1.27: Percents Part 3

Last updated
Save as PDF

Page ID: 56866

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$karim-manjra-07NDEeH3lR4-unsplash.jpg$

You may use a calculator throughout this module.

There is one more situation involving percents that often trips people up: working backwards from the result of a percent change to find the original value.

$\text{Amount}=\text{Rate}\cdot\text{Base}$

$A=R\cdot{B}$

Finding the Base After Percent Increase

Suppose a $12\%$ tax is added to a price; what percent of the original is the new amount?

Well, the original number is $100\%$ of itself, so the new amount must be $100\%+12\%=112\%$ of the original.

As a proportion, $\dfrac{A}{B}=\dfrac{112}{100}$. As an equation, $A=1.12\cdot{B}$.

If a number is increased by a percent, add that percent to $100\%$ and use that result for $R$.

The most common error in solving this type of problem is applying the percent to the new number instead of the original. For example, consider this question: “After a $12\%$ increase, the new price of a computer is $ $1,120$. What was the original price?”

People often work this problem by finding $12\%$ of $ $1,120$ and subtracting that away: $12\%$ of $1,120$ is $134.40$, and $1,120-134.40=985.60$. It appears that the original price was $ $985.60$, but if we check this result, we find that the numbers don’t add up. $12\%$ of $985.60$ is $118.272$, and $985.60+118.272=1,103.872$, not $1,120$.

The correct way to think about this is $1,120=1.12\cdot{B}$. Dividing $1,120$ by $1.12$ gives us the answer $1,000$, which is clearly correct because we can find that $12\%$ of $1,000$ is $120$, making the new amount $1,120$. The original price was $ $1,000$.

To summarize, we cannot subtract $12\%$ from the new amount; we must instead divide the new amount by $112\%$.

Exercises $\PageIndex{1}$

1. A sales tax of $8\%$ is added to the selling price of a lawn tractor, making the total price $ $1,402.92$. What is the selling price of the lawn tractor without tax?

2. The U.S. population in 2018 was estimated to be $327.2$ million, which represents a $7.6\%$ increase from 2008. What was the U.S. population in 2008?

Answer

1. $ $1,299.00$

2. $304.1$ million

Finding the Base After Percent DEcrease

Suppose a $12\%$ discount is applied to a price; what percent of the original is the new amount?

As above, the original number is $100\%$ of itself, so the new amount must be $100\%-12\%=88\%$ of the original.

As a proportion, $\dfrac{A}{B}=\dfrac{88}{100}$. As an equation, $A=0.88\cdot{B}$.

If a number is decreased by a percent, subtract that percent from $100\%$ and use that result for $R$.

As above, the most common error in solving this type of problem is applying the percent to the new number instead of the original. For example, consider this question: “After a $12\%$ decrease, the new price of a computer is $ $880$. What was the original price?”

People often work this problem by finding $12\%$ of $880$ and adding it on: $12\%$ of $880$ is $105.60$, and $880+105.60=985.60$. It appears that the original price was $ $985.60$, but if we check this result, we find that the numbers don’t add up. $12\%$ of $985.60$ is $118.272$, and $985.60-118.272=867.328$, not $880$.

The correct way to think about this is $880=0.88\cdot{B}$. Dividing $880$ by $0.88$ gives us the answer $1,000$, which is clearly correct because we can find that $12\%$ of $1,000$ is $120$, making the new amount $880$. The original price was $ $1,000$.

To summarize, we cannot add $12\%$ to the new amount; we must instead divide the new amount by $88\%$.

Exercises $\PageIndex{1}$

3. A city department’s budget was cut by $5\%$ this year. If this year’s budget is $ $3.04$ million, what was last year’s budget?

4. CCC’s enrollment in Summer 2019 was $9,116$ students, which was a decrease of $2.17\%$ from Summer 2018. What was the enrollment in Summer 2018? (Round to the nearest whole number.)^[1]

5. An educational website claims that by purchasing access for $ $5$, you’ll save $69\%$ off the standard price. What was the standard price? (Use you best judgment when rounding your answer.)

Answer

3. $ $3.20$ million

4. $9,318$ students

5. $ $16.13$, or more likely, just $ $16$.

These enrollment numbers don't match those in Percents Part 2, which makes me wonder how accurate the yearly reports are. Or maybe I inadvertently grabbed data from two different ways that enrollment was being counted. ↵

Search

Exercises \(\PageIndex{1}\)

Exercises \(\PageIndex{1}\)