SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (38 page)

The values for
(C) and (D) don’t work, either:

Only (E) is in anywhere near the proper place
to create a line that passes through (2,1) and is perpendicular to the original line
l
. So (E) must be right:

(I realize that none of the diagrams I supplied is exactly to scale, but the diagrams you would draw for yourself in your test booklet would also be out
-of-scale. During the test, the point isn’t to create a perfectly scaled drawing, but to get a solid idea of where the different elements of the question would be relative to one another.)

So, as we’ve just seen,
it’s possible to solve this question, which is the last one in the section and which was missed by a lot of people, without ever using any kind of formula or consulting a calculator or even adding two single-digit numbers. All we had to do was remember to notice the answer choices, plot a few points (or even just think about plotting some of them), and then realize that only one answer choice was close to working.

Notice that the College Board could have made this question much harde
r by including wrong answers that were closer to the right answer, or by changing the overall scale of the question. But they didn’t. They left a shortcut for alert test-takers to seize. Remember that, and look for these kinds of things on other questions.

People mess this question up all the time, but it’s really just a pretty straightforward application of the concept of slope. We know that for two reasons: the first one is that the diagram shows a diagonal line with its horizontal and vertical changes marked off, and the second reason (perhaps a bit more obvious) is that the question includes the word “slope.”

Remember that slope, by definition, is
the ratio of the vertical change to horizontal change. That means that, in this case, for every 7 units of vertical change there are 16 units of horizontal change. If we only have 3.5 units of vertical change, then we need 8 units of horizontal change to maintain the ratio, because 3.5 is half of 7 and 8 is half of 16. So the correct answer is (A).

Notice that one of the wrong answers is 32, which is what we get if we accidentally double 16 instead of cutting it in half.

I often talk about how the College Board likes to mislead you by making questions seem to be something they’re not. This question might be one of the all-time best examples of that technique. It starts out looking like a classic probabili
ty question, one of those situations where somebody has a certain number of things in a bag and you have to calculate the chance that they’ll pull a certain kind of thing out of the bag at random.

But that’s no
t what it is at all. Instead, it’s much simpler than that, but it’s unlike anything that any test-taker has probably ever seen before.

This is one more example of why you can’t take anything for granted on the SAT. Everything needs to be read carefully.

Ari starts out with 3 red things and 4 green things. If he takes 13 more pieces and we need to end up with more reds than greens, we might imagine that the 13 is made up of 7 reds and 6 greens, just as a place to start; if that's the case, then there would be 10 red and 10 green, which doesn't satisfy the requirements of the question. So in order to end up with more reds than greens, Ari would need to pull out at least 8 reds, so the answer is 8.

A
gain, there's absolutely no formula for this, and it has nothing to do with probability. It's just reading and thinking. Most SAT Math questions are just reading and thinking.

A lot of people who see this question immediately start worrying about the word “tri-factorable,” as though it were a real math term instead of something that the College Board made up specifically for this question.

There are two ways that we can figure out that the word “tri-factorable” was just made up for this question. The first way is that the question tells us what the word means: a tri-factorable number is one that is the product of 3 consecutive numbers. (If this were a real math term, the College Board wouldn’t bother to define it. For slope questions, they don’t say, “What is the slope of this line, if slope is defined as blahblahblah,” because you’re supposed to know what slope is, because slope is a real math concept.)

The second way that we can know this is a made-up word is to be familiar with what the SAT is allowed to ask us about when it comes to math. The SAT can only ask about the concepts in the toolbox in this book, and the toolbox doesn’t cover tri-factorability.

So it must just be that we’re supposed to figure out what the word means from reading the question. If we know what the words "product" and "consecutive" mean, then we know what a tri-factorable number is. Remember, as always, that the most important skill on the SAT is reading carefully.

Now, let’s proceed.

We know that all of these questions can be answered in less than 30 seconds, and we know that it would clearly take a lot longer than 30 seconds to attempt to tri-factor each of the first 1,000 integers.

So there must be another way to go.

It’s important to remember that we can always just list out the answers to these kinds of questions that ask how many numbers in a certain set satisfy a certain condition. Either we’ll start listing them and realize there’s only a small number of them, or we’ll start listing them and realize there’s a certain pattern they all follow, and then we can predict the final number from the pattern.

So how can we figure out a tri-factorable number? Since they’re made by multiplying consecutive integers, I’d start with the smallest positive integer and see what happens:

1 * 2 * 3 = 6

So if we start with 1 as the first positive integer, we arrive at the product 6, which must be the first tri-factorable number.

The next tri-factorable number will start with 2:

2
* 3 * 4 = 24

From there we can basically get on a roll:

3 * 4 * 5 = 60

4 * 5 * 6 = 120

5 * 6 * 7 = 210

6 * 7 * 8 = 336

7 * 8 * 9 = 504

8 * 9 * 10 = 720

9 * 10 * 11 = 990

10 * 11 * 12 = 1320

Oops—notice that that last number is bigger than 1000! So there are 9 numbers that work: the one I get when I multiply three consecutive numbers starting with 1, the one I get when starting with 2, starting with 3, with 4, with 5, 6, 7, 8, and finally 9, and that's it.

Notice that this question, like so many other ‘challenging’ SAT Math questions, involves nothi
ng more than careful reading, basic arithmetic, and a willingness to play around with familiar concepts in strange ways. This is typical for the SAT, as we’ve seen many times by now and will see again.

Notice also that there are many ways to mess this question up. We might miscount, or accidentally overlook one or more of the tri-factorable numbers. Any one of those tiny mistakes will cause us to miss the question completely.

This is another one of those questions that we could try to answer in a few different ways. Many people will attempt an algebraic solution, or you could also try guessing and checking. There’s also a
nother approach we’ll talk about after I go over the algebraic one.

For call A, the
cost will be $1 + $0.07(
t
-20).

For call
B, the cost will be $0.06(
t
).

So just set them equal
and solve:

$1 + $0.07(
t
-20) = $0.06(
t
)              (initial setup)

1 + .07
t
- 1.4 = .06
t
                            (distribute $0.07 on the left)

1 - 1.4 = -.01
t
                                          (combine like terms)

.4 = .01
t
                                          (simplify the expression on the left)

40 =
t
                                                        (isolate
t
)

T
he faster approach is a little more holistic, and would probably not be tried by most test-takers, even trained ones. Still, I thought we should talk about it. To work this problem out without really writing out any algebra, we could realize that the $1 for the first 20 minutes works out to 5 cents per minute, which is one cent per minute less than the other rate. After those 20 minutes are up, the rate goes to 7 cents per minute, which is one cent per minute more than the other rate. So to get everything to equal out, you'll need 20 more minutes of talking at the higher rate after the initial 20 minutes of talking at the lower rate. 20 + 20 is 40, so 40 is the answer.

Just to be clear, both approaches are equally valid, of course. I just wanted to introduce the sec
ond one as an exercise of sorts, to keep calling your attention to the fact that the College Board usually sets up SAT Math questions so that they can be attacked quickly and easily if we know how to look at them.

This question is rated 5 out of 5 for difficulty, but, of course, we know that that doesn’t mean much. All it means is that a lot of untrained people missed it. As we’ll see in this explanation, this is yet another SAT Math question that requires nothing more than careful reading and careful thinking.

We’re told the perimeter is
p
, and we know that each square has a side of
k
. The perimeter consists of 16 sides, so the perimeter’s length is 16
k
.

The area of each square must be
k
2
, and there are 10 squares. So the area is 10
k
2
.

Since the question says that
p
and
a
are equal, we know that

16
k
= 10
k
2

Solving, we get

16 = 10
k
              (divide through by
k
)

1.6 =
k
                            (isolate
k
)

And that’s all. Notice that this question only required us to know the definitions of the word “perimeter” and “area,” and basic algebra. There was no formula involved
, apart from the formula for the area of a square, which the test provides for us at the beginning of the section. There was also no real need for a calculator, since the question only involves dividing by 10, which we can accomplish by moving the decimal point.

Again, these attributes are
pretty typical of the ‘hardest’ SAT Math questions.

This question stumps a lot of people. In my experience, almost everybody who misses it does so because they don’t read it carefully, or they don’t notice the answer choices.