2015 AMC 10B Problems/Problem 3

Problem

Kaashish 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 $y=28$ . $2x+3(28)=100$ , which gives $x=\boxed{\textbf{(A) }8}$ .

Video Solution 1

https://youtu.be/W4eSkC1O-Nk

~Education, the Study of Everything

Video Solution

https://youtu.be/KrlMrXVNKTM

~savannahsolver

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.

Page

Toolbox

Search

2015 AMC 10B Problems/Problem 3

Contents

Problem

Solution

Video Solution 1

Video Solution

See Also