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2015 AMC 10B Problems/Problem 3

Revision as of 21:20, 3 March 2015 by Tkhalid (talk | contribs) (Created page with "Let the first number be <math>x</math> and the second be <math>y</math>. We have <math>2x+3y=100</math>. We are given one of the numbers is 28. If <math>x</math> were to be 28...")
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Let the first number be $x$ and the second be $y$. We have $2x+3y=100$. We are given one of the numbers is 28. If $x$ were to be 28, $y$ would not be an integer, thus $y=28$. $2x+3(28)=100$, solving gives $x=8$, so the answer is $\boxed{\textbf{(A)} 8}$.

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