Ratio of the rates of A and B = 7/4 : 1 = 7 : 4
So, in a race of 7 m, A gains 3m over B.
3 m are gained by A in a race of 7 m.
84 m are gained by A in a race of (7/3 x 84) m = 196 m.
Winning post must be 196 m away from the starting point.

Answer : Option D

Explanation :

P run 1 km in 3 minutes
Q run 1 km in 4 minutes 10 secs
$MF#%\text{=> Q runs 1 km in }\dfrac{25}{6}\text{ minutes}\\\\ \text{=> Q runs }\left(1 \times \dfrac{6}{25} \times 3\right)= \dfrac{18}{25} = 0.72\text{ km in 3 minutes}$MF#%
Hence, in a 1 km race, P can give Q (1 - 0.72)=0.28 km = 280 metre

Answer : Option D

Explanation :

It means, when A convers 200 m, B covers only (200-31)=169 m
and C covers only (200-18)=182 m
=> When C covers 182 m, B covers 169 m
$MF#%\text{=> When C covers 350 m, B covers }\dfrac{169}{182}\times 350\text{ = 325 m}$MF#%
Hence, C beats B by 350-325 = 25 metre

Answer : Option C

Explanation :

$MF#%\text{Speed of A = 5 kmph = }5\times \dfrac{5}{18} = \dfrac{25}{18} \text{m/s}\\\\ \text{Time taken by A to cover 100 m} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{100}{\left(\dfrac{25}{18}\right)}\text{ = 72 seconds}$MF#%
It is given that, A gives B a start of 8 m and still beats him by 8 seconds.
=> B takes (72+8)=80 seconds to cover (100-8)=92 metre
$MF#%\text{Speed of B =} \dfrac{\text{distance}}{\text{time}} = \dfrac{92}{80} \text{m/s} = \dfrac{92}{80} \times \dfrac{18}{5}\text{ kmph}\text{ = 4.14 kmph}$MF#%

A: B = 80 : 75, A : C = 80 : 65
B/C = ( B/ A * A/C) = (75 / 80 * 80 / 65) = 15/13 = 60 /52 = 60: 52
Therfore ,In a game of 60, B can give C 8 points.

Answer : Option B

Explanation :

In a game of 100 points, A scores 100 points, B scores (100-20)=80 points and
C scores (100-28)=72 points
i.e., when B scores 80 points, C scores 72 points
$MF#%\text{=> When B scores 100 points, C scores }\dfrac{72}{80} \times 100\text{ = 90 points}$MF#%
i.e., In a game of 100 points, B can give C (100-90)=10 points

Distance covered by B in 9 sec. = 100 x 9 m = 20 m. 45

A beats B by 20 metres.

Answer : Option C

Explanation :

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Solution 1
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Speed of A : Speed of B = ⁷⁄₃ : 1 = 7 : 3
It means, in a race of 7 m, A gains (7-3)=4 metre
$MF#%\text{If A needs to gain 80 metre, race should be of }\dfrac{7}{4} \times 80\text{ = 140 metre}$MF#%
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Solution 2
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If A is n times as fast as B and A gives B a start of x meters, then the length of the race course , so that A and B reaches the winning post at the same time = $MF#%x\left(\dfrac{n}{n-1}\right)$MF#% metres

B runs
45
m in 6 sec.
2

B covers 300 m in

6 x
2
x 300
sec
= 80 sec.
45

Ratio of the speeds of A and B =
5
: 1 = 5 : 3.
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

5
x 80
m
= 200 m.
2

Winning post is 200 m away from the starting point.