This Algebra lesson teaches how to solve math word problems that deal with two variables that are related to each other in some sort of inverse variation or proportion. (This topic is covered in Saxon Algebra 1, Lesson 113B.) A necessary skill for solving these kinds of problems is being able to read statements in English and represent them as Algebraic equations. For this introduction to this topic we will be working with equations with only two variables. When we have two variables that are related to each other in a direct proportion (or are said to vary directly), we would use the following general equation:

V1 = k*V2

Where V1 and V2 are our variables and k is our constant of proportionality. With inverse variation, however, we say that a variable “varies inversely” as the other variable or the value of one variable is “inversely proportional” and we get a general equation like this:

V1 = (k / V2)

A very important part of dealing with these kind of Algebra word problems is recognizing when we are dealing with a “direct” or an “inverse” variation, and use the appropriate general equation. For this article we will be focusing on inverse variation word problems only. You can see examples of direct variation word problems in my article here.

Here we are given three statements in English. How would we represent them using Algebraic equations?

“The grams of copper were inversely proportional to the grams of bronze”

C = (k / B)

“The cost varied inversely as the number of items”

C = (k / N )

“The number of eggs is inversely proportional to the number of foxes”

E = ( k / F)

In order to solve these kinds of problems we have to first solve for the constant of proportionality. Once we know this constant k we can plug in the value of the variable we know, to find the value of the variable for which we are solving.

Sample Inverse Variation Math Word Problem #1:

The grams of iron were inversely proportional to the grams of copper. If there were 15 grams of iron when there was 3 grams of copper, how many grams of iron will there be when there is 100 grams of copper?

Solution: First write the algebraic equation that describes the relationship between the grams of iron and the grams of copper.

I = ( k / C )

Input the values we know for I and C and solve for k, our constant of proportionality.

15 = ( k / 3 )

k = 45

This gives us the general equation of:

I = (45 / C)

Look at the last part of the word problem and plug in the value we know (in this case, the copper), to find the value we don’t know (the iron).

I = (45 / 100) = 0.45 grams of Iron

Sample Inverse Variation Math Word Problem #2:

The pressure of a perfect gas varies inversely as the volume. When the pressure of the gas is 15 atmospheres the volume is 5 liters. What will be the volume of the gas if the pressure is changed to 10 atmospheres?

Solution: First write the algebraic equation that describes the relationship between the gas pressure and the volume.

P = ( k / V )

Input the values we know for P and V and solve for k, our constant of proportionality.

15 = ( k / 5 )

k = 75

This gives us the general equation of:

P = (75 / V)

Look at the last part of the word problem and plug in the value we know (in this case, the pressure), to find the value we don’t know (the volume).

10 = (75 / V)

10V = 75

V = 7.5 liters

**Source**

John H. Saxon, Jr. Algebra 1. An Incremental Development. Third Edition.