The SAT Prep Black Book
The Math Path In Action
To demonstrate how this approach to SAT Math questions works, and to help you improve your understanding and execution of that approach, we”ll go through an SAT Math section from the second edition of the College Board”s Blue Book, The Official SAT Study Guide. (Remember, the Blue Book is the only book I recommend for practice questions, because it”s the only book whose questions are totally guaranteed to play by the College Board”s rules.) I”ve chosen the first practice-test section that includes both Multiple Choice problems and Student-Produced Response problems. It starts on page 413 of the Blue Book.
Page 413, Question 1
You really have to be careful on a problem like this—it”s the kind of thing that most test-takers will feel they can”t possibly miss because it seems so simple. But it”s very easy to overlook a couple of small details on this question and end up with a wrong answer.
The question asks us to figure out how many new houses were built from 1961 to 1990. To answer that question, we have to look at the relevant years on the chart. The first row shows the houses built from 1961 to 1970; the next row shows the houses from 1971 to 1980; the third row shows us the houses from 1981 to 1990. There are 2 house icons in the first row, 4 in the second row, and 8 in the third row. That makes a total of 14 house icons.
Notice that choice (A) reflects a mistake that would be pretty easy to make here: if we assumed that each house icon represented one house, then we”d think (A) was right. But each icon represents 2,000 houses, so the right answer is 28,000. That makes (E) right.
As always, I”d also try to think about where some of the other wrong answers might have come from. In this case, (B) is what we”d get if we counted all the icons in the entire chart (including 1991 – 2000) and also made the mistake of thinking each icon only represented one house. (C) is what we”d get if we correctly realized that each icon represented 2,000 houses but accidentally only counted the third row. Finally, (D) is what we”d get if we didn”t count the second row for some reason.
I always try to figure out where some of the wrong answers are coming from because it helps me be sure I”ve understood the question. It”s not necessary to figure out every wrong answer, even though we were able to do that in this question. But if you can”t figure out where any of them are coming from at all, then there”s a very good chance you made a mistake, and you should re-check your work.
Also note that, in this case, 3 out of the 5 choices are numbers in the tens of thousands. This strongly suggests (BUT DOES NOT GUARANTEE) that the right answer should be one of those 3. If we had accidentally overlooked the information about each icon counting for 2,000 houses, then noticing that most of the answer choices are huge numbers should have helped us realize we might have missed something.
This is a perfect example of the kind of question that you should absolutely lock down, double-check, and be done with in well under a minute, possibly in as little as 10 seconds. It”s important to work quickly and efficiently through questions like this in order to save time for questions where you might have more difficulty gaining a foothold.
One last thing: note that this question, like many SAT Math questions, is primarily a critical reading question at heart. The major way people will make mistakes here is by misreading the question or the chart, not by failing to multiply the numbers 14 and 2,000.
Page 413, Question 2
This question presents us with a type of drawing that we might not ever have encountered before. But that”s fine: remember that this question, like all SAT Math questions, can only be made out of the same basic math ideas that were in the Math Toolbox earlier in this book. So let”s see what we can figure out.
We know the diagram is drawn to scale, because it doesn”t say that it”s not. But the answer choices are so close to one another that it would probably be hard to answer this question by eyeballing the drawing and using the scale.
So let”s think in terms of geometry. We know that we can fill in the measurements of the angles opposite the labeled angles, because opposite angles are identical. So the angle opposite the 35-degree angle is also 35 degrees, and the one opposite the 45-degree angle is also 45 degrees.
Now it looks like we”re getting somewhere: once we label those opposite angles, we can see that w is the third angle in a triangle, and that the other two angles are 35 degrees and 45 degrees.
That means w + 35 + 45 = 180. So w + 80 = 180, which means w is 100 degrees. So (B) is right.
Notice that (B) is in the middle of a 3-term series in the answer choices, with the other two terms differing by 10. This should alert you to the fact that it would be very easy to be off by 10 in this question if we forgot to carry a digit during the addition or the subtraction. So we should go back and check to make sure we didn”t mess that up.
Still, it”s very reassuring that we like the answer choice in the middle of this range, because typically the College Board likes to make the number in the middle of the range be correct. (As we discussed in the section on patterns, they do this because it gives us the option to be off in either direction and still find an answer choice reflecting our mistake.)
This is another question that probably should take less than 30 seconds to do. We want to answer it quickly, check to make sure we”re right, and move on.
Page 414, Question 3
A lot of students ask me about this question because they can”t figure out how to set it up algebraically.
But this is the SAT—we don”t have to do it algebraically! In fact, in most cases I prefer not to use algebra on the SAT, because it often takes longer and it increases the chance that we”ll make a mistake.
So let”s just think about this. (There”s no magical way to “just think about” a question—all we do is dive in somewhere and then see what adjustments, if any, we need to make.)
If each of the 19 tables had 4 people, then there would be seats for 76 people, because 4*19 = 76. We need to find seats for 84 people, though, which is 8 more than 76.
So 8 of the tables will need to have an extra seat, which makes (E) the right answer.
Notice that this isn”t really the kind of question you would probably ever be asked in a math class, and the solution I just gave isn”t the kind of solution your math teacher would ever accept. This is normal for the SAT, and we need to get used to it.
Also notice that the answer choices form a series in this question, and the correct answer happens to be the largest number in the series, which is something that I said doesn”t happen often. I was right—it doesn”t happen often—but that doesn”t mean it never happens. In this case, when I see that the answer I like is the last number in a series, I”d just double-check my work and make sure I hadn”t made a mistake, and move on.
One more thing—you might ask how I knew to start out by multiplying 4 and 19. But I didn”t “know” that would lead to the answer right away until I tried it. On the SAT Math section, we have to be willing to play with questions like this, and we have to get away from the idea of following established formulas.
Page 414, Question 4
This question often overwhelms test-takers who think in terms of school math, because it has two variables in it, and one of those variables (m) is actually impossible to solve for. In school math, it might be a problem if we couldn”t solve for a variable, but this is SAT Math. The SAT often gives us questions in which it”s impossible to work out the value of a particular variable.
Since the question asks which answer choice would be equal to the original expression when a equals 4, and since every answer choice has 4”s in it instead of a, I would start out by plugging 4 in for a, and determining that the original given expression equals 4m2 + 4m + 4. Then we can go through each answer choice and distribute the 4 in each one, until we end up with a choice that also results in 4m2 + 4m + 4 as an answer.
The answer choice that fits the bill is (D).
Notice that the choices here fit a pattern we discussed earlier in this book: the elements of the right answer appear very frequently in the wrong answers.
In other words, in this case, the correct answer has the parenthetical expression with m2 , m, and 1. Notice that 3 out of the 5 choices have m2 somewhere in them, 4 out of 5 have m, and 4 out of 5 have 1 in them. Also notice that 4 out of 5 don”t have a 4 in the parentheses, and 4 out of 5 don”t involve squaring the expression in parentheses (like choice (B) does).
So if we were going to try to predict the right answer, based purely on the answer choices, we”d probably want the choice with these attributes:
odoesn”t have a 4 in the parentheses
odoesn”t square the expression in parentheses
We”d want that answer choice because those are the most popular features in the field of answer choices. The choice that satisfies all of those conditions is (D), which is the right answer in this case.
Just to be completely clear, let me point out that I would never, ever recommend that you answer a question based only on figuring out which choice has the most popular features. That will work a whole lot of times, but it doesn”t work every time, and my goal is for us to be correct on every single question. So this idea is a pattern to be considered, not a rule to be followed unfailingly.
In this particular question, if we accidentally misread (E) and thought it was correct, being aware of the imitation pattern might let us realize that none of the other answer choices have a 4 inside the parentheses, which is probably a bad sign. If we liked (A) for some reason, then noticing that the other choices were all highly similar to one another and that none of them had m3 might help us realize we needed to re-evaluate our conclusion.
Notice that we answered the question without ever finding out what m represents. This kind of thing is normal for the SAT. On the SAT, we need to get away from the idea that we have to solve for every variable we see in a question.
Page 414, Question 5
This question frustrates a lot of students because they”ve never seen a diagram like this in their math classes. But we have to remember that this kind of thing is standard on the SAT: no matter how much you practice, you”re going to encounter things on test day that don”t look like anything else you”ve ever seen, at least on the surface. We have to learn how to attack these kinds of situations systematically, and confidently.
In this case, they”re asking us for the area of the shaded portion. As trained test-takers we know two things:
They have to give us all the geometry formulas we need.
They didn”t give us a formula for the area of a portion of a square.
That means there must be a way to figure out the area of this shaded region without having a formula uniquely for that purpose.
We do have a formula for the area of a circle, though. And we could use that formula to find the area of the whole circle altogether, and then divide the area of the circle up to find the area of the shaded region.
So now we have to figure out how to find the area of the circle. On the SAT, there”s only one way the College Board can ask us to do that: we have to use A = pi(r2), where A is the area and r is the radius. (Again, this formula is provided at the beginning of the math section if you don”t remember it.)
How can we find the radius? In this case, we use the only number provided anywhere in the diagram: we look at the fact that the side length of the square is 2 units, and we realize that this corresponds to the diameter of the circle. If the diameter is 2, then the radius is 1. So the radius of the circle is 1 unit, which makes the area of the entire circle pi(12), which is just pi.
Now we have to figure out what portion of the circle the shaded area represents. There are two ways to do this. Since the diagram is drawn to scale, we could just eyeball it and realize that the shaded area is one-quarter of the circle. If we want to be more precise, we could realize that the angle at point O must be a 90-degree angle, since O is the center of the circle and the center of the square. And 90 degrees is one-quarter of 360 degrees, so, again, the shaded area must be a quarter of the circle.
That means the correct answer here is pi/4. Again, pi is the area of the entire circle, and the shaded region is 1/4 of the circle, so its area is pi/4. And (A) is correct.
There are a lot of ways to mess this question up, and most of them will involve either thinking that the radius is 2 or accidentally thinking that the shaded portion is 1/2 of the circle (this can happen if a person tries to work out the area mathematically and makes a mistake in the process, instead of just looking at the picture). Notice that both mistakes are reflected in the wrong answer choices. In fact, (E) is what you”d get if you made both mistakes together.
This is a question I would definitely double-check, or even triple-check, for several reasons. The first reason is that the answer choice patterns aren”t really pointing to (A) being right, even though it is. The patterns are really suggesting that (B) would be right (it”s in the middle of a geometric series with (A) and (C) and (E), and there are more 2”s in the answer choices than 4”s). Also, even apart from the answer choices, this is a question where it would be very easy to make a simple mistake and be off by a factor of 2 or 4, and the answer choices are clearly waiting for that.
This is a great example of the kind of question that causes the most trouble for the most test-takers. It”s something the vast, vast majority of test-takers have the skills and knowledge to answer correctly, but it”s also something where a simple mistake or two can easily be made, wrecking the question. If you want to improve your SAT Math score, this kind of question is where you should probably focus your energy first. Most people would significantly increase their scores if they just stopped giving away points on questions like this, rather than focusing on questions that seem harder.
Page 415, Question 6
This question asks about perpendicular lines. We need to know that when two lines are perpendicular to one another, their slopes are opposite reciprocals (this was an idea discussed in the toolbox earlier in this book). This idea is crucial to solving this question (at least when it comes to the way most people will solve it), but notice that the idea is a property of perpendicular lines, not a formula. Remember that challenging math questions on the SAT rely much more heavily on properties and definitions than on formulas.
To solve this question in the most straightforward way, it would probably be best to get the expression for the original line into slope-intercept format (the one that looks like y = mx + b). If we do that, we get this:
x + 3y = 12
3y = -x + 12 (get the x on the right-hand side)
y = -x/3 + 4 (isolate y)
So the slope of the given line is -1/3.
A line perpendicular to this must have a slope that is the opposite reciprocal of -1/3, which is 3. So (C) must be correct, because it gives a coefficient of 3 for x in the y = mx + b format, where m is the slope.
Note that we have the usual kinds of wrong answers we”d expect on an SAT Math question. The slope of (B) is the opposite of the correct slope, and we can probably imagine how a person might make that mistake. The slope of (D) has the right sign (it”s positive) but uses 1/3 instead of 3, and, again, that”s an understandable mistake.
Notice, also, that there”s no other slope in the answer choices that has both its reciprocal and its opposite in the choices. This is a manifestation of the imitation pattern, and it”s a good sign that (C) is, indeed, correct.
A less straightforward way to answer this, but one that wouldn”t require knowing that perpendicular slopes are opposite reciprocals, would be to use your calculator. You could graph the original line on your calculator, and then graph each answer choice. This would let you see on your own which choice generated a perpendicular line. Of course, this would be a little time-consuming, and you”d have to make sure you didn”t key in the graphs wrong. But it could be a viable way to solve the question if you had forgotten perpendicular lines have slopes that are negative reciprocals of each other.
Finally, note that the y-intercepts don”t matter here. The question asks about perpendicular lines, and that idea only requires us to consider the slopes, not the intercepts. This question provides yet another example of a situation in which we must consider the answer choices as part of the question, rather than hoping to predict the correct answer choice completely from scratch—in this case, there are an infinite number of valid y-intercepts that could be part of a line perpendicular to the given line.
Page 415, Question 7
This question bothers a lot of people—but, as we often see on the SAT Math section, it really just comes down to basic properties and definitions. In this case, we need to know what a triangle is. (In case you”ve forgotten, it”s a closed three-sided figure.)
If we go through each choice and imagine trying to draw it, we”d see that they can all be drawn except for (E)—if you try to draw a triangle with two sides of 5 units and a third side of 10 units, you end up with just a line segment of 10 units, because in order to reach the endpoints of the long “side,” the two short “sides” have to open up completely to make a straight line.
We could also think of this in terms of the so-called “triangle inequality,” which says that the length of any side of a triangle must be less than the sum of the other two sides. But I deliberately wanted to talk about it without referring to that idea because, as I keep saying, it”s very important to learn to attack SAT Math in an informal way.
There”s another important element here, too. Notice that the question uses the word “EXCEPT.” I can”t tell you how many times people have missed questions like this because they overlooked that word, and ended up just choosing the first answer choice that would work under normal circumstances. This is one more example of how important it is to read everything on the SAT very, very carefully. It”s also a good example of the importance of checking over the other answer choices before you move on—if you overlook the word “EXCEPT” and choose (A) because it”s the first thing you see that works to make a triangle, you can hopefully catch your mistake if you glance through the other choices and notice that (B), (C), and (D) also work.
Page 415, Question 8
There are a ton of things we can learn from this question, especially. Please, please read this whole description and pay attention!
First, this question is rated 5 out of 5 for difficulty by the College Board. In other words, this is a question that a ton of people miss. But, as we”ll see in a moment, it only involves basic arithmetic. In fact, it involves a specific application of an idea from arithmetic that has possibly never, ever come up in a normal math class before. What it does NOT involve is any kind of formula, or any way that a calculator is likely to be helpful. Remember, this is SAT Math, and it”s likely to involve tricks and misdirection more than formulas.
I”d also like to point out, before we even get started, that this is a question in which several of our reliable answer-choice patterns would have allowed us to predict the correct answer without even seeing the question, just from the answer choices alone. I am NOT saying that you should ever try to answer a question just from the answer choices—I”m just saying that the answer choices are hinting very strongly at (C) being the right answer, if we know how to read them. The fact that so many people missed this question shows us that most people aren”t paying any attention to the answer choices, which is part of the reason why most people have a really hard time on the SAT Math section.
This question, then, is a positively classic example of the way most test-takers throw points away for no reason on the SAT Math section. People don”t miss this question because they don”t know basic arithmetic. They miss it because they don”t pay attention to details and they don”t check over the other answer choices and think about SAT patterns.
In other words, they miss this question because of a lack of SAT-specific skills. Don”t be like them.
Okay, enough preamble. Now let”s actually answer the question.
The question asks what percent of the votes were cast for Candidate 1 given that Candidate 1 received 28,000 more votes. Most people will try to solve this by figuring out that 28,000 is 1% of 2.8 million. So far, so good. Then they”ll assume that this must mean Candidate 1 got 51% of the vote. They”ll choose (D), and they”ll move on to the next question without giving this a second thought.
And they”ll be wrong.
Here”s the mistake: if Candidate 1 pulls 51% of the vote, than Candidate 2 must pull 49%, and 51% is 2 percentage points more than 49%, not 1 percentage point more. So in order for Candidate 1 to have a 1 percentage-point margin over Candidate 2, the correct split isn”t 51 – 49.
It”s 50.5 to 49.5.
That”s why (C) is correct.
Notice that there are 3 answer choices that end in a 5, and only 2 that end in a 1. This goes along with the imitation pattern, and it strongly suggests that the correct answer should end in a 5. Also, notice that 3 out of the 5 choices involve decimals, which also suggests that 51% isn”t correct. We could even say that 50.5% is the middle entry in a series of sorts, where the other numbers are 50.05 and 55—it”s not a traditional series in the mathematical sense, but there”s a clear progression with reference to the decimal places, and (C), the right answer, is in the middle of that progression.
This is yet another great example, then, of the tremendous importance of thinking about the answer choices as part of the question. They”re not just there to take up space—the College Board uses them deliberately, in ways that we can exploit.
Page 416, Question 9
Sometimes people make this question a lot harder than it needs to be, by coming up with a decimal approximation for √18 or by converting it to 3√2. But it”s much easier just to start by squaring both sides, so we get 2p = 18. That means p = 9.
Remember that SAT Math gets a lot easier if we look for ways to keep it simple.
Page 416, Question 10
This question, like many SAT Math questions, is really just a matter of knowing definitions and doing basic arithmetic.
When we round the given number to the nearest whole number, we get 2. When we round it to the nearest tenth, we get 1.8. Since 2 – 1.8 is 0.2, the answer is just 0.2.
Notice that this question is pretty impossible to answer if we don”t understand the concept of rounding, or if we don”t know which decimal place represents tenths. Also notice that the math is incredibly basic, but people will still miss this question because of its strange presentation.
Page 417, Question 11
As is often the case, there are many ways to answer this question. We could do it with pure algebra, letting x be the number of towels and writing 6 = 2x/5. But that would be a little more formal than I like to be on the SAT Math section.
So let”s just think through it instead. If 6 towels represents 2/5 of all the towels, than 3 towels must represent 1/5 (because 3 is half of 6 and 1/5 is half of 2/5). So if 3 towels are 1/5 of the towels, then there must be 15 towels, because 3*5 is 15. So the answer is 15.
Just to be clear, there”s nothing inherently wrong with doing it algebraically, as long as you feel comfortable with that and you set it up correctly and don”t make any mistakes. I”m just more comfortable with mental math, and I find that it tends to work a lot better on the SAT, so I encourage it in my students as much as possible.
Page 417, Question 12
This question is very difficult for a lot of test-takers. But if we just remember to read carefully and pay attention to details, we can figure this out.
The question describes 5 points on a line, like this: A B C D E.
We know that AD is 4.5, and BE is 3.5. Why not try to plot that out?
When we do try to plot it, we realize that we don”t have enough information yet. If AD is 4.5 and BE is 3.5, there”s an infinite number of ways we could draw that. For instance, it might look like this:
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.
Page 417, Question 13
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.
Page 417, Question 14
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 if 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.
Page 418, Question 15
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.)
Page 418, Question 16
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 let 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.
Page 418, Question 17
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.
One thing that we should note right away is that the diagram is not drawn to scale. We”ll want to focus on the part that doesn”t seem to be to scale and see if we can figure out what”s going on there.
Another important feature of this question is that it gives us information about the dimensions of the diagram in the text, but that information isn”t labeled on the diagram itself. This means that the information about those dimensions will be the key to solving the problem.
So let”s see what happens when we put all of that together. If CD is 10, and DE is also 10, and EF is also 10, then we know that each of the overlapping triangles has side lengths of 20. In other words, before they were overlapping, the two triangles were both 20 units on every side. And if DE is 10 units, then all the sides of the small triangle are also 10 units, because the two big triangles are equilateral. So a more accurate representation of the situation would be something like this:
With that in mind, let”s just work our way around the figure, starting with C and going clockwise.
CD is 10 units.
DE is 10 units.
EF is 10 units.
FA is 20 units.
From A to the indented corner on the left is 10 units. (Remember, the diagram in the original question isn”t drawn to scale.)
From the indented corner on the left to B is 10 units.
BC is 20 units.
So if we add that all up, we get 10 + 10 + 10 + 20 + 10 + 10 +20 = 90 units.
So the answer is 90.
Notice, once more, that none of the arithmetic in this question was difficult. The difficulty came from figuring out what was being asked, and we did that by reading carefully and knowing to pay attention to the way the College Board draws diagrams.
Page 418, Question 18
The diagram in this question is drawn to scale, so we want to take note of that. Unfortunately, it doesn”t have any kind of scale labeled on it, but we may be able to work around that if it turns out we need to.
The question is also talking about 3 different variables: x, k, and a. This will give a lot of people trouble, because most test-takers who treat SAT Math like school math will think they need to figure out the values of those variables.
But if we look carefully, we see that there”s not enough information to figure out k. Instead of worrying about that, we just realize that k must not matter in the question. Easy enough.
Let”s focus on the end of the question. It asks for the value of a, but all we know about a from the rest of the question is that it”s positive and that g(a – 1.2) = 0.
A lot of people will want to plug a – 1.2 in for x in the original function equation, because that”s the typical knee-jerk reaction that would be appropriate in a school situation. But if we consider that move for a moment, I think we”ll see that it”s not likely to help us very much. We”d end up with this expression, which would be very ugly:
g(x) = k((a – 1.2) + 3)((a – 1.2) – 3)
This doesn”t really look promising, from an SAT Math standpoint. (Now, it”s true that the last question on the grid-ins is often more complicated than other SAT Math questions, but I”d still be reluctant to go about expanding and condensing that hideous expression unless absolutely necessary.)
So let”s try a different tack.
One thing that”s kind of interesting is the fact that we were told the value of g(a – 1.2) is zero. Zero is a unique number with unique properties, particularly when it comes to functions and graphs. On a graph, the zeros of a function are the points where the function crosses the x-axis; in a function equation, we find the roots of an expression by setting the factors of the expression equal to zero.
So let”s think about that for a second. In the equation, the ways to make g(x) come out to zero are to set either (x + 3) or (x – 3) to zero. (k can”t be zero, if it were, every single g(x) value would also be zero, and that isn”t what the graph shows.)
So if either (x + 3) or (x – 3) is zero, that means the g(x) value would be zero when x is either -3 or 3. And that fits with the graph, because the places where the function equals zero seem to be at x = -3 and x = 3. So we know that g(-3) = 0, and g(3) = 0. And that should be kind of our eureka moment: if g(-3) and g(3) are both zero, and g(a – 1.2) is zero as well, than that must mean that (a – 1.2) is the same thing as either (-3) or (3)!
That means that a is either 4.2, or -1.8. But the question tells us a must be positive, so that means it”s 4.2.
Once again, the College Board has created a misleading question out of very basic ideas. In this question, we needed to know that the x-intercepts of a graph are the places where its y-value is zero; we needed to know that 4.2 is bigger than zero but -1.8 is not; we needed to know that 4.2 – 1.2 is 3. That”s about it, from a math standpoint, and none of those ideas is very complicated on its own. I would bet that over 90% of the people who took this test knew every single one of those facts, but they missed the question because they didn”t realize it was asking about those facts in the first place.
So the challenge came from the sheer weirdness of the question, and we had to use our reading skills to understand what was being asked, and our thinking skills to realize that the idea of the zeroes of the function was very important.
By now, you”re probably beginning to develop a strong appreciation for the strange way the College Board designs SAT Math questions. In the next section, we”ll continue to develop your understanding by exploring a large selection of questions from the Blue Book that most people have some trouble with.
If you”d like to see videos of some sample solutions like the ones in this book, please visit http://www.SATprepVideos.com. A selection of free videos is available for readers of this book.