Question

A runs 123 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) 300m C) 270m D) 160m

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• A can run $1\dfrac { 2 } { 3 }$ as fast as B

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• A gives B a start of 80 m

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$\red{\underline \bold{To \: Find:}}$

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• How far must the winning post be so that A and B might reach it at the same time.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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The ratio of the speed of running of both A and B

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$\purple\longrightarrow$ $\sf \dfrac { A } { B } =\dfrac { \dfrac { 5 } { 3 } } { 1 } = 5 : 3$

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Now as we can see that the total length of the race according to the ratio between their speeds is 5m, where is (5 – 3) is the distance by which A is at front of B.

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Therefore, the distance at which the winning post should be situated from the start should be at  

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$\sf \longmapsto 80 \times \dfrac { 5 } { 3 } = 200$

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The ratio 5 : 2 is the ratio of the total length upon the length by which A is ahead, therefore as seen in the question the distance by which A is ahead is 80 m, therefore multiplying it with the ratio 5 : 2 we get 200 m.  

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Option A is correct

$\rule{200}5$