Difference between revisions of "2009 AMC 8 Problems/Problem 15"
| Line 9: | Line 9: | ||
\textbf{(D)}\ 8 \frac34 \qquad | \textbf{(D)}\ 8 \frac34 \qquad | ||
\textbf{(E)}\ 9\frac78</math> | \textbf{(E)}\ 9\frac78</math> | ||
| + | |||
| + | ==Solution== | ||
| + | Asumming excesses of the other ingredients, the chocolate can make <math>\frac52 \cdot 5=12.5</math> servings, the sugar can make <math>\frac{2}{1/4} \cdot 5 = 40</math> servings, the water can make unlimited servings, and the milk can make <math>\frac74 \cdot 5 = 8.75</math> servings. The Limited by the amount of milk, Jordan can make at most <math>\boxed{\textbf{(D)}\ 8 \frac34}</math> servings. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=14|num-a=16}} | {{AMC8 box|year=2009|num-b=14|num-a=16}} | ||
Revision as of 16:15, 25 December 2012
Problem
A recipe that makes 
servings of hot chocolate requires 
squares of chocolate, 
cup sugar, 
cup water and 
cups milk. Jordan has squares of chocolate, cups of sugar, lots of water and 
cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?
Solution
Asumming excesses of the other ingredients, the chocolate can make 
servings, the sugar can make 
servings, the water can make unlimited servings, and the milk can make 
servings. The Limited by the amount of milk, Jordan can make at most 
servings.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
