Inverse Variation Equation

In this chapter we will learn the methods to solve inverse variation equations with examples.

Let us first review the basic concept of inverse variation.

What is Inverse Variation?

Inverse variation is expressed by below equation.

\mathtt{y\ =\ \frac{k}{x}}

Where, x & y are variables and k is constant.

Sometimes the equation is also expressed as;
k = x . y

The equation states that y is inversely proportional to x.
This means that;

If x increase, value of y decrease proportionally.

If x decrease, value of y increase proportionally.

Now the question arise; with increase in x how much value of y will decrease.

It all depends on value of constant k.

⟹ When k is large, increase in x will cause less decrease in y

⟹ When k is small, increase in x will cause steep decrease in y.

Value of Constant in Inverse Variation

In Inverse Variation, the value of constant is given as;

\mathtt{y\ =\ \frac{k}{x}}\\\ \\ \mathtt{k\ =\ y.x}

Constant k is the multiplication of x and y variable.

We can infer that in indirect variation, the values of x and y changes but the multiplication of x & y always remains constant.

Solving Inverse variation equations

Given below are solved inverse variation questions.

Different types of questions are discussed with full explanation. Make sure you go through each of them and learn the method of solving inverse variation questions.

Example 01
Variable x and y are inversely proportional. The value of k = 10. Find the value of y if x = 4

The expression of inverse variation is written as;

\mathtt{y\ =\ \frac{k}{x}}

Its given that k= 10 & x = 4.
Putting the values in equation we get;

\mathtt{y\ =\ \frac{10}{4}}\\\ \\ \mathtt{y\ =\ 2.5}

Hence, for x = 4, the value of y is 2.5

Example 02
y and x are in inverse variation. When x = 5 then value of y is 20. Find the value of x when y = 40.

Since y and x are in inverse proportion, we can write;

k = x . y

It’s given that when x = 5 then y = 20.

Putting the values in above equation.

k = 5 . 20

k = 100

Hence, the value of constant is 100.

The equation can be written as;
100 = x . y

Now we have to find value of x for y = 40

Putting the values in equation, we get;

100 = x 40

x = 100 / 40

x = 2.5

Hence for y= 40, the value of x is 2.5

Example 03
The weight of man is inversely related to his running capacity. If 60 kg man runs 3 kms then how many kilometers will 90 kg man run?

Here weight and running capacity is inversely proportional.

It means that if weight increases, the running distance will decrease.

Let x be the weight of man.
y be the running distance
k is the constant.

Inverse equation is represented as;
x . y = k

60 . 3 = k

k = 180

The value of constant k = 180

Now find running capacity of 90 kg man.
Here k = 180 & x = 90

The equation is expressed as;
90 . y = 180

y = 180 / 90

y = 2 kms.

Hence running capacity of 90 kg man is 2kms.

Example 04
4 Men can complete the work in 20 days. Find the time taken by 10 men to complete the same work.

Here number of men and time for work are in inverse proportion.

If we increase man power then time taken will reduce.

Decrease man power will increase time taken to complete the work.

Hence, the inverse equation can be expressed as;
x . y = k

x = Number of workers
y = time in days

4 men complete work in 20 days.
Putting values in above equation.

4 x 20 = k

k = 80

Hence, the value of constant is 80.

Now find time taken by 10 men.
Here x = 10 & k = 80

Putting values in equation;
10 x y = 80

y = 80 / 10

y = 8 days

Hence 10 men complete the same work in 8 days.

Example 05
15 kids can drink a jar of juice in 40 minutes. How much time will 20 kids take to empty the same jar?

As the number of kids increases, the time taken to empty the jar will decrease. Hence, It’s a question of inverse variation.

The inverse variation is expressed as;
y . x = k

y = Time taken
x = number of kids
k = constant

Putting the values;
40 x 15 = k

k = 600

Hence, the value of k = 600.

Now we have to find the time taken by 20 kids.
Here, x = 20 & k = 600.

Putting the values in equation.

y . 20 = 600

y = 600 / 20

y = 30 minutes.

Hence, 20 kids will empty the jar in 30 minutes.

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