The SAT Prep Black Book(114)

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.

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 360o. So we can find the value of c by beginning with 360o and subtracting out all the other angle measurements, giving us 360 - 2a - 3b, 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.





Page 598, Question 18


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.