🇺🇸 AMC 8 ⇄ switch contest
1993 AJHSME

Problem 21

Problem 21 · 1993 AJHSME Hard
Fractions, Decimals & Percents percent-area

If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by

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Answer: D — 80%.
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Hint 1 of 2
Percents on area don't ADD — they MULTIPLY. Length and width each scale by their own factor, and area is length × width, so the two factors multiply together. (Adding 20%+50% to get 70% is the trap.)
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Hint 2 of 2
Turn each increase into a multiplier: +20% means ×1.2, +50% means ×1.5. Multiply those, then see how far above 1 the result lands.
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Approach: multiply the two scale factors
  1. +20% on length is ×1.2; +50% on width is ×1.5. Since area = length × width, the new area is 1.2 × 1.5 = 1.8 times the old.
  2. 1.8 times means 0.8 more than the original — an increase of 80%.
  3. No-algebra check: take a 10×10 = 100 rectangle. New sides 12 and 15 give 180 — up 80 out of 100, i.e. 80%. Trap to dodge: percentage changes on a product never just add; choice C (70%) is the bait for 20+50.
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