If $x$ cows give $x+1$ cans of milk in $x+2$ days, how many days will it take $x+3$ cows to give $x+5$ cans of milk?
$\textbf{(A) }\frac{x(x+2)(x+5)}{(x+1)(x+3)}\qquad \textbf{(B) }\frac{x(x+1)(x+5)}{(x+2)(x+3)}\qquad\\ \textbf{(C) }\frac{(x+1)(x+3)(x+5)}{x(x+2)}\qquad \textbf{(D) }\frac{(x+1)(x+3)}{x(x+2)(x+5)}\qquad \\ \textbf{(E) }\text{none of these}$
First, the problem states:

 $x$ cows $\qquad x+1$ cans $\qquad x+2$ days 
Multiply the current number of cows by $\frac{x+3}{x}$ to get the number of cows you want, and divide the number of days by the same amount because an increase in cows will cause a decrease in time.

 $x \cdot \frac{x+3}{x} \qquad x+1 \qquad (x+2) \cdot \frac{x}{x+3}$
Finally, multiply the current amount of cans by $\frac{x+5}{x+1}$ to get the number of cans you want, and multiply the number of days by the same amount because an increase in cans will cause an increase in time.

 $x \cdot \frac{x+3}{x} \qquad (x+1) \cdot \frac{x+5}{x+1} \qquad (x+2) \cdot \frac{x}{x+3} \cdot \frac{x+5}{x+1}$ 
Therefore, our answer is $\boxed{\textbf{(A) }\frac{x(x+2)(x+5)}{(x+1)(x+3)}}$ ~[jiang147369](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Jiang147369)