Difference between revisions of "2015 AMC 10B Problems/Problem 3"
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Isaac has written down one integer two times and another integer three times. The sum of the five numbers is <math>100</math>, and one of the numbers is <math>28</math>. What is the other number? | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is <math>100</math>, and one of the numbers is <math>28</math>. What is the other number? | ||
| − | <math>\textbf{(A)} 8\qquad\textbf{(B)} 11\qquad\textbf{(C)} 14\qquad\textbf{(D)} 15\qquad\textbf{(E)} 18</math> | + | <math>\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18</math> |
==Solution== | ==Solution== | ||
Revision as of 12:51, 5 March 2015
Problem
Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 
, and one of the numbers is 
. What is the other number?
Solution
Let the first number be 
and the second be 
. We have 
. We are given one of the numbers is 28. If were to be 28, would not be an integer, thus 
. 
, solving gives 
, so the answer is 
.
See Also
| 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. 
