Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Given expression ,}} \cr & \left( {1000 - \frac{1}{{1000}}} \right) \times 7 \cr & = \left( {7000 - \frac{7}{{1000}}} \right) \cr & = 6999\frac{{993}}{{1000}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{4 \times {3^3} + {3^2} + 3 + 1}}{{4 \times {3^3}}} \cr & = \frac{{108 + 9 + 3 + 1}}{{108}} \cr & = \frac{{121}}{{108}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{4.5.6 + 5.6 + 2.6 + 2.3}}{{1.2.3.4.5.6}} \cr & = \frac{{120 + 30 + 12 + 6}}{{720}} \cr & = \frac{{168}}{{720}} \cr & = \frac{7}{{30}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {\frac{{0.009 \times 0.036 \times 0.016 \times 0.08}}{{0.002 \times 0.0008 \times 0.0002}}} \, \cr & \Rightarrow \sqrt {\frac{{9 \times 36 \times 16 \times 8}}{{2 \times 8 \times 2}}} \cr & \Rightarrow 3 \times 6 \times 2 \cr & \Rightarrow 36 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & 2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} \cr & = \,6\sqrt 6 - 2\sqrt {\frac{2}{3} \times 9} - 4\sqrt 6 \cr & = 6\sqrt 6 - 2\sqrt 6 - 4\sqrt 6 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right) . . . . + \left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr} $$

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

$$\eqalign{ & {\text{Sum of given fractions}} \cr & {\text{ = }}\frac{7}{4} + \frac{5}{2} + \frac{{67}}{{12}} + \frac{{10}}{3} + \frac{9}{4} \cr & = \left( {\frac{{21 + 30 + 67 + 40 + 24}}{{12}}} \right) \cr & = \frac{{185}}{{12}} \cr & {\text{The whole number just }} \cr & {\text{less than }}\frac{{185}}{{12}}{\text{ is 15}} \cr & {\text{let }}\frac{{185}}{{12}} - x = 15 \cr & {\text{Then,}} \cr & x = \left( {\frac{{185}}{{12}} - 15} \right) \cr & \,\,\,\,\,\, = \frac{5}{{12}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {110 + \frac{1}{4}} \, \cr & \Rightarrow \sqrt {\frac{{441}}{4}} \cr & \Rightarrow \frac{{21}}{2} \cr & \Rightarrow 10.5 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{5}}\,\,{\text{| 675}} \cr & - - - - - - \cr & 5\,\,|\,\,135 \cr & - - - - - - \cr & 5\,\,|\,\,27 \cr & - - - - - - \cr & 3\,\,|\,\,9 \cr & - - - - - - \cr & 3\,\,|\,\,3\quad \cr & - - - - - - \cr & \,\,\,\,\,|\,\,1 \cr & {\text{Factors are}} \cr & = \,5 \times 5 \times 3 \times 3 \times 3 \cr} $$
∴ It must be multiplied by 5 to make a perfect cube.