Practice MCQ Questions and Answer on Simplification

191.

Find the value of $$\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30........\mathrm{\infty }}}}}.$$

• (A) 3
• (B) 4
• (C) 6
• (D) 5

Show Answer

 $$\eqalign{ & x = \sqrt {30 + \sqrt {30 + \sqrt {30 + \sqrt {30\,........\,\infty } } } } \cr & 5 \times 6 = 30 \cr & {\text{Hence }}x = 6 \cr} $$

192.

421 ÷ 35 × 299.97 ÷ 25.05 = ?2

• (A) 22
• (B) 24
• (C) 28
• (D) 12

Show Answer

 $$\eqalign{ & {\text{Let the number be }}x \cr & {\text{Given }} \cr & 421 \div 35 \times 299.97 \div 25.05 = {x^2} \cr & \approx {\left( x \right)^2} = 420 \div 35 \times 300 \div 25 \cr & \approx 12 \times 300 \div 25 = 12 \times 12 \cr & \approx {x^2} = 12 \times 12 \cr & \therefore x = \sqrt {12 \times 12} = 12 \cr & {\text{Hence the number is 12}} \cr} $$

193.

Simplify : $$\text{10}\frac{1}{8}\text{ of }\frac{12}{15}÷\frac{35}{36}\text{ of }\frac{20}{49}$$

• (A) $$17\frac{5}{12}$$
• (B) $$17\frac{8}{17}$$
• (C) $$20\frac{3}{25}$$
• (D) $$20\frac{103}{250}$$

Show Answer

 $$\eqalign{ & {\text{Given expression,}} \cr & \frac{{81}}{8}\,of\,\frac{{12}}{{15}} \div \frac{{35}}{{36}}\,of\,\frac{{20}}{{49}} \cr & = \frac{{81}}{{10}} \div \frac{{25}}{{63}} \cr & = \frac{{81}}{{10}} \times \frac{{63}}{{25}} \cr & = \frac{{5103}}{{250}} \cr & = 20\frac{{103}}{{250}}\, \cr} $$

194.

The value of $$\sqrt{32}$$  - $$\sqrt{128}$$  + $$\sqrt{50}$$ correct to 3 places of decimal is $$\sqrt{32}$$  - $$\sqrt{128}$$  + $$\sqrt{50}$$  = ?

• (A) 1.732
• (B) 1.141
• (C) 1.414
• (D) 1.441

Show Answer

 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {32} {\text{ }} - \sqrt {128} {\text{ + }}\sqrt {50} \cr & \Rightarrow \sqrt {16 \times 2} - \sqrt {64 \times 2} + \sqrt {25 \times 2} \cr & \Rightarrow 4\sqrt 2 - 8\sqrt 2 + 5\sqrt 2 \cr & \Rightarrow \sqrt 2 \cr & \Rightarrow 1.414 \cr} $$

195.

If $$x=\sqrt{3}\text{ + }\sqrt{2}\text{,}$$   then the value of $$x^3-\frac{1}{x^3}$$   is?

• (A) $$10\sqrt{2}$$
• (B) $$14\sqrt{2}$$
• (C) $$22\sqrt{2}$$
• (D) $$8\sqrt{2}$$

Show Answer

 $$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr} $$   $$ = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + $$      $$3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)$$ $$\eqalign{ & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr} $$

196.

$$4\frac{4}{5}÷6\frac{2}{5}=?$$

• (A) $$\frac{3}{4}$$
• (B) $$\frac{5}{7}$$
• (C) $$\frac{5}{8}$$
• (D) $$\frac{7}{11}$$

Show Answer

 $$\eqalign{ & {\text{Given expression}} \cr & \frac{{24}}{5} \div \frac{{32}}{5} \cr & = \frac{{24}}{5} \times \frac{5}{{32}} \cr & = \frac{3}{4} \cr} $$

197.

If $$\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79},$$    where x, y and z are natural numbers, then the value of (2x + 3y - z) is:

• (A) 0
• (B) 4
• (C) 1
• (D) 2

Show Answer

 $$\eqalign{ & \frac{1}{{x + \frac{1}{{y + \frac{2}{{z + \frac{1}{4}}}}}}} = \frac{{29}}{{79}} \cr & \frac{1}{{2 + \frac{1}{{1 + \frac{2}{{5 + \frac{1}{4}}}}}}} = \frac{{29}}{{79}} \cr & x = 2,\,y = 1,\,z = 5 \cr & \left( {2x + 3y - z} \right) \cr & = 2 \times 2 + 3 \times 1 - 5 \cr & = 2 \cr} $$

198.

To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?

• (A) 10
• (B) 35
• (C) 62.5
• (D) Cannot be determined

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{1}}\,{\text{bucket}} = x \cr & {\text{Then,}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{tank}} = 25x \cr & {\text{New}}\,{\text{capacity}}\,{\text{of}}\,{\text{bucket}} = \frac{2}{5}x \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{buckets}} \cr & = \frac{{25x}}{{\left( {2x/5} \right)}} \cr & = {25x \times \frac{5}{{2x}}} \cr & = \frac{{125}}{2} \cr & = 62.5 \cr} $$

199.

Direction: In the question given below the given mathematical symbols are changed from '+' to '÷', '-' to '×', '÷' to '-' and from '×' to '+', then choose your answers from the following options. 67 × 119 + 17 - 27 × 259 = ?

• (A) -13
• (B) -3
• (C) 4
• (D) 7

Show Answer

 67 × 119 + 17 - 27 ÷ 259 ⇒ 67 + 119 ÷ 17 × 27 - 259 ⇒ 67 + 7 × 27 - 259 ⇒ 67 + 189 - 259 ⇒ 256 - 259 ⇒ -3

200.

If x = a + m, y = b + m, z = c + m, then the value of $$\frac{x^2+y^2+z^2-yz-zx-xy}{a^2+b^2+c^2-ab-bc-ca}$$       is = ?

• (A) 1
• (B) $$\frac{x+y+z}{a+b+c}$$
• (C) $$\frac{a+b+c}{x+y+z}$$
• (D) Not possible to find

Show Answer

 $$\eqalign{ & \frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{\text{ }} \cr & = \frac{{{{\left( {a + m} \right)}^2} + {{\left( {b + m} \right)}^2} + {{\left( {c + m} \right)}^2} - \left( {b + m} \right)\left( {c + m} \right) - \left( {c + m} \right)\left( {a + m} \right) - \left( {a + m} \right)\left( {b + m} \right)}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {m^2} + 2am + {b^2} + {m^2} + 2bm + {c^2} + {m^2} + 2cm - bc - bm - cm - {m^2} - ca - cm - am - {m^2} - ab - am - bm - {m^2}}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = 1 \cr} $$