SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (36 page)

Or it might look like this:

Or anything in between.

At first, this uncertainty might seem troubling, but we have to remember that this is an SAT Math question, and that SAT Math questions often deal in uncertainty. So there’s no need to panic—we just need to figure out how it can be possible that the test is allowing there to be multiple arrangements here.

One clue is the phrasing of the question: it asks for “one possible value” of the BC length. As trained test-takers, we know that the College Board only uses that phrase in a question when there are multiple possible values involved. So it’s clearly okay for us not to be sure exactly where A, B, D, and E are relative to one another in every case. We only need to work out one single possible arrangement. So let’s pick the one where D and E are very close together, like this:

Now the question tells us that CD is 2 units. So if we put C in between B and D, and 2 units away from D, we get something like this:

In other words, if we put D and E basically right next to each other, and B is 3.5 units away from E, then let’s say it’s 3.49 units from D. If C is 2 units from D, then B is 1.49 units from C, because
3.49 – 2 = 1.49.

Now, there are other possible answers here, so your work doesn’t have to look exactly like mine. What’s most important, I think, is not to get too thrown off by the fact that more than one outcome can be correct. For a question like this, we don’t have to figure out the entire range of possible solutions, as we might do in a math class in school. We just have to find one possible value and
realize that there are other possible values out there.

Note that this question is fairly challenging for a lot of test-takers even though it only involves the most basic arithmetic (addition and subtraction) and the most basic geometry concepts (points on a line). This is the kind of thing you have to look out for on the SAT: simple concepts presented in strange ways.

This is yet another question that can’t really be set up with a typical algebraic formula. So let’s just think about it.

If the ratio of rainy days to sunny days is 3 to 2, then we can think about the days in terms of “blocks” of 5 days, because every 3 days of rain requires 2 days of sun, and 3 + 2 = 5.

So if we’re dealing with blocks of 5 days, and we have a 30-day month, then there can be 6 blocks of days in that month, because 6*5 = 30.

That means that there will be 6*3 rainy days, and 6*2 sunny days. So there will be 18 rainy days and 12 sunny days. (Just to be sure we haven’t made a mistake, at this point I’d probably quickly add 18 and 12 in my head to confirm that it’s 30, just like it needs to be.)

Now we have to be very careful at this point. The question is asking how many more rainy days there were than sunny days—the question is NOT asking how many rainy days there were, nor how many sunny days there were. It’s asking for the difference between those two numbers.

I’m sure a lot of people either answer 18 or 12 to this question, because they forget what it was asking, but the real answer is 6.

This is yet another example, then, of the critical role that reading comprehension and attention to detail will play in your SAT preparation.

Again, we have to be sure to read carefully. And we should probably try to avoid using traditional, formal series notation on this question. As is usually the case, this question will probably be a lot easier if we just think about it and figure it out using basic reasoning.

The gap between the 3rd term and the 6th term is exactly 3 terms (because 6 – 3 = 3). In terms of units, the same gap is 60 units, because 77 – 17 = 60.

So a difference of 3 terms corresponds to a difference of 60 units. That means each term is a difference of 20 units, because 60/3 = 20. (The fact that the actual arithmetic has been pretty easy so far is a good sign that we’re approaching the question correctly.)

We can confirm that each term is 10 more than the one before it by filling in the gaps in the original sequence: 17, 37, 57, 77.

So i
f each term is 20 units and the 6th term is 77, then we know the 7th term is 97 and the 8th term is 117.

Remember: read carefully, and don’t be afraid to think about basic math concepts in new ways. There’s a reason the College Board made the numbers fairly easy to work with here: they didn’t want you to have to use a formula. They wanted to give you the chance just to sit back for a second and think about the question, and count stuff out if you needed to. They could have asked for the thousandth term, or they could have made the difference between terms equal 13.4 instead of 20, or who knows what, but they didn’t. Instead, they made the arithmetic part fairly easy once we figured out what was going on.

In this question, the phrase “least value” indicates that there are at least two values that
x
might have. We need to keep that in mind as we work through the question.

We could choose to do this in a formal algebraic way. That would look like this:

|
x
– 3| = 1/2

x
– 3 = 1/2              OR             
x
– 3 = -1/2              (create the two absolute value possibilities)

x
= (1/2) + 3              OR             
x
= (-1/2) + 3              (isolate
x
)

x
= 3.5                            OR             
x
= 2.5                            (simplify the right-hand side)

Then we’d know that 2.5 is the correct answer, because it’s the lower of the two possible values.

Instead of the algebra, we could also just look at the expression in the question and realize that it’s basically just telling us that
x
is 1/2 a unit away from 3 on a number line. The two numbers that are 1/2 unit away from 3 are 2.5 and 3.5, and the smaller one of those is 2.5, so that’s the answer.

(If that second approach didn’t make any sense, don’t worry about it. It was just another way to go, that’s all. The algebra works fine too, of course.)

This is a question that many test-takers don’t even understand when they first read it, because it sounds quite odd—usually we don’t describe a “four-digit integer” with a string of capital letters like
WXYZ
. But in these cases we should always trust that the College Board will explain what it means—it has to, or else the question can’t be asked. So let’s keep reading.

The question says that each capital letter represents a digit in the number, and then it tells us some things about the relationships of the values of those digits. Our job is to figure out what the values are.

Many people will try to approach this question like a system of equations. I suppose that might be possible with a lot of effort, but it won’t come easily to anyone.

Another popular (but largely unsuccessful) approach is to try picking
numbers for each digit and play around with it until you hit on a working arrangement.

Let’s try something else, though. Let’s just look at what’s going on and try to think about what it says. And, above all, let’s remember not to panic just because this is a weird question. By now we know that the SAT likes to try to scare us by asking questions that look weird. No big deal.

One thing that I’d notice is that
X
is clearly the biggest number, because it’s the sum of all the other numbers. Okay, that might come in handy.

What’s a little less obvious, but still clear, is that
Z
is the smallest number. We know this because rule 2 tells us that
Y
is 1 less than
W
, but rule 3 tells us
Z
is 5 less than
W
.

So, since
Z
has to be the lowest digit out of these 4 digits, why not see what happens if we make it equal zero, which is the smallest digit possible? If we try setting
Z
equal to zero and it works, great! If it doesn’t work, we’ll probably figure out some more information about the question as we try to use 0 for
Z
. So let’s give it a shot and see what happens.

If
Z
is 0, then rule 3 tells us that
W
is 5.

If
W
is 5, then rule 2 tells us
Y
is 4.

Plugging all of that into rule 1, we would get that
X
equals 5 + 4 + 0, or 9. That makes sense, since we said before that
X
must be the biggest digit.

So the answer will be that
WXYZ
corresponds to 5940.

You might wonder how I knew to start with the idea of
Z
being 0. But I didn’t know beforehand that 0 would work; it was just an informed hunch. If I had started with
Z
as 2, I would have seen my mistake, adjusted, and tried again. And I didn’t even have to start with a value for
Z
. I might have started with
W
being 8 or something. In any case, the important thing isn’t to try to nail the question on the first guess; the important thing is to be willing to play around with the results until you get something to work.

And l
et me make something else extremely clear about this question: the lesson to be learned here is not how to approach future SAT Math questions about mysterious 4-digit numbers whose digits have particular relationships to one another. The chance of you ever seeing a question like this on the test again is basically zero. The much more important thing to try to pick up on is the general thought process—the way we read carefully, think about what the words mean, avoid using formal solutions whenever possible, and just kind of experiment with the question until we find an answer. That underlying skill set is critical on the SAT.

This question, like so many others, manages to be challenging even though it only involves
a simple  idea—in this case, the idea of equilateral triangles. As will often be the case on the SAT, there is no way to apply a formula or use a calculator. Instead, we just have to think about it.