Percentage questions on the Digital SAT are rarely about computing $p\%$ of a number — they are about direction: increases, discounts, and especially recovering an original value after a change. One idea solves all of them: a percent change is a multiplication. A $20\%$ discount multiplies by $0.8$; a $35\%$ increase multiplies by $1.35$. Write the multiplier, and the equation writes itself.
How the question announces itself
“After a $20\%$ discount, the sale price is $48$ dollars. What was the original price?” — reverse problem.
“The population increased by $15\%$ from 2020 to 2021 and by $8\%$ from 2021 to 2022…” — successive changes.
“The value of $x$ is $p\%$ greater than $y$…” — comparison as a multiplier.
Any stem mixing a percent with a before/after pair of values.
The method
Turn the percent into a multiplier. Decrease of $p\%$ → multiply by $\left(1 - \dfrac{p}{100}\right)$. Increase of $p\%$ → multiply by $\left(1 + \dfrac{p}{100}\right)$. Write the relationship as one equation: $(\text{original}) \times (\text{multiplier}) = (\text{final})$.
Reverse problems divide — they never “un-discount”. If $0.8p = 48$, then $p = 48 \div 0.8 = 60$. Adding $20\%$ back to $48$ gives $57.60$ — a different (wrong) number, and always one of the options. The original is recovered by dividing by the multiplier, full stop.
Chain successive changes by multiplying. Two changes compose: $\times 1.15$ then $\times 1.08$ is $\times 1.242$ — a $24.2\%$ total increase, not $23\%$. Percents never add across steps unless they act on the same base.
The named mistakes behind every wrong answer
Wrong base
Applying the percent to the final value instead of the original. The multiplier equation forces the base to be explicit — that is why it works.
Un-discounting by adding the percent back
$48 \times 1.2 = 57.60 \neq 60$. Increasing by $20\%$ does not undo decreasing by $20\%$, because the two percents act on different bases.
Adding successive percents
$+15\%$ then $+8\%$ is not $+23\%$. Compose multipliers instead.
Treating percent as a flat amount
Reading “$20\%$ off” as “20 dollars off.” Distractors are built from this misread — check the units in the stem.
Try one
Question 1 — Percentages (easy)
After a $20\%$ discount, the sale price of a jacket is $48$ dollars. What was the original price of the jacket, in dollars?
$38.40$
$57.60$
$68$
$60$
Answer: D. A $20\%$ discount leaves $80\%$ of the original price $p$, so $0.8p = 48$ and $p = 60$.
The full Percentages lesson inside Satisfied covers percent change over tables, mixture problems, and the two-variable variants — with adaptive drills per difficulty.