Direct and Inverse Proportions
An Introduction to Direct and Inverse Proportion
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 chapterbased 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 –

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: 4litre 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 = 2×200
y2 = (2×200)/4
y2 = 100
Therefore, 2 litre of milk costs Rs.100.
x1/y1 = x3/y3
4/200=3/y3
4y3= 3×200
y3 = (3×200)/4
y3 = 150
Therefore, 3 litre of milk costs Rs.150.
x1/y1 = x4/y4
4/200=5/y4
4y4= 5×200
y4 = (5×200)/4
y4 = 250
Therefore, 4 litre of milk costs Rs.250.
x1/y1 = x5/y5
4/200= 8/y5
4y5= 8×200
y5 = (8×200)/4
y5= 400
Therefore, 5 litre of milk costs Rs.400.

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.
You can learn about the concepts of direct and inverse proportions in detail from subject experts by joining our free live online classes. Our compact study materials, including, exercisebased solutions, will further help you to improve your understanding of the chapter and related concepts much better.
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1. What is the inverse proportion? Give suitable examples.
Ans. Inverse proportion shows the impact brought by the change of value of one object on another. It occurs when an increase in the value of one object decreases the value of another. One of the most common examples of inverse proportion is when the cost of one object increases sharply the demand for it decreases. Similary, as the number of distance covered by a car increases the litre of petrol in the car decreases. One can find more such examples by referring to latest solutions and study materials based on the chapter and improve their understanding of the same more effectively.
2. What is direct variation proportion? Give a suitable example.
Ans. Direct variation is also known as direct proportion and tends to relate the impact of a change of one object on another. In this situation, when the value of one object increases, the value of another object increases. For example, when the number of products in the shopping list increases, the bill amount also increases.
3. What is the difference between direct and inverse variation?
Ans. The fundamental difference between direct and inverse proportions lies in the way two items relate to one another. In case of direct variation, when the value of one object increases, the value of other increases as well. In case of inverse variation, when the value of one object increases the value of the other object decreases. One can learn more about the differences between the two by referring to chapterbased solutions and notes based on direct and inverse variations. Besides providing a fair idea about the expalanations it will also help to understand the different approaches required for solving different types of sums.
4. What is the direct variation? What are the methods of direct proportion?
Ans. Direct variation method can be best described as the situation, wherein an increase in the value of an object leads to an increase in the other object. In this method, the ratio tends to remain constant throughout. Typically, the problems based on direct proportion are solved by using two methods, the tabular method and the unitary method.