Alternative check: We can scale speeds directly. Take vB = 169 units and vC = 182 units. If C covers 350 m, the time factor is proportional to 350 / 182. In that time B would cover (169 / 182) * 350 = 325 m. The difference 350 - 325 again equals 25 m, confirming the previous calculation and ensuring that there is no arithmetic mistake. Why Other Options Are Wrong: - 22.75 m and 19.5 m: These values arise if incorrect ratios are used or if distances are subtracted before forming speed ratios. - 18.5 m: This is smaller than the correct difference and does not match the exact fractional computation. Common Pitfalls: A frequent mistake is to subtract gaps directly, for example by manipulating 31 m and 18 m without converting them into speed ratios. Another error is reversing the ratio, using vC / vB instead of vB / vC at the wrong stage. Always express speeds in terms of one runner (here A), then divide to get vB : vC cleanly, and only then apply this ratio to the new race distance. Final Answer: In a 350 m race, C will beat B by 25 m.

Question 18

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 race question links two separate pairwise results to compare A and C directly. It is a classic exam pattern where one runner beats another by a certain distance, and that second runner beats a third. The challenge is to convert these distance advantages into speed ratios and then combine them to find the final lead. Given Data / Assumptions: - In a 100 m race, when A finishes 100 m, B has covered only 75 m. - In the same type of 100 m race, when B finishes 100 m, C has covered only 96 m. - All runners move at constant speeds throughout the race. - We must find how many metres A is ahead of C when A has completed the 100 m race. Concept / Approach: Distances covered in the same time are proportional to speeds. From the first statement we find the ratio of speeds vB and vA. From the second we get vC and vB. Multiplying these ratios gives vC relative to vA. Once we know vC / vA, we find how far C runs in the time when A covers 100 m and compute the difference. Step-by-Step Solution: Step 1: From A beating B by 25 m in 100 m race, we have vB / vA = 75 / 100 = 3 / 4. Step 2: From B beating C by 4 m in 100 m race, we have vC / vB = 96 / 100 = 24 / 25. Step 3: Multiply the ratios to get vC / vA = (vC / vB) * (vB / vA) = (24 / 25) * (3 / 4) = 72 / 100 = 18 / 25. Step 4: When A runs 100 m, the time taken is T = 100 / vA. Step 5: In the same time, C covers distance = vC * T = vC * 100 / vA = 100 * (vC / vA) = 100 * 18 / 25 = 72 m. Step 6: Therefore A is ahead of C by 100 - 72 = 28 m. Verification / Alternative check: Take vA = 25 units. Then vB = 3/4 of 25 = 18.75 units, and vC is 24/25 of vB, which is 18 units. Time for A to run 100 m is 100 / 25 = 4 time units. In 4 units, C covers 18 * 4 = 72 m. The lead is again 28 m, so the numerical result is consistent and the method is correct. Why Other Options Are Wrong: - 21 m and 26 m: These come from incorrect combination of ratios or partial subtraction of the original leads. - 29 m: Slightly larger than the correct value, usually caused by rounding or arithmetic slips. Common Pitfalls: Many learners simply add the two gaps, 25 m and 4 m, and claim that A beats C by 29 m, which is wrong. Others subtract them and get 21 m. The correct approach is always to convert leads into speed ratios, multiply those ratios, and only then convert back to distances for the new comparison. Remember that race results are about speeds, not about direct addition of distances. Final Answer: In the 100 m race, A can beat C by 28 m.

Question 19

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 in a game of 100 points?

Accepted Answer

Introduction / Context: This is a classic problem on comparative performance in games. When we say that one player can give another a certain number of points in a fixed target game, it means that the stronger player can still reach the target while the weaker player is lagging. The question asks us to deduce the relative strengths of three players and then find the handicap one of them can give to another. Given Data / Assumptions: - Game target is 100 points. - A can give B 20 points, meaning when A scores 100, B scores only 80. - A can give C 28 points, meaning when A scores 100, C scores only 72. - All players score points at constant relative rates in any game length. - We must find how many points B can give C in a 100 point game. Concept / Approach: The idea is similar to relative speed. Here, scoring rate replaces speed. When A scores 100 while B scores 80, their scoring rates are in the ratio 100 : 80. Likewise, A and C have a rate ratio 100 : 72. Once we know these, we can find the rate ratio of B to C and then determine the handicap B can offer to C in a race to 100 points. Step-by-Step Solution: Step 1: From A vs B, scoring rate ratio vA : vB = 100 : 80 = 5 : 4. Step 2: From A vs C, scoring rate ratio vA : vC = 100 : 72 = 25 : 18. Step 3: Express vB and vC in terms of vA. From step 1, vB = (4/5) * vA. From step 2, vC = (18/25) * vA. Step 4: Find vB : vC = (4/5) * vA : (18/25) * vA = (4/5) : (18/25) = (4/5) * (25/18) = 100 / 90 = 10 / 9. Step 5: Let B and C play to 100 points. If B scores 100 points, C will score (9/10) * 100 = 90 points in the same time. Step 6: Therefore B can give C 10 points in a game of 100 points. Verification / Alternative check: We can pick a common unit model. Assume 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 2.5 units, C scores 36 * 2.5 = 90 points. So if B starts from 0 and C starts from 10, both reach 100 together. This confirms that the correct handicap from B to C is 10 points. Why Other Options Are Wrong: - 8 points and 14 points: These values correspond to incorrect manipulation of the rate ratios and do not satisfy the exact proportional relation. - 40 points: This is far too large and would imply C is very weak compared to B, which is not supported by the data. Common Pitfalls: Learners sometimes treat the given points as direct differences and try to subtract or add 20 and 28 without converting them into rate ratios. Others forget that the game length remains 100 points and mistakenly change the target. Always interpret “A can give B k points in 100” as meaning when A reaches 100, B reaches only 100 minus k, then treat points scored per unit time as proportional to these numbers. Final Answer: In a 100 point game, B can give C a start of 10 points.

Question 20

In a race of 1000 m, runner A can beat B by 100 m, and in a race of 800 m, B can beat C by 100 m. By how many metres will A beat C in a 600 m race?

Accepted Answer

Introduction / Context: This is another version of the earlier relative speed question involving three runners A, B, and C. We are given A versus B over 1000 m and B versus C over 800 m and asked to find A's lead over C over 600 m. The structure is the same, but the options are expressed with the word metres written out. It still tests the same ratio and proportional reasoning skills. Given Data / Assumptions: - In a 1000 m race, A finishes when B has covered 900 m. - In an 800 m race, B finishes when C has covered 700 m. - Speeds of all runners remain constant. - We must find A's lead over C in a 600 m race. Concept / Approach: We again convert the information into speed ratios. From each race we get vA relative to vB and vB relative to vC. Combining these gives vA relative to vC. Once we know vC / vA, we can compute how far C runs while A completes 600 m, and then subtract to get the lead. This is a direct proportional application. Step-by-Step Solution: Step 1: From the first race, vA : vB = 1000 : 900 = 10 : 9. Step 2: From the second race, vB : vC = 800 : 700 = 8 : 7. Step 3: Let vB = 9k and vA = 10k. Also write vB = 8t and vC = 7t. Step 4: Since vB is same, 9k = 8t, so t = 9k / 8 and vC = 7 * 9k / 8 = 63k / 8. Step 5: Ratio vC / vA = (63k / 8) / (10k) = 63 / 80. Step 6: Time taken by A to run 600 m is T = 600 / vA. Step 7: In time T, C runs distance = vC * T = 600 * (vC / vA) = 600 * 63 / 80 = 472.5 m. Step 8: Lead of A over C = 600 - 472.5 = 127.5 metres. Verification / Alternative check: Choose vA = 80 and vC = 63 units. Time for A to run 600 m is 600 / 80 = 7.5. In this time C runs 63 * 7.5 = 472.5 m, which produces the same difference of 127.5 m. This confirms the consistency of the computed lead with the combined ratio method. Why Other Options Are Wrong: - 125.5 m and 126.5 m: These are close but incorrect, often caused by premature rounding during calculations. - 128.5 m: Slight overestimate that does not match the exact fractional result 600 * 63 / 80. Common Pitfalls: A common mistake is to average or directly add the differences 100 m and 100 m without translating them into speed ratios. Others treat the new race distance as 1000 m or 800 m instead of 600 m. Always re-build the ratio chain carefully, then apply it only to the exact race length mentioned in the question. Final Answer: A will beat C by 127.5 metres in the 600 m race.

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