Difference between revisions of "2014 AMC 10A Problems/Problem 6"
Flamedragon (talk | contribs) (→See Also) |
Jskalarickal (talk | contribs) (→Solution 1) |
||
| Line 10: | Line 10: | ||
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 2== | ==Solution 2== | ||
Revision as of 21:29, 23 November 2014
- The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6, so both problems redirect to this page.
Contents
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 1
We need to multiply by 
for the new cows and 
for the new time, so the answer is 
, or 
.
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 
.
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 | ||
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
