Practice MCQ Questions and Answer on Simplification

71.

The value of $$\frac{\sqrt{80}-\sqrt{112}}{\sqrt{45}-\sqrt{63}}=?$$

• (A) $$\frac{3}{4}$$
• (B) $$1\frac{3}{4}$$
• (C) $$1\frac{1}{3}$$
• (D) $$1\frac{7}{9}$$

Show Answer

 $$\eqalign{ & {\text{According to the question,}} \cr & \frac{{\sqrt {80} - \sqrt {112} }}{{\sqrt {45} - \sqrt {63} }} \cr & \Rightarrow \frac{{4\sqrt 5 - 4\sqrt 7 }}{{3\sqrt 5 - 3\sqrt 7 }} \cr & \Rightarrow \frac{{4\left( {\sqrt 5 - \sqrt 7 } \right)}}{{3\left( {\sqrt 5 - \sqrt 7 } \right)}} \cr & \Rightarrow \frac{4}{3} \cr & \Rightarrow 1\frac{1}{3} \cr} $$

72.

(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

73.

A sum of Rs. 1360 has been divided among A, B and C such that A gets $$\frac{2}{3}$$ of what B gets and B gets $$\frac{1}{4}$$ of what C gets. B's share is:

• (A) Rs. 120
• (B) Rs. 160
• (C) Rs. 240
• (D) Rs. 300

Show Answer

 $$\eqalign{ & {\text{Let}}\,C's\,{\text{share}} = Rs.\,x \cr & \text{Then,}\,B's\,{\text{share}} = \,Rs.\,\frac{x}{4}, \cr & A's\,{\text{share}} = Rs.\,\left( {\frac{2}{3} \times \frac{x}{4}} \right) = Rs.\,\frac{x}{6} \cr & \therefore \frac{x}{6} + \frac{x}{4} + x = 1360 \cr & \Rightarrow \frac{{17x}}{{12}} = 1360 \cr & \Rightarrow x = \frac{{1360 \times 12}}{{17}} = Rs.\,960 \cr & {\text{Hence}},\,B's\,{\text{share}} = Rs.\,\left( {\frac{{960}}{4}} \right) \cr & = Rs.\,240 \cr} $$

74.

Simplify the following $$\frac{0.75}{10}+\frac{0.31}{100}+\frac{0.09}{1000}$$

• (A) 0.03215
• (B) 0.05621
• (C) 0.07819
• (D) 0.08561

Show Answer

 $$\eqalign{ & \frac{{0.75}}{{10}} + \frac{{0.31}}{{100}} + \frac{{0.09}}{{1000}} \cr & = \frac{{75}}{{1000}} + \frac{{3.1}}{{1000}} + \frac{{0.09}}{{1000}} \cr & = \frac{{78.19}}{{10000}} \cr & = 0.07819 \cr} $$

75.

The value of $$\frac{3}{4}\times 2\frac{2}{3}÷\frac{5}{9}\text{ of }1\frac{1}{5}+\frac{2}{23}\times 3\frac{5}{6}÷\frac{2}{7}\text{ of }2\frac{1}{3}$$        is:

• (A) $$1\frac{5}{6}$$
• (B) $$1\frac{2}{3}$$
• (C) $$3\frac{1}{2}$$
• (D) $$4\frac{5}{6}$$

Show Answer

 $$\eqalign{ & \frac{3}{4} \times 2\frac{2}{3} \div \frac{5}{9}{\text{ of }}1\frac{1}{5} + \frac{2}{{23}} \times 3\frac{5}{6} \div \frac{2}{7}{\text{ of }}2\frac{1}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \div \frac{5}{9}{\text{ of }}\frac{6}{5} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{7}{\text{ of }}\frac{7}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \div \frac{2}{3} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \times \frac{3}{2} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{3} \cr & = 3 + \frac{1}{2} \cr & = 3\frac{1}{2} \cr} $$

76.

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

77.

32 shirt pieces of 120cm each can be cut from a reel of cloth. After cutting these pieces 80 cm of cloth remains. What is the length of reel of cloth in meters?

• (A) 38.70 meters
• (B) 39.20 meters
• (C) 3870 meters
• (D) 3920 meters

Show Answer

 $$\eqalign{ & {\text{Length of reel}} \cr & {\text{ = (32}} \times {\text{120 + 80)}}\,{\text{cm}} \cr & {\text{ = 3920}}\,{\text{cm}} \cr & {\text{ = 39}}{\text{.20}}\,{\text{m}} \cr} $$

78.

The value of $$0.4\bar{6}+0.7\overline{23}-0.3\bar{9}\times 0.\bar{7}$$     is:

• (A) $$0.\overline{57}$$
• (B) $$0.\overline{87}$$
• (C) $$0.\overline{97}$$
• (D) $$0.\overline{77}$$

Show Answer

 $$\eqalign{ & 0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \cr & = \frac{{46 - 4}}{{90}} + \frac{{723 - 7}}{{990}} - \frac{{39 - 3}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{36}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{28}}{{90}} \cr & = \frac{{14}}{{90}} + \frac{{716}}{{990}} \cr & = \frac{{154 + 716}}{{990}} \cr & = \frac{{870}}{{990}} \cr & = 0.\overline {87} \cr} $$

79.

$$\left(x+\frac{1}{x}\right)$$ $$\left(x-\frac{1}{x}\right)$$ $$\left(x^2+\frac{1}{x^2}-1\right)$$  $$\left(x^2+\frac{1}{x^2}+1\right)$$   is equal to ?

• (A) $$x^6-\frac{1}{x^6}$$
• (B) $$x^8-\frac{1}{x^8}$$
• (C) $$x^6+\frac{1}{x^6}$$
• (D) $$x^8+\frac{1}{x^8}$$

Show Answer

 Given expression, $$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$     $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$ $$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$

80.

The simplified value of $$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\text{ is = ?}$$

• (A) 3
• (B) 2
• (C) 6
• (D) 4

Show Answer

 $$\eqalign{ & {\text{According to question,}} \cr & \frac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }} \cr & = \frac{{\sqrt {4 \times 4 \times 2} + \sqrt {4 \times 4 \times 3} }}{{\sqrt {2 \times 2 \times 2} + \sqrt {2 \times 2 \times 3} }} \cr & = \frac{{4\sqrt 2 + 4\sqrt 3 }}{{2\sqrt 2 + 2\sqrt 3 }} \cr & = \frac{{4\left( {\sqrt 2 + \sqrt 3 } \right)}}{{2\left( {\sqrt 2 + \sqrt 3 } \right)}} \cr & = \frac{4}{2} \cr & = 2 \cr} $$