This is a simple question, where we have to find the time taken by A to beat B.
First, we will consider the given data.
Given,
Distance = 1 KM
50 meters = 10 seconds
1 KM = 1000 meters
1m= 10/50
So, we have to now find the Time for 1000 m.
=> 1 m= 10/50
=> 1/5 sec
Now,
Time for 1000 m = 1000×1/5
=200 sec
A can give B (5 min – 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
a:b = 60:40
a:c = 60:45
c/a x a/b = 45/60 x 60/40 = 45/40 = 90/80
So C gives B => 10 points.
When A runs 1000m, B runs 900m.
Hence, when A runs 500m, B runs 450m.
Again, when B runs 400m, C runs 360m.
And, when B runs 450m, C runs = 360 × 450/400 = 405m.
Required distance = 500 - 405 = 95 meter.
Speed of Shiny = 50/ 10 = 5m/s
Time taken by shiny to complete the race is B = 1000/5 = 200 sec.
Time taken by Baley to complete the race is D = 1000/4 = 250 sec.
Hence, D-B = 50 sec
A : B = 200 : 169.
A : C = 200 : 182.
$ \dfrac{C}{B} $=$ \left(\dfrac{C}{A} \times\dfrac{A}{B} \right) $=$ \left(\dfrac{182}{200} \times\dfrac{200}{169} \right) $= 182 : 169.
When C covers 182 m, B covers 169 m.
When C covers 350 m, B covers$ \left(\dfrac{169}{182} \times 350\right) $m= 325 m.
Therefore, C beats B by $\left(350 - 325\right)$ m = 25 m.
Distance covered by B in 9 sec. =$ \left(\dfrac{100}{45} \times 9\right) $m= 20 m.
$\therefore$ A beats B by 20 metres.
A : B = 100 : 80.
A : C = 100 : 72.
$\therefore \dfrac{B}{C} $=$ \left(\dfrac{B}{A} \times\dfrac{A}{C} \right) $=$ \left(\dfrac{80}{100} \times\dfrac{100}{72} \right) $=$ \dfrac{10}{9} $=$ \dfrac{100}{90} $= 100 : 90.
$\therefore$ B can give C 10 points.
B runs 35 m in 7 sec.
$\therefore$ B covers 200 m in$ \left(\dfrac{7}{35} \times 200\right) $= 40 sec.
Bs time over the course = 40 sec.
$\therefore$ As time over the course $\left(40 - 7\right)$ sec = 33 sec.
When B runs 25 m, A runs$ \dfrac{45}{2} $m.
When B runs 1000 m, A runs$ \left(\dfrac{45}{2} \times\dfrac{1}{25} \times 1000\right) $m= 900 m.
$\therefore$ B beats A by 100 m.
B runs$ \dfrac{45}{2} $m in 6 sec.
$\therefore$ B covers 300 m in$ \left(6 \times\dfrac{2}{45} \times 300\right) $sec= 80 sec.
Ratio of the speeds of A and B =$ \dfrac{5}{3} $: 1 = 5 : 3.
Thus, in race of 5 m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
80 m will be gained by A in race of$ \left(\dfrac{5}{2} \times 80\right) $m= 200 m.
$\therefore$ Winning post is 200 m away from the starting point.
A : B = 100 : 75
B : C = 100 : 96.
$\therefore$ A : C =$ \left(\dfrac{A}{B} \times\dfrac{B}{C} \right) $=$ \left(\dfrac{100}{75} \times\dfrac{100}{96} \right) $=$ \dfrac{100}{72} $= 100 : 72.
$\therefore$ A beats C by $\left(100 - 72\right)$ m = 28 m.
While A covers 1000 m, B covers $\left(1000-40\right)$=960 m and C covers $\left(1000-64\right)$=936 m
i.e., when B covers 960 m, C covers 936 m
$\text{When B covers 1000 m, C covers }\dfrac{936}{960} \times 1000\text{ = 975 m}$
i.e., B can give C a start of $\left(1000-975\right)$ = 25 m
While A scores 90 points, B scores $\left(90-15\right)$=75 points and C scores $\left(90-30\right)$= 60 points
i.e., when B scores 75 points, C scores 60 points
$\text{=> When B scores 100 points, C scores }\dfrac{60}{75}\times 100\text{ = 80 points}$
i.e., in a game of 100 points, B can give C $\left(100-80\right)$=20 points