SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (44 page)

This question is a marvelous example of how important it is to be aware of the limitations of what the SAT Math section can ask you. Most people try to approach this question by taking the 6th root of both sides of the equation, but, as trained test-takers, we need to know that the SAT can never ask us to take any root besides the square root. So taking the 6th root isn’t going to be the easiest way to go here (and anyway, the 6th root of 432 is an irrational number).

So now what?

Well, as is often the case, there are two basic ways to go here. We can try a kind of backsolving, brute-force sort of approach, or we can try using algebraic principles. The first kind of approach requires less familiarity with math but will be a bit more time-consuming for most people, since it’s basically just trying a whole bunch of hit-or-miss combinations of
a
and
b
values. The second approach will go a little faster but will be harder for most test-takers to think of or pull off.

If we were going to do the first approach (the hit-or-miss approach), we’d start out by listing all the factors for the quantities in the answer choices. Notice, by the way, that the choices have a lot of factors in common, since they’re all multiples of 6. That would suggest that concentrating on numbers like 2, 3, and 6 might not be a bad way to start.

Anyhow, to start working through our options, we’d take an answer choice like (A), figure out its factors (they’re 1, 2, 3, and 6), and then get out our calculators and start plugging in combinations of those factors for
a
and
b
in the original expression to see if we could come up with a value of 432.

This will be made more challenging and time-consuming by the fact that, to be really thorough, you have to try each factor for
a
and again for
b
, since they aren’t treated the same in the expression.

But it’s made a little easier by the fact that some combinations of factors will prove to be too small or too large, which might help you adjust your guessing and hit on the right answer more quickly. A lot of people who try these kinds of backwards solutions like to start with (C), since it’s the value in the middle, see how that goes, and then adjust up or down. In general I think that’s a fine approach, but in this case it might result in more work because the larger numbers probably also have more pairs of factors to work with. So it’s a toss-up, in my opinion—start where you like.

The second approach, as I mentioned, is a bit more mathematical, and is the one I would probably employ (not that that matters).

I would start out by realizing that I can simplify the expression on the left-hand side of the equation, so that the whole equation reads like this:

a
3
b
2
= 432

Now that’s starting to look like something we might have seen at some point in algebra class—it’s starting to look a bit like a prime factorization, with
a
and
b
being the possible factors of 432.

So at this point, I’d break down 432 into its prime factors. I’d do this by using what my algebra teacher, Mrs. Turner, used to call a “factor tree.” They look something like this:

From this, we can see that the prime factorization of 432 is 2
4
* 3
3
.

That’s almost what we’re looking for, except for one small problem: 432 is the same thing as 2
4
* 3
3
, but we’re looking to express it as
a
3
b
2
. It looks like the
a
can match up with the 3 in our factorization, because they’re both cubed, but how do we get
b
2
to match up with 2
4
? We’ll have to realize that 2
4
is 16, and 16 is 4
2
.

That means that
a
3
b
2
is 432 when
a
is 3 and
b
is 4. And that, in turn, means
ab
is 12, so (B) is correct.

As far as our answer choices are concerned, we see that we have a series of multiples of 6. In this situation we’d expect the right answer not to be a value on either end of the series—maybe that’s not a ton of help here, but at least it doesn’t suggest we’re wrong.

A lot of people get nervous when they see this question, because they’ve never encountered a shape like the one in the diagram. But as trained test-takers, we have to remember that we’re always going to see new stuff on the SAT, and our job is to figure out how to apply basic math knowledge in these new situations.

In this case, as in many other questions, it’s just a matter of reading carefully. The only line segments that will be on edges are the ones to the vertex directly to the left of V, directly to the right of V, and directly underneath V. That means that the lines to the other 8 vertices aren't on the edges of the figure.

Like many of the SAT Math questions that most test-takers struggle with, this question involves no formula and no chance to use a calculator. It’s just a matter of reading carefully
and then subtracting 3 from 11 and arriving at 8 as the answer.

This question challenges many test-takers, usually because the drawing looks so complicated. But if we think carefully and work through the question, we’ll find that it only relies on basic math at its core, just like all SAT Math questions.

The question asks us to find
p
. Since the only mention of
p
in the whole question is in the equation
y
=
px
3
, that’s pretty much where we’ll have to start.

The only way we can get
p
out of
y
=
px
3
is to know the values for
y
and
x
. So that means we need to be able to identify the
x
and
y
values for some point on the graph of the function.

Apart from (0,0), which would be useless for us here, the only clear points that the function passes through are the ones labeled
A
and
C
.

Since
C
has
x
and
y
values that are positive, let’s focus there. We know that
C
is at (1/2,
c
). So now we need to figure out the dimensions of triangle
ABCD.

Since
we know the area of
ABCD
is 4, and the diagram shows its width is 1 unit based on the
x
-coordinates, then we know that
ABCD
has a height of 4.

T
his means that point
C
is at (1/2 , 2). Now we just plug those
x
and
y
values into
y
=
px
3
:

2 =
p
(1/2)
3
              (plug in 1/2 for
x
and 2 for
y
)

2 =
p
(1/8)              (simplify on the right)

16 =
p
                            (isolate
p
)

This question was a bit more involved than many SAT Math questions. We know to expect that some time near the end of the grid-ins. But the solution still only involved a bizarre combination of basic facts—something else we’ve come to expect from SAT Math questions in general.

The real lesson to take away from this question is the idea of tracing the solution back through the concepts in the question. Here, they asked about
p
, so we looked in the question for a statement that was relevant to
p
, then we thought about things that were relevant to that statement, and so on. This general approach will come in handy on other SAT Math questions, even though this particular question will never appear on a real test in the future.

Notice, also, that this question is actually very similar to question 18 on page 530 of the College B
oard’s Blue Book. I’ll repeat here what I said earlier in the explanation for that question: do not be misled by this coincidence into assuming that the College Board frequently repeats a limited number of question types in each SAT Math section. On the contrary, in the entire Blue Book these two questions are one of the only examples of question material being repeated so closely on the SAT Math section.

Many students get nervous over this question because they don’t like things that deal with ranges and inequalities, but it’s important to remember that we don’t have to do these questions in ways that we could defend to a math teacher. We just need to find the answer any way we can.

I would start by multiplying the smallest and greatest values of
x
and
y
together, and then seeing how those things relate back to the answer choices. The smallest and greatest values of
x
are 0 and 8, while the smallest and greatest values of
y
are -1 and 3. So:

0 * -1 = 0

0 * 3 = 0

8 * -1 = -8

8 * 3 = 24

So our range of possible values for
xy
at least needs to include -8 and 24.

That means (E) is the only choice that works.

If we look at the answer choices, it’s a good sign that many of the choices end in 24. It’s not a great sign that only one choice ends in -8, but we could easily pause to double-check the validity of -8 by realizing that 8 * -1 is, indeed, -8.

In fact, from a certain way of looking at it, once we’re positive -8 is a valid option we can actually stop doing any further work, because only one choice includes it.

(Now, if you’re a math teacher or a strong math student you might be shaking your head at my approach here, which is quite informal. In fact, you might shake your head at a lot of what we do in the Black Book because of its informality. But my goal isn’t to show my students how to attack SAT Math questions formally, because formal solutions to SAT Math questions are typically slower and more difficult than the simple, direct approaches I want my students to follow. I’m not teaching math in this book; I’m teaching how to answer SAT Math questions. The distinction is important.)

Many test-takers miss this question, even though it involves nothing more than careful reading and basic arithmetic.

The question says that the new average is equal to the median number of siblings per student. That means we need to figure out what the median number of siblings is, and then figure out how to make the new average number of siblings equal to that median number.

So let’s start by figuring out the median number of siblings.

We see there are 3 students with 0 siblings, 6 students with 1 sibling each, 2 students with 2 siblings each, and 1 student with 3 siblings. So if we arrange the number of siblings for each student in a row from least to greatest, we’ll get this:

0,0,0,1,1,1,1,1,1,2,2,3

So the median number of siblings is 1, because the numbers in the middle of that series are both 1.

So when the new kid joins the class, the average number of siblings that everyone has is going to equal this median number of 1. When the new kid joins the class, there will be 13 total students in the class, which means the total number of siblings for all the kids in the class must also be 13 (because 13/13 is 1).