Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr} $$
Applying componendo and dividendo
$$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$

$$\eqalign{ & \frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}} \cr & {\text{Assume }}x = y = z = 1 \cr & \Leftrightarrow \frac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{1 + 1 + 1}} \cr & \Leftrightarrow \frac{9}{3} \cr & \Leftrightarrow 3 \cr} $$

$$\eqalign{ & {\text{Put }}\left( {x + y + z} \right) = 10 \cr & x = 2 \cr & y = 3 \cr & z = 5 \cr & x\left( {x + y + z} \right) = 20 \cr & \Leftrightarrow 2\left( {10} \right) = 20 \cr & \Leftrightarrow 20 = 20 \cr & {\text{Similarly other will satisfied,}} \cr & {\text{So, value of 2}}\left( {x + y + z} \right) \cr & = 2\left( {10} \right) \cr & = 20 \cr} $$

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

$$\eqalign{ & {x^2} + {y^2} = 14 \cr & x + y = 4\,.............(i) \cr & {\text{Squaring both sides}} \cr & \Rightarrow {x^2} + {y^2} + 2xy = 16 \cr & \Rightarrow 14 + 2xy = 16 \cr & \Rightarrow 2xy = 2 \cr & \Rightarrow xy = 1 \cr & \Rightarrow {x^2} + {y^2} = 14 \cr} $$
Subtracting (2xy) from both sides
$$\eqalign{ & \Rightarrow {x^2} + {y^2} - 2xy = 14 - 2xy \cr & \Rightarrow {\left( {x - y} \right)^2} = 14 - 2 \times 1 \cr & \Rightarrow x - y = \sqrt {12} \cr & \Rightarrow x - y = 2\sqrt 3 \,.........(ii) \cr & {\text{Solve equation (i) and (ii)}} \cr & \Rightarrow y = 2 - \sqrt 3 \cr & \Rightarrow x = 2 + \sqrt 3 \cr} $$

$$\eqalign{ & \left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{1}{9}{a^6}{b^{11}}{c^7} \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{{\frac{1}{9}{a^6}{b^{11}}{c^7}}}{{\frac{{10}}{{27}}{a^2}b{c^3}}} \cr & \Rightarrow \frac{3}{{10}}{a^4}{b^{10}}{c^4} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \because a = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}{\text{, }}b = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{Find }}\frac{{{a^2}}}{b} + \frac{{{b^2}}}{a} = \,? \cr & \Rightarrow \frac{{{a^3} + {b^3}}}{{ab}} = \,? \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - 3ab\left( {a + b} \right)}}{{ab}} = \,? \cr & \Rightarrow a + b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} + \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\sqrt 3 }^2} - {{\sqrt 2 }^2}}} \cr & \Rightarrow \frac{{2\left( {{{\sqrt 3 }^2} + {{\sqrt 2 }^2}} \right)}}{{3 - 2}} \cr & \Rightarrow \frac{{2 \times \left( 5 \right)}}{1} \cr & \Rightarrow a + b = 10 \cr & {\text{Again, }} \cr & \Rightarrow a \times b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow ab = 1 \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^3} - 3ab\left( {a + b} \right)}}{{ab}} \cr & \Rightarrow \frac{{{{10}^3} - 3 \times 1 \times 10}}{1} \cr & \Rightarrow 1000 - 30 \cr & \Rightarrow 970 \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{{25}}{6} \cr & \frac{{{a^2}}}{{{b^2}}} = \frac{{{{25}^2}}}{{{6^2}}} \cr & \Rightarrow \frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}} \cr & \Rightarrow \frac{{{{25}^2} - {6^2}}}{{{{25}^2} + {6^2}}} \cr & \Rightarrow \frac{{625 - 36}}{{625 + 36}} \cr & \Rightarrow \frac{{589}}{{661}} \cr} $$

$$\eqalign{ & 2x + 3y = k\left( {2 = x,{\text{ }}0 = y} \right) \cr & \therefore 2 \times 2 + 3 \times 0 = k \cr & \Leftrightarrow k = 4 \cr} $$

Correct option is C.( becasue x = y)