Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {a^2} + {b^2} + {c^2} + 3 = 2\left( {a - b - c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 3 - 2a + 2b + 2c = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0 \cr & a = 1 \cr & b = - 1 \cr & c = - 1 \cr & \therefore 2a - b + c \cr & = 2 + 1 - 1 \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$ Now the term whose denominator is largest is the smallest term. So, a b c

Solution:

 $$\eqalign{ & \sqrt {0.03 \times 0.3a} = {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times \sqrt b \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{0}}{\text{.03}} \times {\text{0}}{\text{.3a}} = {\left( {0.3} \right)^2} \times {\left( {0.3} \right)^2} \times b \cr & \Rightarrow \frac{3}{{100}} \times \frac{3}{{10}}a = \frac{9}{{100}} \times \frac{9}{{100}} \times b \cr & \Rightarrow 9a = \frac{{81}}{{10}}b \cr & \Rightarrow 90a = 81b \cr & \Rightarrow 10a = 9b \cr & \Rightarrow \frac{a}{b} = \frac{9}{{10}} \cr & \Rightarrow \frac{a}{b} = 0.9 \cr} $$

Solution:

 $$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} \cr & \Rightarrow {\text{A}}:{\text{B}} = 8:6 \cr & \Rightarrow {\text{A}}:{\text{B}} = 4:3 \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} \cr & \Rightarrow {\text{B}}:{\text{C}} = 9:15 \cr & \Rightarrow {\text{B}}:{\text{C}} = 3:5 \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} \cr & \Rightarrow {\text{C}}:{\text{D}} = 20:18 \cr & \Rightarrow {\text{C}}:{\text{D}} = 10:9 \cr & {\text{A}}:{\text{ B }}:{\text{C }}:{\text{D}} \cr & 4{\text{ }}:{\text{ }}3 \cr & \,\,\,\,\,\,\,\,\,\,3{\text{ }}:{\text{ }}5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{\text{ }}:{\text{ }}9 \cr & \overline {\underline {{\text{ }}8{\text{ }}:6{\text{ }}:10{\text{ }}:{\text{ }}9{\text{ }}} } \cr} $$

Solution:

 x = y = 333,     z = 334 ⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2] ⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)2 + (333 - 334)2 + (334 - 333)2 ⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1) ⇒ x3 + y3 + z3 - 3xyz = 1000

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:

 $$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = \frac{{31}}{9} \cr & x + \frac{1}{x} = \sqrt {\frac{{31}}{9} + 2} \cr & x + \frac{1}{x} = \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = {n^3} - 3n \cr & {x^3} + \frac{1}{{{x^3}}} = {\left( {\frac{7}{3}} \right)^3} - 3 \times \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343}}{{27}} - 7 \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343 - 7 \times 27}}{{27}} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{154}}{{27}} \cr} $$

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:

 a - b = 18, a3 - b3 = 324 (a - b)(a2 + b2 + ab) = 324 ⇒ 18[(a - b)2 + 3ab] = 324 ⇒ 324 + 3ab = 18 ⇒ 3ab = 18 - 324 ⇒ 3ab = -306 ⇒ ab = -102

Solution:

 $$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Leftrightarrow x = 3 + 2 + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Leftrightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & {\text{Similarly,}} \cr & \Leftrightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \therefore \left( {\sqrt x + \frac{1}{{\sqrt x }}} \right) \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$