🇺🇸 AMC 8 ⇄ switch contest
1989 AJHSME

Problem 4

Problem 4 · 1989 AJHSME Medium
Fractions, Decimals & Percents estimation

Estimate to determine which of the following numbers is closest to 401.205.

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Answer: E — 2000.
Show hints
Hint 1 of 3
The answer choices jump by factors of 10 (.2, 2, 20, 200, 2000). When choices are that far apart, you don't need an exact answer — a rough size estimate picks the winner.
Still stuck? Show hint 2 →
Hint 2 of 3
Replace the ugly numbers with nearby friendly ones before dividing: 401 → 400, .205 → .2.
Still stuck? Show hint 3 →
Hint 3 of 3
Dividing by .2 means asking 'how many fifths?' — and there are 5 fifths in every 1, so dividing by .2 is the same as multiplying by 5.
Show solution
Approach: round to friendly numbers, then divide
  1. Because the choices differ by whole factors of 10, the problem literally says 'estimate' — so round the messy numbers first: 401 ≈ 400 and .205 ≈ .2.
  2. Now 400 ÷ .2. Dividing by .2 (one-fifth) is the same as multiplying by 5, so 400 × 5 = 2000. Closest choice: 2000.
  3. Why this transfers: when answer choices are spread far apart, estimating is faster and safer than an exact computation — and turning 'divide by .2' into 'times 5' kills the decimal entirely.
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