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| − | ==Problem==
| + | #REDIRECT[[2014 AMC 10A Problems/Problem 6]] |
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| − | Suppose that <math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days. At this rate, how many gallons of milk will <math>d</math> cows give in <math>e</math> days?
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| − | <math> \textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}</math>
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| − | ==Solution 1==
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| − | We need to multiply <math>b</math> by <math>\frac{d}{a}</math> for the new cows and <math>\frac{e}{c}</math> for the new time, so the answer is <math>b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A)}}</math>.
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| − | (Solution by ahaanomegas)
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| − | ==Solution 2==
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| − | We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is <math>\dfrac{ac}{b}</math>.
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| − | Let <math>g</math> be the answer to the question. We have <math>\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}</math>
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