Difference between revisions of "2014 AMC 10A Problems/Problem 6"
(→Solution 4) |
MathDragon2 (talk | contribs) m (→Solution 1) |
||
Line 9: | Line 9: | ||
==Solution 1== | ==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)} \frac{bde}{ac}}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 17:49, 31 December 2017
- The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6, so both problems redirect to this page.
Problem
Suppose that cows give gallons of milk in days. At this rate, how many gallons of milk will cows give in days?
Solution 1
We need to multiply by for the new cows and for the new time, so the answer is , or .
Solution 2
We plug in , , , , and . Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"
If 2 cows give 3 gallons of milk in 4 days, then 2 cows give gallons of milk in 1 day, so 1 cow gives gallons in 1 day. This means that 5 cows give gallons of milk in 1 day. Finally, we see that 5 cows give gallons of milk in 6 days. Substituting our values for the variables, this becomes , which is .
Solution 3
We see that the the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is .
Let be the answer to the question. We have
Solution 4
The problem specifics "rate," so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days
Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days
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.