Difference between revisions of "2014 AMC 10A Problems/Problem 6"
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==Solution== | ==Solution== | ||
| + | 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>. | ||
| + | |||
| + | 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> | ||
==See Also== | ==See Also== | ||
Revision as of 23:32, 6 February 2014
Problem
Suppose that 
cows give 
gallons of milk in 
days. At this rate, how many gallons of milk will 
cows give in 
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
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 
.
Let 
be the answer to the question. We have 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AMC 10 Problems and Solutions | ||
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