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SSC CHSL 2021 PYQ — Question 60

SSC CHSL - 6 Aug 2021 - Shift 3 • Quantitative Aptitude • Q60

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If x+81x=18x+\frac{81}{x}=18 where x > 0, then the value of x2+162x2x^2+\frac{162}{x^2} is:

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 83

Explanation:

According to the given question,

x+81x=18x+\frac{81}{x}=18 ⇒ x2 + 81 = 18x

⇒ x2 - 18x + 81 = 0

⇒ x2 - 9x - 9x + 81 = 0

⇒ x(x - 9) - 9(x - 9) = 0

⇒ (x - 9)(x - 9) = 0

⇒ x = 9; x = 9

Now, the required value = 92+162929^2+\frac{162}{9^2} = 81+1628181+\frac{162}{81} = 81 + 2 = 83.

Hence the required answer is 83.