2014 AMC 10A Problems/Problem 6

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 by ahaanomegas)

Solution 2

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 10A Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also