Question : Evaluate $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, given that $\sqrt{6}=2.45$.
Option 1: 7.7
Option 2: 9.9
Option 3: 8.8
Option 4: 6.6
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Correct Answer: 9.9
Solution :
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Rationalizing the denominator,
= $\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
= $\frac{(3+2+2\sqrt{6})}{(3-2)}$
= $5+2\sqrt6$
= 5 + (2 × 2.45)
= 5 + 4.9
= 9.9
Hence, the correct answer is 9.9.
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