Ratio and Proportion - 01

Explanation:-

And ratio of numbers of boys to the number of girls is 5x : 3x
According to the question, $$\eqalign{ & \Rightarrow 5x + 3x = 48 \cr & \Rightarrow 8x = 48 \cr & \Rightarrow \boxed{x = 6} \cr & {\text{Boys}} = 30 \cr & {\text{Girl}} = 18 \cr & {\text{Now,}}\frac{{30}}{{18 + y}} = \frac{6}{5} \cr & \Rightarrow 25 = 18 + y \cr & \Rightarrow y = 7 \cr}$$

Explanation:-

Let the number of officers and soldiers = 3x, 31x
According to the question, $$\eqalign{ & \Rightarrow \frac{{3x - 6}}{{31x - 22}} = \frac{1}{{13}} \cr & \Rightarrow 39x - 78 = 31x - 22 \cr & \Rightarrow 8x = 56 \cr & \Rightarrow x = 7 \cr}$$ So, number of officer before
= 3x = 3 × 7 = 21

Explanation:-

$$\eqalign{ & {\text{Milk in 1st glass}} \cr & = \frac{1}{2}{\text{ unit;}} \cr & {\text{Milk in 2nd glass}} \cr & = \frac{3}{4}{\text{ unit;}} \cr & {\text{Water in 1st glass}} \cr & = \frac{1}{2}{\text{ unit;}} \cr & {\text{Water in 2nd glass}} \cr & = \frac{1}{4}{\text{ unit}}{\text{.}} \cr & \therefore {\text{Required ratio}} \cr & = \frac{{\frac{1}{2} + \frac{3}{4}}}{{\frac{1}{2} + \frac{1}{4}}} \cr & = \frac{5}{4} \times \frac{4}{3} \cr & = 5:3. \cr}$$

Explanation:-

$$\eqalign{ & {\text{Gold in C}} \cr & = \left( {\frac{7}{9} + \frac{7}{{18}}} \right)\,{\text{units}} \cr & = \frac{{\text{7}}}{{\text{6}}}\,{\text{units}}{\text{.}} \cr & {\text{Copper in C}} \cr & = \left( {\frac{2}{9} + \frac{{11}}{{18}}} \right)\,{\text{units}} \cr & = \frac{5}{6}\,{\text{units}}{\text{.}} \cr & \therefore {\text{Gold}}:{\text{Copper}} \cr & = \frac{7}{6}:\frac{5}{6} \cr & = 7:5. \cr}$$

Explanation:-

let the salaries of A and B last year be Rs. 3x and Rs. 4x respectively.
Then, $$\eqalign{ & {\text{A's present salary}} \cr & = {\text{Rs}}.\left( {\frac{5}{4} \times 3x} \right) \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{15x}}{4}} \right). \cr & {\text{B's present salary}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{3}{2} \times 4x} \right) \cr & = {\text{Rs}}.6x. \cr & \therefore \frac{{15x}}{4} + 6x = 4160 \cr & \Rightarrow 39x = 4160 \times 4 \cr & \Rightarrow x = \frac{{4160 \times 4}}{{39}}. \cr & {\text{So,}} \cr & {\text{A's present salary}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{15}}{4} \times \frac{{4160 \times 4}}{{39}}} \right) \cr & = {\text{Rs}}.1600. \cr}$$

Explanation:-

$$\eqalign{ & \frac{{{m_1} \times {h_1}}}{{{w_1}}} = \frac{{{m_2} \times {h_2}}}{{{w_2}}} \cr & \Rightarrow \frac{{5 \times 5}}{5} = \frac{{10 \times 10}}{{{w_2}}} \cr & \Rightarrow {w_2} = 20 \cr}$$

Explanation:-

Let the initial fares of 1st, 2nd and 3rd class be Rs. 8x, Rs. 6x, and Rs. 3x respectively. $$\eqalign{ & {\text{Revised fare of 1st class}} \cr & {\text{ = }}\frac{5}{6}{\text{of Rs}}.8x \cr & = {\text{Rs}}.\left( {\frac{{20x}}{3}} \right). \cr & {\text{Revised fare of }}2{\text{nd class}} \cr & {\text{ = }}\frac{{11}}{{12}}{\text{of Rs}}.6x \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{11x}}{2}} \right). \cr}$$ Let the number of passengers of 1st, 2nd and 3rd class be 9y, 12y and 26y respectively.
Then, $$\eqalign{ & = \frac{{20x}}{3} \times 9y + \frac{{11x}}{2} \times 12y + 3x \times 26y = 1088 \cr & \Rightarrow 60xy + 66xy + 78xy = 1088 \cr & \Rightarrow 204xy = 1088 \cr & \Rightarrow xy = \frac{{1088}}{{204}} = \frac{{16}}{3}. \cr}$$ ∴ Collection from passengers of 1st class = 60xy $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {60 \times \frac{{16}}{3}} \right) \cr & = {\text{Rs}}.320. \cr}$$

Explanation:-

Let the number of coins with Lalita, Palita and Salita in the end be 24x, 21x and 16x respectively.
Then,
Number of coins received by Lalita, Palita and Salita from their mother are
(24x + 50), (21x + 85) and (16x + 39) respectively.
So, (24x + 50) + (21x + 85) + (16x + 39) = 3224
⇒ 61x = 3050
⇒ x = 50.
Hence, number of coins received by Lalita from her mother
= (24 × 50 + 50)
= 1250.

Explanation:-

Let the initial total weight of A and B be x kg.
Then, $$\eqalign{ & = 125\% {\text{ of }}x = 80 \cr & \Rightarrow x = 80 \times \frac{{100}}{{125}} = 64{\text{kg}}. \cr & {\text{A's initial weight}} \cr & = \left( {64 \times \frac{3}{8}} \right){\text{kg}} \cr & = 24{\text{kg}}{\text{.}} \cr & {\text{B's initial weight}} \cr & = \left( {64 \times \frac{5}{8}} \right){\text{kg}} \cr & = 40{\text{kg}}. \cr & {\text{A's new weight}} \cr & = 120\% {\text{ of }}24{\text{kg}} \cr & = 28.8{\text{kg}}{\text{.}} \cr & {\text{B's new weight}} \cr & = \left( {80 - 28.8} \right){\text{kg}} \cr & = 51.2{\text{kg}} \cr & {\text{Increase in B's weight}} \cr & = \left( {51.2 - 40} \right){\text{kg}} \cr & = 11.2{\text{kg}}. \cr & \therefore {\text{Increase }}\% \cr & = \left( {\frac{{11.2}}{{40}} \times 100} \right)\% \cr & = 28\% . \cr}$$

Explanation:-

Let the earnings of A and B Rs. 8x and Rs. 9x respectively.
Then, $$\eqalign{ & = \frac{{150\% {\text{ of }}8x}}{{75\% {\text{ of }}9x}} = \frac{{16}}{9} \cr & \Rightarrow \frac{{\frac{3}{2} \times 8x}}{{\frac{3}{4} \times 9x}} = \frac{{16}}{9} \cr & \Rightarrow \frac{{16}}{9} = \frac{{16}}{9}. \cr}$$ Hence, A's earnings cannot be determined.