Difference between revisions of "2006 AMC 10B Problems/Problem 13"
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== Problem == | == Problem == | ||
+ | Joe and JoAnn each bought <math>12</math> ounces of coffee in a <math>16</math> ounce cup. Joe drank <math>2</math> ounces of his coffee and then added <math>2</math> ounces of cream. JoAnn added <math>2</math> ounces of cream, stirred the coffee well, and then drank <math>2</math> ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? | ||
+ | |||
+ | <math> \textbf{(A) } \frac{6}{7}\qquad \textbf{(B) } \frac{13}{14}\qquad \textbf{(C) }1 \qquad \textbf{(D) \ } \frac{14}{13}\qquad \textbf{(E) } \frac{7}{6} </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | After drinking and adding cream, Joe's cup has <math>2</math> ounces of cream. | ||
+ | |||
+ | After adding cream to her cup, JoAnn's cup had <math>14</math> ounces of liquid. By stirring and then drinking <math>2</math> ounces out of the <math>14</math> ounces of liquid, she drank | ||
+ | <math>\frac{2}{14}=\frac{1}{7}</math>th of the cream. So there are <math>2\cdot\frac{6}{7}=\frac{12}{7}</math> ounces of cream left. | ||
+ | |||
+ | So the desired ratio is: <math>2 \div \frac{12}{7}= \boxed{\textbf{(E) }\frac{7}{6}}</math>. | ||
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== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=12|num-a=14}} | |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:11, 26 January 2022
Problem
Joe and JoAnn each bought ounces of coffee in a
ounce cup. Joe drank
ounces of his coffee and then added
ounces of cream. JoAnn added
ounces of cream, stirred the coffee well, and then drank
ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?
Solution
After drinking and adding cream, Joe's cup has ounces of cream.
After adding cream to her cup, JoAnn's cup had ounces of liquid. By stirring and then drinking
ounces out of the
ounces of liquid, she drank
th of the cream. So there are
ounces of cream left.
So the desired ratio is: .
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 |
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