1. ## (SAT) Proportionality

If 500 pounds of mush will feed 20 pigs for a week, for how many days will 200 pounds of mush feed 14 pigs?

Inversely proportional.

$\displaystyle 500\ \times\ \frac{20}{7} = \frac{10 000}{7}$
$\displaystyle = 1428\frac{4}{7} \implies 1428\frac{4}{7}$ pig-days

$\displaystyle 200\ \times\ y\ days = 1428\frac{4}{7}$
$\displaystyle y = \frac{1428\frac{4}{7}}{200} = 7\frac{1}{7}$

2. $n_{pigs} \cdot n_{days} \cdot n_{lb/(day \cdot pig)} = n_{lb}$

Since the amount of mush eaten per pig per day is assumed to be constant then $20 \cdot 7 \cdot n_{lb/(day \cdot pig)} = 500$

$n_{lb/(day \cdot pig)} = \dfrac{500}{20 \cdot 7} = \dfrac{25}{7}$

Using this info:

$14 \cdot n_{days} \cdot \dfrac{25}{7} = 200$

Solve for n. I get 4 days which is around what we'd expect from intuition (not like 0.4, 40 and 25 which I got when trying to solve!)

3. First, you need to know how much a pig will eat each day.

If 500 lbs feeds 20 pigs for 7 days, then if we set up the ratio mush : day

$\displaystyle 500 : 7$

$\displaystyle \frac{500}{7} : 1$.

So 20 pigs will eat $\displaystyle \frac{500}{7}$ lbs of mush per day.

Now we can set up another ratio of mush : pigs

$\displaystyle \frac{500}{7} : 20$

$\displaystyle \frac{25}{7} : 1$

$\displaystyle 50:14$.

So 14 pigs will eat 50 lbs of mush per day.

Finally, setting up the ratio mush : days

$\displaystyle 50 : 1$

$\displaystyle 200 : 4$.

So 14 pigs will eat 200 lbs of mush in 4 days.

4. This is a tough question for the SAT. I would do it with a ratio (direct proportion) followed by an inverse proportion as follows:

First I'll answer this question: "If 500 pounds of mush will feed 20 pigs for a week, then 200 pounds of mush will feed how many pigs for a week?"

$\rm{mush}\ \ 500 \ 200$

$\rm{pigs} \ \ \ \ 20 \ \ \ x$

So $\frac{500}{20}=\frac{200}{x}$

$500x=4000$

$x=8$

Now I'll answer this question: "If you can feed 8 pigs for 7 days, then for how many days can you feed 14 pigs?"

$8\cdot 7=14x$

$56=14x$

$x=4$

The problem is particularly confusing because there are three active variables here. I don't believe I've ever seen this in an actual SAT problem (if it ever did occur you could probably get it wrong and still get an 800). I have seen an additional inactive variable by which I mean the third variable wasn't needed to solve the problem (it was just there to confuse you). I'm curious what book this problem came from.