SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (37 page)

One thing that we should note right away is that the diagram is not drawn to scale. We’ll want to focus on the part that doesn’t seem to be to scale and see if we can figure out what’s going on there.

Another important feature of this question is that it gives us information about the dimensions of the diagram in the text, but that information isn’t labeled on the diagram itself. This means that the information about those dimensions will be the key to solving the problem.

So let’s see what happens when
we put all of that together. If CD is 10, and DE is also 10, and EF is also 10, then we know that each of the overlapping triangles has side lengths of 20. In other words, before they were overlapping, the two triangles were both 20 units on every side. And if DE is 10 units, then all the sides of the small triangle are also 10 units, because the two big triangles are equilateral. So a more accurate representation of the situation would be something like this:

With that in mind, let’s just work our way around the figure, starting with C and going clockwise.

CD is 10 units.

DE is 10 units.

EF is 10 units.

FA is 20 units.

From A to the indented corner on the left is 10 units. (Remember, the diagram in the original question isn’t drawn to scale.)

From the indented corner on the left to B is 10 units.

BC is 20 units.

So if we add that all up, we get 10 + 10 + 10 + 20 + 10 + 10 +20 = 90 units.

So the answer is 90.

Notice, once more, that none of the arithmetic in this question was difficult. The difficulty came from figuring out what was being asked, and we did that by reading carefully and knowing to pay attention to the way the College Board draws diagrams.

The diagram in this question is drawn to scale, so we want to take note of that. Unfortunately, it doesn’t have any kind of scale labeled on it, but we may be able to work around that if it turns out we need to.

The question is also talking about 3 different variables:
x
,
k
, and
a
. This will give a lot of people trouble, because most test-takers who treat SAT Math like school math will think they need to figure out the values of those variables.

But if we look carefully, we see that there’s not enough information to figure out
k
. Instead of worrying about that, we just realize that
k
must not matter in the question. Easy enough.

Let’s focus on the end of the question. It asks for the value of
a
, but all we know about
a
from the rest of the question is that it’s positive and that
g
(
a
– 1.2) = 0.

A lot of people will want to plug
a
– 1.2 in for
x
in the original function equation, because that’s the typical knee-jerk reaction that would be appropriate in a school situation. But if we consider that move for a moment, I think we’ll see that it’s not likely to help us very much. We’d end up with this expression, which would be very ugly:

g
(
x
) =
k
((
a
– 1.2) + 3)((
a
– 1.2) – 3)

This
doesn’t really look promising, from an SAT Math standpoint. (Now, it’s true that the last question on the grid-ins is often more complicated than other SAT Math questions, but I’d still be reluctant to go about expanding and condensing that hideous expression unless absolutely necessary.)

So let’s try a different tack.

One thing that’s kind of interesting is the fact that we were told the value of
g
(
a
– 1.2) is
zero
. Zero is a unique number with unique properties, particularly when it comes to functions and graphs. On a graph, the zeros of a function are the points where the function crosses the
x
-axis; in a function equation, we find the roots of an expression by setting the factors of the expression equal to zero.

So let’s think about that for a second. In the equation, the ways to make
g
(
x
) come out to zero are to set either (
x
+ 3) or (
x
– 3) to zero. (
k
can’t be zero, if it were, every single
g
(
x
) value would also be zero, and that isn’t what the graph shows.)

So if either (
x
+ 3) or (
x
– 3) is zero, that means the
g
(
x
) value would be zero when
x
is either -3 or 3. And that fits with the graph, because the places where the function equals zero seem to be at
x
= -3 and
x
= 3. So we know that
g
(-3) = 0, and
g
(3) = 0. And that should be kind of our eureka moment: if
g
(-3) and
g
(3) are both zero, and
g
(
a
– 1.2) is zero as well, than that must mean that (
a
– 1.2) is the same thing as either (-3) or (3)!

That means that
a
is either 4.2, or -1.8. But the question tells us
a
must be positive, so that means it’s 4.2.

Once again, the College Board has created a misleading question out of very basic ideas. In this question, we needed to know that the
x
-intercepts of a graph are the places where its
y
-value is zero; we needed to know that 4.2 is bigger than zero but -1.8 is not; we needed to know that 4.2 – 1.2 is 3. That’s about it, from a math standpoint, and none of those ideas is very complicated on its own. I would bet that over 90% of the people who took this test knew every single one of those facts, but they missed the question because they didn’t realize it was asking about those facts in the first place.

So the challenge came from the sheer weirdness of the question, and we had to use our reading skills to understand what was being asked, and our thinking skills to realize that the idea of the zeroes of the function was very important.

By now, you’re probably beginning to develop a strong appreciation for the strange way the College Board designs SAT Math questions. In the next section, we’ll continue to develop your understanding by exploring a large selection of questions from the Blue Book that most people have some trouble with.

If you’d like to see videos of some sample solutions like the ones in this book, please visit
http://www.SATprepVideos.com
.
A selection of free videos is available for readers of this book. 

At this point we’ve taken a look at a variety of Math questions, but you’re probably interested in seeing more. I understand where you’re coming from with that, and I’m about to show you some more Math solutions, but . . . before I do, I want to make sure I reiterate what you should be learning from each solution.

I am NOT doing these solutions in an effort to show you something formal that you should memorize and then expect to repeat on future SAT Math questions, as we would normally do for a math class in school. Remember that the SAT Math section will show you very strange combinations of very basic ideas, and specific questions and questions ‘types’ are not repeated for the most part. The odds are that when you take the SAT for real, you’ll see a whole test full of math questions that don’t look like any other SAT Math questions you’ve ever seen before, at least not on the surface. And this will be the case no matter how many practice SAT Math questions you work on beforehand. Every test combines the same basic stuff in new ways.

You may have noticed I’m repeating this idea a lot. There’s a reason for that:

It’s
very important, and most people ignore it.

So if you want to get better at SAT Math, the goal is to learn the underlying process, and to practice using it against real SAT Math questions written by the College Board. That’s why I’ve
included these solutions: to give you an idea of the right general approach so you can continue to refine your instincts, not to give you cookie-cutter instructions for specific question types, because specific question types basically don’t exist on the SAT Math section, practically speaking.

With that important reminder out of the way, let’s take a look at some of the SAT Math questions that students have typically asked about.

As with other question explanations in this book, you’ll need a copy of the second edition of the College Board’s
Official SAT Study Guide
(otherwise known as the Blue Book) to follow along. Let’s get started.

As with many SAT Math questions, t
here are basically two ways to do this: we can pick a concrete number that satisfies the requirements for
k
and then look to see what happens to
k
+ 2, or we can think in the abstract about properties relating to the concept of remainders. In general, abstract solutions will be faster to reach but harder for many students to execute, while concrete solutions will give most students added confidence but end up taking more time. For this particular question, though, the amount of time spent on each solution is likely to be roughly the same.

If we go the concrete route of
picking a number to be
k
, we have to make sure that it satisfies the setup. In this case, a number like 13 would work, because 13 divided by 7 gives a remainder of 6. So then we'd look to see what happens when we divide 15 by 7, since 15 is 2 more than 13; the result is a remainder of 1.

The other way to do this is to think in the abstract: if
k
has a remainder of 6 when it's divided by 7, then
k
is 6 more than some multiple of 7, and
k
+ 2 will be 8 more than that multiple of 7. We know that 8 is more than 7, so one more 7 will "fit in" when
k
+ 2 is divided by 7, and we’ll be left with a remainder of 8 - 7, or 1.

Either way, (B) is the right answer.

Of course, you wouldn’t need to do both solutions on test day. I’m just doing both of them to show that there are a variety of ways to attack this question successfully, as will be the case for most SAT Math questions.

This is one of those questions that lets us work on identifying a whole variety of the patterns and rules we talk about in this book.

First, we’d want to notice that the diagram is drawn to scale. That means that we might be able to figure out the answer by eyeballing the diagram—or, at the very least, we want to make sure that our final answer makes sense in the context of the scale of the diagram.

Also notice that the diagram and the answer choices have a lot of expressions with √2 in them. We know that √2 relates to 45
o
-45
o
-90
o
triangles, which seems relevant to the question since ABC, ADB, and BDC are all 45
o
-45
o
-90
o
triangles.

There are a variety of ways we could try to figure out the area of the shaded region. The most straigh
tforward way is probably to work out the distance of EF, along with the distance from F to the base of the figure, and then multiply those two things together, since they would represent the length and the height of the rectangle. We might do that by realizing that BE and BF are each 5√2 units long, and that they are each the legs of a 45
o
-45
o
-90
o
triangle with EF as its hypotenuse. Since the ratio of the sides of a 45
o
-45
o
-90
o
triangle is 1:1:√2, the distance of EF is 5√2(√2), which is 10.

Since the shaded region is divided into two sq
uares, we know that the distance from E or F to the base of the figure must be 5. So the shaded area is 50, because we multiply 10 (the length) by 5 (the height) to find the area. So (C) is correct.

There are other ways to get this answer, as well. We could also realize that AC must be 10
√2(√2), or 20 (again because of the 1:1:√2 ratio for sides of a 45
o
-45
o
-90
o
triangle). Then we could realize that the height of triangle ABC must be 10 (because BD is one of the legs for which BC is a hypotenuse, and this is another 45
o
-45
o
-90
o
triangle for which the 1:1:√2 ratio applies). That means the area of the entire triangle ABC must be 100 units. The shaded region represents 1/2 of the area of ABC (we know this either by eyeballing the scaled diagram, or by realizing that the four unshaded small triangles are the same total area as the shaded rectangle).

Now let’s turn to the answer choices. An awareness of the patterns that frequently appear in answer choices on the SAT Math section would help us to realize that 50 looks like a very likely option to be the correct answer. It fits the halves-and-doubles pattern and it’s also the middle number in a series (the series is 25, 50, 100). It also doesn’t have
√2 in it, which is probably good because 3 out of the 5 choices don’t have √2.

Like most SAT Math questions, this one rewards us for reading very carefully and thinking about the definitions and properties of basic terms.

We’re told that 2 of the faces are black and the rest are white; this means there must be 4 white faces, since the total number of faces on any cube is 6.

If the total area of the white faces of the cube is 64 square inches, and if there are 4 white faces, then the total area per face is 64/4, or 16.

That means each face is 16 square inches. And since the dimensions of a cube are all identical, that means each face is a 4 x 4 square, which means the cube is 4 inches in each dimension.

The volume of the cube, then, is 4 x 4 x 4, or 64.
So the correct answer is (A).

Notice that some of the answer choices differ from the correct answer by a factor of 2 or 4. This is a strong, strong reason to go back over your work and check it for small mistakes. If we misread the setup, or if we accidentally mis-multiplied or mis-divided, we can easily be off from the right answer in a way that will be accounted for in the answer choices. Also notice that (B) is 5 x 5 x 5, and (D) is 6 x 6 x 6. This reinforces our belief that we should find the correct answer by cubing something, but it also means we have to make sure we were right to cube 4 instead of 5 or 6.

As will often be the case on the SAT, there are multiple valid approaches to this question. If we wanted to use a simple permutation solution, we could realize that there are 5 options for one color, and 4 options for the other color (there are only 4 options for the second color because we’re not allowed to repeat the selected color—otherwise there would be 5 options for both). 5
x 4 = 20, so there would be 20 possible arrangements.

But the SAT doesn’t make us do things like this in the formal way—the SAT doesn’t care how we get the answer, as long as it’s right. So, if we want, we can just list out the different possible arrangements and then count them up. (To be clear, this list-and-count approach will take a good bit longer than we would normally like to spend on a question, but it can be a very concrete way to arrive at the answer if you don’t feel comfortable with permutations. Remember, too, that the reason we try to go through most questions as quickly and efficiently as possible is so that we can have more time if we need it on questions like this.)

So if we call the colors a, b, c, d, and e, and then list things out, here are the different arrangements:

zone 1 / zone 2
a/b
a/c
a/d
a/e
b/a
b/c
b/d
b/e
c/a
c/b
c/d
c/e
d/a
d/b
d/c
d/e
e/a
e/b
e/c
e/d

That means
there are 20 different arrangements, so (B) is correct.

Most people try to approach this using some variation on the slope formula (which is
(
y
2
–
y
1
)/(
x
2
–
x
1
),  not that I recommend it here).

But we have to remember that this is the SAT, and the College Board often likes to leave short cuts hidden in the questions for people who think to look for them.

Instead of going through the hassle of working out the slope of the original line, the slope of the perpendicular line, and the missing
y
-value that would make the whole thing work out correctly, let’s take a second and actually notice the answer choices for a minute.

Remember that we always want to check out the answer choices before we commit to a course of action. They’ll help us understand a lot of stuff in a lot of questions if we just pay attention to them a little.

In this case, I’m noticing that the answer choices are kind of falling into 3 different groups, in a sense. We have negative numbers, then we have 2 and 3, and then we have 5. Hmmmm.

Let’s try actually plotting line
l
and see what it looks like. Here’s a rough approximation of the situation described in the question:

If we
plot the points (0,0) and (2,1), and then think about the options offered by each of the answer choices, we might start to notice something. (A) and (B) clearly don't work if you're trying to draw line that would go through either value and (2,1), and be perpendicular to the existing line: