2008 AMC 8 Problems/Problem 13: Difference between revisions

Latest revision as of 23:57, 27 October 2024

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 1

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

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

Solution 2

Using variables $a$ , $b$ , and $c$ to denote the boxes, with $a \le b \le c$ . It is obvious that $a+b=122$ and $a+c=125$ , since these are the two smallest pairs. Subtracting the former equation from the latter results in $c-b=3$ . Additionally, $c+b=127$ . Solving for $c$ and $b$ gives $c=65$ and $b=62$ , so we can thus find $a=60$ . Solving $60+62+65=\boxed{\textbf{(C) }\ 187}$

~megaboy6679

Video Solution

https://www.youtube.com/watch?v=LKhOV9p4WiY ~David

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 8: / Line 8: @@
 \textbf{(E)}\ 354</math>
+==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:
+<cmath>\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}.</cmath>
+==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>
+~megaboy6679
 ==Video Solution==
 https://www.youtube.com/watch?v=LKhOV9p4WiY   ~David
-==Solution==
-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>
 ==See Also==
 {{AMC8 box|year=2008|num-b=12|num-a=14}}
 {{MAA Notice}}

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