Difference between revisions of "2008 AMC 8 Problems/Problem 13"

m (Solution)
Line 11: Line 11:
 
Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures
 
Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures
  
<math></math>\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}$
+
<cmath>\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=12|num-a=14}}
 
{{AMC8 box|year=2008|num-b=12|num-a=14}}

Revision as of 23:20, 24 December 2012

Problem

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$, $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?

$\textbf{(A)}\ 160\qquad \textbf{(B)}\ 170\qquad \textbf{(C)}\ 187\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 354$

Solution

Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures

\[\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}\]

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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