Difference between revisions of "2015 AMC 10B Problems/Problem 3"
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==Problem== | ==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? | + | 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== | ||
| − | 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, <math>y</math> would not be an integer, thus <math>y=28</math>. <math>2x+3(28)=100</math>, | + | 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 <math>28</math>. If <math>x</math> were to be <math>28</math>, <math>y</math> would not be an integer, thus <math>y=28</math>. <math>2x+3(28)=100</math>, which gives <math>x=\boxed{\textbf{(A) }8}</math>. |
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/W4eSkC1O-Nk | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/KrlMrXVNKTM | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | {{AMC10 box|year=2015|ab=B| | + | {{AMC10 box|year=2015|ab=B|num-b=2|num-a=4}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:37, 26 March 2024
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 
. If were to be , would not be an integer, thus 
. 
, which gives 
.
Video Solution 1
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
