2014 AMC 10A Problem 6

Find the output of a single cow in a single day. Since $a$ cows produce $b$ gallons in $c$ days, one cow produces $\frac{b}{a}$ gallons in $c$ days, hence $\frac{b}{ac}$ gallons per day.

Milk output scales with both the number of cows and the number of days, so $d$ cows in $e$ days produce
$$d \cdot e \cdot \frac{b}{ac} = \frac{bde}{ac}.$$

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

"At this rate" signals a joint proportion: the quantity produced is proportional to each of the inputs (cows, days). Normalizing to one unit of each input gives a rate you can rescale to any other combination. The same template handles workers-and-hours and machines-and-widgets problems.

Plug in numbers: let $a = 2$ cows give $b = 6$ gallons in $c = 3$ days (so each cow gives $1$ gallon per day). Then $d = 4$ cows in $e = 5$ days give $20$ gallons. Only choice (A) yields $\frac{6 \cdot 4 \cdot 5}{2 \cdot 3} = 20$.

• Inverting the rate (choosing ac/(bde)) or putting c in the numerator because 'more days' feels like it should multiply.
• Testing the choices with values like $a = b = c = 1$, which cannot distinguish the formulas.