Difference between revisions of "2012 AMC 10B Problems/Problem 4"

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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
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== Solution ==
  
  
 
In total, there were <math>3+4=7</math> marbles left from both Ringo and Paul. <math>7/6</math>=1R1. <math>\text{This means that that there is}</math>                <math> \boxed{1}</math> <math>\text{marbles left}</math> or  
 
In total, there were <math>3+4=7</math> marbles left from both Ringo and Paul. <math>7/6</math>=1R1. <math>\text{This means that that there is}</math>                <math> \boxed{1}</math> <math>\text{marbles left}</math> or  
  
['''A''']
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<math> \textbf{(A)}</math>

Revision as of 23:27, 23 February 2012

Problem 4

When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$


Solution

In total, there were $3+4=7$ marbles left from both Ringo and Paul. $7/6$=1R1. $\text{This means that that there is}$ $\boxed{1}$ $\text{marbles left}$ or

$\textbf{(A)}$