Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & x*y = {\left( {x + 3} \right)^2}\left( {y - 1} \right) \cr & \Leftrightarrow 5*4 = {\left( {5 + 3} \right)^2}\left( {4 - 1} \right) \cr & \Leftrightarrow 5*4 = 64 \times 3 \cr & \Leftrightarrow 5*4 = 192 \cr} $$

Solution:

 $$\eqalign{ & x = 5 - \sqrt {21} \cr & 2x = 10 - 2\sqrt {21} \,......(i) \cr & \Rightarrow 2x = {\left( {\sqrt 7 } \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2\left( {\sqrt 7 } \right)\left( {\sqrt 3 } \right) \cr & \Rightarrow 2x = {\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow x = \frac{1}{2}{\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr & \therefore \frac{{\sqrt x }}{{\sqrt {32 - 2x} - \sqrt {21} }} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {32 - \left( {10 - 2\sqrt {21} } \right)} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {22 + 2\sqrt {21} } - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {{{\left( {\sqrt {21} + 1} \right)}^2}} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {21} + 1 - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr} $$

Solution:

 $$\eqalign{ & 3a = 4b = 6c \cr & \Rightarrow \frac{{3a}}{{12}} = \frac{{4b}}{{12}} = \frac{{6c}}{{12}} \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{c}{2} = k \cr & \Rightarrow a = 4k,\,b = 3k,\,c = 2k \cr & a + b + c = 27\sqrt {29} \cr & 9k = 27\sqrt {29} \cr & k = 3\sqrt {29} \cr & a = 4 \times 3\sqrt {29} ,\,b = 3 \times 3\sqrt {29} ,\,c = 2 \times 3\sqrt {29} \cr & \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {29\left( {144 + 81 + 36} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$

Solution:

 a = 55 b = 17 c = -72 a + b + c = 55 + 17 - 72 = 0 ∴ a3 + b3 + c3 - 3abc = 0 (a + b + c) = 0 Answer = 0

Solution:

 $$\eqalign{ & {a^3} - {b^3} = 56 \cr & \Rightarrow a - b = 2 \cr & \,\,\,\,\left( {{\text{By cubing}}} \right) \cr & \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr & \Rightarrow 56 - 3ab \times 2 = 8 \cr & \Rightarrow - 6ab = 8 - 56 \cr & \Rightarrow 6ab = 48 \cr & \Rightarrow ab = 8 \cr & \left( {a - b} \right) = 2 \cr & \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr & \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr & \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr & \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr & \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr & \Rightarrow {a^2} + {b^2} = 20 \cr} $$

Solution:

 $$\eqalign{ & \left( {2 + \sqrt 3 } \right)a = \left( {2 - \sqrt 3 } \right)b = 1 \cr & \Rightarrow \frac{1}{a} = \left( {2 + \sqrt 3 } \right) \cr & {\text{By rationals}} \cr & \Rightarrow \frac{1}{b} = \left( {2 - \sqrt 3 } \right) \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 4 \cr} $$

Solution:

 $$\eqalign{ & x:y = 3:4 \cr & \therefore \frac{{5x - 2y}}{{7x + 2y}} \cr & = \frac{{5 \times 3 - 2 \times 4}}{{7 \times 3 + 2 \times 4}} \cr & = \frac{{15 - 8}}{{21 + 8}} \cr & = \frac{7}{{29}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{2x - y}}{{x + 2y}}\, \times \frac{1}{2}\left( {{\text{Cross multiply}}} \right) \cr & \Rightarrow 4x - 2y = x + 2y \cr & \Rightarrow 3x = 4y \cr & \Rightarrow x:y = 4:3 \cr & \therefore \frac{{3x - y}}{{3x + y}}{\text{ }} \cr & {\text{ = }}\frac{{3 \times 4 - 3}}{{3 \times 4 + 3}} \cr & {\text{ = }}\frac{{12 - 3}}{{12 + 3}} \cr & {\text{ = }}\frac{9}{{15}}{\text{ }} \cr & {\text{ = }}\frac{3}{5} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given}} \cr & x + y + z = 6 \cr & xy + yz + zx = 10 \cr & {\text{To find }}{x^3} + {y^3} + {z^3} - 3xyz{\text{ = ?}} \cr & \Rightarrow {\text{ Using formula,}} \cr} $$   $$ \Rightarrow {\left( {x + y + z} \right)^2} = $$   $${x^2} \,+$$ $$\, {y^2} \,+ $$ $$\, {z^2} \,+ $$ $$\, 2\left( {xy + yz + zx} \right)$$ $$\eqalign{ & \Rightarrow {6^2} = {x^2} + {y^2} + {z^2} + 2 \times 10 \cr & \Rightarrow 36 = {x^2} + {y^2} + {z^2} + 20 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 16 \cr} $$   $$ \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = $$     $$\left( {x + y + z} \right)$$  $$\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$$ $$\eqalign{ & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6\left[ {16 - 10} \right] \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6 \times 6 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 36 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given }}m = - 4,n = - 2, \cr & {\text{Find }}{m^3} - 3{m^2} + 3m + 3n + 3{n^2} + {n^3} = ? \cr & {\text{Putting value of }}m{\text{ and }}n \cr & \Rightarrow {\left( { - 4} \right)^3} - 3{\left( { - 4} \right)^2} + 3\left( { - 4} \right) + 3\left( { - 2} \right) + 3{\left( { - 2} \right)^2} + {\left( { - 2} \right)^3} \cr & \Rightarrow - 64 - 48 - 12 - 6 + 12 - 8 \cr & \Rightarrow - 64 - 60 - 2 \cr & \Rightarrow - 126 \cr} $$