Test 1 Section 7 - #17 (page 418)

They tell us these are overlapping equilateral triangles, and that CD, DE, and EF are all 10 inches long, so let's just fill in what we know first (the fact that they're equilateral and thus have angles that all equal 60° is very important here).

(This is totally not to scale.  Sorry.  Drawing with a mouse is hard!)  

Note that the little triangle in the middle is also equilateral.  We can use that to fill in all our distances.  The bases of the big triangles (CE and DF on your diagram) are both 20, so all the sides of the big triangles will be 20.  The smaller triangle will have sides of 10 inches each.  That means we're cutting the overlapping sides in half:

To find our perimeter and finish the question, just count up all the sides that are bolded.  They add up to 90.

Test 3 Section 5 - #15 (page 530)

This one's gonna get a little messy.  Here's our starting point:

Note that I've just gone ahead and put 40° in for x and 30° in for y.  If we just work in the order they tell us things, we're going to scribble all over our diagram before we can fill in any actual NUMBERS for our angle measures.  Instead, let's start with the last thing they tell us, since that'll allow us to start putting numbers in right away.  If OB bisects angle AOD, then:

Next, we can use the first thing they told us: OD bisects angle AOF.  That means angle DOF is going to be 80 as well:

And now we're almost there. Since y was 30°, we know angle DOE is 50°, so fill that in.  At this point, we kinda have everything we need.  The question asked us for angle BOE, which has to be the sum of DOE (which we just figured out is 50°) and BOD (which we already know is 40°):

In this last diagram, I've also gone ahead and filled in all the other angles just for practice (based on what they tell us about OC bisecting AOE, which we actually don't need).  Angle BOE is 90°.

Test 3 Section 5 - #16 (page 530)

Well, according to the way the problem is set up, we're going to have one "1", two "2"'s, three "3"'s, etc.  We want to know when the first "12" will come.  Well, that's going to come after the eleventh "11".  Let's just count up how many terms will come first:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66

So if 66 terms come before the first "12", the first "12" will be the 67th term.

Test 3 Section 5 - #14 (page 529)

When the SAT spells out an equation in words instead of just giving it to you, it's usually going to be a very simple equation once you translate it.  Most of the translating is pretty easy:

• "Five times a number" can be translated to "5n"
• "the number added to 5" can be translated to "n + 5"
• "is the same as" has to mean just "="

So we rewrite what they spelled out, and solve it:

5n = n + 5

4n = 5

n = 5/4 or 1.25

Test 3 Section 5 - #8 (page 527)

You can BACKSOLVE this one.  Start with (C), which says m = 24.  If that's true, then:

(x - 8)(x - k) = x² - 5kx + 24

You know from all the factoring/FOILing you've ever done in school that k has to be 3, because (-8)(-3) = 24.  But does that work with the rest of the equation?  Test, substituting 3 in for k: 

(x - 8)(x - 3) = x² - 5(3)x + 24

x² - 11x + 24 = x² - 15x + 24

EEK!  When you FOIL out the left hand side, it doesn't work!

So do the same process now, using (B)* instead, which says x = 16:

(x - 8)(x - k) = x² - 5kx + 16

Using the same logic we just used above, we know k has to be 2, since (-8)(-2) = 16.  Test it, using 2 for k:

(x - 8)(x - 2) = x² - 5(2)x + 16

x² - 10x + 16 = x² - 10x + 16

Everything matches up this time!  (B) must be our answer!

* How do we know to try (B) next instead of (D), you ask?  Well, basically looking at how -5(3) = 15 was BIGGER than -8 - 3 = -11, and realizing that those numbers are only going to get further away from each other if our answer gets bigger.  This is a bit persnickety to explain, but the good news is if you try (D), you'll see very quickly that it didn't work and you got further away, so you'll know to try (B) next.  

Test 3 Section 5 - #5 (page 526)

To start, just look at triangle ACD.  Since ABCD is a square, AD = CD, so we're looking at an isosceles right triangle (a.k.a. a 45°-45°-90° triangle).

If you don't know it by heart, take a look at the beginning of the section for the Reference Information table, which tells you that a triangle like this with legs s will have a hypotenuse of s√2.

Here's the tricky bit: the hypotenuse of triangle ACD is just plain old 4.  There's no radical to be found.  All that means though, is that the legs already had a √2 in them.

Think of it this way: what do we have to do to s√2 to make it be just s?  We divide by √2!  So we do the same thing with our hypotenuse of 4: we divide it by √2.  So here's what we can get from that:

which can be rewritten as:

The area of the square is going to be (2√2)². That's 2²(√2)², or (4)(2) = 8. Answer choice (A).

Test 3 Section 5 - #18 (page 530)

Ok.  They tell us the equation of that parabola is y = ax².  Whenever you get a question like this (which is not all that uncommon in the hard math questions) you're looking to find a point on the graph!  Since they also tell us that the area of the square is 64, we know the sides of the square must each be 8:

Since parabolas are symmetrical and this one has its minimum at (0,0), we know now what the actual points are at each corner of our square:

And that, friends, is where this gets fun.  Let's use (4,8) since both numbers are positive.  Now that we have a point on our parabola, we just need to plug values in for x and y!

y = ax²

8 = a(4)²

8 = 16a

0.5 = a

This would, of course, also work with (-4,8). Either way, you're going to get 0.5 (or 1/2) for a.