Question

The value of $\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}$ $+\left(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}\right) \div \sqrt{\frac{16}{81}}$

• A. $\frac{69}{16}$
• B. $\frac{31}{3}$
• C. $\frac{54}{7}$
• D. $\frac{39}{16}$

Answer 1

Answer: (C)

$ \sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}} +\left(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}\right) \div \sqrt{\frac{16}{81}} $
$ =(\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+15}}}}) +\left(\frac{15}{27}-\frac{5}{12}\right) \div\left(\frac{4}{9}\right) $
$\left[\mathrm{As}, \sqrt{225}=\sqrt{3^2 \times 5^2} \sqrt{729}=\sqrt{3^6}=3^3=27\right]$
$ =(\sqrt{10+\sqrt{25+\sqrt{108+13}}})+\left(\frac{20-15}{36}\right) \times \frac{9}{4} $
$ {\left[\text { As, } \sqrt{169}=\sqrt{13^2}=13\right]} $
$ =\sqrt{10+\sqrt{25+11}}+\frac{5}{16}\left[\text { As, } \sqrt{121}=\sqrt{11^2}=11\right] $
$ =\sqrt{10+\sqrt{36}}+\frac{5}{16} $
$ =\sqrt{10+6}+\frac{5}{16} $
$ =4+\frac{5}{16} $
$ =\frac{69}{16} $