# Quant Questions – Ratio and Proportion Problems Set 3

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Dear Readers, Important Ratio and Proportion Problems for upcoming competitive Exams with Solutions. Ratio and Proportion is an important topic from banking exam point of view. We could solve Ratio & Proportion problems easily if we understand the basic concepts. Aspirants preparing for SBI Clerk, SBI PO, IBPS Clerk, IBPS PO, Insurance, RBI and other competitive examination can make of use these ratio and proportion problems. We have included some important ratio & proportion problems that are repeatedly asking in competitive exams. Practice lot of questions to become an expert in solving ratio & proportion problems and learn to use shortcuts.

## Ratio and Proportion Problems – Set 3

1).In a class of 60 students, where the girls are twice that of boys. Kamal ranked seventeenth from the top. If there are 9 girls ahead of Kamal, the number of boys in rank after him is:
a)  3
b)  7
c)  12
d)  13
e)  None of these

2).Two varieties of rice at Rs. 10 per kg and Rs. 12 per kg are mixed together in the ratio 1 : 2. What is the price of the resulting mixture?
a)  Rs. 10.50 per kg
b)  Rs. 10.67 per kg
c)  Rs. 11.20 per kg
d)  Rs. 11.33 per  kg
e)  None of these

3).If A varies directly proportional to C, and B also varies directly proportional to C, which one of the following is not correct?
a)  (A + B) ∝ C
b)  (A – B) ∝ (1/C)
c)  √(AB) ∝ C
d)  (A / B) = constant
e)  None of these

4).If the ratio of the areas of two squares is 9 : 1, the ratio of their perimeters is:
a)  9 : 1
b)  3 : 1
c)  3 : 4
d)  1 : 3
e)  None of these

5).The ratio of the number of boys and girls in a school is 3 : 2. If 20% of the boys and 25% of the girls are scholarship holders, then the percentage of the students who did not get scholarship is:
a)  68%
b)  78%
c)  82%
d)  72%
e)  None of these

6).In a mixture of 60 litres, the ratio of acid and water is 2 : 1. If the ratio of acid and water is to be 1 : 2, then the amount of water (in litres) to be added to the mixture is
a)  50
b)  45
c)  55
d)  60
e)  None of these

7).Salaries of Akash, Babloo and Chintu are in the ratio of 2 : 3 : 5. If their salaries were increased by 15%, 10% and 20% respectively, what will be the new ratio of their salaries?
a)  3 : 3 : 10
b)  23 : 33 : 60
c)  20 : 22 : 40
d)  None of these
e)  Cannot be determined

8).If x : y = 3 : 4 then
(7x + 3y) : (7x – 3y) is equal to :
a)  5 : 2
b)  4 : 3
c)  11 : 3
d)  None of these
e)  Cannot be determined

9).If A : B = 5 : 7 ; C : D = 2A : 3B then AC : BD is:
a)  20 : 38
b)  50 : 147
c)  10 : 21
d)  None of these
e)  Cannot be determined

1). c) 2). d) 3). b) 4). b) 5). b) 6). d) 7). b) 8). c) 9). b)

Solution:

1).Girls : Boys = 2 : 1. Total no.of students is 60
Girls 60 × (2 / 3) = 40 and boys = 20. Kamal has been ranked 17 which means there are 16 students before him of which 9 are girlsàremaining 7 are boys.
7 boys are ahead of Kamal + Kamal is the 8th boy, hence, there are (20 – 8) = 12 boys after him.

2).Let the amount of rice be 1 kg and 2 kg. (The given ratio is 1 : 2). So, total cost = 10 × 1 + 12 × 2 = 10 + 24 = 34
Rs. 34 is the cost of (1 + 2 = 3) kg of rice.
Cost per kg = (34 / 3) = 11.33 per kg

3).A ∝ C = A = K1C
B ∝ C = B = K2C
Where K1and K2 are constants
(A + B) = (K1 +K2)C
(K1 +K2) is again a constants
(A + B) ∝ C √(A.B) = √[ (K1C) × (K2C)] = √(K1 K2C)
Here √(K1 K2) is a constantà√(A . B)∝ C
(A / B) = (K1C / K2C) = (K1 / K2)
(K1/ K2) is (1/ K2) a constant
But, (A – B) = (K1C – K2C) = (K1 – K2) C
(K1– K2) is a constantà(A – B) ∝ C

4).(x2 / y2) = (9 / 1) where x and y are the sides of the two squares
= ratio of perimetersà(x /y) = (3 / 1)à(4x / 4y) = (12 / 4) = (3 / 1)

5).Number of students in school = 100 (let)
Boys = (3 / 5) × 100 = 60, Girls = 40
Students who did not get scholarship
Boys = 60 × (80 / 100) = 48
Girls = 40 × (75 / 100) = 30
Students who do not get scholarship = 78
Required percentage = 78

6).In 60 litres of mixture,
Acid = (2/3) × 60 = 40 litres, Water = 20 litres
If x litres of water be mixed, then [40 / (20+ x)] = 1/2à 20 + x = 80
x = 80 – 20 = 60 litres

7).Required ratio  = [(2 × 115) / 100] : [(3 × 110) / 100] : [(5 × 120) / 100]
230 : 330 : 600à23 : 33 : 60

8).(x / y) = (3 / 4)
(7x + 3y) / (7x – 3y) = [7 (x /y) + 3] / [7(x /y) – 3]
[7 × (3 / 4) + 3] / [7 × (3 / 4) – 3] = (21 + 12) / (21 – 12)
33 / 9 = 11 / 3 = 11 : 3