\(3\sqrt{5}+\sqrt{125} = 17.88\)

\(\Rightarrow 3\sqrt{5}+\sqrt{25\times5} = 17.88\)

\(\Rightarrow 3\sqrt{5}+5\sqrt{5} = 17.88\)

\(\Rightarrow8\sqrt{5} = 17.88\)

\(\Rightarrow \sqrt{5}=2.235\)

\(\therefore \sqrt{80}+6\sqrt{5} =\sqrt{16\times5}+6\sqrt{5}\)

\(= 4\sqrt{5}+6\sqrt{5}\)

\(= 10\sqrt{5} = (10\times2.235)=22.35\)

\( \sqrt{4a^{2}-4a+1}+3a = \sqrt{(a^{2})+(2a)^{2}-2\times1\times2a}+3a\)

= \( \sqrt{(1-2a)^{2}}+3a\)

   = (1 - 2a) + 3a

   = (1 + a)

   = (1 + 0.1039)

   = 1.1039

\(x =\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)}{(3-1)}^{2} = \frac{3+1+2\sqrt{3}}{2} = 2+\sqrt{3}.\)

\(y =\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}\times\frac{(\sqrt{3}-1)}{(\sqrt{3}-1)} = \frac{(\sqrt{3}-1)}{(3-1)}^{2} = \frac{3+1-2\sqrt{3}}{2} = 2-\sqrt{3}\)

\(\therefore a^{b}+a^{b} = (2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}\)

= 2(4 + 3)

   = 14

Money collected = (59.29 x 100) paise = 5929 paise.

\(\therefore Number of members = \sqrt{5929} = 77.\)

\( \sqrt{(7+3\sqrt{5)}(7-3\sqrt{5)}}=\sqrt{(7)^{2}-(3\sqrt{5})^{2}} = \sqrt{49-45} = \sqrt{4} = 2\)

\(\frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}}+\sqrt{125} = \frac{(\sqrt{5})^{2}-20+2\sqrt{5}\times5\sqrt{5}}{2\sqrt{5}}\)

= \(\frac{5-20+50}{2\sqrt{5}}\)

= \(\frac{35}{2\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}\)

= \(\frac{35\sqrt{5}}{10}\)

= \(\frac{7\times2.236}{2}\)

= 7 x 1.118

= 7.826

Given Exspression = \( \frac{25}{11}\times\frac{14}{5}\times\frac{11}{14} = 5\)

Let \(\sqrt{0.0169\times x} = 1.3\)

Then, 0.0169x = (1.3)² = 1.69

\(\Rightarrow x= \frac{1.69}{0.0169} = 100\)

\((\sqrt{3}-\frac{1}{\sqrt{3}})^{2} = (\sqrt{3})^{2}+(\frac{1}{\sqrt{3}})^{2}-2\times\sqrt{3}\times\frac{1}{\sqrt{3}}\)

\(= 3+\frac{1}{3}-2\)

\(= 1+\frac{1}{3}\)

\(= \frac{4}{3}\)

A number ending in 8 can never be a perfect square.