Worked out Problems on Ratio and Proportion

Worked out problems on ratio and proportion are explained here in detailed description using step-by-step procedure. Solved examples involving different questions related to comparison of ratios in ascending order or descending order, simplification of ratios and also word problems on ratio proportion.

Sample questions and answers are given below in the worked out problems on ratio and proportion to get the basic concepts of solving ratio proportion.

1. Arrange the following ratios in descending order.

2 : 3, 3 : 4, 5 : 6, 1 : 5

Solution:

Given ratios are 2/3, 3/4, 5/6, 1/5

The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60

Now, 2/3 = (2 × 20)/(3 × 20) = 40/60

3/4 = (3 × 15)/(4 × 15) = 45/60

5/6 = (5 × 10)/(6 × 10) = 50/60

1/5 = (1 × 12)/(5 × 12) = 12/60

Clearly, 50/60 > 45/60 > 40/60 > 12/60

Therefore, 5/6 > 3/4 > 2/3 > 1/5

So, 5 : 6 > 3 : 4 > 2 : 3 > 1 : 5

2. Two numbers are in the ratio 3 : 4. If the sum of numbers is 63, find the numbers.

Solution:

Sum of the terms of the ratio = 3 + 4 = 7

Sum of numbers = 63

Therefore, first number = 3/7 × 63 = 27

Second number = 4/7 × 63 = 36

Therefore, the two numbers are 27 and 36.

3. If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y)

Solution:

x : y = 1 : 2 means x/y = 1/2

Now, (2x + 3y) : (x + 4y) = (2x + 3y)/(x + 4y) [Divide numerator and denominator by y.]

= [(2x + 3y)/y]/[(x + 4y)/2] = [2(x/y) + 3]/[(x/y) + 4], put x/y = 1/2

We get = [2 (1/2) + 3)/(1/2 + 4) = (1 + 3)/[(1 + 8)/2] = 4/(9/2) = 4/1 × 2/9 = 8/9

Therefore the value of (2x + 3y) : (x + 4y) = 8 : 9

More solved problems on ratio and proportion are explained here with full description.

4. A bag contains $510 in the form of 50 p, 25 p and 20 p coins in the ratio 2 : 3 : 4. Find the number of coins of each type.

Solution:

Let the number of 50 p, 25 p and 20 p coins be 2x, 3x and 4x.

Then 2x × 50/100 + 3x × 25/100 + 4x × 20/100 = 510

x/1 + 3x/4 + 4x/5 = 510

(20x + 15x + 16x)/20 = 510

⇒ 51x/20 = 510

x = (510 × 20)/51

x = 200

2x = 2 × 200 = 400

3x = 3 × 200 = 600

4x = 4 × 200 = 800.

Therefore, number of 50 p coins, 25 p coins and 20 p coins are 400, 600, 800 respectively.

5. If 2A = 3B = 4C, find A : B : C

Solution:

Let 2A = 3B = 4C = x

So, A = x/2 B = x/3 C = x/4

The L.C.M of 2, 3 and 4 is 12

Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12

= 6x : 4x : 3x

= 6 : 4 : 3

Therefore, A : B : C = 6 : 4 : 3

6. What must be added to each term of the ratio 2 : 3, so that it may become equal to 4 : 5?

Solution:

Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5

⇒ (2 + x)/(5 + x) = 4/5

5(2 + x) = 4(3 + x)

10 + 5x = 12 + 4x

5x - 4x = 12 - 10

x = 2

7. The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now?

Solution:

Length of ribbon originally = 30 cm

Let the original length be 5x and reduced length be 3x.

But 5x = 30 cm

x = 30/5 cm = 6 cm

Therefore, reduced length = 3 cm

= 3 × 6 cm = 18 cm

More worked out problems on ratio and proportion are explained here step-by-step.

8. Mother divided the money among Ron, Sam and Maria in the ratio 2 : 3 : 5. If Maria got $150, find the total amount and the money received by Ron and Sam.

Solution:

Let the money received by Ron, Sam and Maria be 2x, 3x, 5x respectively.

Given that Maria has got $ 150.

Therefore, 5x = 150

or, x = 150/5

or, x = 30

So, Ron got = 2x

= $ 2 × 30 = $60

Sam got = 3x

= 3 × 60 = $90

Therefore, the total amount $(60 + 90 + 150) = $300

9. Divide $370 into three parts such that second part is 1/4 of the third part and the ratio between the first and the third part is 3 : 5. Find each part.

Solution:

Let the first and the third parts be 3x and 5x.

Second part = 1/4 of third part.

= (1/4) × 5x

= 5x/4

Therefore, 3x + (5x/4) + 5x = 370

(12x + 5x + 20x)/4 = 370

37x/4 = 370

x = (370 × 4)/37

x = 10 × 4

x = 40

Therefore, first part = 3x

= 3 × 40

= $120

Second part = 5x/4

= 5 × 40/4

= $50

Third part = 5x

= 5 × 40

= $ 200

10. The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term.

Solution:

Let the fourth term be x.

Thus 42, 36, 35, x are in proportion.

Product of extreme terms = 42 ×x

Product of mean terms = 36 X 35

Since, the numbers make up a proportion

Therefore, 42 × x = 36 × 35

or, x = (36 × 35)/42

or, x = 30

Therefore, the fourth term of the proportion is 30.

More worked out problems on ratio and proportion using step-by-step explanation.

11. Set up all possible proportions from the numbers 8, 12, 20, 30.

Solution:

We note that 8 × 30 = 240 and 12 × 20 = 240

Thus, 8 × 30 = 12 × 20 ………..(I)

Hence, 8 : 12 = 20 : 30 ……….. (i)

We also note that, 8 × 30 = 20 × 12

Hence, 8 : 20 = 12 : 30 ……….. (ii)

(I) can also be written as 12 × 20 = 8 × 30

Hence, 12 : 8 = 30 : 20 ……….. (iii)

Last (I) can also be written as

12 : 30 = 8 : 20 ……….. (iv)

Thus, the required proportions are 8 : 12 = 20 : 30

8 : 20 = 12 : 30 12 : 8 = 30 : 20 12 : 30 = 8 : 20

12. The ratio of number of boys and girls is 4 : 3. If there are 18 girls in a class, find the number of boys in the class and the total number of students in the class.

Solution:

Number of girls in the class = 18

Ratio of boys and girls = 4 : 3

According to the question,

Boys/Girls = 4/5

Boys/18 = 4/5

Boys = (4 × 18)/3 = 24

Therefore, total number of students = 24 + 18 = 42.

13. Find the third proportional of 16 and 20.

Solution:

Let the third proportional of 16 and 20 be x.

Then 16, 20, x are in proportion.

This means 16 : 20 = 20 : x

So, 16 × x = 20 × 20

x = (20 × 20)/16 = 25

Therefore, the third proportional of 16 and 20 is 25.