Simplification — Part 2 MCQs with Explanations

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

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

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

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{8}}\frac{1}{2} - \left[ {3\frac{1}{4} + \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{{9 - 2 - 1}}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2} \times 1} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{{5 - 2}}{4}} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \frac{3}{4}} \right] \cr & \Rightarrow \frac{{17}}{2} - \frac{{16}}{4} \cr & \Rightarrow \frac{{34 - 16}}{4} \cr & \Rightarrow \frac{9}{2} \cr & \Rightarrow 4\frac{1}{2} \cr} $$

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

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

$$\eqalign{ & {\frac{{41}}{6} \times \frac{{16}}{3} + \frac{{53}}{3} \times \frac{9}{2}} \cr & = {\frac{{328}}{9} + \frac{{159}}{2}} \cr & = \frac{{656 + 1431}}{{18}} \cr & = \frac{{2087}}{{18}} \cr & = 115\frac{{17}}{{18}} \cr} $$

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

$$\eqalign{ & x = \frac{1}{{2 + \frac{1}{2}}} = \frac{2}{5} \cr & \Rightarrow \frac{1}{x} = \frac{5}{2} \cr} $$

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

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

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

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{81}}{8}\,of\,\frac{{12}}{{15}} \div \frac{{35}}{{36}}\,of\,\frac{{20}}{{49}} \cr & = \frac{{81}}{{10}} \div \frac{{25}}{{63}} \cr & = \frac{{81}}{{10}} \times \frac{{63}}{{25}} \cr & = \frac{{5103}}{{250}} \cr & = 20\frac{{103}}{{250}}\, \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {0.003} \right)}^2} + {{\left( {0.021} \right)}^2} + {{\left( {0.0065} \right)}^2}}}} \cr & = \sqrt {\frac{{0.0009 + 0.0441 + 0.004225}}{{0.000009 + 0.000441 + 0.00004225}}} \cr & = \sqrt {\frac{{0.049225}}{{0.00049225}}} \cr & = \sqrt {100} \cr & = 10 \cr} $$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{6}{{5 - \frac{5}{3}}} \div \frac{{4 - \frac{2}{{4 - \frac{1}{2}}}}}{{5 - \frac{3}{2}}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{2}{3} \times \frac{1}{2}} \right\}{\text{ }} \cr & = \frac{6}{{\left( {\frac{{10}}{3}} \right)}} \div \frac{{4 - \frac{2}{{\left( {\frac{7}{2}} \right)}}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{1}{3}} \right\} \cr & = \frac{{6 \times 3}}{{10}} \div \frac{{4 - \frac{{2 \times 2}}{7}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \times 1 \cr & = \frac{9}{5} \div \frac{{\left( {4 - \frac{4}{7}} \right)}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \cr & = \frac{9}{5} \div \left( {\frac{{24}}{7} \times \frac{2}{7}} \right) - \frac{2}{5} \cr & = \frac{9}{5} \times \frac{{49}}{{48}} - \frac{2}{5} \cr & = \frac{{147}}{{80}} - \frac{2}{5} \cr & = \frac{{147 - 32}}{{80}} \cr & = \frac{{115}}{{80}} \cr & = \frac{{23}}{{16}} \cr & = 1\frac{7}{{16}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{I = }}\frac{3}{5} \div \frac{5}{6} \cr & {\text{I = }}\frac{3}{5} \times \frac{6}{5} = \frac{18}{{25}} \cr & \cr & {\text{II = 3}} \div \left[ {\frac{4}{5} \times \frac{1}{6}} \right] \cr & {\text{II = 3}} \times \frac{{30}}{4} \cr & {\text{II = }}\frac{{45}}{2} \cr & \cr & {\text{III = }}\left[ {3 \div \left( {4 \div 5} \right)} \right] \div 6\, \cr & {\text{III = }}\left[ {3 \times \frac{5}{4}} \right] \times \frac{1}{6}\, \cr & {\text{III = }}\frac{5}{8} \cr & \cr & {\text{IV = 3}} \div {\text{4}}\left( {5 \div 6} \right) \cr & {\text{IV = 3}} \div {\text{4}} \times \frac{5}{6} \cr & {\text{IV = 3}} \times \frac{6}{{20}} \cr & {\text{IV = }}\frac{9}{{10}} \cr } $$

$$\eqalign{ & {\text{ = }}\, - 7m - \left[ {3n - \left\{ {8m - \left( {4n - 10m} \right)} \right\}} \right]\,\,\, \cr & = \, - 7m - \left[ {3n - \left\{ {8m - 4n + 10m} \right\}} \right]\, \cr & = - 7m - \left[ {3n - 18m + 4n} \right] \cr & = - 7m - \left[ {7n - 18m} \right] \cr & = - 7m - 7n + 18m\,\,\, \cr & = \,11m - 7n\,\, \cr} $$

$$\eqalign{ & \frac{2}{a} + \frac{1}{b} \cr & = \frac{{2b + a}}{{ab}} \cr & = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$

According to question, $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}}$$ As we know that $${a^3} + {b^3} + {c^3} - 3abc = $$ $$\left( {a + b + c} \right)$$ $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$ $$ = \frac{{{{(0.9)}^3} + {{(0.2)}^3} + {{(0.3)}^3} - 3 \times 0.9 \times 0.2 \times 0.3}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$ $$ = \frac{{\left( {0.9 + 0.2 + 0.3} \right)\left[ {{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9} \right]}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$ $$\eqalign{ & = 0.9 + 0.2 + 0.3 \cr & = 1.4 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}} \div \frac{4}{7}\left( {\frac{2}{5} + \frac{3}{{10}}} \right){\text{of }}\frac{{\frac{1}{2} + \frac{1}{3}}}{{\frac{1}{2} - \frac{1}{3}}} \cr & = \frac{{\frac{3}{2}}}{{\frac{1}{2}}} \div \frac{4}{7}\left( {\frac{{20 + 15}}{{50}}} \right) \times \frac{{\frac{5}{6}}}{{\frac{1}{6}}} \cr & = \frac{3}{2} \times \frac{2}{1} \div \frac{4}{7} \times \frac{{35}}{{50}} \times \frac{5}{6} \times \frac{6}{1} \cr & = \frac{3}{2} \times \frac{2}{1} \div 2 \cr & = \frac{3}{2} \times 2 \div \frac{1}{2} \cr & = \frac{3}{2} \cr} $$

Given expression, $$ = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ {2{a^3} + \left( { - 7{a^3}} \right)} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right]$$ $$\eqalign{ & = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ { - 5{a^3}} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {3{a^3} + 4{b^3}{\text{ + 5}}{a^3}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {13{a^3} - 3{b^3}} \right] \cr & = 2{a^3} - 13{a^3} + 3{b^3} \cr & = - 11{a^3} + 3{b^3} \cr} $$

$$\left[ {1 + \frac{1}{{x + 1}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 2}}} \right]$$ $$\,\left[ {1 + \frac{1}{{x + 3}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 4}}} \right]$$ $$ = \left[ {\frac{{\left( {x + 1} \right) + 1}}{{x + 1}}} \right]$$ $$\left[ {\frac{{\left( {x + 2} \right) + 1}}{{x + 2}}} \right]$$ $$\left[ {\frac{{\left( {x + 3} \right) + 1}}{{x + 3}}} \right]$$ $$\left[ {\frac{{\left( {x + 4} \right) + 1}}{{x + 4}}} \right]$$ $$\eqalign{ & = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left( {\frac{{x + 3}}{{x + 2}}} \right)\left( {\frac{{x + 4}}{{x + 3}}} \right)\left( {\frac{{x + 5}}{{x + 4}}} \right) \cr & = \frac{{x + 5}}{{x + 1}} \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{4}{5}{\text{ and }}\frac{b}{c} = \frac{{15}}{{16}} \cr & \Rightarrow \left( {\frac{a}{b} \times \frac{b}{c}} \right) = \left( {\frac{4}{5} \times \frac{{15}}{{16}}} \right) \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \,\frac{{{c^2} - {a^2}}}{{{c^2} + {a^2}}} \cr & = \frac{{1 - \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}}{{1 + \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{1 - {{\left( {\frac{a}{c}} \right)}^2}}}{{1 + {{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & = \frac{{\left( {\frac{7}{{16}}} \right)}}{{\left( {\frac{{25}}{{16}}} \right)}} \cr & = \frac{7}{{25}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{1}{8} + \frac{1}{{{8^2}}} + \frac{1}{{{8^3}}} + \frac{1}{{{8^4}}} + \frac{1}{{{8^5}}} \cr & \Rightarrow \frac{1}{8} + \frac{1}{{64}} + \frac{1}{{512}} + \frac{1}{{4096}} + \frac{1}{{32768}} \cr} $$ ⇒ 0.125 + 0.015625 + 0.00195313 + 0.00024414 + 0.0000305175 ⇒ 0.143

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {0.00005746} \cr & \Rightarrow \sqrt {\frac{{5746}}{{100000000}}} \cr & \Rightarrow \frac{{75.8}}{{10000}} \cr & \Rightarrow 0.00758 \cr} $$