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

Revision as of 18:54, 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

@@ Line 9: / Line 9: @@
 ==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}}
 {{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 18:54, 4 March 2015

Problem

Solution

See Also