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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 7"
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<math>a+b+c+d+e=4</math>
 
<math>a+b+c+d+e=4</math>
 
Adding all the equations gives us <math>\displaystyle 5 (a + b + c + d + e) = 28 \Longrightarrow a + b + c + d + e = 5.6 </math>.
 
Adding all the equations gives us <math>\displaystyle 5 (a + b + c + d + e) = 28 \Longrightarrow a + b + c + d + e = 5.6 </math>.
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 6|Previous Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 8|Next Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 8|Next Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 12:27, 23 July 2006

Problem

Each card below covers up a number. The number written below each card is the sum of all the numbers covered by all of the other cards. What is the sum of all of the hidden numbers?

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

$\mathrm{(A) \ }4.2 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ }5.6 \qquad \mathrm{(D) \ }6.2 \qquad \mathrm{(E) \ }6.8$

Solution

If we call the squares $a,b,c,d,e,f$ (in order from left to right), we have: $b+c+d+e+f=3$ $a+c+d+e+f=8$ $a+b+d+e+f=5$ $a+b+c+e+f=6$ $a+b+c+d+f=2$ $a+b+c+d+e=4$ Adding all the equations gives us $\displaystyle 5 (a + b + c + d + e) = 28 \Longrightarrow a + b + c + d + e = 5.6$.

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