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

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{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #10]] and [[2002 AMC 10A Problems|2002 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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===Solution 1===
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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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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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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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===Solution 2===
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Let's consider this in steps.
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We have 4 ounces of coffee in the first cup.
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We hace 4 ounces of cream in the second cup.
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We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).
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We then take half of the second cup and pour it into the first cup. <math>6/2=3</math>, so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.
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Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces.
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In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is <math>\boxed{\text{(D) } \frac 25}</math>.
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~MathKatana
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==Video Solution==
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https://www.youtube.com/watch?v=3Zri7R40wFM    ~David
  
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==Video Solution by Daily Dose of 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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https://youtu.be/H8A6o_O6SRY
  
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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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~Thesmartgreekmathdude
  
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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]]
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{{MAA Notice}}

Latest revision as of 17:40, 15 October 2024

The following problem is from both the 2002 AMC 12A #10 and 2002 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

Solution 1

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 $\boxed{\text{(D) } \frac 25}$.

Solution 2

Let's consider this in steps. We have 4 ounces of coffee in the first cup. We hace 4 ounces of cream in the second cup. We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely). We then take half of the second cup and pour it into the first cup. $6/2=3$, so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture. Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{\text{(D) } \frac 25}$.

~MathKatana

Video Solution

https://www.youtube.com/watch?v=3Zri7R40wFM ~David

Video Solution by Daily Dose of Math

https://youtu.be/H8A6o_O6SRY

~Thesmartgreekmathdude

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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