# Aptitude Questions with Answers- Ratio and Proportion Solved Examples 2

## IBPS Aptitude - Ratio and Proportion Solved Questions 2

**16.**Rs 12200 divided among 5 women, 4 boys and 7 men , such that the share a woman, a boy and a man are in the ratio of 7:3:2. Find the share of a girl?

**Explanation :**Let the share of a woman , a boy and that of man are 7x ,3x and 2x.

Total
amount paid = (5 x 7x) + (4 x 3x) +(7 x 2x) =61x = 12200

=>35x+12x+14x=12200

; =>61x=12200
=>x=200

The
share of a girl= 3x = 3 x Rs 200 = Rs 600

**17.**A bag consists of 1 rupee, 50 paise and 25p coins and their total value in rupees is Rs 1120 .These three types of coins are in the ratio 3:20:4. How may 25 paisa coins are there?

**Explanation :**The coins are 1 rupee, 50 paisa and 25 paisa.

The
ratio is 100p: 50 p : 25 p =4 : 2: 1

The
coins are in the ratio = 3 :20: 4

The ratio of their values in rupees=> 12 :40 :4

The value of 25 p coins in total Rs 1120 = 1120 x $\frac{4}{56}$= 80 rupees

Number
of 25 p coins are = 80x4 = 320 coins

**18**. A bag contains one rupee, 50 paisa and 10 paisa coins in the ratio of 3:2:1. How many 50 paisa coins are there, if the total number of money in the bag is Rs 820?

**Explanation**:

Let the number of 100p, 50p and 10p coins be 3x ,2x and x .

The
ratio of the values of coins = 100 p: 50p :10 p = 10:5:1

Total
amount = (10 x 3x ) + (5 x 2x) +( 1 x x) = 30x+10x+x=41x

The
ratio of their values in rupees= 30x:10x:x

Given
total amount is Rs 820 => 41x = 820 => x=Rs 20

Total value of 50p coins (in rupees) = 10 x 20 =Rs 200

Total
50 paisa coins are = 200 x 2 = 400 coins

**19**. Two numbers are in the ratio 2:5. If each number is increased by 12, the ratio becomes 4:7.Find the two numbers?

**Explanation**:

Let the numbers be 2x and 5x.

2x+12
5x+12

=
4
7

ð 7(2x+12) = 4(5x+12)

ð 14x+84 = 20x+48

ð 6x=36

ð x=6

The numbers are 2x and 5x
= 2 x 6 and 5 x 6 =12 and 30

**20.**Two numbers are in the ratio 3:2. If 20 is subtracted from each of them, the ratio becomes 5:3. Find the numbers

**Explanation**: Let two numbers be 3x and 2x.

3x-20
/ 2x-20 = 5:3

ð 3 (3x-20) = 5 (2x-20)

ð 9x -60 = 10x- 100

ð X= 40

Therefore, the numbers are 3x and 2x -= 3x40 and 2x40=120 and 80

**21**. Two vessels of equal capacity are $\frac{8}{11}$ and $\frac{3}{5}$ full of wine. They are then filled up with water and the contents are poured in another vessel. Find the ratio of the wine and water in the final mixture?

**Explanation:**

8
11

th of first
vessel and
3
5

th of second vessel have wine.
Quantity of wine in 1

^{st}vessel =
8
11

Quantity of water in 1

^{st}vessel =
3
11

Quantity of wine in 2

^{nd}vessel =
3
5

Quantity of water in 2

^{nd}vessel =
2
5

Total quantity of wine =

8
11

+
3
5

=
73
55

Total quantity of water =

3
11

+
2
5

=
37
55

Ratio of wine and water in final mixture= 73 :
37

**22**. In a mixture of 90 litres wine and water are in the ratio 3:2. If 12 litres of water is added to the mixture, what will be the ratio of wine and water in the final mixture?

**Explanation :**Quantity of wine in mixture = 90 x

3
5

= 54
litres
Quantity of water in the mixture = 90 x

2
5

=
36 lifters
When 12 liters of water is added to the
mixture, quantity of water = 36 +12 = 48 litres

Ratio of wine and water in the final mixture=
54 : 48 =9:8

**23.**The marks of a and b are in the ratio of 3:2 and marks of b and c are in the ratio of 7:1 and sum of the marks of a,b and c is 1480. Find the marks of A?

**Explanation:**A : B = 3 :2

B: C
= 7 :1

Equating the common term B, the ratio of A,B and C = 21 :14:2

Total
marks obtained by A,B and C= 1480

Marks of A = 1480 x

21
37

= 840**24**. What is the number which when subtracted from the terms of the ratio 5:9 makes it equal to 1:3 ?

**Explanation:**Let the number to be subtracted = x

(5-x)
(9-x)

=
1
3

ð 3(5-x) =1(9-x)

ð 15-3x = 9-x=>
x=3

The number to be subtracted from the terms of the ratio 5:9 to
make it ratio 1:3 is 3

**25**. The incomes of A and B are in the ratio of 3:2 and their expenditures are in the ratio of 5:3. If each of A and B saves Rs 2000 , the find the income of A?

**Explanation :**Income – Savings = Expenditures

Let
the incomes of A and B be 3x and 2x respectively.

Each
of A and B saves Rs 2000.

3x-2000
2x -2000

= 5:3
ð 3(3x-2000)= 5(2x-2000)

ð 9x-6000=
10x-10000

ð X=4000

Income of A =3x = 3 x
4000 = Rs 12000

**26**. An organization reduces the number of employees in the ratio of 9:7 and increases their salaries in the ratio of 10:13. Find in what ratio the expenditure of that organization on salaries is changed and also find the difference between the present and previous expenditures on salaries, if it was previously Rs 5400 lacs.

**Explanation:**Ratio of the employees = 9:7

Ratio of the salaries = 10 :13.

Ratio of previous and present expenditures on salaries

= 9 x 10: 7 x 13 = 90 :91

Therefore, salaries will
increase in the ratio of 90:91.

Expenditure on salaries previously => 90x = 5400 lacs

Expenditure on salaries now => 91x =

5400
90

x 91 =5460
Therefore, the difference
is = Rs 5460 –Rs 5400 = 60 lacs

**27.**Ratio of the fares of first, second and third class tickets of a train is 3:2:1 and ratio of number of passengers travel in first , second and third classes is 2:3:4 . If the total collection from 2

^{nd}class tickets is Rs 12000, find the total collection from 3

^{rd}class tickets?

**Explanation:**Ratio of the fare of 1

^{st}, 2

^{nd}and 3

^{rd}class tickets = 3 : 2 :1

Ratio of the passengers travel in these classes = 2 : 3 :4

The ratio of total collection from 1

= 3x2 + 2x3 + 1x4 =6:6:4= 3:3:2

^{st}, 2^{nd}and 3^{rd}class= 3x2 + 2x3 + 1x4 =6:6:4= 3:3:2

Collection from 2

^{nd}class tickets = 3x=12000 =>x=4000
Total collection from 3

= Rs 8000

^{rd}class tickets = 2x = 2 x 4000= Rs 8000

**28**. 48 litres of mixture contains milk and water in the ratio 5:1. How much water must be added to this mixture so as to have milk in the ratio 2:1?

**Explanation :**The quantity of milk in 48 litres of mixture = 48 x

5
6

=40 litres
Quantity of water in the mixture = 48- 40 = 8 litres

We have
to add water . So there is no change in the quantity of milk.

Milk
Water

=
40
8+x

=
2
1

ð 1 x 40 =2( 8+x)

ð 40= 16+x => x=
12

**29.**A vessel of 270 litres capacity is full of pure wine. On the first day $\frac{1}{3}$ of the milk is drawn and filled with water on the second day $\frac{1}{3}$ of the mixture is drawn and filled with water and again the same thing done on the third day. Find the quantity of wine and water at the end?

**Solution:**Total pure wine = 270 litres

When $\frac{1}{3}$ is drawn, we will have

2
3

of the
previous wine.
Quantity of wine at the end = 270 x

2
3

x
2
3

x
2
3

= 80 litres

Quantity of water = 270-80= 190 litres

**30.**A sum of Rs 45 is made up of 100 coins partly of 50 paisa and partly of 25 paise. How many 25 paisa coins are there?

**Solution**: Let the number of 50 paisa coins and 25 paisa coins be x and y respectively.

Number of coins => x+ y = 100
------- (1)

Value of coins => 50x +25y = 4500 ----(2)

Solving
equations 1 and 2 , we get x=80 and y=20

So the number of 25 paisa coins are 20.