Exams › SSC CGL (Prelims) › General

Suppose $p = \sqrt{3} + \sqrt{7}$, $q = \sqrt{8} + \sqrt{2}$, and $r = 2\sqrt{3} + 2$. Which of the following is true?

1. p < q < r
2. r < q < p
3. q < p < r
4. p < r < q

Correct answer: q < p < r

Solution

Approximating gives $q = \sqrt{8}+\sqrt{2} = 2\sqrt{2}+\sqrt{2} = 3\sqrt{2} \approx 4.24$, $p \approx 1.73+2.65 = 4.38$, and $r \approx 3.46+2 = 5.46$. Hence the order is $q < p < r$.

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