Question 64

In racing over a distance d at uniforgn speed, A can beat B by 20 metres, B can beat C by 10 metres, and A can beat C by 28 metres. Then d, in metres, is

Solution

Let speed of A, B and C be $$a,b,c$$ m/s respectively.

Time taken by A to cover $$d$$ metres = Time taken by B to cover $$(d-20)$$ metres

=> $$\frac{s_a}{s_b}=\frac{d}{d-20}$$ ---------------(i) [Since, speed is directly proportional to distance]

Similarly, Time taken by A to cover $$d$$ metres = Time taken by C to cover $$(d-28)$$ metres

=> $$\frac{s_a}{s_c}=\frac{d}{d-28}$$ --------------(ii)

and Time taken by B to cover $$d$$ metres = Time taken by C to cover $$(d-10)$$ metres

=> $$\frac{s_b}{s_c}=\frac{d}{d-10}$$ -------------(iii)

Equating equations (i) and (ii), => $$\frac{s_b}{s_c}=\frac{d-20}{d-28}$$

Comparing it with equation (iv), => $$\frac{d}{d-10}=\frac{d-20}{d-28}$$

Solving above equation, we get : $$d=100$$ m

=> Ans - (C)

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