Alternatively, the question stem asks for an approximate answer, so you
can also try estimating from the start. Draw the triangle carefully and start
with the same perpendicular line as before. This line is a little shorter than
the side of length
 (which is about 1.4). Call the two shorter legs 1.2
and calculate the hypotenuse. It equals 1.2 multiplied by 1.4, or
approximately 1.7. (Bonus question: How can you estimate that math
quickly? Answer below.)
Now, examine the answer choices using 1.4 for
 and 1.7 for
 :
They’re all close, but you can pretty confidently eliminate answers (A) and
(E). Furthermore, the answer needs to be less than 2, so (B) can’t be it.
Answer (C) is closer than (D), so (C) is probably it. Unfortunately, you might
guess wrong at this point. But the odds are much better than they were
at the outset.
It is worthwhile to look for multiple solution paths as you practice. Your
top-down brain will become faster, more organized, and more flexible,
enabling your bottom-up brain to have more flashes of insight.
That was a substantial introduction. Now, on to Chapter 1!
3.4(A)
2(B)
1.7(C)
1.4(D)
1(E)


PART ONE
Problem Solving and Data
Suﬀiciency Strategies
In This Part
Chapter 1: Problem Solving: Advanced Principles
Chapter 2: Problem Solving: Strategies & Tactic
Chapter 3: Data Suﬀiciency: Principles
Chapter 4: Data Suﬀiciency: Strategies & Tactics


CHAPTER 1
Problem Solving: Advanced
Principles


In This Chapter...
Principle #1: Understand the Basics
Principle #2: Build a Plan
Principle #3: Solve—and Put Pen to Paper
Principle #4: Review Your Work


Chapter 1
Problem Solving: Advanced
Principles
Chapters 1 and 2 of this book focus on the more fundamental of the two
types of GMAT math questions: Problem Solving (PS). Some of the content
applies to any kind of math problem, including Data Suﬀiciency (DS).
However, Chapters 3 and 4 deal specifically with DS issues.
This chapter outlines broad principles for solving advanced PS problems.
You’ve already seen very basic versions of the first three principles in the
introduction, in the dialogs between the top-down and the bottom-up
brain.
As mentioned earlier, these principles draw on the work of George Pólya,
who was a brilliant mathematician and teacher of mathematics. Pólya was
teaching future mathematicians, not GMAT test-takers, but what he said
still applies. His little book How to Solve It has been in print since 1945—it’s
worth getting a copy.
In the meantime, keep reading.


Principle #1: Understand the Basics
Take time to think and plan before you start solving a diﬀicult problem. If
Quant is your strength, you may want to dive straight into every problem
as soon as you see it, without pausing to consider all of the angles. There
are two good reasons to slow down:
To remind yourself to slow down and plan, understand each problem by
taking three steps:
Glance at the entire problem: is it PS or DS? If it’s PS, glance at the answer
choices. If it’s DS, glance at the statements. Knowing what type of problem
you’re dealing with will help you read more eﬀectively.
Pólya recommended that you ask yourself simple questions as you read a
problem. Here are some great Pólya-style questions that can help you
understand:
You need to manage your time and mental energy across the entire
GMAT. If you pause briefly to find a more eﬀicient solution, you’ll save
time and energy for other problems.
1.
If you start doing math without thinking first, you might have to change
your approach later in the problem, which takes time that you don’t
have. You also risk falling for traps.
2.


You don’t need to meticulously go through every one of these questions
whenever you solve a problem. (However, that’s a good thing to do when
you review a problem!) They’re here to help you consider how you might
read more productively.
As you read the problem, jot down any given numbers or formulas on your
scrap paper. That doesn’t mean you should start doing math while you’re
trying to read. If you start trying to solve the problem when you haven’t
even finished reading it, you’re getting ahead of yourself.
Start the following problem by taking these three steps: Glance, Read, and
Jot.
What exactly is the problem asking for?
What information would I need in order to find the answer?
What information do I already have?
What information don’t I have?
Sometimes you care about something you don’t know. This could be
an intermediate unknown quantity that you didn’t think of earlier.
Other times, you don’t know something, and you don’t care. For
instance, if a problem includes the quantity 11! (11 factorial), you will
practically never need to know the exact value of that quantity.
What is this problem testing? In other words, why is this problem on the
GMAT? What aspect of math are they testing? What kind of reasoning do
they want me to demonstrate?
Have I seen a problem like this before? Have you already solved a similar
problem? What approach worked best?


Try-It #1-1
x = 910 − 317 and
 is an integer. If n is a positive integer that has
exactly two factors, how many different values for n are possible?
Glance. This is a PS problem. The answers are numbers but in written form;
this format is reserved for problems that ask for the number of numbers or
number of possibilities for something. The numbers are small.
Read. Dive into the text. Here are some possible answers to the Pólya
questions:
What
exactly is
the
problem
asking
for?
The number of possible values for n.
This means that n might have multiple possible values. In fact, it probably can
take on more than one value.
I may not need these actual values. I just need to count them.
What are
the
quantities
I care
about?
I’m given x and n as variables. These are the quantities I care about.
One(A)
Two(B)
Three(C)
Four(D)
Five(E)


What do I
know?
x = 910 − 317
That is, x = a specific large integer, expressed in terms of powers of 9 and 3.
 is an integer.
That is, x is divisible by n, or n is a factor of x.
Finally, n is a positive integer that has exactly two factors.
Prime numbers have exactly two factors. So I can rephrase the information: n is
prime. (Primes are always positive.)
What
don’t I
know?
Here’s something I don’t know: I don’t know the value of x as a series of digits.
Using a calculator or Excel, I could find out that x equals 3,357,644,238. But I don’t
know this number at the outset. Moreover, because this calculation is far too
cumbersome, it must be the case that I don’t need to find this number.
What is
this
problem
testing?
From the foregoing, I can infer that this problem is testing Divisibility & Primes. I
will probably also need to manipulate exponents, since I see them in the
expression for x.
You can ask these questions in whatever order is most helpful for the
problem. For instance, you might not look at what the problem is asking
for until you’ve understood the given information.
Jot. As you decide that a piece of information is important, jot it down on
your scrap paper. At this point, your scrap paper might look something like
this:


Principle #2: Build a Plan
Next, think about how you will solve the problem:
Reflect. Here are some Pólya questions that help you think about what you
know and come up with a plan:
Is a good
approach
already
obvious?
From your answers above, you may already see a way to reach the answer. If
you can envision the rough outlines of the correct path, then go ahead and
get started.
If not, what in
the problem
can help me
figure out a
good
approach?
If you are stuck, look for particular clues to tell you what to do next. Revisit
your answers to the basic questions. What do those answers mean? Can you
rephrase or reword them? Can you combine two pieces of information in any
way, or can you rephrase the question, given everything you know?
Can I
remember a
similar
problem?
Try relating the problem to other problems you’ve faced. This can help you
categorize the problem or recall a solution process.


Organize. For the Try-It #1-1 problem, some of the information is already
rephrased (reorganized). Go further now, combining information and
simplifying the question:
Given: n is a prime number    AND    n is a factor of x.
Combined: n is a prime factor of x.
Question: How many diﬀerent values for n are possible?
Combined: How many diﬀerent values for n, a prime factor of x, are possible?
Rephrased: How many distinct prime factors does x have?
You need the prime factorization of x. Notice that n is not even in the
question anymore. The variable n just gave you a way to ask this
underlying question.
Consider the other given fact: x = 910 − 317. It can be helpful initially to put
certain complicated facts to the side. At this stage, however, you know that
you need the prime factors of x. So now you have the beginning of a plan:
factor this expression into its primes.


Principle #3: Solve—and Put Pen to
Paper
The third step is to do the work: solve.
You’ll want to execute that solution in an error-free way—it would be terrible to
get all the thinking correct, then make a careless computational mistake. That’s
why we say you should put pen to paper.
In the expression 910 − 317, the 3 is prime but the 9 is not. Since the problem is
asking about prime factors, rewrite the equation in terms of prime numbers:
Next, pull out a common factor from both terms. The largest common factor is
317:
Now, you have what you need: the prime factorization of x. The number x has
three distinct prime factors: 2, 3, and 13. The correct answer is (C).
The idea of putting pen to paper also applies when you get stuck anywhere
along the way on a monster problem.


Think back to those killer Try-It problems in the introduction. Those are not the
kinds of problems you can figure out just by looking at them.
When you get stuck on a tough problem, take action. Do not just stare, hoping
that you suddenly get it.
Instead, ask yourself the Pólya questions again and write down whatever you
can:
This way, your top-down brain can help your bottom-up brain find the correct
leads—or help it let go. In particular, it’s almost impossible to abandon an
unpromising line of thinking without writing something down.
Think back to the sequence problem in the introduction. You’ll keep seeing 64
as 82 unless you try writing it in another way.
Do not try to juggle everything in your head. Your working memory has limited
capacity, and your bottom-up brain needs that space to work. A multistep
problem cannot be solved in your brain as quickly, easily, and accurately as it
can be on paper.
As you put pen to paper, there are three themes you'll want to keep in mind.
1. LOOK FOR PATTERNS
Reinterpretations of given information or of the question
Intermediate results, whether specific or general
Avenues or approaches that didn’t work


Every GMAT Quant problem has a two-minutes-or-faster solution path, which
may depend upon a pattern that you’ll need to extrapolate. You’ll know a
pattern is needed when a problem asks something that would be impossible to
calculate (without a calculator) in two minutes. When this happens, write out
the first five to eight items in the sequence or list in order to try to spot the
pattern.
Try-It #1-2
 for all integer values of n greater than 1. If S1 = 1,
what is the sum of the first 61 terms in the sequence?
Nobody is going to write out all 61 terms and then add them up in two minutes.
There must be a pattern. The recursive definition of Sn doesn’t yield any secrets
upon first glance. So write out the early cases in the sequence, starting at n = 1
and looking for a pattern:
−48(A)
−31(B)
−29(C)
 1(D)
 30(E)


etc.
The terms of the sequence are
 . . . . Three terms
repeat in this cyclical pattern forever; every third term is the same. Note: If you
don’t spot a pattern within the first five to eight terms, stop using this approach
and see whether there’s another way (including guessing!).
The problem asks for the sum, so find the sum of each group of three
consecutive terms:
 . There are 20 groups in
the first 61 terms, and one additional term that hasn’t been counted yet. So the
sum of the first 61 terms is as follows:


The correct answer is (C).
It is almost impossible to stare at the recursive definition of this sequence and
discern the resulting pattern.
The best way to identify the pattern is to calculate a few values of the sequence
and look for the pattern. You will learn more about pattern recognition in
Chapter 5.
2. DRAW IT OUT
Some problems are much easier to solve if you draw out what’s happening in
the problem. Whenever a story problem describes something that could
actually happen in the real world, you could try to draw out the solution. For
instance, if a problem involves motion, you can draw snapshots representing
the problem at diﬀerent points in time.
Try-It #1-3
Truck A is on a straight highway heading due south at the same time
Truck B is on a different straight highway heading due east. At 1:00
p.m., Truck A is exactly 14 miles north of Truck B. If both trucks are
traveling at a constant speed of 30 miles per hour, at which of the
following times will they be exactly 10 miles apart?
1:10 p.m.(A)
1:12 p.m.(B)
1:14 p.m.(C)
1:15 p.m.(D)
1:20 p.m.(E)


Represent Truck A and Truck B as of 1:00 p.m. How does the
distance between Truck A and Truck B change as time goes
by?
Try another point in time. Since the answers are all a
matter of minutes a er 1:00 p.m., try a convenient
increment of a few minutes. A er 10 minutes, each truck
will have traveled 5 miles (30 miles per 60 minutes = 5 miles in 10 minutes). How
far apart will the trucks be then? On the diagram to the right, the distance is
represented by x.
Because Truck A is traveling due south and Truck B
is traveling due east, the triangle must be a right
triangle.
Therefore, x2 = 92 + 52.
At this point, you could solve the problem in one of
two ways. The first is to notice that once both trucks
travel 6 miles, the diagram will contain a 6 : 8 : 10
triangle. Therefore,
 of an hour later, at
1:12 p.m., the trucks will be exactly 10 miles apart.
Alternatively, you could set up an algebraic equation
and solve for the unknown number of miles
traveled, such that the distance between the trucks
is 10. Call that distance y:


Therefore, y could equal
6 or 8 miles. In other
words, the trucks will be
exactly 10 miles apart at
1:12 p.m. and at 1:16
p.m. Either way, the
correct answer is (B).
Notice how instrumental
these diagrams were for
the solution process. You
may already accept that Geometry problems require diagrams. However, many
other kinds of problems can benefit from visual thinking. You will learn more
about advanced visualization techniques in Chapter 7.
3. SOLVE AN EASIER PROBLEM
A problem may contain large numbers or complicated expressions that actually
distract you from the task at hand: finding a solution path.
When this happens, one tactic is to simplify part of the problem and solve that.
Once you understand how the math works, return to the more complex problem
and apply the same solution path.


Try-It #1-4
If x and y are positive integers and
 is the square of an odd
integer, what is the smallest possible value of xy ?
As you read, jot down the given information.
Note that you might not immediately write down the square of an odd integer
info if you still have to puzzle out what it means:
What does the square of an odd integer look like? List out a few examples, on
paper or in your head:
 1(A)
 8(B)
10(C)
15(D)
28(E)


Are there any patterns or commonalities? All of the numbers are odd. All of the
numbers are perfect squares. Therefore,
 is an odd perfect square.
Add that to your notes.
The question asks for the smallest possible value of xy. What do you need to
figure out in order to find that?
If the
 expression is distracting you, try figuring out what this would
mean for a simpler version of the expression.
Simpler problem: What if
 is an odd perfect square?
In order for the number to be odd, you have to get rid of the even
number 20 (because an even number times any number equals an even
number). The only way to get rid of the even part is to divide it out by y. If
y is 4, then the expression would become
 . As long as x is
an odd number, 5x will be odd, too.
Interesting. How can you apply that thinking to the real problem?
It’s still true that, in order for
 to be odd, you have to get rid of the
even factors in the numerator.


In other words, y2 must cancel out all the even factors in
1,620. The y2 must contain at least two 2’s, so y itself has to
contain at least one 2.
Okay, that takes care of y: at minimum, y must be 2. If so,
then the expression becomes
 .
Now, what about x? If you’re not sure, return to your simpler problem thinking.
Simpler problem: In the last step, in order to make 5x odd, x has to be
odd. 5x also has to be a perfect square. If you make x = 5, then 5x = 25, an
odd perfect square.
Why does that work? A perfect square must contain two of each factor: 5
and 5, for example. The expression 5x, therefore, needs a second 5 to
make this a perfect square.
Back to the real problem. Make sure that 405x has two pairs
of every factor. 405 contains only one 5, so x must contain
another 5. Also contained in 405 is 81, which is 92. That set
of factors already represents a perfect square, so the
minimum requirement is that x equals 5.
If y must be 2, at minimum, and x must be 5, at minimum, then the smallest -
possible value of xy is 10.
You can generalize this approach. If a problem has many complexities, you can
attack it by ignoring some of the complexities at first. Solve a simpler problem.


Then, see whether you can adjust the solution to the simpler problem in order
to solve the original.
To recap, put your work on paper. Don’t try to solve hard problems in your head.
Instead, do the following:
In general, jot down intermediate results as you go. You may see them in a new
light and consider how they fit into the solution.
Also, try to be organized. For instance, make tables to keep track of cases. The
more organized you are, the more insights you will have into diﬀicult problems.
Find a pattern: Write out the first few cases.
Visualize a scene: Draw it out!
Solve an easier problem, then apply your method to the harder problem.


Principle #4: Review Your Work
When you are done with a test or practice set, you are not really done.
When you first do a problem under timed conditions, your brain is too busy
solving the problem to eﬀectively learn and remember. What you learn
from a new problem comes a er you’ve finished it and picked your answer,
when you look at it with a clear head and no timer. Give yourself twice as
much time to review each problem as you spent doing the problem in the
first place.
Here are some things you might consider as you review a problem. Most of
these questions are useful even if you got the problem correct. Don’t
restrict yourself to reviewing problems you got wrong. Review any problem
you might learn something from and ask yourself:
What are all of the pathways to the answer? Which is the best? What is
the easiest and fastest way to implement it?
What clues in the problem told me to use a certain approach or take a
certain step? If I see one of those clues in a diﬀerent problem, what
should I do?
What traps or tricks are built into this problem?
Where could I have made a mistake?
If I did make a mistake, what went wrong in my problem-solving
process? Do I need to change how I approach similar problems?


What could I take from this problem to help me solve other problems in
the future?


When you do the following problem set, apply the first three principles
from this chapter to each problem: Understand, Plan, and Solve. Then,
review each problem in depth. As you review, do two things:
The solutions include our own responses to these two tasks. Yours might
look diﬀerent, and that's fine.
Problem Set
Identify exactly what the problem is asking for and what that means in
the simplest possible terms.
1.
Note at least one general takeaway that might be useful on other
problems in the future.
2.
1. Each factor of 210 is inscribed on its own plastic ball and all of the
balls are placed in a jar. If a ball is randomly selected from the jar,
what is the probability that the ball is inscribed with a multiple of
42 ?


(A)
(B)

(C)
(D)

(E)
2. If x is a positive integer, what is the units digit of (24)5 + 2x(36)6(17)3 ?
2(A)
3(B)
4(C)
6(D)
8(E)
3. A baker makes a combination of chocolate chip cookies and peanut
butter cookies for a school bake sale. His recipes only allow him to
make chocolate chip cookies in batches of 7 and peanut butter
cookies in batches of 6. If he makes exactly 95 cookies for the bake
sale, what is the minimum number of chocolate chip cookies that
he could make?


7(A)
14(B)
21(C)
28(D)
35(E)
4. A rectangular solid is changed such that the width and length are
each increased by 1 inch and the height is decreased by 9 inches.
Despite these changes, the new rectangular solid has the same
volume as the original rectangular solid. If the width and length of
the original rectangular solid are equal and the height of the new
rectangular solid is 4 times the width of the original rectangular
solid, what is the volume of the rectangular solid?
  18(A)
  50(B)
100(C)
200(D)
400(E)
5. The sum of all distinct solutions for x in the equation x2 − 8x + 21 = |x
− 4| + 5 is equal to which of the following?


−7(A)
  7(B)
 10(C)
 12(D)
 14(E)


Solutions
Each solution addresses the two steps from the instructions:
Identify exactly what the problem is asking for, and what that means in
the simplest possible terms.
1)
Note at least one general takeaway that might useful on other problems
in the future.
2)
1. (C)
1. What it’s asking: The problem is asking for the probability that the
selected ball is a multiple of 42.
The quantities you care about are the factors of 210.
What you
know:
There are many balls, each with a diﬀerent factor of 210. Each factor of 210 is
represented. One ball is selected randomly. Some balls have a multiple of 42
(e.g., 42 itself ); some do not (e.g., 1).
What you
don’t
know:
How many factors of 210 there are
How many of these factors are multiples of 42
What the
problem
is testing:
Probability; Divisibility & Primes


The real question:
Plan: 210 to primes → build full list of factors from prime components →
distinguish between multiples of 42 and non-multiples → count factors →
compute probability.
Alternatively, you could list all the factors of 210 using factor pairs.
1 210
2 105
3 70
5 42
6 35
7 30
10 21
14 15


There are 16 factors of 210, and two of them (42 and 210) are multiples
of 42.
You can also count the factors using 210’s prime factorization: (2)(3)(5)
(7) = (21)(31)(51)(71).
Here’s a shortcut to determine the number of distinct factors of 210. Add
1 to the power of each prime factor and multiply:
There are 16 diﬀerent factors of 210: 2 x 2 x 2 x 2 = 16.
How many of these 16 factors are multiples of 42? 42 itself is a multiple
of 42, of course. To find any others, divide 210 by 42 to get 5. This
number is a prime, so the only other possible factor is 42 × 5, or 210.
There are two multiples of 42 out of a total of 16 factors, so the
probability is
 .
The correct answer is (C).
2. At least one takeaway: The problem is straightforward in one sense: it
says the word factor explicitly. Listing all the factors is feasible in two
minutes, but you do need to be going down that solution path fairly
quickly because it will take some time. It may be slightly faster to use


the factor-counting shortcut, but only if you do know how to deal with
the multiples of 42.
2. (A) 2:
1. What it’s asking: The problem asks for the units digit. Because the
problem talks about a product, you care only about the units digits, not
the overall values. Furthermore, the problem provides crazy numbers;
you are absolutely not going to multiply these out. There must be some
kind of pattern at work. Use the Last Digit Shortcut (discussed in the All
the Quant guide).
What jumps out? If x is a positive integer, then 2x must be even and 5 +
2x must be odd.
Units digit of (24)5 + 2x = units digit of (4)odd. The pattern for the units
digit of 4integer = [4, 6]. Thus, the units digit is 4.
Units digit of (36)6 must be 6, as every power of 6 ends in 6.
Units digit of (17)3 = units digit of (7)3. The pattern for the units digit of
7integer = [7, 9, 3, 1]. Thus, the units digit is 3.
The product of the units digits is (4)(6)(3) = 72, which has a units digit of
2. The correct answer is (A).
2. At least one takeaway: Patterns were very important on this one! If
you forget any of the units digit patterns, start listing out the early cases.
At most, you’ll need to list four cases to find the pattern.


3. (E) 35:
1. What it’s asking: The problem asks for the minimum number of
chocolate chip cookies.
Given: The baker only makes chocolate chip (C) or peanut butter (P) cookies. He can only
make chocolate chip cookies in batches of 7 and peanut butter cookies in batches
of 6. He makes exactly 95 cookies total.
What jumps out? C and P must be integers. Therefore:
The answer choices are small multiples of 7, so work backwards from
the answers on this problem. Because the problem asks you to minimize
the number of chocolate chip cookies, start with the smallest answer
choice.
Make a chart:
7C 6P = 95 − 7C Is 6P a multiple of 6?
(i.e., Is P an integer?)
7 88 N
14 81 N
21 74 N


7C 6P = 95 − 7C Is 6P a multiple of 6?
(i.e., Is P an integer?)
28 67 N
35 60 Y
Use the answer choices to calculate the value of 6P. Cross oﬀ an answer
choice if 6P is not a multiple of 6. The first answer choice that works is
the last one. The correct answer is (E).
2. At least one takeaway: The two competing constraints made testing
choices the most eﬀicient method.
4. (E) 400:
There are three equations and three variables. What is the easiest way to
solve for the volume?
What it’s asking: The question asks for the volume of the box. Draw
out the scenarios:
1.


The width, w, appears in all three constraint equations, so solve for the
other variables in terms of w and substitute into the longest constraint:
Substitute:
Since w can’t be zero, you can divide it
out safely.
Solve for all variables:
The correct answer is (E).
At least one takeaway: The question is complex enough that you
could check your work at the end by also calculating the volume of
the new solid (w + 1)(l + 1)(h − 9). As always, you have to decide
whether to spend that time here versus elsewhere.
2.


Notice that, though the initial volume formula seemed long and
annoying, the calculations canceled out nicely in the end. This is
common on the GMAT—common enough, in fact, to suspect that you
may be doing something wrong if the algebra becomes very messy.
5. (D) 12:
The question implies that there may be multiple solutions, as does the
nonlinear given equation. What is the most eﬀicient way to find those
solutions?
There are actually two good approaches; choose the one that is easier
for you.
Approach #1: Do the algebra. Split it into two equations, the “positive”
version and the “negative” version:
Scenario 1
x − 4 ≥ 0
Scenario 2
x − 4 ≤ 0
What it’s asking: The question asks for the sum of the distinct
solutions. In other words, if the number 2 were to show up twice as a
solution, you would count it only once.
1.


Sum of the diﬀerent solutions: 5 + 4 + 3 = 12. The correct answer is (D).
Approach #2: simplify the equation and use theory to finish it oﬀ. Isolate
the absolute value:
Think it through. You square a number and get the absolute value of
that same number (not squared!). Only a few numbers can make that
true: 1 squared equals |1|, 0 squared equals |0|, −1 squared equals |−1|.
That’s it!
Sum of the diﬀerent solutions: 5 + 4 + 3 = 12.
At least one takeaway: When a problem asks for distinct solutions,
count each unique solution; ignore multiple instances of the same
value. Consider whether you prefer the pure algebraic approach or
the theoretical approach. Both are valid solution methods. Which one
do you think you will be able to remember and use more easily?
2.


CHAPTER 2
Problem Solving: Strategies &
Tactics


In This Chapter...
Advanced Strategies
Advanced Guessing Tactics


Chapter 2
Problem Solving: Strategies & Tactics
Sometimes you will encounter a Problem Solving (PS) problem that you
can’t answer—either because its content is diﬀicult or obscure or because
you don’t have enough time to solve completely in two minutes.
This chapter describes a series of diﬀerent methods you might try in these
circumstances. Here, we make the distinction between solution strategies
and guessing tactics.
Solution strategies are broad: they apply to a wide variety of problems,
they provide a complete approach, and they can be used safely in most
circumstances.
In contrast, guessing tactics can help you eliminate a few answer choices,
but o en leave a fair amount of uncertainty. Moreover, a particular tactic
may only be useful in special situations or for parts of a problem.
The first section of this chapter outlines four Problem Solving strategies:
PS Strategy 1: Choose Smart Numbers


The rest of the chapter is devoted to four specialized tactics that can knock
out answer choices or provide clues about how to approach the problem
more eﬀectively:
One of the most productive strategies on the GMAT is to pick good
numbers and plug them into unknowns. Try this when the concepts are
especially complex or when conditions are placed on key inputs that are
otherwise unspecified (e.g., n is a prime number).
PS Strategy 2: Work Backwards
Another common approach is to work backwards from the answer
choices, testing to see which one fits. Doing so can o en help you avoid
demanding calculations or the need to set up and solve complicated
algebraic expressions.
PS Strategy 3: Test Cases
In certain circumstances, a problem allows multiple possible scenarios,
or cases. On PS questions, the problem usually asks you to find
something that must be true or could be true. On these problems, you
can test diﬀerent numbers to eliminate answers until only one remains.
PS Strategy 4: Avoid Needless Computation
The GMAT rarely requires you to carry out intensive calculations to arrive
at an answer. Look for opportunities to avoid tedious computation by
factoring, simplifying, or estimating.


PS Tactic 1: Look for Answer Pairs
Some PS questions have answer choices that pair with each other in
some way. The correct answer may be part of one of these pairs.
PS Tactic 2: Apply Cutoﬀs
Sometimes a back-of-the-envelope estimation can help you eliminate
any answer choice above or below a certain cutoﬀ.
PS Tactic 3: Look at Positive/Negative
Some PS questions include both positive and negative answer choices.
In such cases, look for clues as to the correct sign of the correct answer.
PS Tactic 4: Draw to Scale
Many Geometry problems allow you to eliminate some answer choices
using visual estimation, as long as you draw the diagram accurately
enough.


Advanced Strategies
1. CHOOSE SMART NUMBERS
Some types of problems allow you to pick real numbers and solve the
problem arithmetically rather than algebraically. For instance, almost any
Problem Solving problem that has variables in the answer choices gives you
this opportunity. Likewise, you can o en pick a smart number for a fraction or
percent problem without specified absolute value amounts.
Other problem types allow this strategy as well. For instance, a problem may
put specific conditions on the inputs but not give you exact numbers. In this
case, you can go ahead and just pick inputs that fit the conditions. If a
problem specifies that “x must be a positive even integer” but does not
specify the value of x, picking 2 for x will probably get you to a solution
quickly and easily.
The GMAT Oﬀicial Guide contains many problems that are diﬀicult to solve
algebraically but much easier to solve with real numbers. However, as an
advanced test-taker, you might consider it a point of pride not to plug in a
number. You might want to prove a “theoretically correct” answer. Overcome
your pride! The GMAT is not a math test; the GMAT tests you on your flexibility
of thinking and your ability to manage a very limited amount of time. Use the
easiest and most eﬀicient solution path, not the textbook math solution path!
Try-It #2-1


Andra, Elif, and Grady each invested in a certain stock. Andra
invested q dollars, which was 40% more than Elif invested. If Elif
invested 25% less than Grady invested, what was the total amount
invested by all three, in terms of q ?
This problem can be solved algebraically: write a couple of equations and
solve for all three variables, then add them up. A glance at the answers,
though, indicates that the algebra is likely to get messy. Instead, choose a real
number and solve the problem arithmetically.
If you’ve made it to the GMAT Advanced Quant book, then you have likely
used this strategy before (or at least learned about it). At times, you may have
been frustrated because this technique didn’t actually seem easier than doing
the math algebraically. If so, here’s the missing piece: you need to learn how
to choose smart numbers in the best possible way.
Most of the time, you’re going to choose for the variable given (in this case, q).
In some cases, though, starting with the given variable doesn’t make your life
any easier. The problem above actually has three unknowns: one for Andra,
one for Elif, and one for Grady. Take a look at the relationship between those
unknowns before you decide which one is the best starting point.
2q(A)
(B)
(C)
(D)
(E)


Andra invests 40% more than Elif. For just these two, it would be easier to pick
a number for Elif and then calculate Andra’s amount.
Elif invests 25% less than Grady. For these two, it is easier to start with Grady
and then calculate Elif. As a result, start with Grady, then find Elif, then find
Andra.
If Grady invests $100, then Elif invests 25% less, or $75. Andra invests 40%
more than Elif, or $75 + $30 = $105. Make sure to note on your scrap paper
that q = 105.
Collectively, the three invest $100 + $75 + $105 = $280. Find the answer choice
that matches $280:
The correct answer is (E).
Note a few important aspects that will help you to choose smart numbers
eﬀiciently and eﬀectively.
First, if there are multiple unknowns and you have to choose where to start,
pause to think about how to make the math as easy as possible. In the case of
the problem above, if you had picked q = 100 for Andra, your next step would
2q = 2(105) = $210. Eliminate.(A)
 Eliminate.(B)
Eliminate.
(C)
. Eliminate.(D)
 Correct!(E)


have been to figure out Elif’s amount. It is not the case that Elif would be $60,
or 40% less than Andra.
Rather, Andra is 40% more than Elif: 1.4e = $100, so
 . Elif would
actually equal approximately $71.42857. Nobody’s going to want to go down
that path! At this stage, you have two choices: you can go back and pick for
someone else or you can think about what numbers would make this
particular path easier. For example, a value of q = 140 instead of 100 would
result in an easier calculation,
 , and a value of 100 for Elif. (In this case,
that would still result in a messy number for the next calculation; on another
problem that didn’t have so many numbers, though, making a one-number
adjustment can still leave you well within the two-minute time frame.)
Second, you do not need to calculate the value of every answer choice. You
can stop and eliminate a choice whenever you can tell that it will not equal
$280. When you plug in 105 for q, answer (B) isn’t an integer and answer (C)’s
units digit isn’t 0, so neither can be the correct answer.
Third, practice this strategy extensively in order to expose yourself to these
little variations and possible sticking points. As you become proficient with
the strategy, you’ll be amazed at how much time and mental eﬀort it can save
you on the GMAT.
2. WORK BACKWARDS
In a number of cases, the easiest way to solve a GMAT problem is to start from
the answer choices and work backwards. Don’t be too proud to try this
technique either. The GMAT doesn’t reward perfect math technique; it


rewards finding the correct answer as eﬀiciently as possible or guessing when
needed.
Try-It #2-2
If
 , which of these integers could be the value of z ?
Look at all of those fractions! Solving for z algebraically in this problem would
not be easy. Instead, notice two important clues: the problem asks for the
value of a single variable and the answer choices oﬀer nice-and-easy integers.
Work backwards! People o en start at the beginning, with choice (A), but
actually start with (B) or (D); you’ll learn why in a moment:
(B)
 INCORRECT
The right side is smaller than the le . The numerators are always 2, so what
needs to happen to bring the two halves of the equation together?
The le side needs to be made smaller, so the denominator needs to be
bigger; eliminate answer choice (A) as well. This is why you start with choice
(B) or (D)—you can eliminate other answer choices that are too small or too
large without having to test them. Try answer (D) next:
0(A)
1(B)
2(C)
3(D)
4(E)


(D)
 CORRECT
You can stop when you find an answer that works. If you are paying attention
to how the math works as you solve, you can o en get away with trying just
two or three answer choices on these problems.
Try-It #2-3
A certain college party is attended by both male and female students.
The ratio of male-to-female students is 3 to 5. If 5 of the male
students were to leave the party, the ratio would change to 1 to 2.
How many total students are at the party?
Of course, you could set up equations for the unknowns in the problem and
solve them algebraically. However, the numbers in the answers are pretty
straightforward integers. Try Working Backwards.
Again, begin with answer (B) or answer (D), whichever number looks easier
for the problem:
24(A)
30(B)
48(C)
80(D)
90(E)


What characteristic must be true of the correct answer? It must be a multiple
of 8, so eliminate answer (E). Try the remaining answers:
Once you have found the answer using this technique, you can stop; you
don’t need to test any remaining answers. Also, you don’t have to translate
equations or figure out how to eliminate variables and solve. You just work
each number through the problem until you’re done.
As you work through an answer, think about how the math is playing out.
Again, you will usually be able to eliminate some answers without actually
having to try them.
3. TEST CASES
In the original ratio, you can use an unknown multiplier to represent
the total number of students: 3x + 5x = 8x. If there are 30 students
total, then the unknown multiplier is
 = 3.75, leaving you with a
non-integer number of students. This is impossible, so this answer
must be incorrect.
(B)
If there are 24 total students, then the unknown multiplier is 3: there
are 9 male students and 15 female students at the party. If 5 males
leave, there will be 4 le , leaving a ratio of 4 : 15. Eliminate answer (A).
This ratio is also pretty far away from the correct ratio of 1 : 2,
so consider trying the larger remaining answer, 80, next.
(A)
If there are 80 students total, then the unknown multiplier is 10, so
there are 30 males and 50 females. If 5 males leave, the new ratio is 25
: 50, or 1 : 2. This is the correct answer!
(D)


Some problems allow you to choose real numbers to solve, but you can’t
choose just one set of numbers as you do when you choose smart numbers.
Rather, you have to test multiple scenarios to get yourself to the one correct
answer. On Problem Solving problems, this tends to occur with must be true
or could be true problems.
Try It #2-4
If n is a positive integer, what must be true of n3 − n ?
This problem is asking about a theoretical concept: what mathematical
characteristic must be true of this expression? In order to solve, you can test
allowable cases to narrow down the answers until only one remains.
Note that in this PS problem, one of the answers must be correct. In this case,
any positive integer n will help you get to the answer, so start with the
simplest possible positive integer: 1. In general, when Testing Cases on PS or
DS, start with the simplest number that fits the problem’s parameters.
Try n = 1:
  n n3 − n
It is divisible by 4.(A)
It is odd.(B)
It is a multiple of 6.(C)
It is a prime number.(D)
It has, at most, two distinct prime factors.(E)


n n3 − n
Case 1 1 0
Try your next case, ignoring answers (B) and (D) from now on:
  n  n3 − n
Case 2 2 6
You’re down to (C) and (E). Because one of the statements talks about having
more than a certain number of prime factors, try a larger number next:
  n  n3 − n
Yes, 0 is divisible by 4. (0 is divisible by any number except 0.)(A)
No, 0 is not odd. Eliminate.(B)
Yes, 0 is a multiple of 6. (0 is a multiple of any number.)(C)
No, 0 is not a prime number. Eliminate.(D)
Yes, 0 does not have more than two prime factors.(E)
No, 6 is not divisible by 4. Eliminate.(A)
Yes, 6 is a multiple of 6.(C)
Yes, 6 does not have more than two distinct prime factors. (The prime
factors are 2 and 3.)
(E)


n  n3 − n
Case 3 5 120
The correct answer is (C). You could save yourself some time on this one by
recognizing that the expression n3 − n can be rewritten as:
In other words, the expression represents three consecutive integers. Test
some cases to discover what must be true about the product of three
consecutive integers.
Case 1: If n = 1, then the three consecutive integers are 0, 1, 2. This product is
even and not prime, so eliminate answers (B) and (D).
Case 2: If n = 2, then the three consecutive integers are 1, 2, and 3. This
product does not contain two 4’s, so it is not a multiple of 4. Eliminate (A).
Case 3: If n = 3, then the three consecutive integers are 3, 4, and 5. You could
multiply out the consecutive integers, but only go down the computation
Yes, 120 is a multiple of 6.(C)
No, 120 does have more than two distinct prime factors. (The prime
factors are 2, 3, and 5.) Eliminate.
(E)


path if you must. In this case, you’re trying to find factors, and the three
consecutive integers tell you that directly; you don’t need to multiply them
out. (You’ll learn more about avoiding unnecessary computation in the next
section.)
Instead, if possible, try to notice a pattern. In every set of three consecutive
numbers, you will always have at least one even number. You will also always
have a multiple of 3. (In Case 1, 0 is a multiple of 3.) As a result, the product
will always be a multiple of 6.
You may also notice that the problem contains no upper limit. By choosing a
large enough number, you’re going to be able to create a number that
contains more than two distinct prime numbers.
You can test cases directly to eliminate the four incorrect answers, or you can
use a few cases to help you figure out the theory underlying the problem,
which will also get you to the correct answer.
4. AVOID NEEDLESS COMPUTATION
You won’t see many GMAT problems that require substantial calculation to
arrive at a precise answer. Rather, correct answers on diﬀicult problems will
generally be relatively easy to compute once the diﬀicult concept or trick in
the problem has been correctly identified and addressed.
On several types of GMAT problems, a significant amount of computation can
be avoided. A simple rule of thumb is this: if it seems that calculating the
answer is going to take a lot of work, there’s a good chance that a shortcut
exists. Look for the back door!


Estimation
Intelligent estimation can save you time and eﬀort on many problems. For
example, you can round to nearby benchmarks or be ready to switch a
fraction to a decimal or percent, and vice versa.
Try-It #2-5
The percent change from 29 to 43 is approximately what percent of
the percent change from 43 to 57 ?
In this case, the question stem straight up tells you that you can estimate. Any
time you see the word approximately (or a synonym), definitely do not try to
solve for the exact answer.
The question stem is pretty complex. Use the percent change formula:
A direct translation would look like this:
50%(A)
66%(B)
110%(C)
133%(D)
150%(E)


You can simplify things by solving the question in parts. First, it talks about
the percent change from 29 to 43.
The percent change for the top half of the fraction, is
 , but that number is
cumbersome. Estimate! This is approximately
 .
The percent change from 43 to 57 is
 . This is about
 . The question is
really asking:
The fraction
 is equivalent to 150%. The correct answer is (E).
The wording of the question can sometimes provide a strong clue that
estimation should be used. Look for phrases such as these:
The test doesn’t have to tell you that you can estimate, though. You can
usually estimate when the answer choices are far apart.
Heavy Long Division
 is what percent of
 ?
. . . the number is approximately equal to which of the following?
. . . this result is closest to which of the following?
Which of the following is most nearly equal to . . .?


Very few problems on the GMAT truly require long division, even though it
might appear otherwise. You can almost always approximate the answer or
reduce the division by taking out common factors.
Try-It #2-6
 is equivalent to which of the following?
At first glance, it appears that precise long division is necessary. The answer
choices are very close together, making estimation diﬀicult. However, with
some manipulation and factoring, the solution is much more straightforward.
The key to factoring this fraction is to move the decimals of both the
numerator and denominator three places to the right so that you’re dealing
with integers. Then you might notice that 3,507 is divisible by 7 (3,500 and 7
are both divisible by 7, providing a clue that you may be able to factor out 7).
Moreover, 10,020 is divisible by 10 and by 2 (10,020 ends in a 0, and 1,002 is
even):
The correct answer is (A).
0.35(A)
0.3505(B)
0.3509(C)
0.351(D)
0.3527(E)


Alternatively, you might observe that 10.02 is very slightly larger than 10.
Therefore, the correct answer will be slightly smaller than
 Guess between choices (A) and (B).
Try-It #2-7
What is the value of
Since 102 is extremely small compared to 105, and the choices are somewhat
spread out, estimate:
Notice that by ignoring the 102 term, you made the denominator slightly
larger than it originally was. Therefore, 0.81918 is slightly smaller than the
correct answer, 0.82. The correct answer is (D).
Quadratic Expressions in Word Problems
Some Word Problems result in a quadratic equation. You are probably pretty
good at solving quadratic equations, so your natural bias would be to set up
and solve the equation. However, if the coeﬀicients are huge, the equation
8.19(A)
8.02(B)
0.89(C)
0.82(D)
0.81(E)


may be very diﬀicult to solve. In these cases, try testing the answer choices in
the original problem (not in the translated and manipulated quadratic),
especially when the answer choices contain easier numbers.
Try-It #2-8
A shoe cobbler charges n dollars to repair a single pair of loafers.
Tomorrow, he intends to earn $240 repairing loafers. If he were to
reduce his fee per pair by $20, he would have to repair an additional
pair of the loafers to earn the same amount of revenue. How many
pairs of loafers does he intend to repair tomorrow?
The problem may not seem too bad . . . until you try to set it up algebraically.
Assign x to represent the number of pairs of loafers the cobbler intends to
repair tomorrow. Using the equation for revenue gives you the following:
Furthermore, reducing his fee by $20 would result in the need to repair an
additional pair of shoes for the same amount of revenue gives you the
following:
1(A)
2(B)
3(C)
4(D)
5(E)


From here, the algebra gets complicated very quickly. (The algebra is shown
later, if you want to see!)
Switch to a diﬀerent approach—Working Backwards from the answer choices.
Start with (B) or (D).
The revenue in answer (B), $300, is too large. Do you need to make x larger or
smaller? If you’re not sure, try (D) next, and notice something very important.
The revenue in (D), $200, is too small. The correct revenue, therefore, should
fall between (B) and (D). The answer must be (C)!
  x n x × n x + 1 n − 20 (x + 1) × (n − 20)
(B) 2 $120 $240 3 $100 $300
(D) 4 $60 $240 5 $40 $200
If you’re not sure, try that answer:
  x n x × n x + 1 n − 20 (x + 1) × (n − 20)
(C) 3 $80 $240 4 $60 $240
Here’s the algebra in case you want to see it (but it’s not recommended to do
this):


Advanced Guessing Tactics
To repeat, the tactics below are less universally useful than the strategies
we just covered. However, when all else fails, “break the glass” and try one
or more of these tactics. They’re almost always better than guessing
completely randomly. If you’re way behind on time and you have to
sacrifice a couple of problems, though, guess immediately and move on.
The examples below will not be too diﬀicult in order to illustrate the tactic
clearly and not distract you with other issues. Of course, if you can solve
the problem directly, do so. But also study the tactic, so you’re ready to use
it on a harder problem of the same type.
1. Look for Answer Pairs            Certainty: Moderate
GMAT answer choices are sometimes paired in a mathematically relevant
way. Pairs of answers may:
The correct answer is sometimes part of such a pair. Why? The GMAT likes
to put in a final obstacle. Say you do everything correct, except you solve
Add up to 1 on a probability or fraction question
Add up to 100% on questions involving percents
Add up to 0 (be opposites of each other)
Multiply to 1 (be reciprocals of each other)


for the wrong unknown or forget to subtract from 1. Under the pressure of
the exam, people make this sort of penultimate error all the time
(penultimate means next to last).
In order to catch folks in this trap, the GMAT has to make an answer choice
that’s paired to the correct answer—it’s correct except for that one last
step.
This means that you can o en eliminate unpaired answer choices. Also, the
way in which the answers are paired may provide clues about the correct
solution method and/or traps in the problem.
Try-It #2-9
At a certain high school, the junior class is twice the size of the
senior class. If
 of the seniors and
 of the juniors study
Japanese, what fraction of the students in both classes do not study
Japanese?
Note that two pairs of answers each add up to 1:
 , and
. Answer (A) is not likely to be correct, because it is not part of an
(A)
(B)
(C)
(D)
(E)


answer pair. The fact that these pairs sum to 1 also provides a clue to
double-check the wording of the question: do vs. do not study Japanese
(the sum of the fractions of the students that do study Japanese and those
who do not will equal 1).
To solve, you could use a double-set matrix that shows juniors versus
seniors and Japanese studiers versus non-Japanese studiers. However,
let’s go back to the first strategy in this chapter and choose smart numbers.
Pick a smart number that is a multiple of the denominators in the problem:
3 × 4 = 12.
If the junior class has 12 people, then the senior class thus has 6. If
 of
the seniors study Japanese, then
 seniors study Japanese. If
 of the juniors study Japanese, then
 juniors study
Japanese. There are 12 + 6 = 18 students total, and 2 + 3 = 5 of them study
Japanese. Thus,
 of the students do study Japanese, so the fraction of
the students that do not study Japanese is
 . The correct answer is (E).
2. Apply Cutoﬀs           Certainty: High
You may be able to eliminate answers above or below some easily
calculated threshold value. You may have to imagine that the problem is
slightly diﬀerent (and easier) to come up with that threshold value, but
once you do, you can o en get rid of two or three answer choices.


This strategy can sometimes be used in combination with an answer pairs
strategy, as pairs of answers are o en composed of a high and a low value.
In the previous problem, for example,
 of the seniors and
 of the
juniors study Japanese. Therefore, somewhere between
   and
 of the
students overall, or less than half, must study Japanese. This implies that
the fraction of students who do not study Japanese must be more than
half. You could eliminate answer choices (A), (B), and (C) because each of
these answer choices is smaller than
 .
Try-It #2-10
The eSoroban device is available in two colors, orange and green.
In 2013, 60% of the eSoroban devices sold were purchased by
women,
 of whom purchased the orange device. If an equal
number of orange and green eSoroban devices were sold in 2013,
what fraction of men who purchased an eSoroban in
2013 purchased the green device?
(A)
(B)
(C)
(D)
(E)


More than half of the people purchasing this device are women, and more
than half of them buy the green version. If an equal number of green and
orange devices are sold, then more than half of the men must buy the
orange version. Therefore, less than half of men buy the green version:
eliminate answers (D) and (E).
Furthermore, women represent 60% of purchases, while men represent
just 40%. Pretend for a moment that there were equal numbers of men and
women. If that were the case, then
 of women would buy orange and
 would buy green. On the flip side,
 of men would buy green and
would buy orange.
In fact, there are more women than men, so the proportion of men buying
orange devices has to be higher than
 . Fewer than
 must buy the
green version. Eliminate answer (C).
From here, if you have a strong number sense, you might guess that
answer (B) is more likely the correct answer, because a
 split would
have made the answer
 . The actual split is
 , so the correct value is
not that much lower than
 . But here’s the actual solution, using a
double-set matrix and a smart number of 100 total people:
  Men Women Total
Orange
 50


Men Women Total
Green 50 − 35 = 15 60 − 25 = 35 50
Total 40 60 100
Because 15 men purchased a green device out of 40 men total, the
proportion is
 =
 . The correct answer is (B).
3. Look at Positive/Negative            Certainty: High
A special case of the cutoﬀ strategy occurs when some of the answer
choices are positive and others are negative. In this case, focus on figuring
out the sign of the correct answer, then eliminate any answer choices of
the opposite sign. (Again, this is something to do when you have run out of
direct approaches or you are short on time!)
Try-It #2-11
If x □ y is defined to equal
 for all x and y, then (−1 □ 2) □ 3 is
equivalent to which of the following?

(A)

(B)

(C)
−
(D)


The negative sign in the term −1 will be extinguished, because the term
before the □ symbol is squared when this function is calculated. Therefore,
the correct answer will be positive. Eliminate (D) and (E). By the way, the
correct answer is (C).
4. Answer Properties                Certainty: Moderate
You may not have time, while doing a Problem Solving problem, to figure
out the correct answer. However, it might take less time to figure out
something about the correct answer. You saw two examples of this in the
previous two guessing strategies: you can sometimes figure out whether
the correct answer is above or below a certain value, or whether it’s
positive or negative. Here are some other things you may be able to
determine:
5. Draw to Scale            Certainty: Moderate
The estimation technique can be extended to some Geometry problems.
You can approximate the length of a line segment, size of an angle, or area
of an object by drawing it as accurately as possible on your scrap paper.
−
(E)
Is the correct answer even or odd?
Is the correct answer divisible by a certain value?
What is the units digit of the correct answer?
Does the correct answer include square roots?
Does the correct answer include decimals?


When you take the GMAT, the laminated scratch booklet is printed with a
light grid. This grid can help you draw very accurate scale pictures, and
o en the picture alone is enough to answer the question.
Try-It #2-12
In the diagram to the right, equilateral
triangle ADE is drawn inside square
BCDE. A circle is then inscribed inside
triangle ADE. What is the ratio of the area
of the circle to the area of the square?
If a Problem Solving question does not say that the diagram is not drawn to
scale, then the diagram is drawn to scale. The circle is about
 the height
of the square and about
 the width. Therefore, the area of the circle
should be approximately
 of the area of the square. Eliminate any answer
choices that are far away from that estimate:
Ok
On the high side
Too high
(A)
(B)
(C)
(D)
(E)
(A)
(B)
(C)


Too high
Way too high
Only answers (A) and (B) are remotely reasonable, and only (A) is really
close to the estimate.
The most eﬀicient way to solve this problem fully is to assign a radius of 1
to the circle. This way, the circle has an area of π. Now work outwards.
If the center of the circle is O, then OF = 1.
Because ADE is an equilateral triangle, angle ADE
= 60°. OD bisects angle ADE, so angle ODE = 30°.
Therefore, triangle OFD is a 30 : 60 : 90 triangle,
and
 . DE must be twice the length of
DF, meaning that
 , and the area of
square BCDE =
 .
So the ratio of the area of the circle to the area of the square is
 .
Notice how much easier the draw to scale tactic was! Will it work every
time? No. But it works o en enough that you want to think about using it.
(D)
(E)


Problem Set
Solve problems 1–12. In each case, identify whether you could use
advanced strategies (Choose Smart Numbers, Work Backwards, Test Cases,
or Avoid Needless Computation) and guessing tactics (Look for Answer
Pairs, Apply Cutoﬀs, Look at Positive-Negative, or Draw to Scale) in any
beneficial way. Of course, there are textbook ways to solve these problems
directly; instead of trying the textbook method first, focus on applying the
strategies and tactics described in this chapter.
1. A popular yoga studio that is always filled to capacity moves to a
new location that is able to serve 45% more students.
Unfortunately, 20% of the current students will no longer attend
classes at the new location. If classes at the new location are also
filled to capacity, what fraction of the students at the new location
will be new students?


(A)
(B)
(C)
(D)
(E)
2. If x < 10, y < 8, and y < x, what must be true?
xy < 80I.
II.
x 2 + y 2 > 1III.
None(A)
II only(B)
III only(C)
I and II only(D)
II and III only(E)
3. If
 , which of the following could be the value of x ?


6(A)
  8(B)
  9(C)
12(D)
15(E)
4. If
 , which of the following could be the value of x  ?
−2(A)
  0(B)
  1(C)
  3(D)
  5(E)
5. If 154 is
 of x, approximately what is the value of 2x  ?
104(A)
114(B)
208(C)
228(D)
416(E)
6.
 is equivalent to which of the following?


0.6(A)
0.603(B)
0.606(C)
0.615(D)
0.66(E)
7. In a certain clothing store, the most expensive pair of socks sells for
$1 less than twice the price of the cheapest pair of socks. A
customer notices that for exactly $18, she can buy three fewer pairs
of the most expensive socks than the cheapest socks. What could
be the number of pairs of the cheapest socks she could have
purchased?
  3(A)
  5(B)
  6(C)
12(D)
36(E)
8. If
 , then m must equal which of the following?
−2(A)
−1(B)
  0(C)
  1(D)
  2(E)


9. Simplify
10. If 3x + 3x + 3x = 1, what is x ?
−1(A)
−
(B)
  0(C)

(D)
  1(E)
11. If a and b are integers and a is a factor of b, what must be true?
a < bI.
The distinct prime factors of a2 are also factors of b.II.
III.


Without solving problems 13–16, which answers could you confidently
eliminate and why?
None(A)
II only(B)
III only(C)
I and II only(D)
II and III only(E)
12. The integer k is positive but less than 400. If 21k is a multiple of 180,
how many unique prime factors does k have?
One(A)
Two(B)
Three(C)
Four(D)
Five(E)
13.
0.693(A)
0.7(B)
0.71(C)
6.93(D)
7.1(E)


14.
 is equivalent to which of the following?
−1.8(A)
−0.18(B)
  0(C)
  0.18(D)
  1.8(E)
15. In the 7-inch square to the right, another
square is inscribed. What fraction of the
larger square is shaded?
(A)
(B)
(C)
(D)
(E)
16. If Mason is now twice as old as Gunther was 10 years ago, and G is
Gunther’s current age in years, which of the following represents
the sum of Mason’s and Gunther’s ages 4 years from now?


(A)
3G + 28(B)
3G − 12(C)
8 − G(D)
(E)


Solutions
1. (D)
 : (Choose Smart Numbers) Both the 45% figure and the answer
choices look cumbersome, but the problem never provides a real value
for the number of students, either before the move or a er. Choose a
smart number of 100 for this Percents problem.
If the new studio is filled to capacity, at 145 students, but only 80 old
students continue to attend, then the studio will have 65 new students.
The fraction of new students at the new location, then, is
 .
The correct answer is (D).
Old capacity: 100 students
New capacity: 145 students
Old students who will stay with the studio: 80
2. (A) None: (Test Cases) Test real numbers to prove or disprove the
statements. Make sure to choose only values that are allowed by the
problem: x < 10, y < 8, and y < x.
Statement I: If x = 9 and y = 7, then xy = 63, which is less than 80. If, on
the other hand, x = −10 and y = −20, then xy = 200, which is greater than
80. Statement I does not have to be true. Eliminate answer (D).
Statement II: Positive values for both x and y will make
 , but if x
= 9 and y = −1, then
 , which is not greater than 1. Statement II


does not have to be true. Eliminate answers (B) and (E).
Statement III: If x = 3 and y = 2, then x2 + y2 is 13, which is greater than 1.
If x = −2 and y = −3, then x2 + y2 is still 13, which is still greater than 1.
Don’t forget about fractions! If x = 0.3 and y = 0.2, then x2 + y2 = 0.09 +
0.04, which is not greater than 1. Fractions between 0 and 1 get smaller
when squared. Eliminate answers (C) and (E).
None of the statements must be true. The correct answer is (A).
3. (D) 12: (Work Backwards) This problem can be solved algebraically, but
not easily. You’d actually need to use the quadratic formula, and the
equation would turn out to be really nasty. No thank you!
Instead, notice that the problem asks for a single variable and the
answer choices are decently small integers. Start with answer (B) or (D).
That math is a pain. Is answer (D) any easier? If so, start there.
Yes! Much easier: 122 = 144, and that goes very nicely with 288.
This equation simplifies to 5 + 2 = 7, so the correct answer is (D). Note
that if (D) had not been the correct answer, you would have been able to
(B)
(D)


tell whether to try a larger or smaller answer next based on whether (D)
was too small or too large.
4. (D) 3: (Work Backwards) Typically, when Working Backwards, you start
with answer (B) or (D). If you can tell that a problem isn’t likely to have a
consistent pattern, though, as with this absolute value equation, then
just start at one end—answer (A) or answer (E)—and stop when you find
the correct answer.
  x x2 − 6 | x2 − 6 |
(A) −2 (−2)2 − 6 = 4 − 6 = −2 2
(B) 0 (0)2 − 6 = 0 − 6 = −6 6
(C) 1 (1)2 − 6 = 1 − 6 = 5 5
(D) 3 (3)2 − 6 = 9 − 6 = 3 3
The correct answer is (D).
5. (E) 416: (Avoid Needless Computation) Not only do the numbers in the
problem look ugly, but the problem asks for an approximate answer.
Estimate.
It would be much easier if 154 were 150. What about that fraction? First
of all, put it over 100:
 .
Much better! This is almost
 , so use that fraction instead.


Translate the equation with the estimated values:
Finally, make sure to answer the right question that was asked! The
answer is not (C), 208, because the question asks for the value of 2x.
The correct answer is (E).
6. (A) 0.6: (Avoid Needless Computation)
Alternatively, note that 2.010 is very slightly larger than 2. Therefore, the
fraction is very slightly smaller than
 Only (A) is
smaller than 0.603. The correct answer is (A).
7. (D) 12: (Work Backwards) This problem can be solved algebraically, but
the math gets pretty challenging (see end of this explanation). It’s easier
to work from the answer choices. Start with (B) or (D).
Call the number of cheap pairs c and the number of expensive pairs e.
Call the cost of a cheap pair $c and the cost of an expensive pair $e. First,
try (B).
# Cheap Pairs
(c)
$c Pairs
 # e Pairs (c −
3)
$e Pairs
 Match?
$e = 2($c) − 1?


# Cheap Pairs
(c)
$c Pairs
 # e Pairs (c −
3)
$e Pairs
 Match?
$e = 2($c) − 1?
(B) 5
 2 9
No ✗
Answer (B) is incorrect. Note that you don’t actually have to figure out
the value of the right-hand side of the equation, as long as you know
that it will not equal 9. The two sides of the equation are pretty far apart,
so (A) and (C) are also probably not correct answers. Try answer (D) next.
# Cheap Pairs (c) $c Pairs
 # e Pairs (c − 3) $e Pairs
 Match?
$e = 2($c) − 1?
(D) 12 $1.50 9 $2 2 = 2(1.5) – 1
Yes ✓
Answer (D) makes the final equation work, so it is the correct answer.
Here’s how the algebra would have to be set up. Let c = the number of
pairs of cheap socks, e = the number of pairs of expensive socks, x = the
cost for one pair of cheap socks, and y = the cost for one pair of
expensive socks.
From sentence 1: y = 2x − 1


From sentence 2: 18 = cx and 18 = ey = (c − 3)y
From here, you would solve the equations from sentence 2 for x and y,
respectively, and plug them into the equation from sentence 1:
Plug in to equation 1:
With the correct manipulation (over multiple, complicated steps!), that
equation would eventually become c2 − 21 + 108 = 0 and you could solve
for the two solutions, c = 9 and c = 12. Only 12 is in the answer choices,
so choice (D) is the correct answer.
8. (D) 1: (Work Backwards) In order for the le -hand side to equal 1, m has
to be positive. Try only the three positive answer choices. In this
circumstance, start with the middle number of the remaining choices. If
m = 1, then the le -hand side simplifies to
 .
The correct answer is (D).
9.
 (Avoid Needless Computation)
10. (A) −1: (Work Backwards) You might solve this one by inspection: three
identical “somethings” sum to 1, so one of those “somethings” equals


, or
 . Therefore, x = −1.
Testing choices is fast, too. In this case, (B) and (D) are both fractions,
but the other answers are integers, so try testing (A), (C), and (E) instead.
Stop when you find the correct answer.
Many people would try (C) first because 0 is an easy number. Notice that
it’s too big. Which number should you try next, −1 or 1? Since 0 leads to a
number that’s too big, try −1 next.
  x 3x + 3x + 3x
(C) 0 1 + 1 + 1 = 3
(A) −1
The correct answer is (A).
11. (B) II Only: (Test Cases) First, consider what the given information a is a
factor of b tells you:
Next, test the statements.
Statement I: a < b. A factor can be smaller than the main number, but a
factor can also equal the main number: 6 is a factor of 6. Statement I
2 is a factor of 6
6 is not a factor of 2
1, 2, 3, and 6 are all factors of 6


does not have to be true. Eliminate answer (D).
Statement II: The distinct prime factors of a2 are also factors of b. If a = 2,
then a2 = 4. There is just one distinct prime factor of a2: 2. In this case, a
is a factor of b, yes. If a = 6, then a2 = 36. There are two distinct prime
factors of a2: 2 and 3. In this case, a is still a factor of b.
Notice any patterns? The distinct prime factors of a2 are the same as the
distinct prime factors of a, since a2 is made up of a multiplied by itself.
So, first, this statement is really saying that the distinct prime factors of
a are also factors of b.
If a is a factor of b, then by definition all of a’s prime factors also have to
be factors of b. No matter what numbers you try for statement II, a will
be a factor of b. Statement II must be true, so eliminate answers (A) and
(C).
Statement III:
  If b is 6, then a could be 6, in which case
. Alternatively, if b is 6, then a could be 2, in which case
 .
So far, this statement looks good.
Don’t forget about negative numbers! The problem doesn’t specify
positive integers. What if b is −6? In this case, a could still be 2, so
. Statement III does not have to be true. Eliminate answer (E)
and choose correct answer (B).
Note: The GMAT doesn’t o en test the factors of negative numbers, but
the definitions provided in the math review of the Oﬀicial Guide do


allow for the possibility.
12. (C) Three: (Choose Smart Numbers) A laborious way to solve this
problem would be to determine all the possible values for k and take a
prime factorization of each value, counting the number of diﬀerent
prime factors that each value has. Ugh!
An easier technique is to pick a smart number: one value of k that
satisfies the constraints. Any value of k that fits the constraints must
have the same number of diﬀerent prime factors as any other legal value
of k. Otherwise, the problem could not exist as written. There would be
more than one correct answer.
The problem states that 21k is a multiple of 180, so
 must be
an integer. In other words, k must be divisible by 60. The easiest number
to choose is k = 60.
The prime factorization of 60 is 2 × 2 × 3 × 5, so 60 has the unique prime
factors 2, 3, and 5. Thus, k has three unique prime factors. The correct
answer is (C).
13. (A), (D), and (E) can be eliminated: If you ignore the 103 in the
denominator, the division is
 . This is an
approximation of the answer, not an exact computation of it, so
eliminate (A). You have slightly overstated the denominator, thus slightly
understated the result. Answers (B) and (C) are possibilities, but (D) and
(E) are both too large. The correct answer turns out to be (B).
14. (A), (C), (D), and (E) can be eliminated: The numerator is negative, so
eliminate (C), (D), and (E). The numbers in the numerator will add to


more than –25 (i.e., closer to 0), so eliminate (A).
The correct answer must be (B).
15. (A), (C), (D), and (E) can be eliminated: Ignoring the dimensions 3 and 4
for a moment, think about the types of squares that might be inscribed
in the larger square:
The shaded area can be at most
 of the larger square, which occurs
when the smallest possible square is inscribed in the larger square. This
gives you a great cutoﬀ.
The larger the inscribed square, the smaller the shaded area and the
more the inscribed square must be rotated from the vertical orientation
of the minimum inscribed square.
Eliminate (D) and (E), since they are larger than
 . Eliminate (C), since
the labeled lengths of 3 and 4 are not equal, indicating that the
inscribed square is rotated from the minimum square position.
Answer (B) is paired with (D) to sum to 1. Answer (A) is unpaired, so
between (A) and (B), the more likely answer is (B). Choice (B) is in fact
the correct answer.


There is a way to arrive at the exact answer: compute the relevant areas
and take the ratio. The point of this exercise, though, is to practice
guessing tactics.
16. (D) and (E) can be eliminated: Gunther must be at least 10 years old for
him to have had a non-negative age “10 years ago” and for Mason to
have a non-negative age now. Therefore, eliminate (D) and (E), as both
choices will give a negative result when G > 10. The correct answer is in
fact (C).


CHAPTER 3
Data Suﬀiciency: Principles


In This Chapter...
Principle #1: Follow a Consistent Process
Principle #2: Never Rephrase Yes/No as Value
Principle #3: Work from Facts to Question
Principle #4: Be a Contrarian
Principle #5: Assume Nothing


Chapter 3
Data Suﬀiciency: Principles
The goal of every Data Suﬀiciency (DS) problem is the same: determine
what information will let you answer the given question. This gives you a
significant advantage. Once you know whether a piece of information lets
you answer the given question, you can stop calculating. You do not have
to waste time finishing that calculation.
However, this type of problem presents its own challenges. DS answer
choices are never numbers, so you can’t plug them back into the question
to check your work. Also, answer choice (E)—that the two statements
combined are NOT suﬀicient—leaves open the possibility that the
embedded math question is not solvable even with all the information
given. That is, unlike Problem Solving (PS), DS may contain math problems
that cannot be solved! This aspect of DS is unsettling. (Note: If you don’t
already have the five DS answer choices memorized, then you are not yet
ready for this chapter. Practice with other material and then return here
when you feel fully comfortable with how DS works.)
On DS problems, the issue being tested is “answer-ability” itself—can the
given question be answered, and if so, with what information? So the GMAT
disguises “answer-ability” as best it can. The given facts and the question


itself are generally presented in ways that make this determination
diﬀicult.
For instance, information that seems to be suﬀicient may actually be
insuﬀicient if it permits an alternative scenario that leads to a diﬀerent
answer. Likewise, information that seems to be insuﬀicient may actually be
suﬀicient if all the possible scenarios lead to the same answer for the
question.
Advanced DS problems require you to step up your game. You will have to
get really good at Testing Cases, which is even more important for
advanced DS than for advanced PS problems.
You will also have to get really good at simplifying the given facts and the
given question. The GMAT increases the trickiness of the phrasing of the
question and/or statements even more than the complexity of the
underlying concepts.
So how do you approach advanced DS problems? Unfortunately, there is
no “one-size-fits-all” approach:
All that said, there are a few guiding principles you can follow.
The best approach may involve precise application of theory, or it may
involve a “quick and dirty” approach.
The statements may be easy or diﬀicult to interpret.
The question may require no rephrasing or elaborate rephrasing. In fact,
the crux of the problem may rest entirely on a careful rephrasing of the
question.


Principle #1: Follow a Consistent
Process
This is the most important principle. A consistent process will prevent the
most common errors. It will focus your eﬀorts at any stage of the problem.
Perhaps most importantly, it will reduce your stress level, because you will
have confidence in the approach that you’ve practiced.
Follow these rules of the road:
Do your work on paper, not in your head. This fits with the put pen to
paper concept discussed in Chapter 1. Writing down each thought as it
occurs helps you keep track of the work you’ve done. Your mind is also
freed up to think ahead. Many DS questions are explicitly designed to
confuse you if you do all the work in your head.
Label everything and separate everything physically on your paper. If
you mix up the elements of the problem, you will o en mess up the
problem itself. Keep these four elements straight:
Facts given in the question stem. You can leave them unlabeled or
you can put them with each statement.
Question. Label the question with a question mark. Obvious, right?
Amazingly, many people fail to take this simple step. Without a
question mark, you might think the question is a fact—and you will


Here is one sample schema for setting up your work on a Data Suﬀiciency
question:
It may be worth rewriting the facts from the question stem alongside
each statement. Although this may seem redundant, the time is well
spent if it prevents you from forgetting to use those facts.
get the problem wrong. Keep the question mark as you rephrase. It’s
also helpful to keep the helping verb in a Yes/No question:
Original question:
You write down:
You could just write down:
Do NOT just write down:
Statement (1). Label this with a (1).
Statement (2). Label this with a (2).
Rephrase the question and statements whenever possible. Question
stems and statements are o en more complex than they need to be; if
you can simplify the information up front, you will save yourself time
and eﬀort later in the problem. In particular, try to rephrase the question
before you dive into the statements.
Evaluate the easier statement first. If the second statement looks much
easier to you than the first, then start with the second statement.
Physically separate the work that you do on the individual statements.
Doing so can help reduce the risk of statement carryover—
unintentionally letting one statement influence you as you evaluate the
other.


Scrapwork for Rephrasing: ~~~~

Constraints: ~~~~

AD
BCE

(1)

FACT from question stem: ~~~~
FACT from (1): ~~~~
. . . and any work you do to combine
these facts
(2)

FACT from question stem: ~~~~
FACT from (2): ~~~~
. . . and any work you do to combine
these facts
QUESTION: .
. . ?
ANSWER ANSWER
Notice the physical separation between statement (1) and statement (2).
You might even consider going so far as to cover up the statement (1) work
when evaluating statement (2). Also observe that this schema explicitly
parses out the facts given in the question stem and evaluates those facts
alongside each statement.


Try this example problem, and then take a look at what the scrap paper
would look like according to this schema.
Try-It #3-1
If x and y are integers and 4xy = x2y + 4y, what is the value of xy ?
Because the question stem contains an equation, simplify it before
considering the statements:
Therefore, either x = 2 or y = 0, or both. One of the following scenarios must
be true:
x y xy = ?
2 Not 0 2y
Not 2 0 0
2 0 0
y − x = 0(1)
x3 < 0(2)


Interesting. It turns out that you only need to know the value of y. If y = 0,
then xy = 0. Otherwise, x must equal 2, in which case the value of xy is still
determined by the value of y.
(1) INSUFFICIENT: Rather than trying to combine this algebraically with the
equation in the question stem, try a couple of the possible scenarios that
fit the statement y − x = 2. Construct scenarios using the earlier table as
inspiration:
Since there are two possible answers, this statement is not suﬀicient.
(2) SUFFICIENT: If x 3 < 0, then x < 0. If x does not equal 2, then y must equal
0, according to the fact from the question stem. Therefore, xy = 0.
Notice how valuable it was to evaluate the fact in the question stem first
and to use it to rephrase the question. Then, when you reach the
statements, your work is made much easier. Here is approximately how
your paper could look:
Scrapwork for
Rephrasing:
Does x = 2 or does y = 0?
x = 2, y = 0 or both
If y = 0, value of xy = 0.
If x = 2, value of xy
depends on y.
If x = 2, then y = 2 + x = 4, so xy = (2)(4) = 8.
If y = 0, then x = y − 2 = −2, so xy = (−2)(0) = 0.


Constraints:                            x and y integers

(1)

Case 1: If x = 2, y = 4, so xy = 8.
Case 2: If y = 0, x = −2, so xy = 0.
(2)

If x is negative, y must equal 0.
QUESTION:
Does y = 0 ?
What is y ?
INSUFFICIENT y = 0 SUFFICIENT
The correct answer is (B).
If you don’t rephrase the question, it is easy to fall into the trap of thinking
that statement (2) alone is not suﬀicient.
You can lay out your paper in many other ways. For instance, you might go
with this version:
Stem:                                           Q: . . . ?


1)
2)
In this second layout, facts go on the le , while the question and any
rephrasing go on the right. Then the process is always to see whether you
can bridge the gap, going from le to right.
The important thing is that you develop a consistent layout that you
always use. Don’t give away points on Data Suﬀiciency because your work
is sloppy or you mix up the logic.


Principle #2: Never Rephrase Yes/No
as Value
All DS questions can be divided into two types: Value questions (such as
“What is a?”) and Yes/No questions (such as “Is x an integer?”). Value
questions and Yes/No questions are fundamentally diﬀerent: they require
diﬀerent levels of information to answer the question. Therefore, never
rephrase a Yes/No question as a Value question. Value questions usually
require more information than Yes/No questions.
Try-It #3-2
Is the integer n odd?
You don’t need to know which value n might be, just whether n is odd.
Therefore, do not rephrase this question to “What is integer n?” Doing so
unnecessarily increases the amount of information you need to answer the
question. Of course, if you happen to know what n is, then great, you can
answer any Yes/No question about n. But you generally don’t need to know
the value of n to answer Yes/No questions about n, and the GMAT loves to
exploit that truth at your expense.
n2 − 2n is not a multiple of 4.(1)
n is a multiple of 3.(2)


(1) SUFFICIENT: n2 − 2n = n(n − 2). If n is even, both terms in this product
will be even and the product will be divisible by 4. Since n2 − 2n is not a
multiple of 4, integer n cannot be even—it must be odd.
(2) INSUFFICIENT: Multiples of 3 can be either odd or even.
The correct answer is (A).
Rephrasing a Yes/No question into a Value question makes the question
unnecessarily picky. Yes/No questions can o en be suﬀiciently answered
despite having multiple possible values for the answer. In the last question,
for example, n could be any odd integer. If you rephrased this question to
“What is n?” you would incorrectly conclude that the answer is (E).
Note that the converse of this principle is not always true. Occasionally, it’s
okay to rephrase a Value question as a Yes/No question—specifically, when
it turns out that there are only two possible values.
Try-It #3-3
If x is a positive integer, what is the remainder of
Some quick analysis will show that x2 − 1 can be factored into (x + 1)(x − 1).
If x is odd, then both of these terms will be even and the product will be
divisible by 4, yielding a remainder of 0 when divided by 4. If x is even, then
x2 will be divisible by 4, so the remainder of x2 − 1 will be 3.
There are only two possible values of the remainder: 0 and 3. So this Value
question can be rephrased as the Yes/No question, “Is x odd?” or similarly,


“Is x even?” Although this Value question seemed at first to have several
diﬀerent potential outcomes, only two are possible, so you are able to
change the question to a Yes/No format by suitable rephrasing.


Principle #3: Work from Facts to
Question
Especially for simple Yes/No questions, people o en assume the answer to
the question is Yes before looking at the statements. They only test cases
for which the answer is Yes, rather than testing all of the possible cases
allowed by the statements.
This line of thinking is backwards—and tempting, because of the order in
which things are presented.
Instead ask, “If I start with the applicable facts and consider all
possibilities, do I get a definitive answer to the question?”
Always work from the given facts to the question—never the reverse! This
is why you have to keep the facts separated from the question and why you
should always clearly mark the question on your paper.
Try-It #3-4
If x ≠ 0, is x = 1 ?
(1)
(2)


If you work from the question to the facts, you would assume a Yes for the
question, then plug this information into the statements. For instance, you
would plug x = 1 into each statement. You would see that the value fits the
equations in both statements, and you would pick (D) incorrectly. That’s
especially easy to do in this case because this particular question is much
simpler to think about than the statements (which are nasty little
equations).
Don’t start from the question! No matter what, when you are judging
suﬀiciency, always proceed from the facts to the question. It doesn’t
matter how easy or hard the question is at that point. A er you’ve
rephrased, put the question on hold and work from the statements and
any other given facts to the question:

AD
BCE

(1)
 (2)
Question:
Is x = 1 ?
Maybe Yes
The correct answer is (B).
In a Yes/No question, when evaluating the statements, always try to determine whether
the question can be answered the same way under any possibility that is consistent with


those facts. A Yes answer means Always Yes, for all allowed scenarios. Likewise, a No
answer means Always No.


Principle #4: Be a Contrarian
To avoid statement carryover and to gain insight into the nature of a
problem, deliberately try to violate one statement as you evaluate the
other statement. This will make it much harder for you to make a faulty
assumption that leads to an incorrect answer. Think outside the first
statement’s box.
Try-It #3-5
If x ≠ 0, is xy > 0 ?
The question is actually asking whether x and y have the same sign.
(1) INSUFFICIENT: This indicates nothing about the sign of y.
In evaluating statement (2), you might be tempted to assume that x must
be positive. A er all, you just read information in statement (1) that
indicates that x is positive. Besides, it is natural to assume that a given
variable will have a positive value, because positive numbers are much
more intuitive than negative numbers.
(1)
(2)


Instead, follow Principle #4: actively try to violate statement (1), helping
you to expose the trick in this question.
(2) INSUFFICIENT: Consider the possibility that x is negative. In this case, it
is necessary to flip the sign of the inequality when you cross-multiply. That
is, if x < 0, then
 means that 1 > xy, and the answer to the question
is MAYBE.
(1) & (2) SUFFICIENT: If x is positive, then statement (2) says that 1 < xy (do
not flip the sign when cross-multiplying). Thus, xy > 0.
The correct answer is (C).
When evaluating individual statements, deliberately trying to violate the other statement
can help you see the full pattern or trick in the problem. You will be less likely to fall victim
to statement carryover.


Principle #5: Assume Nothing
This principle is a corollary of the previous principle: avoid assuming
constraints that aren’t actually given in the problem—particularly
assumptions that seem natural to make.
Try-It #3-6
Is z an even integer?
The wording of this question has a tendency to bias people toward
integers. A er all, the “opposite” of even is odd, and odd numbers are
integers, too. However, the question does not state that z must be an
integer in the first place, so do not assume that it is.
(1) SUFFICIENT: The fact that
 is an even integer implies that z = 2 × (an
even integer), so z must be an even integer. (In fact, according to statement
(1), z must be divisible by 4.)
(2) INSUFFICIENT: The fact that 3z is an even integer implies that z =
, so z might not be an integer at all. For example, z could
 is an even integer.(1)
3z is an even integer.(2)


equal
 .
One way to avoid assuming is to invoke Principle #3: work from facts to
question. Statement (2) indicates that 3z = even integer = −2, 0, 2, 4, 6, 8, 10,
etc. No even integers have been skipped over, nor have you allowed the
question to suggest z values. That is how assumptions sneak in.
Next, divide the numbers in your list by 3:
, etc. Is z an even integer?
Maybe. Note that you could have stopped computing a er the second
value, since you already achieved a Yes and a No.
The correct answer is (A).
If you had assumed that z must be an integer, you might have evaluated
statement (2) with two cases:
You would have incorrectly concluded that statement (2) was suﬀicient and therefore
incorrectly selected answer (D).
Another common assumption is that a variable must be positive. Do not
assume that any unknown is positive unless it is stated as such in the
information given (or if the unknown counts physical things or measures
some other positive-only quantity).
3 × even = even, so z could be even.
3 × odd = odd, so z is definitely not odd.


Problem Set
For problems 1–3, apply Principle #1 (follow a consistent process) from this
chapter to arrive at a solution to each problem. Note that the solutions
presented for problems 1–3 are specific examples. Your process may be
diﬀerent. Also note that the Data Suﬀiciency answer choices are not listed
but all DS problems throughout this guide use the standard DS answers in
the standard order.
1. There are 19 batters on a baseball team. Every batter bats either
right-handed only, le -handed only, or both right-handed and le -
handed. How many of the 19 batters bat le -handed?
Seven of the batters bat right-handed but do not bat le -
handed.
(1)
Four of the batters bat both right-handed and le -handed.(2)
2. If a is a positive integer and 81 divided by a results in a remainder of
1, what is the value of a ?
The remainder when a is divided by 40 is 0.(1)
The remainder when 40 is divided by a is 40.(2)


Solve problems 4 and 5. Apply Principle #2 (Never Rephrase Yes/No as
Value); describe why these Yes/No questions cannot be rephrased as Value
questions.
3. If a, b, c, d, and e are positive integers such that
 is
 an integer?
d − e ≥ 4(1)
d − e > 4(2)
4. If a, b, and c are each integers greater than 1, is the product abc
divisible by 6 ?
The product ab is even.(1)
The product bc is divisible by 3.(2)
5. If n is a positive integer, is n − 1 divisible by 3 ?
n2 + n is not divisible by 6.(1)
3n = k + 3, where k is a positive multiple of 3(2)
6. Revisit problems #4 and #5 above, this time deliberately violating
Principle #3 (Work from Facts to Question). Determine the incorrect
answer you might have selected if you had reversed the process
and worked from the question to the facts.


First, attempt to solve problem 7 by evaluating statement (1), and then
evaluating statement (2) without violating the information in statement (1).
Then, re-solve the problem by applying Principle #4 (be a contrarian). Do
you get the same answer? Verify that applying Principle #4 leads to the
correct answer, whereas not following the principle could lead to an
incorrect answer.
For problems 8 and 9, apply Principle #5 (assume nothing) by identifying
the explicit constraints given in the problem. What values are still
permissible? Next, solve using these constraints. Verify that diﬀerent
(incorrect) answers are attainable if incorrect assumptions about the
variables in the problem are made, and identify examples of such incorrect
assumptions.
7. Is m ≠ 0, is m3 > m2 ?
m > 0(1)
m2 > m(2)
8. If yz ≠ 0, is 0 < y < 1 ?
y = z2                                     (1)
(2)
9. Is x > y ?
|x – y| < |x|(1)


|x| > |y|(2)


Solutions
(Note that the solutions presented for problems 1–3 are specific examples.
Your notes may be diﬀerent.)
1. (A):
Rephrasing:
19 batters
total →
integers
only!
Some R
only
Some L
only
Some R & L
0 “neither”
How many
le -handed
batters?
L Not L Total
R
Not R 0
Total x 19
What is the integer x?
(1)


L Not L Total
R 7
Not R 0
Total x 7 19
Question:
Integer x = ?
Can find x.
(SUFFICIENT)
2. (B):
Rephrasing: 81 divided by a → remainder of 1.
                          a goes evenly into 80, and a ≠ 1.
                         a is one of the factors of 80 other than 1.
                         a = 2, 4, 5, 8, 10, 16, 20, 40, or 80.
Which of the numbers listed above is the value of a ?
(1)
 → remainder of 0.
(2)
 → remainder of 40.


40 goes evenly into a.
a is a multiple of 40.
a must be larger than 40.
Question:
Which of the listed numbers is a ?
a could be 40 or 80.
(INSUFFICIENT)
a must be 80.
(SUFFICIENT)
3. (D):
Rephrasing:
Is
 an integer?
Is
 an integer?
Is
Is
(1)
 (2)
Question:
Is d − e ≥ 4 ?
Yes
(SUFFICIENT)
Yes
(SUFFICIENT)


4. (C): Each of the below is an accurate Yes/No rephrase:
Alternatively, you could ask “Is there an even integer and a multiple of 3
among a, b, and c ?”
(1) INSUFFICIENT: ab is divisible by 2, but it’s unclear whether it is
divisible by 3 (or whether c is divisible by 3).
(2) INSUFFICIENT: bc is divisible by 3, but it’s unclear whether it is
divisible by 2 (or whether a is divisible by 2).
(1) AND (2) SUFFICIENT: Statement (1) indicates that a or b is even, and
statement (2) indicates that b or c is divisible by 3. Therefore abc is
divisible by both 2 and 3.
The correct answer is (C).
Is
 ?
Is abc = 6, 12, 18, 24, 30, 36, 42, 48, etc.?
Is abc divisible by 2 and by 3 ?
5. (A): An accurate Yes/No rephrase is the following:


This narrows down the values of interest to a certain type of number,
which follows a pattern: 1, 4, 7, 10, etc.
(1) SUFFICIENT: If n2 + n = n (n + 1) is not divisible by 6, you can rule out
certain values for n.
n n + 1 n (n + 1) not divisible by 6
1 2 ✓
2 3 ✗
3 4 ✗
4 5 ✓
5 6 ✗
6 7 ✗
7 8 ✓
The pattern from the rephrasing is apparent here: n can only be 1, 4, 7,
10, etc., all integers that are one greater than a multiple of 3.


Alternatively, use theory. The integers n − 1, n , and n + 1 must be
consecutive. If n( n + 1) is not divisible by 3, then n − 1 must be divisible
by 3, since in any set of three consecutive integers, one of the integers
must be divisible by 3.
(2) INSUFFICIENT: If 3n = 3 × pos integer + 3, then n = pos integer + 1 .
Therefore, n is an integer such that n ≥ 2. This does not resolve whether
n is definitely one greater than a multiple of 3.
The correct answer is (A).
6. Revisiting #4, working incorrectly from the Question to the Facts:
Manipulating the question to abc = 6 × integer (and losing track of the
question mark), someone might be tempted to check whether it is
possible for abc to be a multiple of 6, instead of whether abc is definitely
a multiple of 6:
The incorrect thinking would lead someone to wrong answer (D). Be
sure to revisit how to do this problem correctly!
Revisiting #5, working incorrectly from the Question to the Facts:
Someone working incorrectly might try multiples of 3 for n − 1 to see
whether they “work” with the statements:
(1) abc = 6 × integer, so ab is even. ✓
(2) abc = 6 × integer, so bc is divisible by 3. ✓


(1) SUFFICIENT:
n − 1 n n + 1 n2 + n = n(n+ 1) not divis by 6
3 4 5 20 ✓
6 7 8 56 ✓
9 10 11 110 ✓
(2) SEEMS SUFFICIENT (incorrectly): If 3n = k + 3, then k = 3n − 3 = 3(n −
1).
n − 1 k = 3n − 3 = 3(n − 1) pos mult of 3
3 9 ✓
6 18 ✓
9 27 ✓
This incorrect work would lead you to wrong answer (D). Be sure to
revisit how to do this problem correctly, so you are certain how to do so
for the future!
7. (C):


Non-Contrarian Approach Contrarian Approach
(1) INSUFFICIENT:
m > 0 or m = pos, so “Is pos3 >
pos2?”
(1) INSUFFICIENT:
m > 0 or m = pos, so “Is pos3 > pos2?”
(2) SEEMS SUFFICIENT:
m2 > m implies that m is not a
fraction or 1, therefore m3 >
m2.
Or, if you assume that m > 0,
carrying over from (1), you
might do the following:
m2 > m
m > 1 (dividing by m)
Therefore, m3 > m2.
(2) INSUFFICIENT:
m2 > m implies that m is not a fraction or 1, so if m > 1
then the answer is Yes.
BUT, contradicting (1), what if m is negative? That is
possible according to (2), since
neg2 > neg.
Is m3 > m2? → Is neg3 > neg2? No!
Yes, if m > 1.
No, if m is a proper fraction
or 1.
(That is, 0 < m ≤ 1.)
Yes, if m > 1.
No, if m is a proper fraction or 1.
(That is, 0 < m ≤ 1.)


Non-Contrarian Approach Contrarian Approach
(1) AND (2) SUFFICIENT:
Combined, the statements eliminate negative, zero,
positive proper fractions, and 1 for the value of m. If m > 1,
then m3 > m2. Definite Yes answer.
The (incorrect) answer is (B). The correct answer is (C).
Note that both solutions were hampered by inadequate rephrasing.
Ideally, you would first rephrase as follows:
Takeaway: With proper rephrasing, other errors are less likely. But even
with inadequate rephrasing, taking a contrarian approach can save you
from a wrong answer.
Is m3 > m2?
Is m > 1? (It’s okay to divide by m2, which must be positive: a square is
never negative and m is also not 0, according to the question stem.)
8. (C): The explicit constraint is yz  ≠ 0, which indicates that y ≠ 0 and z ≠ 0.
Both y and z could be any nonzero value, including positive integers,
negative integers, positive fractions, negative fractions, etc.
Making faulty assumptions:


For (1), a faulty assumption could be made by those who plug in values
for z.
For example, if you plug in z = −2, −1, 1, 2, 3, 4, etc., you would get y = 1,
4, 9, 16, etc. That would yield a definite No answer to the question, as all
the y values are at least as great as 1. The (unverbalized) assumption is
that z is an integer, but that’s not necessarily so.
For (2), most people would want to cross-multiply, so a potential false
assumption is that y is positive (and this is reinforced by the fact that, in
statement (1), y is in fact positive):
Conclusion: SEEMS SUFFICIENT (incorrect)
These faulty assumptions would lead to the incorrect answers (A) or (D).
Correct solution:
(1) INSUFFICIENT: y must be positive, but is it a fraction or an integer?
If z = 2, then y = 4, and the answer is No.


(2) INSUFFICIENT:
Since you’re assuming y < 0,
the rephrasing is y < −1.
Is 0 < y < 1 ? YES NO
(1) AND (2) SUFFICIENT: Since y is positive, statement (1) indicates that 0
< y < 1.
The correct answer is (C).
If z =
 , then y =
 , and the answer is Yes.
9. (E):


(1) INSUFFICIENT: Test cases for x and y. Avoid limiting yourself to cases
in which x is greater than y. Since “Is x > y?” is a question and not a
statement, x might be greater than or less than y. Actively seek cases
that give a No answer.
x y |x – y| < |x| Is x > y ?
3 4 |–1| < |3|    (valid) No
4 3   |1| < |3|    (valid) Yes
Since x could be greater than y or less than y, given the information in
the statement, this statement is insuﬀicient.
(2) INSUFFICIENT: The second case tested above also fits this
statement and yields an answer of Yes. Find another case in which |x| >
|y|, but x is not greater than y.
x  y |x| > |y| Is x > y ?
4 3   |4| > |3|    (valid) Yes
–4 3 |–4| > |3|    (valid) No
(1) AND (2) INSUFFICIENT: A case that fits both statements and gives an
answer of Yes has already been found. Try to find another case for which
both statements are true, but which gives an answer of No. As you test
diﬀerent values for x and y, you may find certain cases for which one or
both statements are not true. Cross these oﬀ on your paper, and do not
take them into consideration when choosing your answer.


x y (1): |x – y| < |x| (2): |x| > |y| Is x > y ?x y (1): |x – y| < |x| (2): |x| > |y| Is x > y ?
4 3 |1| < |3|    (valid) |4| > |3|    (valid) Yes
–4 3 |–7| < |–4|: (not valid) n/a n/a
–4 –3 |–1| < |–4| |–4| > |–3| No
The answer to the question can be either Yes (if x = 4 and y = 3) or No (if
x = –4 and y = –3), so the statements together are insuﬀicient.
The correct answer is (E).


CHAPTER 4
Data Suﬀiciency: Strategies & Tactics


In This Chapter...
Advanced Strategies
Advanced Guessing Tactics
Summary
Common Wrong Answers


Chapter 4
Data Suﬀiciency: Strategies & Tactics
Sometimes you will encounter a Data Suﬀiciency (DS) problem that you
can’t answer—either because its content is diﬀicult or obscure, or because
you don’t have enough time to solve completely in two minutes.
Like Chapter 2, this chapter describes a series of diﬀerent methods you
might try in these circumstances. Again, this chapter distinguishes
between solution strategies and guessing tactics.
Strategies are broad: they apply to a wide variety of problems, they provide
a complete approach, and they can be used safely in most circumstances.
In contrast, tactics can help you eliminate a few answer choices, but o en
leave a fair amount of uncertainty. Moreover, a particular tactic may only
be useful in special situations or for parts of a problem.
The first section of this chapter outlines six DS strategies:
DS Strategy 1: Compute to Completion
For some problems, you won’t necessarily be able to tell whether the
answer can be calculated until you follow through on the calculations all