Ratio - Aptitude MCQ Questions and Solutions with Explanations

Let income in the first year be Rs. x
And
expenses in the second year be Rs. y
Then,
$$\eqalign{ & = \frac{x}{{45000}} = \frac{2}{3}\,{\text{and }}\,\frac{{25000}}{y} = \frac{5}{9} \cr & \Rightarrow x = \frac{{2 \times 45000}}{3} = 30000\,{\text{and}} \cr & \,\,\,\,\,\,\,y = \frac{{25000 \times 9}}{5} = 45000 \cr} $$
∴ Total savings for 2 years
= Rs. [(30000 - 25000) + (45000 - 45000)]
= Rs. 5000

Clearly, A and B save 20% and 30% of their respective salaries
Let the salaries of A and B be x and y respectively
Then,
$$\eqalign{ & {\text{ = }}\frac{{{\text{20}}\% {\text{ of }}x}}{{{\text{30}}\% {\text{ of }}y}} = \frac{4}{3} \cr & \Rightarrow \frac{x}{5} \times \frac{{10}}{{3y}} = \frac{4}{3} \cr & \Rightarrow \frac{x}{y} = 2 \cr & \Rightarrow x = 2y \cr & \therefore x + y = 2100 \cr & \Rightarrow 2y + y = 2100 \cr & \Rightarrow 3y = 2100 \cr & \Rightarrow y = 700 \cr & {\text{A's salary}} = x = 2y \cr & = {\text{Rs}}{\text{. }}\left( {2 \times 700} \right) \cr & = {\text{Rs}}{\text{. }}1400 \cr} $$

According to the question,

Sn	:	Fe
A → 1	:	2 = 3) × 5
B → 2	:	3 = 5) × 3

Making quantity equal

Sn	:	Fe
A → 5	:	10 = 15) × 3
B → 6	:	9 = 15) × 4

Sn	:	Fe
A →15	:	30
B →24	:	36
39	:	66

In final mixture Sn : Fe
= 39 : 66
= 13 : 22

Let the 2 number be (2x, 3x)
According to the question,
$$\eqalign{ & \Rightarrow \frac{{2x + 8}}{{3x + 8}} = \frac{3}{4} \cr & \Rightarrow 8x + 32 = 9x + 24 \cr & \Rightarrow x = 8 \cr} $$
Therefore, the sum of the two numbers is
= 2x + 3x = 5x
= 5 × 8 = 40

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

Let the incomes of the two persons be 5x and 3x
and
their expenditures be 9y and 5y respectively.
Then,
= 5x - 9y = 2600.....(i)
= 3x - 5y = 1800.....(ii)
Multiplying (i) by 3 and (ii) by 5, we get:
= 15x - 27y = 7800.....(iii)
= 15x - 25y = 9000.....(iv)
Subtracting (iii) from (iv), we get : 2y = 1200 or y = 600
Putting y = 600 in (i), we get : 5x = 8000 or x = 1600.
∴ Their incomes are Rs. (5 × 1600) and Rs. (3 × 1600)i.e.
Rs. 8000 and Rs. 4800 respectively

Let the number of students in Arts and Commerce faculties be 4x and 5x respectively.
Then,
$$\eqalign{ & = \frac{{4x}}{{5x + 65}} = \frac{8}{{11}} \cr & \Rightarrow 44x = 8\left( {5x + 65} \right) \cr & \Rightarrow 4x = 520 \cr & \Rightarrow x = 130 \cr} $$
∴ Number of students in Arts faculty
= (4 × 130)
= 520

Let the numbers be 7x and 11x respectively
Then,
$$\eqalign{ & = \frac{{7x + 7}}{{11x + 7}} = \frac{2}{3} \cr & \Rightarrow 21x + 21 = 22x + 14 \cr & \Rightarrow x = 7 \cr & \therefore {\text{Smaller number}} \cr & = 7 \times 7 \cr & = 49 \cr} $$

Let their income be 4x and 3x
Their savings = Rs. 3200 each
According to the question,
$$\eqalign{ & \Rightarrow \frac{{4x - 3200}}{{3x - 3200}} = \frac{{12}}{7} \cr & \Rightarrow \frac{{x - 800}}{{3x - 3200}} = \frac{3}{7} \cr & \Rightarrow 7x - 5600 = 9x - 9600 \cr & \Rightarrow 2x = 4000 \cr & \Rightarrow x = 2000 \cr & \Rightarrow {\text{A}} = 2000 \cr & \Rightarrow {\text{Income of A}} = 4x \cr & \Rightarrow 4 \times 2000 = {\text{Rs}}{\text{. }}8000 \cr} $$

According to question,
$$\eqalign{ & \frac{{{x^3} - {y^3}}}{{{x^2} + xy + {y^2}}} = \frac{5}{1} \cr & \frac{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}{{{x^2} + xy + {y^2}}} = \frac{5}{1} \cr & x - y = 5.....({\text{i}}) \cr & \frac{{{x^2} - {y^2}}}{{x - y}} = \frac{7}{1} \cr & \frac{{\left( {x + y} \right)\left( {x - y} \right)}}{{x - y}} = \frac{7}{1} \cr & x + y = 7.....({\text{ii}}) \cr & {\text{Solve equations (i) and (ii)}} \cr & x = 6 \cr & y = 1 \cr & \therefore \frac{{2x}}{{3y}} = \frac{{2 \times 6}}{{3 \times 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{12}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{4}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4:1 \cr} $$