Decimals, Percents, and Real Numbers
7.4 Percents
Percents
Percents are very useful in conveying information. Most people have some intuitive understanding of percents and how they work. However, it is important to understand the formal definition of percents, because they can be used in dubious ways. Mathematics do not lie, but people do. We see it all the time in advertising and product labels. "Now with 50% more!" More than what?
Definition: Percent
A percent is a rational number that is a part of the whole, where the whole is $100$. Percent literally means "per $100$."
$$n \% =\frac{n}{100}$$
where $n$ is any nonnegative number.
The utility in percents is that they allow us to compare different quantities to a very familiar one. If one were to write $\frac{4897}{7300}$, it is a bit hard to understand what that means. However, if we reduce it to a fraction of $100$,
$$\frac{4897}{7300}=\frac{4897}{7300}\cdot\frac{100}{100}\approx67.08\%$$
we can see that if the whole were $100$, this number would account for about $67$ of them, or $67\%$. A bit more than two thirds.
Example: Writing numbers as percents
Write each of the following as a percent.
- $0.03$
- $0.\overline{3}$
- $1.2$
- $0.00042$
- $1$
- $\frac{3}{5}$
- $\frac{2}{3}$
- $1\frac{2}{7}$
Show Solution
Multiply each number by $100%$.
- $0.03=3\%$
- $0.\overline{3}=33.\overline{3}\%$
- $1.2=120\%$
- $0.00042=0.042\%$
- $1=100%$
- $\frac{3}{5}=60\%$
- $\frac{2}{3}=66.\overline{6}\%$
- $1\frac{2}{7}=\frac{9}{7}\approx128.571428\%$
Example: Writing percents as decimals
Write each of the following percents as a decimal.
- $5\%$
- $6.3\%$
- $100\%$
- $250\%$
- $\frac{2}{3}\%$
- $33\frac{1}{3}\%$
Show Solution
Divide each number by $100$.
- $5\%=0.05$
- $6.3\%=0.063$
- $100\%=1$
- $250\%=2.5$
- $\frac{2}{3}\%=\frac{2}{300}=\frac{1}{150}$
- $33\frac{1}{3}\%=\frac{1}{3}=0.\overline{3}$
MyOpenMath: Percent conversions
Percent Applications
Once you understand the definition, the next step is learning to identify what the percent statement is really telling you: the part, the whole, or the rate.
Example: Part of a class
In a class of $30$ students, $60\%$ of them are girls. How many students are girls? How many are boys?
Show Solution
The number of girls is
$$0.60\cdot 30=18.$$
So there are $18$ girls. The remaining students are boys:
$$30-18=12.$$
So there are $12$ boys.
Example: Percentage score
Alice scored $24$ out of $28$ correct on a test. What is her percentage score?
Show Solution
$$\frac{24}{28}=\frac{6}{7}\approx0.857142\ldots$$
Converting to a percent gives
$$85.7142\ldots\%$$
so her score is about $85.71\%$.
Example: Total from a percent part
Isabell went to her local zoo, which featured $84$ felines. If the felines make up $21\%$ of the animals at the zoo, how many animals does the zoo have in total?
Show Solution
Let the total number of animals be $T$. Then
$$0.21T=84$$
so
$$T=\frac{84}{0.21}=400.$$
The zoo has $400$ animals in total.
Example: Percent discount
Skis at a sports store are on sale for $\$476$. If the original price was $\$560$, what percent were they discounted?
Show Solution
The amount of the discount is
$$560-476=84.$$
To find the percent discount, compare the discount to the original price:
$$\frac{84}{560}=0.15=15\%.$$
So the skis were discounted by $15\%$.
MyOpenMath: Percent applications
Mental Math with Percents
Every few months or so, someone claims to have found a percent trick that they wish their math teacher had taught them in school. The trick is usually something like
$$8\%\text{ of }50\text{ is the same as }50\%\text{ of }8.$$
The reality is that someone probably tried very hard to teach them this trick. That trick is usually the commutative property of multiplication for rational numbers:
$$x\%\text{ of }y=\frac{x}{100}\cdot y=\frac{x}{100}\cdot\frac{y}{1}=\frac{y}{100}\cdot\frac{x}{1}=y\%\text{ of }x.$$
It follows that having a little practice with mental math and percents can be very useful. Some difficult problems can be made quite easy with a clever use of what we know about rational numbers.
Example: Estimating a sale price
Laura wants to buy a blouse originally priced at $\$26.50$ but now on sale at $40\%$ off. She has $\$20$ in her wallet. How might she use mental math and estimation to determine if she has enough money?
Show Solution
Since $40\%$ is $10\%$ but $4$ times, we can find $10\%$ pretty easily, and then take it away $4$ times.
$$0.10\cdot 26.50=2.65.$$
Multiplying that by $4$ to get $40\%$
$$4(2+0.65)=8+2.60=10.60$$
$$26.50-10.60=15.90.$$
She does have enough money, and she will have $\$4.10$ left.
Example: Percent reasoning
Which of the following statements could be true and which are false? Explain your reasoning.
- Leonardo got a $10%$ raise at the end of his first year on the job and a $10%$ raise after another year. His total raise was $20%$ of his original salary.
- Jung and Dina paid $45\%$ of their first department store bill of $\$620$ and $48\%$ of their second bill of $\$380$. They paid $45+48=93\%$ of their total bill.
- In a town, $65\%$ of the adult population work in town, $25\%$ work out of town, $15\%$ are unemployed, and everyone is in exactly one of these categories.
- In Clean City, the fine for smoking in public places is $40\%$ of monthly income, for driving a polluting car is $50\%$, and for littering is $30\%$. Mr. Schmutz committed all three in one day and paid a fine of $120\%$ of his monthly income.
Show Solution
- False. Let the original salary be $S$. After two raises of $10\%$ each, the new salary is
$$S(1.10)(1.10)=1.21S,$$
so the total raise is $21\%$, not $20\%$.
- False. They paid
$$0.45(620)+0.48(380)=279+182.4=461.4.$$
Their total bill was
$$620+380=1000,$$
so they paid
$$\frac{461.4}{1000}=0.4614=46.14\%,$$
not $93\%$.
- False. The categories are disjoint and include everyone, so the percentages must add to $100\%$. But
$$65+25+15=105,$$
which is impossible.
- Could be true. If each fine is assessed as the stated percent of monthly income and then added, the total fine is
$$40\%+50\%+30\%=120\%$$
of monthly income.
MyOpenMath: Mental math with percents
Computing Simple Interest
Consider the idea of simple interest. Say a person puts $\$100$ in an account that draws $5\%$ interest each month. The table below shows the amount of money in the account at the end of each month.
| Month | Current Value | Interest | Total |
|---|---|---|---|
| 1 | $\$100$ | $\$5$ | $\$105$ |
| 2 | $\$105$ | $\$5$ | $\$110$ |
| 3 | $\$110$ | $\$5$ | $\$115$ |
| 4 | $\$115$ | $\$5$ | $\$120$ |
It does not matter what the starting amount, or principal, is. The formula remains the same. If $P$ is the principal, $r$ is the interest rate, and $t$ is the number of time periods for which interest is accrued, then the amount of money in the account at the end of the interest period is
You might recognize this as our good ol' arithmetic sequence.
Fact: Simple Interest
Given the principal $P$, the interest rate $r$, and the time $t$, the amount of money in an account drawing simple interest is given by
$$A=P(1+rt).$$
Example: Simple interest over two time periods
Vera opened a savings account that pays simple interest at the rate of $5\frac{1}{4}\%$ per year. If she deposits $\$2000$ and makes no additional deposits, find the interest and the final amount for the following time periods.
- $1$ year
- $90$ days
Show Solution
Use the simple-interest model
$$I=Prt,\qquad A=P+I,$$
with $P=2000$ and $r=5\frac{1}{4}\%=0.0525$.
- For $t=1$ year:
$$I=2000(0.0525)(1)=105.$$
So the interest is $\$105$, and the final amount is
$$A=2000+105=2105,$$
so $\$2105$.
- For $90$ days (using a $365$-day year, so $t=\frac{90}{365}$):
$$I=2000(0.0525)\left(\frac{90}{365}\right)\approx25.89.$$
So the interest is about $\$25.89$, and the final amount is
$$A\approx2000+25.89=2025.89,$$
so about $\$2025.89$.
Example: Finding the annual interest rate
Find the annual interest rate if a principal of $\$10{,}000$ increased to $\$10{,}900$ at the end of $1$ year.
Show Solution
Using the simple interest formula,
$$10900=10000(1+r\cdot1).$$
Divide both sides by $10000$:
$$1.09=1+r.$$
So
$$r=0.09=9\%.$$
The annual simple interest rate is $9\%$.