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.

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

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.

If A scores 100 points while B scores only 80 points, then 'A can give B 20 points'.

- 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.

- 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.

- 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

- 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

- 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.

- 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.

- 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.

- 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.

- 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).