AMC 8 · 1999 · #3
Grade 7 arithmeticPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question hands you five labeled candidates and asks which one fails a single, easy-to-check property (sum equals 1). That is exactly Tool #3 (Eliminate Possibilities): walk through the candidates, compute the sum of each, and cross off any whose sum is 1. The one that survives the elimination is the answer. No algebra or pattern-hunting is needed — every triplet is just three numbers to add.
Test (A): the three fractions add to 1, so cross it off.
Adding fractions with unlike denominators by rewriting with a common denominator is the Grade 5 standard.
5.NF.A.1Eliminate PossibilitiesTest (B): the signed integers add to 1, so cross it off.
Adding a positive and its opposite gives 0 — the Grade 7 additive-inverse idea.
7.NS.A.1Eliminate PossibilitiesTest (C): the three decimals add to 1, so cross it off.
Adding decimals to the hundredths is the Grade 5 standard; here all three numbers already line up at the tenths place.
5.NBT.B.7Eliminate PossibilitiesTest (D): the signed decimals add to 0, not 1, so (D) survives.
1.1 and -2.1 differ in size by exactly 1.0, so adding them gives -1.0, which then cancels the last +1.0 to leave 0.
7.NS.A.1Eliminate PossibilitiesCheck (E): the last triplet also adds to 1, so (D) is the only survivor.
Both fractions share denominator 2, so they combine directly to -4, and -4 + 5 = 1.
7.NS.A.1Eliminate PossibilitiesWhen the question is "which one is different?", just test each candidate and cross off the ones that pass. The single triplet that fails is your answer — here, (D) sums to 0, not 1.