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

Revision as of 12:07, 17 April 2021

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

Solution 6

In the right formula, plugging in $d=a$ and $e=c$ should simplify to $b$ , as if it doesn't, we'd essentially be saying " $a$ cows give $b$ gallons of milk in $c$ days, but $a$ cows don't give $b$ gallons of milk in $c$ days." The only one of the answer choices that simplifies like this is $\boxed{\textbf{(A)}}$

Solution 7

Call the rate at which a single cow produces milk $r$ . Using the given information we have $rac=b$ , and we want to find $rde$ . We solve for $r$ in the first equation to get $r=b*\tfrac{1}{ac}$ . Plugging in this value for $r$ into the expression we want, we get that $rde=\frac{bde}{ac}$ , or

Video Solution

https://youtu.be/OW4rHUTPgPA

~savannahsolver

@@ Line 36: / Line 36: @@
 ==Solution 7==
-Call the rate at which one cow produces milk <math>r</math>. Using the given information we have <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\frac{ac}</math>
+Call the rate at which a single cow produces milk <math>r</math>. Using the given information we have <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*\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
 ==Video Solution==

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

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