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Which value is not possible for the given condition – CAT 2004 Question

This post is a Question of CAT 2004 Question.It is to find out Which value is not possible for the given condition. It uses the concept of ratio.
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Which value is not possible for the given condition – CAT 2004 Question


(Q.)If $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = r , then r can not take any value except [CAT 2004 Question]

(a)$\frac{1}{2}$

(b)-1

(c)$\frac{1}{2}$ or -1

(d)$-\frac{1}{2}$ or -1

Solution :
Given Condition is
$\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = r

This can also be written as
$\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a+b+c}{(a+b)+(b+c)+(c+a)}$ = $\frac{1}{2}$

Now Here, there are two possibilities ,

(I) When a+b+c $\neq$ 0 , then

$\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a+b+c}{(a+b)+(b+c)+(c+a)}$ = $\frac{1}{2}$

(II) If a+b+c = 0 , then

a = -(b+c)
b = -(c+a)
c = -(a+b)

Hence , $\frac{a}{b+c}$ = $\frac{a}{-a}$ = -1

Similarly for others , $\frac{b}{a+c}$ = $\frac{c}{a+c}$ = -1

Hence, r can not take values $\frac{1}{2}$ and -1.
Hence, OPTION(c)

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 1.)  $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a}{(b+c)}$ = $\frac{a+b+c}{(a+b)+(b+c)+(c+a)}$ = $\frac{1}{2}$

 2.)   $\frac{a}{b}$ = $\frac{c}{d}$ = $\frac{e}{f}$ = $\frac{Sum_Of_All_Numerators}{Sum_Of_All_Denominators}$ = $\frac{a+c+e}{b+d+f}$