Practice MCQ Questions and Answer on Algebra

161.

If $$x=7-4\sqrt{3}\text{,}$$    then $$\sqrt{x}\text{ + }\frac{1}{\sqrt{x}}$$   is equal to?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Show Answer

 $$\eqalign{ & x = 7 - 4\sqrt 3 \cr & \Rightarrow x = 4 + 3 - 4\sqrt 3 \cr & \Rightarrow x = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2 \times 2\sqrt 3 \cr & \Rightarrow {\left( {2 - \sqrt 3 } \right)^2} \cr & \therefore \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] \cr & \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + \sqrt 3 \cr & \therefore \sqrt x {\text{ + }}\frac{1}{{\sqrt x }}{\text{ }} \cr & = 2 - \sqrt 3 + 2 + \sqrt 3 \cr & = 4 \cr} $$

162.

If $${\left(\sqrt{5}\right)}^7÷{\left(\sqrt{5}\right)}^5={\text{5}}^{\text{P}}\text{,}$$     then the value of P is?

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

Show Answer

 $$\eqalign{ & {\left( {\sqrt 5 } \right)^7} \div {\left( {\sqrt 5 } \right)^5}{\text{ = }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \frac{{{{\left( {\sqrt 5 } \right)}^7}}}{{{{\left( {\sqrt 5 } \right)}^5}}} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {\left( {\sqrt 5 } \right)^2} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {{\text{5}}^{\text{1}}} = {\text{ }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \boxed{{\text{P}} = 1} \cr} $$

163.

If x = 2015, y = 2014, z = 2013, then the value of x² + y² + z² - xy - yz - zx is?

• (A) 3
• (B) 4
• (C) 6
• (D) 2

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 $$\eqalign{ & x = 2015 \cr & y = 2014 \cr & z = 2013 \cr & \therefore {x^2} + {y^2} + {z^2} - xy - yz - zx \cr & = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {2015 - 2014} \right)}^2} + {{\left( {2014 - 2013} \right)}^2} + {{\left( {2013 - 2015} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr & = 3 \cr} $$

164.

If $$\sqrt{0.03\times 0.3a}$$   = $$\text{0}\text{.3}\times \text{0}\text{.3}\times \sqrt{b}\text{,}$$    value of $$\frac{a}{b}\text{is?}$$

• (A) 0.009
• (B) 0.03
• (C) 0.9
• (D) 0.08

Show Answer

 $$\eqalign{ & \sqrt {0.03 \times 0.3a} = {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times \sqrt b \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{0}}{\text{.03}} \times {\text{0}}{\text{.3a}} = {\left( {0.3} \right)^2} \times {\left( {0.3} \right)^2} \times b \cr & \Rightarrow \frac{3}{{100}} \times \frac{3}{{10}}a = \frac{9}{{100}} \times \frac{9}{{100}} \times b \cr & \Rightarrow 9a = \frac{{81}}{{10}}b \cr & \Rightarrow 90a = 81b \cr & \Rightarrow 10a = 9b \cr & \Rightarrow \frac{a}{b} = \frac{9}{{10}} \cr & \Rightarrow \frac{a}{b} = 0.9 \cr} $$

165.

If $$p+\frac{1}{p}=112,$$   find the value of $${\left(p-112\right)}^{15}+\frac{1}{p^{15}}=?$$

• (A) 10
• (B) 0
• (C) 15
• (D) 1

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 $$\eqalign{ & p + \frac{1}{p} = 112,\,\,{\left( {p - 112} \right)^{15}} + \frac{1}{{{p^{15}}}} = ? \cr & p - 112 = - \frac{1}{p} \cr & {\left( {p - 112} \right)^{15}} + \frac{1}{{{p^{15}}}} \cr & = {\left( { - \frac{1}{p}} \right)^{15}} + \frac{1}{{{p^{15}}}} \cr & = - \frac{1}{{{p^{15}}}} + \frac{1}{{{p^{15}}}} \cr & = 0 \cr} $$

166.

If a = 1 + √3, b = 1 - √3, then what is the value of a + b?22

• (A) 4
• (B) 8
• (C) 0
• (D) 2

Show Answer

 a = 1 + √3 ∴ a2 = 1 + 3 + 2√3 b = 1 - √3 ∴ b2 = 1 + 3 - 2√3 ∴ a2 + b2 = 8

167.

If (3a + 1)² + (b - 1)² + (2c - 3)² = 0 then the value of (3a + b + 2c) is equal to?

• (A) 3
• (B) -1
• (C) 2
• (D) 5

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 $$\eqalign{ & {\left( {3a + 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c - 3} \right)^2} = 0 \cr & {\left( {3a + 1} \right)^2} = 0 \cr & \Rightarrow 3a = - 1 \cr & \Rightarrow a = - \frac{1}{3} \cr & {\left( {b - 1} \right)^2} = 0 \cr & \Rightarrow b - 1 = 0 \cr & \Rightarrow b = 1 \cr & {\left( {2c - 3} \right)^2} = 0 \cr & \Rightarrow c = \frac{3}{2} \cr & \therefore 3a + b + 2c \cr & = 3 \times - \frac{1}{3} + 1 + \frac{3}{2} \times 2 \cr & = - 1 + 1 + 3 \cr & = 3 \cr} $$

168.

If $${\left(a+\frac{1}{a}\right)}^2=3$$    then $$a^3+\frac{1}{a^3}=?$$

• (A) $$2\sqrt{3}$$
• (B) 2
• (C) $$3\sqrt{3}$$
• (D) 0

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 $$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & a + \frac{1}{a} = \sqrt 3 \cr & {\text{Take cube on both sides}} \cr & {\left( {a + \frac{1}{a}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$

169.

If $$\frac{x}{y}=\frac{4}{5}\text{,}$$   then the value of $$\left(\frac{4}{7}+\frac{2y-x}{2y+x}\right)$$   is?

• (A) $$\frac{3}{7}$$
• (B) $$\text{1}\frac{1}{7}$$
• (C) 1
• (D) 2

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 $$\eqalign{ & \frac{x}{y} = \frac{4}{5} \cr & \frac{4}{7} + \frac{{2y - x}}{{2y + x}} \cr & = \frac{4}{7} + \frac{{y\left( {2 - \frac{x}{y}} \right)}}{{y\left( {2 + \frac{x}{y}} \right)}} \cr & = \frac{4}{7} + \frac{{\left( {2 - \frac{4}{5}} \right)}}{{\left( {2 + \frac{4}{5}} \right)}} \cr & = \frac{4}{7} + \frac{{10 - 4}}{{10 + 4}} \cr & = \frac{4}{7} + \frac{6}{{14}} \cr & {\text{ = }}\frac{{8 + 6}}{{14}} \cr & {\text{ = }}\frac{{14}}{{14}} \cr & {\text{ = 1}} \cr} $$

170.

If a, b, c are non - zero $$a+\frac{1}{b}=1$$   and $$b+\frac{1}{c}=1,$$   then the value of abc is?

• (A) -1
• (B) 3
• (C) -3
• (D) 1

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 $$\eqalign{ & a + \frac{1}{b} = 1,{\text{ }}b + \frac{1}{c} = 1 \cr & {\text{Values of }}a,{\text{ }}b,{\text{ }}c{\text{ assume}} \cr & a = \frac{1}{2} \cr & b = 2 \cr & c = - 1 \cr & \therefore abc \cr & = \frac{1}{2} \times 2 \times - 1 \cr & = - 1 \cr} $$