Percentages and Percent Change, SAT math lesson

A jacket costs 80 dollars, goes up to 100, and later drops back to 80. The rise was 25 percent, yet the fall was only 20 percent. Both changes are 20 dollars, but they are measured against different starting prices. Many percent questions on the SAT turn on this one point, which amount is the base of the percentage.

Key terms

Percentage

Definition For a real number $p$ , the expression $p\%$ means $\frac{p}{100}$ . A percentage of a quantity $A$ is

p\% \text{ of } A = \frac{p}{100} \cdot A.

The number $p$ may be larger than $100$ , and $100\%$ of $A$ is $A$ itself. To ask what percent a part $B$ is of a quantity $A$ with $A \ne 0$ , compute $\frac{B}{A} \cdot 100$ .

In plain words Percent means “per hundred.” Turn the percentage into a decimal and multiply.

Example Write $35\%$ as $0.35$ , so $35\%$ of $60$ is $0.35 \cdot 60 = 21$ . Percentages above $100$ work the same way. For instance $140\%$ of $35$ is $1.4 \cdot 35 = 49$ , which is more than $35$ . In the other direction, the number $18$ is $75\%$ of $24$ , because

\frac{18}{24} \cdot 100 = 75.

Original value

Definition In a percent change from one value to another, the original value (or reference value) is the value before the change, the one the change is compared with. It must be nonzero. In “what percent of $A$ is $B$ ” the reference value is $A$ , the quantity after “of.”

In plain words The original value is the base you divide by. Pick it first, before any arithmetic.

Example For a price that goes from $80$ to $100$ dollars, the original value is $80$ . For the later move from $100$ back to $80$ , the original value is $100$ . The same $20$ dollar change is $\frac{20}{80} = 25\%$ of the first base and $\frac{20}{100} = 20\%$ of the second.

Percent change

Definition If a quantity changes from an original value $O \ne 0$ to a new value $N$ , the percent change is

\text{percent change} = \frac{N - O}{O} \cdot 100\%.

A positive result is a percent increase and a negative result is a percent decrease. Percent change is undefined when the original value is $0$ .

In plain words Find the change, divide by where you started, and write the answer as a percent.

Example A town grows from $12{,}500$ to $13{,}400$ people. The percent change is

\frac{13{,}400 - 12{,}500}{12{,}500} \cdot 100\% = 7.2\%,

an increase of $7.2$ percent. Had you divided by the new value $13{,}400$ instead, you would get about $6.7$ percent, a typical wrong choice.

Multiplier

Definition Increasing a quantity by $p\%$ multiplies it by $1 + \frac{p}{100}$ . Decreasing it by $p\%$ , for $0 \le p \le 100$ , multiplies it by $1 - \frac{p}{100}$ . Both cases fit one rule if a decrease is written as a negative change, so a change of $p\%$ with $p \ge -100$ multiplies by $1 + \frac{p}{100}$ . After a change of $p\%$ and then a change of $q\%$ , an amount $A$ becomes

A \left(1 + \frac{p}{100}\right)\left(1 + \frac{q}{100}\right),

with the second change applied to the new amount. For that reason an increase and a decrease by the same percentage do not cancel unless the percentage is $0$ . For $p > 0$

\begin{aligned} &\left(1 + \frac{p}{100}\right)\left(1 - \frac{p}{100}\right) \\[0.4em] &\quad = 1 - \frac{p^2}{10{,}000} < 1. \end{aligned}

In plain words One multiplication does the whole change. Two changes in a row mean two multipliers, and the order of the factors does not matter.

Example The jacket goes up $25\%$ and then down $20\%$ ,

80(1.25)(0.80) = 100(0.80) = 80,

so it ends where it started. With a $20\%$ increase followed by a $20\%$ decrease instead, the multiplier is $1.2 \cdot 0.8 = 0.96$ , and an $80$ dollar jacket ends at $76.80$ dollars, a net decrease of $4\%$ .

Reverse percentage

Definition If a quantity after a change of $p\%$ equals $N$ , and the multiplier $m = 1 + \frac{p}{100}$ is nonzero, then the original value is

O = \frac{N}{m}.

Here $p$ is negative for a decrease. Finding $O$ this way is called a reverse percentage problem. The original value is not found by applying the opposite percentage to $N$ .

In plain words Undo a multiplier by dividing by it. Taking $15\%$ off and then adding $15\%$ back does not return you to the start.

Example After a $15\%$ discount a pair of shoes costs $68$ dollars. The multiplier is $0.85$ , so

O = \frac{68}{0.85} = 80 \text{ dollars.}

Check it, $80 \cdot 0.85 = 68$ . The tempting answer $68 \cdot 1.15 = 78.20$ dollars is wrong, because the $15\%$ was taken from $80$ , not from $68$ .

Percentage point

Definition When two quantities are themselves percentages, the difference between them is measured in percentage points. If a rate goes from $a\%$ to $b\%$ , it changes by $b - a$ percentage points. Its relative percent change is

\frac{b - a}{a} \cdot 100\%, \qquad a \ne 0.

These are different numbers in general, so a statement must say which one it means.

In plain words Percentage points are a plain subtraction of two percentages. Percent change compares that difference with the starting rate.

Example A savings rate rises from $4\%$ to $5\%$ . That is an increase of $1$ percentage point. Relative to the original rate it is

\frac{5 - 4}{4} \cdot 100\% = 25\%,

so the rate rose by $25$ percent. Saying “the rate rose by $1$ percent” would be wrong, since $1\%$ of $4\%$ is only $0.04$ percentage points.

Check yourself

The price of a bicycle is increased by $10\%$ , and then the new price is decreased by $10\%$ . How does the final price compare with the original price? (A) It is equal to the original price. (B) It is $1\%$ less than the original price. (C) It is $1\%$ more than the original price. (D) It is $20\%$ less than the original price.

Show the answer

The answer is (B). The combined multiplier is $1.1 \cdot 0.9 = 0.99$ , so the final price is $99\%$ of the original. Choice (A) assumes the two changes cancel, but the decrease is taken from the larger, already increased price. Choice (C) gets the direction wrong, and choice (D) adds the two percentages as if both were decreases.

After an increase of $30\%$ , the population of a town is $6{,}890$ . What was the population before the increase?

Show the answer

The multiplier is $1.3$ , so the original population is

\frac{6{,}890}{1.3} = 5{,}300.

Enter 5300 without the comma. Taking $30\%$ off $6{,}890$ would give $4{,}823$ , which is wrong because the increase was $30\%$ of the original.

A survey rate rose from $5\%$ to $6\%$ . Which statement is true? (A) The rate rose by $1$ percent. (B) The rate rose by $1$ percentage point, which is a $20$ percent increase. (C) The rate rose by $20$ percentage points. (D) The rate rose by $6$ percent.

Show the answer

The answer is (B). The difference is $6 - 5 = 1$ percentage point, and relative to the original rate it is

\frac{6 - 5}{5} \cdot 100\% = 20\%.

Choice (A) mixes up percentage points with percent change. Choice (C) uses the relative change but labels it as percentage points, and choice (D) just repeats the new rate.

Percentages and Percent Change

Theory, part 5

Practice, part 6

Key terms

Percentage

Original value

Percent change

Multiplier

Reverse percentage

Percentage point

Check yourself

How this lesson connects

Builds on

Used later in