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- In a game of 200m race, A can win B by 50 m and B can win C by 8 m.
Question 1:
In a game of 200m race, A can win B by 50 m and B can win C by 8 m. In the same race, A can win C by: A. 52m
B. 58m
C. 56m
Feedback Given that, A can win B by 50 m.
That is, When A runs 200 m then B can run (200 - 50 = )150 m.
And, B can win C by 8 m means, When B runs 200 m then C can run (200 - 8 =) 192 m.
Therefore, A:B = 200:150 and B:C = 200:192
We have to find A: C.
A:C = A/C = A/B x B/C = (200 / 150) x ( 200 / 192) = (4 / 3) x (200 / 192) = 200 / (3 x 48) = 200 /1 44
That is, A:C = 200:144.
Hence, A wins C by 200-144 m = 56 m.
That is, When A runs 200 m then B can run (200 - 50 = )150 m.
And, B can win C by 8 m means, When B runs 200 m then C can run (200 - 8 =) 192 m.
Therefore, A:B = 200:150 and B:C = 200:192
We have to find A: C.
A:C = A/C = A/B x B/C = (200 / 150) x ( 200 / 192) = (4 / 3) x (200 / 192) = 200 / (3 x 48) = 200 /1 44
That is, A:C = 200:144.
Hence, A wins C by 200-144 m = 56 m.
D. 54m
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