The SAT Prep Black Book(112)

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.

oSeventeen 1-point tokens

oTwelve 1-point tokens and one 5-point token

oSeven 1-point tokens and two 5-point tokens

oSeven 1-point tokens and one 10-point token

oTwo 1-point tokens and three 5-point tokens

oTwo 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.





Page 548, Question 16


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 45o-45o-90o 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√22 and 22. 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.