Applications of Proportions
Exercises
Answers
Example 6.40
| (a) |
|
| \(3\) is to \(7\) as \(15\) is to \(35 .\) | |
| Write as a proportion. | \(\dfrac{3}{7}=\dfrac{15}{35}\) |
| (b) |
|
| \(5\) hits in \(8\) at-bats is the same as \(30\) hits in \(48\) atbats. | |
| Write each fraction to compare hits to at-bats. | \(\dfrac{\text { hits }}{\text { at-bats }}=\dfrac{\text { hits }}{\text { at-bats }}\) |
| Write as a proportion. | \(\dfrac{5}{8}=\dfrac{30}{48}\) |
| (c) |
|
| \(\$ 1.50\) for \(6\) ounces is equivalent to \(\$ 2.25\) for \(9\) ounces. | |
| Write each fraction to compare dollars to ounces. | \(\dfrac{\$}{\text { ounces }}=\dfrac{\$}{\text { ounces }}\) |
| Write as a proportion. | \(\dfrac{1.50}{6}=\dfrac{2.25}{9}\) |
Example 6.41
To determine if the equation is a proportion, we find the cross products. If they are equal, the equation is a proportion.
| (a) |
|
| \(\dfrac{4}{9}=\dfrac{12}{28}\) | |
| Find the cross products. | \(28 \cdot 4=112\) \[9 \cdot 12=108\]  |
Since the cross products are not equal, \(28 \cdot 4 \neq 9 \cdot 12\), the equation is not a proportion.
| (b) |
|
| \(\dfrac{17.5}{37.5}=\dfrac{7}{15}\) | |
| Find the cross products. | \(15 \cdot 17.5=262.5 \quad 37.5 \cdot 7=262.5\)  |
Since the cross products are equal, \(15 \cdot 17.5=37.5 \cdot 7\), the equation is a proportion.
Example 6.42
| \(\dfrac{x}{63}=\dfrac{4}{7}\) | ||
| To isolate \(x\), multiply both sides by the LCD, \(63\). | \(\color{red}{63}\left(\dfrac{x}{63}\right)=\color{red}{63}\left(\dfrac{4}{7}\right)\) | |
| Simplify. | \(x=\dfrac{9 \cdot \not{7} \cdot 4}{\not{7}}\) | |
| Divide the common factors. | \(x=36\) | |
| Check: To check our answer, we substitute into the original proportion. | ||
| \(\dfrac{x}{63}=\dfrac{4}{7}\) | ||
| Substitute \(x=\color{red}{36}\) | \(\dfrac{\color{red}{36}}{63} \stackrel{?}{=} \dfrac{4}{7}\) | |
| Show common factors. | \(\dfrac{4 \cdot 9}{7 \cdot 9} \stackrel{?}{=} \dfrac{4}{7}\) | |
| Simplify. | \(\dfrac{4}{7}=\dfrac{4}{7}✓\) | |
Example 6.43
Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.
 |
||
| Find the cross products and set them equal. | \(4 \cdot 144=a \cdot 9\) | |
| Simplify. | \(576=9 a\) | |
| Divide both sides by 9. | \(\dfrac{576}{9}=\dfrac{9 a}{9}\) | |
| Simplify. | \(64=a\) | |
| Check your answer. | ||
| \(\dfrac{144}{a}=\dfrac{9}{4}\) | ||
| Substitute \(a=\color{red}{64}\) | \(\dfrac{144}{64} \stackrel{?}{=} \dfrac{9}{4}\) | |
| Show common factors.. | \(\dfrac{9 \cdot 16}{4 \cdot 16} \stackrel{?}{=} \dfrac{9}{4}\) | |
| Simplify. | \(\dfrac{9}{4}=\dfrac{9}{4}✓\) | |
Another method to solve this would be to multiply both sides by the LCD, \(4a\)Try it and verify that you get the same solution.
Example 6.44
| Find the cross products and set them equal. |  |
|
| \(y \cdot 52=91(-4)\) | ||
| Simplify. | \(52 y=-364\) | |
| Divide both sides by \(52\). | \(\dfrac{52 y}{52}=\dfrac{-364}{52}\) | |
| Simplify. | \(y=-7\) | |
| Check: | ||
| \(\dfrac{52}{91}=\dfrac{-4}{y}\) | ||
| Substitute \(y=\color{red}{-7}\) | \(\dfrac{52}{91} \stackrel{?}{=} \dfrac{-4}{\color{red}{-7}}\) | |
| Show common factors. | \(\dfrac{13 \cdot 4}{13 \cdot 4} \stackrel{?}{=} \dfrac{-4}{\color{red}{-7}}\) | |
| Simplify. | \(\dfrac{4}{7}=\dfrac{4}{7}\checkmark\) | |