Practice MCQ Questions and Answer on Algebra

1.

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

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

Show Answer

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

2.

If $$a - \frac{1}{{a - 5}} = 10$$   , then the value of $${\left( {a - 5} \right)^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}$$    is:

• (A) 140
• (B) 70
• (C) 100
• (D) 120

Show Answer

 $$a - \frac{1}{{a - 5}} = 10$$ We can write it is as (subtracting 5 from both sides) $$\eqalign{ & a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr & {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$ So, take the cube of this equation, $$\eqalign{ & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$

3.

If $$x + \frac{1}{x} = 99{\text{,}}$$   find the value of   $$\frac{{100x}}{{2{x^2} + 2 + 102x}}$$    is?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{2}$$
• (C) $$\frac{1}{3}$$
• (D) $$\frac{1}{4}$$

Show Answer

 $$\eqalign{ & x + \frac{1}{x} = 99 \cr & \therefore {x^2} + 1 = 99x \cr & \Rightarrow 2\left( {{x^2} + 1} \right) = 2 \times 99x \cr & \Rightarrow 2{x^2} + 2 = 198x \cr & = \frac{{100x}}{{2{x^2} + 2 + 102x}} \cr & = {\text{ }}\frac{{100x}}{{198x + 102x}} \cr & = \frac{{100x}}{{300x}} \cr & = \frac{1}{3} \cr} $$

4.

If the equation 2x² - 7x + 12 = 0 has two roots $$\alpha$$ and $$\beta$$, then the value of $$\frac{\alpha }{\beta }{\text{ + }}\frac{\beta }{\alpha }\,{\text{is?}}$$

• (A) $$\frac{7}{2}$$
• (B) $$\frac{1}{{24}}$$
• (C) $$\frac{7}{{24}}$$
• (D) $$\frac{{97}}{{24}}$$

Show Answer

 $$\eqalign{ & {\text{2}}{x^2} - 7x + 12 = 0 \cr & {\text{roots are }}\alpha {\text{ , }}\beta \cr & \therefore \alpha \beta = + \frac{c}{a},\alpha + \beta = \frac{{ - b}}{a} \cr & \therefore \alpha + \beta = + \frac{7}{2},\alpha \beta = \frac{{12}}{2} = 6 \cr & \therefore \frac{\alpha }{\beta } + \frac{\beta }{\alpha } \cr & = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} \cr & = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr & = \frac{{{{\left( {\frac{7}{2}} \right)}^2} - 2 \times 6}}{6} \cr & = \frac{{\frac{{49}}{4} - 12}}{6} \cr & = \frac{{49 - 48}}{{6 \times 4}} \cr & = \frac{1}{{24}} \cr} $$

5.

If a + b = 1, c + d = 1 and a - b = $$\frac{d}{c}{\text{,}}$$  then the value of c² - d² = ?

• (A) $$\frac{a}{b}$$
• (B) $$\frac{b}{a}$$
• (C) 1
• (D) -1

Show Answer

 $$\eqalign{ & a + b = 1 \cr & c + d = 1 \cr & a - b = \frac{d}{c} \cr & or\,\,\frac{1}{{a - b}} = \frac{c}{d} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{c}{d}\left( {\therefore a + b = 1} \right) \cr & {\text{By C & D rule }} \cr & \Rightarrow \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{a}{b} = \frac{{c + d}}{{c - d}} \cr} $$ Now multiply & divide by (c + d) $$\eqalign{ & \frac{a}{b} = \frac{{\left( {c + d} \right)}}{{\left( {c - d} \right)}} \times \frac{{\left( {c + d} \right)}}{{\left( {c + d} \right)}} = \frac{{{{\left( {c + d} \right)}^2}}}{{{c^2} - {d^2}}} \cr & \Rightarrow \frac{a}{b} = \frac{{{{\left( {c + d} \right)}^2}}}{{\left( {{c^2} - {d^2}} \right)}} \cr & \Rightarrow c + d = 1 \cr & \Rightarrow \frac{a}{b} = \frac{1}{{{c^2} - {d^2}}} \cr & \Rightarrow {c^2} - {d^2} = \frac{b}{a} \cr} $$

6.

If $$x + \frac{1}{x} = 5{\text{,}}$$   then the value of $$\frac{{5x}}{{{x^2} + 5x + 1}}$$   is?

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

Show Answer

 $$\eqalign{ & x + \frac{1}{x} = 5{\text{ then, }}\frac{{5x}}{{{x^2} + 5x + 1}} \cr & \Rightarrow \frac{5}{{x + 5 + \frac{1}{x}}} \cr & \Rightarrow \frac{5}{{x + \frac{1}{x} + 5}} \cr & \Rightarrow \frac{5}{{5 + 5}} \cr & \Rightarrow \frac{1}{2} \cr} $$

7.

If a² + b² = 4b + 6a - 13, then what is the value of a + b?

• (A) 3
• (B) 2
• (C) 5
• (D) 10

Show Answer

 a2 + b2 = 4b + 6a - 13 a2 + 9 - 3a + b2 + 4 - 4b = 0 (a - 3)2 + (b - 2)2 = 0 a = 3, b = 2 a + b = 5

8.

If $$a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$$   & $$b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{,}}$$    then the value of $$\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}{\text{ is?}}$$

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

Show Answer

 $$\eqalign{ & a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{ }} \cr & \therefore a = \frac{1}{b} \cr & a + b = a + \frac{1}{a} \cr & \Rightarrow \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 1 \right)}^2}}} \cr & \Rightarrow \frac{{6 + 2\sqrt 5 + 6 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow \frac{{12}}{4} \cr & \Rightarrow 3 \cr & \therefore \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & \Rightarrow \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} \cr & \Rightarrow 9 - 2 \cr & \Rightarrow 7\left( {ab = 1} \right) \cr & \therefore \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

9.

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

• (A) $$\frac{{43}}{{23}}$$
• (B) $$\frac{{47}}{{21}}$$
• (C) $$\frac{{41}}{{23}}$$
• (D) $$\frac{{45}}{{21}}$$

Show Answer

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & \left( {{\text{By squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = 23 \cr & {\text{Now,}} \cr & \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & {\text{Divided by }}{x^2}, \cr & \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr & \Rightarrow \frac{{43}}{{23}} \cr} $$

10.

If x + y + z = 17, xyz = 171 and xy + yz + zx = 111, then the value of $$\root 3 \of {\left( {{x^3} + {y^3} + {z^3} + xyz} \right)} $$     is:

• (A) -64
• (B) 0
• (C) 4
• (D) -4

Show Answer

 x3 + y3 + z3 - 3xyz = (x + y + z)[x2 + y2 + z2 - (xy + yz + zx)] (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) 289 = x2 + y2 + z2 + 2 × 111 x2 + y2 + z2 = 67 Now, x3 + y3 + z3 - 3xyz = 17(67 - 111) x3 + y3 + z3 - 3xyz = -44 × 17 x3 + y3 + z3 - 3xyz + 4xyz = -748 + 4xyz x3 + y3 + z3 + xyz = -748 + 4 × 171 x3 + y3 + z3 + xyz = -748 + 684 x3 + y3 + z3 + xyz = -64 $$\eqalign{ & \root 3 \of {\left( {{x^3} + {y^3} + {z^3} + xyz} \right)} \cr & = \root 3 \of { - 64} \cr & = - 4 \cr} $$