SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (34 page)

So here’s the bottom line: you should ignore the idea of an order of difficulty, because it serves no purpose. Instead, treat each question as a separate event. Don’t be afraid of any questions, and don’t take them for granted either.
 

To demonstrate how this approach to SAT
Math questions works, and to help you improve your understanding and execution of that approach, we’ll go through an SAT Math section from the second edition of the College Board’s Blue Book,
The Official SAT Study Guide
. (Remember, the Blue Book is the only book I recommend for practice questions, because it’s the only book whose questions are totally guaranteed to play by the College Board’s rules.) I’ve chosen the first practice-test section that includes both Multiple Choice problems and Student-Produced Response problems. It starts on page 413 of the Blue Book.

You really have to be careful on a problem like this—it’s the kind of thing that most test-takers will feel they can’t possibly miss
because it seems so simple. But it’s very easy to overlook a couple of small details on this question and end up with a wrong answer.

The question asks us to figure out how many new houses were built from 1961 to 1990. To answer that question, we have to look at the relevant years on the chart. The first row shows the houses built from 1961 to 1970; the next row shows the houses from 1971 to 1980; the third row shows us the houses from 1981 to 1990. There are 2 house icons in the first row, 4 in the second row, and 8 in the third row. That makes a total of 14 house icons.

Notice that choice (A) reflects a mistake that would be pretty easy to make here: if we assumed that each house icon represented one house, then we’d think (A) was right. But each icon represents 2,000 houses, so the right answer is 28,000. That makes (E) right.

As always, I’d also try to think about where some of the other wrong answers might have come from. In this case, (B) is what we’d get if we counted all the icons in the entire chart (including 1991 – 2000)
and
also made the mistake of thinking each icon only represented one house. (C) is what we’d get if we correctly realized that each icon represented 2,000 houses but accidentally only counted the third row. Finally, (D) is what we’d get if we didn’t count the second row for some reason.

I always try to figure out where some of the wrong answers are coming from because it helps me be sure I’ve understood the question. It’s not necessary to figure out
every
wrong answer, even though we were able to do that in this question. But if you can’t figure out where any of them are coming from at all, then there’s a very good chance you made a mistake, and you should re-check your work.

Also note that, in this case, 3 out of the 5 choices are numbers in the tens of thousands. This strongly suggests (BUT DOES NOT GUARANTEE) that the right answer should be one of those 3. If we had accidentally overlooked the information about each icon counting for 2,000 houses, then noticing that most of the answer choices are huge numbers should have helped us realize we might have missed something.

This is a perfect example of the kind of question that you should absolutely lock down, double-check, and be done with in well under a minute, possibly in as little as 10 seconds. It’s important to work quickly and efficiently through questions like this in order to save time for questions where you might have more difficulty gaining a foothold.

One last thing: note that this question, like many SAT Math questions, is primarily a critical reading question at heart. The major way people will make mistakes here is by misreading the question or the chart, not by failing to multiply the numbers 14 and 2,000.

This question presents us with a type of drawing that we might not ever have encountered before.  But that’s fine: remember that this question, like all SAT Math questions, can only be made out of the same basic ma
th ideas that were in the Math Toolbox earlier in this book. So let’s see what we can figure out.

We know the diagram is drawn to scale, because it doesn’t say that it’s not. But the answer choices are so close to one another that it would probably be hard to answer this question by eyeballing the drawing and using the scale.

So let’s think in terms of geometry. We know that we can fill in the measurements of the angles opposite the labeled angles, because opposite angles are identical. So the angle opposite the 35-degree angle is also 35 degrees, and the one opposite the 45-degree angle is also 45 degrees.

Now it looks like we’re getting somewhere: once we label those opposite angles, we can see that
w
is the third angle in a triangle, and that the other two angles are 35 degrees and 45 degrees.

That means
w
+ 35 + 45 = 180. So
w
+ 80 = 180, which means
w
is 100 degrees. So (B) is right.

Notice that (B) is in the middle of a 3-term series in the answer choices, with the other two terms differing by 10. This should alert you to the fact that it would be very easy to be off by 10 in this question if we forgot to carry a digit during the addition or the subtraction. So we should go back and check to make sure we didn’t mess that up.

Still, it’s very reassuring that we like the answer choice in the middle of this range, because typically the College Board likes to make the number in the middle of the range be correct. (As we discussed in the section on patterns, they do this because it gives us the option to be off in either direction and still find an answer choice reflecting our mistake.)

This is another question that probably should take
less than 30 seconds to do. We want to answer it quickly, check to make sure we’re right, and move on.

A lot of students ask me about this question because they can’t figure out how to set it up algebraically.

But this is the SAT—we don’t have to do it algebraically! In fact, in most cases I prefer not to use algebra on the SAT, because it often takes longer and it increases the chance that we’ll make a mistake.

So let’s just think about this.
(There’s no magical way to “just think about” a question—all we do is dive in somewhere and then see what adjustments, if any, we need to make.)

If each of the 19 tables had
4 people, then there would be seats for 76 people, because 4*19 = 76. We need to find seats for 84 people, though, which is 8 more than 76.

So 8 of the tables will need to have an extra seat, which makes (E) the right answer.

Notice that this isn’t really the kind of question you would probably ever be asked in a math class, and the solution I just gave isn’t the kind of solution your math teacher would ever accept. This is normal for the SAT, and we need to get used to it.

Also notice that the answer choices form a series in this question, and the correct answer happens to be the largest number in the series, which is something that I said doesn’t happen often. I was right—it doesn’t happen often—but that doesn’t mean it never happens. In this case, when I see that the answer I like is the last number in a series, I’d just double-check my work and make sure I hadn’t made a mistake, and move on.

One more thing—you might ask how I knew to start out by multiplying 4 and 19. But I didn’t “know” that would lead to the answer right away until I tried it. On the SAT Math section, we have to be willing to play with questions like this, and we have to get away from the idea of following established formulas.

This question often overwhelms test-takers who think in terms of school math, because it has two variables in it, and one of those variables (
m
) is actually impossible to solve for. In school math, it might be a problem if we couldn’t solve for a variable, but this is SAT Math. The SAT often gives us questions in which it’s impossible to work out the value of a particular variable.

Since the question asks which answer choice would be equal to the original expression when
a
equals 4, and since every answer choice has 4’s in it instead of
a
, I would start out by plugging 4 in for
a
, and determining that the original given expression equals 4
m
2
+ 4
m
+ 4. Then we can go through each answer choice and distribute the 4 in each one, until we end up with a choice that also results in 4
m
2
+ 4
m
+ 4 as an answer.

The answer choice that fits the bill is (D).

Notice that the choices here fit a pattern we discussed earlier in this book: the elements of the right answer appear very frequently in the wrong answers.

In other words, in this case, the correct answer has the parenthetical expression with
m
2
,
m
, and 1. Notice that 3 out of the 5 choices have
m
2
somewhere in them, 4 out of 5 have
m
, and 4 out of 5 have 1 in them. Also notice that 4 out of 5 don’t have a 4 in the parentheses, and 4 out of 5 don’t involve squaring the expression in parentheses (like choice (B) does).

So if we were going to try to predict the right answer, based purely on the answer choices, we’d probably want the choice with these attributes:

o
includes
m
2

o
includes
m

o
includes 1

o
doesn’t have a 4 in the parentheses

o
doesn’t square the expression in parentheses

We’d want that answer choice because those are the most popular features in the field of answer choices. The choice that satisfies all of those conditions is (D), which is the right answer in this case.

Just to be completely clear, let me point out that I would never, ever recommend that you answer a question based only on figuring out which choice has the most popular features. That will work a whole lot of times, but it doesn’t work every time, and my goal is for us to be correct on every single question. So this idea is a pattern to be considered, not a rule to be followed unfailingly.

In this particular question, if we accidentally misread (E) and thought it was correct, being aware of the imitation pattern might let us realize that none of the other answer choices have a 4 inside the parentheses, which is probably a bad sign. If we liked (A) for some reason, then noticing that the other choices were all highly similar to one another and that none of them had
m
3
might help us realize we needed to re-evaluate our conclusion.

Notice that we answered the question without ever finding out what
m
represents. This kind of thing is normal for the SAT. On the SAT, we need to get away from the idea that we have to solve for every variable we see in a question.

This question frustrates a lot of students because they’ve never seen a diagram like this in their math classes. But we have to remember that this kind of thing is standard on the SAT: no matter how much you practice, you’re going to encounter things on test day that don’t look like anything else you’ve ever seen, at least on the surface. We have to learn how to attack these kinds of situations systematically, and confidently.

In this case, they’re asking us for the area of the shaded portion. As trained test-takers we know two things:

They have to give us all the geometry formulas we need.

They didn’t give us a formula for the area of a portion of a square.

That means there must be a way to figure out the area of this shaded region without having a formula uniquely for that purpose.

We do have a formula for the area of a circle, though. And we could use that formula to find the area of the whole circle altogether, and then divide the area of the circle up to find the area of the shaded region.

So now we have to figure out how to find the area of the circle. On the SAT, there’s only one way the College Board can ask us to do that: we have to use
A
= pi(
r
2
), where A is the area and
r
is the radius. (Again, this formula is provided at the beginning of the math section if you don’t remember it.)

How can we find the radius? In this case, we use the only number provided anywhere in the diagram: we look at the fact that the side length of the square is 2 units, and we realize that this corresponds to the diameter of the circle. If the diameter is 2, then the radius is 1. So the radius of the circle is 1 unit, which makes the area
of the entire circle
pi(1
2
), which is just pi.

Now we have to figure out what portion of the circle the shaded area represents. There are two ways to do this. Since the diagram is drawn to scale, we could just eyeball it and realize that the shaded area is one-quarter of the circle. If we want to be more precise, we could realize that the angle at point
O
must be a 90-degree angle, since
O
is the center of the circle and the center of the square. And 90 degrees is one-quarter of 360 degrees, so, again, the shaded area must be a quarter of the circle.