3 TIPS on cracking Aptitude Questions on Races and Games

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Tip #1: Acquaint yourself with the terms

Dead Heat Race: A race in which all the contestants reach the Goal at the same time.

Start: If A and B are two contestants in a race, such that before the start of the race, A is at the starting point and B is ahead of A by 12 meters, then we say that ‘A gives B a start of 12 meters’.

Game: A game of 100, means that the person among the contestants who scores 100 points first is the winner. If A scores 100 points while B scores only 80 points, then we say that 'A can give B 20 points’. This implies that if A actually gave B a start of 20 points, then the contest would result in a dead heat.

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Tip #2: Assume that the speed or the scoring rate for each player is constant

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.

Tip #3: If A runs x times faster than B, A’s speed is actually 1+x the speed of B

Question: A runs 1⅔ 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 1⅔ 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.

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