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

Revision as of 20:26, 11 December 2014

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?

$\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}$ (Error compiling LaTeX. Unknown error_msg)

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 $a=2$ , $b=3$ , $c=4$ , $d=5$ , $e=6$ . 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?"

Well if 2 cows give 3 gallons of milk in 4 days. Then 2 cows give $\frac{3}{4}$ gallons of milk in 1 day. And then 1 cow gives $\frac{3}{4\cdot2}$ in 1 day. This means that 5 cows give $\frac{5\cdot3}{4\cdot2}$ gallons of milk in 1 day. Finally we get 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 $\boxed{\textbf{(A)}}$ .

Solution 3

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 $\dfrac{ac}{b}$ .

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

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.

@@ Line 18: / Line 18: @@
 Well if 2 cows give 3 gallons of milk in 4 days. Then 2 cows give <math>\frac{3}{4}</math> gallons of milk in 1 day. And then 1 cow gives <math>\frac{3}{4\cdot2}</math> in 1 day.  This means that 5 cows give <math>\frac{5\cdot3}{4\cdot2}</math> gallons of milk in 1 day. Finally we get that 5 cows give <math>\frac{5\cdot3\cdot6}{4\cdot2}</math> gallons of milk in 6 days. Substituting our values for the variables this becomes <math>\frac{dbe}{ac}</math> which is <math>\boxed{\textbf{(A)}}</math>.
-==Solution 2==
+==Solution 3==
 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 <math>\dfrac{ac}{b}</math>.

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