Practice MCQ Questions and Answer on Simplification

Solution:

 $$\eqalign{ & \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} - \frac{1}{5} \times \left( {\frac{7}{8} - \frac{5}{4}} \right)} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} - \frac{1}{5} \times \left( {\frac{{7 - 10}}{8}} \right)} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} + \frac{1}{5} \times \frac{3}{8}} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} + \frac{3}{{40}}} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{{32 + 3}}{{40}}} \right] \cr & = \frac{{33}}{{40}} + \frac{7}{{40}} \cr & = \frac{{40}}{{40}} \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & \frac{{0.75}}{{10}} + \frac{{0.31}}{{100}} + \frac{{0.09}}{{1000}} \cr & = \frac{{75}}{{1000}} + \frac{{3.1}}{{1000}} + \frac{{0.09}}{{1000}} \cr & = \frac{{78.19}}{{10000}} \cr & = 0.07819 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{price}}\,{\text{of a}}\,{\text{saree}}\,{\text{and}}\,{\text{a}}\,{\text{shirt}} \cr & \,{\text{be}}\,Rs.\,x\,{\text{and}}\,Rs.\,y\,{\text{respectively}}. \cr & {\text{Then,}}\,2x + 4y = 1600\,....(i) \cr & \,\,\,\,\,\,\,\,\,\,\,{\text{and}}\,x + 6y = 1600\,....(ii) \cr & {\text{Divide}}\,{\text{equation}}\,(i)\,by\,2, \cr & {\text{we}}\,{\text{get}}\,{\text{the}}\,{\text{below}}\,{\text{equation}}. \cr & = > \,x + 2y = 800\,\, - - - (iii) \cr & {\text{Now}}\,{\text{subtract}}\,(iii)\,{\text{from}}\,(ii) \cr & x\,\, + \,\,\,\,6y\,\, = \,\,1600\,\,\,( - ) \cr & x\,\, + \,\,\,\,2y\,\, = \,\,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4y\,\, = \,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & {\text{Therefore}},\,y = 200 \cr & {\text{Nowapply}}\,{\text{value}}\,{\text{of}}\,y\,{\text{in}}\,(iii) \cr & = > x + 2 \times 200 = 800 \cr & = > x + 400 = 800 \cr & {\text{Therefore}}\,x = 400 \cr & {\text{Solving}}\,(i)\,{\text{and}}\,(ii)\,{\text{we}}\,{\text{get}}\, \cr & x = 400,\,y = 200 \cr & \therefore {\text{Cost}}\,{\text{of}}\,{\text{12}}\,{\text{shirts}} \cr & = Rs.\,\left( {12 \times 200} \right) \cr & = Rs.\,2400 \cr} $$

Solution:

 Let the number of men in the 1st, 2nd, 3rd and 4th battalions be x, y, z and t respectively. Then, $$\eqalign{ & \frac{1}{2}x = \frac{2}{3}y = \frac{3}{4}z = \frac{4}{5}t \cr & \Rightarrow x = \frac{4}{3}y, \cr & \,\,\,\,\,\,\,\,z = \frac{8}{9}y, \cr & \,\,\,\,\,\,\,\,\,t = \frac{5}{6}y{\text{ }} \cr & {\text{Now,}} \cr & x + y + z + t = 7300 \cr & \Rightarrow \frac{4}{3}y + y + \frac{8}{9}y + \frac{5}{6}y = 7300 \cr & \Rightarrow \frac{{24y + 18y + 16y + 15y}}{{18}} = 7300 \cr & \Rightarrow 73y = 7300 \times 18 \cr & \Rightarrow y = 1800 \cr} $$

Solution:

 $$\eqalign{ & x = \frac{{5\frac{3}{4} - \frac{3}{7} \times 15\frac{3}{4} + 2\frac{2}{{35}} \div 1\frac{{11}}{{25}}}}{{\frac{3}{4} \div 5\frac{1}{4} + 5\frac{3}{5} \div 3\frac{4}{{15}}}} \cr & x = \frac{{\frac{{23}}{4} - \frac{3}{7} \times \frac{{63}}{4} + \frac{{72}}{{35}} \div \frac{{36}}{{25}}}}{{\frac{3}{4} \div \frac{{21}}{4} + \frac{{28}}{5} \div \frac{{49}}{{15}}}} \cr & x = \frac{{\frac{{23}}{4} - \frac{{27}}{4} + \frac{{10}}{7}}}{{\frac{1}{7} + \frac{{12}}{7}}} \cr & x = \frac{{ - \frac{4}{4} + \frac{{10}}{7}}}{{\frac{{13}}{7}}} \cr & x = \frac{{\frac{{ - 7 + 10}}{7}}}{{\frac{{13}}{7}}} \cr & x = \frac{3}{{13}} \cr & x + y = \frac{7}{{13}} \cr & y = \frac{7}{{13}} - \frac{3}{{13}} \cr & y = \frac{4}{{13}} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & 11760 \times \frac{1}{3} \times \frac{1}{5} \times \frac{3}{8} \cr & = \frac{{1176}}{4} \cr & = 294 \cr} $$

Solution:

 $$\eqalign{ & 0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \cr & = \frac{{46 - 4}}{{90}} + \frac{{723 - 7}}{{990}} - \frac{{39 - 3}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{36}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{28}}{{90}} \cr & = \frac{{14}}{{90}} + \frac{{716}}{{990}} \cr & = \frac{{154 + 716}}{{990}} \cr & = \frac{{870}}{{990}} \cr & = 0.\overline {87} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & \frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}} \cr & {\text{let }}a = 0.73{\text{ and }}b = 0.27 \cr} $$  $$ = \frac{{\left( {0.73 + 0.27} \right)\left[ {{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27} \right]}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}}$$ $$\eqalign{ & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.73 + 0.27 \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\, \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + \sqrt {49} } } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + 7} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {36} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + 6} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {25} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + 5} } \cr & \Rightarrow \sqrt {5 + \sqrt {16} } \cr & \Rightarrow \sqrt {5 + 4} \cr & \Rightarrow \sqrt 9 \cr & \Rightarrow 3 \cr} $$