A race or a games of skill includes the contestants in a contest and their skill in the concerned contest/game.
Note: For a winner, finishing point is as same as the winning point/goal.
To cover a race of 100 metres in this case, A will have to cover 100 metres while B will have to cover only (100 - 12) = 88 metres.
In a 100 race, 'A can give B 12 m' or 'A can give B a start of 12 m' or 'A beats B by 12 m' means that while A runs 100 m, B runs (100 - 12) = 88 m.
If A scores 100 points while B scores only 80 points, then we say that 'A can give B 20 points'.
Question: In a game of 100 points, A can give B 20 points and C 28 points. How many points can B give C?
Solution:
By the time A scores 100 points, B scores only 80 and C scores only 72 points.
Let the Scoring Rate of A be Sa. (Scoring Rate = score/ time)
Scoring Rate of B, Sb = 80/100 x Sa = 0.8 Sa
Scoring Rate of C, Sc = 72/100 x Sa = 0.72 Sa
Time taken for B to get 100 points = 100/Sb = 100/ (0.8 x Sa)
Score taken by C in this time period = Sc x 100/ (0.8 x Sa) = 72/0.8 = 90
Thus, B can give C 10 points.
Question: In a 200 m race A beats B by 35 m or 7 sec. Find A's time over the course.
Solution:
By the time A completes the race, B is 35m behind A and would take 7 more seconds to complete the race.
=> B can run 35 m in 7 s. Thus, B's speed = 35 / 7 = 5 m/s.
Time taken by B to finish the race = 200 / 5 = 40 s.
Thus, A's time over the course = (40 - 7)s = 33 s.
Question: A runs 1? 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?
Solution:
Speed of A, Sa = 5/3 x Sb
Let the distance of the course be 'd' meters
Time taken by A to cover distance 'd' = Time taken by B to cover distance'd-80'
d/[5/3 x Sb] = (d-80)/Sb
3d = 5d - 400
⇒ 2d = 640 ⇒ d = 200m
Question: A runs 1? times faster than 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?
Solution:
Speed of A, Sa = (1 + 5/3) x Sb = 8/3 x Sb
Let the distance of the course be 'd' meters
Time taken by A to cover distance 'd' = Time taken by B to cover distance 'd-80'
d/[8/3 x Sb] = (d-80)/Sb
3d = 8d - 640
⇒ 5d = 640 ⇒ d = 128m
Note: Here, A 5/3 times faster than B, i.e., A's speed = B's speed + 5/3 times B's speed = 8/3 times B's speed.
| 1. What are the key concepts to understand when solving problems related to races and games? | ![]() |
| 2. How can I approach a question involving two runners starting at the same point but running different distances? | ![]() |
| 3. What strategies can I use to solve problems that involve multiple participants in a race? | ![]() |
| 4. In games of skill, how do I determine the winner if players have different probabilities of winning? | ![]() |
| 5. What role do time and speed play in solving race-related aptitude questions? | ![]() |