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

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==Problem==
 
==Problem==
  
A recipe that makes <math> 5</math> servings of hot chocolate requires <math> 2</math> squares of chocolate, <math> \frac{1}{4}</math> cup sugar, <math> 1</math> cup water and <math> 4</math> cups milk. Jordan has <math> 5</math> squares of chocolate, <math> 2</math> cups of sugar, lots of water and <math> 7</math> cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?
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A recipe that makes <math> 5</math> servings of hot chocolate requires <math> 2</math> squares of chocolate, <math> \frac{1}{4}</math> cup sugar, <math> 1</math> cup water and <math> 4</math> cups milk. Jordan has <math> 5</math> squares of chocolate, <math> 2</math> cups of sugar, lots of water, and <math> 7</math> cups of milk. If he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he can make?
  
  
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\textbf{(D)}\  8 \frac34  \qquad
 
\textbf{(D)}\  8 \frac34  \qquad
 
\textbf{(E)}\  9\frac78</math>
 
\textbf{(E)}\  9\frac78</math>
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==Solution==
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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.
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==See Also==
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{{AMC8 box|year=2009|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 18:53, 20 August 2020

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 he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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