Practice MCQ Questions and Answer on Algebra

1.

If 5√5x + 2√2y = (Ax + √2y)(Bx + 2y + Cxy), then the value of (A + B - C) is:3322222

• (A) 15
• (B) 20
• (C) 40
• (D) 30

Show Answer

 $$\eqalign{ & 5\sqrt 5 {x^3} + 2\sqrt 2 {y^3} = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\left( {\sqrt 5 x} \right)^3} + {\left( {\sqrt 2 y} \right)^3} = \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) \cr & \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\text{Comparison both side:}} \cr & A = \sqrt 5 \cr & B = 5 \cr & C = - \sqrt {10} \cr & {A^2} + {B^2} + {C^2} = 5 + 25 - 10 = 20 \cr} $$

2.

If p(x + y)² = 5 and q(x - y)² = 3 then the simplified value of p²(x + y)² + 4pqxy - q²(x - y)² is:

• (A) 2(p + q)
• (B) -(p + q)
• (C) -2(p +q)
• (D) p + q

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 $$\eqalign{ & p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr & {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr & p{\left( {2 + 1} \right)^2} = 5{\text{ and }}q{\left( {2 - 1} \right)^2} = 3 \cr & p = \frac{5}{9},\,q = 3 \cr & \to {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr & = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times 2 \times 1 - {\left( 3 \right)^2}{\left( {2 - 1} \right)^2} \cr & = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr & = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr & = \frac{{25 + 120 - 81}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Put the value of }}p{\text{ and }}q{\text{ in option A}} \cr & {\text{option A }} \to {\text{2}}\left( {p + q} \right) \cr & = 2\left( {\frac{5}{9} + 3} \right) \cr & = 2 \times \frac{{32}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Option A is satisfied}} \cr & {\text{So, 2}}\left( {p + q} \right)\,{\text{is answer}} \cr} $$

3.

If $$3\sqrt{\frac{1-a}{a}}+9=19-3\sqrt{\frac{a}{1-a}},$$      then what is the value of a?

• (A) $$\frac{3}{10},\frac{7}{10}$$
• (B) $$\frac{1}{10},\frac{9}{10}$$
• (C) $$\frac{2}{5},\frac{3}{5}$$
• (D) $$\frac{1}{5},\frac{4}{5}$$

Show Answer

 $$\eqalign{ & {\text{Put }}\sqrt {\frac{{1 - a}}{a}} = x \cr & \Rightarrow 3x + 9 = 19 - \frac{3}{x} \cr & \Rightarrow 3x + \frac{3}{x} = 10 \cr & \Rightarrow x = 3,\,\frac{1}{3} \cr & {\text{Now, }}\sqrt {\frac{{1 - a}}{a}} = 3 \cr & \Rightarrow \frac{{1 - a}}{a} = 9 \cr & \Rightarrow 10a = 1 \cr & \Rightarrow a = \frac{1}{{10}} \cr & {\text{and }}\sqrt {\frac{{1 - a}}{a}} = \frac{1}{3} \cr & \Rightarrow \frac{{1 - a}}{a} = \frac{1}{9} \cr & \Rightarrow 10a = 9 \cr & \Rightarrow a = \frac{9}{{10}} \cr & \therefore \,a = \frac{1}{{10}}\,{\text{and }}\frac{9}{{10}} \cr} $$

4.

If p = 124, then the value of $$\sqrt[3]{p\left(p^2+3p+3\right)+1}=?$$

• (A) 5
• (B) 7
• (C) 123
• (D) 125

Show Answer

 $$\eqalign{ & p = 124 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {125} \right)}^3}} \cr & = 125 \cr} $$

5.

What is the value of $$\frac{\left(a^2+b^2\right)\left(a-b\right)-\left(a^3-b^3\right)}{a^{2}b-a{b}^2}?$$

• (A) 0
• (B) 1
• (C) -1
• (D) 3

Show Answer

 $$\eqalign{ & \frac{{\left( {{a^2} + {b^2}} \right)\left( {a - b} \right) - \left( {{a^3} - {b^3}} \right)}}{{{a^2}b - a{b^2}}} \cr & = \frac{{\left( {{a^2} + {b^2}} \right)\left( {a - b} \right) - \left( {{a^3} - {b^3}} \right)}}{{ab\left( {a - b} \right)}} \cr & = \frac{{\left( {a - b} \right)\left( {{a^2} + {b^2} - {a^2} - {b^2} - ab} \right)}}{{ab\left( {a - b} \right)}} \cr & = - 1 \cr} $$

6.

If $$\frac{a}{b}=\frac{7}{9},\frac{b}{c}=\frac{3}{5}\text{,}$$    then the value of a : b : c is?

• (A) 7 : 9 : 15
• (B) 7 : 9 : 5
• (C) 21 : 35 : 45
• (D) 7 : 3 : 15

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 $$\eqalign{ & {\text{ }}\frac{a}{b} = \frac{7}{9},{\text{ }}\frac{b}{c} = \frac{3}{5} \cr & \therefore a\,\,\,:\,\,\,b\,\,\,:\,\,\,c \cr & \,\,\,\,\,7\,\,\,:\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{3_{ \times 3}}:\,\,\,{5_{ \times 3}} \cr & \overline {{\text{ }}\,\,{\text{ }}7\,\,\,:\,\,\,\,9\,\,:\,\,15{\text{ }}} \cr} $$

7.

If x + y + z = 19, x² + y² + z² = 133, and xz = y², x > z > 0, what is the value of (x - z)?

• (A) 0
• (B) 5
• (C) -2
• (D) -5

Show Answer

 x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2 (x - z) = ? (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) 361 = 133 + 2(xy + yz + y2) 228 = 2(x + y + z)y $$y = \frac{{114}}{{19}} = 6$$ x + z = 13 xz = 36 x - z = ? (x + z)2 - (x - z)2 = 4xz 169 - (x - z)2 = 144 x - z = 5

8.

The value of $${\left(x^{b+c}\right)}^{b-c}$$   $$\times$$ $${\left(x^{c+a}\right)}^{c-a}$$   $$\times$$ $${\left(x^{a+b}\right)}^{a-b}$$   is?

• (A) 1
• (B) 2
• (C) -1
• (D) 0

Show Answer

 $$\eqalign{ & {\left( {{x^{b + c}}} \right)^{b - c}} \times {\left( {{x^{c + a}}} \right)^{c - a}} \times {\left( {{x^{a + b}}} \right)^{a - b}}\left( {x \ne 0} \right) \cr & = {x^{{b^2} - {c^2}}} \times {x^{{c^2} - {a^2}}} \times {x^{{a^2} - {b^2}}} \cr & = {x^{{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}}} \cr & = {x^0} \cr & = 1 \cr} $$

9.

If x + y = 15, then the value of (x - 10)³ + (y - 5)³ is?

• (A) 25
• (B) 125
• (C) 625
• (D) 0

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 $$\eqalign{ & x + y = 15 \cr & \Rightarrow x - 10 = 5 - y \cr & \Rightarrow x - 10 = - \left( {y - 5} \right) \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr & \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$

10.

If $$x+\frac{1}{x}=2\text{,}$$   then the value of $$x^{12}-\frac{1}{x^{12}}$$   is?

• (A) -4
• (B) 4
• (C) 2
• (D) 0

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 $$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 2\ , . . . . . . (i) \cr & {\text{The value of }}{x^{12}} - \frac{1}{{{x^{12}}}} =\, ? \cr & \Rightarrow {\text{If }}x = 1 \cr & \Rightarrow x + \frac{1}{x} = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & {\text{Then, }}{x^{12}} - \frac{1}{{{x^{12}}}} \cr & \Rightarrow {1^{12}} - \frac{1}{{{1^{12}}}} \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$