A 25% rise followed by a 25% fall does not return to the start. One rule (a percent is always a percent of something) plus three original practice items with trap analysis.
When a price rises 25% and then falls 25%, it does not return to where it started. It ends up lower. The SAT counts on you to forget that, because the trap answer always looks like the number that "makes sense." The fix is one rule: a percent is always a percent of something, and the something changes.
For two values, old and new:
The denominators are different by design. That is the entire trap. Every wrong answer in a percent-change question is a right answer to a different question, usually one where you used the wrong base.
These are original SAT-style examples written for practice, not official SAT questions.
Example 1. A store raises the price of a $40 jacket by 25%. One week later, it lowers the new price by 25%. What is the final price?
A. $37.50 B. $40 C. $36 D. $45
Raise: $40 × 1.25 = $50. Then lower: $50 × 0.75 = $37.50. The answer is A.
Choice B is the trap. Twenty-five percent up and twenty-five percent down feel like they should cancel, but the second 25% is taken from $50, which is larger than $40. The drop ($12.50) is bigger than the raise ($10). Whenever a percent decrease follows a percent increase of the same number, the result is always below the start, because the decrease acts on the larger base.
Example 2. A town's population grew from 8,000 to 12,000 over ten years. By what percent did the population increase?
A. 33 1/3% B. 50% C. 66 2/3% D. 150%
The increase is 12,000 − 8,000 = 4,000, and the base is the starting value, 8,000. So 4,000 ÷ 8,000 × 100 = 50%. The answer is B.
Choice D is the trap. 150% is 12,000 ÷ 8,000 × 100: the percent of the original. The question asks for the percent increase, which is the percent of minus 100. Every time you compute a percent of, stop and ask whether the question wanted the change. Choice A is a subtler trap: it is the percent the population would drop if it went from 12,000 back to 8,000, using the wrong base. Choice C is 8,000 ÷ 12,000 × 100, a percent of built on the new value instead of the old.
Example 3. A price drops by 20%. By what percent must it then increase to return to its original value?
A. 16% B. 20% C. 25% D. 80%
Start with 100 for concreteness. After the drop: 100 × 0.80 = 80. To get back to 100, the price must rise 20, and 20 ÷ 80 × 100 = 25%. The answer is C.
Choice B is the trap. A 20% decrease followed by a 20% increase leaves you at 96, not 100, because the increase acts on the smaller base of 80. The general shortcut: after a drop of p%, the increase needed to recover is p ÷ (100 − p) × 100. For p = 20, that is 20 ÷ 80 × 100 = 25%.
On every percent question, write the base next to the percent before you compute anything. If the problem says "increase," the base is the starting number. If the problem describes a second change, the base is the number after the first change. That one label kills the trap.
Practice the habit on TaroPrep's free Digital SAT math questions, where percent problems show up in both Algebra and Problem Solving and Data Analysis.
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