Practice MCQ Questions and Answer on Algebra

131.

If $$a+\frac{1}{a}+1=0\left(a\ne 0\right)\text{,}$$     then the value of $$\left(a^4-a\right)\text{is?}$$

• (A) 0
• (B) 1
• (C) 2
• (D) -1

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 $$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$

132.

If $$p^2+q^2=7pq\text{,}$$    then the value of $$\frac{p}{q}\text{ + }\frac{q}{p}$$   is equal to?

• (A) 9
• (B) 5
• (C) 7
• (D) 3

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 $$\eqalign{ & {p^2} + {q^2} = 7pq \cr & \frac{p}{q}{\text{ + }}\frac{q}{p} = 7 \cr} $$ (Divide whole equation by pq)

133.

If x² - 3x + 1 = 0, then the value of $$\frac{x^6+x^4+x^2+1}{x^3}$$     will be?

• (A) 18
• (B) 15
• (C) 21
• (D) 30

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 $$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{\text{ }} \cr & = \frac{{{x^6}}}{{{x^3}}} + \frac{{{x^4}}}{{{x^3}}} + \frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}} \cr & = {x^3} + \frac{1}{{{x^3}}} + \frac{1}{x} + x \cr & = 18 + 3 \cr & = 21 \cr} $$

134.

If $$x+\frac{4}{x}-4=0,$$   then the value of x² - 4 is equal to:

• (A) 2
• (B) 4
• (C) 0
• (D) 1

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 $$\eqalign{ & x + \frac{4}{x} = 4 \cr & x = 2 \cr & {\text{So, }}{x^2} - 4 = ? \cr & {x^2} - 4 = 4 - 4 = 0 \cr} $$

135.

a + b + c = 0, then the value of $$\frac{a^2+b^2+c^2}{ab+bc+ca}$$   is?

• (A) 2
• (B) -2
• (C) 0
• (D) 4

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 $$\eqalign{ & {\text{If }}a + b + c = 0 \cr & {\text{Put }}a = 1,{\text{ }}b = 1{\text{ and }}c = - 2 \cr & \therefore \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} \cr & = \frac{{1 + 1 + 4}}{{1 - 2 - 2}} \cr & = \frac{6}{{ - 3}} \cr & = - 2 \cr} $$

136.

If x⁴ - 83x² + 1 = 0, then a value of x³ - x^-3 can be:

• (A) 758
• (B) 756
• (C) 739
• (D) 737

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 $$\eqalign{ & {x^4} - 83{x^2} + 1 = 0 \cr & {x^2} - 83 + \frac{1}{{{x^2}}} = 0 \cr & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & x - \frac{1}{x} = 9 \cr & {\left( {x - \frac{1}{x}} \right)^3} = {9^3} \cr & {x^3} - \frac{1}{{{x^3}}} - 3 \times 9 = 729 \cr & {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

137.

If x * y = (x + 3)² (y - 1), then the value of 5 * 4 is?

• (A) 192
• (B) 182
• (C) 180
• (D) 172

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 $$\eqalign{ & x*y = {\left( {x + 3} \right)^2}\left( {y - 1} \right) \cr & \Leftrightarrow 5*4 = {\left( {5 + 3} \right)^2}\left( {4 - 1} \right) \cr & \Leftrightarrow 5*4 = 64 \times 3 \cr & \Leftrightarrow 5*4 = 192 \cr} $$

138.

If x - √7x + 1 = 0, then what is the value of $$x^5+\frac{1}{x^5}?$$2

• (A) 19√7
• (B) 25√7
• (C) 27√7
• (D) 21√7

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 $$\eqalign{ & {x^2} - \sqrt 7 x + 1 = 0 \cr & {x^2} + 1 = \sqrt 7 x \cr & x + \frac{1}{x} = \sqrt 7 \cr & {x^5} + \frac{1}{{{x^5}}} = \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = \left( {{{\left( {\sqrt 7 } \right)}^2} - 2} \right)\left( {{{\left( {\sqrt 7 } \right)}^3} - 3 \times \sqrt 7 } \right) - \sqrt 7 \cr & = 5 \times 4\sqrt 7 - \sqrt 7 \cr & = 19\sqrt 7 \cr} $$

139.

If $$x+\frac{1}{x}=3\text{,}$$   then the value of $$\frac{3x^2-4x+3}{x^2-x+1}$$   is?

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

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 $$\eqalign{ & x + \frac{1}{x} = 3 \cr & \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr & = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr & = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr & = \frac{{9 - 4}}{2} \cr & = \frac{5}{2} \cr} $$

140.

Simplify $$\frac{x^2+2x+y^2}{x^3-5x^2}\text{if }x+\frac{y^2}{x}=5.$$

• (A) $$\frac{5}{y^2}$$
• (B) $$\frac{7}{y^2}$$
• (C) $$-\frac{5}{y^2}$$
• (D) $$-\frac{7}{y^2}$$

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 $$\eqalign{ & {\text{Given:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {\text{Put }}y = 2,\,x = 1 \cr & {\text{Now,}} \cr & \frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{{{\left( 1 \right)}^2} + 2 \times 1 + {{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2}}} \cr & = \frac{7}{{ - 4}} \cr & {\text{Put }}y = 2{\text{ in option and option D will satisfy the condition}} \cr & \cr & {\bf{Alternate \,solution:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {x^2} + {y^2} = 5x \cr & {x^2} - 5x = - {y^2} \cr & {\text{Now, }}\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{5x + 2x}}{{x\left( {{x^2} - 5x} \right)}} \cr & = \frac{{7x}}{{x\left( { - {y^2}} \right)}} \cr & = - \frac{7}{{{y^2}}} \cr} $$