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 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:

Accepted Answer

A : B = 100 : 90. A : C = 100 : 87. B/C = (B/A*A/C) = (90/100*100/87) = 30/29 When B runs 30 m, C runs 29 m. When B runs 180 m, C runs (29/30*180)m =174m B beats C by (180 - 174) m = 6 m.

Question 8

In a 1 km race, A beats B by 28 meters in 7sec. Find A's time over the course?

Accepted Answer

B covers 28 meters in 7sec. So, B's time over the course = (7/28)*100 =250 sec. Whereas A's time over the course = 250 -7 = 243 sec. i.e A's time over the course is 4min , 3 sec.

Question 9

A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:

Accepted Answer

When B runs 25 m, A runs 45/2 m When B runs 1000 m, A runs (45/2*1/25*1000)m = 900 m. B beats A by 100 m.

Question 10

A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

Accepted Answer

A's speed = (5*5/15)m/sec = (25/18)m/sec Time taken by A to cover 100 m = (100*18/15)sec = 72sec Time taken by B to cover 92 m = (72 + 8) = 80 sec. B's speed =(92/80*18/5)kmph =4.14kmph

Question 11

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

Accepted Answer

Distance covered by B in 9 sec. = (100/45)*9m = 20m A beats B by 20 metres.

Question 12

In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:

Accepted Answer

To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m. While A covers 3 m, B covers 4 m. While A covers 360 m, B covers (4/3)*360 = 480 m. Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind. A wins by 20 m.

Question 13

In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:

Accepted Answer

B runs 35 m in 7 sec. B covers 200 m in (7/35*200) = 40 sec. B's time over the course = 40 sec. A's time over the course (40 - 7) sec = 33 sec.

Question 14

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

Accepted Answer

A : B = 60 : 45. A : C = 60 : 40. B/C = (B/A*A/C) = (45/60*60/45) = 45/40 = 90/80 = 90:80 B can give C 10 points in a game of 90.

Question 15

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

Accepted Answer

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m. When B runs 900 m, distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 m. In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C. In a race of 600 m, the number of meters by which A beats C = (600 x 212.5)/1000 = 127.5 m.

Question 16

A runs 123 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Accepted Answer

Ratio of the speeds of A and B =5 : 1 = 5 :3 Thus, in race of 5 m, A gains 2 m over B. 2 m are gained by A in a race of 5 m. 80 m will be gained by A in race of 5 2 * 80 m =200m => Winning post is 200 m away from the starting point.

Question 17

In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:

Accepted Answer

A : B = 200 : 169. A : C = 200 : 182. C/B = (C/A*A/B) = (182/200*200/169) = 182:169 When C covers 182 m, B covers 169 m. When C covers 350 m, B covers (169/180*350)m = 325 m Therefore, C beats B by (350 - 325) m = 25 m.

Question 18

In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by

Accepted Answer

28m

Question 19

In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C is

Accepted Answer

10 points

Question 20

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

Accepted Answer

When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters. Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters. So, in a race of 1000 meters, A beats C by (1000 - 787.5) = 212.5 meters to C. So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.

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