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Difference between revisions of "2002 AMC 12A Problems/Problem 10"

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{{duplicate|[[2002 AMC 12A Problems|2009 AMC 12A #10]] and [[2002 AMC 10A Problems|2009 AMC 10A #17]]}}
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== Problem ==
 
== Problem ==
 
Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
 
Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
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== Solution ==
 
== Solution ==
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We take this problem step by step:
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We will simulate the process in steps.
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In the beginning, we have:
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* <math>4</math> ounces of coffee in cup <math>1</math>
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* <math>4</math> ounces of cream in cup <math>2</math>
  
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In the first step we pour <math>4/2=2</math> ounces of coffee from cup <math>1</math> to cup <math>2</math>, getting:
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* <math>2</math> ounces of coffee in cup <math>1</math>
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* <math>2</math> ounces of coffee and <math>4</math> ounces of cream in cup <math>2</math>
  
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Step 0: We begin with 4 ounces of coffee in cup 1 and 4 ounces of cream in cup 2
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In the second step we pour <math>2/2=1</math> ounce of coffee and <math>4/2=2</math> ounces of cream from cup <math>2</math> to cup <math>1</math>, getting:
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* <math>2+1=3</math> ounces of coffee and <math>0+2=2</math> ounces of cream in cup <math>1</math>
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* the rest in cup <math>2</math>
  
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Step 1: We now have 2 ounces of coffee in cup 1 and 4 ounces of cream and 2 ounces of coffee in cup 2.
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Hence at the end we have <math>3+2=5</math> ounces of liquid in cup <math>1</math>, and out of these <math>2</math> ounces is cream. Thus the answer is <math>\boxed{\text{(D) } \frac 25}</math>.
  
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Step 2: We now have <math>2+\frac22=3</math> ounces of coffee and <math>\frac42=2</math> of cream in cup 1 (and two ounces of cream and one of coffee in cup 2, though this does not matter).
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== See Also ==
  
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{{AMC12 box|year=2002|ab=A|num-b=9|num-a=11}}
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{{AMC10 box|year=2002|ab=A|num-b=16|num-a=18}}
  
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So 2 out of the 5 ounces of liquid in cup 1 is cream, so the answer is <math>\frac25</math>, or <math>\mathrm{D}</math>.
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[[Category:Introductory Algebra Problems]]

Revision as of 07:04, 18 February 2009

The following problem is from both the 2009 AMC 12A #10 and 2009 AMC 10A #17, so both problems redirect to this page.


Problem

Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$

Solution

We will simulate the process in steps.

In the beginning, we have:

  • $4$ ounces of coffee in cup $1$
  • $4$ ounces of cream in cup $2$

In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$, getting:

  • $2$ ounces of coffee in cup $1$
  • $2$ ounces of coffee and $4$ ounces of cream in cup $2$

In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$, getting:

  • $2+1=3$ ounces of coffee and $0+2=2$ ounces of cream in cup $1$
  • the rest in cup $2$

Hence at the end we have $3+2=5$ ounces of liquid in cup $1$, and out of these $2$ ounces is cream. Thus the answer is $\boxed{\text{(D) } \frac 25}$.

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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