1999 AMC 8 Problem 3 — Solution | LIVE by Po-Shen Loh

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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 $1?$

$(\frac12,\frac13,\frac16)$

$(2,-2,1)$

$(0.1,0.3,0.6)$

$(1.1,-2.1,1.0)$

$(-\frac32,-\frac52,5)$

Answer: D

Concepts:fraction decimal

Difficulty rating: 560

Solution:

We have that A is the same as $\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.$

B reduces to $2 - 2 + 1 = 1.$

C adds up to $.1 + .3 + .6 = 1.$

D however adds to $1.1 - 2.1 + 1.0 = 0.$

To make sure, we check that E has a sum of $-\dfrac{8}{2} + 5 = 5 - 4 = 1.$

Thus, D is the correct answer.

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