In this chapter we will learn methods to solve inverse variation problems using ratio and proportion concept.

Let us first review the basics of inverse variation.

## What is Inverse Variation?

Inverse variation is represented by following expression;

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

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

The expression states that;

⟹ **Increase in value of x results in proportional decrease in value of y.**

⟹ **Decrease in value of y results in proportional increase in value of x.**

Hence there is inverse relationship between the variables of x and y.

### Proportion used in Inverse variation

Let’s rearrange the expression of inverse variation.

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

From the expression we can infer that the product of variable x and y will remain same.

Even if the value of x and y changes, the value of constant k will remain unchanged.

Let there are two set of values x1, y1 and x2, y2 for inverse variation.

As the value of product x and y remains the same, we can write the expression as;**x1 . y1 = x2. y2**

This expression is useful to solve problems related to inverse variation. So make sure to remember the expression for future use.

I hope you have understood the above concept. Let us now move to solve the related problems.

## Solved Inverse Variation Problem using Proportion Method

**Example 01**

6 Men can plough the field in 15 days. How many days will 9 men take to plough the same field?

**Solution**

Here number of men & time taken are inversely related.

If more men are present, less time will be taken to plough the field and vice versa.

Let;

x1 = 6 men

y1 = 15 days

x2 = 9 men

y2 = ?

Using Inverse proportion expression;

x1 . y1 = x2 . y2

Putting the values;

\mathtt{6.\ 15\ \ =\ 9\ .\ y2}\\\ \\ \mathtt{y2\ =\ \frac{6\ .\ 15}{9}}\\\ \\ \mathtt{y2\ =\ 10\ days}

**Hence, 9 men can plough the field in 10 days.**

**Example 02**

A car travelling at speed of 40 km/hr takes 3 hours to complete the journey. If the speed of car is increased to 60 km/hr, find the time to complete the same journey.**Solution**

Speed and journey time are inversely related.

If the speed increase the journey time will decrease and vice versa.

Let;

x1 = 40 km/hr

y1 = 3 hours

x2 = 60 km/hr

y2 = ?

Putting the values in inverse variation expression.

\mathtt{x1\ .\ y1\ =\ x2\ .\ y2}\\\ \\ \mathtt{40.\ 3\ \ =\ 60\ .\ y2}\\\ \\ \mathtt{y2\ =\ \frac{40\ .\ 3}{60}}\\\ \\ \mathtt{y2\ =\ 2\ hours}

Hence **at 60 km/hr, the journey will be completed in 2 hours**.

**Example 03**

20 soldiers can complete the ration in 50 days. If the number of soldiers increased to 25, in how many days the ration get empty?

**Solution**

Number of soldiers and days to complete the ration are inversely proportional.

If number of soldiers increase, the days to complete ration will decrease and vice-versa.

Let;

x1 = 20 soldiers

y1 = 50 men

x2 = 25 soldiers

y2 = ?

Putting the values in Inverse Variation expression.

\mathtt{x1\ .\ y1\ =\ x2\ .\ y2}\\\ \\ \mathtt{20.\ 50\ \ =\ 25\ .\ y2}\\\ \\ \mathtt{y2\ =\ \frac{20\ .\ 50}{25}}\\\ \\ \mathtt{y2\ =\ 40\ days}

Hence **with 25 soldiers, the ration will get over in 40 days.**

**Example 04**

6 pipes can fill the drum in 18 hours. How many pipes are needed to fill the drum in 6 hours.

**Solution**

Number of pipes and time to fill drum are inversely proportion.

If pipes are increased, the time taken to fill the jar will be reduced and vice-versa.

Let;

x1 = 6 pipes

y1 = 18 hours

x2 = ?

y2 = 6 hours

Putting the values in inverse variation equation.

\mathtt{x1\ .\ y1\ =\ x2\ .\ y2}\\\ \\ \mathtt{6.\ 18\ \ =\ x2\ .\ 8}\\\ \\ \mathtt{x2\ =\ \frac{6\ .\ 18}{6}}\\\ \\ \mathtt{x2\ =\ 18\ pipes}

Hence, **18 pipes can fill the drum in 6 hours.**