SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (45 page)

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So now we add up the numbers of siblings of existing students, so we can see how far we are from a total of 13 students:

0 + 0 + 0 + 1 + 1 + 1 + 1 + 1 + 1 + 2 + 2 + 3 = 13

So actually, it turns out that the 12 students already in the class have a total of 13 siblings already, which means that the 13th student needs to have zero siblings in order for the average number of siblings in the class to be 1.

That means (A) is correct.

Now let’s review the math ideas that were necessary to answer this question. We had to know how to read a relatively simple table. We had to know what the words “median” and “average” meant. We had to arrange a lot of single-digit numbers from least to greatest. We had to figure out that a lot of single-digit numbers added up to 13. We had to know that 13/13 is 1. And we had to know that 13 + 0 is 13.

None of that would probably qualify as advanced mathematical knowledge at your school, but test-takers miss this question very, very often. We should ask ourselves why that is, so we can make sure we don’t make their mistakes.

In my experience, people who miss this question (or just give up on it in the first place) do so because they don’t understand what it’s asking when they read it. This is why I often say that critical reading skills are the most important thing on the entire SAT. And it’s because of the great number of questions like this—questions that present bizarre combinations of simple
math ideas—that test-takers who try to prepare for the SAT by learning advanced math usually don’t see any improvement.

Page 773, Question 17

For this question, as for many abstract questions on the SAT Math section, there are two ways to go: we can come up with concrete values for the variables described in the question and run a test, or we can think about the question in the abstract. The concrete approach tends to take more time but will often feel more comfortable, while the abstract approach is faster but will be harder for many test-takers to manage.

In this explanation, I’ll start with a concrete example and then show the abstract one. You could use either on the test, of course, without bothering to think about the other.

Let’s assume that
p
is 3,
r
is 5, and
s
is 7, as those are three different prime numbers greater than 2, like the question asks us for.

In that case,
n
is 3 * 5 * 7, or 105.

Now let’s figure out what the factors of
n
are:

1, 3, 5, 7, 15, 21, 35, 105

There are 8 of them, so the answer to the question is 8.

Now, if we wanted to think about this in the abstract, we’d note that since
n
is the same as
prs
, and since
p
,
r
, and
s
don’t have any other factors besides 1 (since they’re prime), then the factors of
prs
will include every possible combination of
p
,
r
, and
s
, like this:

p
,
r
,
s
,
pr
,
ps
,
rs
,
prs

. . . plus, of course, 1.

That also makes for a total of 8 factors.

(Notice that the concrete example we did follows our abstract thinking perfectly if we substitute 3 for
p
, 5 for
r
, and 7 for
s
.)

Page 785, Question 2

People often miss this question, even though it’s fairly early in the section and doesn’t really involve any math. When they miss it, it’s usually as a result of not having read it carefully. Most people choose (A) because the figure in that choice has a line of symmetry, and they’re rushing through the question. They don’t notice that 3 other choices also have a line of symmetry, which might help them realize their mistake.

If we read carefully
and realize we’re looking for a figure with 2 lines of symmetry, then we realize that (D) has one line of symmetry going vertically straight down its center and another line going horizontally straight through its center.

It’s a real waste to miss a question like this! Make sure you pay proper attention to every single question on the entire test. Even the so-called ‘easy’ ones can trip you up if you rush.

Page 789, Question 18

This is one of those questions in which knowing the unwritten rules of the SAT Math section comes in especially handy.

Most people who try to answer this question will either use some version of a summation formula or they’ll actually try to add up every integer starting from -22 until they hit a running total of 72.

Either approach could conceivably work, but neither is very fast, nor very easy.

Instead, we want to remember that there must be some way to do this question in under 30 seconds relying only on some combination of basic arithmetic, algebra, and/or geometry.

One thing that would jump out at me right away is that we’re looking for a
positive
sum, but we’re starting by adding up some
negative
numbers in the beginning of our series. That means we’ll have to add quite a few numbers before our running total becomes positive, because at the beginning of our series (-22, -21, -20, . . .), the running total will just become increasingly negative.

So wait a second . . . when does the running total start to creep towards a positive balance?

Well, that won’t happen until we start adding in some positive numbers, of course. So our series will have to go past zero, for sure.

And when it passes zero, for a while, each positive number added to the running total will only serve to erase one of the negative numbers already added to it.
Positive 1 undoes -1, for instance, and positive 2 undoes -2, and so on.

If we think about it, then, we’ll see that every number from -22 to positive 22, added together, gives a sum of zero, because every positive number in the series is outweighed by a negative number and vice-versa.

So all the numbers from -22 to positive 22 add up to zero. Then what happens?

We add in positive 23, positive 24, and positive 25, and we’ve reached our grand total of 72.

That means that all the numbers from -22 to positive 25 add up to positive 72. So (B) is the answer.

Notice some of the wrong answers here. (C) is the sum of the two numbers in the original question, while (E) is the difference. (A) is the first positive number that produces a positive value when added to the running total, so some test-takers will probably accidentally choose it because they’ve realized it marks the beginning of the positive running total without remembering what the question was actually asking.

Page 799, Question 14

People often get frustrated by this question because they can’t figure out the values of
b
and
c
. In school math classes, we’re often programmed to find the values for every variable in a question, but on the SAT Math section we often have to let go of that. Sometimes there’s not enough information in a question to figure out the value of every variable, but the question can still be answered anyway. One of the clearest signs of that situation is when the question involves multiple variables but doesn’t actually ask you for each one of their values.

For questions about function graphs, I often like to look at the intercepts. In this case, we know that
c
is positive, so the
y
-intercept must also be positive, because the
y
-intercept is what we get when
x
= 0.

O
nly (E) has a graph with a positive
y
-intercept, so only (E) works.

Page 800, Question 15

I often resist the idea of identifying ‘types’ of math questions on the SAT Math section, because I want you to realize that you’ll basically have to figure out stuff you’ve never seen before when you take the SAT. But if there were going to be a type of question that I might specifically train somebody for, it would be the kind of thing described in this question. (By the way, this comes up less than once per test in the College Board’s Blue Book. The odds that it will ever come up for you on a real test are less than 50%. And, even if it does come up, it’ll probably only count for one question. These are the kinds of reasons why I generally avoid thinking in terms of question types.)

Anyway, for these questions that ask you about the distance from one point of a 3-dimensional figure to another point, you basically have to
apply the Pythagorean theorem twice.

Let's
label the corner underneath point
B
as point
Q
, and let’s label the corner to the right of point
A
as point
T
.

So then triangle
ATQ
has legs of 1 and 2 units, which means that
AQ
has a distance of √5.

So
now we can use
AQ
, which was the hypotenuse of
ATQ
, as a leg in the triangle
ABQ
. Using the Pythagorean theorem again, we get this:

AB
2
= √5
2
+ 1
2

AB
2
= 5+ 1

AB
2
= 6

AB
= √6

So the answer is (D).

Note that one of the wrong answer choices is √5, for people who forget to finish the question.

Page 800, Question 16

This is a question with a made-up function. For these kinds of things, it’s always very important to read and follow directions very carefully.

In this case, people often fail to realize that the r
ight-hand side of the equation should be
(
a
-2)
2
- (
a
-2). We have to apply the oval function to the whole expression inside the oval, and the whole expression inside the second oval is (
a
- 2).

Once we do that correctly, all that remains is to solve:

a
2
-
a
= (
a
-2)
2
- (
a
-2)                            (substitute the starting values from the question)

a
2
-
a
= (
a
2
- 4
a
+ 4) -
a
+2              (FOIL the expression on the right)

a
2
-
a
=
a
2
- 5
a
+ 6                            (combine like terms on the right)

-
a
= - 5
a
+ 6                                          (subtract
a
2
from both sides)

0 = - 4
a
+ 6                                          (combine
a
terms)

4
a
= 6                                                        (separate
a
term from constant)

a
= 6/4                                                        (isolate
a
)

a
= 3/2                                                        (simplify)

So (C) is correct.

As we might expect, one of the wrong answers is twice as much as the right answer. The other wrong answers seem to reflect what we might have ended up with if we had handled the algebra incorrectly.

By the way, another approach to this question, which would probably take longer but would avoid so much algebra, would be to take each answer choice and plug it in to the original expression until you find one choice that results in a true mathematical statement. But even if you do the question in this way, you still need to make sure you handle the substitution properly by reading the question very carefully.

Page 835, Question 18

To answer this one we have to think about what we're given and what we're asked for. We're given the averages of the two groups and asked for the ratio of their sizes. Note that it's impossible to know the actual number of people in each group--one major clue to this is the fact that the test asks for the ratio.

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