Test 8 Section 3 - #16 (page 835)

In a shaded region problem like this, it's almost always simpler to find the area of the entire figure, and then subtract the areas of the unshaded regions.  In this question, doing so is going to save us a bit of time, and keep us working with simpler numbers.

So, the area of the larger square (our entire figure) is 3², = 9

All four of the triangles around the shaded square are the same*, so we just need to find the area of one of them, and then multiply by 4.  The area of a triangle is: A = ½bh.  In our triangles, the base, b, is 1, and the height, h, is 2.

A = ½(1)(2) = 1

We've got 4 of them, so when we're subtracting the areas of the unshaded regions from the areas of the whole regions, it looks like this.

9 - 4(1) = 5

The area of the shaded square is 5.

Note: it's also possible to find any side of the square directly using the Pythagorean Theorem (see image below) since they're all hypotenuses of the right triangles.  I just like my way better (and it's a lot more universally useful).

* They have to be the same, because we know the shaded thing is a square, which means all its sides are the same.  So we have 4 right triangles with equal hypotenuses.  We can prove the congruency of their acute angles by extending like every single line in the diagram (a huge mess, but it works) and using the rules about parallel lines and transversals.  The right angles are also congruent (of course), so we have Angle-Angle-Side triangle congruency.  

Or, you could just guesstimate because this isn't the kind of thing the SAT tries to trick you about.  :)

Test 8 Section 3 - #18 (page 835)

Now, normally we'd love to use Plug-In on a question like this, what with the variables and all.  Unfortunately, because the question is asking us, in the end, for p/n, we can't exactly just make up p and n, you know?  This one is supposed to be hard, and it is.  That's why this is a #18!  Anyway, to solve, we're going to have to use the Average Table (yay!), with variables in it (yuck!).  Here goes:

[# of Numbers]	[Average]	[Sum]
p	70	70p
n	92	92n
p + n	86	70p + 92n

Remember, the Average Table is based on the fact that [# of Numbers] x [Average] = [Sum].  We can add and/or subtract up and down the outer columns, but we can only fill in the middle column based on what we know from the question, or what we can figure out from the outer columns.

We get the sums (with variables) to be 70p, and 92n.  We know that the total number of kids (both classes) is going to be p + n.  We know that their average is going to be 86.  And filling in from the chart, we know the sum that gets us there is 70p + 92n.

That's as far as we can get w/ the average table, but it's far enough.  Here's what I've got written down at this point:

(p + n) * 86 = (70p + 92n)

Remember, [# of Numbers] x [Average] = [Sum]

Then, we just do some algebra.  Distribute:

86p + 86n = 70p + 92n

combine the p's and n's:

6n = 16p

and...go for the gold.  They want p/n.  Let's give it to them.

p/n = 6/16

that simplifies to

p/n = 3/8, which is the answer.  Phew.