SSC CHSL 6 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

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Question : 52 of 100

Marks: +1, -0

If

a+b-c=0

, then what is the value of

\;{(b-c)^{2}}/{4 b c} \;{(c-a)^{2}}/{4 c a} \;{(a+b)^{2}}/{4 a b} ?

$\;{1}/{2}$
$\;{1}/{64}$
$-\;{1}/{2}$
$\;{-1}/{64}$

Solution:

Given,

a+b-c=0 ⇒ b-c=-a

,

c-a=b

and

a+b=c

∴ \;{(b-c)^{2}}/{4 b c} × \;{(c-a)^{2}}/{4 c a} × \;{(a+b)^{2}}/{4 a b}

=\;{(-a)^{2}}/{4 b c} × \;{(b)^{2}}/{4 c a} × \;{(c)^{2}}/{4 a b}

=\;{a^{2} b^{2} c^{2}}/{64 a^{2} b^{2} c^{2}}=\;{1}/{64}

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