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Ratio - Aptitude MCQ Questions and Solutions with Explanations

Ratio MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Ratio with easy and logical explanations. Ratio MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-2 section-2

Each section contains 100 MCQs on Ratio .
11
If x : y = 3 : 1, then x3 - y3 : x3 + y3 = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & = \frac{x}{y} = \frac{3}{1} \cr & \Rightarrow \frac{{{x^3}}}{{{y^3}}} = \frac{{27}}{1} \cr & \Rightarrow \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} = \frac{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) - 1}}{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) + 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{27 - 1}}{{27 + 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{26}}{{28}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{13}}{{14}} \cr & = \left( {{x^3} - {y^3}} \right):\left( {{x^3} + {y^3}} \right) \cr & = 13:14 \cr} $$
12
If 5a + 3b : 2a - 3b = 23 : 5, then the value of a : b is
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & = \frac{{5a + 3b}}{{2a - 3b}} = \frac{{23}}{5} \cr & \Rightarrow 5\left( {5a + 3b} \right) = 23(2a - 3b) \cr & \Rightarrow 25a + 15b = 46a - 69b \cr & \Rightarrow 21a = 84b \cr & \Rightarrow \frac{a}{b} = 4. \cr & {\text{Hence}},\,a:b = 4:1 \cr} $$
13
If x : y = 7 : 3, then the value of $$\frac{{xy + {y^2}}}{{{x^2} - {y^2}}}$$   is
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & = \frac{x}{y} = \frac{7}{3} \cr & = \frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{\left( {\frac{x}{y}} \right) + 1}}{{\left( {\frac{{{x^2}}}{{{y^2}}}} \right) - 1}} \cr & = \frac{{\frac{7}{3} + 1}}{{{{\left( {\frac{7}{3}} \right)}^2} - 1}} \cr & = \frac{{10}}{3} \times \frac{9}{{40}} \cr & = \frac{3}{4} \cr} $$
14
The salaries of A, B and C are in the ratio 1 : 3 : 4. If the salaries are increased 5%, 10% and 15% respectively. Then the increased salaries will be in the ratio = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
  A : B : C
  1 : 3 : 4
Let 100x : 300x : 400x
    ↓+5%       ↓+10%       ↓+15%  
Let 105x : 330x : 460x
  105 : 330 : 460
  21 : 66 : 92
15
There are three numbers A, B, C such that twice A is equal to thrice B and four times B is equal to five times C. Then the ratio between A and C is = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Given 2A = 3B and 4B = 5C
i.e.
A : B = 3 : 2
B : C = 5 : 4
A : B = 3 : 2 (multiply with 5) and
B : C = 5 : 4 (multiply with 2)
i.e. A : B = 15 : 10 and
B : C = 10 : 8
∴ A : B : C = 15 : 12 : 8
∴ A : C = 15 : 8
16
If a : b = b : c, then a4 : b4 would be equal to
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & = \frac{a}{b} = \frac{b}{c} \cr & \Rightarrow {b^2} = ac \cr & \Rightarrow \frac{{{a^4}}}{{{b^4}}} = \frac{{{a^4}}}{{{{\left( {ac} \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^4}}}{{{a^2}{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^2}}}{{{c^2}}} \cr & \Rightarrow {a^4}:{b^4} = {a^2}:{c^2} \cr} $$
17
If (4x2 - 3y2) : (2x2 + 5y2) = 12 : 19, then x : y is
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{ = }}\frac{{4{x^2} - 3{y^2}}}{{2{x^2} + 5{y^2}}} = \frac{{12}}{{19}} \cr & \Rightarrow 19\left( {4{x^2} - 3{y^2}} \right) = 12\left( {2{x^2} + 5{y^2}} \right) \cr & \Rightarrow 76{x^2} - 57{y^2} = 24{x^2} + 60{y^2} \cr & \Rightarrow 52{x^2} = 117{y^2} \cr & \Rightarrow 4{x^2} = 9{y^2} \cr & \Rightarrow \frac{{{x^2}}}{{{y^2}}} = \frac{9}{4} \cr & \Rightarrow {\left( {\frac{x}{y}} \right)^2} = {\left( {\frac{3}{2}} \right)^2} \cr & \Rightarrow \frac{x}{y} = \frac{3}{2} \cr & \therefore x:y = 3:2 \cr} $$
18
If x : y = 3 : 4 and a : b = 1 : 2, then the value of $$\frac{{2xa + yb}}{{3yb - 4xa}}$$   is
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & = \frac{x}{y} = \frac{3}{4}{\text{ and }}\frac{a}{b} = \frac{1}{2} \cr & \Rightarrow \frac{{xa}}{{yb}} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \cr & \frac{{2xa + yb}}{{3yb - 4xa}} \cr & = \frac{{2\left( {\frac{{xa}}{{yb}}} \right) + 1}}{{3 - 4\left( {\frac{{xa}}{{yb}}} \right)}} \cr & = \frac{{2 \times \frac{3}{8} + 1}}{{3 - 4 \times \frac{3}{8}}} \cr & = \frac{7}{4} \times \frac{2}{3} \cr & = \frac{7}{6} \cr} $$
19
Income of A and B are in the ratio 4 : 3 and their annual expenses are in the ratio 3 : 2. If each save Rs. 60000 at the end of the year, the annual income of A is = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
  A : B
Income 4x : 3x
  Expenses   3y : 2y
Saving 6000   :   6000

Income = Expenses + Saving
$$\therefore \frac{{4x - 60000}}{{3x - 60000}} = \frac{3}{2}$$
⇒ 8x - 120000 = 9x - 180000
⇒ x = 60000
∴ Income of A = 4 × 60000 = 240000
20
The weight of Mr. Gupta and Mrs. Gupta are in the ratio 7 : 8 and their total weight is 120 kg. After taking a dieting course Mr. Gupta reduces by 6 kg and the ratio between their weights change to 5 : 6, so Mrs. Gupta has reduced by = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
  Mr.   :   Mrs.
Before 7x : 8x
After 5y : 6y

Before 7x + 8x = 120
⇒ 15x = 120
⇒ x = 8
∴ Mr. Gupta = 7 × 8 = 56
Mrs. Gupta = 8 × 8 = 64
∴ After losing 6kg by Mr. Gupta the ratio becomes 5 : 6
⇒ Let Mrs. Gupta loss x kg
$$\eqalign{ & \therefore \frac{{56 - 6}}{{64 - x}} \times \frac{5}{6} \cr & \left( {{\text{Cross multiplication}}} \right) \cr} $$
⇒ 300 = 320 - 5x
⇒ 5x = 20
∴ x = 4 kg