**Ratio & Proportion Questions for RRB NTPC Set-2 PDF**

Download RRB NTPC Ratio & Proportion Questions and Answers Set-2 PDF. Top 15 RRB NTPC Ratio & Proportion questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Download Ratio & Proportion Questions for RRB NTPC Set-2 PDF

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**Question 1: **The average earning of 3 people goes up by 15% after promotion but their individual earning goes up by 10%,4% and 15%. What is the ratio of their original earnings ?

a) 1

b) 2

c) 3

d) 4

**Question 2: **A solution containing milk and water in ratio 2:3 has total volume A. Volume B is taken out of this solution and replaced with same amount of another solution containing milk and water in ratio 4:5. If the final solution has milk and water in ratio 3:4. Find the ratio of volumes A and B.

a) 14:9

b) 8:3

c) 9:5

d) 11:3

**Question 3: **Two alloys contain gold and silver. In what ratio should alloy A having 40% silver be mixed with alloy B having 40% gold to obtain a mixture having 55% silver ?

a) 1:4

b) 2:7

c) 1:3

d) 4:5

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**Question 4: **Find the average weight of class if it is given that the ratio of the number of girls to the total students in the class is 3 : 10 and the average weight of boys and girls are 40 kg and 30 kg respectively.

a) 34 kg

b) 36 kg

c) 35 kg

d) 37 kg

**Question 5: **Ram’s investment grew by 25%, 30% in the first year and second year respectively. If after three years his net return on the investment for the entire period is 50%, then find out the return on his investment in the third year alone?

a) -7.69 percent

b) 7.14 percent

c) 7.69 percent

d) -14.28 percent

**Question 6: **In a class, a student weighing 70 kg is replaced by another student weighing 35 kg. This results in the average age of the entire class going down by 0.25 kg. Find the number of students in the class.

a) 100

b) 110

c) 125

d) 140

e) 150

**Question 7: **The average age of 4 children born at intervals of 4 years is 8 years. Find the age of the youngest child?

a) 6 years

b) 8 years

c) 4 years

d) 2 years

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**Question 8: **Average age of rahul and his wife is 27 years.If after 4 years a son is born to them then what is the average age of the family 6 years from now ?

a) 21.66 years

b) 20.66 years

c) 23.66 years

d) 22.66 years

**Question 9: **What is the fourth proportional of 12,36 and 108 ?

a) 144

b) 162

c) 324

d) 216

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**Question 10: **Find the fourth number that makes the series 16, 24, 96 to form a continuous proportion.

a) 100

b) 120

c) 144

d) 180

**Question 11: **Find the fourth proportional to 8, 18 and 24.

a) 36

b) 54

c) 72

d) 90

**Question 12: **The ratio of two numbers is 2:5. If the difference between the squares of the numbers is 1029, then find the sum of the two numbers.

a) 35

b) 56

c) 49

d) 28

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**Question 13: **Ratio of ages of son and father 5 years ago was 2:5 and the ratio of their ages 9 yrs from now will be 8:13. What is the ratio of ages of son to father after 15 years from now ?

a) 3:5

b) 4:7

c) 5:6

d) 2:3

**Question 14: **Ratios of incomes of M and N is 5:3 and N got a hike of Rs 500 and the ratio of their incomes become 3:2. What is the initial income of M ?

a) 2200

b) 2800

c) 2000

d) 2500

**Question 15: **Akhi distributes 4600 rupees among 35 boys and 45 girls in such a way that each boy gets double of what each girl gets then what is the total amount received by boys?

a) Rs 2800

b) Rs 3500

c) Rs 2500

d) Rs 3000

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**Answers & Solutions:**

**1) Answer (C)**

Let the original earnings of the people be $x$,$y$ and $z$.

Total earning = $x+y+z$

**2) Answer (A)**

After taking out volume B, volume of solution = A-B

Before mixing composition will not change. It will remain 2:3.

Before mixing it with the other solution, Amount of milk = $\frac{2}{(2+3)}\times(A-B)$ = $\frac{2}{(5)}\times(A-B)$

Amount of water = $\frac{3}{2+3}\times(A-B)$ = $\frac{3}{5}\times(A-B)$

After replacing with B amount of the solution with ratio 4:5,

Final ratio = $\frac{Amount of milk in first solution + Amount of milk added}{Amount of water in first solution + Amount of water added}$ = $\frac{{\frac{2}{5}\times(A-B)+\frac{4}{4+5}\times{B}}}{{\frac{3}{5}\times(A-B)+\frac{5}{4+5}\times{B}}}$ = $\frac{3}{4}$

⇒ $({\frac{2}{5}\times(A-B)+\frac{4}{9}\times{B}}) \times 4 = ({\frac{3}{5}\times(A-B)+\frac{5}{9}\times{B}}) \times 3$

⇒ $\frac{8}{5} \times A + \frac{8}{45} \times B = \frac{9}{5} \times A -\frac{6}{45} \times B$

⇒ $\frac{A}{B}$ = $\frac{14}{9}$

**3) Answer (C)**

For 100 gms of alloy A :

Silver = 40 gms, Gold = 100 – 40 = 60 gms

For 100 gms of alloy B :

Silver = 100 – 40 = 60 gms, Gold = 40 gms

Let the two alloys be mixed in the ratio $k$ ie

Volume of alloy A = $k \times $ Volume of alloy B

Thus, in the mixture,

Volume of silver = 40k + 60 gms

Volume of Gold = 60k + 40 gms

Since we know that the mixture has 55% silver, this means that gold = 100 – 55 = 45%

Thus, ratio of volume of silver and gold can be expressed as :

$\dfrac{40k+60}{60k+40} = \dfrac{55}{45} $

$ \Rightarrow \dfrac{2k+3}{3k+2} = \dfrac{11}{9} $

$ \Rightarrow 18k + 27 = 33k + 22 $

$ \Rightarrow k = \dfrac{5}{15} = \dfrac{1}{3} $

Thus the required ratio is 1:3

**4) Answer (D)**

Let ‘100x’ be the total number of students in the class.

Number of girls in the class = 30x

Number of boys in the class = 70x

Therefore, the average weight of the class = $\dfrac{30x*30+70x*40}{100x}$ = 37 kg

**5) Answer (A)**

Let us assume that Ram’s initial investment = Rs. 100x. This amount will grow by 50% in three years, therefore the total amount that Ram will get after 3 years = 1.5*100x = Rs. 150x.

It is given that his investment grew by 25%, 30% in the first year and second year respectively. Therefore, we can say that the total amount that Ram will get after 2 years = 100x*1.25*1.3 = Rs. 162.5x

Hence, we can say that return on investment in the third year = $\frac{150x – 162.50x}{162.50}\times 100 = -7.69$ percent.

**6) Answer (D)**

Let ‘x’ kg be the average weight of the class before the stated operation.

Let ‘n’ be the number of students in the class.

Total weight of the entire class before the operation = xn

Total weight of the entire class after the operation = xn – 70 + 35 = xn – 35

Hence, the average weight of the class

$\Rightarrow$ $\dfrac{xn – 35}{n} = x – 0.25$

$\Rightarrow$ $35 = 0.25n$

$\Rightarrow$ $n = 140$

Therefore, option D is the correct answer.

**7) Answer (D)**

Let the age of the youngest child be x years

Then, Ages of 4 children = x, x+4, x+8, x+12 years

Given, $\dfrac{x+x+4+x+8+x+12}{4} = 8$

⇒ $\dfrac{4x+24}{4} = 8$

⇒ 4x+24 = 32

⇒ 4x = 8

⇒ x = 2

Hence, The age of the youngest child = 2 years.

**8) Answer (D)**

let the ages of rahul and his wife be r and w respectively

Given r+w=54

Age of son after 6 years will be 6-4=2 years

So the average age of the family becomes (54+2+6+6)/3=68/3=22.66 years

**9) Answer (C)**

For fourth proportional we have 12:36::108:x

Therefore 12x=36*108

x=324

**10) Answer (C)**

a, b, c, d in continued proportion => ad = bc

16 * d = 24 * 96

=> d = 24 * 6 = 144

**11) Answer (B)**

For a, b, c and d to be proportional, a:b :: c:d must hold true.

=> 8 : 18 :: 24 : x

=> 8x = 24 * 18

=> x = 3 * 18

=> x = 54

**12) Answer (C)**

Let the two numbers be 2x and 5x.

Given, $(5x)^2 – (2x)^2 = 1029

⇒ $25x^2 – 4x^2 = 1029

⇒ $21x^2 = 1029

⇒ x^2 = 49

⇒ x = 7.

Then, the two numbers will be,

2x = 2*7 = 14

5x = 5*7 = 35

Then, Sum of two numbers = 14+35 = 49.

**13) Answer (D)**

Let the present age of son be ‘s’ and father’s be ‘f’

(s-5)/(f-5)=2:5

5s-25=2f-10

5s-2f=15

(s+9)/(f+9)=8:13

13s+117=8f+72

8f-13s=45

20s-8f=60

7s=105

s=15

f=30

15 years from now their ages will be s=15+15 and f=30+15

Ratio =30:45

=2:3

**14) Answer (D)**

let the incomes of M and N be 5x and 3x.

Now (5x)/(3x+500)=3:2

10x=9x+1500

x=500

Income of M =5*500

=2500

**15) Answer (A)**

let the amount received by each girl be x then amount receive by each boy becomes 2x.

Therefore 35*2x+45*x=4600

70x+45x=1100

115x=4600

x=40

So each boy gets Rs 80 and each girl gets Rs 40

Total amount received by boys is 35*80=2800

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