Solving Inverse Variation using Ratio and Proportion

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?

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.

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.

Speed and journey time are inversely related.

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

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?

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.

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.

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.

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.

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