1999 AMC 8 Problem 1 — Solution | LIVE by Po-Shen Loh

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1999 AMC 8 Problem 1

Attempt Problem 1 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).

1.

$(6 \ ? \ 3) + 4 - (2 - 1) = 5.$ To make this statement true, the question mark between the $6$ and the $3$ should be replaced by

$\div$

$\times$

$+$

$-$

None of these

Answer: A

Concepts:order of operations

Difficulty rating: 370

Solution:

We have that $4 - (2 - 1) = 4 - 1 = 3.$

Then $(6 \ ? \ 3) + 3 = 5$ $6 \ ? \ 3 = 2.$

The only choice that works is $\div.$

Thus, A is the correct answer.

Problem 2#2 →

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