Example 3.5.3

Suppose we want to know the total cost for a box of cereal that weighs \(18\) ounces, assuming it costs the same per ounce as the \(21\)-ounce box. Letting \(C\) be this unknown cost (in dollars), we could set up the following proportion:

\begin{align*} \frac{\text{cost in dollars}}{\text{weight in oz}}\amp=\frac{\text{cost in dollars}}{\text{weight in oz}}\\ \frac{\$3.99}{21\,\text{oz}}\amp=\frac{\$C}{18\,\text{oz}} \end{align*}

To solve this proportion, we will first note that it will be easier to solve without units:

\begin{equation*} \frac{3.99}{21}=\frac{C}{18} \end{equation*}

Next we want to recognize that each side contains a fraction. Our usual approach for solving this type of equation is to multiply each side by the least common denominator (LCD). In this case, the LCD of \(21\) and \(18\) is \(126\text{.}\) As with many other proportions we solve, it is often easier to just multiply each side by the common denominator of \(18\cdot 21\text{,}\) which we know will make each denominator cancel:

\begin{align*} \frac{3.99}{21}\amp=\frac{C}{18}\\ \multiplyleft{18\cdot21}\frac{3.99}{21}\amp=\frac{C}{18}\multiplyright{18\cdot21}\\ 18\cdot\cancelhighlight{21}\frac{3.99}{\cancelhighlight{21}}\amp=\frac{C}{\cancelhighlight{18}}\cdot\cancelhighlight{18}\cdot21\\ 71.82\amp=21C\\ \divideunder{71.82}{21}\amp=\divideunder{21C}{21}\\ C\amp=3.42 \end{align*}

So assuming the cost is proportional to the cost of the \(21\)-ounce box, the cost for an \(18\)-ounce box of cereal would be \(\$3.42\text{.}\)

in-context