Difference between revisions of "2008 AMC 8 Problems/Problem 13"
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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 | ||
| − | <cmath>\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}</cmath> | + | <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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:23, 8 August 2021
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 
pounds or more than 
pounds. So the boxes are weighed in pairs in every possible way. The results are 
, 
and 
pounds. What is the combined weight in pounds of the three boxes?
Solution
Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures
![\[\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}.\]](http://latex.artofproblemsolving.com/2/0/c/20c6d8379377b5f2b3ec4a4e6155ba77b4a867d2.png)
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
