Quantitative Aptitude - SPLessons

Race – Game Problems

Chapter 29

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Race – Game Problems

shape Introduction

The chapter, Races and Games of Skill deals with races, race course, starting point, goal, winner, dead heat race, start, games. The problems are related to time over course, distance between starting and ending point, speed, level points etc.


shape Concepts

Races: A race is a contest of speed in running, riding, driving, sailing or rowing.


Race course: A race course is the ground or path on which contests are made.


Starting point: A starting point is the point from which a race begins.


Winning point: Winning point is the point set to bound a race. It is also known as goal.


Winner: Winner is the first person who reaches the winning goal.


Dead heat race: Dead heat race is defined as the race in which all the persons contesting a race reach the goal exactly at the same time.


Start: If before the start of the race, A is at the starting point and B is ahead of A by 12 meters, then ‘A gives B, a start of 12 meters’.

Here, A and B are two contestants in a race.


Example: To cover a race of 100 meters in this case, A will have to cover 100 meters while B will have to cover only (100 – 12) = 88 meters.

In a 100 meters race, ‘A can give 12 meters’ or ‘A can give B a start of 12 meters’ or ‘A beats B by 12 meters’ means that while A runs 100 meters, B runs (100 – 12) = 88 meters.


Games: ‘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 ‘A can give B 20 points’.


Example 1:
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 = \(\frac{score}{time}\))


    Scoring Rate of B, Sb = \(\frac{80}{100}\) x Sa = 0.8 Sa


    Scoring Rate of C, Sc = \(\frac{72}{100}\) x Sa = 0.72 Sa


    Time taken for B to get 100 points = \(\frac{100}{sb}\) = \(\frac{100}{(0.8 \times Sa)}\)


    Score taken by C in this time period = Sc x \(\frac{100}{(0.8 \times Sa)}\) = \(\frac{72}{0.8}\) = 90


    Thus, B can give C 10 points.


Example 2:
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 = \(\frac{35}{7}\) = 5 m/s.


    Time taken by B to finish the race = \(\frac{200}{5}\) = 40 s.


    Thus, A’s time over the course = (40 – 7)s = 33 s.


Example 1:
A runs \(1\frac{2}{3}\) 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 = \(\frac{5}{3}\) 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’


    \(\frac{d}{[\frac{5}{3} \times Sb]}\) = \(\frac{(d – 80)}{Sb}\)


    3d = 5d – 400


    => 2d = 640 => d = 200m


Example 2:
A runs \(1\frac{2}{3}\) 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 + \(\frac{5}{3}\)) x Sb = \(\frac{8}{3}\) 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’


    \(\frac{d}{[\frac{8}{3} \times Sb]}\) = \(\frac{(d – 80)}{Sb}\)


    3d = 8d – 640


    => 5d = 640 => d = 128m

shape Samples

1. P, Q, and R are three contestants in a km race. If P can give Q a start of 40 meters and P can give S a start of 64 meters, how many meters start can Q give S?

Solution:

    Given,

    P, Q and S contestants took a race of 1 km

    So, while P covers 1000 meters, Q covers (1000 – 40) meters = 960 meters and S covers (1000 – 64) meters = 936 meters.

    When Q covers 960 meters, S covers 936 meters.

    When Q covers 1000 meters, S covers ( \(\frac{936}{960} * 1000\) ) = 975 meters.

    Therefore, Q can give S a start of (1000 – 975) = 25 meters.


2. In a 100 meters race, S runs at 8 km per hour. If S gives T a start of 6 meters and still beats by 15 seconds. What is the speed of T?

Solution:

    Given,

    The race is for 100 meters

    Time taken by A to cover 100 meters = (\(\frac{60 * 60}{8000} * 100\)) seconds = 45 seconds.

    Therefore, T covers (100 – 4) meters = 96 meters in (45 + 15) seconds = 60 seconds.

    Therefore, B’s speed = ( \( \frac{96 * 60 * 60}{60 * 1000} \)) km/hr = 5.76 km/hr.


3. In a kilo metre race, A beats B by 38 meters or 8 seconds. Find A’s time over the course?

Solution:

    Given that,

    B covers 38 meters in 8 seconds.

    Therefore, B’s time course = \(\frac{8}{38} * 1000\) seconds = 210.52 seconds.

    Therefore, A’s time over the course = 210.52 – 8 = 202.52 seconds = 3 minutes 4 seconds.


4. C can run 1 km in 3 min. 10 sec. and D can cover the same distance in 3 min. 20 sec. By what distance can C beat D?

Solution:

    Given that,

    C beats D by 10 sec.

    Distance covered by D in 10 sec. = \( \frac{1000}{200} * 100 \) m = 50 m

    Therefore, C beats D by 50 meters.


5. In a game of 80 points, S can give T 5 points and U 15 points. Then how many points T can give U in a game of 60?

Solution:

    Given that,

    S : T = 80 : 75,

    S : U = 80 : 65

    Therefore,

    \(\frac{T}{U}\) = \(\frac{T}{S} * \frac{S}{U}\) = (\(\frac{75}{80} * \frac{80}{65}\)) = \(\frac{15}{13}\) = \(\frac{60}{52}\) = 60 : 52

    Therefore, In a game of 60, T can give U 8 points (since 60 -52 = 8).