Difference between revisions of "2015 AMC 10B Problems/Problem 3"

Revision as of 17:54, 4 March 2015

Problem

Isaac has written down one integer two times and another integer three times. The sum of the five numbers is $100$ , and one of the numbers is $28$ . What is the other number?

$\textbf{(A)} 8\qquad\textbf{(B)} 11\qquad\textbf{(C)} 14\qquad\textbf{(D)} 15\qquad\textbf{(E)} 18$

Solution

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 $\boxed{\textbf{(A)} 8}$ .

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC10 box|year=2015|ab=B|before=Problem 2|num-a=4}}
+{{AMC10 box|year=2015|ab=B|num-b=2|num-a=4}}
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Difference between revisions of "2015 AMC 10B Problems/Problem 3"

Revision as of 17:54, 4 March 2015

Problem

Solution

See Also