Why can't I just add and subtract the percents in a chain?

Because each percent applies to a different base. The second change acts on the amount left after the first change, not on the original. Adding percents ignores this, so you must multiply the multipliers instead. For example, +20\% then -20\% gives 1.2× 0.8 = 0.96, a 4\% loss, not zero.

How do I know whether to divide or multiply in a percentage problem?

If you are applying a change to a known starting value, multiply by the multiplier. If you are given the value after a change and need the original, divide by the multiplier. The key clue is whether the number you have is the before amount or the after amount.

What's the fastest way to handle percent change on the Digital SAT?

Convert every percent to a decimal multiplier right away: increases become 1+p/100 and decreases become 1-p/100. Then the whole problem becomes multiplication or division, which is quick with the on-screen calculator and less error-prone than tracking percents.

Does the order of percent changes change the final answer?

No. Since the changes are multiplications and multiplication is commutative, the final value is the same regardless of order. Order only matters if a question asks for an intermediate value between the two changes.

DSAT-3.2

Percentages

Master Digital SAT percentages: percent of, percent change, chained changes, and reverse percentage to recover original values. Methods, examples, and practice.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Percentages, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Percentages free

What this lesson covers

Percentages show up all over the Digital SAT — in word problems, data interpretation, and multi-step algebra. The good news is that almost every percentage question reduces to a few reliable moves: taking a percent of a number, measuring how much something grew or shrank, applying several changes in a row, and working backward from a final amount to the original.

This lesson gives you a clean toolkit for each of those moves and shows you the traps the test loves to set, especially the difference between a percent and a multiplier. Learn to translate every percentage into a decimal multiplier and most questions become one careful equation.

Percent of a number

A percent is just a fraction out of 100, so

p\%

means

\frac{p}{100}

. To take a percent of a quantity, convert to a decimal and multiply.

p\% \text{ of } N = \frac{p}{100}\cdot N

For example,

30\%

80

0.30\times 80 = 24

. The word "of" almost always signals multiplication, and the word "is" signals equals. So "18 is what percent of 40?" translates to

18 = \frac{p}{100}\cdot 40

, giving

p = 45

.

A huge time-saver is knowing that percent statements are reversible:

p\%

N

equals

N\%

p

. That means

18\%

50

equals

50\%

18 = 9

. The Digital SAT rewards students who see these shortcuts instead of reaching for the calculator every time.

Phrase	Translation
$p\%$ of $N$	$\frac{p}{100}\cdot N$
$A$ is $p\%$ of $B$	$A = \frac{p}{100}\cdot B$
$A$ is what percent of $B$	$p = \frac{A}{B}\cdot 100$

The most common misconception is treating a percent as if it were a raw number — writing

30

instead of

0.30

. Always convert before you multiply.

Percent change and the multiplier method

Percent change measures how much a quantity grew or shrank relative to its starting value:

\text{percent change} = \frac{\text{new} - \text{old}}{\text{old}}\times 100

The denominator is always the original amount, not the new one. A positive result is a percent increase; a negative result is a percent decrease.

The fastest way to apply a change is the multiplier. Increasing by

p\%

multiplies by

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

; decreasing by

p\%

multiplies by

\left(1-\frac{p}{100}\right)

. So a

25\%

increase means

\times 1.25

, and a

40\%

decrease means

\times 0.60

Change	Multiplier
increase 20%	$1.20$
increase 5%	$1.05$
decrease 15%	$0.85$
decrease 100%	$0$

Multipliers are powerful because they turn awkward wording into arithmetic. "A price rises from 40 to 50" gives

\frac{50-40}{40}=0.25

, a

25\%

increase. Watch the direction: going from 40 to 50 is a

25\%

increase, but going from 50 to 40 is a

20\%

decrease, because the base changed. The test frequently exploits this asymmetry.

Chained percent changes

When several percent changes happen in sequence, you multiply the multipliers together — you do not add the percents. This is the single most tested percentage trap on the Digital SAT.

Suppose a value increases

20\%

and then decreases

20\%

. The combined multiplier is

1.20\times 0.80 = 0.96

, a net

4\%

decrease — not zero. The decrease acts on a larger base, so it removes more than the increase added.

For a value

V

hit by changes of

a\%

then

b\%

V_{\text{final}} = V\left(1+\tfrac{a}{100}\right)\left(1+\tfrac{b}{100}\right)

Order does not matter for the final amount because multiplication commutes, but it matters for any intermediate value the question might ask about.

For repeated identical changes, use an exponent. A quantity growing

8\%

per year for

t

years is multiplied by

(1.08)^t

. This connects directly to exponential-growth questions elsewhere on the test.

A common misconception: a

50\%

increase followed by a

50\%

decrease returns you to the start. It does not —

1.5\times 0.5 = 0.75

, a

25\%

net loss. Always convert each step to a multiplier and multiply.

Reverse percentage: recovering the original

Reverse-percentage problems give you a final value that already includes a change and ask for the original. The mistake students make is applying the percent to the final number. Instead, set up the multiplier equation and divide.

If a price after a

20\%

increase is 90, then

\text{original}\times 1.20 = 90 \quad\Rightarrow\quad \text{original} = \frac{90}{1.20} = 75.

Notice you divide by the multiplier, not by

0.20

. Subtracting

20\%

of 90 would wrongly give 72, because the

20\%

was based on the original 75, not on 90.

The general form: if a final value

F

results from an

x\%

change to an unknown

N

N = \frac{F}{1 \pm \frac{x}{100}}

using

+

for an increase and

-

for a decrease. So a sale price of 68 after a

15\%

discount means

N = \frac{68}{0.85} = 80

.

The Digital SAT loves to phrase these as "after tax," "after a discount," or "including a markup." Whenever the number you are given is the after-change amount, divide by the multiplier to undo it.

How the Digital SAT tests percentages

Percentage questions appear in both easy and hard slots, and the difficulty usually comes from wording, not arithmetic. Expect these formats.

Straight computation questions ask for a percent of a value or a percent change and can often be entered directly with the built-in calculator. Do not let the calculator make you sloppy about which number is the base.

Multi-step word problems bury a percent inside a larger scenario — a salary after two raises, a population after decline and rebound, a bill with tax added to a discounted price. Translate each phrase into a multiplier and chain them.

Reverse questions give the after amount and ask for a before amount. Recognizing the direction is half the battle.

Data and table questions, which connect to the statistics lessons in this unit, ask what percent one cell is of a row total or how a total changed between two columns. The skill is the same: identify the base, then divide.

A reliable habit for every percentage problem: write down what counts as the base (the "100%") before doing anything else. Most wrong answers come from using the wrong base.

Key terms

Percent.: A ratio expressed out of 100; $p\%$ equals the fraction $\frac{p}{100}$ or the decimal $\frac{p}{100}$ .
Percent of.: The result of multiplying a quantity by a percent written as a decimal: $p\%$ of $N = \frac{p}{100}\cdot N$ .
Percent change.: The difference between new and old values divided by the old value, times 100; positive for increase, negative for decrease.
Multiplier.: The factor that applies a percent change in one step: $1+\frac{p}{100}$ for an increase, $1-\frac{p}{100}$ for a decrease.
Base.: The reference quantity that counts as 100% in a percentage calculation; always the original value in a percent-change problem.
Chained percent change.: Two or more successive percent changes applied by multiplying their multipliers rather than adding percents.
Reverse percentage.: Finding an original value from a post-change final value by dividing the final value by the multiplier.

Worked example

A jacket's price is first increased by 25% for the holiday season, then marked down 30% in a clearance sale. The final clearance price is $52.50. What was the original price of the jacket before any changes?

Start by turning each change into a multiplier. A

25\%

increase is

\times 1.25

, and a

30\%

decrease is

\times 0.70

.

Let the original price be

P

. Applying both changes in order gives the final price:

P\times 1.25\times 0.70 = 52.50.

Multiply the two multipliers first:

1.25\times 0.70 = 0.875

. So

0.875\,P = 52.50.

Because 52.50 is the after-change amount, recover the original by dividing by the combined multiplier:

P = \frac{52.50}{0.875} = 60.

The original price was 60 dollars. As a check,

60\times 1.25 = 75

after the markup, and

75\times 0.70 = 52.50

after the clearance discount — matching the given final price.

Notice the net effect was

0.875

, a

12.5\%

decrease overall, even though the stated percents were 25 and 30. Adding or subtracting the percents would have been wrong.

Practice questions

A store raises the price of a lamp by 40%, then later lowers the new price by 40%. Compared with the original price, the final price is:

16% higher
the same
16% lower
20% lower

Answer: 16% lower

Convert each change to a multiplier: a

40\%

increase is

\times 1.40

and a

40\%

decrease is

\times 0.60

. The combined multiplier is

1.40\times 0.60 = 0.84

, which is a

16\%

decrease from the original. The increase and decrease do not cancel because the second change acts on a larger base.

After a 15% discount, a pair of shoes costs $68. To the nearest dollar, what was the original price before the discount?

Answer: $80

15\%

discount multiplies the original by

0.85

. So

\text{original}\times 0.85 = 68

, which gives

\text{original} = \frac{68}{0.85} = 80

. Dividing by the multiplier undoes the change; subtracting

15\%

of 68 would incorrectly give about 57.80 because that uses the wrong base.

A town's population was 12,000 in 2020. It grew 10% in 2021 and then fell 5% in 2022. What was the population at the end of 2022?

Answer: 12,540

Apply successive multipliers:

12000\times 1.10 = 13200

after the growth, then

13200\times 0.95 = 12540

after the decline. You cannot just net the percents to

+5\%

, since the

5\%

drop applies to the larger 13,200 base.

FAQ

Why can't I just add and subtract the percents in a chain?: Because each percent applies to a different base. The second change acts on the amount left after the first change, not on the original. Adding percents ignores this, so you must multiply the multipliers instead. For example, $+20\%$ then $-20\%$ gives $1.2\times 0.8 = 0.96$ , a $4\%$ loss, not zero.
How do I know whether to divide or multiply in a percentage problem?: If you are applying a change to a known starting value, multiply by the multiplier. If you are given the value after a change and need the original, divide by the multiplier. The key clue is whether the number you have is the before amount or the after amount.
What's the fastest way to handle percent change on the Digital SAT?: Convert every percent to a decimal multiplier right away: increases become $1+\frac{p}{100}$ and decreases become $1-\frac{p}{100}$ . Then the whole problem becomes multiplication or division, which is quick with the on-screen calculator and less error-prone than tracking percents.
Does the order of percent changes change the final answer?: No. Since the changes are multiplications and multiplication is commutative, the final value is the same regardless of order. Order only matters if a question asks for an intermediate value between the two changes.

Learn this with a teacher, not a page

The Crimsora tutor teaches Percentages live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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