\textbf{Q.} A runs times as fast as B. If A gives B a start of , how far must the winning post be so that A and B might reach it at the same time?

200 m Explanation: Correct Option - 1 \textbf{Q.} A runs times as fast as B. If A gives B a start of , how far must the winning post be so that A and B might reach it at the same time? \textbf{Given:} A gives B a start of \textbf{Formula Used:} \textbf{Calculation:} A runs times as fast as B So the ratio of speeds of A and B is . Let the speeds of A and B be and respectively. Let the distance covered by B be . Then the distance covered by A . They both reach at the same time, hence Simplifying, A will cover the total distance

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Explanation

Correct Option - 1

\textbf{Q.} A runs $1 \frac{2}{3}$ 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?

\textbf{Given:}
A gives B a start of $80 m$

\textbf{Formula Used:}
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > Speed = \frac{Distance}{Time} < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

\textbf{Calculation:}

A runs $1 \frac{2}{3}$ times as fast as B
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > A = \frac{5}{3} B \Rightarrow \frac{A}{B} = \frac{5}{3} < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

So the ratio of speeds of A and B is $5 : 3$ .

Let the speeds of A and B be $5 x$ and $3 x$ respectively.
Let the distance covered by B be $L$ .
Then the distance covered by A $= L + 80$ .

They both reach at the same time, hence
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > \frac{L}{3 x} = \frac{L + 80}{5 x} < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

Simplifying,
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > 5 xL = 3 xL + 240 x < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > 5 L - 3 L = 240 < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > 2 L = 240 < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > L = 120 < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

A will cover the total distance
$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > L + 80 = 120 + 80 = 200 < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

$< / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " > ∴ The winning post will be 200 meters long. < / s p an >< s p an s t y l e = " f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; f o n t - s i z e : 14 pt; " >$

Explanation

Correct Option - 1

NIMCET 2018 — LOGICAL REASONING PYQ

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