Ratio - Aptitude MCQ Questions and Solutions with Explanations

Required ratio
= 120% of Rs. 30 : 110% of Rs. 20
= 36 : 22
= 18 : 11

Ratio of the monthly incomes of families A and B
= Ratio of the areas of rectangles A and B
= 5 × 2 : 4 × 1
= 10 : 4
= 5 : 2
Let the monthly income of family A be Rs. x
Then,
5 : 2 :: x : 12000
⇒ 2x = 5 × 12000 = 60000
$$\eqalign{ & \Rightarrow x{\text{ = }}\frac{{60000}}{2} \cr & = 30000 \cr} $$

	Box	:	Paper bundle
Weight	3	:	22
Let	3x	:	22x

Total weight = 3x + 22x = 25x
25x = 36kg
$$\eqalign{ & x = \frac{{36}}{{25}} \times 1000 \cr & \,\,\,\,\, = 1440{\text{ grams}} \cr} $$
(∴ 1 kg = 1000 grams)
∴ Weight of paper bundles = 22x
= 22 × 1440 grams
= 31680 grams

$$\eqalign{ & {\text{Let number be }}x\,\&\, y \cr & {\text{Given,}} \cr & \,\,x:y \cr & \,\,\,\,3:4 \cr & \,3a:4a \cr & {\text{Now , given that}} \cr & \Rightarrow {\text{8}}{\left( {3a} \right)^2} = {\left( {4a} \right)^2} + 224 \cr & \Rightarrow 72{a^2} = 16{a^2} + 224 \cr & \Rightarrow 56{a^2} = 224 \cr & \Rightarrow {a^2} = 4 \cr & \Rightarrow a = 2 \cr & {\text{Numbers are}} \cr & x = 3 \times 2 = 6 \cr & y = 4 \times 2 = 8 \cr} $$

(a + b) : (b + c) : (c + a)
6 : 7 : 8
Le t 6x : 7x : 8x
∴ a + b + b + c + a = 6x + 7x + 8x
$$\eqalign{ & \Rightarrow 2a + 2b + 2c = 21x \cr & \Rightarrow 2\left( {a + b + c} \right) = 21x \cr & \Rightarrow a + b + c = \frac{{21}}{2}x \cr & \Rightarrow a + b + c = 14{\text{ given}} \cr & \Rightarrow \frac{{21x}}{2} = 14 \cr & \Rightarrow x = \frac{{28}}{{21}} = \frac{4}{3} \cr & \therefore a + b = 6 \times \frac{4}{3} = 8 \cr & \therefore a + b + c = 14 \cr & \Rightarrow 8 + c = 14 \cr & \Rightarrow c = 14 - 8 = 6 \cr} $$

Let the number of boys be x
$$\eqalign{ & {\text{Then,}} \cr & {\text{Number of girls}} \cr & = 120\% {\text{ of }}x = \frac{{6x}}{5} \cr & \therefore x + \frac{{6x}}{5} = 66 \cr & \Rightarrow \frac{{11x}}{5} = 66 \cr & \Rightarrow x = \frac{{66 \times 5}}{{11}} \cr & = 30 \cr & {\text{So, number of boys}} \cr & = 30 \cr & {\text{And, new number of girls}} \cr & = \left( {\frac{{6 \times 30}}{5}} \right) + 4 = 40 \cr & \therefore {\text{Required ratio}} \cr & = 30:40 \cr & = 3:4 \cr} $$

$$\eqalign{ & {\text{Let the third number be }}x \cr & {\text{Then,}} \cr & {\text{first number}} \cr & = 120\% {\text{ of }}x = \frac{{120}}{{100}}x = \frac{{6x}}{5} \cr & {\text{And, second number}} \cr & = 150\% {\text{ of }}x = \frac{{150}}{{100}}x = \frac{{3x}}{2} \cr & \therefore {\text{Required ratio}} \cr & = \frac{{6x}}{5}:\frac{{3x}}{2} \cr & = 12:15 \cr & = 4:5 \cr} $$

Let the terms of the ratio be x and x + 40
$$\eqalign{ & {\text{Then,}} \cr & = \frac{x}{{x + 40}} = \frac{2}{7} \cr & \Rightarrow 7x = 2x + 80 \cr & \Rightarrow 5x = 80 \cr & \Rightarrow x = 16 \cr & \therefore {\text{Required ratio}} \cr & = 16:56 \cr} $$

$$\eqalign{ & 5.5{\text{a}} = 0.65{\text{b}} \cr & \Rightarrow \frac{{55}}{{10}}{\text{a}} = \frac{{65}}{{100}}{\text{b}} \cr & \Rightarrow 55{\text{a}} = \frac{{65}}{{10}}{\text{b}} \cr & \Rightarrow 550{\text{a}} = 65{\text{b}} \cr & {\text{a}}:{\text{b}} = 65:550 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 13:110 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,{25^{2.5}}:{5^{3{\text{ }}}} \cr & {(5)^{2.5 \times 2}}:{5^{3{\text{ }}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{5^5}:{5^3} \cr & \,\,\,\,\,\,{5^3}{.5^2}:{5^{3{\text{ }}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{5^2}:1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,25:1 \cr} $$