# Word Problems: Inverse Variation

While direct variation describes a linear relationship between two variables, inverse variation describes another kind of relationship. For two quantities with inverse variation, as one quantity increases, the other quantity decreases.

For example, when you travel to a particular location, as your speed increases, the time it takes to arrive at that location decreases. When you decrease your speed, the time it takes to arrive at that location increases. So, the quantities are inversely proportional.

An inverse variation can be expressed by the equation $xy=k$ or $y=\frac{k}{x}$ .

That is, $y$ varies inversely as $x$ if there is some nonzero constant $k$ such that, $xy$ = $k$ or $y=\frac{k}{x}$ where $x\ne 0$ and $y\ne 0$ .

Some word problems require the use of inverse variation. Here are the ways to solve inverse variation word problems.

- Understand the problem.
- Write the formula.
- Identify the known values and substitute in the formula.
- Solve for the unknown.

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Example 1:
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The volume $V$ of a gas varies inversely as the pressure $P$ on it. If the volume is $240\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{3}$ under pressure of $30\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{kg}/{\text{cm}}^{2}$ , what pressure has to be applied to have a volume of $160\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{3}$ ?

The volume $V$ varies inversely as the pressure $P$ means when the volume increases, the pressure decreases and when the volume decreases, the pressure increases.

Now write the formula for inverse variation.

$\text{PV}=$ $k$

Substitute $240$ for $V$ $30$ for $P$ in the formula and find the constant

$\left(240\right)\left(30\right)=k$

$7200=k$

Now write an equation and solve for the unknown.

We have to find the pressure when the volume is $160\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{3}$ .

So,

$\left(160\right)\left(P\right)=7200$ .

Solve for $P$ .

$\begin{array}{l}P=\frac{7200}{160}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=45\end{array}$

Therefore, pressure $45\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{kg}/{\text{cm}}^{2}$ be applied to have a volume of $160\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{3}$ .

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Example 2:
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The length of a violin string varies inversely as the frequency of its vibrations. A violin string $14$ inches long vibrates at a frequency of $450$ cycles per second. Find the frequency of a $12$ -inch violin string.

The length( $l$ ) varies inversely as the frequency( $f$ ), when the length increases, the frequency decreases and when the length decreases, the frequency increases.

Now write the formula for inverse variation.

$lf=k$ .

Substitute $450$ for $f$ $14$ for $l$ in the formula and find the constant.

$\left(450\right)\left(14\right)=k$

$6300=k$

Now write an equation and solve for the unknown.

We have to find the frequency of $12$ -inch violin string.

So,

$\left(12\right)\left(f\right)=6300$ .

Solve for $f$ .

$\begin{array}{l}f=\frac{6300}{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=525\end{array}$

Therefore, $12$ -inch violin string vibrates at a frequency of $525$ cycles per second.

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