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Difference between revisions of "2009 AMC 8 Problems/Problem 15"

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==Solution==
 
==Solution==
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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. Limited by the amount of milk, Jordan can make at most <math>\boxed{\textbf{(D)}\ 8 \frac34}</math> servings.
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Assuming 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. 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:10, 3 September 2019

Problem

A recipe that makes $5$ servings of hot chocolate requires $2$ squares of chocolate, $\frac{1}{4}$ cup sugar, $1$ cup water and $4$ cups milk. Jordan has $5$ squares of chocolate, $2$ cups of sugar, lots of water, and $7$ cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?


$\textbf{(A)}\ 5 \frac18 \qquad \textbf{(B)}\ 6\frac14 \qquad \textbf{(C)}\ 7\frac12 \qquad \textbf{(D)}\ 8 \frac34 \qquad \textbf{(E)}\ 9\frac78$

Solution

Assuming excesses of the other ingredients, the chocolate can make $\frac52 \cdot 5=12.5$ servings, the sugar can make $\frac{2}{1/4} \cdot 5 = 40$ servings, the water can make unlimited servings, and the milk can make $\frac74 \cdot 5 = 8.75$ servings. Limited by the amount of milk, Jordan can make at most $\boxed{\textbf{(D)}\ 8 \frac34}$ 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

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