Table of contents | |
Races and Games | |
Important Terms | |
TIPS to Crack Questions | |
Tip 1 | |
Tip 2 | |
Tip 3 |
Races and games in government exams test quantitative skills, requiring candidates to apply mathematical concepts such as speed, distance, and time. They evaluate analytical thinking and decision-making abilities, essential for data analysis and logistical tasks in government roles.
A race or a games of skill includes the contestants in a contest and their skill in the concerned contest/game.
Acquaint yourself with the terms
Assume that the speed or the scoring rate for each player is constant
Q1: In a game of 100 points, A can give B 20 points and C 28 points. How many points can B give C?
Sol:
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.
Q2: In a 200 m race A beats B by 35 m or 7 sec. Find A's time over the course.
Sol:
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.
If A runs x times faster than B, A’s speed is actually 1+x the speed of B
Q1: 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?
Sol:
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
Q2: 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?
Sol:
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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1. What are races and games in the context of aptitude questions? |
2. What are some important terms related to races and games in aptitude questions? |
3. What are some tips for solving aptitude questions on races and games? |
4. What are some common types of questions on races and games in aptitude exams? |
5. How can I improve my problem-solving skills for races and games in aptitude exams? |
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