Ratio - Aptitude MCQ Questions and Solutions with Explanations

a   :   b   ::   c   :   d
    a × d = b × c
So, go through options
(A). 9 × 16 = 12 × 12 (right)
(B). 13 × 4 = 11 × 5 (wrong)
(C). 30 × 24 = 45 × 13 (wrong)
(D). 3 × 5 = 5 × 2 (wrong)
So answer is 12 : 9 :: 16 : 12

Given, A : B = 7 : 9
B : C = 3 : 5
Equal the value of B in both equation
B : C = 3 : 5 (multiply with 3)
i.e. B : C = 9 : 15
∴ A : B : C = 7 : 9 : 15

Length : Width
    5x   :: 2x
Width ⇒ 2x ⇒ 40m
                 x ⇒ 20m
∴ Length ⇒ 5 × 20 = 100m

$$\eqalign{ & {\text{First}}:{\text{Second}} = 3:2, \cr & {\text{Second}}:{\text{Third}} = 3:2 \cr & = \left( {3 \times \frac{2}{3}} \right):\left( {2 \times \frac{2}{3}} \right) \cr & = 2:\frac{4}{3} \cr & \therefore {\text{Ratio between the numbers}} \cr & = 3:2:\frac{4}{3} \cr & = 9:6:4. \cr} $$
Let the numbers be 9x, 6x and 4x
Then,
= (9x)² + (6x)² + (4x)² = 532
⇒ 81x² + 36x² + 16x² = 532
⇒ 133x² = 532
⇒ x² = 4
⇒ x = 2
So, third number = 4x = 4 × 2 = 8

Let the numbers be 17x and 45x
Then,
$$\eqalign{ & = \frac{1}{5}{\text{of }}45x - \frac{1}{3}{\text{of }}17x = 15 \cr & \Leftrightarrow 9x - \frac{{17}}{3}x = 15 \cr & \Leftrightarrow \frac{{10x}}{3} = 15 \cr & \Leftrightarrow x = 15 \times \frac{3}{{10}} = \frac{9}{2} \cr & \therefore {\text{Smaller number}} \cr & = 17x = \left( {17x \times \frac{9}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{153}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{76}}\frac{1}{2} \cr} $$

Let the incomes of A, B, C be 7x, 9x and 12x and their expenditures be 8y, 9y and 15y respectively
Then, A's saving = (7x - 8y)
$$\eqalign{ & \therefore 7x - 8y = \frac{1}{4}{\text{of }}7x \cr & \Rightarrow 8y = 7x - \frac{{7x}}{4} \cr & \Rightarrow 8y = \frac{{21}}{4}x \cr & \Rightarrow y = \frac{{21}}{{32}}x \cr & {\text{So, A's expenditure}} \cr & = \left( {8 \times \frac{{21}}{{32}}x} \right) = \frac{{168}}{{32}}x \cr & {\text{B's expenditure}} \cr & = \left( {9 \times \frac{{21}}{{32}}x} \right) = \frac{{189}}{{32}}x \cr & {\text{C's expenditure}} \cr & = \left( {15 \times \frac{{21}}{{32}}x} \right) = \frac{{315}}{{32}}x \cr & \therefore {\text{A's saving}} \cr & = \left( {7x - \frac{{168}}{{32}}x} \right) = \frac{{56}}{{32}}x \cr & \therefore {\text{B's saving}} \cr & = \left( {9x - \frac{{189}}{{32}}x} \right) = \frac{{99}}{{32}}x \cr & \therefore {\text{C's saving}} \cr & = \left( {12x - \frac{{315}}{{32}}x} \right) = \frac{{69}}{{32}}x \cr & {\text{Hence, required ratio}} \cr & = \frac{{56}}{{32}}x:\frac{{99}}{{32}}x:\frac{{69}}{{32}}x \cr & = 56:99:69 \cr} $$

Ratio of the age of two person A : B = 4 : 7 = 4x : 7x
Difference of their ages is = 7x - 4x = 30
i.e. x = 10
Age of A = 4 × 10 = 40
Age of A = 7 × 10 = 70
Sum of their ages = 40 + 70 = 110

	Rs. 1	:	50 P	:	25 p
Values of coins	8x	:	11x	:	3x
Number of coins	8x×1	:	4x×2	:	3x×4
	8x	:	8x	:	12x

$$\eqalign{ & \therefore {\text{Total coins}} \cr & \Rightarrow 8x + 8x + 12x = 28x \cr & \Rightarrow 28x = 280{\text{ (given)}} \cr & \Rightarrow x = \frac{{280}}{{28}} \cr & \Rightarrow x = 10 \cr} $$
∴ Number of 50 P coins are 8x
= 8 × 10 = 80

The distance covered by policeman in 5 steps is equal to that of thief in 7 steps
⇒ 5P = 7T
P : T
7 : 5 (distance covered in each step)
⇒ And policeman goes 8 steps while thief moves 10 steps

		Policeman	:	Thief
Steps		8	:	10
Distance in each step		7	:	5
		56	:	50
Speed	=	28	:	25

Let the basic salaries of A and B be Rs. x and Rs. y respectively.
Then,
= x + 65% of x = y + 80% of y
$$\eqalign{ & \Rightarrow \frac{{165}}{{100}}x = \frac{{180}}{{100}}y \cr & \Rightarrow 165x = 180y \cr & \Rightarrow \frac{x}{y} = \frac{{180}}{{165}} = \frac{{12}}{{11}} \cr & \therefore {\text{Required ratio}} \cr & = 12:11 \cr} $$