2014 AMC 10A Problem 6

The simplest amount to work with is what one cow gives in one day. Find that first, then scale up.

1. $a$ cows give $b$ gallons in $c$ days.
2. One cow gives one $a$-th of that: $\frac{b}{a}$ gallons in $c$ days.
3. In one day that cow gives one $c$-th of that: $\frac{b}{ac}$ gallons.
4. Now scale up. $d$ cows give $d$ times as much, and $e$ days give $e$ times as much:
$$d \cdot e \cdot \frac{b}{ac} = \frac{bde}{ac}.$$

To check, try numbers. Say $2$ cows give $6$ gallons in $3$ days, so one cow gives $1$ gallon a day. Then $4$ cows in $5$ days give $20$ gallons, and the formula gives $\frac{6 \cdot 4 \cdot 5}{2 \cdot 3} = \frac{120}{6} = 20$.

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

"At this rate" means the milk grows in step with both the number of cows and the number of days. Bringing everything down to one cow for one day gives a single rate, and from there any number of cows and days is one multiplication. The same method works for workers and hours, or machines and parts.

Skip the algebra and test numbers against the choices. With $a = 2$, $b = 6$, $c = 3$, $d = 4$, $e = 5$, the true answer is $20$ gallons (as worked out above). Choice (A) gives $20$. Choice (B) gives $\frac{6}{120} = \frac{1}{20}$, choice (C) gives $\frac{240}{3} = 80$, choice (D) gives $\frac{360}{2} = 180$, and choice (E) gives $\frac{36}{20}$. Only (A) matches.