SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (41 page)

When you have a question like this where the answer choices are all relatively small numbers, that's a sign from the test that counting the options is the best way to go. On the other hand, if the number of options was much higher, that would be a sign that there's a solution where you don't actually have to count things.

From the beginning, though, we want to notice that each answer choice is only one more than the choice before it, so if we make a very small mistake and end up overlooking one of the possible combinations, or accidentally counting a combination twice, there will be a wrong answer choice waiting for us. Remember that it’s always very important to pay attention to small details on the SAT, but it’s especially important on questions like this.

So at this point I would go ahead and list the possibilities systematically, to make sure I don't skip anything.
In this case, I would start with the smallest-value tokens and then work up from there.

o
Seventeen 1-point tokens

o
Twelve 1-point tokens and one 5-point token

o
Seven 1-point tokens and two 5-point tokens

o
Seven 1-point tokens and one 10-point token

o
Two 1-point tokens and three 5-point tokens

o
Two 1-point tokens, one 5-point token, and one 10-point token

So the answer is (E), six.

Normally I would be suspicious of liking the largest answer choice in a series, because the College Board usually likes to make the correct answer be somewhere in the middle of a series if one appears in the answer choices. So I would double-check my counting to make sure that everything I listed was really valid.

I’d also like to reiterate
that this question is a great example of how SAT math can be "tricky" rather than really difficult. There's no advanced math concept or common formula involved here or anything, just a simple arithmetic issue where it's very easy to get mixed up and overlook some options.

This question involves a lot of geometric concepts, but as long as we think through them carefully we should have no problem navigating it—after all, since this is the SAT Math section, all of the individual concepts involved must be fairly simple on their own, even if they’re combined in ways that might be strange.

The first idea mentioned in the question is that of a cube with volume 8. If the volume is 8, then the side length of the cube must be 2, because 2 cubed is 8.

Then we’re told that the cube is inscribed in a sphere. That’s a hard thing to represent with a diagram on a two-dimensional page, so let’s describe it with words instead: think of a throwing die stuck inside a ping-pong ball so that it can’t move.

Now the question asks for the diameter of the sphere. Since we were only given one numerical measurement in the entire question (the volume of the cube), it must be possible to figure out the diameter of the sphere from some measurement related to the cube.

At this point, we need to realize that the distance from one corner of the cube to the very opposite corner of the cube (in other words, the distance to the corner on the other side of the center of the cube) is the same as the diameter of the sphere. I’ll make a diagram and leave the sphere out of it so you can see the distance I’m talking about. The dashed line is the distance we want to find:

To find this distance, we have to use the Pythagorean theorem twice. First, we’ll use it to find the distance of the diagonal across one of the faces of the cube; next, we’ll use that diagonal as a leg to find the actual corner-to-opposite-corner distance we’re looking for
. So here’s step 1:

That diagonal across the bottom face is the hypotenuse of a 45
o
-45
o
-90
o
triangle with legs of length 2, so its length is 2√2.

Notice that the diagonal of 2
√2 now forms the leg of another right triangle whose hypotenuse is the distance we’re looking for, and whose other leg is one of the vertical edges of the cube.

This new triangle then has legs of 2
√2 and 2, which means the hypotenuse is the square root of the sum of 2√2
2
and 2
2
. That sum is 12, so the hypotenuse is the square root of 12, which we can simplify like this:

√
12 = √4 * √3

= 2
√3

So (D) is correct.

I would like to add that I have always felt like there is a much simpler way to approach this question, but I don’t quite see it. Note that the values in the answer choices are all pretty well spread out from one another, for the most part—I’m pretty sure there’s something we’re supposed to be able to notice from the setup that might let us realize that the correct answer must be a value between 3 and 4 or something. The reasoning would go something like this: 2 is much too small to be the distance across the cube because it’s already the side-length of the cube, and 4 is much too big, because it’s twice the side-length. I feel like there must somehow be a way to tell that 2.5 is also too small. Otherwise, I don’t see any reason for the College Board to have included the decimal approximations in choice (B) and choice (D), because the College Board only seems to include these approximations when there’s some kind of rough reckoning that can be used to rule out certain choices.

So if you’d like a little mental exercise, see if you can figure out a way to eliminate values of 2.5 or smaller in this question without doing the actual math. I’m pretty sure that will leave you with the fastest possible solution to the question.

(If, like me, you can’t see a way to do that, there’s no real problem—we should still have plenty of time left over from answering most of the other questions as quickly and efficiently as possible.)

This question offers a classic example of the kinds of information we can glean from a question by thinking of the answer choices from the very beginning.

Most test-takers will try to answer this question by coming up with an algebraic expression on their own and then looking in the answer choices to find a match. That approach can work if you do it perfectly, but it’s very challenging for some people.

What I would recommend instead is to think about the similarities and differences in the answer choices and how they might be relevant to the concepts in the question.

If we look at the elements in the answer choices, we see that each choice is a fraction, with either
n
or 100
n
on the top, and either
n
+ 75 or 2
n
+ 75 on the bottom (choice (C) also has 100 on the bottom).

So we basically only need to figure out the answers to these questions:

1. Should 100 be involved in the correct expression? If so, in the numerator or in the denominator?

2.
Should the denominator contain the expression
n
+ 75 or 2
n
+ 75?

Let’s think about that. As for the issue with 100, we’d want to realize that the question is asking for a percentage but the answer choices are all fractions. Ideally, thinking about the percentages-versus-fractions issue along with the idea of 100 should remind us that we have to multiply fractions by 100 to turn them
into percentages (because fractions describe a portion of a single unit, and percentages describe a portion of 100 units). So we should have 100 in the correct expression, and it should be in the numerator because we’re multiplying by it.

Now, should the denominator have
n
or 2
n
? For a lot of people, the temptation is to say that 2
n
doesn’t make any sense, because it doesn’t appear in the text of the question. But there are a couple of clues in the answer choices that should make us re-examine that assumption. For one thing, if 2
n
were just some pointless, random mistake, we wouldn’t expect to see it repeated across multiple answer choices; instead, we’d expect that other choices would have other random values like 3
n
or 4
n
. On top of that, 2
n
+ 75 actually appears
more often
in the denominator than just
n
+ 75, which would suggest, according to the answer choice patterns we talked about for the SAT Math section, that 2
n
is actually the correct version. (Just to be clear, there’s no guarantee that 2
n
is correct just because it shows up more often, but in general the elements that show up more often will tend to be part of the correct expression.)

So we really want to revisit this idea of 2
n
. Is there any way it could make sense—either as the correct answer, or as an understandable mistake?

Actually, it makes sense as the correct answer when we realize the question is asking us to compare the number of male students to the entire number of students in the whole college. So the number in the denominator needs to reflect both the male and female students together. We know that the number of female students is given by
n
+ 75, and the number of male students is given by
n
. So the sum of the female and male students will be
n
+ 75 +
n
, or 2
n
+ 75.

That means (E) must be correct.

(Notice, by the way, that 100 appears in 3 of the 5 answer choices, which strongly suggests it’s part of the right answer. In those 3 appearances, it shows up twice in the numerator, which suggests—but, again, does NOT guarantee—that it ought to be in the numerator in the correct answer.)

I remember being asked about this question for the first time in a live class I was teaching literally the first weekend after the second edition of the Blue Book came out. I had never seen the question before and for some reason I completely panicked in front of my students (which I almost never do). I had to admit sheepishly that I wasn’t seeing whatever I needed to be seeing, and ask the students to allow me to give them the solution over lunch break when I would have time to think more clearly. (This, by the way, is the only time in my life I ever needed to do that. But sometimes these things happen. As it would turn out, all of my troubles were caused by straying from my normal game plan because I let myself get stressed out. More on that in a moment.)

The difficulty in this question, for most people, arises from the fact that it looks like the small triangle at the top of the figure has a nearly horizontal base. We really want that base to be horizontal, because then it will be parallel to the base of the large triangle, and then
c
would just be 180 - a - b, like choice (C) says.

Unfortunately, the base of that top triangle just isn’t horizontal, no matter how we look at it, which means (C) can’t be right. And this is where panic mode might start to set in.

When we panic on the SAT Math section, our normal reaction is to try to make things as complicated and advanced as we can, because that kind of thing usually works on math in school. But on the SAT we want to try to have the opposite reaction, actually—we want to try to look at the question in a simpler, more basic way.