Difference between revisions of "2015 AMC 10B Problems/Problem 3"
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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>. | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/W4eSkC1O-Nk | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== |
Revision as of 17:10, 2 August 2022
Problem
Kaashish 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.