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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 1"
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Let <math>a</math> be the length and <math>b</math> be the width. We have that <math>3ab=2b(a+3) \Longrightarrow ab=6</math>. <math>a=6, b=1</math> works, so <math>6</math> is our answer.
 
Let <math>a</math> be the length and <math>b</math> be the width. We have that <math>3ab=2b(a+3) \Longrightarrow ab=6</math>. <math>a=6, b=1</math> works, so <math>6</math> is our answer.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 2|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 12:31, 23 July 2006

Problem

If the width of a particular rectangle is doubled and the length is increased by 3, then the area is tripled. What is the length of the rectangle?

$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 6 \qquad \mathrm{(E) \ } 9$

Solution

Let $a$ be the length and $b$ be the width. We have that $3ab=2b(a+3) \Longrightarrow ab=6$. $a=6, b=1$ works, so $6$ is our answer.

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