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 = [latex]\frac{score}{time}[/latex])
Scoring Rate of B, Sb = [latex]\frac{80}{100}[/latex] x Sa = 0.8 Sa
Scoring Rate of C, Sc = [latex]\frac{72}{100}[/latex] x Sa = 0.72 Sa
Time taken for B to get 100 points = [latex]\frac{100}{sb}[/latex] = [latex]\frac{100}{(0.8 \times Sa)}[/latex]
Score taken by C in this time period = Sc x [latex]\frac{100}{(0.8 \times Sa)}[/latex] = [latex]\frac{72}{0.8}[/latex] = 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 = [latex]\frac{35}{7}[/latex] = 5 m/s.
Time taken by B to finish the race = [latex]\frac{200}{5}[/latex] = 40 s.
Thus, A’s time over the course = (40 – 7)s = 33 s.

Example 1: A runs [latex]1\frac{2}{3}[/latex] 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 = [latex]\frac{5}{3}[/latex] 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’
[latex]\frac{d}{[\frac{5}{3} \times Sb]}[/latex] = [latex]\frac{(d - 80)}{Sb}[/latex]
3d = 5d – 400
=> 2d = 640 => d = 200m

Example 2: A runs [latex]1\frac{2}{3}[/latex] 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 + [latex]\frac{5}{3}[/latex]) x Sb = [latex]\frac{8}{3}[/latex] 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’
[latex]\frac{d}{[\frac{8}{3} \times Sb]}[/latex] = [latex]\frac{(d - 80)}{Sb}[/latex]
3d = 8d – 640
=> 5d = 640 => d = 128m

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 ( [latex]\frac{936}{960} * 1000[/latex] ) = 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 = ([latex]\frac{60 * 60}{8000} * 100[/latex]) 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 = [latex]\frac{8}{38} * 1000[/latex] 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,
[latex]\frac{T}{U}[/latex] = [latex]\frac{T}{S} * \frac{S}{U}[/latex] = ([latex]\frac{75}{80} * \frac{80}{65}[/latex]) = [latex]\frac{15}{13}[/latex] = [latex]\frac{60}{52}[/latex] = 60 : 52
Therefore, In a game of 60, T can give U 8 points (since 60 -52 = 8).