SSC CGL 2011 Question Paper Tier 1 - Shift 2 (June 19th)

Question 139

$$\frac{4\sqrt{3} + 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} = a + b\sqrt{6}$$

Solution

First multiplying numerator and denominator with $$\sqrt{2}$$
than it will be $$\frac{4\sqrt{6} + 10}{4\sqrt{6} + 6}$$
now rationalising it with $${4\sqrt{6} - 6}$$, then it will be $$\frac{16\sqrt{6} + 36}{60}$$
or $$\frac{6}{10} + \frac{4\sqrt{6}}{15}$$

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