7.6 Applications of percents  (Page 2/4)

A student who makes $125 a month working part-time receives a 4% salary raise. What is the student's new monthly salary?

With a 4% raise, this student will make 100% of the original salary + 4% of the original salary. This means the new salary will be 104% of the original salary. We need to determine what part of $125 is to be taken. What part indicates percentage .

$\begin{array}{cccccc}\underbrace{\text{What number}}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{104\%}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{125}& \text{Missing product statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \\ P& =& 104\%& \cdot & 125& \text{Convert to decimals.}\hfill \\ P& =& 1.04& \cdot & 125& \text{Multiply.}\hfill \\ P& =& 130& & & \end{array}$

Thus, this student's new monthly salary is $130.

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An article of clothing is on sale at 15% off the marked price. If the marked price is $24.95, what is the sale price?

Since the item is discounted 15%, the new price will be $\text{100\%}-\text{15\%}=\text{85\%}$ of the marked price. We need to determine what part of 24.95 is to be taken. What part indicates percentage .

$\begin{array}{cccccc}\underbrace{\text{What number}}& \underset{\text{↓}}{\text{is}}& \underset{\text{↓}}{85\%}& \underset{\text{↓}}{\text{of}}& \text{\$}\underset{\text{↓}}{24.95}\text{.}& \text{Missing product statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \\ P& =& 85\%& \cdot & 24.95& \text{Convert to decimals.}\hfill \\ P& =& .85& \cdot & 24.95& \text{Multiply.}\hfill \\ P& =& 21.2075& & & \begin{array}{c}\text{Since this number represents money,}\hfill \\ \text{we'll round to 2 decimal places}\hfill \end{array}\\ P& =& 21.21& & & \end{array}$

Thus, the sale price of the item is $21.21.

$\begin{array}{cccccc}\underset{\text{↓}}{15}& \underset{\text{↓}}{\text{is}}& \underbrace{\text{what percent}}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{50}\text{?}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \text{[(product) = (factor) ⋅ (factor)]}\\ 15& =& Q& \cdot & 50& \end{array}$

Recall that (missing factor) = (product) ÷ (known factor).

$\begin{array}{cccc}Q& =& 15÷50& \text{Divide.}\hfill \\ Q& =& 0.3& \text{Convert to a percent.}\hfill \\ Q& =& 30\%\end{array}$

Thus, 15 is 30% of 50.

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$\begin{array}{cccccc}\underset{\text{↓}}{4.32}& \underset{\text{↓}}{\text{is}}& \underbrace{\text{what percent}}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{72}\text{?}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \text{[(product) = (factor) ⋅ (factor)]}\\ 4.32& =& Q& \cdot & 72& \end{array}$

$\begin{array}{cccc}Q& =& 4.32÷72& \text{Divide.}\hfill \\ Q& =& 0.06& \text{Convert to a percent.}\hfill \\ Q& =& 6\%\end{array}$

Thus, 4.32 is 6% of 72.

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On a 160 question exam, a student got 125 correct answers. What percent is this? Round the result to two decimal places.

We need to determine the percent.

$\begin{array}{cccccc}\underset{\text{↓}}{125}& \underset{\text{↓}}{\text{is}}& \underbrace{\text{what percent}}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{160}\text{?}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \text{[(product) = (factor) ⋅ (factor)]}\\ 125& =& Q& \cdot & 160& \end{array}$

$\begin{array}{cccc}Q& =& 125÷160\hfill & \text{Divide.}\hfill \\ Q& =& 0.78125\hfill & \text{Round to two decimal places.}\hfill \\ Q& =& .78\hfill \end{array}$

Thus, this student received a 78% on the exam.

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A bottle contains 80 milliliters of hydrochloric acid (HCl) and 30 milliliters of water. What percent of HCl does the bottle contain? Round the result to two decimal places.

We need to determine the percent. The total amount of liquid in the bottle is

$\text{80 milliliters}+\text{30 milliliters}=\text{110 milliliters}$ .

$\begin{array}{cccccc}\underset{\text{↓}}{80}& \underset{\text{↓}}{\text{is}}& \underbrace{\text{what percent}}& \underset{\text{↓}}{\text{of}}& \underset{\text{↓}}{110}\text{?}& \text{Missing factor statement.}\hfill \\ \underset{\text{↓}}{\text{(percentage)}}& \underset{\text{↓}}{\text{=}}& \underset{\text{↓}}{\text{(percent)}}& \underset{\text{↓}}{\cdot }& \underset{\text{↓}}{\text{(base)}}& \text{[(product) = (factor) ⋅ (factor)]}\\ 80& =& Q& \cdot & 110& \end{array}$

$\begin{array}{cccc}Q& =& 80÷110\hfill & \text{Divide.}\hfill \\ Q& =& 0.727272\text{. . .}\hfill & \text{Round to two decimal places.}\hfill \\ Q& \approx & 73\%\hfill & \text{The symbol "≈" is read as "approximately."}\end{array}$

Thus, this bottle contains approximately 73% HCl.

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