Practice MCQ Questions and Answer on Simplification

91.

Simplify the following:
$$\frac{{\left(0.1\right)}^2+{\left(0.01\right)}^2+{\left(0.009\right)}^2+2\left[0.1\times 0.01+0.01\times 0.009+0.009\times 0.1\right]}{0.01+0.001+0.0009}$$

• (A) 0.19
• (B) 0.01
• (C) 1.01
• (D) 1.19

Show Answer

 $$\eqalign{ & \frac{{{{\left( {0.1} \right)}^2} + {{\left( {0.01} \right)}^2} + {{\left( {0.009} \right)}^2} + 2\left[ {0.1 \times 0.01 + 0.01 \times 0.009 + 0.009 \times 0.1} \right]}}{{0.01 + 0.001 + 0.0009}} \cr & = \frac{{10{{\left( {0.1 + 0.01 + 0.009} \right)}^2}}}{{\left( {0.1 + 0.01 + 0.009} \right)}} \cr & = 10 \times 0.119 \cr & = 1.19{\text{ Answer}} \cr} $$

92.

The value of x in the equation $$\frac{113\times 4-x\times 2}{13\times 9-5\times 7}\text{ = 5}$$     is?

• (A) 21
• (B) 27
• (C) 35
• (D) 42

Show Answer

 $$\eqalign{ & \frac{{113 \times 4 - x \times 2}}{{13 \times 9 - 5 \times 7}} = 5 \cr & \Rightarrow \frac{{452 - 2x}}{{117 - 35}} = 5 \cr & \Rightarrow \frac{{452 - 2x}}{{82}} = 5 \cr & \Rightarrow 452 - 2x = 410 \cr & \Rightarrow 2x = 452 - 410 = 42 \cr & \Rightarrow x = 21 \cr} $$

93.

The value of $$3\frac{1}{3}÷2\frac{1}{2}\text{ of }1\frac{3}{5}+\left(\frac{3}{8}+\frac{1}{7}\times 1\frac{3}{4}\right)$$      is:

• (A) $$\frac{55}{24}$$
• (B) $$\frac{35}{24}$$
• (C) $$\frac{25}{24}$$
• (D) $$\frac{5}{24}$$

Show Answer

 $$\eqalign{ & 3\frac{1}{3} \div 2\frac{1}{2}{\text{ of }}1\frac{3}{5} + \left( {\frac{3}{8} + \frac{1}{7} \times 1\frac{3}{4}} \right) \cr & = \frac{{10}}{3} \div \frac{5}{2} \times \frac{8}{5} + \left( {\frac{3}{8} + \frac{1}{7} \times \frac{7}{4}} \right) \cr & = \frac{{10}}{3} \times \frac{1}{4} + \frac{5}{8} \cr & = \frac{5}{6} + \frac{5}{8} \cr & = \frac{{35}}{{24}} \cr} $$

94.

The number of pairs of natural numbers the difference of whose squares is 45 will be ?

• (A) 2
• (B) 3
• (C) 6
• (D) 5

Show Answer

 Let the numbers be x and y According to question, $$\eqalign{ & \left( {x > y} \right) \cr & {x^2} - {y^2} = 45 \cr & \left( {x + y} \right)\left( {x - y} \right) = 45 \cr & {\text{Make factor of 45}} \cr & {\text{15}} \times 3 \cr & \,\,9 \times 5\,\,\,\,\,\,\,\,\,\,(3\,{\text{pairs}}) \cr & 45 \times 1 \cr} $$ These pairs gives the value of x and y which satisfy the given condition.

95.

If $$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$$     and $$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$$     where a, b, c, p, q, r are non-zero real numbers, then $$\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$$    is equal to = ?

• (A) 0
• (B) 1
• (C) 3
• (D) 9

Show Answer

 $$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr} $$

96.

Evaluated : $$\frac{9\left|3-5\right|-5\left|4\right|÷10}{-3\left(5\right)-2\times 4÷2}$$

• (A) $$\frac{9}{10}$$
• (B) $$-\frac{8}{17}$$
• (C) $$-\frac{16}{19}$$
• (D) $$\frac{4}{7}$$

Show Answer

 $$\eqalign{ & {\text{According to question}} \cr & \frac{{9|3 - 5| - 5|4| \div 10}}{{ - 3\left( 5 \right) - 2 \times 4 \div 2}} \cr & \Rightarrow \frac{{9 \times 2 - 20 \div 10}}{{ - 15 - 2 \times 2}} \cr & \Rightarrow \frac{{18 - 2}}{{ - 15 - 4}} \cr & \Rightarrow - \frac{{16}}{{19}} \cr} $$

97.

Let 0 x 1, then the correct inequality is = ?

• (A) $$x\sqrt{x}{x}^2$$  
• (B) $$\sqrt{x}x{x}^2$$  
• (C) $$x^{2}x\sqrt{x}$$  
• (D) $$\sqrt{x}{x}^{2}x$$  

Show Answer

 $$\eqalign{ & 0 x 1, \cr & {\text{Let }}x = \frac{4}{{10}} \cr & {\text{So}},\sqrt x = \frac{2}{{\sqrt {10} }}\,\& \, \cr & \,\,{x^2} = \frac{{16}}{{100}} = 0.16 \cr & {\text{Now}}, \cr & \because 0.16 \frac{4}{{10}} \frac{2}{{\sqrt {10} }} \cr & \therefore {x^2} x \sqrt x \cr} $$

98.

If $$\frac{a}{b}=\frac{1}{3},$$ $$\frac{b}{c}=2,$$ $$\frac{c}{d}=\frac{1}{2},$$ $$\frac{d}{e}=3$$   and $$\frac{e}{f}=\frac{1}{4},$$   then what is the value of $$\frac{abc}{def}=?$$

• (A) $$\frac{1}{4}$$
• (B) $$\frac{3}{4}$$
• (C) $$\frac{3}{8}$$
• (D) $$\frac{27}{4}$$

Show Answer

 $$\eqalign{ & \frac{a}{b} = \frac{1}{3}\,\, \Rightarrow b = 3a; \cr & \frac{b}{c} = 2\,\,\, \Rightarrow c = \frac{b}{2} = \frac{{3a}}{2}; \cr & \frac{c}{d} = \frac{1}{2}\,\,\, \Rightarrow d = 2c = 2\left( {\frac{{3a}}{2}} \right) = 3a; \cr & \frac{d}{e} = 3\,\,\, \Rightarrow e = \frac{d}{3} = \left( {\frac{{3a}}{3}} \right) = a; \cr & \frac{e}{f} = \frac{1}{4}\,\,\, \Rightarrow f = 4e = 4a \cr & \therefore \frac{{abc}}{{def}} = \frac{{\left( a \right)\left( {3a} \right)\left( {\frac{{3a}}{2}} \right)}}{{\left( {3a} \right)\left( a \right)\left( {4a} \right)}} \cr & \frac{{abc}}{{def}} = \frac{9}{2}{a^3} \times \frac{1}{{12{a^3}}} \cr & \frac{{abc}}{{def}} = \frac{3}{8} \cr} $$

99.

The value of $$\frac{5}{1\frac{7}{8}\text{of 1}\frac{1}{3}}\times \frac{2\frac{1}{10}}{3\frac{1}{2}}\text{ of 1}\frac{1}{4}=?$$

• (A) $$1\frac{1}{2}$$
• (B) 0.05
• (C) 1
• (D) 2

Show Answer

 $$\eqalign{ & {\text{According to the question}} \cr & \frac{5}{{1\frac{7}{8}{\text{of 1}}\frac{1}{3}}} \times \frac{{2\frac{1}{{10}}}}{{3\frac{1}{2}}}{\text{ of 1}}\frac{1}{4} \cr & \Rightarrow \frac{5}{{\frac{{15}}{8} \times {\text{ }}\frac{4}{3}}} \times {\text{ }}\frac{{21}}{{10}} \times \frac{2}{7} \times \frac{5}{4} \cr & \Rightarrow 2 \times \frac{3}{4} \cr & \Rightarrow \frac{3}{2} \cr & \Rightarrow 1\frac{1}{2} \cr} $$

100.

421 ÷ 35 × 299.97 ÷ 25.05 = ?2

• (A) 22
• (B) 24
• (C) 28
• (D) 12

Show Answer

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