Practice MCQ Questions and Answer on Simplification

111.

The difference of two numbers is 1564. After dividing the larger number by the smaller, we get 6 as quotient and 19 as remainder. What is the smaller number?

• (A) 456
• (B) 287
• (C) 623
• (D) 309

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 $$\eqalign{ & {\text{Let two numbers}} = a{\text{ and }}b \cr & a > b \cr & a - b = 1564{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & 6b + 19 = a \cr & a - 6b = 19{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & a - b = 1564 \cr & a - 6b = 19 \cr & - \,\,\,\,\, + \,\,\,\,\, - \cr & \overline {\,\,\,\,\,\,\,5b = 1545\,} \cr & b = 309 \cr & {\text{Smaller number}} = 309 \cr} $$

112.

The value of $$\frac{16-0.002}{8+0.4+0.02}$$   is:

• (A) 2.1
• (B) 1.5
• (C) 3.2
• (D) 1.9

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 $$\eqalign{ & \frac{{16 - 0.002}}{{8 + 0.4 + 0.02}} \cr & = \frac{{16000 - 2}}{{8000 + 400 + 20}} \cr & = \frac{{15098}}{{8420}} \cr & = 1.9{\text{ Answer}} \cr} $$

113.

The value of 18% of 15% of $$\frac{25}{9}$$ of 3800 is:

• (A) 285
• (B) 582
• (C) 583
• (D) 385

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 $$\eqalign{ & 3800 \times \frac{{25}}{9} \times 15\% \times 18\% \cr & = 3800 \times \frac{{25}}{9} \times \frac{3}{{20}} \times \frac{{18}}{{100}} \cr & = 19 \times 15 \cr & = 285 \cr} $$

114.

The value of $$\frac{{\left(a+b\right)}^2}{\left(a^2-b^2\right)}\text{is}=\text{?}$$

• (A) $$\frac{ab}{a+b}$$
• (B) $$\frac{2ab}{a-b}$$
• (C) $$\frac{a+b}{a-b}$$
• (D) None of these

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 $$\eqalign{ & \frac{{{{\left( {a + b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)}} \cr & = \frac{{\left( {a + b} \right)\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{a + b}}{{a - b}}\, \cr} $$

115.

(98764 + 89881 + 99763 + 66342) ÷ (1186 + ? + 1040 + 1870) = 55

• (A) 2254
• (B) 2354
• (C) 2368
• (D) 2404

Show Answer

 Let the missing number is x. then, $$\frac{{354750}}{{4096 + x}}$$   = 55 ⇒ 55( 4096 + x ) = 354750 ⇒ 55x = 354750 - 225280 ⇒ 55x = 129470 ⇒ x = 2354

116.

Given that $$\sqrt{574.6}$$  = 23.97, $$\sqrt{5746}$$  = 75.8 then $$\sqrt{0.00005746}$$   = ?

• (A) 0.002394
• (B) 0.0002397
• (C) 0.0007580
• (D) 0.00758

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 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {0.00005746} \cr & \Rightarrow \sqrt {\frac{{5746}}{{100000000}}} \cr & \Rightarrow \frac{{75.8}}{{10000}} \cr & \Rightarrow 0.00758 \cr} $$

117.

$$\frac{3}{2}\times \frac{11}{5}÷\left(\frac{25}{44}\times \frac{11}{5}\right)÷\frac{33}{15}=?$$

• (A) $$\frac{1}{2}$$
• (B) $$\frac{2}{3}$$
• (C) $$\frac{126}{125}$$
• (D) $$5\frac{101}{125}$$

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & \frac{3}{2} \times \frac{{11}}{5} \div \left( {\frac{{25}}{{44}} \times \frac{{11}}{5}} \right) \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \div \frac{5}{4} \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \times \frac{4}{5} \times \frac{{15}}{{33}} \cr & = \frac{6}{5} \cr & = 1\frac{1}{5} \cr} $$

118.

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

• (A) 10
• (B) 6
• (C) 4
• (D) 0

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 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

119.

The sum of the first 99 terms of the series $$\frac{3}{4}+\frac{5}{16}+\frac{7}{144}+\frac{9}{400}+.....$$

• (A) $$\frac{99}{100}$$
• (B) $$\frac{999}{1000}$$
• (C) $$\frac{9999}{10000}$$
• (D) 1

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

120.

Simplify : $$\frac{1+\frac{1}{2}}{1-\frac{1}{2}}÷\frac{4}{7}\left(\frac{2}{5}+\frac{3}{10}\right)$$     $$\text{of}$$ $$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}$$

• (A) $$\frac{2}{3}$$
• (B) $$37\frac{1}{2}$$
• (C) $$\frac{3}{2}$$
• (D) $$13\frac{3}{8}$$

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 $$\eqalign{ & {\text{According to question,}} \cr & \frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}} \div \frac{4}{7}\left( {\frac{2}{5} + \frac{3}{{10}}} \right){\text{of }}\frac{{\frac{1}{2} + \frac{1}{3}}}{{\frac{1}{2} - \frac{1}{3}}} \cr & = \frac{{\frac{3}{2}}}{{\frac{1}{2}}} \div \frac{4}{7}\left( {\frac{{20 + 15}}{{50}}} \right) \times \frac{{\frac{5}{6}}}{{\frac{1}{6}}} \cr & = \frac{3}{2} \times \frac{2}{1} \div \frac{4}{7} \times \frac{{35}}{{50}} \times \frac{5}{6} \times \frac{6}{1} \cr & = \frac{3}{2} \times \frac{2}{1} \div 2 \cr & = \frac{3}{2} \times 2 \div \frac{1}{2} \cr & = \frac{3}{2} \cr} $$