Algebra — Part 2 MCQs with Explanations

$$\eqalign{ & x + y = 7 \cr & \left( {{\text{cubing both sides}}} \right){\text{ }} \cr & \Rightarrow {\left( {x + y} \right)^3} = {\left( 7 \right)^3} \cr & \Rightarrow {x^3} + {y^3} + 3\left( {x + y} \right)xy = 343 \cr & \Rightarrow {x^3} + {y^3} + 21xy = 343 \cr} $$

$$\eqalign{ & {\text{Given, }}a \otimes b = 2a{\text{ }} \cr & {\text{Where }}a > b \cr & a \otimes b = a + b \cr & {\text{Where }}a < b \cr & a \otimes b = {a^2} \cr & {\text{Where }}a = b \cr & \frac{{\left( {5 \otimes 7} \right) + \left( {4 \otimes 4} \right)}}{{3\left( {5 \otimes 5} \right) - \left( {15 \otimes 11} \right) - 3}}{\text{ }} \cr & {\text{ = }}\frac{{\left( {5 + 7} \right) + {{\left( 4 \right)}^2}}}{{3{{\left( 5 \right)}^2} - \left( {2 \times 15} \right) - 3}} \cr & {\text{ = }}\frac{{12 + 16}}{{75 - 30 - 3}} \cr & {\text{ = }}\frac{{28}}{{42}} \cr & {\text{ = }}\frac{2}{3} \cr} $$

$$\eqalign{ & {\left( {125} \right)^x} = 3125 \cr & \Rightarrow {\left( {{5^3}} \right)^x} = {\left( 5 \right)^5} \cr & \Rightarrow {\left( 5 \right)^{3x}} = {\left( 5 \right)^5} \cr & \Rightarrow 3x = 5 \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\eqalign{ & {\text{ }}n + \frac{2}{3}n + \frac{1}{2}n + \frac{1}{7}n = 97 \cr & \Rightarrow \frac{{42n + 28n + 21n + 6n}}{{42}} = 97 \cr & \Rightarrow \frac{{97n}}{{42}} = 97 \cr & \Rightarrow n = 42 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 4 \cr & \left( {{\text{on squaring }}} \right) \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 - 2 = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 20 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 20 \cr & \Rightarrow x + \frac{1}{x} = \sqrt {20} \cr & \Rightarrow x + \frac{1}{x} = \sqrt {4 \times 5} \cr & \Rightarrow x + \frac{1}{x} = 2\sqrt 5 \cr} $$

$$\eqalign{ & 4{b^2} + \frac{1}{{{b^2}}} = 2 \cr & \Rightarrow {\left( {2b} \right)^2} + {\left( {\frac{1}{b}} \right)^2} + 4 - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow 2b + \frac{1}{b} = \sqrt 6 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3 \times 2b \times \frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 6\sqrt 6 - 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 0 \cr} $$

$$\eqalign{ & {\left( {\frac{3}{5}} \right)^3} \times {\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\left( {\frac{3}{5}} \right)^{3 - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow - 3 = 2x - 1 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & \frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = \frac{2}{1}{\text{ }} \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1} \cr & \left[ {\frac{{\text{A}}}{{\text{B}}} = \frac{{\text{C}}}{{\text{D}}}} \right] \cr & \left[ {\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = \frac{{{\text{C}} + {\text{D}}}}{{{\text{C}} - {\text{D}}}}} \right] \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = 3{\text{ }} \cr & {\text{Squaring both sides}} \cr & \Rightarrow \frac{{3 + x}}{{3 - x}} = 9 \cr & \Rightarrow 3 + x = 27 - 9x \cr & \Rightarrow 10x = 24 \cr & \Rightarrow x = \frac{{24}}{{10}} \cr & \Rightarrow x = \frac{{12}}{5} \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & = \frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }} \times \frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }} \cr & = \frac{{{{\left( {\sqrt {1 + x} + \sqrt {1 - x} } \right)}^2}}}{{{{\left( {\sqrt {1 + x} } \right)}^2} - {{\left( {\sqrt {1 - x} } \right)}^2}}} \cr & = \frac{{1 + x + 1 - x + 2\sqrt {1 - {x^2}} }}{{1 + x - 1 + x}} \cr & = \frac{{2 + 2\sqrt {1 - {x^2}} }}{{2x}} \cr & = \frac{{1 + \sqrt {1 - {x^2}} }}{x} \cr & = \frac{{1 + \sqrt {1 - \frac{3}{4}} }}{{\sqrt 3 }} \times 2 \cr & = \frac{{\left( {1 + \frac{1}{2}} \right)}}{{\sqrt 3 }} \times 2 \cr & = \frac{{\frac{3}{2} \times 2}}{{\sqrt 3 }} \cr & = \sqrt 3 \cr} $$

$$\eqalign{ & {4^{4x + 1}} = \frac{1}{{64}} \cr & \Rightarrow {4^{4x + 1}} = \frac{1}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow {4^{4x + 1}} = {\left( 4 \right)^{ - 3}} \cr & \Rightarrow 4x + 1 = - 3 \cr & \Rightarrow 4x = - 4 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\left( {\sqrt 5 } \right)^7} \div {\left( {\sqrt 5 } \right)^5}{\text{ = }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \frac{{{{\left( {\sqrt 5 } \right)}^7}}}{{{{\left( {\sqrt 5 } \right)}^5}}} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {\left( {\sqrt 5 } \right)^2} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {{\text{5}}^{\text{1}}} = {\text{ }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \boxed{{\text{P}} = 1} \cr} $$

$$\eqalign{ & 1.5a = 0.04b \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} = \frac{4}{{100}} \times \frac{{10}}{{15}} = \frac{2}{{75}} \cr & {\text{Let }}a = 2x,{\text{ }}b = 75x \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{75x - 2x}}{{75x + 2x}} = \frac{{73}}{{77}} \cr & \cr & {\bf{Alternate:}} \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{1.5 - 0.04}}{{1.5 + 0.04}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.46}}{{1.54}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{73}}{{77}} \cr} $$

$$\eqalign{ & \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{x\left\{ {\left( {x + \frac{1}{x}} \right) - 1} \right\}}}{{x\left\{ {\left( {x + \frac{1}{x}} \right) + 1} \right\}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\left( {x + \frac{1}{x}} \right) - 1}}{{\left( {x + \frac{1}{x}} \right) + 1}} = \frac{3}{2} \cr & \,\,\,\,\,\,\,\,\,\,{\text{Let }}\left( {x + \frac{1}{x} = y} \right) \cr & \Rightarrow \frac{{y - 1}}{{y + 1}} = \frac{3}{2} \cr & \Rightarrow 2\left( {y - 1} \right) = 3\left( {y + 1} \right) \cr & \Rightarrow 2y - 2 = 3y + 3 \cr & \Rightarrow y = - 2 - 3 \cr & \Rightarrow y = - 5 \cr & \therefore x + \frac{1}{x} = - 5 \cr & {\text{ }} \cr} $$

$$\eqalign{ & \Rightarrow x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 2 \times 3\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 6\sqrt 8 \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 6\sqrt 8 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} \cr & = 17 + 6\sqrt 8 + 17 - 6\sqrt 8 \cr & = 34 \cr} $$

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

$$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \frac{{{x^3} + \frac{1}{x}}}{{{x^2} - x + 1}}{\text{ }}\left( {{\text{Divide by }}x} \right) \cr & \Rightarrow \frac{{\frac{{{x^3}}}{x} + \frac{1}{{{x^2}}}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - 1 + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} \cr & \therefore x + \frac{1}{x} = 3 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 9 - 2 = 7 \cr & \therefore \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} = \frac{7}{{3 - 1}} = \frac{7}{2} \cr} $$

$$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$

$$\eqalign{ & x = a + \frac{1}{a} \cr & y = a - \frac{1}{a} \cr & \therefore \left( {x + y} \right) = a + \frac{1}{a} + a - \frac{1}{a} = 2a \cr & \therefore \left( {x - y} \right) = a + \frac{1}{a} - a + \frac{1}{a} = \frac{2}{a} \cr & \therefore {x^4}{\text{ + }}{{\text{y}}^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2a \times \frac{2}{a}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

$$\eqalign{ & p = 101 \cr & \root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} \cr & = \root 3 \of {{p^3} - 3{p^2} + 3p - 1} \cr & \therefore \left[ {{{\left( {p - 1} \right)}^3} = {p^3} - {{\left( 1 \right)}^3} - 3p\left( {p - 1} \right)} \right] \cr & = \root 3 \of {{{\left( {p - 1} \right)}^3}} \cr & = p - 1 \cr & = 101 - 1 \cr & = 100{\text{ }} \cr} $$

$$\eqalign{ & 50\% {\text{ of }}\left( {p - q} \right) = 30\% {\text{ of }}\left( {p + q} \right) \cr & \frac{{p - q}}{2} = \frac{3}{{10}}\left( {p + q} \right) \cr & \left[ {50\% = \frac{1}{2}} \right] \cr & \Rightarrow 5\left( {p - q} \right) = 3\left( {p + q} \right) \cr & \Rightarrow 5p - 5q = 3p + 3q \cr & \Rightarrow 2p = 8q \cr & \Rightarrow 1p = 4q \cr & \Rightarrow p:q = 4:1 \cr} $$

$$\eqalign{ & x:y = 2:1 \cr & {\text{then, }}5{x^2} - 13xy + 6{y^2} \cr & \Rightarrow 5 \times 4 - 13 \times 2 \times 1 + 6 \times {1^2} \cr & \Rightarrow 20 - 26 + 6 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{a}{3} = \frac{b}{2} \Rightarrow \frac{a}{b} = \frac{3}{2} \cr & \therefore \frac{{2a + 3b}}{{3a - 2b}} \cr & = \frac{{2 \times 3 + 3 \times 2}}{{3 \times 3 - 2 \times 2}} \cr & = \frac{{6 + 6}}{{9 - 4}} \cr & = \frac{{12}}{5} \cr} $$

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

$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

$$\eqalign{ & \sqrt {1 - \frac{{{x^3}}}{{100}}} = \frac{3}{5} \cr & \Rightarrow 1 - \frac{{{x^3}}}{{100}} = {\left( {\frac{3}{5}} \right)^2} \cr & \Rightarrow 1 - \frac{9}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow \frac{{16}}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow {x^3} = \frac{{16 \times 100}}{{25}} \cr & \Rightarrow {x^3} = 16 \times 4 \cr & \Rightarrow {x^3} = 64 \cr & \Rightarrow \boxed{x = 4} \cr} $$

$$\eqalign{ & \sqrt {1 + \frac{x}{9}} = \frac{{13}}{3} \cr & {\text{By option }} \cr & {\text{Put }}x = 160 \cr & \sqrt {1 + \frac{{160}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \sqrt {\frac{{169}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \frac{{13}}{3} = \frac{{13}}{3} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Squaring both sides}} \cr & {\left( {\sqrt {1 + \frac{{x}}{9}} } \right)^2} = {\left( {\frac{{13}}{3}} \right)^2} \cr & \Rightarrow 1 + \frac{x}{9} = \frac{{169}}{9} \cr & \Rightarrow \frac{{9 + x}}{9} = \frac{{169}}{9} \cr & \Rightarrow 9 + x = 169 \cr & \Rightarrow \boxed{x = 160} \cr} $$

$$\eqalign{ & \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{\left( {4\sqrt 3 + 5\sqrt 2 } \right)\left( {4\sqrt 3 - 3\sqrt 2 } \right)}}{{48 - 18}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 + 18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 }}{{30}} + \frac{{18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{4}{{15}}\sqrt 6 + \frac{3}{5} = a + b\sqrt 6 \cr & \Rightarrow \frac{3}{5} + \frac{4}{{15}}\sqrt 6 = a + b\sqrt 6 \cr} $$ By comparing coefficients of rational and irrational parts. $$\eqalign{ & \Rightarrow a = \frac{3}{5}{\text{ , }}b = \frac{4}{{15}} \cr & \therefore \left( {\frac{3}{5},\frac{4}{{15}}} \right) \cr} $$

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

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

$$\eqalign{ & \sqrt {1 + \frac{x}{{961}}} = \frac{{32}}{{31}} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow 1 + \frac{x}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow \frac{{961 + x}}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow x = 1024 - 961 \cr & \Rightarrow x = 63 \cr} $$

$$\eqalign{ & {\text{ }}\sqrt {0.04 \times 0.4 \times a} = 0.004 \times 0.4 \times \sqrt b \cr & \Rightarrow \sqrt {\frac{4}{{100}} \times \frac{4}{{10}} \times a} = \frac{4}{{1000}} \times \frac{4}{{10}} \times \sqrt b \cr & \Rightarrow \frac{4}{{10}}\sqrt {\frac{a}{{10}}} = \frac{4}{{10}} \times \frac{4}{{1000}} \times \sqrt b \cr & \Rightarrow \frac{a}{{10}} = \frac{{16}}{{1000000}} \times b \cr & \Rightarrow \frac{a}{b} = \frac{{16 \times 10}}{{1000000}} \cr & \Rightarrow \frac{a}{b} = \frac{{16}}{{100000}} \cr & \Rightarrow \frac{a}{b} = 16 \times {10^{ - 5}} \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{3}{1} \cr & \Rightarrow \frac{{2 \times 9 + 3 \times 9 + 4 \times 9}}{{2 \times 1 + 3 \times 1 + 4 \times 1}} \cr & \Rightarrow \frac{{18 + 27 + 36}}{{2 + 3 + 4}} \cr & \Rightarrow \frac{{81}}{9} \cr & \Rightarrow 9 \cr} $$

$$\eqalign{ & x \propto \frac{1}{{{y^2} - 1}}{\text{ }}\left( {{\text{Given}}} \right) \cr & x = k \times \frac{1}{{{y^2} - 1}}\left( {k{\text{ is constant}}} \right) \cr & {\text{Now }}x = 24{\text{ when }}y = 10{\text{ given}} \cr & \Rightarrow 24 = k \times \frac{1}{{{{\left( {10} \right)}^2} - 1}} \cr & \Rightarrow 24 = \frac{k}{{99}} \cr & \Rightarrow k = 24 \times 99 \cr & x = ? \cr & y = 5 \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{25 - 1}} \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{24}} \cr & \Rightarrow x = 99 \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {x^2} + 2x + 1 + {y^2} = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr} $$ Hence both terms are squares and there addition is zero. So, it can be possible only when both terms are zeros. $$\eqalign{ & \therefore x + 1 = 0 \cr & \Rightarrow x = - 1 \cr & y = 0 \cr & \therefore {x^{31}} + {y^{35}} \cr & \Rightarrow {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & \Rightarrow - 1 \cr} $$

a 2 + b 2 + c 2 = 2(a - b - c) - 3 ⇒ a 2 + b 2 + c 2 = 2a - 2b - 2c - 3 ⇒ a 2 + b 2 + c 2 - 2a + 2b +2c + 1 + 1 + 1 = 0 ⇒ (a 2 - 2a + 1) + (b 2 + 2b + 1) + (c 2 + 2c + 1) = 0 ⇒ (a - 1) 2 + (b + 1) 2 + (c + 1) 2 = 0 a = 1 b = -1 c = -1 ∴ 2a - 3b + 4c = 2 × 1 - 3 × (-1) + 4 × (-1) = 2 + 3 - 4 = 1

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

$$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$ Now, $$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} \times \frac{{\left( {a - b} \right)}}{{a - b}}$$ Multiply divide by (a - b) in 1 st term Multiply divide by (b - c) in 2 nd term Multiply divide by (c - a) in 3 rd term $$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {a - b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$$ $$\eqalign{ & {\text{Let, }}a - b = x \cr & b - c = y \cr & c - a = z \cr & \therefore x + y + z = 0 \cr & \therefore {x^3} + {y^3} + {z^3} = 3xyz \cr & \therefore {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \cr & = 3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) \cr & \therefore \frac{{3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \cr & = 3 \cr} $$

$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} = 11 \cr & \Leftrightarrow a = {11^3} = 1331 \cr & {a^2} - 331a \cr & = a\left( {a - 331} \right) \cr & = 1331\left( {1331 - 331} \right) \cr & = 1331\left( {1000} \right) \cr & = 1331000 \cr} $$

$$\eqalign{ & {a^2} + {b^2} = 2{\text{ }} \cr & {c^2} + {d^2} = 1 \cr & {\text{Put values of }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \cr & {\text{Take ,}}a = b = 1 \cr & c = 1 \cr & d = 0 \cr & \Rightarrow {\left( {ad - bc} \right)^2}{\text{ + }}{\left( {ac - bd} \right)^2} \cr & \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {1 + 0} \right)^2} \cr & \Rightarrow {\left( { - 1} \right)^2} + {\left( 1 \right)^2} \cr & \Rightarrow 2 \cr} $$