$$\eqalign{ & \frac{{3a + 5b}}{{3a - 5b}} = 5 \cr & \Rightarrow 3a + 5b = 15a - 25b \cr & \Rightarrow 12a = 30b \cr & \Rightarrow 2a = 5b \cr & a:b \cr & 5:2 \cr} $$

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

$${\left( {a + b - c} \right)^2}{\text{ + }}{\left( {b + c - a} \right)^2}$$     $${\text{ + }}{\left( {c + a - b} \right)^2}$$
$$\eqalign{ & \Rightarrow a + b + c = 0{\text{ }}\left( {{\text{ Given}}} \right) \cr & \Rightarrow a + b = - c \cr & \Rightarrow b + c = - a \cr & \Rightarrow a + c = - b \cr} $$
$$ \Rightarrow {\left( {a + b - c} \right)^2} + {\left( {b + c - a} \right)^2}$$      $$ + {\left( {c + a - b} \right)^2}$$
$$\eqalign{ & \Rightarrow {\left( { - c - c} \right)^2}{\text{ + }}{\left( { - a - a} \right)^2}{\text{ + }}{\left( { - b - b} \right)^2} \cr & \Rightarrow {\left( { - 2c} \right)^2}{\text{ + }}{\left( { - 2a} \right)^2}{\text{ + }}{\left( { - 2b} \right)^2} \cr & \Rightarrow 4{c^2} + 4{a^2} + 4{b^2} \cr & \Rightarrow 4\left( {{a^2} + {b^2} + {c^2}} \right) \cr} $$

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

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

$$\eqalign{ & x = 2015 \cr & y = 2014 \cr & z = 2013 \cr & \therefore {x^2} + {y^2} + {z^2} - xy - yz - zx \cr & = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {2015 - 2014} \right)}^2} + {{\left( {2014 - 2013} \right)}^2} + {{\left( {2013 - 2015} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr & = 3 \cr} $$

$$\eqalign{ & 3{a^2} = {b^2}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{ }}\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}} \cr} $$
  $$ = \frac{{{a^3} + {b^3} + 3ab\left( {a + b} \right)\, - \,\left( {{a^3} - {b^3} - 3ab\left( {a - b} \right)} \right){\text{ }}}}{{{a^2} + {b^2} + 2ab + {\text{ }}{a^2} + {b^2} - 2ab}}$$
$$\eqalign{ & = \frac{{2{b^3}{\text{ + 6}}{{\text{a}}^2}{\text{b }}}}{{2{a^2} + 2{b^2}{\text{ }}}} \cr & = \frac{{{b^3}{\text{ + 3}}{{\text{a}}^2}{\text{b }}}}{{{a^2} + {b^2}{\text{ }}}} \cr & = \frac{{{b^3} + {b^3}{\text{ }}}}{{\frac{{{b^2}}}{3} + {b^2}{\text{ }}}} \cr & = \frac{{2{b^3}}}{{{b^2}\left( {\frac{1}{3} + 1} \right)}} \cr & = \frac{{2b}}{{\frac{4}{3}}} \cr & = \frac{{3b}}{2} \cr} $$

$$\eqalign{ & {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr & = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr & = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr & {\text{Put }} t = a - 2b \cr & = {t^2} - 2t - 8 \cr & = {t^2} - 4t + 2t - 8 \cr & = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr & = \left( {t + 2} \right)\left( {t - 4} \right) \cr & = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr & \left( {{\text{Put the value of assume }}t} \right) \cr} $$

$$\frac{1}{{{a^2} + ax + {x^2}}}$$   $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$   $$ + $$ $$\frac{1}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$       $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$\eqalign{ & = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax + 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} \cr & = 0 \cr} $$

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

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

$$\eqalign{ & x:y \cr & 7:3{\text{ }} \cr & \therefore {\text{ }}\frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{21 + 9}}{{49 - 9}} \cr & = \frac{{30}}{{40}} \cr & = \frac{3}{4} \cr} $$

Given [p] means the greatest positive integer less than or [p] equal to p
$$\eqalign{ & \Rightarrow \left[ {\text{p}} \right] = {\text{ p}} \cr & \Rightarrow \left[ -{\text{p}} \right] = {\text{ p}} \cr & \Rightarrow \left[ { - \frac{1}{4}} \right]{\text{ + }}\left[ {4 - \frac{1}{4}} \right] + \left[ 3 \right] \cr & \Rightarrow \frac{1}{4} + 4 - \frac{1}{4} + 3 \cr & \Rightarrow 7 \cr} $$

$$\eqalign{ & x = 9999{\text{ }}\left( {{\text{ Given}}} \right) \cr & \frac{{4{x^3} - x}}{{\left( {2x + 1} \right)\left( {6x - 3} \right)}} \cr & = \frac{{x\left( {4{x^2} - 1} \right)}}{{3\left( {2x + 1} \right)\left( {2x - 1} \right)}} \cr & = \frac{{x\left( {4{x^2} - 1} \right)}}{{3\left( {4{x^2} - 1} \right)}} \cr & = \frac{x}{3} \cr & \therefore \frac{{9999}}{3} = 3333 \cr} $$

$$\because $$ x = 11
x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1
= x⁵ - 11x⁴ - x⁴ + 11x³ + x³ - 11x² - x² + 11x + x - 1
= 11⁵ - 11.11⁴ - 11⁴ + 11.11³ + 11³ - 11.11² - 11² + 11.11 + 11 - 1
= 0 - 0 + 0 + 0 + 11 - 1
= 10

$$\eqalign{ & {x^2} + {y^2} + {z^2} - xy - yz - zx \cr & = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr} $$
  $$ = \frac{1}{2}$$ $$\left[ {{{\left( {997 - 998} \right)}^2} + {{\left( {998 - 999} \right)}^2} + {{\left( {999 - 997} \right)}^2}} \right]$$
$$\eqalign{ & = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr & = 3 \cr} $$

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

$$\eqalign{ & \frac{{{{\left( {243} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{9^n}{{.3}^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{3^{2n}}{{.3}^{n - 1}}}} \cr & = \frac{{{3^{n + 2n + 1}}}}{{{3^{2n + n - 1}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{3n + 1 - 3n + 1}} \cr & = {3^2} \cr & = 9 \cr} $$

$$\eqalign{ & x = 0.5 \cr & y = 0.2 \cr & \sqrt {0.6} \times {\left( {3y} \right)^x}{\text{ }} \cr & = \sqrt {0.6} \times {\left( {3 \times 0.2} \right)^{0.5}}{\text{ }} \cr & = \sqrt {0.6} \times \sqrt {0.6} \cr & = 0.6 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr & = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr & = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr & = \frac{{9 - 4}}{2} \cr & = \frac{5}{2} \cr} $$

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

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

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

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

$$\eqalign{ & {x^{x\sqrt x }} = {\left( {x\sqrt x } \right)^x} \cr & {x^{x\sqrt x }} = {\left( {{x^{\frac{3}{2}}}} \right)^x} \cr & {x^{x\sqrt x }} = {x^{\frac{3}{2}x}} \cr} $$
(If bases are same then their power is also same)
$$\eqalign{ & \therefore x\sqrt x = \frac{3}{2}x \cr & \Rightarrow \sqrt x = \frac{3}{2} \cr & \Rightarrow x = {\left( {\frac{3}{2}} \right)^2} \cr & \Rightarrow x = \frac{9}{4} \cr} $$

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

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

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

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

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