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?
![$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 6 \qquad \mathrm{(E) \ } 9$](http://latex.artofproblemsolving.com/b/f/3/bf3cf45fe611d0698538a185c14d41a0d91220e5.png)
Solution
Let be the length and
be the width. We have that
.
works, so
is our answer.