Difference between revisions of "2006 AMC 12B Problems/Problem 11"

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Joe has 2 ounces of cream, as stated in the problem.
 
Joe has 2 ounces of cream, as stated in the problem.
  
JoAnn had 14 ounces of liquid, and drank <math>\frac{1}{7}</math> of it. Therefore, she drank <math>\frac{1}{7}</math> of her cream, giving her <math>2*\frac{6}{7}</math>.
+
JoAnn had 14 ounces of liquid, and drank <math>\frac{1}{7}</math> of it. Therefore, she drank <math>\frac{1}{7}</math> of her cream, leaving her <math>2*\frac{6}{7}</math>.
  
 
<math>\frac{2}{2*\frac{6}{7}}=\frac{7}{6} \Rightarrow \boxed{\text{(E)}}</math>
 
<math>\frac{2}{2*\frac{6}{7}}=\frac{7}{6} \Rightarrow \boxed{\text{(E)}}</math>
 +
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|num-b=10|num-a=12|ab=B}}
 
{{AMC12 box|year=2006|num-b=10|num-a=12|ab=B}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 10:55, 3 December 2007

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?

$\text {(A) } \frac 67 \qquad \text {(B) } \frac {13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac {14}{13} \qquad \text {(E) } \frac 76$

Solution

Joe has 2 ounces of cream, as stated in the problem.

JoAnn had 14 ounces of liquid, and drank $\frac{1}{7}$ of it. Therefore, she drank $\frac{1}{7}$ of her cream, leaving her $2*\frac{6}{7}$.

$\frac{2}{2*\frac{6}{7}}=\frac{7}{6} \Rightarrow \boxed{\text{(E)}}$

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 12 Problems and Solutions
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