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SSC CGL 1 Sep 2016 Paper

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Question : 66 of 100

Marks: +1, -0

If

\frac{a^2}{b+c}

\frac{b^2}{c+a}

\frac{c^2}{a+b}

\frac{1}{1+a}

\frac{1}{1+b}

\frac{1}{1+c}

is

1
2
3
4

Solution:

\frac{a^2}{b+c}

=

a^2

=

a^2

= 1

⇒

\frac{1}{a+1}

= 1

⇒

\frac{a}{a+b+c}

= b + c

⇒

\frac{b^2}{c+a}

+ a = a + b + c

⇒ a (a + 1) = a + b + c

⇒

b^2

=

b^2

Similarly,

\frac{1}{b+1}

= 1 ⇒

\frac{b}{a+b+c}

= c + a

⇒

\frac{c^2}{a+b}

+ b = a + b + c

⇒ b (b + 1) = a + b + c

⇒

c^2

=

c^2

and

\frac{1}{1+a}

= 1 ⇒

\frac{1}{1+b}

= a + b

⇒

\frac{1}{1+c}

+ c = a + b + c

⇒ c (c + 1) = a + b + c

⇒

\frac{a}{a+b+c}

+

\frac{b}{a+b+c}

+

\frac{c}{a+b+c}

=

\frac{a+b+c}{a+b+c}

+

b/{a+b+c}

+

c/{a+b+c}

=

{a+b+c}/{a+b+c}

= 1

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