Problem Solving

[GMAT math practice question]

$$If\ a\ =\sqrt{\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{2}}},\ b=\ \sqrt{\sqrt{\frac{1}{5}}-\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}},\ c=\sqrt{-\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}},\ which\ of\ the\ following\ is\ true?$$

A. a < c < b
B. a < b < c
C. c < b < a
D. c < a < b
E. b < a < c

Since $$\sqrt{\frac{1}{5}}<\sqrt{\frac{1}{3}}<\sqrt{\frac{1}{2}}$$

We have $$\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{2}}<\sqrt{\frac{1}{5}}-\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}<\ -\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}$$

So $$\sqrt{\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{2}}}<\sqrt{\sqrt{\frac{1}{5}}-\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}}<\sqrt{-\sqrt{\frac{1}{5}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}}$$

Thus c < b < a.

Therefore, C is the answer.

Answer: C