Difference between revisions of "2014 AMC 10A Problems/Problem 6"

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==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>.
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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==
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==Solution 3==
 
==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 <math>\dfrac{ac}{b}</math>.
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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>
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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>
 
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\dfrac{bde}{ac}\implies\boxed{\textbf{(A)}}</cmath>
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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>
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==Solution 5==
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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{(A)\frac{bde}{ac}}</math>
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==Video Solution==
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https://youtu.be/OW4rHUTPgPA
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 14:47, 30 June 2020

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

$\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}$

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 $\boxed{\textbf{(A)} \frac{bde}{ac}}$.

Solution 2

We plug in $a=2$, $b=3$, $c=4$, $d=5$, and $e=6$. 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 $\frac{3}{4}$ gallons of milk in 1 day, so 1 cow gives $\frac{3}{4\cdot2}$ gallons in 1 day. This means that 5 cows give $\frac{5\cdot3}{4\cdot2}$ gallons of milk in 1 day. Finally, we see that 5 cows give $\frac{5\cdot3\cdot6}{4\cdot2}$ gallons of milk in 6 days. Substituting our values for the variables, this becomes $\frac{dbe}{ac}$, which is $\boxed{\textbf{(A)}\ \frac{bde}{ac}}$.

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 $\dfrac{ac}{b}$.

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

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 \[\implies\text{rate}=\dfrac{b}{ac}\]

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 \[\implies\boxed{\textbf{(A)} \dfrac{bde}{ac}}\]

Solution 5

If $a$ cows give $b$ gallons of milk in $c$ days, that means that one cow will give $\frac{b}{a}$ gallons of milk in $c$ days. Also, we want to find the number of gallons of milk will $d$ cows give in $e$ days, so in $\frac{e}{c}$ days $d$ cows give $\frac{bd}{a}$ gallons of milk. Multiplying with the formula $d=rt$, we get $\boxed{(A)\frac{bde}{ac}}$

Video Solution

https://youtu.be/OW4rHUTPgPA

~savannahsolver

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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