$$\eqalign{ & {\text{Given expression}} \cr & \frac{{24}}{5} \div \frac{{32}}{5} \cr & = \frac{{24}}{5} \times \frac{5}{{32}} \cr & = \frac{3}{4} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\frac{1}{3} + \frac{1}{4}\left[ {\frac{2}{5} - \frac{1}{2}} \right]}}{{1\frac{2}{3}\,{\text{of }}\frac{3}{4} - \frac{3}{4}\,{\text{of }}\frac{4}{5}}} \cr & \Rightarrow \frac{{\frac{1}{3} + \frac{1}{4}\left[ {\frac{{4 - 5}}{{10}}} \right]}}{{\frac{5}{3}\, \times \frac{3}{4} - \frac{3}{4}\, \times \frac{4}{5}}} \cr & \Rightarrow \frac{{\frac{1}{3} - \frac{1}{4} \times \frac{1}{{10}}}}{{\frac{5}{4} - \frac{3}{5}}} \cr & \Rightarrow \frac{{\frac{{40 - 3}}{{120}}}}{{\frac{{25 - 12}}{{20}}}} \cr & \Rightarrow \frac{{37}}{{120}} \times \frac{{20}}{{13}} \cr & \Rightarrow \frac{{37}}{{78}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{0.3555 \times 0.5555 \times 2.025}}{{0.225 \times 1.7775 \times 0.2222}}\, \cr & = \frac{{3555 \times 5555 \times 2025}}{{225 \times 17775 \times 2222}}\, \cr & = 4.5 \cr} $$

$$\eqalign{ & {\text{Green portion}} \cr & = \left[ {1 - \left( {\frac{1}{{10}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{40}} + \frac{1}{{50}} + \frac{1}{{60}}} \right)} \right] \cr & = \left[ {1 - \frac{1}{{10}}\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}} \right)} \right] \cr & = 1 - \frac{1}{{10}} \times \frac{{147}}{{60}} \cr & = 1 - \frac{{147}}{{600}} \cr & = \frac{{453}}{{600}} \cr & {\text{let the length of the pole be }}x{\text{ metres}} \cr & {\text{Then, }}\frac{{453}}{{600}}x = 12.08 \cr & \Leftrightarrow x = \left( {\frac{{12.08 \times 600}}{{453}}} \right) = 16 \cr} $$

Let the capacity of the drum be x liters
Then,
$$\eqalign{ & \frac{3}{4}x - \frac{7}{{12}}x = 30 \cr & \Leftrightarrow 9x - 7x = 12 \times 30 \cr & \Leftrightarrow 2x = 360 \cr & \Leftrightarrow x = 180 \cr} $$

Suppose the taller tree grows x feet in 1 year.
Then the shorter tree grows
$$\eqalign{ & = \frac{{3x}}{5}{\text{ft in 1 year}} \cr & \therefore 4x + 4 \times \frac{3}{5}x = 8 \cr & \Leftrightarrow 4x + \frac{{12}}{5}x = 8 \cr & \Leftrightarrow \frac{{32}}{5}x = 8 \cr & \Leftrightarrow x = \frac{{8 \times 5}}{{32}} = \frac{5}{4} \cr & {\text{Growth of shorter tree in}} \cr & {\text{2 years = }}\left( {2 \times \frac{{3x}}{5}} \right){\text{ft}} \cr & {\text{ = }}\left( {2 \times \frac{3}{5} \times \frac{5}{4}} \right){\text{ft}} \cr & {\text{ = }}\frac{3}{2}{\text{ft}} \cr & {\text{ = 1}}\frac{1}{2}{\text{ft}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{28 + 14 + 7 + 4 + 2 + 1}}{{28}} \cr & = \frac{{56}}{{28}} \cr & = 2 \cr} $$

$$\eqalign{ & \frac{1}{x} = 4 - \left( {\frac{1}{3} + \frac{1}{2}} \right) \cr & \,\,\,\,\,\, = 4 - {\frac{{2 + 3}}{6}} \cr & \,\,\,\,\,\, = 4 - \frac{5}{6} \cr & \,\,\,\,\,\, = \frac{{24 - 5}}{6} \cr & \,\,\,\,\,\, = \frac{{19}}{6} \cr & \Rightarrow x = \frac{6}{{19}} \cr} $$

$$\eqalign{ & {\text{I = }}\frac{3}{4} \div \frac{5}{6} \cr & \,\,\,\, = \frac{3}{4} \times \frac{6}{5} \cr & \,\,\,\, = \frac{9}{{10}} \cr & {\text{II = 3}} \div \left[ {\left( {4 \div 5} \right) \div 6} \right] \cr & \,\,\,\,\,\, = 3 \div \left( {\frac{4}{5} \times \frac{1}{6}} \right) \cr & \,\,\,\,\,\, = 3 \div \frac{4}{{30}} \cr & \,\,\,\,\,\, = 3 \times \frac{{30}}{4} \cr & \,\,\,\,\,\, = \frac{{45}}{2} \cr & {\text{III = }}\left[ {3 \div \left( {4 \div 5} \right)} \right] \div {\text{6}} \cr & \,\,\,\,\,\,\,\,{\text{ = }}\left[ {3 \div \frac{4}{5}} \right] \div 6 \cr & \,\,\,\,\,\,\,\, = \left[ {3 \times \frac{5}{4}} \right] \div 6 \cr & \,\,\,\,\,\,\,\, = \frac{{15}}{4} \times \frac{1}{6} \cr & \,\,\,\,\,\,\,\, = \frac{5}{8} \cr & {\text{IV = 3}} \div 4 \div \left( {5 \div 6} \right) \cr & \,\,\,\,\,\,\,\, = 3 \div 4 \div \frac{5}{6} \cr & \,\,\,\,\,\,\,\, = \frac{3}{4} \times \frac{6}{5} \cr & \,\,\,\,\,\,\,\, = \frac{9}{{10}} \cr & {\text{So, I and IV are equal}}{\text{.}} \cr} $$

According to the question,
As we know that the square of 252 is which that is near to the value of 63522.
∴ 63522 - x = 63504
⇒ x = 18

$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {0.00060516} {\text{ }} \cr & \Rightarrow \sqrt {\frac{{60516}}{{100000000}}} \cr & \Rightarrow \frac{{246}}{{10000}} \cr & \Rightarrow 0.0246 \cr} $$

$$\eqalign{ & {\text{Given equation is :}} \cr & \frac{{\left( {5x + 1} \right)}}{x} \times \frac{{\left( {4y + 3} \right)}}{4} = 20 \cr & \Leftrightarrow \left( {5x + 1} \right)\left( {4y + 3} \right) = 80x \cr & {\text{Clearly, }}x = 3{\text{ and}} \cr & {\text{ }}y = 3{\text{ satisfy}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {248 + \sqrt {52 + \sqrt {144} } } \cr & \Rightarrow \sqrt {248 + \sqrt {52 + 12} } \cr & \Rightarrow \sqrt {248 + \sqrt {64} } \cr & \Rightarrow \sqrt {256} \cr & \Rightarrow \pm 16 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{5}{{3 + \frac{3}{{1 - \frac{2}{3}}}}}\,\, \cr & \Rightarrow \frac{5}{{3 + \frac{3}{{\frac{1}{3}}}}}\, \cr & \Rightarrow \frac{5}{{3 + 9}} \cr & \Rightarrow \frac{5}{{12}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}}{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}} \cr & {\text{let ,}}1 + \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{111}}{{101}} = a \cr & \,\,\,\,\,\,\,\,1 - \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{91}}{{101}} = b \cr & \Rightarrow \frac{{{a^2} - {b^2}}}{{a + b}}\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{a + b}} \cr & \Rightarrow \left( {a - b} \right) \cr & \Rightarrow \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow \frac{{20}}{{101}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{5}{2}of\frac{3}{4} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \frac{3}{2}\left[ {\frac{2}{3} - \frac{1}{3}} \right] \cr & = \frac{5}{2}of\frac{3}{4} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \left( {\frac{3}{2} \times \frac{1}{3}} \right) \cr & = \frac{{15}}{8} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \frac{1}{2} \cr & = \frac{{15}}{8} \times \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times 2 \cr & = \frac{5}{8} + 1 \cr & = \,1\frac{5}{8} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} \cr & \Rightarrow \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} \cr & \Rightarrow \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}} \cr & \Rightarrow 6.4 + 0.64 + 0.064 + 0.0064 \cr & \Rightarrow 7.11 \cr} $$

$$\eqalign{ & {\text{Required differnce}} \cr & {\text{ = }}\frac{{19}}{{16}} - \frac{{16}}{{19}} = \frac{{{{19}^2} - {{16}^2}}}{{304}} \cr & = \frac{{\left( {19 + 16} \right)\left( {19 - 16} \right)}}{{304}} \cr & = \frac{{35 \times 3}}{{304}} \cr & = \frac{{105}}{{304}} \cr} $$

$$\eqalign{ & \,a = \left( {4 \div 3} \right) \div {\text{3}} \div {\text{4}} \cr & \,\,\,\,\,\,{\text{ = }}\frac{4}{3} \times \frac{1}{3} \times \frac{1}{4} \cr & \,\,\,\,\,\, = \frac{1}{9} \cr & b = 4 \div \left( {3 \div 3} \right) \div {\text{4}} \cr & \,\,\,\,\,{\text{ = 4}} \div 1 \div 4 \cr & \,\,\,\,\, = 1 \cr & c = 4 \div 3 \div \left( {3 \div 4} \right) \cr & \,\,\,\,\, = 4 \div 3 \div \frac{3}{4} \cr & \,\,\,\,\, = \frac{4}{3} \times \frac{4}{3} \cr & \,\,\,\,\, = \frac{{16}}{9} \cr & {\text{Clearly, c is the greatest}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {0.01} + \sqrt {0.81} + \sqrt {1.21} + \sqrt {0.0009} \cr & = 0.1 + 0.9 + 1.1 + 0.03 \cr & = 2.13 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {\frac{{0.342 \times 0.684}}{{0.000342 \times 0.000171}}} \cr & = \sqrt {\frac{{342 \times 684 \times 1000000}}{{342 \times 171}}} \cr & = \sqrt {4 \times 1000000} \cr & = 2 \times 1000 \cr & = 2000 \cr} $$

$$\eqalign{ & {\text{According to the question}} \cr & \frac{5}{{1\frac{7}{8}{\text{of 1}}\frac{1}{3}}} \times \frac{{2\frac{1}{{10}}}}{{3\frac{1}{2}}}{\text{ of 1}}\frac{1}{4} \cr & \Rightarrow \frac{5}{{\frac{{15}}{8} \times {\text{ }}\frac{4}{3}}} \times {\text{ }}\frac{{21}}{{10}} \times \frac{2}{7} \times \frac{5}{4} \cr & \Rightarrow 2 \times \frac{3}{4} \cr & \Rightarrow \frac{3}{2} \cr & \Rightarrow 1\frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{225}}{{836}} \times \frac{{152}}{{245}} \div 1\frac{{43}}{{77}} \cr & = \frac{{225}}{{836}} \times \frac{{152}}{{245}} \div \frac{{120}}{{77}} \cr & = \frac{{225}}{{836}} \times \frac{{152}}{{245}} \times \frac{{77}}{{120}} \cr & = \frac{3}{{28}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{\left( { - \frac{2}{3} - \frac{1}{2}} \right) + \left( {\frac{4}{5} + \frac{1}{5}} \right) + \left( {\frac{3}{4} - \frac{1}{2}} \right)}}{{\left( {\frac{2}{3} - \frac{4}{3} + \frac{1}{3}} \right) - \left( {\frac{1}{5} + \frac{4}{5}} \right) + \frac{1}{2}}} \cr & = \frac{{ - 1 + 1 + \frac{1}{4}}}{{ - \frac{1}{3} - 1 + \frac{1}{2}}} \cr & = \frac{{\frac{1}{4}}}{{\frac{{ - 2 - 6 + 3}}{6}}} \cr & = \frac{{\frac{1}{4}}}{{ - \frac{5}{6}}} \cr & = \frac{1}{4} \times \left( { - \frac{6}{5}} \right) \cr & = - \frac{3}{{10}} \cr} $$

$$\eqalign{ & \frac{3}{2} \times \frac{{11}}{5} \div \left( {\frac{{25}}{{44}} \times \frac{{11}}{5}} \right) \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \div \frac{5}{4} \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \times \frac{4}{5} \times \frac{{15}}{{33}} \cr & = \frac{6}{5} \cr & = 1\frac{1}{5} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{\sqrt {80} - \sqrt {112} }}{{\sqrt {45} - \sqrt {63} }} \cr & \Rightarrow \frac{{4\sqrt 5 - 4\sqrt 7 }}{{3\sqrt 5 - 3\sqrt 7 }} \cr & \Rightarrow \frac{{4\left( {\sqrt 5 - \sqrt 7 } \right)}}{{3\left( {\sqrt 5 - \sqrt 7 } \right)}} \cr & \Rightarrow \frac{4}{3} \cr & \Rightarrow 1\frac{1}{3} \cr} $$

$$\eqalign{ & {\text{According to the question}} \cr & x + y = 22\,......(i) \cr & {x^2} + {y^2} = 404\,......(ii){\text{ }} \cr & \therefore {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \cr & \Rightarrow {\left( {22} \right)^2} = 404 + 2xy \cr & \Rightarrow 484 = 404 + 2xy \cr & \Rightarrow 2xy = 80 \cr & \Rightarrow xy = 40 \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & \frac{5}{3} \div \frac{2}{7} \times \frac{x}{7} = \frac{5}{4} \times \frac{2}{3} \div \frac{1}{6} \cr & {\text{Then,}} \cr & \frac{5}{3} \times \frac{7}{2} \times \frac{x}{7} = \frac{5}{4} \times \frac{2}{3} \times 6 \cr & \Leftrightarrow \frac{5}{6}x = 5 \cr & \Leftrightarrow x = \left( {\frac{{5 \times 6}}{5}} \right) \cr & \Leftrightarrow x = 6 \cr} $$

According to the question,
$$\eqalign{ & \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2\left( {5 - 3} \right) + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2 * 2 + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 4 + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3 * 3 - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {11 - 10} \right]}}{4} \cr & = 25 - \frac{5}{4} \cr & = \frac{{100 - 5}}{4} \cr & = \frac{{95}}{4} \cr & = 23.75 \cr} $$

$$\eqalign{ & {\text{Given Expression,}} \cr & {\frac{3}{7} \times 455 + \frac{5}{8} \times 456} \cr & = \left( {3 \times 65 + 5 \times 57} \right) \cr & = 195 + 285 \cr & = 480 \cr} $$