7.6 Applications of percents  (Page 3/4)

Five years ago a woman had an annual income of $19,200. She presently earns $42,000 annually. By what percent has her salary increased? Round the result to two decimal places.

We need to determine the percent.

$\begin{array}{cccccc}\underset{\text{↓}}{\mathrm{42,000}}& \underset{\text{↓}}{\text{is}}& \underbrace{\text{what percent}}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{\mathrm{19,200}}\text{?}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \\ \mathrm{42,000}& =& Q& \cdot & \mathrm{19,200}& \end{array}$

$\begin{array}{cccc}Q& =& \mathrm{42,000}÷\mathrm{19,200}\hfill & \text{Divide.}\hfill \\ Q& =& 2.1875\hfill & \text{Round to two decimal places.}\hfill \\ Q& =& 2.19\hfill & \text{Convert to a percent.}\hfill \\ Q& =& \mathrm{219\%}\hfill & \text{Convert to a percent.}\hfill \end{array}$

Thus, this woman's annual salary has increased 219%.

$\begin{array}{cccccc}\underset{\text{↓}}{15}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{30\%}& \underset{\text{↓}}{\text{of}}& \underbrace{\text{what number?}}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \text{[(percentage) = (factor) ⋅ (factor)]}\hfill \\ 15& =& 30\%& \cdot & B& \text{Convert to decimals.}\hfill \\ 15& =& .30& \cdot & B& \text{[(missing factor) = (product) ÷ (known factor)]}\hfill \end{array}$

$\begin{array}{ccc}B& =& 15÷.30\hfill \\ B& =& 50\hfill \end{array}$

Thus, 15 is 30% of 50.

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$\begin{array}{cccccc}\underset{\text{↓}}{56.43}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{33\%}& \underset{\text{↓}}{\text{of}}& \underbrace{\text{what number?}}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}\\ 56.43& =& 33\%& \cdot & B& \text{Convert to decimals.}\hfill \\ 56.43& =& .33& \cdot & B& \text{Divide.}\hfill \end{array}$

$\begin{array}{ccc}B& =& 56.43÷.33\hfill \\ B& =& 171\hfill \end{array}$

Thus, 56.43 is 33% of 171.

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Fifteen milliliters of water represents 2% of a hydrochloric acid (HCl) solution. How many milliliters of solution are there?

We need to determine the total supply. The word supply indicates base .

$\begin{array}{cccccc}\underset{\text{↓}}{15}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{2\%}& \underset{\text{↓}}{\text{of}}& \underbrace{\text{what number?}}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}\\ 15& =& 2\%& \cdot & B& \text{Convert to decimals.}\hfill \\ 15& =& .02& \cdot & B& \text{Divide.}\hfill \end{array}$

$\begin{array}{ccc}B& =& 15÷.02\hfill \\ B& =& 750\hfill \end{array}$

Thus, there are 750 milliliters of solution in the bottle.

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In a particular city, a sales tax of $6\frac{1}{2}$ % is charged on items purchased in local stores. If the tax on an item is $2.99, what is the price of the item?

We need to determine the price of the item. We can think of price as the starting place . Starting place indicates base . We need to determine the base.

$\begin{array}{cccccc}\underset{\text{↓}}{2.99}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{6\frac{1}{2}\%}& \underset{\text{↓}}{\text{of}}& \underbrace{\text{what number?}}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}\\ 2.99& =& 6\frac{1}{2}\%& \cdot & B& \text{Convert to decimals.}\hfill \\ 2.99& =& 6.5\%& \cdot & B\\ 2.99& =& .065& \cdot & B& \text{[(missing factor) = (product) ÷ (known factor)]}\hfill \end{array}$

$\begin{array}{cccc}B& =& 2.99÷.065\hfill & \text{Divide.}\\ B& =& 46\hfill & \end{array}$

Thus, the price of the item is $46.00.

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A clothing item is priced at $20.40. This marked price includes a 15% discount. What is the original price?

We need to determine the original price. We can think of the original price as the starting place . Starting place indicates base . We need to determine the base. The new price, $20.40, represents $\text{100\%}-\text{15\%}=\text{85\%}$ of the original price.

$\begin{array}{cccccc}\underset{\text{↓}}{20.40}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{85\%}& \underset{\text{↓}}{\text{of}}& \underbrace{\text{what number?}}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}\\ 20.40& =& 85\%& \cdot & B& \text{Convert to decimals.}\hfill \\ 20.40& =& .85& \cdot & B& \text{[(missing factor) = (product) ÷ (known factor)]}\hfill \end{array}$

$\begin{array}{cccc}B& =& 20.40÷.85\hfill & \text{Divide.}\\ B& =& 24\hfill & \end{array}$

Thus, the original price of the item is $24.00.

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