Difference between revisions of "2014 AMC 12A Problems/Problem 4"

(Solution)
(Solution)
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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>
 
<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>
  
==Solution==
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==Solution 1==
  
 
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>.
 
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>.
  
 
(Solution by ahaanomegas)
 
(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>

Revision as of 19:02, 7 February 2014

Problem

Suppose that $a$ cows give $b$ gallons of milk in $c$ days. At this rate, how many gallons of milk will $d$ cows give in $e$ days?

$\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}$ (Error compiling LaTeX. Unknown error_msg)

Solution 1

We need to multiply $b$ by $\frac{d}{a}$ for the new cows and $\frac{e}{c}$ for the new time, so the answer is $b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}$, or $\boxed{\textbf{(A)}}$.

(Solution by ahaanomegas)

Solution 2

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 $\dfrac{ac}{b}$.

Let $g$ be the answer to the question. We have $\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}$