Practice MCQ Questions and Answer on Simplification

101.

The value of $$\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5}\times \left(\frac{7}{8}-\frac{5}{4}\right)\right]$$      is:

• (A) 5
• (B) 0
• (C) 10
• (D) 1

Show Answer

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

102.

$$\sqrt{110+\frac{1}{4}}$$   is equal to = ?

• (A) 12.0
• (B) 11.5
• (C) 11.0
• (D) 10.5

Show Answer

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

103.

5 - [4 - {3 - (3 - 3 - 6)}] is equal to:

• (A) 10
• (B) 6
• (C) 4
• (D) 0

Show Answer

 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

104.

Supply the two missing figures in order indicated by x and y in the given equation, the fractions being in their lowest terms. $$\text{5}\frac{1}{x}\times y\frac{3}{4}=20$$

• (A) 3, 1
• (B) 3, 3
• (C) 4, 1
• (D) 5, 3

Show Answer

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

105.

The value of $$2a^3-\left[3a^3+4b^3-\left\{2a^3+\left(-7a^3\right)\right\}\text{ + 5}{a}^3-\text{7}{\text{b}}^3\right]\text{ is - }$$

• (A) $$-11a^3\text{ + 3}{\text{b}}^3$$
• (B) $$\text{7}{\text{b}}^3\text{ + 3}{a}^3$$
• (C) $$\text{11}{a}^3-3{\text{b}}^3$$
• (D) $$-\left(11a^3\text{ + 3}{\text{b}}^3\right)$$

Show Answer

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

106.

Find the simplest value of $$\frac{6.25-1.96}{1.1}.$$

• (A) 3.8
• (B) 3.9
• (C) 4
• (D) 2.9

Show Answer

 $$\eqalign{ & \frac{{6.25 - 1.96}}{{1.1}} \cr & = \frac{{{{\left( {2.5} \right)}^2} - {{\left( {1.4} \right)}^2}}}{{\left( {1.1} \right)}} \cr & = \frac{{\left( {1.1} \right) \times 3.9}}{{\left( {1.1} \right)}} \cr & = 3.9{\text{ Answer}} \cr} $$

107.

The simplified value of $$\frac{\left(1+\frac{1}{1+\frac{1}{100}}\right)\left(1+\frac{1}{1+\frac{1}{100}}\right)-\left(1-\frac{1}{1+\frac{1}{100}}\right)\left(1-\frac{1}{1+\frac{1}{100}}\right)}{\left(1+\frac{1}{1+\frac{1}{100}}\right)+\left(1-\frac{1}{1+\frac{1}{100}}\right)}=?$$

• (A) 100
• (B) $$\frac{200}{101}$$
• (C) 200
• (D) $$\frac{202}{100}$$

Show Answer

 $$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr} $$

108.

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}$$     is equal to = ?

• (A) 2
• (B) 2.5
• (C) 3
• (D) 3.5

Show Answer

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

109.

The value of $$0.\overline{57}-0.4\overline{32}+0.3\bar{5}$$    is:

• (A) $$0.4\overline{98}$$
• (B) $$0.4\overline{94}$$
• (C) $$0.\overline{498}$$
• (D) $$0.\overline{494}$$

Show Answer

 $$\eqalign{ & 0.\overline {57} - 0.4\overline {32} + 0.3\overline 5 \cr & = 0.\overline {57} - 0.4\overline {32} + 0.3\overline {55} \cr & = \frac{{57}}{{99}} - \frac{{432 - 4}}{{990}} + \frac{{355 - 3}}{{990}} \cr & = \frac{{570 - 428 + 352}}{{990}} \cr & = 0.4\overline {94} \cr} $$

110.

Simplify the value of $$\frac{\text{0}\text{.9}\times \text{0}\text{.9}\times \text{0}\text{.9 + 0}\text{.2}\times \text{0}\text{.2}\times \text{0}\text{.2 + 0}\text{.3}\times \text{0}\text{.3}\times \text{0}\text{.3}-\text{3}\times 0.9\times \text{0}\text{.2}\times \text{0}\text{.3}}{\text{0}\text{.9}\times \text{0}\text{.9 + 0}\text{.2}\times \text{0}\text{.2 + 0}\text{.3}\times \text{0}\text{.3}-0.9\times \text{0}\text{.2}-\text{0}\text{.2}\times \text{0}\text{.3}-0.3\times 0.9}=?$$

• (A) 1.4
• (B) 0.054
• (C) 0.8
• (D) 1.0

Show Answer

 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 $${a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$   $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$ $$ = \frac{{{{(0.9)}^3} + {{(0.2)}^3} + {{(0.3)}^3} - 3 \times 0.9 \times 0.2 \times 0.3}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$ $$ = \frac{{\left( {0.9 + 0.2 + 0.3} \right)\left[ {{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9} \right]}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$ $$\eqalign{ & = 0.9 + 0.2 + 0.3 \cr & = 1.4 \cr} $$