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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>
  
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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's <math>\textit{Introduction to Algebra}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 20:57, 4 November 2017

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's $\textit{Introduction to Algebra}$

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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