Question

Which of the following irrational number lies between $\frac{3}{4}$ and $\frac{7}{8}$?

• A. $\sqrt{\frac{38}{64}}$
• B. $\sqrt{\frac{100}{128}}$
• C. $\frac{6}{8}$
• D. None of these

Answer 1

The correct option is A $\sqrt{\frac{38}{64}}$

Option $A$ $\Rightarrow \sqrt{\frac{38}{64}}=\frac{\sqrt{38}}{8}\Rightarrow$ an irrational number

$\because 6<\sqrt{38}<7$

$\therefore \frac{6}{8}<\sqrt{\frac{38}{64}}<\frac{7}{8}$

$\therefore \frac{3}{4}<\sqrt{\frac{38}{64}}<\frac{7}{8}$.

Option $B$ $\Rightarrow \sqrt{\frac{100}{128}}\Rightarrow \sqrt{\frac{10\times 10}{2^7}}=\frac{10}{8\sqrt{2}}=\frac{5\sqrt{2}}{8}$

But $5\sqrt{2}>7$

$\therefore \sqrt{\frac{100}{128}}>\frac{7}{8}$

Option $C$ $\Rightarrow \frac{6}{8}\Rightarrow$ rational number

Hence, only option $A$ is the right answer.