Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 1"
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== Solution == | == Solution == | ||
− | 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> | + | 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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