Direct and Inverse Proportions

Direct and Inverse Proportions introduces students to the possibility of relating two things or situations with one another. It helps to understand the impact of change in one item (x) on another (y) more efficiently. 

Being one of the introductory chapters, students benefit from their understanding of concepts and theories and can apply the same in real life situations effectively. Several study material and chapter-based solutions are available these days for students to access and further improve their understanding of these vital concepts.

CBSE Class 8 chapter 13 Direct and Inverse Proportions – Variation

Suppose two objects, ‘x’ and ‘y’ depend on each other in a way such that an increase or decrease in the value of either of them affects the other. In such a case, the objects will be considered to be in variation.

CBSE Class 8 Chapter 13 Direct and Inverse Proportions – Direct variation 

It can be best described as the situation, wherein –

• An increase in ‘x’ leads to an increase in ‘y’.

• A decrease in ‘x’ leads to a decrease in ‘y’.

Notably, the ratio of respective values has to be same. 

To elaborate -

‘x’ and ‘y’ will be in direct proportion if only 

• x/y =k(constant),

or

• x=ky.

Also, in such a condition, if y1, y2 represent the values of y corresponding to the values x1, x2 respectively then x1/y2 = x2/y1.

The direct proportion between two objects is represented by the sign – 

∝

Notably, there are several methods which can be used to solve problems based on direct proportion.

CBSE Class 8 Chapter 13 Direct and Inverse Proportions – Methods of Direct Proportion

There are 2 distinct methods for solving problems based on direct proportion, namely –

1. Tabular method

In this method, the ratio is constant. It means if one ratio is mentioned, the value of the others can also be found.

x1/y1 = x2/y2 = x3/y3 = xn/yn

Example: 4-litre milk costs Rs.200. Tabulate the cost of milk of 2l, 3l, 5l and 8l.

Sol: Suppose, x litre of milk costs Rs.Y

It is a given that as the volume increases, the cost will increase too. 

x1/y1 = 4/200

x1/y1 = x2/y2

4/200=2/y2

4y2 = 2x200

y2 = (2x200)/4

y2 = 100

Therefore, 2 litre of milk costs Rs.100.

x1/y1 = x3/y3

4/200=3/y3

4y3= 3x200

y3 = (3x200)/4

y3 = 150

Therefore, 3 litre of milk costs Rs.150.

x1/y1 = x4/y4

4/200=5/y4

4y4= 5x200

y4 = (5x200)/4

y4 = 250

Therefore, 4 litre of milk costs Rs.250.

  x1/y1 = x5/y5

4/200= 8/y5

4y5= 8x200

y5 = (8x200)/4

y5= 400

Therefore, 5 litre of milk costs Rs.400.

2. Unitary Method

When two quantities ‘x’ and ‘y’ are said to be in direct proportion, the relation would be expressed as –

k= x/y or,

x = ky

Example: If Sam gets Rs.2000 for 4 hours of work, how many hours will he have to work to earn Rs.60,000.

Sol: 

k= Number of hours/salary of a worker 

= 4/2000 

= 1/500

By using this relation, x = ky

x = 1/500 x 60000 = 12

Therefore, Sam has to work for 12 hours to earn Rs.60000.

CBSE Class 8 Chapter 13 – Inverse Proportion

Typically, two quantities, says, ‘x’ and ‘y’ are said to be in inverse proportion when –

• An increase in ‘x’ leads to a decrease in ‘y’.

• A decrease in ‘y’ leads to an increase in ‘x’.

It must be noted that the respective values of the ratio must be the same.

‘x’ and ‘y’ will be inversely proportional when k=xy

In such a condition, y1, y2 are values of y corresponding to values of x1, x2. Notably, when two quantities x and y are considered to be in inverse proportion, they are expressed as ∝ 1/y.

Example: If it takes 15 artists to make a statue in 48 hours, how many artists would be required to complete the same in 30 hours?

Sol: Let y be the number of required artists.

It is a given that as the number of artists will be increased, the time taken to complete the work will decrease. Resultantly, the number of hours and artists are in an inverse proportion.

48 x 15 = 30 x y (x1y1 = x2y2)

Therefore, 

y = (48 x 15)/30 = 24

Hence, the statue will be completed in 30 hours by 24 artists.

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