1999 AMC 8 Problems/Problem 1

Revision as of 16:23, 16 July 2018 by Bestaops (talk | contribs) (Solution)

Problem

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

$\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}$

Solution

Simplifying the given expression, we get: $(6?3)=2$

At this point, it becomes clear that it should be $\div,$ so the answer is $\boxed{A}$.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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