Middle School Mathematics
Middle school mathematics

Decimals and Percentages

Learning objectives

Fractions, decimals and percents

Definition

A percent is a fraction with denominator 100: \(p\%=\frac{p}{100}\). The word means "per hundred".

Move the slider to shade \(p\) of the 100 squares. The page shows the same number as a percent, a fraction and a decimal, and finds that percent of a whole.

Example · Conversions

\(\frac34=0.75=75\%\) and \(\frac18=0.125=12.5\%\).

To turn a decimal into a percent, multiply by 100: \(0.07=7\%\).

Percent of a quantity

Worked example

Find \(35\%\) of 80.

\[35\%\cdot 80=0.35\cdot 80=28\]

Worked example

What percent is 18 of 24?

\[\frac{18}{24}=\frac34=75\%\]

Worked example · Finding the whole

15 is \(30\%\) of a number. Find the number.

We have \(0.30\cdot x=15\), so \(x=\frac{15}{0.30}=50\).

Percent change

Theorem · Percent change

If a quantity changes from \(a\) to \(b\), the percent change is

\[\frac{b-a}{a}\cdot 100\%\]

A positive result is an increase and a negative result is a decrease.

Worked example

A price goes from 40 to 50, and later back from 50 to 40. Find both percent changes.

\[\frac{50-40}{40}\cdot 100\%=25\%,\qquad \frac{40-50}{50}\cdot 100\%=-20\%\]

Remark

The two changes are not opposites, because each percent is taken of a different starting value.

Worked example · Discount and tax

A jacket costs 60. It is discounted by \(20\%\), then \(10\%\) tax is added. What is the final price?

\[60\cdot 0.80=48,\qquad 48\cdot 1.10=52.8\]

Remark · Successive changes

Percent changes do not simply add up. An increase of \(20\%\) followed by a decrease of \(20\%\) gives \(1.2\cdot 0.8=0.96\), which is a \(4\%\) decrease overall.

Exercises

Write \(\frac58\) as a decimal and as a percent.

Show solution

\(\frac58=0.625=62.5\%\)

Find \(12\%\) of 250.

Show solution

\(0.12\cdot 250=30\)

What percent is 45 of 60?

Show solution

\(\frac{45}{60}=\frac34=75\%\)

After a \(25\%\) discount a jacket costs 45. What was the original price?

Show solution

The price is \(75\%\) of the original: \(0.75x=45\), so \(x=60\).

A population grows from 8000 to 9200. Find the percent increase.

Show solution

\(\frac{9200-8000}{8000}\cdot 100\%=\frac{1200}{8000}\cdot 100\%=15\%\)

A price rises by \(10\%\) and then falls by \(10\%\). Is the final price the same as the start?

Show solution

No: \(1.10\cdot 0.90=0.99\), so the final price is \(1\%\) lower than the start.

Summary