Practice MCQ Questions and Answer on Simplification

231.

A tree grows only $$\frac{3}{5}$$ as fast as the one beside it. In four years the combined growth of the trees is eight feet. How much does the shorter tree grow in 2 years?

• (A) Less than 2 feet
• (B) 2 feet
• (C) $${\text{2}}\frac{1}{2}$$ feet
• (D) 3 feet

Show Answer

 Suppose the taller tree grows x feet in 1 year. Then the shorter tree grows $$\eqalign{ & = \frac{{3x}}{5}{\text{ft in 1 year}} \cr & \therefore 4x + 4 \times \frac{3}{5}x = 8 \cr & \Leftrightarrow 4x + \frac{{12}}{5}x = 8 \cr & \Leftrightarrow \frac{{32}}{5}x = 8 \cr & \Leftrightarrow x = \frac{{8 \times 5}}{{32}} = \frac{5}{4} \cr & {\text{Growth of shorter tree in}} \cr & {\text{2 years = }}\left( {2 \times \frac{{3x}}{5}} \right){\text{ft}} \cr & {\text{ = }}\left( {2 \times \frac{3}{5} \times \frac{5}{4}} \right){\text{ft}} \cr & {\text{ = }}\frac{3}{2}{\text{ft}} \cr & {\text{ = 1}}\frac{1}{2}{\text{ft}} \cr} $$

232.

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

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

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

233.

A canteen requires 798 bananas for a week. Total how many bananas did it require for the months of January, February and March 2008?

• (A) 10277
• (B) 10374
• (C) 10480
• (D) 10586

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 $$\eqalign{ & {\text{Number of bananas required daily}} \cr & {\text{ = }}\frac{{798}}{7} \cr & = 144 \cr & \therefore \,\,\,{\text{Required number}} \cr & {\text{ = }}\left[ {114 \times \left( {\mathop {31}\limits^{{\text{Jan}}} + \mathop {29}\limits^{{\text{Feb}}} + \mathop {31}\limits^{{\text{Mar}}} } \right)} \right] \cr & = \left( {114 \times 91} \right) \cr & = 10374 \cr} $$

234.

The value of $$0.\overline {45} \times 1.\overline {22} $$   is:

• (A) $$0.\overline 3 $$
• (B) $$0.\overline 4 $$
• (C) $$0.\overline 6 $$
• (D) $$0.\overline 5 $$

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 $$\eqalign{ & 0.\overline {45} \times 1.\overline {22} \cr & = \frac{{45}}{{99}} \times \frac{{122 - 1}}{{99}} \cr & = \frac{{45}}{{99}} \times \frac{{121}}{{99}} \cr & = \frac{5}{9} \cr & = 0.\overline 5 {\text{ Answer}} \cr} $$

235.

If $$\left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154,}}$$     then the value of $$\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$$   is = ?

• (A) 198
• (B) 200
• (C) 216
• (D) None of these

Show Answer

 $$\eqalign{ & \left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154}} \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = 1156 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = 36 \cr & \Rightarrow a + \frac{1}{a} = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^3} = {6^3} = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 6 = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 216 - 18 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 198 \cr} $$

236.

Simplify : $$\frac{{ - \frac{1}{2} - \frac{2}{3} + \frac{4}{5} - \frac{1}{3} + \frac{1}{5} + \frac{3}{4}}}{{\frac{1}{2} + \frac{2}{3} - \frac{4}{3} + \frac{1}{3} - \frac{1}{5} - \frac{4}{5}}}$$

• (A) $$ - \frac{3}{{10}}$$
• (B) $$ - \frac{{10}}{3}$$
• (C) $$ - 2$$
• (D) $${\text{1}}$$

Show Answer

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

237.

The value of $$\sqrt {400} $$  + $$\sqrt {0.0400} $$   + $$\sqrt {0.000004} $$   = ?

• (A) 0.222
• (B) 20.22
• (C) 20.202
• (D) 2.022

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 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {400} + \sqrt {0.0400} + \sqrt {0.000004} \cr & \Rightarrow 20 + 0.2 + 0.002 \cr & \Rightarrow 20.202 \cr} $$

238.

$$\left( {? - 968} \right) \div 79 \times 4 = 512$$

• (A) 10185
• (B) 10190
• (C) 11075
• (D) 11080

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 $$\eqalign{ & {\text{Let,}} \cr & \left( {x - 968} \right) \div 79 \times 4 = 512 \cr & {\text{Then,}} \cr & \frac{{x - 968}}{{79}} \times 4 = 512 \cr & \Rightarrow x - 968 = \frac{{512 \times 79}}{4} \cr & \Rightarrow x - 968 = 10112 \cr & \Rightarrow x = 10112 + 968 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 11080 \cr} $$

239.

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

240.

The value of $$\frac{4}{5} \div 3\frac{1}{4}{\text{ of }}\frac{8}{{13}} - \frac{{\frac{1}{5} - \frac{1}{8}}}{{\frac{1}{5} + \frac{1}{8}}} \times 5\frac{1}{5} + \frac{5}{6}\,{\text{is:}}$$

• (A) $$\frac{2}{{15}}$$
• (B) $$\frac{7}{{30}}$$
• (C) $$\frac{1}{{15}}$$
• (D) $$\frac{1}{{30}}$$

Show Answer

 $$\eqalign{ & \frac{4}{5} \div 3\frac{1}{4}{\text{ of }}\frac{8}{{13}} - \frac{{\frac{1}{5} - \frac{1}{8}}}{{\frac{1}{5} + \frac{1}{8}}} \times 5\frac{1}{5} + \frac{5}{6} \cr & = \frac{4}{5} \div \frac{{13}}{4} \times \frac{8}{{13}} - \frac{{\frac{1}{5} - \frac{1}{8}}}{{\frac{1}{5} + \frac{1}{8}}} \times 5\frac{1}{5} + \frac{5}{6} \cr & = \frac{2}{5} - \frac{3}{{13}} \times \frac{{26}}{5} + \frac{5}{6} \cr & = \frac{2}{5} - \frac{6}{5} + \frac{5}{6} \cr & = \frac{{12 - 36 + 25}}{{30}} \cr & = \frac{1}{{30}} \cr} $$