Difference between revisions of "2008 AMC 8 Problems/Problem 13"
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− | ==Solution== | + | ==Solution 1== |
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> | ||
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+ | ==Solution 2== | ||
+ | Using variables <math>a</math>, <math>b</math>, and <math>c</math> to denote the boxes, with <math>a \le b \le c</math>. It is obvious that <math>a+b=122</math> and <math>a+c=125</math>, since these are the two smallest pairs. Subtracting the former equation from the latter results in <math>c-b=3</math>. Additionally, <math>c+b=127</math>. Solving for <math>c</math> and <math>b</math> gives <math>c=65</math> and <math>b=62</math>, so we can thus find <math>a=60</math>. Solving <math>60+62+65=\boxed{\textbf{(C) }\ 187}</math> | ||
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+ | ~megaboy6679 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 23:57, 27 October 2024
Contents
[hide]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 1
Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures:
Solution 2
Using variables , , and to denote the boxes, with . It is obvious that and , since these are the two smallest pairs. Subtracting the former equation from the latter results in . Additionally, . Solving for and gives and , so we can thus find . Solving
~megaboy6679
Video Solution
https://www.youtube.com/watch?v=LKhOV9p4WiY ~David
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.