Practice MCQ Questions and Answer on Ratio

Solution: $$\eqalign{ & {\text{A}}:{\text{B}} = {\text{3}}:{\text{4}}, \cr & {\text{B}}:{\text{C}} = {\text{5}}:{\text{6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {5 \times \frac{4}{5}} \right):\left( {6 \times \frac{4}{5}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 4:\frac{{24}}{5} \cr & {\text{C}}:{\text{D}} = {\text{7}}:{\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {7 \times \frac{{24}}{{35}}} \right):\left( {5 \times \frac{{24}}{{35}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{24}}{5}:\frac{{24}}{7} \cr & \therefore {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & = 3:4:\frac{{24}}{5}:\frac{{24}}{7} \cr & = 105:140:168:120 \cr} $$ Clearly, C will get the maximum

Solution: Let Maya's and Chhaya's present age is 6x and 5x respectively And, $$\frac{{6{\text{x}} + 15}}{{5{\text{x}} + 15}} = \frac{9}{8}$$ Or, 48x + 120 = 45x = 135 Or, 3x = 15 Or, x = 5 Present age of Maya = 6x = 30

Solution: Let the number are 5x, 7x and 12x According to the question, ⇒ 5x + 12x = 7x + 50 ⇒ 17x - 7x = 50 ⇒ 10x = 50 ⇒ x = 5 Sum of all three numbers = 5x + 7x + 12x = 24x = 24 × 5 =120

Solution: $$\eqalign{ & {\text{Second son's share}} \cr & = {\text{Rs}}{\text{.}}\left( {10800 \times \frac{4}{{12}}} \right) \cr & = {\text{Rs}}{\text{. }}3600 \cr} $$ Money distributed between the two daughters = Rs. [3600 - (1000 + 600)] = Rs. 2000 $$\eqalign{ & {\text{First daughter's share}} \cr & = {\text{Rs}}{\text{.}}\left( {2000 \times \frac{{11}}{{20}}} \right) \cr & = {\text{Rs}}.1100. \cr & {\text{Second daughter's share}} \cr & = {\text{Rs}}{\text{.}}\left( {2000 \times \frac{9}{{20}}} \right) \cr & = {\text{Rs}}{\text{. 9}}00 \cr} $$

Solution: $$\eqalign{ & {\frac{A}{B}} \times {\frac{B}{C}} = {\frac{3}{4}} \times {\frac{{12}}{{13}}} \cr & Or,\,\frac{A}{C} = \frac{{36}}{{52}} = 9:13 \cr} $$

Solution: $$\eqalign{ & a + b\sqrt 3 = \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} \cr & \Rightarrow 2 + \sqrt 3 \cr} $$ By rationalisation of denominator ⇒ a + b$$\sqrt 3 $$ = 2 + $$\sqrt 3 $$ ⇒ Now compare the rational & irrational parts ∴ a = 2    b = 1 ∴ a : b    2 : 1

Solution: $$\eqalign{ & {\text{Given }} \cr & \frac{x}{y} = \frac{3}{4} \cr & \Rightarrow \frac{{2x + 3y}}{{3y - 2x}} \cr & = \frac{{2 \times 3 + 3 \times 4}}{{3 \times 4 - 2 \times 3}} \cr & = \frac{{6 + 12}}{{12 - 6}} \cr & = \frac{{18}}{6} \cr & = 3:1 \cr} $$ This is the required ratio 3 : 1

Solution: $$\eqalign{ & = \frac{2}{5}{\text{A}} + 40 = \frac{2}{7}{\text{B}} + 20 \cr & \Rightarrow \frac{2}{7}{\text{B}} = \frac{2}{5}{\text{A}} + 20 \cr & \Rightarrow {\text{B}} = \frac{7}{2}\left( {\frac{2}{5}{\text{A}} + 20} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{7}{5}{\text{A}} + 70} \right) \cr & {\text{And,}} \cr & = \frac{2}{5}{\text{A}} + 40 = \frac{9}{{17}}{\text{C}} + 10 \cr & \Rightarrow \frac{9}{{17}}{\text{C}} = \frac{2}{5}{\text{A}} + 30 \cr & \Rightarrow {\text{C}} = \frac{{17}}{9}\left( {\frac{2}{5}{\text{A}} + 30} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{34}}{{45}}{\text{A}} + \frac{{170}}{3} \cr & = {\text{A}} + {\text{B}} + {\text{C}} = 600 \cr & \Rightarrow {\text{A}} + \left( {\frac{7}{5}{\text{A}} + 70} \right) + \left( {\frac{{34}}{{45}}{\text{A}} + \frac{{170}}{3}} \right) = 600 \cr & \Rightarrow \frac{{142{\text{A}}}}{{45}} = 600 - \frac{{380}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1420}}{3} \cr & \Rightarrow {\text{A}} = \frac{{1420}}{3} \times \frac{{45}}{{142}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 150 \cr} $$

Solution: Let the monthly salary of Pia and Som be 5a and 4a respectively. Then, Money given by Pia to her mother $$\eqalign{ & = 5a \times \frac{3}{5} = 3a \cr } $$ Money given by Pia as sister's tuition fees $$\eqalign{ & = 15\% {\text{ of }}5a \cr & = \frac{{15 \times 5a}}{{100}} \cr & = \frac{{75a}}{{100}} = 0.75a \cr} $$ Money given by Pia towards loan $$\eqalign{ & = 18\% {\text{ of }}5a \cr & = \frac{{18 \times 5a}}{{100}} \cr & = \frac{{90a}}{{100}} = 0.9a \cr & \therefore {\text{Total money given}} \cr & = 3a + 0.75a + 0.90a \cr & = 4.65a \cr & \therefore {\text{Remaining amount}} \cr & = 5a - 4.65 \cr & = 0.35a \cr & {\text{Pia have remaining amount}} \cr & = {\text{Rs}}{\text{. }}2100 \cr & 0.35a = 2100 \cr & \therefore \frac{{35a}}{{100}} = 2100 \cr & \Rightarrow a = \frac{{2100 \times 100}}{{35}} \cr & a = {\text{Rs}}{\text{. }}6000 \cr & \therefore {\text{Monthly salary of Som}} \cr & = 4a = 4 \times 6000 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}24000 \cr} $$

Solution: $$\eqalign{ & {\text{A}}:{\text{B}} = 3:7, \cr & {\text{B}}:{\text{C}} = 6:5, \cr & = \left( {6 \times \frac{7}{6}} \right):\left( {5 \times \frac{7}{6}} \right) \cr & = 7:\frac{{35}}{6} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:7:\frac{{35}}{6} \cr & = 18:42:35. \cr & {\text{Sum of ratio terms}} \cr & = \left( {18 + 42 + 35} \right) \cr & = 95 \cr & \therefore {\text{B's share}} = \cr & {\text{Rs}}{\text{.}}\left( {33630 \times \frac{{42}}{{95}}} \right) \cr & = {\text{Rs}}{\text{. }}14868 \cr} $$