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 <math>100</math>, and one of the numbers is <math>28</math> | + | 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> | ||
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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>. | 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>. | ||
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| + | ==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|num-b=2|num-a=4}} | {{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. 
