WRITE INVERSE VARIATION EQUATIONS

If an increase () [or decrease ()] in one quantity produces a proportionate decrease ([or increase ()] in another quantity, then we say that the two quantities are in inverse variation.

Equation of Inverse Variation :

An inverse variation can be represented by the equation

xy = k or y = k/x

That is, y varies inversely x and k (≠ 0) is known as the constant of variation, it cannot be equal to zero.

Example 1 :

In an inverse variation, y = 1 when x = 6.Write an inverse variation equation that shows the relationship between x and y.

Solution :

Equation of inverse variation :

y = k/x ----(1)

Substitute x = 6 and y = 1.

1 = k(6)

1 = 6k

Divide both sides by 6.

1/6 = k

Substitute k = 1/6 in (1).

y = 6/x

Example 2 :

In an inverse variation, y = 50 when x = 40. Write an inverse variation equation that shows the relationship between x and y.

Solution :

Equation of inverse variation :

y = k/x ----(1)

Substitute x = 40 and y = 50.

50 = k/40

Multiply both sides by 40.

2000 = k

k = 2000

Substitute k = 2000 in (1)

y = 2000/x

Example 3 :

In an inverse variation, y = 50 when x = 8.Write an inverse variation equation that shows the relationship between x and y.

Solution :

Equation of inverse variation :

y = k/x ----(1)

In order to find the value of "k" in the equation, we need to apply the values of x and y in the equation.

50 = k/8

Multiply both sides by 8.

400 = k

Substitute k = 400 in (1).

y = 400/x

Example 4 :

In an inverse variation, y = 2 when x = 2.Write an inverse variation equation that shows the relationship between x and y.

Solution :

Equation of inverse variation :

y = k/x ----(1)

Substitute x = 2 and y = 2.

2 = k/2

Multiply both sides by 2. 

4 = k

Substitute k = 4 in (1).

y = 4/x

Example 5 :

In an inverse variation, y = 3 when x = 8.Write an inverse variation equation that shows the relationship between x and y.

Solution :

Equation of inverse variation :

y = k/x ----(1)

Substitute x = 8 and y = 3.

3 = k/8

Multiply both sides by 8.

24 = k

Substitute k = 24 in (1).

y = 24/x

Solving Word Problems Using Inverse Variation Equations

Example 6 :

A man can finish a piece of work, working 8 hours a day in 5 days. If he works now 10 hours daily, in how many days can he finish the same work ?

Solution :

8 hours/day ----> 5 days

10 hours/day ----> ?

When number of hours per day increases, number of days required to complete the work will decrease. So, it is inverse variation.

Let x be the number of hours work done per day and y be the number of days required to complete the work.

Then,

y = k/x ----(1)

Substitute x = 8 and y = 5.

5 = k/8

Multiply both sides by 8.

40 = k

Substitute k = 40 in (1).

y = 40/x

Substitute x = 10.

y = 40/10

y = 4

When the man works 10 hours daily, he will complete the work in 4 days.

Example 7 :

In a factory, 10 men can complete a work in 30 days. In how many days will 20 men complete the same work ?

Solution :

10 men ----> 30 days

20 men ----> ?

When number of men increases, number of days required to complete the work will decrease. So, it is inverse variation.

Let x be the number of men and y be the number of days required to complete the work.

Then,

y = k/x ----(1)

Substitute x = 10 and y = 30.

30 = k/10

Multiply both sides by 10.

300 = k

Substitute k = 300 in (1).

y = 300/x

Substitute x = 20.

y = 300/20

y = 15

20 men will complete the work in 15 days.

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