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Difference between revisions of "2015 AMC 10B Problems/Problem 3"
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==Problem==
 
==Problem==
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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?
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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?
  
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<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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<math>\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18</math>
  
 
==Solution==
 
==Solution==
  
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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 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>, solving gives <math>x=8</math>, so the answer is <math>\boxed{\textbf{(A)} 8}</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>.
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2015|ab=B|before=Problem 2|num-a=3}}
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{{AMC10 box|year=2015|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:52, 5 March 2015

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?

$\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18$

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 $x=\boxed{\textbf{(A) }8}$.

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

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