5.2: Proportions

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Page ID: 20042

Kelly Brooks
Eastern Gateway Community College

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A proportion determines the relationship between two variables as it relates to the magnitude or size of the variables. A proportional relationship shows that two ratios are equivalent such that as one amount changes, another amount changes by a relative amount.

To solve proportion problems, we set up an equation such that two ratios are equivalent:

\[\dfrac{\text{One characteristic from one item}}{\text{A second characteristic from the same item}}=\dfrac{\text{the same first characteristic from a second item}}{\text{the same second characteristic from the second item}} \label{ratio} \]

Example $\PageIndex{1}$

Seven containers of a chemical costs $125. If the cost of the containers is proportional, meaning the unit price does not change, determine the cost of 12 containers of the same chemical.

Solution

Let’s begin by setting up the fractions like in Equation \ref{ratio}:

\[\begin{align*} \dfrac{\text{Number of containers}}{\text{Cost of those containers}} &=\dfrac{\text{Number of another set of containers}}{\text{Cost of those containers}} \\[4pt] \dfrac{7\, \text{containers}}{\$125} &=\dfrac{12\, \text{containers}}{x} \end{align*}\]

To solve for $x$ we can use one of two methods:

Method 1: Solve for $x$ by isolating the $x$ (using information learned in Unit 3)

Multiply both sides of the equation by $x$: \[\dfrac{7}{125}x=12 \nonumber\]

Multiply both sides of the equation by $125: \[\begin{align*} 7x &=12\cdot 125 \\[4pt] 7x&=1500 \end{align*}\]

Divide both sides of the equation by 7 containers: \[x=\dfrac{1500}{7}=\$214.29 \nonumber\]

Method 2: Use cross multiplication: Multiply the numerator of the first fraction by the denominator of the second fraction and set it equal to the multiplication of the numerator of the second fraction by the denominator of the first fraction.

\[\begin{align*} \left(7 \, \text{containers}\right)\cdot \left(\$x\right) &=\left(12 \, \text{containers} \right)\cdot \left(\$125\right) \\[4pt] 7x &=1500 \\[4pt] x&=\dfrac{1500}{7} \\[4pt] &=\$214.29 \end{align*}\]

Hence, the cost of the 12 containers is $214.29.

Example $\PageIndex{2}$

We want to estimate the number of fish in a pond. Suppose we capture 280 fish, tab them, and throw them back into the pond. After a couple of days, we go back to the pond and capture 420 fish, of which 28 are tagged. Estimate the number of fish in the pond.

Example $\PageIndex{3}$

We wish to mix a solution of hypochlorite and water by dissolving 4.8 pounds of hypochlorite in 70 gallons of water. For the same concentration, how many pounds of hypochlorite should we dissolve in 20 gallons of water?

Solution

Let’s begin by setting up the fractions like in Equation \ref{ratio}:

\[\begin{align*} \dfrac{\text{Number of pounds of hypochlorite}}{\text{Number of gallons of water}}&=\dfrac{\text{Number of pounds of hypochlorite}}{\text{Number of gallons of water}} \\[4pt] \dfrac{4.8\, \text{lbs}}{70\, \text{gal}} &=\dfrac{x\, \text{lbs}}{20\, \text{gal}} \\[4pt] \left(4.8 \, \text{lbs}\right)\left(20\, \text{gal}\right) &=\left(70\, \text{gal}\right)\left(x \, \text{lbs}\right) \\[4pt] 96 &=70x \\[4pt] \dfrac{96}{70} &=x \\[4pt] x &=1.4\, \text{lbs} \end{align*}\]

Hence, we need 1.4 pounds of hypochlorite to dissolve in 20 gallons of water to create the same concentration.

Example \(\PageIndex{1}\)

Example \(\PageIndex{2}\)

Example \(\PageIndex{3}\)