# 2014 AMC 10A Problems/Problem 6

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 $\boxed{\textbf{(A)}}$.

## 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 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 3

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 $$\implies\dfrac{bde}{ac}\implies\boxed{A}$$

## See Also

 2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 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 byProblem 3 Followed byProblem 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.

Invalid username
Login to AoPS