SSC CGL 2021 PYQ — Question 74

SSC CGL - 16 Aug 2021 - Shift 2 • Quantitative Aptitude • Q74

If a - 12a\frac{12}{a} = 1, where a > 0, then the value of a2 + 16a2\frac{16}{a^2} is :

Show answer & explanation

Correct answer: B

Explanation

The correct answer is Option 2 i.e. 17.

Given, a - 12/a = 1

⇒ a2 - 12 = a

⇒ a2 - a - 12 = 0

⇒ a2 - 4a + 3a - 12 = 0 

⇒ a(a - 4) + 3(a - 4) = 0

⇒ a = 4, -3

but as given a > 0,

⇒ a = 4

Putting in  a2 + 16a2\frac{16}{a^2} ,

⇒ 16 + 1 = 17