ABC26GN1394 · Square Roots and Cube Roots

Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:

$\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}=$ ?
(a)$2 \sqrt{6}$
(b)2
(c)$6 \sqrt{2}$
(d)$\frac{2}{\sqrt{6}}$
Answer
Answer (as printed): B
Explanation
$\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}=\frac{\sqrt{4 \times 6}+\sqrt{36 \times 6}}{\sqrt{16 \times 6}}=\frac{2 \sqrt{6}+6 \sqrt{6}}{4 \sqrt{6}}=\frac{8 \sqrt{6}}{4 \sqrt{6}}=2$.

Explanation as extracted from the printed page; notation may be imperfect.

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