SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (42 page)

On the day that I panicked in front of my students, for some reason I completely abandoned this principle and started writing out extremely complicated algebraic expressions trying to relate the different measurements of the various angles to one another. It was a mess. In general, if you find yourself writing out gigantic algebraic expressions to solve an SAT Math question, you’re probably doing it wrong. At the very least, you’re doing it in a much, much more complicated way than necessary.

So anyhow, when I looked back at the question a second time over my lunch break, I immediately saw the solution—and it was much, much easier than anything I had previously thought of. In fact, it took about 5 seconds to do, and didn’t even involve picking up a pencil.

The fast solution comes from noticing that the 3 small triangles with angle measurements marked in them all combine to form a quadrilateral, and the angle measurements in a quadrilateral add up to 360
o
. So we can find the value of
c
by beginning with 360
o
and subtracting out all the other angle measurements, giving us 360 - 2
a
- 3
b
, which is what (E) says.

There’s a very big lesson in the mistake I made on this question when I first saw it—it’s one I apparently needed to be reminded of at the time, and one you need to learn now if you want to do well on the SAT Math section. Remember that questions on this test can be answered quickly and usually pretty simply. Remember that having to write out long algebraic expressions or go through 15 steps to get an answer means you’re not looking at the question in the best way. Remember to look at the answer choices (I would have seen that 360 appears twice in the choices, which should have been a dead giveaway that a 4-sided figure might be involved, since there are clearly no circles in the question). Remember, above all, that this is the SAT, and the typical school approaches to math just don’t work here
most of the time.

At first glance, this question might look like a normal permutation question. But when we read closely, we realize that this question has a unique aspect that makes it different from typical permutation questions. In this question, one of the cards is forbidden to appear in two of the five positions.

So we can’t use a standard factorial calculation here, because the factorial wouldn’t take into account the limitations on the gray card. The factorial would tell us the total number of arrangements if every card could appear in every position.

But we can use the underlying logic that makes the factorial work, and modify how we apply it so that it fits the current situation. I can think of two ways to do that.

For the first way, we’ll start with the idea that we’re going to figure out the number of cards that can appear in each position, and then we’ll multiply all the possibilities for each position together so we arrive at the total number of possible outcomes for all five positions together.

Since we have to make sure we don’t end up with the gray card last, I’d figure out the number of possibilities for the end positions first, and then the number of possibilities for the middle positions.

For the first end position, we can put any one of 4 cards (any of the non-gray cards).

For the second end position, we can put any one of 3 cards (there are 3 non-gray cards left after we use the first one for the first end position).

For the first non-end position, we can put any one of 3 cards (there are 2 non-gray cards left, plus the possibility of the gray card).

For the second non-end position, we can put any one of 2 cards.

Finally, for the last non-end position, there will only be one card remaining.

So multiplying the number of possibilities at each position gives us  4 * 3 * 2 * 1 * 3, or 72.

Another way to go would be to figure out the number of possible arrangements if the gray card could go anywhere, and then subtract out the possibilities with the gray card at either end.

If we didn't care about where the gray card went, there would be 120 possible arrangements of the cards, because there would be 5 possible cards for the first slot, 4 for the second slot, 3 for the third, 2 for the fourth, and 1 for the fifth, and 5 * 4 * 3 * 2 * 1 = 120.

Now we have to subtract out the situations in which the gray card is at the first position or the last one. Since there are 5 cards, it stands to reason that each card is in each of the five positions for 1/5 of the 120 arrangements. We don't want to count the 1/5 of 120 where the gray card is first, and we don't want to count the 1/5 of
that 120 where it's last. So we want 120 - 2/5(120), or 120 – 2(24), or 120 – 48, or 72.

Either of these approaches is perfectly valid, and there are probably other ways you could choose to tackle this question as well. I would probably advise against trying to list out all of the possible outcomes here, for two reasons. First of all, there are too many things involved, so it would probably take
too long. Secondly, there are no answer choices, so if you miscounted by even one possible arrangement you’d end up with the wrong answer (if there were answer choices and you ended up differing from one of the choices by only one, you could probably assume that you had miscounted, but without answer choices that will be harder to catch).

This question is one more great example of why we don’t really need to use formulas very often on the SAT. There’s no formulaic way to attack this question
that the average high school student will be familiar with in advance; instead, we have to think about the bizarre situation it presents us with, and then figure out a way to respond to that.

Students ask about this question all the time. It’s probably one of the most time-consuming questions in the entire Blue Book.

For this one, we have to realize that we need the equation for lin
el
, so we can plug
t
and
t
+1 in for
x
and
y
, and then solve.

In order to figure out the equation, we need two things: the
y
-intercept, and the slope.

It may seem hard to figure out the
y
-intercept for lin
el
, because we don’t seem to have enough information. But, as always, we need to remember that the SAT gives us enough information to answer questions, and we need to remember that careful reading is very important. We actually can figure out the
y
-intercept for lin
el
, because we're told that the line goes through the origin, which means that its
y
-intercept is 0. So now we just need the slope.

We can find the slope because it
must be the negative reciprocal of the slope of the other line, since the two lines are perpendicular. Since the other line has a slope of -4,lin
el
has a slope of 1/4.

Now we know that lin
el
has the equation
y
= (1/4)x, so we just plug in
t
and
t
+ 1 and solve:

t
+ 1 = (1/4)
t
              (original equation with
t
and (
t
+ 1) subbed for
x
and
y
)

4
t
+ 4 =
t
              (multiply both sides by 4)

3
t
= -4                            (combine like terms)

t
= -4/3                            (isolate
t
)

So (A) is correct.

Looking at the other answer choices, I would be a little worried because it looks like the choices are trying to distract me from choice (E)—choice (E) is the only one with both its reciprocal and its opposite in the answer choices. So I would double-check my work to make sure that I hadn’t accidentally switched my signs or something.

Test-takers miss this question all the time, even though it only involves basic arithmetic. This is just one more example of how important it is to pay attention to details on the SAT, and really make sure we lock down every question we can.

If we’re familiar with the definitions and properties of words like “remainder” and “factor,” we can work this out
logically. If we were going to think about it in the abstract, we’d realize that if the remainder is going to be 3, then we know that we’re looking for factors of 12 that aren’t also factors of 13, 14, and 15, because 15 - 12 is 3. The factors of 12 are 1, 2, 3, 4, 6, and 12. But 1, 2, and 3 don't work, because they’re all factors of either 14 or 15, so we're left with 4, 6, and 12, and the answer is that there are three possible values for
k
.

The more concrete way to do this would be to list all the numbers of from 1 to 15, and then write down the remainders when 15 is divided by each of them:

15/1 = 15 r 0

15/2 = 7 r 1

15/3 = 5 r 0

15 / 4 = 3 r 3

15/5 = 3 r 0

15/6 = 2 r 3

15/7 = 2 r 1

15/8 = 1 r 7

15/9 = 1 r 6

15/10 = 1 r 5

15/11 = 1 r 4

15/12 = 1 r 3

15/13 = 1 r 2

15/14 = 1 r 1

15/15 = 1 r 0

So we can see that there are three numbers that produce a remainder of 3
, and they are 4, 6, and 12.

So (C) is correct.

When people get this wrong, it’s either because they’ve forgotten the meaning of the word “remainder,” or because they’ve miscounted the number of things that produce a remainder of 3. Both of these are elementary math mistakes—exactly the kind of thing that you can’t let happen on the SAT Math section.

Most test-takers will try to apply the distance formula here (or else they’ll fail to remember the distance formula and then just take a guess at the answer based on nothing).

The distance formula would ultimately work here, but I prefer to think in terms of the Pythagorean theorem—and, after all, the distance formula is just one specific application of the Pythagorean theorem anyway.

(The Pythagorean theorem says that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a
2
+
b
2
=
c
2
, where
c
is the hypotenuse and
a
and
b
are the legs. If you forget the Pythagorean theorem, you can always go to the beginning of any SAT Math section and find it in the box of provided formulas.)

Some people would choose to diagram this question. Let’s work it out mathematically first, and then we’ll see what a diagram might look like.

If we’re thinking in terms of the Pythagorean theorem, the distance between the two points will be the hypotenuse in the theorem, and the horizontal and vertical changes between the two points will be the legs in the theorem.

We know that the hypotenuse, or the distance, is 17.

We know that the separation in the
y
axis is 15, because point
B
has a
y
value of 18 and point
A
has a
y
value of 3, and 18 - 3 is 15. We’re looking for one possible value of
x
, so we need to know how big the separation in the
x
-axis must be in order for it to be true that
a
2
+
b
2
=
c
2
when
a
is 15 and
c
is 17. So let’s figure it out. 15
2
is 225, and 17
2
is 289. So that gives us this:

225 +
b
2
= 289              (Pythagorean theorem)

b
2
= 64                            (combine like terms)

b
= 8 or -8              (isolate
b
)

So we know that the separation in the
x
-axis must be 8. (Don’t make the mistake of thinking that 8 is the answer to the question! 8 is just the length of the separation in one axis, but the question is asking for an actual coordinate in the
x
-axis.)

Since the
x
-axis separation is 8 units and the
x
-value for
B
is 10, we know that the
x
-value for point
A
must be either 2 or 18. So the answer can be either 2 or 18.