# Run Race MCQ1( One Runs faster than the other)

Run Race MCQ1: A runs 1 – 2/3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

a) 200 m               b) 300 m        c) 270 m          d) 160 m

Solution:

Let B runs at speed ‘x’

By given condition,

A will run with speed ​$$1\frac{2}{3}x=\frac{5}{3}x$$

We know the formula of distance;

$s=vt$

Where,

S= distance

V= speed

t= time

From there we can say;

$t=\frac{s}{v}$

or

$Time= \frac {Distance}{Speed}$

In the statement “If A gives B a start of 80 m” means that B is 80 meter closer to the winning pole than A

Let the winning pole is at distance ‘y’ from B. Then by given condition it will be at distance ‘y+80’ from A.

Using above formula,

Time taken by B to reach winning pole;

$t=\frac{y}{x}$

Time by A to reach winning pole;

$t= (y+80)\frac {3}{5x}$

If bot A and B reach at the same time then times taken by both should be equal

$(y+80)\frac {3}{5x}=\frac{y}{x}$

or

$5y=3(y+80)$

$2y=240$

$y=120$

So, the winning pole must be erected at distance of 120 from B

or

y+80 = 120+80=200 from A

Note:

This question has been solved on other websites as well —  but in a confusing style and not in a  proper and easy way

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