Difference between revisions of "1999 AMC 8 Problems/Problem 1"

m (Solution)
m (Solution)
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Simplifying the given expression, we get: <math>(6?3)=2.</math>
 
Simplifying the given expression, we get: <math>(6?3)=2.</math>
  
At this point, it becomes clear that it should be <math> \div,</math> so the answer is <math> \boxed{A} </math>.
+
At this point, it becomes clear that it should be <math> \div (A) </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|before=First<br />Question|num-a=2}}
 
{{AMC8 box|year=1999|before=First<br />Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:24, 16 July 2018

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 $\div (A)$.

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

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