# Ratios and Proportions – Quiz 2

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# Ratios and Proportions – Quiz 2

### Introduction

Ratios and Proportions is one of important topic in Quantitative Aptitude Section. In Ratios and Proportions – Quiz 1 article candidates can find a question with an answer. By solving this question candidates can improve and maintain, speed, and accuracy in the exams. Ratios and Proportions - Quiz 1 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

The salaries A, B, C are in the ratio 2:3:5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
A. 3 : 3 : 10 B. 10 : 11 : 20 C. 23 : 33 : 60 D. Cannot be determined

C
Let A = 2k, B = 3k and C = 5k
A's new salary = $(\frac {115}{100} \times 2k) = \frac {23 k}{10}$
B's new salary = $(\frac {120}{100} \times 5k) = \frac {33 k}{10}$
C's new salary = $(\frac {115}{100} \times 2k) = 6k$
New ratio = ($\frac {23 k}{10}$ : $\frac {33 k}{10}$ : 6k) = 23 : 33 : 60

### Q2

A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs.2.40 the number of girls is
A. 35 B. 40 C. 45 D. 50

B
Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x + y = 100
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x + 2.4y =312
Step (iii)
Solving (i) and (ii)
3.6x +3.6y = 360 multiplying by 3.6
3.6x+2.4y1.20yy=312=48=40
The number of girls is 40.

### Q3

In a mixture 60 litres, the ratio of milk and water 2:1. If the this ratio is to be 1:2, then the quantity of water to be further added is:
A. 20 litres B. 30 litres C. 40 litres D. 60 litres

D
Quantity of milk = 60 $\times \frac{2}{3}$ = 40 litres
Quantity of water in it =(60−40) litres =20 litres.
New ratio =1:2
Let, the quantity of water to be added further be x litres.
Then milk : water = $\frac{40}{20 + x}$
Now, $\frac{40}{20 + x} = \frac{1}{2}$
20 + x = 80
$\Rightarrow$x = 60

### Q4

The ratio of the number of boys and girls in a college is 7:8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
A. 8 : 9 B. 17 : 18 C. 21 : 22 D. 5 : 4

C
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120 % of 7x) and (110 % of 8x).
$\Rightarrow (\frac{120 }{100} \times 7x)$ and $(\frac{110}{100}\times 8x)$
$\Rightarrow \frac{42 x}{5}$ and $\frac{44x}{5}$
So, the required ratio = ($\frac{42 x}{5}$ : $\frac{44x}{5}$) = 21 : 22

### Q5

A sum of Rs. 36.90 is made up of 180 coins which are either 10 paise coins or 25 p coins. The number of 10 p coins is:
A. 48 B. 54 C. 56 D. 60

B
Step (i) Total number of coins = 180
Let x be number of 10p coins and y be number of 25p coins
x + y = 180 --------(i)
Step (ii) Given 10p coins and 25p coins make the sum = Rs. 36.90
$\frac{10 x}{100} + \frac{25 y}{100} = 36.9$
$\Rightarrow$10x + 25y = 3690-----(ii)
Step (iii)
Solving (i) and (ii),
10x + 10y = 1800
10x + 25y = 3600
-15y = 1890
y = 126
Substitute y value in equation (i)
x = 180 – 126 = 54
Number of 10p coins = 54