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SSC CGL 2024 PYQ — Question 72

SSC CGL - 13 Sep 2024 - Shift 3 • Quantitative Aptitude • Q72

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Let a + b = 1, then the value of (1a2+1b2+2ab)({1\over a^2} + {1\over b^2} + {2\over ab}) is:

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Correct answer: A

Explanation

The correct answer is Option 1 i.e. 1a2b21\over a^2 b^2.

(1a2+1b2+2ab)({1\over a^2} + {1\over b^2} + {2\over ab})

⇒ b2+a2+2aba2b2{b^2+a^2+2ab\over a^2b^2}

⇒ (b+a)2a2b2{(b+a)^2\over a^2b^2}  [(b + a) = 1]

⇒ 1a2b2{1\over a^2b^2}