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.(6 \ ? \ 3) + 4 - (2 - 1) = 5. To make this statement true, the question mark between the 66 and the 33 should be replaced by
÷\div
×\times
++
−-
None of these
Answer: A
Concepts:order of operations
Small hint:
Simplify the part without the question mark first
Big hint:
Find the operation that makes 6 ? 3=26\ ?\ 3=2
Solution:
We have that 4−(2−1)=4−1=3. 4 - (2 - 1) = 4 - 1 = 3.
Then (6 ? 3)+3=5 (6 \ ? \ 3) + 3 = 5 6 ? 3=2. 6 \ ? \ 3 = 2.
The only choice that works is ÷.\div.
Thus, A is the correct answer.
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