SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 66 of 100

Marks: +1, -0

If

a^{2}+49 b^{2}+c^{2}+18=2

(28 b-c-a)

, then the value of

(a+7 b-c)

is

2
4
6
$-1$

Solution:

Given,

a^{2}+49 b^{2}+c^{2}+18

=2(28 b-c-a)

⇒ a^{2}+49 b^{2}+c^{2}+18=56 b-2 c-2 a

⇒ a^{2}+1+2 a+(7 b)^{2}+16-56 b

+c^{2}+1+2 c=0

⇒(a+1)^{2}+(7 b-4)^{2}+(c+1)^{2}=0

Now,

\;\; a+1=0

⇒ \;\; a=-1

,

7 b-4=0

⇒ \;\; b=\;{4}/{7}

and

c+1=0

⇒ \;\; c=-1

∴

The value of

(a+7 b-c)

=[-1+7 × \;{4}/{7}-(-1)]

=-1+4+1=4

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