Problem 133

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