Practice MCQ Questions and Answer on Algebra

411.

If $$\sqrt{3}+\frac{1}{\sqrt{3}}\text{,}$$   then the value of $$\left(x-\frac{\sqrt{126}}{\sqrt{42}}\right)$$ $$\left(x-\frac{1}{x-\frac{2\sqrt{3}}{3}}\right)$$   is?

• (A) $$\text{5}\frac{\sqrt{3}}{6}$$
• (B) $$\frac{2\sqrt{3}}{3}$$
• (C) $$\frac{5}{6}$$
• (D) $$\frac{2}{3}$$

Show Answer

 $$\eqalign{ & \because x = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{{3 + 1}}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{4}{{\sqrt 3 }} \cr & {\text{So, }}\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right) \cr & = \left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{4}{{\sqrt 3 }} - \frac{2}{{\sqrt 3 }}}}} \right) \cr & = \left( {\frac{{4\sqrt {42} - \sqrt {126} \times \sqrt 3 }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{2}{{\sqrt 3 }}}}} \right) \cr & = \frac{{4\sqrt {42} - 3\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}\left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2}} \right) \cr & = \left( {\frac{{\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{{8 - 3}}{{2\sqrt 3 }}} \right) \cr & = \frac{1}{{\sqrt 3 }} \times \frac{5}{{2\sqrt 3 }} \cr & = \frac{5}{6} \cr} $$

412.

If x, y are two positive real number and $$x^{\frac{1}{3}}=y^{\frac{1}{4}},$$   then which of the following relations is true?

• (A) x3 = y4
• (B) x3 = y
• (C) x = y4
• (D) x20 = y15

Show Answer

 $$\eqalign{ & {\text{ }}{x^{\frac{1}{3}}} = {y^{\frac{1}{4}}}, \cr & \Rightarrow {\text{LCM of 3, 4}} = 12{\text{ }} \cr & \therefore {\text{ }}{\left( {{x^{\frac{1}{3}}}} \right)^{12}} = {\left( {{y^{\frac{1}{4}}}} \right)^{12}} \cr & \Rightarrow {x^4} = {y^3} \cr & {\text{take power '5' on both sides}} \cr & \Rightarrow {\left( {{x^4}} \right)^5} = {\left( {{y^3}} \right)^5} \cr & \Rightarrow {x^{20}} = {y^{15}} \cr} $$

413.

If a^2x+2 = 1, where is a positive real number other than 1, then x is equal to?

• (A) -2
• (B) -1
• (C) 0
• (D) 1

Show Answer

 $$\eqalign{ & \Rightarrow {a^{2x + 2}} = 1 \cr & \Rightarrow {a^{2x + 2}} = {a^0} \cr & \Rightarrow 2x + 2 = 0 \cr & \Rightarrow x = - \frac{2}{2} \cr & \Rightarrow x = - 1 \cr} $$

414.

If $$x+\frac{1}{x}=3\text{,}$$   where $$x\ne 0\text{,}$$  then the value of $$\frac{x^4+3x^3+5x^2+3x+1}{x^4+1}$$     = ?

• (A) 3
• (B) 5
• (C) 7
• (D) 2

Show Answer

 $$\eqalign{ & x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + 1 = 3x\,.....(i) \cr & \Rightarrow {\left( {{x^2} + 1} \right)^2} = {\left( {3x} \right)^2} \cr & \Rightarrow {x^4} + 1 + 2{x^2} = 9{x^2} \cr & \Rightarrow {x^4} + 1 = 7{x^2}\,.....(ii) \cr & \therefore \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{7{x^2} + 3{x^3} + 5{x^2} + 3x}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{12{x^2} + 3{x^3} + 3x}}{{7{x^2}}} \cr & {\text{From equation (i)}} \cr & \Rightarrow \frac{{12x + 3\left( {{x^2} + 1} \right)}}{{7x}} \cr & \Rightarrow \frac{{12x + 3 \times 3x}}{{7x}} \cr & \Rightarrow \frac{{21x}}{{7x}} \cr & \Rightarrow 3 \cr} $$

415.

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

• (A) $$\frac{2}{3}$$
• (B) 2
• (C) 1
• (D) 4

Show Answer

 $$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{2}{{{x^2} - x + 2}} = ? \cr & x + \frac{1}{x} = 1 \cr & {x^2} + 1 = x \cr & \left( {{x^2} - x} \right) = - 1 \cr & {\text{Putting value in,}} \cr & = \frac{2}{{{x^2} - x + 2}} \cr & = \frac{2}{{ - 1 + 2}} \cr & = 2 \cr} $$

416.

If pq(p + q) = 1, then the value of $$\frac{1}{p^3{q}^3}$$  - $$p^3$$ - $$q^3\text{,}$$  is equal to?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Show Answer

 $$\eqalign{ & pq\left( {p + q} \right) = 1 \cr & \Rightarrow p + q = \frac{1}{{pq}} \cr & \,\,\,\,\,\,\, \text{Cubing both side} \cr & \Rightarrow {p^3} + {q^3} +3 pq\left( {p + q} \right) = \frac{1}{{{p^3}{q^3}}} \cr & \,\,\,\,\,\,\, \text{Puting the value of } pq = \frac{1}{\left({p + q}\right)} \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = \left( {\frac{{3}}{{p + q}}} \right)\left( {p + q} \right) \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = 3 \cr} $$

417.

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

Show Answer

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

418.

The third proportional of the following numbers (x - y)², (x² - y²) = ?

• (A) (x + y3) (x + y4)
• (B) (x + y)4 (x - y)2
• (C) (x - y) (x + y)2
• (D) (x + y)2 (x - y)3

Show Answer

 $$\eqalign{ & {\text{Let,}} \cr & a = {\left( {x - y} \right)^2}, \cr & b = \left( {{x^2} - {y^2}} \right){\text{and}} \cr & c\,\,{\text{be}}\,{\text{the}}\,{\text{third}}\,{\text{proportional}} \cr & {\text{Therefore}}\,\,\,a:b::b:c \cr & i.e.\,\,\,c = \frac{{{b^2}}}{a} \cr & \Rightarrow c = \frac{{{{\left( {{x^2} - {y^2}} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {x - y} \right)}^2}{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {x - y} \right){\left( {x + y} \right)^2} \cr} $$

419.

If ab + bc + ca = 8 and a² + b² + c² = 20, then a possible value of $$\frac{1}{2}$$(a + b + c)[(a - b)² + (b - c)² + (c - a)²] is:

• (A) 72
• (B) 56
• (C) 84
• (D) 80

Show Answer

 ab + bc + ca = 8 a2 + b2 + c2 = 20 a2 + b2 + c2 - ab - bc - ca = $$\frac{1}{2}$$[(a - b)2 + (b - c)2 + (c - a)2] 20 - 8 = $$\frac{1}{2}$$[(a - b)2 + (b - c)2 + (c - a)2)] 12 = $$\frac{1}{2}$$[(a - b)2 + (b - c)2 + (c - a)2] (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) a + b + c = $$\sqrt {20 + 2 \times 8} = 6$$ $$\frac{1}{2}$$(a + b + c)[(a - b)2 + (b - c)2 + (c - a)2] = 12(a + b+ c) = 12 × 6 = 72

420.

x is a negative number such that k + k^-1 = -2, then what is the value of $$\frac{k^2+4k-2}{k^2+k-5}?$$

• (A) 7
• (B) 1
• (C) -7
• (D) -1

Show Answer

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