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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 1"

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== Solution ==
 
== Solution ==
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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=6b</math>.  Dividing by <math>b</math> give <math>a=6</math> so <math>\mathrm{(D)}</math> is our answer.
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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=6b</math>.  Dividing by <math>b</math> yields <math>a=6</math> so <math>\mathrm{(D)}</math> is our answer.
  
 
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Latest revision as of 03:03, 1 November 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 $3ab=2b(a+3) \Longrightarrow ab=6b$. Dividing by $b$ yields $a=6$ so $\mathrm{(D)}$ is our answer.

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