Reverse percentage
SAT glossary · Course Foundations
Definition
Recovering the original from a post-change value: "after a 20% discount a coat costs $48" means 0.80x = 48, so x = 60 — the 48 is 80% of the ORIGINAL, not of itself. To undo a multiplier you DIVIDE by it; the distractor 48 × 1.2 = 57.60 adds the percentage back and is the single most common percentage error, because the two bases are different numbers.
Reverse percentage in plain English
A 'reverse percentage' problem gives you the value AFTER some percentage change happened, and asks you to work backward to find the ORIGINAL value before the change. For example: 'after a 20% discount, a coat costs $48' means the $48 represents 80% of the original price (since 20% was taken off), so you'd set up 0.80x=48 and solve to get x=60. The critical insight is that $48 is 80% of the ORIGINAL price, not of ITSELF — to undo a multiplier, you DIVIDE by it, you don't multiply. The classic wrong answer here is calculating 48×1.2=57.60, which mistakenly tries to 'add the 20% back' onto the discounted price — but that 20% was a percentage of a DIFFERENT (bigger) starting number, so simply adding it back onto the smaller number gives the wrong result.
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