2002 AMC 12A Problems/Problem 10
- 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?
Solution
Solution 1
We will simulate the process in steps.
In the beginning, we have:
- ounces of coffee in cup
- ounces of cream in cup
In the first step we pour ounces of coffee from cup to cup , getting:
- ounces of coffee in cup
- ounces of coffee and ounces of cream in cup
In the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:
- ounces of coffee and ounces of cream in cup
- the rest in cup
Hence at the end we have ounces of liquid in cup , and out of these ounces is cream. Thus the answer is .
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. , 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 .
~MathKatana
Video Solution
https://www.youtube.com/watch?v=3Zri7R40wFM ~David
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.