Question

If $\frac{\frac{1}{\sqrt{9}}-\frac{1}{\sqrt{11}}}{\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{11}}}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$ then x equals

• A. 2
• B. 3
• C. 10
• D. $\frac{1}{10}$

Answer 1

Answer: (A)

2

$\frac{\frac{1}{\sqrt{9}}-\frac{1}{\sqrt{11}}}{\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{11}}}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{\frac{\sqrt{11}-\sqrt{9}}{\sqrt{99}}}{\frac{\sqrt{11}+\sqrt{9}}{\sqrt{99}}}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}-\sqrt{9}}\times \frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{11+9-2\sqrt{11}\sqrt{9}}{2}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\frac{20-2\sqrt{99}}{2}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{2(10-\sqrt{99})}{2}\times \frac{10+\sqrt{99}}{x}=\frac{1}{2}$

$\Rightarrow \frac{100-99}{x}=\frac{1}{2}$

$\Rightarrow x=2$