Difference between revisions of "2014 AMC 10A Problems/Problem 6"
m (→Solution 3) |
Mathfun1000 (talk | contribs) m (Minor Fixes to a few solutions) |
||
(13 intermediate revisions by 6 users not shown) | |||
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== | ||
Line 19: | Line 19: | ||
==Solution 3== | ==Solution 3== | ||
− | We see that | + | We see that the number of cows is inversely proportional to the number 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> | 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> | ||
+ | |||
+ | ==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 <cmath>\implies\text{rate}=\dfrac{b}{ac}</cmath> | ||
+ | |||
+ | 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 <cmath>\implies\boxed{\textbf{(A) } \dfrac{bde}{ac}}</cmath> | ||
+ | ==Solution 5== | ||
+ | If <math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days, that means that one cow will give <math>\frac{b}{a}</math> gallons of milk in <math>c</math> days. Also, we want to find the number of gallons of milk will <math>d</math> cows give in <math>e</math> days, so in <math>\frac{e}{c}</math> days <math>d</math> cows give <math>\frac{bd}{a}</math> gallons of milk. Multiplying with the formula <math>d=rt</math>, we get <math>\boxed{\textbf{(A) } \dfrac{bde}{ac}}</math> | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | In the right formula, plugging in <math>d=a</math> and <math>e=c</math> should simplify to <math>b</math>, as if it doesn't, we'd essentially be saying "<math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days, but <math>a</math> cows don't give <math>b</math> gallons of milk in <math>c</math> days." The only one of the answer choices that simplifies like this is <math>\boxed{\textbf{(A)}}</math> | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | Call the rate at which a single cow produces milk <math>r</math>. Using the given information we get that <math>rac=b</math>, and we want to find <math>rde</math>. We solve for <math>r</math> in the first equation to get <math>r=b \cdot \tfrac{1}{ac}</math>. Plugging in this value for <math>r</math> into the expression we want, we get that <math>rde=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A)}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/OW4rHUTPgPA | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 10:03, 7 September 2021
- 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?
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 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
Solution 5
If cows give gallons of milk in days, that means that one cow will give gallons of milk in days. Also, we want to find the number of gallons of milk will cows give in days, so in days cows give gallons of milk. Multiplying with the formula , we get
Solution 6
In the right formula, plugging in and should simplify to , as if it doesn't, we'd essentially be saying " cows give gallons of milk in days, but cows don't give gallons of milk in days." The only one of the answer choices that simplifies like this is
Solution 7
Call the rate at which a single cow produces milk . Using the given information we get that , and we want to find . We solve for in the first equation to get . Plugging in this value for into the expression we want, we get that , or .
Video Solution
~savannahsolver
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.