Practice MCQ Questions and Answer on Algebra

61.

If $$x^2+\frac{1}{x^2}=7,$$   then the value of $$x^3+\frac{1}{x^3}$$  where x > 0 is equal to:

• (A) 18
• (B) 12
• (C) 15
• (D) 16

Show Answer

 $$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & x + \frac{1}{x} = 3 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 = 18 \cr} $$

62.

If a + b + c = 26 and ab + bc + ca = 109, find the value of a² + b² + c² = ?

• (A) 458
• (B) 472
• (C) 452
• (D) 476

Show Answer

 $$\eqalign{ & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr} $$

63.

If $$x=p+\frac{1}{p}$$   and $$y=p-\frac{1}{p}$$   then the value of x⁴ - 2x²y² + y⁴ = ?

• (A) 24
• (B) 4
• (C) 16
• (D) 8

Show Answer

 $$\eqalign{ & x = p + \frac{1}{p}{\text{ }} \cr & y = p - \frac{1}{p} \cr & \therefore x + y = p + \frac{1}{p} + p - \frac{1}{p} \cr & \Leftrightarrow x + y = 2p \cr & \therefore x - y = p + \frac{1}{p} - p + \frac{1}{p} \cr & \Leftrightarrow x - y = \frac{2}{p} \cr & \therefore {x^4} - 2{x^2}{y^2} + {y^4} \cr & = {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2p \times \frac{2}{p}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

64.

If x = 2, y = 1 and z = -3, then x³ + y³ + z³ - 3xyz is equal to?

• (A) 6
• (B) 0
• (C) 2
• (D) 8

Show Answer

 $$\eqalign{ & {\text{According to the question,}} \cr & x = 2,{\text{ }}y = 1,{\text{ }}z = - 3 \cr & {x^3} + {y^3} + {z^3} - 3xyz = ? \cr & {\text{As we know that}} \cr & a + b + c = 0{\text{ then }}{a^3} + {b^3} + {c^3} - 3abc = 0 \cr & 2 + 1 - 3 = 0 \cr & \therefore {x^3} + {y^3} + {z^3} - 3xyz = 0 \cr} $$

65.

If $${\text{7}}^x\text{ = }\frac{1}{343}\text{,}$$   then the value of x is?

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

Show Answer

 $$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$ (If bases are equal then their power are also equal)

66.

If (135√5x - 2√2y) ÷ (3√5x - √2y) = Ax + By + $$\sqrt{10}$$ Cxy, then the value of (A + B - 9C) is:2322

• (A) 18
• (B) 12
• (C) 20
• (D) 10

Show Answer

 $$\eqalign{ & \left( {135\sqrt 5 {x^3} - 2\sqrt 2 {y^3}} \right) \div \left( {3\sqrt 5 x - \sqrt 2 y} \right) = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & \Rightarrow \frac{{{{\left( {3\sqrt 5 x} \right)}^3} - {{\left( {\sqrt 2 y} \right)}^3}}}{{3\sqrt 5 x - \sqrt 2 y}} = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & \Rightarrow \frac{{\left( {3\sqrt 5 x - \sqrt 2 y} \right)\left( {45{x^2} + 2{y^2} + 3\sqrt {10} xy} \right)}}{{\left( {3\sqrt 5 x - \sqrt 2 y} \right)}} = A{x^2} + B{y^2} + \sqrt {10} Cxy \cr & {\text{Comparison both side,}} \cr & A = 45,\,B = 2,\,C = 3 \cr & A + B - 9C = \left( {45 + 2 - 27} \right) \cr & A + B - 9C = 20 \cr} $$

67.

The value of $$\frac{1}{a^2+ax+x^2}$$   $$-$$ $$\frac{1}{a^2-ax+x^2}$$   $$+$$ $$\frac{2ax}{a^4+a^2{x}^2+x^4}$$    is?

• (A) 2
• (B) 1
• (C) -1
• (D) 0

Show Answer

 $$\frac{1}{{{a^2} + ax + {x^2}}}$$   $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$   $$ + $$ $$\frac{2ax}{{{a^4} + {a^2}{x^2} + {x^4}}}$$ $$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$       $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$ $$\eqalign{ & = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = 0 \cr} $$

68.

If $$x+\frac{1}{x}=2,$$   x â‰Â  0, then the value of $$x^2\text{ + }\frac{1}{x^3}$$   is equal to?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Show Answer

 $$\eqalign{ & {\text{ }}x + \frac{1}{x} = 2{\text{, }}\,\,\,x \ne 0 \cr & {\text{Put }}x = 1 \cr & 1 + 1 = 2 \cr & \therefore {x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

69.

If $$x-\frac{1}{x}=5,$$   x â‰Â  0, then what is the value of $$\frac{x^6+3x^3-1}{x^6-8x^3-1}?$$

• (A) $$\frac{3}{8}$$
• (B) $$\frac{13}{12}$$
• (C) $$\frac{4}{9}$$
• (D) $$\frac{11}{13}$$

Show Answer

 $$\eqalign{ & {\text{Given, }}x - \frac{1}{x} = 5 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = {5^3} + 3 \times 5 = 140 \cr & \frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}} \cr & = \frac{{{x^3} + 3 - \frac{1}{{{x^3}}}}}{{{x^3} - 8 - \frac{1}{{{x^3}}}}} \cr & = \frac{{{x^3} - \frac{1}{{{x^3}}} + 3}}{{{x^3} - \frac{1}{{{x^3}}} - 8}} \cr & = \frac{{140 + 3}}{{140 - 8}} \cr & = \frac{{143}}{{132}} \cr & = \frac{{13}}{{12}} \cr} $$

70.

If x + y = 2a, then the value of $$\frac{a}{x-a}$$  $$+$$ $$\frac{a}{y-a}$$  is?

• (A) 0
• (B) -1
• (C) 1
• (D) 2

Show Answer

 $$\eqalign{ & {\text{Given, }}x + y = 2a \cr & {\text{Find, }}\frac{a}{{x - a}} + \frac{a}{{y - a}}{\text{ = ?}} \cr & \mathop {\mathop x\limits_ \downarrow \,}\limits_3 + \mathop {\mathop y\limits_ \downarrow }\limits_1 = \mathop {\mathop {2a}\limits_ \downarrow }\limits_2 \cr & {\text{Let }}x = 3,{\text{ }}y = 1,{\text{ }}a = 2 \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{2}{{\left( {3 - 2} \right)}} + \frac{2}{{\left( {1 - 2} \right)}} \cr & = \frac{2}{1} + \frac{2}{{ - 1}} \cr & = 0 \cr} $$