1999 AMC 8 Problems/Problem 1

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

Simplify the given expression: $(6?3)+4-(2-1)=5$

$(6?3)+4-1=5$

$(6?3)+3=5$

$(6?3)=2$

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

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png