Difference between revisions of "2000 AMC 12 Problems/Problem 13"
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We also know that the amount Angela drank, which is <math>\frac{c}{6} + \frac{m}{4},</math> is equal to <math>8</math> ounces, thus <math>\frac{c}{6} + \frac{m}{4} = 8.</math> Rearranging gives <cmath>24p - c = 96.</cmath> Now notice that <math>c > 0</math> (by the problem statement). In addition, <math>m > 0,</math> so <math>c = 8p-m < 8p.</math> Therefore, <math>0 < c < 8p,</math> and so <math>24p > 24p-c > 16p.</math> We know that <math>24p-c = 96,</math> so <cmath>24p > 96 > 16p.</cmath> From the leftmost inequality, we get <math>p > 4,</math> and from the rightmost inequality, we get <math>p < 6.</math> The only possible value of <math>p</math> is <math>p = 5\ \mathrm{(C)}</math>. | We also know that the amount Angela drank, which is <math>\frac{c}{6} + \frac{m}{4},</math> is equal to <math>8</math> ounces, thus <math>\frac{c}{6} + \frac{m}{4} = 8.</math> Rearranging gives <cmath>24p - c = 96.</cmath> Now notice that <math>c > 0</math> (by the problem statement). In addition, <math>m > 0,</math> so <math>c = 8p-m < 8p.</math> Therefore, <math>0 < c < 8p,</math> and so <math>24p > 24p-c > 16p.</math> We know that <math>24p-c = 96,</math> so <cmath>24p > 96 > 16p.</cmath> From the leftmost inequality, we get <math>p > 4,</math> and from the rightmost inequality, we get <math>p < 6.</math> The only possible value of <math>p</math> is <math>p = 5\ \mathrm{(C)}</math>. | ||
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+ | ==Sidenote== | ||
+ | If we now solve for <math>c</math> and <math>m</math>, we find that <math>m=16</math> and <math>c=24</math>. Thus in total the family drank <math>16</math> ounces of milk and <math>24</math> ounces of coffee. Angela drank exactly <math>4</math> ounces of milk and <math>4</math> ounces of coffee. | ||
== See also == | == See also == |
Revision as of 16:30, 28 December 2015
- The following problem is from both the 2000 AMC 12 #13 and 2000 AMC 10 #22, so both problems redirect to this page.
Contents
Problem
One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
Solution
Solution 1:
Let be the total amount of coffee, of milk, and the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so Regrouping, we get . Since both are positive, it follows that and are also positive, which is only possible when .
Solution 2 (less rigorous):
One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has ounces. The "fair share" is So,
Which requires that be since is a whole number.
Solution 3:
Again, let and be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is and also Thus, so and
We also know that the amount Angela drank, which is is equal to ounces, thus Rearranging gives Now notice that (by the problem statement). In addition, so Therefore, and so We know that so From the leftmost inequality, we get and from the rightmost inequality, we get The only possible value of is .
Sidenote
If we now solve for and , we find that and . Thus in total the family drank ounces of milk and ounces of coffee. Angela drank exactly ounces of milk and ounces of coffee.
See also
2000 AMC 12 (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 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.