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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− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 2|Next Problem]] | ||
+ | * [[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?
Solution
Let be the length and be the width. We have that . works, so is our answer.