Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} = - 1 \cr & \Rightarrow \left( {{a^6} + 1} \right) = 0 \cr & {\text{So,}}{a^{18}} + {\text{ }}{a^{12}} + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( {{a^6} + 1} \right) + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$

According to the question,
Divisor = 2 × remainder
= 2 × 80
= 160
Again, 4 × quotient = 160
⇒ Quotient = $$\frac{{160}}{4}$$ = 40
∴ x = Divisor × Quotient + Remainder
x = 160 × 40 + 80 = 6480

$$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$

$$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$

$$\eqalign{ & \left( {{x^4} + 64} \right) \cr & = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 \cr & = {\left( {{x^2} + 8} \right)^2} - \left( {16{x^2}} \right) \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & = \left( {{x^2} + 4x + 8} \right)\left( {{x^2} - 4x + 8} \right) \cr} $$

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

$$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr & {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr & {y^2} = 0 \cr & x = - 1,y = 0 \cr & {\text{Now, }}{x^{31}} + {y^{35}} \cr & = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & = - 1 \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \therefore x = \frac{1}{y} \cr & \Leftrightarrow xy = 1 \cr & x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow x + y = \frac{{12}}{4} \cr & \Rightarrow x + y = 3 \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr & = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & {x^4} + 2{x^3} + a{x^2} + bx + 9 \cr & {\text{Put }}x = 1 \cr & = 1 + 2 \times 1 + a + b + 9 \cr & = 1 + 2 + a + b + 9 \cr & = 13 + a + b \cr} $$
To make a perfect square numbers value of a + b must be either 3 or 13
Now, option (B) a = 6, b = 7
$$\eqalign{ & \therefore a + b = 13 \cr & {\text{make perfect square}} \cr & \left( {25 = {5^2}} \right) \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} = 16,{\text{ }}{x^2} + {y^2} + {z^2} = 25{\text{ }} \cr & {\text{But }}b = c = 0,{\text{ But }}y = z = 0 \cr & {\text{}}a = 4{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 5 \cr & {\text{Now, }} \cr & ax + by + cz = 20 \cr & 4 \times 5 + 0 + 0 = 20 \cr & 20 = 20{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{Now, }}\frac{{a + b + c}}{{x + y + z}} \cr & = \frac{{4 + 0 + 0}}{{5 + 0 + 0}} \cr & = \frac{4}{5} \cr} $$