Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {\text{ }}x = \root 3 \of {2 + \sqrt 3 } \cr & {x^3} = 2 + \sqrt 3 \cr & \frac{1}{{{x^3}}} = \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = 2 - \sqrt 3 \cr & \therefore {x^3}{\text{ + }}\frac{1}{{{x^3}}} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let }}\frac{a}{{q - r}} = \frac{b}{{r - p}} = \frac{c}{{p - q}} = k \cr & \frac{a}{{q - r}} = k{\text{ }}\left( {{\text{On multiplying by }}p} \right) \cr & pa = k\left( {pq - pr} \right)\,......(i) \cr & {\text{In the same way we can write}} \cr & {\text{qb = k}}\left( {qr - qp} \right)\,........(ii) \cr & {\text{And }}rc = k\left( {rp - rp} \right)\,.....iii) \cr & {\text{On adding equation (i), (ii) and (iii) }} \cr & pa + qb + rc \cr & = k\left( {pq - pr + qr - qp + rp - rq} \right) \cr & = 0 \cr} $$

Solution:

 $${x^3} + {y^3} + {z^3} - 3xyz$$ $$ = \frac{1}{2}\left( {x + y + z} \right)$$  $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$ $$ = \frac{1}{2}\left( {222 + 223 + 225} \right)$$    $$\left[ {{{\left( {222 - 223} \right)}^2} + {{\left( {223 - 225} \right)}^2} + {{\left( {225 - 222} \right)}^2}} \right]$$ $$\eqalign{ & = \frac{1}{2}\left( {670} \right)\left( {1 + 4 + 9} \right) \cr & = \frac{1}{2} \times 670 \times 14 \cr & = 4690 \cr} $$

Solution:

 $$\eqalign{ & {a^2} + {b^2} = ab\,........(i) \cr & {a^2} + {b^2} - ab = 0 \cr & \because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( 0 \right) \cr & \Rightarrow {a^3} + {b^3} = 0 \cr} $$

Solution:

 $$\eqalign{ & a + b = 2c \cr & {\text{Taking }}a = 2,{\text{ }}b = 4{\text{ and }}c = 3 \cr & {\text{So,}}2 + 4 = 2 \times 3 \cr & \boxed{6 = 6} \cr & {\text{Now,}}\frac{a}{{a - c}} + {\text{ }}\frac{c}{{b - c}} \cr & = \frac{2}{{2 - 3}} + {\text{ }}\frac{3}{{4 - 3}} \cr & = \frac{2}{{ - 1}} + \frac{3}{1} \cr & = 1 \cr} $$

Solution:

 x = 1 + √2 + √3 ⇒ x - 1 = √2 + √3 On squaring both sides, (x - 1)2 = (√2 + √3)2 ⇒ x2 - 2x + 1 = 2 + 3 + 2√6 Add 3 both side ⇒ x2 - 2x + 4 = 5 + 2√6 + 3 ⇒ x2 - 2x + 4 = 8 + 2√6 ⇒ x2 - 2x + 4 = 2(4 + √6)

Solution:

 $$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr} $$

Solution:

 $$\eqalign{ & {x^m} \times {x^n} = 1 \cr & {x^{m + n}} = {x^0}\left[ {{x^0} = 1} \right] \cr & m + n = 0 \cr & m = - n \cr} $$

Solution:

 $$\eqalign{ & 3x + \frac{1}{{2x}} = 5 \cr & \Rightarrow {\text{Multiply both sides by }}\frac{2}{3} \cr & \therefore 3x \times \frac{2}{3} + \frac{1}{2}x \times \frac{2}{3} = 5 \times \frac{2}{3} \cr & \Rightarrow 2x + \frac{1}{{3x}} = \frac{{10}}{3} \cr & \therefore {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 3.2x.\frac{1}{{3x}}\left( {{\text{ 2}}x{\text{ + }}\frac{1}{{3x}}} \right) = {\left( {\frac{{10}}{3}} \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 2\left( {\frac{{10}}{3}} \right) = \left( {\frac{{1000}}{{27}}} \right) \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000}}{{27}} - \frac{{20}}{3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000 - 180}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{820}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = 30\frac{{10}}{{27}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr & \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr & \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr & {\text{Now,}} \cr & {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr & \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr & \Rightarrow x + y = 8\,.......(i) \cr & Again, \cr & x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr & \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr & {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}xy = 1 \cr & {\text{Substitutes values in the question}}{\text{.}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr & \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr & \Rightarrow \frac{{63}}{{61}} \cr} $$