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Question: In racing over a given distance d at a uniform speed, A can beat B by 30 meters, B can beat C by 20 ...

In racing over a given distance d at a uniform speed, A can beat B by 30 meters, B can beat C by 20 meters and A can beat C by 48 meters. Find ‘d’ in meters.
A. 300
B. 350
C. 450
D. 400

Explanation

Solution

We use the concept of ratio here using the fixed distance and uniform speed of each A, B and C. Write ratio of speed of each two pairs using the given information and form the ratio of third pair by multiplying ratio of first two pairs of ratios.

Complete step by step answer:
We are given a fixed distance ‘d’ meters.
Since A, B and C have uniform speed, let their speeds be a, b and c respectively.
We are given A can beat B by 30 meters
If A covers ‘d’ distance then B covers d−30d - 30 distance
Ratio of distance covered by A to distance covered by B is
⇒ab=dd−30\Rightarrow \dfrac{a}{b} = \dfrac{d}{{d - 30}} … (1)
We are given B can beat C by 20 meters
If B covers ‘d’ distance then C covers d−20d - 20 distance
Ratio of distance covered by B to distance covered by C is
⇒bc=dd−20\Rightarrow \dfrac{b}{c} = \dfrac{d}{{d - 20}} … (2)
We are given A can beat C by 48 meters
If A covers ‘d’ distance then C covers d−48d - 48 distance
Ratio of distance covered by A to distance covered by C is
⇒ac=dd−48\Rightarrow \dfrac{a}{c} = \dfrac{d}{{d - 48}} … (3)
We know ac=ab×bc\dfrac{a}{c} = \dfrac{a}{b} \times \dfrac{b}{c}
Substitute values in RHS form (1) and (2) and in LHS from equation (3)
⇒dd−48=dd−30×dd−20\Rightarrow \dfrac{d}{{d - 48}} = \dfrac{d}{{d - 30}} \times \dfrac{d}{{d - 20}}
Cancel same factor from numerator on both sides of the equation
⇒1d−48=1d−30×dd−20\Rightarrow \dfrac{1}{{d - 48}} = \dfrac{1}{{d - 30}} \times \dfrac{d}{{d - 20}}
Multiply terms in denominator of RHS
⇒1d−48=dd2−30d−20d+600\Rightarrow \dfrac{1}{{d - 48}} = \dfrac{d}{{{d^2} - 30d - 20d + 600}}
⇒1d−48=dd2−50d+600\Rightarrow \dfrac{1}{{d - 48}} = \dfrac{d}{{{d^2} - 50d + 600}}
Cross multiply the values from both sides of the equation
⇒d2−50d+600=d2−48d\Rightarrow {d^2} - 50d + 600 = {d^2} - 48d
Cancel same factor from both sides of the equation
⇒−50d+600=−48d\Rightarrow - 50d + 600 = - 48d
Shift -50d to RHS of the equation
⇒600=50d−48d\Rightarrow 600 = 50d - 48d
⇒600=2d\Rightarrow 600 = 2d
Cancel same factor from both sides of the equation
⇒300=d\Rightarrow 300 = d
∴\therefore Distance is 300 meters

∴\therefore Option A is correct.

Note: Many students make the mistake of writing the equation of speed in terms of distance and time and then try to write equations in terms of speed or time. Keep in mind we need not do all unnecessary calculations, instead we use the fact that distance is the same and speeds are not varying.