1999 AMC 8 Problem 3

Attempt Problem 3 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

Which triplet of numbers has a sum NOT equal to 1?1?
(12,13,16)(\frac12,\frac13,\frac16)
(2,−2,1)(2,-2,1)
(0.1,0.3,0.6)(0.1,0.3,0.6)
(1.1,−2.1,1.0)(1.1,-2.1,1.0)
(−32,−52,5)(-\frac32,-\frac52,5)
Answer: D
Concepts:fractiondecimal
Small hint:
Compute the easiest-looking sums first
Big hint:
Only one listed triplet fails to total 11
Solution:
We have that A is the same as 36+26+16=1. \dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.
B reduces to 2−2+1=1. 2 - 2 + 1 = 1.
C adds up to 0.1+0.3+0.6=1. 0.1 + 0.3 + 0.6 = 1.
D however adds to 1.1−2.1+1.0=0. 1.1 - 2.1 + 1.0 = 0.
To make sure, we check that E has a sum of −82+5=5−4=1. -\dfrac{8}{2} + 5 = 5 - 4 = 1.
Thus, D is the correct answer.
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