Algebra — Part 1 MCQs with Explanations

$$\eqalign{ & 3*5 + 5*3 \cr & \Rightarrow 3*5 = 2 \times 3 - 3 \times 5 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 - 15 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 \cr & \Rightarrow 5*3 = 2 \times 5 - 3 \times 3 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10 - 9 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \cr & \therefore 3*5 + 5*3 \cr & \Rightarrow 6 + 16 = 22 \cr} $$

$$\eqalign{ & {\text{8}} \times {\text{2}} \cr & = 8 + 2 + \frac{8}{2} \cr & = 10 + 4 \cr & = 14 \cr} $$

$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$ Taking L.C.M of each term $$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$ $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$ $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$ $$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{2}{3}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{4}{5}\,\,\left( {{\text{Given}}} \right) \cr & or\,\frac{c}{b} = \frac{5}{4} \cr & \frac{{a + b}}{{b + c}} \cr & = \frac{{b\left( {\frac{a}{b} + 1} \right)}}{{b\left( {\frac{c}{b} + 1} \right)}} \cr & = \frac{{\frac{a}{b} + 1}}{{\frac{c}{b} + 1}} \cr & = \frac{{\left( {\frac{2}{3} + 1} \right)}}{{\left( {\frac{5}{4} + 1} \right)}} \cr & = \frac{{\frac{2 + 3}{3}}}{{\frac{{5 + 4}}{4}}} \cr & = \frac{{5 \times 4}}{{3 \times 9}} \cr & = \frac{{20}}{{27}} \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{20}}{{27}} \cr & {\bf{Alternate:}} \cr & a{\text{ }}\,\,\,{\text{ }}\,{\text{ }}\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,{\text{ }}\,\,\,\,{\text{ }}\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,{\text{ }}\,{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}{\text{ }}\,\,\,\,\,\,{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {8{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}15\,\,\,\,} } \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{8 + 12}}{{12 + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{27}} \cr} $$

$$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr} $$

$$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr} $$

$$\eqalign{ & a:b = 2:3{\text{ and }}b:c = 4:5 \cr & a{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}\,\,\,\,\,\,\,\,{\text{ }}{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {\,\,8{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}15{\text{ }}\,\,{\text{ }}} } \cr & \therefore {a^2}:{b^2}:bc \cr & \Rightarrow {\left( 8 \right)^2}:{\left( {12} \right)^2}:\left( {12 \times 15} \right) \cr & \Rightarrow 64:144:180 \cr & \Rightarrow 16:36:45 \cr} $$

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

$$\eqalign{ & {\text{A}}:{\text{B}}:{\text{C}} \cr & {\text{ }}2:\,\,3\,\,:4 \cr & \therefore \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} \cr & \left( {{\text{multiply with ABC}}} \right) \cr & \therefore \frac{{{\text{A}} \times {\text{ABC}}}}{{\text{B}}}{\text{:}}\frac{{{\text{B}} \times {\text{ABC}}}}{{\text{C}}}{\text{:}}\frac{{{\text{C}} \times {\text{ABC}}}}{{\text{A}}} \cr & \Rightarrow {{\text{A}}^2}{\text{C}}:{{\text{B}}^2}{\text{A}}:{\text{B}}{{\text{C}}^2} \cr & \Rightarrow {\left( 2 \right)^2} \times 4:{\left( 3 \right)^2} \times 2:{\left( 4 \right)^2} \times 3 \cr & \Rightarrow 16:18:48 \cr & \Rightarrow 8:9:24 \cr & \cr & {\bf{Alternate:}} \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:\frac{4}{2} \cr & \Rightarrow \frac{2}{3} \times 12:\frac{3}{4} \times 12:\frac{4}{2} \times 12 \cr & \Rightarrow 8:9:24 \cr} $$

$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr} $$

$$\eqalign{ & 1 < x < 2 \cr & \sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {3 - x} \right)}^2}} \cr & \left( {{\text{Square root cancel with square}}} \right) \cr & \therefore x - 1 + 3 - x \cr & = 2 \cr} $$

$$\eqalign{ & a \oplus b = \left( {a \times b} \right) + b \cr & \Rightarrow 5 \oplus 7 \cr & = \left( {5 \times 7} \right) + 7 \cr & = 35 + 7 \cr & = 42 \cr} $$

Given 47.2506 = 4A + 7B + 2C + $$\frac{5}{{\text{D}}}$$ + 6E . . . . . . . . (1) Now, we can write 47.2506 = 40 + 7 + 0.2 + 0.05 + 0.0006 . . . . . . . . (2) Comparing (1) and (2) we get 4A = 40 ⇒ A = 10 $7B = 7 ⇒ B = 1 2C = 0.2 ⇒ C = 0.1 $$\frac{5}{{\text{D}}}$$ = 0.05 ⇒ D = $$\frac{5}{{0.05}}$$ = 100 and, 6E = 0.0006 ⇒ E = 0.0001 So velue of 5A + 3B + 6C + D + 3E = (5 × 10) + (3 × 1) + (6 × 0.1) + 100 + (3 × 0.0001) = 50 + 3 + 0.6 + 100 + 0.0003 = 153.6003

$$\eqalign{ & x*y = {x^2} + {y^2} - xy \cr & {\text{ }}9*11 = {\left( 9 \right)^2} + {\left( {11} \right)^2} - 11 \times 9 \cr & \Rightarrow 9*11 = 81 + 121 - 99 \cr & \Rightarrow 9*11 = 103 \cr} $$

$$\eqalign{ & {3^{x + 3}} + 7 = 250 \cr & \Rightarrow {3^{x + 3}} = 250 - 7 \cr & \Rightarrow {3^{x + 3}} = 243 \cr & \Rightarrow {3^{x + 3}} = {3^5} \cr & \Rightarrow x + 3 = 5 \cr & \Rightarrow x = 2 \cr} $$

$$\frac{1}{4} \times $$ $$\frac{2}{6} \times $$ $$\frac{3}{8} \times $$ $$\frac{4}{{10}} \times $$ $$\frac{5}{{12}} \times $$ . . . . . $$ \times \frac{{31}}{{64}}$$ $$ = \frac{1}{{{2^x}}}$$ Denominator of first term is cut by the numerator of next terms, second term to the next one and so on. $$\eqalign{ & \Rightarrow {\left( {\frac{1}{2}} \right)^{30}} \times {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{{2^x}}} \cr & \Rightarrow {\left( {\frac{1}{2}} \right)^{30 + 6}} = \frac{1}{{{2^x}}} \cr & \Rightarrow \frac{1}{{{2^{36}}}} = \frac{1}{{{2^x}}} \cr & \Rightarrow x = 36 \cr} $$

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

$$\eqalign{ & \because p = 999 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {1000} \right)}^3}} \cr & = 1000 \cr} $$

$$\eqalign{ & {\text{ }}\frac{a}{b} = \frac{7}{9},{\text{ }}\frac{b}{c} = \frac{3}{5} \cr & \therefore a\,\,\,:\,\,\,b\,\,\,:\,\,\,c \cr & \,\,\,\,\,7\,\,\,:\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{3_{ \times 3}}:\,\,\,{5_{ \times 3}} \cr & \overline {{\text{ }}\,\,{\text{ }}7\,\,\,:\,\,\,\,9\,\,:\,\,15{\text{ }}} \cr} $$

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

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

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

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

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

$$\eqalign{ & {a^2} + {b^2} + {c^2} - ab - bc - ca \cr & = \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {7 - 5} \right)}^2} + {{\left( {5 - 3} \right)}^2} + {{\left( {3 - 7} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {4 + 4 + 16} \right) \cr & = \frac{{24}}{2} \cr & = 12 \cr} $$

$$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$ (If bases are equal then their power are also equal)

$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}$$ A 3 + B 3 + C 3 - 3ABC = (A + B + C)(A 2 + B 2 + C 2 - AB - BC - CA) $$\therefore \frac{{{{\left( {\frac{1}{3}} \right)}^3} + {{\left( {\frac{1}{4}} \right)}^3} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + {{\left( {\frac{1}{5}} \right)}^3}}}{{{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}}}$$ $$ = \frac{{\left( {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right)\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}{{\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}$$ $$\eqalign{ & = \frac{{20 + 15 + 12}}{{60}} \cr & = \frac{{47}}{{60}} \cr} $$

$$\eqalign{ & {\text{0}}{\text{.13}} \times {{\text{p}}^2} = 13 \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{0.13}} \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{13}} \times 100 \cr & \Rightarrow {{\text{p}}^2} = 100 \cr & \Rightarrow {\text{p}} = 10 \cr} $$

$$\eqalign{ & x = 7 - 4\sqrt 3 \cr & \Rightarrow x = 4 + 3 - 4\sqrt 3 \cr & \Rightarrow x = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2 \times 2\sqrt 3 \cr & \Rightarrow {\left( {2 - \sqrt 3 } \right)^2} \cr & \therefore \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] \cr & \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + \sqrt 3 \cr & \therefore \sqrt x {\text{ + }}\frac{1}{{\sqrt x }}{\text{ }} \cr & = 2 - \sqrt 3 + 2 + \sqrt 3 \cr & = 4 \cr} $$

$$\eqalign{ & a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{ }} \cr & \therefore a = \frac{1}{b} \cr & a + b = a + \frac{1}{a} \cr & \Rightarrow \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 1 \right)}^2}}} \cr & \Rightarrow \frac{{6 + 2\sqrt 5 + 6 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow \frac{{12}}{4} \cr & \Rightarrow 3 \cr & \therefore \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & \Rightarrow \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} \cr & \Rightarrow 9 - 2 \cr & \Rightarrow 7\left( {ab = 1} \right) \cr & \therefore \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr} $$

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

$$\eqalign{ & x:y = 3:4 \cr & \frac{{\left( {7x + 3y} \right)}}{{\left( {7x - 3y} \right)}} \cr & = \frac{y}{y}\left( {\frac{{7\frac{x}{y} + 3}}{{7\frac{x}{y} - 3}}} \right) \cr & = \frac{{7 \times \frac{3}{4} + 3}}{{7 \times \frac{3}{4} - 3}} \cr & = \frac{{\frac{{21}}{4} + 3}}{{\frac{{21}}{4} - 3}} \cr & = \frac{{\frac{{21 + 12}}{4}}}{{\frac{{21 - 12}}{4}}} \cr & = \frac{{11}}{3} \,\,or\,\, 11 : 3 \cr} $$

$$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$

$$\eqalign{ & {\text{ }}{x^{\frac{1}{3}}} = {y^{\frac{1}{4}}}, \cr & \Rightarrow {\text{LCM of 3, 4}} = 12{\text{ }} \cr & \therefore {\text{ }}{\left( {{x^{\frac{1}{3}}}} \right)^{12}} = {\left( {{y^{\frac{1}{4}}}} \right)^{12}} \cr & \Rightarrow {x^4} = {y^3} \cr & {\text{take power '5' on both sides}} \cr & \Rightarrow {\left( {{x^4}} \right)^5} = {\left( {{y^3}} \right)^5} \cr & \Rightarrow {x^{20}} = {y^{15}} \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & {\text{or }}1 + x = 1 + \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2 + \sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2\left( {2 + \sqrt 3 } \right)}}{{2 \times 2}} \cr & \left( {{\text{Divided and multiply by 2}}} \right) \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{1 + 3 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.1.\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{4} \cr & \therefore \sqrt {1 + x} = \frac{{1 + \sqrt 3 }}{2} \cr & \cr & {\bf{Similarly:}} \cr & \sqrt {1 - x} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr & \therefore \frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }} \cr & = \frac{{\frac{{1 + \sqrt 3 }}{2}}}{{1 + \frac{{1 + \sqrt 3 }}{2}}} + \frac{{\frac{{\sqrt 3 - 1}}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}} \cr & = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{3 - \sqrt 3 }} \cr & = \frac{{1 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} + \frac{{1 - \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} \cr & = \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \Leftrightarrow {x^2} - 4x - 1 = 0 \cr & \Leftrightarrow {x^2} - 1 = 4x \cr & \left( {{\text{divide }}x{\text{ both sides}}} \right) \cr & \Leftrightarrow x - \frac{1}{x} = 4 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr} $$

$$\eqalign{ & \Rightarrow {a^{2x + 2}} = 1 \cr & \Rightarrow {a^{2x + 2}} = {a^0} \cr & \Rightarrow 2x + 2 = 0 \cr & \Rightarrow x = - \frac{2}{2} \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & \frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b \cr & {\text{L}}{\text{.H}}{\text{.S }}\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} \times \frac{{\sqrt 7 - 2}}{{\sqrt 7 - 2}} \cr & {\text{ }}\left( {{\text{Rationalisation}}} \right) \cr & = \frac{{{{\left( {\sqrt 7 - 2} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - \left( 4 \right)}} \cr & = \frac{{7 + 4 - 4\sqrt 7 }}{{7 - 4}} \cr & = \frac{{11 - 4\sqrt 7 }}{3} \cr & = \frac{{11}}{3} - \frac{4}{3}\sqrt 7 \cr & = - \frac{4}{3}\sqrt 7 + \frac{{11}}{3} \cr & = a\sqrt 7 + b{\text{ }}\left( {{\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}} \right) \cr} $$ (Compare the coefficients of $$\sqrt 7 $$ and constant term) $$\eqalign{ & {\text{a}} = - \frac{4}{3} \cr & b = \frac{{11}}{3} \cr} $$

$$\eqalign{ & a + \frac{1}{b} = 1{\text{, }}b + \frac{1}{c} = 1{\text{, }}c + \frac{1}{a} = ? \cr & {\text{Put values, }} \cr & a = \frac{1}{2},b = 2,{\text{ }}c = - 1 \cr & c + \frac{1}{a} \cr & = - 1 + \frac{1}{{\left( {\frac{1}{2}} \right)}} \cr & = - 1 + 2 \cr & = 1 \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow a + \frac{1}{b} = 1 \cr & \Rightarrow a = 1 - \frac{1}{b} = \boxed{\frac{{b - 1}}{b}} \cr & \frac{1}{a} = \frac{b}{{b - 1}} \cr & \Rightarrow b + \frac{1}{c} = 1 \cr & \Rightarrow \frac{1}{c} = 1 - b,\boxed{c = \frac{1}{{1 - b}}} \cr & \therefore c + \frac{1}{a} \cr & = \frac{1}{{1 - b}} + \frac{b}{{b - 1}} \cr & = \frac{1}{{1 - b}} - \frac{b}{{1 - b}} \cr & = \frac{{1 - b}}{{1 - b}} \cr & = 1 \cr} $$