Difference between revisions of "2006 AMC 10B Problems/Problem 13"

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So the desired ratio is: <math> \frac{2}{\frac{12}{7}} = \frac{7}{6} \Rightarrow E </math>
 
So the desired ratio is: <math> \frac{2}{\frac{12}{7}} = \frac{7}{6} \Rightarrow E </math>
  
This is problem 6.2.5 in "the Art of Problem Solving, Introduction to Algebra".
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This is problem 6.2.5 in ''the Art of Problem Solving, Introduction to Algebra''
  
 
== See Also ==
 
== See Also ==

Revision as of 17:09, 29 December 2013

Problem

Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?

$\mathrm{(A) \ } \frac{6}{7}\qquad \mathrm{(B) \ } \frac{13}{14}\qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ } \frac{14}{13}\qquad \mathrm{(E) \ } \frac{7}{6}$

Solution

After drinking and adding cream, Joe's cup has $2$ ounces of cream.

After adding cream to her cup, JoAnn's cup had $14$ ounces of liquid. By drinking $2$ ounces out of the $14$ ounces of liquid, she drank $\frac{2}{14}=\frac{1}{7}$th of the cream. So there is $2\cdot\frac{6}{7}=\frac{12}{7}$ ounces of cream left.

So the desired ratio is: $\frac{2}{\frac{12}{7}} = \frac{7}{6} \Rightarrow E$

This is problem 6.2.5 in the Art of Problem Solving, Introduction to Algebra

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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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