The SAT Prep Black Book(115)

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.





Page 642, Question 17


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 linel, 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 linel, 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 linel, 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,linel has a slope of 1/4.

Now we know that linel 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)

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

3t = -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.





Page 643, Question 20


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.





Page 655, Question 18


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: a2 + b2 = c2, 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.)