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mcq on ratio and proportion for competitive exams

MCQ on ratio and proportion for competitive exams

Solve the following MCQ on ratio and proportion:

The respective ratio between the present ages of father, mother and daughter is 7 : 6 : 2. The difference between mother’s and the daughter’s age is 24 years. What is the father’s age at present ?

(a) 43 years

(b) 42 years

(c) 39 years

(d) 38 years

(e) None of these

Solution: (e)

Let present age of father, mother and daughter be

7x, 6x, 2x

Given, 6x – 2x = 24

\(\displaystyle \Rightarrow \)4x = 24

\(\displaystyle \Rightarrow \) x = 6

Father age = 7x = 42 years.

When X is subtracted from the numbers 9,15 and 27, the remainders are in continued proportion. What is the value of X ?

(a) 8

(b) 6

(c) 4

(d) 5

(e) None of these

Solution: (e)

9, 15, 27

9 – x , 15 – x, 27 – x

\(\displaystyle \frac{{15-x}}{{9-x}}=\frac{{27-x}}{{15-x}}\)

\(\displaystyle {{(15-x)}^{2}}=(27-x)(9-x)\)

\(\displaystyle 225+{{x}^{2}}-30x=243-9x-27x+{{x}^{2}}\)

\(\displaystyle \Rightarrow \)  –30x + 9x + 27x = 243 –225

\(\displaystyle \Rightarrow \) 6x = 18

\(\displaystyle \Rightarrow \) x = 3

Alternate method:

If, \(\displaystyle \frac{a}{b}=\frac{b}{c}\) be the continued fraction

so, \(\displaystyle \frac{{9-x}}{{15-x}}=\frac{{15-x}}{{27-x}}\)

[9, 15 and 27 are divisible by 3]

We can see that, x = 3, satisfies the above condition.

\(\displaystyle \frac{{9-3}}{{15-3}}=\frac{{15-3}}{{27-3}}\)

\(\displaystyle \Rightarrow \frac{6}{{12}}=\frac{{12}}{{24}}\)

\(\displaystyle \Rightarrow \frac{1}{2}=\frac{1}{2}\)

Therefore, x=3

The respective ratio of the present ages of a mother and daughter is 7 : 1. Four years ago the respective ratio of their ages was 19 : 1. What will be the mother’s age four years from now? 

(a) 42 years

(b) 38 years

(c) 46 years

(d) 36 years

(e) None of these

Solution: (c)

Let the ages of the mother and daughter be 7x and x years respectively.

Therefore, Four years ago, \(\displaystyle \frac{{7x-4}}{{x-4}}=\frac{{19}}{1}\)

\(\displaystyle \Rightarrow \) 19x – 76 = 7x – 4

\(\displaystyle \Rightarrow \) 12x = 72 = x = 6

 Mother’s age after four years = 7x + 4 = 7 × 6 + 4 = 46 years

The ratio of the monthly income of Trupti, Pallavi and Komal is 141 : 172 : 123. If Trupti’s annual income is ₹ 3,38 400, what is Komal’s annual income? (In some cases, monthly income is used, while in others, annual income is used.)

(a) ₹ 4,12,800

(b) ₹ 3,63,500

(c) ₹ 3,17,600

(d) ₹ 2,95,200

(e) None of these

Solution: (d)

Let monthly income to Trupati, Pallavi and Komal is 141x, 172x and 123x respectively.

Trupti’s annual income = 141x × 12 = 338400

     x = \(\displaystyle \frac{{338400}}{{141\times 12}}\)

Komal’s annual income = 123 × 12x

= \(\displaystyle 123\times 12\times \frac{{338400}}{{141\times 12}}=295200\)

The ratio of the number of boys to that of girls was 1 : 2 but when 2 boys and 2 girls left, the ratio became 1 : 3. How many people were at the party originally?

(a) 12

(b) 15

(c) 10

(d) 18

(e) None of these

Solution: (a)

2 b = g … … . (1)

b − 2 / g − 2 = 1/3 … … . (2)

Solving (1) and (2),we get b = 4 , g = 8

Therefore, b + g = 12