CAT 2003 - Retake — QA Question 9

$\begin{array}{rl}x& =\frac{n^2+2\sqrt{n}(n+4)+16}{n+4\sqrt{n}+4}\\ & =\frac{n^2+2n\sqrt{n}+8\sqrt{n}+16}{(\sqrt{n}+2{)}^2}\\ & =\frac{n\sqrt{n}(\sqrt{n}+2)+8(\sqrt{n}+2)}{(\sqrt{n}+2{)}^2}\\ & =\frac{(n\sqrt{n}+8)(\sqrt{n}+2)}{(\sqrt{n}+2{)}^2}\\ \therefore x& =\frac{n\sqrt{n}+8}{\sqrt{n}+2}\end{array}$

When n = 36,

$x=\frac{36\times 6+8}{6+2}=\frac{8(27+1)}{8}=28$

When n = 72,

$x=\frac{432\sqrt{2}+8}{6\sqrt{2}+2}=\frac{4(54\sqrt{2}+1)}{3\sqrt{2}+1}=\frac{309.42}{5.24}\approx 59$

Only the option 3 includes both these values.

Hence, option (c).