Practice MCQ Questions and Answer on Simplification

Solution:

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

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:

 $$\eqalign{ & {\text{Given expression,}} \cr & {\text{ = }}\frac{{4 - 1}}{{4 \times 1}} + \frac{{9 - 4}}{{9 \times 4}} + \frac{{16 - 9}}{{16 \times 9}} + ..... \cr & = \left( {1 - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{9}} \right) + \left( {\frac{1}{9} - \frac{1}{{16}}} \right) + ..... \cr} $$   $$ = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + $$   $$\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + $$   $$\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + $$   $$.....$$ $$\eqalign{ & \therefore \,{99^{{\text{th}}}}{\text{ term of the series}} \cr & {\text{ = }}\left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right) \cr & \therefore \,\,\,{\text{Given expression,}} \cr} $$ $$\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + $$   $$\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + $$   $$\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + $$   . . . . . $$ + $$ $$\left( {\frac{1}{{{{98}^2}}} - \frac{1}{{{{99}^2}}}} \right) + $$   $$\left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right)$$ $$\eqalign{ & = \left( {1 - \frac{1}{{{{100}^2}}}} \right) \cr & = \left( {1 - \frac{1}{{10000}}} \right) \cr & = \frac{{9999}}{{10000}} \cr} $$

Solution:

 $$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr} $$

Solution:

 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {\frac{{0.25}}{{0.0009}}} \times \sqrt {\frac{{0.09}}{{0.36}}\,\,} \cr & \Rightarrow \sqrt {\frac{{25}}{9} \times 100} \times \sqrt {\frac{9}{{36}}\,\,} \cr & \Rightarrow \frac{{5 \times 10}}{3} \times \frac{3}{6} \cr & \Rightarrow \frac{{25}}{3} \cr & \Rightarrow 8\frac{1}{3} \cr} $$

Solution:

 Given, 5 - [4 - {3 - (3 - 3 - 6)}] = 5 - [4 - {3 - (-6)}] = 5 - [4 - {3 +6}] = 5 - [4 - {9}] = 5 - [4 - 9] = 5 - [-5] = 5 + 5 = 10

Solution:

 2.1 + 2.25 ÷ [63 - {7.5 × 8 + (13 - 2.5 × 5)}] = 2.1 + 2.25 ÷ [63 - {60 + 0.5}] = 2.1 + 2.25 ÷ [63 - 60.5] = 2.1 + $$\frac{{2.25}}{{2.50}}$$ = 2.1 + 0.9 = 3.0

Solution:

 As we know a3 + b3 + c3 + 3abc = (a2 + b2 + c2 - ab - bc - ca)(a + b + c) When, a + b + c = 0 Then a3 + b3 + c3 - 3abc = 0 When, x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz ⇒ x3 + y3 + z3 + 3xyz = 3xyz + 3xyz ⇒ x3 + y3 + z3 + 3xyz = 6xyz

Solution:

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

Solution:

 $$\eqalign{ & {\text{A}} = 3\frac{1}{4} \times 4\frac{1}{4} \div 34 - \frac{{47}}{{32}} + \frac{{47}}{{16}} \cr & {\text{B}} = 2\frac{1}{2} + 5\frac{1}{2} \div 55 - \frac{{11}}{{10}} \cr & \therefore {\text{A}} - {\text{B}} = \left( {3\frac{1}{4} \times 4\frac{1}{4} \div 34 - \frac{{47}}{{32}} + \frac{{47}}{{16}}} \right) - \left( {2\frac{1}{2} + 5\frac{1}{2} \div 55 - \frac{{11}}{{10}}} \right) \cr & = \left( {\frac{{13}}{4} \times \frac{{17}}{4} \times \frac{1}{{34}} - \frac{{47}}{{32}} + \frac{{47}}{{16}}} \right) - \left( {\frac{5}{2} + \frac{{11}}{2} \times \frac{1}{{55}} - \frac{{11}}{{10}}} \right) \cr & = \left( {\frac{{13}}{{32}} - \frac{{47}}{{32}} + \frac{{47}}{{16}}} \right) - \left( {\frac{5}{2} + \frac{1}{{10}} - \frac{{11}}{{10}}} \right) \cr & = \left( {\frac{{13 - 47 + 94}}{{32}}} \right) - \left( {\frac{{25 + 1 - 11}}{{10}}} \right) \cr & = \frac{{60}}{{32}} - \frac{{15}}{{10}} \cr & = \frac{{15}}{8} - \frac{3}{2} \cr & = \frac{{15 - 12}}{8} \cr & = \frac{3}{8} \cr} $$