🇺🇸 AMC 8 ⇄ switch contest
1985 AJHSME

Problem 8

Problem 8 · 1985 AJHSME Medium
Arithmetic & Operations substitute-and-compare

If a = −2, the largest number in the set −3a, 4a, 24⁄a, a², 1 is

Show answer
Answer: A — −3a.
Show hints
Hint 1 of 2
Before plugging in, ask: when a is NEGATIVE, which expressions come out positive? Anything that multiplies or divides a single negative a by a positive stays negative; the ones that flip to positive are your only candidates for 'largest'.
Still stuck? Show hint 2 →
Hint 2 of 2
You don't even need all five values. The biggest must be positive, and only −3a (negative times negative) and a² (a negative squared) can be positive — so just compare those two.
Show solution
Approach: substitute and pick the max
  1. Spot the signs first: with a = −2, the expressions 4a and 24⁄a stay negative, so they can't be the largest. The candidates are −3a and a², both positive.
  2. −3(−2) = 6 versus (−2)² = 4. The largest is 6 = −3a.
  3. Why this transfers: sign-sorting before computing saves work — knowing what's positive, negative, or zero often eliminates most options without a single full calculation.
Mark: · log in to save