Ratio - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

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
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {60 \times \frac{{16}}{3}} \right) \cr & = {\text{Rs}}.320 \cr} $$

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

A : B = 3 : 4
B : C = 3.5 : 3 = 7 : 6

A	:	B	:	C
3	:	4
		7	:	6
21	:	28	:	24

A + B + C → 21 + 28 + 24
73 → 730
1 → 10
Then share of B exceeds that of C by (28 - 24) → 4
= 4 × 10 = 40

Total Number of student = 132
The ratio of the pass : fail = 9 : 2
Number of student pass = $$\frac{9}{11} \times 132 = 108$$
Number of student fail = $$\frac{2}{11} \times 132 = 24$$
Now 4 student more passed.
∴ Number of student pass = 108 + 4 = 112 and,
Number of student fail = 24 - 4 = 20
New ratio of pass : fail = 112 : 20 = 28 : 5

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

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

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