Practice Test on Ratio and Proportion | Ratio and Proportion Word Problems with Answers

Practice Test on Ratio and Proportion helps students to get knowledge on different levels. The Ratio and Proportion Questions and Answers provided range from beginner, medium, hard levels. Practice the Questions here and get to know how to solve different problems asked. All the Ratio and Proportion Word Problems covered are as per the latest syllabus. Master the topic of Ratio and Proportion by practicing the Problems on a consistent basis and score better grades in your exam.

Ratio and Proportion Questions and Answers

1. The ratio of monthly income to the savings in a family is 5 : 3 If the savings be $6000, find the income and the expenses?

Solution:

Let us assume the Income be 5x

whereas savings be 3x

Given Savings = $6000

3x = 6000

x = 6000/3

= 2000

Income = 5x

= 5*2000

= $10,000

Expenses = Income – Savings

= 5x – 3x

= 2x

= 2*2000

= $4000

Therefore, Income and Expenses are $10,000 and $ 4000.

2. Two numbers are in the ratio 7: 4. If 3 is subtracted from each of them, the ratio becomes 5 : 2. Find the numbers?

Solution:

Let us consider the number be x

so 7x:4x

If 3 is subtracted the ratio becomes 5:2 then we have

7x-3:4x-3 = 5:2

equating them ad solving we get the values as

7x-3/4x-3 = 5/2

(7x-3)2 = 5(4x-3)

14x-6 = 20x-15

-6+15 = 20x-14x

9 = 6x

6x = 9

x = 9/6

= 3/2

Therefore the numbers are 7(3/2) and 4(3/2)

= 21/2, 6

3. Two numbers are in the ratio 3 : 5. If their sum is 720, find the numbers?

Solution:

Let us consider the number be x

Therefore two numbers become 3x:5x

Since their Sum = 720

3x+5x = 720

8x = 720

x = 720/8

= 90

Numbers are 3x and 4x

Thus, they become 3(90) and 4(90) i.e. 270 and 360.

4. A sum of money is divided among Rohan and Anand in the ratio 4 : 6. If Anand’s share is $600, find the total money?

Solution:

Let the money be x

Rohan and Anand’s Share = 4x:6x

Anand’s Share = $600

6x = $600

x = $100

Rohan’s Share = 4x

= 4*100

= $400

Total Money = Rohan’s Share + Anand’s Share

= $400+$600

= $1000

Therefore, the Sum of Money is $1000

5. The difference between the two numbers is 33 and the ratio between them is 5 : 2. Find the numbers?

Solution:

Let the number be x

From the given data

we have 5x-2x = 33

3x = 33

x = 11

Numbers are 5x, 2x

thus, they become 5*11 and 2*11

= 55, 22

Therefore, the numbers are 55 and 22.

6.  The ages of A and B are in the ratio 3 : 6. Four years later, the sum of their ages is 53. Find their present ages?

Solution:

Let the Present Ages be 3x and 6x

After four Years Age of A and B Becomes 3x+4 and 6x+4

We know sum of their ages after 4 years = 53

3x+4+6x+4 = 53

9x+8 = 53

9x = 53-8

9x = 45

x =5

Present Ages of A and B is 3x and 6x

thus 3*5 and 6*5 i.e. 15 and 30

Therefore, the Present Ages of A and B are 15 and 30.

7. If 3A = 4B = 5C, find the ratio of A : B : C?

Solution:

Let us assume that 3A = 4B = 5C = k

Equating them we have A = k/3, B = k/4, C = k/5

Therefore, Ratio becomes = k/3:k/4:k/5

LCM of 3, 4, 5 is 60

Thus expressing them in terms of least common multiple we have

A:B:C = 20:15:12

Therefore, Ratio of A:B:C is 20:15:12

8. A certain sum of money is divided among a, b, c in the ratio 3:4:5. of a share is $300, find the share of b and c?

Solution:

Let us consider the sum of money as x

Since it is shared among the ratio of 3:4:5 we have 3x:4x:5x

We know a’s share is 3x = $300

x =$100

Share of b = 4x

= 4*100

= $400

Share of C = 5x

= 5*100

= $500

9. Divide $900 among A, B, C in the ratio 3: 4 ∶ 5?

Solution:

Let us assume the total money as x

Since the sum is to be shared among A, B, C in the ratio of 3:4:5 we have

3x+4x+5x = $900

12x = $900

x = $900/12

=$75

Share of A = 3x

= 3*75

= $225

Share of B = 4x

= 4*75

= $300

Share of C = 5x

= 5*75

= $375

10. Find the first term, if second, third, and fourth terms are 21, 80, 120?

Solution:

Let the Terms be a, a+d, a+2d, a+3d

Given Second Term = 21

a+d = 21

Third Term = 80

a+2d = 80

Fourth Term = 120

a+3d = 120

Using the Eliminating Method

a+d = 21

a+2d = 80

_______

Subtracting them we get the value of d as

-d = -59

d= 59

Substitute the value of d in any of the terms

a+d = 21

a+59 =21

a =21-59

= -38

 

 

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