x x2
 = remainder of 4
 = remainder of?
2 4 Yes
 = 0 remainder 2
3 9 Yes
 = 0 remainder 3
There are at least two possible values for the remainder.
(2) SUFFICIENT:
x x3
 = remainder of 2
 = remainder of?
3 27 Yes
 = 0 remainder 3
There is only one possible remainder: 3.
If you aren’t sure that you only need to test these five cases, try the next
five values for x. Note the repeat of the remainder pattern. Only every
third term (3, 8, 13, and so on) will be valid based on the information
from statement (2) and will always have a remainder of 3.
0 0 No (invalid)
1 1 No (invalid)
0 0 No (invalid)
1 1 No (invalid)
2 8 No (invalid)
4 64 No (invalid)


The correct answer is (B).
8. (C): 5x will always end in 5.
(1) INSUFFICIENT: If x is even, then 5x = 25, 625, and so on. This
statement provides no information about y, though. For example if y = 5,
then
 remainder 0. If y = 4, then
 remainder 1.
(2) INSUFFICIENT: Test some diﬀerent values for x.
x 5x Remainder of
1 5 2
2 25 1
3 125 2
4 625 1
A er testing the first two numbers, it’s clear that statement (2) is not
suﬀicient. Because you will have to test the two statements together,
continue testing another couple of numbers to see whether there is a
pattern. When 5 is raised to an odd power, the remainder is 2, but when
5 is raised to an even power, the remainder is 1.


(1) AND (2) SUFFICIENT: When 5even integer is divided by 3, the remainder
is always 1.
The correct answer is (C).
9. 110: The remainder must always be smaller than the divisor. Thus, b
must be at least 10, and d must be at least 11. Therefore, bd must be at
least 110. The purpose of this problem is to remind you of these
constraints on remainders.
10. 63: Each column contains a series of evenly spaced integers. To avoid
calculation, cancel out as many values as possible before summing the
remaining values.
In each row, the sum of the integers in the first two columns is −1, and
the sum of the integers in the third and fourth columns is 1. Therefore,
the sum of the integers in the first four columns in each row is −1 + 1 = 0.
Only the integers in the fi h column need to be considered.
The integers in the fi h column can be rewritten as follows:
The sum of the integers is therefore 3(21) = 63.
11. 165: Write out the first few terms of the sequence to find the pattern.


To simplify the calculation, these terms can be rewritten as follows:
Therefore, the nth term in the sequence can be found by adding 10 a
total of n times, then adding the sum of the integers from 1 to n – 1,
inclusive. For n = 11, this value is equal to 10(11) + (1 + 2 + 3 + ... + 10),
which equals 10(11) + 55 = 165.
12. (B) Between
 and 0: Compute the first few elements of Sn.
n Sn
1
2


n Sn
3
4
5
Use a number line to track the sum:
Place the first term,
 , on the number line. The second term is +
 ,
so the sum will move to the right (closer to 0) on the number line. The
third term is
  so the sum will move to the le , but it can’t go as
far as
 again, because you’re only subtracting
 this time—the
distance is smaller than the first
 hop that you made.
Each subsequent hop flips back and forth between positive and
negative but also keeps getting smaller and smaller, so you’ll never
“break out” of the range
 to 0.
The correct answer is (B).


13. 9: Calculate the first several terms of the sequence to find the pattern:
The pattern should continue, so 8 will be the units digit 9 out of the first
10 times.
14. (B) 24: This is a counting pattern problem. In order for ♦ K♦ to be divisible
by 411, it must be divisible by (22)11 = 222. Thus,  ♦ K♦ must contain 22
twos in its prime factorization.
If K = 10, for example, then ♦ K♦ equals 2 × 4 × 6 × 8 × 10. The number 2
has one 2 in its prime factorization; 4 has two 2’s; 6 has one 2 (and a 3); 8
has three 2’s; 10 has one 2 (and a 5). This amounts to a total of only eight
2’s:
Number 2 4 6 8 10
Prime Factor(s) 2 2, 2 2, 3 2, 2, 2 2, 5
Total 2’s in PF 1 2 1 3 1
Cumulative 2’s 1 3 4 7 8


Keep adding even numbers to the result until you get to 22 twos in total:
Number 2 4 6 8 10 12 14 16 18 20 22 24
Prime
Factor(s)
2 2,
2
2,
3
2,
2,
2
2,
5
2,
2,
3
2,
7
2, 2,
2, 2
2,
3,
3
2,
2,
5
2,
11
2, 2,
2, 3
Total 2’s in
PF
1 2 1 3 1 2 1 4 1 2 1 3
Cumulative
2’s
1 3 4 7 8 10 11 15 16 18 19 22
Thus, the smallest possible value for K is 24. Notice the pattern in the
number of 2’s in each even number: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1 …
The correct answer is (B).
15. $255 (or 28 − 1): At the end of the first week, Mitchell receives $1. At the
end of the second week, he gets $1, plus $1 for the total he had been
paid up to that point, for a total of $2. At the end of the third week, he
gets $1, plus ($1 + $2), or $3, for the total he had been paid up to that
point, so this third week’s total is $4. Put this in a table:
Week # Paid This Week($) Cumulative Pay Including This Week ($)
1 1 1


Week # Paid This Week($) Cumulative Pay Including This Week ($)
2 1 + 1 = 2     1 + 2 = 3
3 1 + 3 = 4     3 + 4 = 7
4 1 + 7 = 8     7 + 8 = 15
5 1 + 15 = 16 15 + 16 = 31
6 1 + 31 = 32 31 + 32 = 63
7 1 + 63 = 64 63 + 64 = 127
8 1 + 127 = 128 127 + 128 = 255
This calculation is not so bad, but you may notice that this payment
schedule is a geometric sequence, 2n − 1, where n is the number of the
week in which Mitchell is being paid. Summing that sequence is
equivalent to 2t − 1, where t is the total number of weeks. In other
words, the cumulative pay is one less than the next power of 2.
The correct answer is 28 − 1 = $255.


CHAPTER 6
Common Terms & Quadratic
Templates


In This Chapter...
Common Terms
Quadratic Templates
Quadratic Templates in Disguise


Chapter 6
Common Terms & Quadratic
Templates
You are probably already familiar with the mechanics of algebraic
manipulations—what is allowed and what is not:
But of all the steps you could take, how do you decide which steps you
should take?
Two indicators can o en help you on the GMAT:
You can substitute one expression for another if they are equal.
You can add some number to one side of an equation as long as you do
the same on the other side of the equation.
You can cross-multiply to simplify equations with fractions on each side
of the equals sign. And so on.
Common Terms1.
Quadratic Templates2.


Common Terms
If you spot common terms, you can o en spot the path all the way to the
solution. Common terms appear on the GMAT in three typical ways.
1. ALGEBRA
Look for terms that appear in the same form more than once. Those
recurring expressions might also appear in slightly modified form such as
reciprocal, negative, or raised to a power:
 is the reciprocal of
 .
 is negative
.
Is
 ? where
 and 3y appear both
squared and multiplied
together.
This expression is of the form:
(More on Quadratic Templates
later in this chapter)


Try-It #6-1
If y ≠ 3, simplify as much as possible:
Spot the common term (3 − y). Note that (−3 + y) is −(3 − y), or −1 × (3 − y).
Factor out the common term (3 − y) and cancel:
.
By the way, the condition that y could not equal 3 just prevented you from
dividing by 0 and ending up with an undefined number.
2. EXPONENTS
Exponents can be manipulated when either bases or exponents are
common. Also look for bases that have common factors, such as 3 and 12
(common factor of 3). You can o en create a common base. For example:
(16x)(42) =
256
4 is a factor of 4, 16, and 256.
Similarly, 4, 16, and 256 are all powers
of 4.


Try-It #6-2
If 3x + 243 = 2(3x), what is the value of x ?
Note the common term 3x, and note the fact that 243 = 35:
3. FACTORS AND MULTIPLES
When many terms share a factor, pull that shared factor out to the side.
These can appear in algebraic or numerical expressions:
→ get common denominators, then cross
them all oﬀ →
x18 + 2x16 + x14 → x14 is a factor of each term → x14(x4 + 2x2 + 1) = x14(x2 +
1)2


Factorials are particularly noteworthy, as they o en have an abundance of
shared factors. For any integer n, the factorial n! is calculated as follows: n!
= n(n − 1)(n − 2)(n − 3) . . . 1. Thus, all the terms in 4! = (4)(3)(2)(1) are also
common factors of 6! = (6)(5)(4)(3)(2)(1) = (6)(5)(4!).
More generally, factorials are “super multiples.” Without ever computing
their precise value, you can tell that they’re divisible by all sorts of
numbers. For example:
If x is an integer
between
7! + 2 and 7! + 4,
inclusive, is x prime?
x is one of the following
integers:
7! + 2 = (7)(6)(5)(4)(3)(2)(1) + 2
7! + 3 = (7)(6)(5)(4)(3)(2)(1) + 3
7! + 4 = (7)(6)(5)(4)(3)(2)(1) + 4
x has one of the following
factors, if x is:
7! + 2 = 2 × Integer
7! + 3 = 3 × Integer
7! + 4 = 4 × Integer…so x is
not prime!
Sometimes a common factor is just a random number buried inside a
couple of larger numbers. Find it and pull it out:
10’s in the numerator
line up with 5’s in the
corresponding digit
place of the
denominator:


Try-It #6-3
If n is a positive integer and
 is a positive
integer, what is the value of n ?
What would need to be true in order for the square root to be a positive
integer? The number under the square root symbol would have to be a
perfect square. Rearrange the expression to determine whether there are
any restrictions that could help narrow down the possibilities before going
to the statements. Try to break the numbers down into primes to locate
and pull out any existing perfect squares:
n is prime.(1)
n < 3(2)


Pull out common terms:
It turns out that you can also pull out the term 7n − 1. The term 7n + 1 = (7n − 1)
(72):
The first two terms are both perfect squares and can be pulled out of the
square root sign:
What would need to be true in order for 7n − 1 to be pulled out of the square
root sign? It would also have to be a perfect square, so n – 1 must be even
and n itself must be odd.


The question is: “If n is odd, what is the value of n?”
(1) INSUFFICIENT: This statement allows multiple possible odd values of n.
(2) SUFFICIENT: The question stem indicates that n is a positive integer,
and this statement specifies that n < 3. The only odd, positive integer less
than 3 is the number 1.
The correct answer is (B).


Quadratic Templates
On the GMAT, quadratic expressions take three common forms called the
Quadratic Templates. Memorize these templates, and get comfortable
transforming back and forth between factored and distributed form:
  Factored
 Distributed
Square of a Sum (a + b)2 = a2 + 2ab + b2
Square of a Diﬀerence (a − b)2 = a2 − 2ab + b2
Diﬀerence of Two Squares (a + b) (a − b) = a2 − b2
QUICK MANIPULATION
Expressions with both squared and non-squared common terms should
make you suspect that you are looking at a Quadratic Template.
Try-It #6-4
Factor
 .


This problem requires you to manipulate a rather complicated expression.
However, by using the common terms, you can put the problem in the
more basic template form to solve:
Once you are comfortable with Quadratic Templates, you can manipulate
even complicated expressions quickly, as in the middle box above. Until
then, write down the templates and the substitution of the common terms,
as in the box on the right.
The very same problem could have been presented in disguise:
Factor
 .
The common terms are slightly harder to spot in this form. In such a case,
start with the squared terms,
 and 25y2. Then, try to untangle their
square roots,
 and 5y, from the remaining term. The factored form is:


Consider all three of the common Quadratic Templates before deciding
which one or ones are most convenient to use.
THE MIDDLE TERM: 2AB
The square of a sum and square of a diﬀerence templates have something
in common: the middle term is ±2ab. The only diﬀerence is the sign of that
middle term.
When you add these two templates, the middle terms cancel, leaving the
end terms:
  Factored
 Distributed
Square of a Sum (a + b)2 = a2 + 2ab + b2
+ Square of a Diﬀerence (a − b)2 = a2 − 2ab + b2

Addition: (a + b)2 + (a − b)2 = 2a2 + 0 + 2b2
2(a2 + b2)


In contrast, when you subtract these two templates, the end terms cancel,
leaving the middle term:
  Factored
 Distributed
Square of a Sum (a + b)2 = a2 + 2ab + b2
+ Square of a Diﬀerence (a − b)2 = a2 − 2ab + b2

Subtraction: (a + b)2 − (a − b)2 = 0 + 4ab + 0
This is handy for simplification. Also, whenever you see the sum of two
squares (a2 + b2), which is not itself a Quadratic Template, remember that it
can be derived from this sum of two templates.
Try It #6-5
What is the sum of 9,9992 and 10,0012?
99,980,001(A)
199,999,998(B)
200,000,002(C)
399,999,996(D)
400,000,004(E)


These numbers all look seriously cumbersome. If only they had given the
easier number of 10,000 instead . . .
When wishful thinking pops up, try to use it to make the problem easier. If
you changed both of these numbers to the form 10,000 plus or minus a
number, what would that be?
These are the Quadratic Templates! In this case, a = 10,000 and b = 1. If
you’ve memorized the sum term, plug these in:
Now, notice something: the part in the parentheses is going to have a units
digit of 1. Multiply the number by 2, and the end result will have a units
digit of 2. Only answer choice (C) fits.
Not sure about that? Go ahead and do the math:
The correct answer is (C).
Even if you don’t memorize the sum term of the Quadratic Templates, the
math is still far easier to do in this rewritten form:


Quadratic Templates in Disguise
Quadratic Templates can be disguised in arithmetic computations.
Try-It #6-6
What is 198 × 202 ?
You can round each number and quickly estimate the result to be about
2002. Or, you could laboriously multiply two 3-digit numbers by hand to get
an exact result. But if you need an exact result quickly, you can use a
Quadratic Template, as shown in the previous section. You just need to
turn 198 into (200 − 2) and 202 into (200 + 2) as shown here:
Another place to hide a Quadratic Template is in an advanced right-triangle
problem:


You know that
 and d2 = b2 + h2 (by the Pythagorean
theorem). Do the common terms b2, h2, and bh look familiar? Use the
Square of a Sum template:
Likewise, there is a similar relationship based on the Square of a Diﬀerence
template:
An advanced GMAT problem can draw on these complicated relationships.
For instance, you can compute the area of a right triangle directly from the
sum of the shorter sides and the hypotenuse:


You should absolutely not memorize these particular formulas. Rather, be
able to recognize when the GMAT is indirectly testing these generic
Quadratic Templates.


Problem Set
For the following problems, use the Pattern Recognition techniques
discussed in this chapter to solve.
For problems 2–4, if x < –1, which of the following inequalities must be
true?
1. If xy > 0, is
(1)
x > 2y(2)
2. Is x4 > x2 ?
3. Is x3 + x4 > x3 + x2 ?
4. Is x6 − x7 > x5 − x6 ?
5. If,
 and |x| ≠ |y|, what is the ratio of x to y ?


4 : 1(A)
3 : 1(B)
2 : 1(C)
3 : 2(D)
1 : 3(E)
6. If
 , what is the value of
7. If n is an integer and (−3)4n = 37n – 3, then n = ?
8. If (ax + by)2 = (bx + ay)2, what is the value of a – b ?
x2 > y2(1)
a and b are positive integers.(2)
9. If x and k are both integers, x > k, and x–k = 625, what is x ?
|k| is a prime number.(1)
x + k > 20(2)


Distribute the expression in problems 10–14 without FOILing (doing the
math the long way). Use the Quadratic Templates.
Factor problems 15–19 according to the Quadratic Templates.
10.
11.
12.
13.
14.
15.
16. 4 + 4a + a2


For problems 20–23, simplify the expressions completely.
17. 81 − x4
18.
19. 4x2 − 12xy + 9y2
20.
21.
22. (111)(89)
23. 3502 − 3202


24.
In the right triangle below, side a is 7 inches longer than side b. If
the area of the triangle is 30 inches2, what is the length of
hypotenuse c ?


Solutions
1. (C): From xy > 0, you know that neither x nor y equals 0, and they must
have the same sign (++ or − −). Combine the exponent in the question to
get
What is the significance of the inequality in the question stem? If the
exponent equals 2, then 52 = 25. In order to be greater than 25, the
exponent has to be greater than 2. The question can be rephrased as “Is
?”
You may want to start with statement (2).
(2) INSUFFICIENT: If y is positive, then
 . If, on the other hand, y is
negative, then
 .
(1) INSUFFICIENT: Simplify the equation:


Therefore, y3 must equal 8, so y = 2. This provides no information about
x, however.
(1) AND (2) SUFFICIENT: If y = 2, then you can plug into statement (2) to
find the range of values for x. Since x > 2y and y is positive, you can
divide the expression by y, giving
 .
The correct answer is (C).
For questions 2–4: since x ≠ 0, divide by the common terms, making sure
to flip the inequality sign if the common term is negative.
2.
TRUE:
 ?
True
Note: x2 is positive.
Divide both sides by x2, leaving the sign as it is.
The square of a number smaller than −1 will be greater than
positive 1.
3.
TRUE:

True
Note: x2 is positive.
Divide both sides by x2, leaving the sign as it is.
Subtract x from both sides.
(Or, you might have subtracted x3 immediately.)
TRUE:

True
Note: x5 is negative.
Divide both sides by x5, flipping the inequality sign.
Group like terms.


4.
5. (C) 2 : 1: Try to find like terms in order to simplify the le -hand side of
the equation:
The ratio of x to y is 2 : 1.
The correct answer is (C).
6.
7. n = 1: Since n is an integer, 4n is even. An even exponent “hides the sign”
of the base, so you can treat the (−3) base as a (3):


8. (C): Begin by simplifying the equation in the question stem, making it
look as similar as possible to the question itself:
Do not divide by x2 – y2, since it may equal 0. Instead, subtract b2(x2 – y2)
from both sides of the equation, then factor:
So either a2 – b2 = 0 (in which case a2 = b2), or x2 – y2 = 0 (in which case x2
= y2). One or both of these must be true.


(1) INSUFFICIENT: If x2 > y2, then x2 cannot equal y2. Therefore, a2 = b2.
Case 1: It is possible that a and b are equal, in which case a – b = 0, so
the answer is 0.
Case 2: It is also possible that a and b are equal in absolute value but
have opposite signs. For example, a = –2 and b = 2. In this case, the
answer is a – b = –4.
More than one answer is possible, so this statement is not suﬀicient.
(2) INSUFFICIENT: This statement only says that a and b are positive
integers, but does not give further indication of what they could be.
There are many possible answers for the value of a – b, so this statement
is insuﬀicient.
(1) AND (2) SUFFICIENT: According to statement (1), a2 = b2. Thus, either
a = b or a = –b. According to statement (2), a and b are both positive, so it
is not possible that a = –b. Therefore, a = b, and the answer to the
question is a – b = 0. The statements together are suﬀicient.
The correct answer is (C).
9. (A): The fact that x and k are both integers (and that x > k) significantly
limits the possible values for x and k. The possible pairings are as
follows:
x k


x k
5 −4
25 −2
625 −1
(1) SUFFICIENT: If the absolute value of k is prime, then only the second
possibility works: x = 25 and k = −2.
(2) INSUFFICIENT: The second and third possibilities both make this
statement true, so it isn’t possible to determine a single value for x.
10.
11.
12.
13. 4:
14.


15.
16.
17.
18.
19.
20.
21. 4:
22. 9,879:
23. 20,100:
24. (a − b) = 7 and
 : From the Pythagorean theorem, a2 + b2 =
c2.
Use the Square of a Diﬀerence template:


Plug in the values:


CHAPTER 7
Visual Solutions


In This Chapter...
Representing Objects with Pictures
Rubber Band Geometry
Baseline Calculations for Averages
Number Line Techniques for Statistics Problems


Chapter 7
Visual Solutions
Visual interpretations—good pictures, essentially—can help you solve
certain types of GMAT problems. This chapter highlights some of these
types of problems and demonstrates how you can use visual techniques to
solve these problems more confidently, accurately, and quickly.
Many problems discussed in this chapter can be solved with other
techniques. Still, visual thinking is a powerful tool. It can expand your
comprehension of a topic. It may enable you to solve particular problem
types more easily or “break through” on a diﬀicult problem. In fact,
visualization is the only realistic way to approach certain problems. So it’s
worth trying your hand with visual approaches.
In this chapter, we will discuss the following Visual Solution techniques:
Representing Objects with Pictures: Many Word Problems and Geometry
problems do not provide a diagram alongside the problem. Drawing a
good picture will make the problem-solving process easier and less
error-prone.
Rubber Band Geometry: For Geometry questions involving both
constraints and flexibility (especially in Data Suﬀiciency), drawing


diﬀerent “rubber band” scenarios according to those constraints and
freedoms can o en help you solve the problem without doing any
computation.
Baseline Calculations for Averages: Visual techniques can help you
compute averages (both basic and weighted) and can also foster a
better understanding of those calculations.
Number Line Techniques for Statistics Problems: You can solve a variety
of common problems involving statistics by using a number line to
visualize and manipulate the problem.


Representing Objects with Pictures
For Word Problems that describe a physical object, or for Geometry
problems that do not give a diagram alongside the problem, drawing a
picture is o en the best approach. Sometimes it’s the only viable approach!
Even if you are good at visualizing objects in your head, draw the picture
anyway. It’s just too easy to make a mistake on many of these questions.
Try-It #7-1
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch.
If the dowel is painted on all surfaces and then cut into
-inch
cubes, what fraction of the resulting cube faces are painted?
If you draw a picture, this problem becomes a matter of counting:
(A)
(B)
(C)
(D)
(E)


Total cubes = (4 inches × 2 cubes per inch) × (1 × 2) × (1 × 2) = 32 cubes
Total cube faces = 32 cubes × 6 faces per cube = 192 faces total
Now, consider the faces that were painted on the front and back of the
dowel, the top and bottom of the dowel, and the ends of the dowel. In the
diagram above, you can see 16 faces on the front, 16 faces on the top, and 4
faces on the end shown. Of course, there are other sides: the back, the
bottom, and the other end. Now you can find the number of painted cubes:
Therefore, the fraction of faces that are painted is
 .
The correct answer is (B).
Notice that there is no shortcut to solving this kind of problem, so don’t
waste time looking for one—just draw the diagram and count.
Even if you can easily picture 3-D shapes and objects in your head, it is still better to draw a
picture on your scrap paper.


The test will usually include among the incorrect answers numbers that you might get by
losing track of your progress as you process the object in your mind.
This kind of process can also help you with questions that deal with the
relative size of diﬀerent objects.
Try-It #7-2
Bucket A has twice the capacity of Bucket B, and Bucket A has
 the
capacity of Bucket C. Bucket B is full of water and Bucket C is half
full of water. When the water from Bucket B is poured into Bucket C,
Bucket C will be filled to what fraction of its capacity?
You could attempt to solve this problem algebraically, but the equations get
messy very quickly. Instead, try drawing buckets A, B, and C in correct
proportion to one another. Then think through the problem:


Algebra Picture
The algebra and the picture say the same thing, but the picture has several
advantages:
Based on this picture, you might pick a capacity of 1 for Bucket B, yielding a
capacity of 2 for Bucket A and 6 for Bucket C. Bucket B would contain 1 unit
of water and Bucket C, 3 units. When the contents of B are poured into C,
Bucket C would then be
 full.
It’s much easier to comprehend at a glance.
It’s harder to mistake relative sizes (e.g., accidentally thinking A is
smallest).
You can easily represent both total capacity and amount of water visually.
It prompts you to pursue the smartest, easiest solution: picking numbers
for the capacities of the buckets.


Notice that the buckets are not labeled in alphabetical order, even though
that would be easy to incorrectly assume. The GMAT frequently adds little
layers of disguise and complexity such as this to induce you to make a
mistake. By drawing the buckets carefully, you minimize the chance that you
will fall into a trap on a problem such as this one.


Rubber Band Geometry
Many Geometry problems—particularly of the Data Suﬀiciency (DS) variety
—describe objects for which only partial information is known. We call
these questions Rubber Band Geometry problems, because they
simultaneously involve constraints and flexibility. Some parts of the
diagram can stretch like a rubber band as you open or close angles.
Your job is to figure out what specifics in the problem are constrained and
what specifics are flexible.
For example, if a problem specifies that a line has a slope of 2, it will be
steep and upward-sloping. In fact, it will always “rise” 2 units for every unit
of “run.” However, you don’t know where the line will appear. The line is
constrained in its slope, but it is flexible in that it can be moved up or down,
right or le . You can draw many diﬀerent lines with a slope of 2 (these lines
will all be parallel, of course).
If, however, the problem only specifies that a line must go through the point
(4, 0) in the coordinate plane, then the line is “fixed” at that point. However,
the slope of the line would now be flexible. You could draw many diﬀerent
lines with diﬀerent slopes that run through that point.
If you knew both of these specifications—that the line must have a slope of
2 and must run through the point (4, 0)—then you would be able to


calculate the exact line that is being described. The slope of the line and a
point that the line goes through specify the line precisely—there is no
remaining flexibility for either the slope or the location of the line. (Note
that in this example, the line is described by the equation y = 2x − 8.)
Given these specifications, every other feature of this line is also known: its
x-intercept, its y-intercept, whether it goes through some fixed point, which
quadrants it crosses, etc. In a DS problem, you could answer any such
questions about this line without actually calculating the answer.
Therefore for these types of problems, your goal is to figure out what
combination of information “cements” the problem in place—in other
words, what combination of information removes all of the remaining
flexibility. No flexibility means suﬀiciency. And by using rubber band
geometry thinking, you can o en do this without using any calculation or
algebra at all.
For each piece of information that you’re given in this type of problem,
think about what is fixed and what is flexible. Try to draw multiple versions
of each object (if possible), testing the boundaries of this flexibility. The
following everyday objects may be useful as analogies in your thinking:
Rubber band: Determines a straight line segment. Can stretch
between any two points.
Drinking straw: Determines a straight line segment, but with fixed
length.
Thumbtack: Fixes a point, but can allow rotations through that point
in many cases.
Wedge: Fixes an angle.


Here are some common examples of how these objects can be used to help
you think through these problems:
Constrained Flexible Analogous Object(s) Mental
Picture/Simplified
Sketch
A line passes
through a
specified
point.
Slope of
the line
Drinking straw = line
Thumbtack = point
Line free to spin about a
point.
Two lines
intersect at a
specified
point.
Slope of
the lines
Drinking straw = line
Thumbtack = point
Both lines free to spin
about a point. The angle
is free to change.
A line passes
through two
points.
Nothing
flexible
Drinking straw = line
Thumbtacks = points
Two points pin down a
line—no flexibility.


Constrained Flexible Analogous Object(s) Mental
Picture/Simplified
Sketch
Specified
distance
between two
points
Absolute
or
relative
location
of the
points
Thumbtacks = points
Drinking straw =
distance between points
Fix one point
temporarily. Line (straw)
free to spin about one
point, tracing the circle
of possible locations of
the other point.
Slope of a line Location
of the
line, or
points
the line
may pass
through
Drinking straw = line
Line is free to “float
around” but not rotate.
Points are on
a line (either
in the
coordinate
plane or on a
basic number
line).
Distance
between
the
points
Rubber band = stretched between
points
 “Stretchy” distance
between points


Constrained Flexible Analogous Object(s) Mental
Picture/Simplified
Sketch
Points on a
line are a
specified
distance
apart.
Order of
the
points
(le -to-
middle-
to-right)
Drinking straws = fixed lengths
between points
Manipulation will be determined by
other stated constraints, but
thinking of the points as the
endpoints of rigid straws will ensure
that you do not forget the distance
constraint.
Lines could be laid end-
to-end, overlapping,
separated, and flipped
right-to-le .
Triangle with
a fixed area
and a fixed
base (and
therefore a
fixed height)
Position
of the
third
vertex
along a
line
parallel
to the
base
Straw = fixed base
Rubber bands = other sides of the
triangle
Thumbtacks = endpoints of the base
This is not intended to be an exhaustive list. The idea is to show you a new
way of thinking through some diﬀicult Geometry problems.


Is B inside the circle?
Try-It #7-3
A circle in a coordinate plane has a center at point A and a diameter
of 6. If points B and C also lie in the same coordinate plane, is point
B inside the circle?
The exact locations of points A, B,
and C do not matter—only the
relative locations of the points
matter. Therefore, you can
arbitrarily assign point A to a
specific location (when possible,
choose the origin of the coordinate
plane) and draw a circle with a
radius of 3 units around it.
(1) INSUFFICIENT: Statement (1)
does not indicate anything about
point B, so it is not suﬀicient.
However, it does indicate that A and C are 2 units apart, so statement (1)
enables you to place point C anywhere along the gray circle.
(2) INSUFFICIENT: Statement (2) does not indicate anything about point B
relative to point A, so it is not suﬀicient. However, it does constrain point B
to be exactly 2 units away from wherever point C is. You could imagine
point C at the center of a circle of size 2, with point B somewhere on the
The distance between point A and point C equals 2.(1)
The distance between point B and point C equals 2.(2)


(1) The distance between A and C is
2.
(2) The distance between B and C is
2.
circle around it (and this circle could
pick up and move anywhere on the
coordinate plane).
(1) AND (2) INSUFFICIENT: Finally,
combine these two statements to
see that depending on where point
C is drawn, point B may be inside
the dotted circle, and it may not be.
The correct answer is (E).
As long as you can achieve a visual
proof of the answer, you don’t need
to prove it algebraically.
That’s what rubber band geometry
is all about: testing scenarios for
Geometry problems without the
need to plug in numbers or use
algebra. All you need is a visual
environment that can be
manipulated—one that preserves all
key constraints and freedoms in the
problem and allows you to see and
test them.


Baseline Calculations for Averages
BASIC AVERAGES
Try-It #7-4
What is the average (arithmetic mean) of 387, 388, and 389 ?
Without even calculating, you may be able to see that the average is 388. How did you
arrive at that answer?
It’s very unlikely that you calculated the average the classical way:
You probably noticed that the numbers are very close together and evenly spaced: 387 is
1 less than 388, and 389 is 1 greater than 388. Thus, the average must be 388—right in the
middle.
Whether you realize it or not, you’re using a relatively advanced technique to solve this
problem: a baseline calculation. The baseline in this case is 388—the middle number.
This concept can be applied to more diﬀicult calculations of averages, making the
calculation process much easier.
Try-It #7-5
A consumer finds that five bags of popcorn contain 257, 261, 273, 280, and 259
corn kernels per bag, respectively. What is the average (arithmetic mean)
number of corn kernels per bag of popcorn?


First, note that all of the bags have at least 257 kernels, so the average must be greater
than 257. How much greater than 257? First, consider how much each term diﬀers from
257. Represent every number with a column rising above the baseline value (in this
example, 257). The biggest numbers rise the highest; a number equal to the baseline has
no height. The height of the column thus represents the diﬀerence between the number
and the baseline value:
Calculate the sum of the diﬀerences: 0 + 4 + 16 + 23 + 2 = 45.
Divide by the number of terms: the average diﬀerence is 45 ÷ 5 = 9.
Therefore, the average number of kernels per bag equals baseline + average diﬀerence:
257 + 9 = 266.
Simply put, a baseline picture is a column chart. The columns don’t show the actual
value of any number—rather, they show the diﬀerence between the baseline and the
number.
!
The baseline can be any convenient number. Consider the following when choosing a baseline:
The smallest term in the set
The largest term in the set
The median term in the set
A round number near the range of values


For sets with apparent symmetry, choosing a baseline in the middle is a good way not
only to confirm the symmetry, but also to compute the average. In this scenario,
represent numbers lower than the baseline with columns that drop below the baseline.
As before, the size of the column represents the diﬀerence between the number and the
baseline.
Use trial and error to pick a possible average baseline, then adjust the drawing and
calculations if necessary.
Try-It #7-6
If a small business paid quarterly taxes last year of $10,079, $10,121, $10,112,
and $10,088, what was the average (arithmetic mean) quarterly tax payment
last year?
In this case, some of the numbers are below $10,100 and others are above $10,100, so
$10,100 is a natural first guess:
If the baseline is the average, then the sum of the diﬀerences from the baseline will be
zero. Since the diﬀerences from the baseline do in fact sum to zero, $10,100 is indeed the
average of this set.
WEIGHTED AVERAGES


Some sets may have many terms, but each of those terms has one of only two possible
values. Rather than add each individual term together, simplify the calculations by using
a weighted average calculation.
For weighted averages, use visualization to advance your understanding of the math.
Two real-life analogies can make it easier to remember how the relative weights of high
and low values determine where the weighted average falls.
Teeter-Totter Hanging Scale
These pictures represent the values in a set as horizontal positions—le -to-right, as on a
number line, not as vertical columns. (It may help to imagine the balance beam or lever
marked oﬀ in equal units like a number line.) Each pin or weight corresponds to the
presence of a value in a set. The X in each picture marks the equilibrium point—in other
words, the weighted average.
Both visual interpretations regard weighted averages as a kind of balancing act: the
weighted average will be closer to the end of the range that has more weight. In the
pictures, there are more instances of the le -hand side number in the set, so the X is
relatively closer to the le -hand side. You can think of this point as the point where the
weight would be “balanced” between the two sets.
Try-It #7-7
A convenience store stocks soda in 12-ounce and 24-ounce bottles. If the
average capacity of all the bottles in the store is 22 ounces, then what fraction of
the bottles in the store are 12 ounces?


Note that 22 is much closer to 24 than to 12. This implies that there will be many more
24-ounce bottles than 12-ounce bottles. Because the question asks about 12-ounce
bottles, you could strategically eliminate any answer greater than or equal to
 .
Use your understanding of weighted averages as a balancing act to work backwards from
the weighted average to the ratio of high-to-low terms:
The total range between the high and low values (24 and 12) is 12 units. Mark oﬀ the
distance from each end to the average of 22. Because the weighted average is closer to
24, that side of the teeter-totter is assigned the greater weight of 10 out of 12. The other
side, 12, is assigned the smaller weight of 2 out of 12.
Therefore, the 24-ounce bottles constitute
 of the total number of bottles. The
12-ounce bottles constitute
 of the total number of bottles.
When using this technique, it is important to remember that the weighted average is
closer to (i.e., fewer units away from) the side that has greater weight, so that side should
always be assigned the higher fraction. It is easy to reverse this logic accidentally when
solving a weighted average problem with this technique, so be very careful! Just
remember: the weighted average point will be closer to the side with more weight.
Try-It #7-8
Darla decides to mix lemonade with limeade to make a new drink called
citrusade. The lemonade is 50% water, 30% lemon juice, and 20% sugar. The
limeade is 40% water, 28% lime juice, and 32% sugar. If the citrusade is more
than 45% water and more than 24% sugar, which of the following could be the
ratio of lemonade to limeade?
3 : 1(A)


The citrusade is more than 45% water. The lemonade is 50% water and the limeade is
40% water, so the lemonade must be more heavily weighted:
Therefore, the lemonade must make up more than 50% of the mixture. Eliminate
answers (D) and (E).
Next, the citrusade is more than 24% sugar. The lemonade is 20% sugar and the limeade
is 32% sugar, so what else can you figure out about the relative weighting?
Since lemonade is more heavily weighted (the mixture is to the right of 50/50), the
weighting of the mixture is somewhere between 24% and the 50/50 weighting of the
sugar, 26%. What weighting does the 24% figure represent?
The weighting is at most
 lemonade, or
 . The weighting, then, must be between
 and
 lemonade and the rest limeade.
7 : 3(B)
3 : 2(C)
4 : 5(D)
3 : 4(E)


Answer (A) represents a weighting of
 lemonade. Answer (B) represents a weighting of
 lemonade. Only answer (C) oﬀers a weighting in the correct range:
 or 3 : 5
lemonade.
The correct answer is (C).


Number Line Techniques for Statistics
Problems
Several other common types of questions involving statistics can be solved with
visualization. Specifically, using a number line can help simplify the work for many of
these problems.
MEDIAN RELATIVE TO MEAN
Most questions involving the term median are really asking about the order of terms in
a set: you line up the terms in a set in order of size, then select the middle term. By
contrast, the average, or arithmetic mean, is the sum of all of the terms divided by the
number of terms. It can be visualized as the balancing point of all the terms laid out on
the number line, as in the discussion of the balancing point for weighted averages in
the previous section. For Data Suﬀiciency questions involving median, you generally
need to picture the placement of the unknown terms relative to the given terms in the
problem.
This technique is similar to rubber band geometry discussed earlier in this chapter,
except this technique applies to problems involving sets rather than problems involving
the coordinate plane. In this technique, you must place fixed terms in order from least
to greatest (as you would on a number line), then move variable terms around
according to the constraints. By doing so, you can visualize what impact these changes
have on the answer.
Try-It #7-9
If set S consists of the numbers n, −2, and 4, is the mean of set S greater than
the median of set S ?


The mean of set S is
 . The median depends on where n
falls relative to −2 and 4: below −2, between −2 and 4, or above 4. One approach to this
question is to think through the potential answers for all possible n values in the likely
range (you can glance at the statements and other values in the list to get a sense of the
relevant range) and draw out the scenarios on a number line. Try numbers around the
relevant numbers, including a smaller and larger one at the far ends of the range; note
that, by definition, all numbers in a set are diﬀerent, so n cannot be −2 or 4:
This requires a fair amount of up-front work, but evaluating the statements is fast as a
result. Statement (1) indicates that n > 2, which is not suﬀicient. If n = 5, for example,
the mean would equal
 and the median would equal 4. By contrast, if n = 10, the
mean would equal 4 and the median would still equal 4.
Similarly, statement (2) indicates that n < 3, which is not suﬀicient. If n = −3, for
example, the mean would equal
 and the median would equal −2. By contrast, if n
= −8, the mean would equal −2 and the median would still equal −2.
Taken together, however, any number in the range of 2 < n < 3 would feature a median
greater than the mean.
n > 2(1)
n < 3(2)


The correct answer is (C).
Notice from this problem that as you move the variable terms, the mean always
changes when the value of the variable terms change, but the median typically changes
in jumps. The median can get stuck while the number you’re changing doesn’t aﬀect
which number is in the middle.
CHANGES IN STANDARD DEVIATION
The GMAT will rarely (if ever) ask you to calculate the standard deviation of a list of
numbers. However, the exam will expect you to have some intuition about standard
deviations.
One way in which the GMAT might test your intuitive knowledge of standard deviations
is by changing numbers within a list and asking you what the impact on standard
deviation would be. The relationship is relatively straightforward:
You might also see the term variance, which is also a measure of the spread of numbers
in a set or list. Variance and standard deviation indicate the same information. A
variance of 0 indicates that all of the numbers are identical, as does a standard
deviation of 0. The larger the variance, or standard deviation, the more the numbers are
spread out. (Variance and standard deviation are never negative.)
Try-It #7-10
Last Year 9 9.5 10 10 11 11 11 11 11 12.5 13 13
This Year 9 x 10 10 11 11 11 11 11 y 13 13
Moving terms away from the mean increases the standard deviation of the list.
Moving terms toward the mean decreases the standard deviation of the list.


The monthly sales (in thousands of $) at a certain restaurant for the past two
years are given in the chart above. If the standard deviation of the monthly
sales is greater this year than last year, which of the following are possible
values for x and y ?
Except for x and y, the two lists of monthly sales numbers are identical, so focus
exclusively on those terms that changed: 9.5 and 12.5 from last year were replaced by x
and y this year. If this year’s standard deviation is greater, then this year’s numbers
must be more spread out from the mean than last year’s. The numbers are close
enough together to indicate that the average should be somewhere around 11.
Visually, here are the interesting terms from last year:
This problem does not require actual computation of the standard deviation using the
new x and y values. The math would be too complex to complete in two minutes.
Instead, determine visually which x and y values increase the standard deviation: the
pair of x and y values that are farther from the mean than are 9.5 and 12.5 will increase
the standard deviation.
9 and 12.5(A)
10 and 11(B)
10 and 12.5(C)
11 and 11(D)
11 and 12.5(E)


All of the sets have either one point or both points shi ed in toward the mean EXCEPT
(A), which has one of the points shi ed away from the mean while the other is
unchanged. Only in this case is the deviation greater.
The correct answer is (A).
The GMAT may also test you on standard deviations by adding numbers to a list. When
new terms are added, the GMAT will o en ask you to compare the old list to the new
list, or to compare various options for the new list, or to do both. You must have a
technique to evaluate the standard deviation of diﬀerent lists relative to one another.
Again, pictures make for great comparison tools!
Try-It #7-11
A list of 12 test scores has an average (arithmetic mean) of 500 and a standard
deviation of 50. Which of the following lists of additional test scores, when
combined with the original list of 12 test scores, must result in a combined list
with a standard deviation less than 50 ?
6 test scores with average of 450 and standard deviation of 50(A)
6 test scores with average of 500 and standard deviation of 25(B)
6 test scores with average of 550 and standard deviation of 25(C)
12 test scores with average of 450 and standard deviation of 25(D)
2 test scores with average of 550 and standard deviation of 50(E)


It is not generally true that all of the terms in a list are within 1 standard deviation of the
mean. However, standard deviation is a measure of the spread of the terms of a list, so
you could represent the original list of scores this way:
The oval spans ±1 standard deviation from the mean, where many of the scores will
likely be. This simplification is acceptable as long as you represent all of the other lists
the same way so that you can compare the relative eﬀects of the new test scores
systematically.
For each of the answer choices, overlay the representative ovals for the new data on top
of the oval for the original data:
(A)
(B)
(C)
(D)
(E)


Only answer (B) concentrates the list of scores closer to the original average of 500.
Thus, adding the data in answer (B) will result in a smaller standard deviation than that
found in the original data. The correct answer is (B).
In general, these are the rules for adding a single term to a list:
Note that mathematically this is a slight oversimplification, but for the purpose of
adding terms to a list of numbers on the GMAT, you can accept this simplification as
true.
FLOATING TERMS IN A SET
On GMAT Statistics problems involving elements (i.e., terms) in a list, you can usually
focus your attention on a single term or two. These terms could be considered the
floating terms—the terms that are unknown or not completely defined among a list of
more clearly defined terms.
As you approach a question of this type, try to rephrase the question quickly so that
you focus on the unknown, or floating, terms rather than on the known terms.
Try-It #7-12
List A contains 5 positive integers, and the average (arithmetic mean) of the
integers in the list is 7. If the integers 6, 7, and 8 are in List A, what is the range
of List A ?
Adding a new term more than 1 standard deviation from the mean generally
increases the standard deviation.
Adding a new term less than 1 standard deviation from the mean generally
decreases the standard deviation.
The integer 3 is in List A.(1)


The average of all five integers in the list is 7. Three of the integers in the list are given
(6, 7, and 8), and they all have an average of 7. Therefore, the floating terms in this
problem must also have an average of 7. Assign x and y to represent these terms:
The rephrased question is thus, “Given that x + y = 14, what is either x or y?” Once you
know one of the values, you can solve for the other and thereby determine the range of
the list.
(1) SUFFICIENT: If 3 is one of the unknown integers, the other must be 11. The range is
thus 11 − 3 = 8.
(2) SUFFICIENT: This statement might seem a little too vague to be suﬀicient, but by
visually listing the possible pairs that add up to 14, you can rule out pairs that don’t fit
the constraint from this statement:
Notice that the pairings represent the constraint x + y = 14. Visually, this means that x
and y are balanced around 7.
The largest term in List A is greater than 3 times and less than 4 times the
size of the smallest term.
(2)


Among these pairs:
Only one pair of integers results in a ratio strictly between 3 and 4. The unknown terms
must therefore be 3 and 11, and the range is 11 − 3 = 8.
The correct answer is (D).
In this problem, the constraint x + y = 14 is a fixed sum. Another common constraint is a
fixed diﬀerence, such as a − b = 2. A fixed diﬀerence can be represented visually as a
fixed distance between a and b on the number line, with a to the right because it is
larger. That distance could move le or right:
MAXIMIZING (OR MINIMIZING) ONE TERM
Another visual technique for statistics involves maximizing (or minimizing) the value of
a term in a set or list of numbers, subject to some constraints. Such problems will
usually employ the word maximum or minimum. For these problems, you o en should
maximize (or minimize) the term by minimizing (or maximizing) the other terms,
because the constraints usually involve mathematical trade-oﬀs.
Try-It #7-13
8 is 1.33 times the size of 6 (the ratio is too low).
9 is 1.8 times the size of 5 (the ratio is too low).
10 is 2.5 times the size of 4 (the ratio is too low).
11 is 3.66 times the size of 3 (an acceptable ratio).
12 is 6 times the size of 2 (the ratio is too high).


In a certain lottery drawing, five balls are selected from a tumbler in which
each ball is printed with a different two-digit positive integer. If the average
(arithmetic mean) of the five numbers drawn is 56 and the median is 60, what
is the greatest value that the lowest number selected could be?
The goal is to maximize the value of the lowest-numbered ball. All balls contain a two-
digit positive integer, and none of the balls have the same number. The problem
provides enough information to calculate the sum and to lay out a visual listing of the
numbers:
In order to maximize the value of the first (lowest) number in the set, what do you need
to do to the other numbers?
You’d want to minimize them. Select the smallest numbers that you can for the
remaining slots:
In some problems, you might actually make diﬀerent slots equal to each other (e.g., the
three largest numbers could be 60, 60, and 60). This problem, though, specifies that the
43(A)
48(B)
51(C)
53(D)
56(E)


integers are all diﬀerent.
The five numbers must sum to 280, so you can set up an algebraic equation and solve.
Can you think of a way to minimize the arithmetic needed to solve in that way?
The three numerical values, 60, 61, and 62, are all larger than the average of 56.
Specifically, they are +4, +5, and +6 away from that average.
The other two numbers, then, need to make up for that overage of 4 + 5 + 6 = 15:
The numbers x and x + 1 are also consecutive, so they need to be −8 and −7 away from
the average of 56. The two remaining numbers are 48 and 49. (We call this the
over/under approach, by the way.)
The correct answer is (B).
Try-It #7-14
The average (arithmetic mean) of six numbers is 18 and the median of the six
numbers is 16. What is the minimum possible value for the greatest number in
the list?
19(A)
20(B)
21(C)
22(D)
23(E)


This time, the goal is to minimize the largest number:
In order to minimize the final term, you’d want to maximize all of the other terms. This
time, though, the list contains an even number of terms, so you can’t just set the
median to the middle number. The two middle numbers average to 16.
The pair (16, 16) averages to 16, as does the pair (15, 17). The first pair, though, is better
when the goal is to minimize the final term, since the terms to the right have to be
equal to or greater than the terms to the le .
In other words, when trying to minimize the final term, it’s true that you want to
maximize the earlier terms, but you also have to think about how to do so in a way that
doesn’t make the final term too large (since it has to be larger than the earlier terms). In
this case, the first four terms are all 16:
In order to minimize the final term, set the last two terms equal to each other and solve
algebraically:
Or you can use the over/under approach:


Those last two terms have to make up for the −8 on the other side, so each x must be 4
over the average, or 22.
TAKING AND GIVING
Another common scenario involves taking value from one term in a list and giving it to
another term. The relative value of the terms in the list will change, leading to some
interesting results.
Try-It #7-15
Jake 51
Keri 63
Luke 15
Mia 38
Nora 22
The table above shows the number of points held by five players of a certain
game. If an integer number of Keri’s points were taken from her and given to
Luke, and the median score of the five players increased, how many points
were transferred from Keri to Luke?
The key to this problem is that by taking enough points from Keri and giving them to
Luke, the median of the list can change.
23(A)
24(B)
25(C)
26(D)
27(E)


Set it up visually. Order the scores from low to high on a number line, and represent the
change in Luke’s score with x:
The current median is Mia’s 38, circled in the diagram. In order for the median to
change, Luke’s score must leap-frog those of Nora and Mia, pushing Mia into the bottom
two scores and making Luke’s score the median. But be careful! You don’t want to
decrease Keri’s score so much that Luke and Mia surpass her, leaving Mia once again in
the median score position.
If 15 + x = 38, Luke would match the current median score. That is x = 23, and Keri’s new
score would be 63 − 23 = 40. However, the median score would remain 38, with both
Luke and Mia having that score. Therefore, x must be greater than 23.
Try x = 24. Luke’s new point value is 15 + 24 = 39. Keri’s new point value is 63 − 24 = 39.
Both Nora and Mia are below Luke (and Keri), so the new median is 39:
The correct answer is (B).


Problem Set
Use the Visual Solutions techniques discussed in this chapter to solve the
following problems.
1. Does a rectangular mirror have an area greater than 10 square
centimeters?
The perimeter of the mirror is 24 cm.(1)
The diagonal of the mirror is less than 11 cm.(2)
2. A number line is numbered with the integers from 0 to 50 inclusive.
An ant walks along the number line as follows: First, it walks in the
positive direction until it reaches a multiple of 5 that it hasn’t
previously reached. Then, it walks in the negative direction for at
least 1 unit, stopping when it reaches any multiple of 2. The ant
starts at 0 on the number line and repeats this process until it first
reaches the point marked 50. How many units does it travel in
total?


60(A)
65(B)
76(C)
80(D)
90(E)
3. If a, b, and c are positive, is
On the number line, a is closer to b than it is to c.(1)
b > c(2)
4. The length of one edge of a cube equals 4. What is the distance
between the center of the cube and one of its vertices?
5. A test is taken by 100 people and possible scores are the integers
between 0 and 50, inclusive. For each of the following scenarios,
determine whether the average (arithmetic mean) score would be
greater than 30 (answer Yes, No, or Uncertain).
More than 70 people scored 40 or higher.a)
75 people scored 40 or higher.b)
Fewer than 10 people scored 50.c)
No more than 2 test-takers scored any given score.d)


6.
As part of an experiment, a student repeatedly tests the
temperature of a light bulb. The bar graph below displays the
number of readings the student recorded at various temperatures,
measured in degrees Fahrenheit. What was the (arithmetic mean)
temperature reading of the light bulb?
7. In a certain dance troupe, there are 55 women and 33 men. If all of
the women are 62 inches tall and all of the men are 70 inches tall,
what is the average height of the dancers in the troupe?
8.


Anton and Lena start at the same point on a circular track,
measuring 10 meters in diameter, and begin walking
counterclockwise at the same time, with Anton walking more
quickly than Lena. When Anton has traveled exactly halfway around
the track, they both stop walking. They then observe that the
distance between them along a straight line measures exactly
 meters, as shown above. What fraction of the track has Lena
covered?
(A)
(B)
(C)
(D)
(E)
9. Eddy, Mario, and Peter have $32, $72, and $98, respectively. They
pool their money and redistribute the entire amount among
themselves. If Eddy now has the median amount of money in the
group, what is the greatest amount of money that Eddy could now
have?


$72(A)
  $85(B)
  $98(C)
$101(D)
$202(E)
10.
A square is drawn in the coordinate plane with its vertices at the
points (–2, –2), (–2, 6), (6, 6), and (6, –2), and a non-vertical line is
drawn that passes through the point (0, 4). The portion of the
coordinate plane that lies within the square, but above the line, is
then shaded as shown above. If A is the area of the shaded region in
square units, which of the following specifies all the possible values
of A ?
8 ≤ A ≤ 16(A)
8 ≤ A < 48(B)
16 ≤ A< 48(C)
8 < A ≤ 32(D)
16 < A ≤ 32(E)


Solutions
1. (C): Because the formula for the area of a rectangle is A = lw, rephrase the
question. Is lw > 10 ?
(1) INSUFFICIENT: The perimeter is 24 cm. The area of a quadrilateral is
maximized when the quadrilateral is a square, so first try l = w = 6. In this case,
the area is 36 cm2 and the answer to the question is Yes.
If, on the other hand, l = 11.5 and w = 0.5, the perimeter is still 24, but the area
is (11.5)(0.5) = 5.75 cm2. In this case, the answer to the question is No.
(2) INSUFFICIENT: The diagonal of the rectangle is less than 11. If the diagonal
is less than 11, then the sides must also be less than 11. If l = w = 6, then the
diagonal is shorter than 11 and, as last time, the area is 36. In this case, the
answer to the question is Yes.
If, on the other hand, l = 3 and w = 1, the diagonal is less than 11, but the area
is (3)(1) = 3 cm2. In this case, the answer to the question is No.
(1) AND (2) SUFFICIENT: The sides must be less than 11 and the perimeter
must be 24. The case of the square still maximizes the area: l = w = 6 and the
area is 36.
The largest possible length is just under 11, making the width just over 1. (<11)
(>1) = something larger than 10. The area must be greater than 10.


The correct answer is (C).
2. (C) 76: The problem describes an ant walking along a number line in a
complex way. It starts at 0, then walks to the right until it reaches a multiple of
5:
Then, it walks to the le until it reaches a multiple of 2. The problem specifies
that the ant must walk to the le at least one unit. Therefore, even if it is
already standing on a multiple of 2, it must walk to the next one. In this case, it
walks le until it reaches 4:
Then it walks to the right again until it reaches a new multiple of 5:
The question asks how far the ant walks, in total, to reach the number 50 for
the first time. Since this is too many units to realistically count by hand, there
must be an underlying pattern.
Write down the first few distances that the ant walks, using your number line:
5, 1, 6, 2, 7, 1, 6, 2, 7, 1 ...


There is a repeating cycle of four numbers. Starting at 5 on the number line,
the ant moves le 1, then right 6, then le 2, then right 7, which takes it to 15
on the number line. That is, the ant walks a total of 1 + 6 + 2 + 7 = 16 units to
move from 5 to 15. Then it walks another 16 units to move from 15 to 25, from
25 to 35, and from 35 to 45. In total, the ant walks 4(16) = 64 units to move
from 5 to 45.
To get from 45 to 50, the ant first walks one unit le to 44, then six units right
to 50. In total, the ant walks 64 + 1 + 6 = 71 units to move from 5 to 50.
Finally, add the five units it walks from the beginning: 71 + 5 = 76.
It is also possible to estimate the answer. As established above, traveling 10
points on the number line (from 15 to 25, from 25 to 35, and so on) requires
the ant to travel 16 units. To begin at 0 and end at 50, the ant would travel
approximately 5(16) = 80 units. This is enough to let you eliminate answers (A),
(B), and (E). By estimation alone, the answer must be either (C) or (D).
The correct answer is (C).
3. (C): The problem indicates that a, b, and c are positive and asks whether a is
greater than
 , which is the average (or arithmetic mean) of b and c.
Draw a picture and rephrase the question: “On the number line, is a
positioned to the right of the midpoint between b and c?”


(1) INSUFFICIENT: It isn’t clear whether b or c is the larger value—the point on
the right—so a could be closer to the smaller number (making a less than the
average) or the larger number (making a greater than the average).
(2) INSUFFICIENT: This statement indicates nothing about a, so it can’t be
suﬀicient.
(1) AND (2) SUFFICIENT: Together, the two statements indicate that b is the
point on the right, so a must be on the right side of the midpoint.
The correct answer is (C).
4.
 : Draw it out!
The length of any side of the cube is 4, and the problem asks for the distance
between the center of the cube and any of its vertices (corners). Chop up the
cube into eight smaller cubes to see that the distance from the center of the 4
× 4 × 4 cube to any corner is the diagonal of a 2 × 2 × 2 cube.


You can find the diagonal of a cube in a variety of ways. Probably the fastest
(besides applying a memorized formula) is to use the “Super Pythagorean”
theorem, which extends to three dimensions:
In the special case when the three sides of the box are equal, as they are in a
cube, use this equation:
Since s = 2,
 .
5. To visualize this set of weighted averages problems, imagine a teeter-totter
that is 50 meters long, marked oﬀ from 0 to 50 to represent scores. One
hundred people of equal weight sit on the teeter-totter at their respective
scores. The weighted average is the position where the teeter-totter would
balance. Thus, to answer whether the average score is greater than 30, take
extremes according to the given conditions and see whether you can swing
the balance to either side of 30 (or whether you are forced to balance on one
side of 30 only).


(A) Uncertain: More than 70 people scored 40 or higher. If all 100 scored 40 (or
higher), then the average is above 30.
If, on the other hand, 71 people scored 40 and the other 29 scored 0, does the
average drop to 30 or lower?
Match up the 29 people who scored 0 with 29 of the people who scored 40.
These 58 people together have an average score of 20. The other 42 people
have a score of 40. If you average these two groups, the average must be
below 30, since 58 is larger than 42.
(B) Uncertain: Seventy-five people scored 40 or higher. If all 100 scored 40,
then the average is higher than 30. If 75 scored 40 and the other 25 scored 0,
then what?
Match the 25 people who scored 0 with 25 people who scored 40. These 50
people have an average score of 20. The remaining 50 people have an average
score of 40, so the overall average is exactly 30. It is possible, therefore, to
have an average that is higher than 30 or an average that is not higher than 30:


(C) Uncertain: Fewer than 10 people scored 50. Say that 1 person scored 50
and the other 99 scored 0. The average is definitely below 30. If on the other
hand, 1 person scored 50 and the other 99 scored 40, then the average is
definitely above 30.
(D) No: Each score was achieved by no more than 2 people. There are 51
integers between 0 and 50, inclusive. In other words, there are 102 possible
scores to spread among the test-takers. Since there are 100 test-takers, almost
every score is taken.
The highest possible average will occur when nobody scores 0 points. If 2 test-
takers score 1, 2 test-takers score 2, and so on up to 50, then the average score
will be approximately 25. As this is the highest allowable average for this
scenario, it’s impossible for the average to be greater than 30.
6. 112°: Eyeball the graph. The average looks like it’s in the 112° range. Calculate
the over/under with 112° as the assumed baseline.
The four 110° readings are each 2° below the baseline, and the three 111°
readings are each 1° below the baseline, for a total of 11° below baseline.
On the other side, the 113°, 114°, and 115° readings are a total of 11° above the
baseline. They balance perfectly! The average is exactly 112°.
7. 65 inches: Algebraic solution:


Visual solution:
8. (B)
 : Draw a diagram showing the track, and label all of the given
information. Because Anton has walked exactly halfway around the track, the
distance between Anton and the starting point is the diameter.
The answer choices are fractions that are fairly far apart from each other,
suggesting that visual estimation could be eﬀective. If you use this approach,
carefully draw your diagram to scale.  Since
  is approximately 5(1.7) =
8.5, the line between Anton and Lena is only slightly shorter than the diameter
of the circle. Therefore, Lena is closer to the starting point than she is to
Anton, so Lena has traveled less than one-quarter of the circle: eliminate
answers (D) and (E).
At the other extreme, choice (A) is quite small, suggesting that the distance
between Anton and Lena would be closer to 10 than it actually is. The only
reasonable answer choices are (B) and (C).


To calculate the exact answer, observe that Anton, Lena, and the starting line
form the three vertices of a triangle. Because one side of the triangle is the
diameter of the circle, the triangle is a right triangle with hypotenuse 10. The
known leg is equal to
 .
You can either use the Pythagorean theorem to calculate the missing leg (the
straight line between Lena and the starting point) or you can note that the
known leg and the hypotenuse have a ratio of
 , so the triangle
must be a 30–60–90 triangle and the remaining side must equal 5.
The arc of the circle that Lena traveled has a corresponding inscribed angle of
30 degrees (the vertex at point A). The corresponding central angle is twice the
inscribed angle, or 60 degrees.
Alternatively, the distance between Lena and the starting line is equal to the
radius of the circle. Therefore, the triangle with vertices at Lena, the starting
line, and the center of the circle is an equilateral triangle, and all of its angles
equal 60 degrees. So Lena has covered 60 degrees out of the 360 degrees in a
full circle, or a fraction of
 .
The correct answer is (B).


9. (D) $101: The pool of money is $32 + $72 + $98 = $202. A er the redistribution,
each person will have an amount between $0 and $202, inclusive. Call the
amounts L, M, and H (low, median, high). To maximize M (Eddy’s share),
minimize L and H.
The correct answer is (D).
Minimum L = $0
Minimum H = M
Maximum M = Total pool of money − Minimum L − Minimum H
M = $202 − $0 − M
2M = $202
M = $101
10. (B) 8 ≤ A < 48: On your paper, draw the square in the coordinate plane. The line
passes through the point (0, 4), so label that point as well. Since the slope of
the line could vary, sketch several possible lines that pass through the point,
being certain to test extreme cases.
If the slope is very positive, then the line is nearly vertical, and the area above
the line will be very close to 2(8) = 16:


As the slope decreases, the area above the line will decrease. If the slope is 1,
the area above the line is the area of a triangle with base 4 and height 4, which
is equal to
 (4)(4) = 8.
Then, as the slope continues to decrease, the area above the line will once
again increase steadily. The greatest area is found when the slope is very
negative, since the area will be slightly less than (6)(8) = 48.


The area of the shaded region can be a minimum of exactly 8 and can be a
maximum of almost, but not equal to, 48 (since the line cannot be vertical).
The correct answer is (B).


CHAPTER 8
Hybrid Problems


In This Chapter...
Pop Quiz!
Hybrid Problems
Identify and Sequence the Parts
Where to Start
Minor Hybrids


Chapter 8
Hybrid Problems
POP QUIZ!
Try these four hybrid problems.
Try-It #8-1
Set A consists of four consecutive positive integers. Set B is
constructed as follows: each integer in set A is randomly either
increased by 10% or decreased by 10%, and the four resulting
values compose set B. What is the range of set B ?
Try-It #8-2
Set S contains 100 consecutive integers. If the range of the negative
elements of set S equals 80, what is the average (arithmetic mean)
of the positive numbers in the set?
The smallest integer in set A was increased by 10% when set B
was constructed.
(1)
The greatest integer in set A was increased by 10% when set B
was constructed.
(2)


Try-It #8-3
If a and b are consecutive positive integers, is ab divisible by 30 ?
Try-It #8-4
A carnival card game gives the contestant a one in three probability
of successfully choosing the correct card and thereby winning the
game. If a contestant plays the game repeatedly, what is the
minimum number of times that he must play the game so that the
probability that he never loses is less than 1% ?
a2 is divisible by 25.(1)
63 is a factor of b2.(2)


Hybrid Problems
Hybrid problems blend topics together. They contain two or more
qualitatively diﬀerent kinds of obstacles that you must surmount on the
way to the answer.
Some hybrid problems feature content areas that are fairly closely related.
Other hybrid problems feature content areas that share little in common;
these problems must be solved in separate steps.
The diﬀiculty of a hybrid problem is related to the following questions:
Problems with minor additional content areas are generally easier to solve
than hybrids that blend topics together in an unusual, fundamental, and
How closely related are the subjects being tested? Are the content
areas covered in the same section of the Manhattan Prep
All the Quant guide? The more closely related the subjects, the
easier it will be to navigate the problem.
How important is each of the subject areas? Are each of the subject
areas fundamental to the problem or is one of them just a low-level
disguise that can be quickly disposed of? The more important each
topic is in solving the problem, the more diﬀicult the problem will
usually be.


clever way. The best hybrids are one of a kind. You will have to bring your A
game to solve them.
But solve them you can. That’s why you’re reading this chapter!


Identify and Sequence the Parts
When you encounter a hybrid problem, first pick out all of the topics
tested. Take a look at the first problem from the pop quiz:
Try-It #8-1
Set A consists of four consecutive positive integers. Set B is
constructed as follows: each integer in set A is randomly either
increased by 10% or decreased by 10%, and the four resulting
values compose set B. What is the range of set B ?
This problem refers to consecutive integers, but it involves percents and
also asks for the range of a set of numbers, which is a statistics concept.
Which aspect of the problem should you tackle first?
In this problem, a set begins as four consecutive integers and is then
changed. So you might start your approach by jotting down some sets of
consecutive integers—either as numbers or in terms of a variable. Then,
The greatest integer in set A was increased by 10% when set B
was constructed.
(1)
The smallest integer in set A was increased by 10% when set B
was constructed.
(2)


apply the percent change described in the problem and, finally, check the
range of the resulting set.
In general, start with wherever you feel the logical beginning is or
whichever part you feel is an easier, cleaner starting point.


Where to Start
STARTING AT THE BEGINNING
As you contemplate the logical order of steps, you might feel less confident
with the second stage than with the first. If so, go ahead and start at the
beginning. Just articulate very clearly, “What intermediate result will I get
once I’m finished with the first part of the problem?”
Begin by defining set A. Set A consists of four consecutive integers, such as
1, 2, 3, 4, or 10, 11, 12, 13.
Next, choose a statement to work with first. Statement (1) says that the
smallest integer in set A is increased by 10%. The other integers might have
been either increased or decreased. Using one of your cases, calculate
what might happen to the set:
Set A Scenarios Set B
1, 2, 3, 4 All values increased by 10% 1.1, 2.2, 3.3, 4.4
1, 2, 3, 4 Smallest value increased by 10%, other values decreased by 10% 1.1, 1.8, 2.7, 3.6
Finally, calculate the range of the resulting set. The range of a set is the
diﬀerence between the greatest number and the smallest number in the
set.


Set A Scenarios Set B Range
1, 2,
3, 4
All values increased by 10% 1.1, 2.2, 3.3,
4.4
4.4—1.1 =
3.3
1, 2,
3, 4
Smallest value increased by 10%, other values
decreased by 10%
1.1, 1.8, 2.7,
3.6
3.6—1.1 =
2.7
Since diﬀerent ranges are possible, the statement is insuﬀicient. Eliminate
answers (A) and (D).
STARTING AT THE END
You might decide that it is easier to start at the end and work backwards.
That’s fine. Just ask yourself, “What information do I need to have as a last
step before arriving at a solution to the question?”
In this problem, you might first consider what would cause set B to have a
diﬀerent range. The range of a set is the diﬀerence between the greatest
and smallest numbers in the set. If the diﬀerence between these two
numbers can vary, the range can vary as well.
According to statement (2), the greatest integer in set A was increased by
10%. However, the smallest integer might have increased by 10%, bringing
it closer to the greatest integer, or it might have decreased by 10%,
bringing it farther from the greatest integer. Therefore, the range of the set
can vary, so statement (2) is insuﬀicient. Eliminate answer (B).


Now, put the two statements together. One option is to try a few diﬀerent
cases, making sure to increase the smallest integer in set A and decrease
the greatest integer in set A. One of the two cases used for statement (1)
works here as well:
Set A Scenario Set B Range
1, 2, 3, 4 All values increased by 10% 1.1, 2.2, 3.3, 4.4 4.4—1.1 = 3.3
Try another case, using diﬀerent numbers:
Set A Scenario Set B Range
10, 11, 12, 13 All values increased by 10% 11, 12.1, 13.2, 14.3 14.3—11 = 3.3
The range is the same, but this doesn’t mean that the statements are
suﬀicient together. To see why, use a diﬀerent approach. Set A always
contains the numbers x, x + 1, x + 2, and x + 3. When the smallest value in
set A is increased by 10%, it goes from x to 1.1x. When the greatest value is
increased by 10%, it goes from x + 3 to 1.1x + 3.3. So these two values diﬀer
by 3.3.
However, you can’t assume that the smallest and greatest values in set
A always become the smallest and greatest values in set B. If two of the
values switch places, the range might change. Here’s an example:
Set A Scenario Set B Range


Set A Scenario Set B Range
10, 11,
12, 13
Smallest and largest values increased by 10%, other
values decreased by 10%
11, 9.9,
10.8, 14.3
14.3—9.9
= 4.4
The range of the set might vary, even when both statements are used. The
correct answer is (E).
Let’s look at another example:
Try-It #8-2
Set S contains 100 consecutive integers. If the range of the negative
elements of set S equals 80, what is the average (arithmetic mean)
of the positive numbers in the set?
In this problem, information is given about the range of numbers in a set.
Thus, knowing how to work with statistics techniques will be important in
solving the problem.
Solving this problem also requires using consecutive integers techniques—
namely, counting consecutive integers and computing their average. These
two topics are closely related but still cover diﬀerent ideas.
If you start with the statistics piece of the problem, then you’ll be able to
find the highest and lowest negative numbers in the set of consecutive
integers. This will act as an input to the formulas for computing the largest
and smallest positive integers in the set, and subsequently the average of
the positive integers in the set.


First, determine what set of consecutive negative integers will result in a
range of 80. Range is defined as the diﬀerence between the highest and
lowest numbers in a set:
The high number among the negative terms is the largest negative integer,
−1:
Therefore, the lowest number in set S is −81. Use this result to jump to the
consecutive integers portion of the solution. In this case, −81 is the
smallest or “first” element in the set:
Therefore, the highest or “last” number in set S is 18.
Finally, calculate the average of the positive terms in the set using 1 as the
smallest positive integer (“first pos”) and 18 as the largest  positive integer
(“last pos”):


The average of the positive numbers in the set is 9.5.
If instead you decided to start with the second step, you would write the
formula for the average of consecutive integers first, focusing on the
positive integers:
The two unknowns you need to solve for are the greatest positive
integer and the least positive integer because the question asks only about
the positive integers in the set.
Some of the integers in set S are negative and some are positive, so clearly
the least positive integer will be 1. Therefore, you only need to figure out
what the greatest integer in the set will be. This is the information needed
in the last step of solving the problem.
In order to calculate this number, you would now need to apply the
definition of range:
The greatest negative integer (the “high” number) is −1:


Therefore, the least number in the set is −81; plug this into the formula for
counting consecutive integers:
Therefore, the highest number in the set is 18.
Finally, plug 18 into the average formula:
Whether you start solving this problem from the beginning or from the end,
study how the steps hook together: the output of one step becomes the
input to another. These are the “turns” you have to make in solving any
hybrid problem. O en, a number that plays one role in a particular formula
or calculation plays a completely diﬀerent role in the next step.


Minor Hybrids
In minor hybrid problems, one of the following conditions applies:
These problems can be easier to solve than major hybrids. However, you
will still benefit greatly from paying close attention to the turns as you
move through the stages of the solution.
Try-It #8-3
If a and b are consecutive positive integers, is ab divisible by 30 ?
This problem tests your skill with both Divisibility & Primes and
Consecutive Integers. Consecutive integer concepts o en lend themselves
well to questions about divisibility.
The content areas in the problem are closely related. For instance,
they are covered in the same section of the Manhattan Prep
All the Quant guide.
1.
One of the content areas is a low-level disguise or some other minor
feature.
2.
a2 is divisible by 25.(1)
63 is a factor of b2.(2)


To be divisible by 30, a number must have 2, 3, and 5 as prime factors.
Thus, the question becomes: “Does ab have 2, 3, and 5 as prime factors?”
Next, use a concept from consecutive integers: a and b are consecutive
positive integers. Thus, either a or b is an even number, which means that
the product ab is automatically divisible by 2. The question can be further
simplified: “Does the product ab have 3 and 5 as prime factors?”
(1) INSUFFICIENT: Statement (1) indicates that 25 is a factor of a2, meaning
that 5 and 5 are prime factors of a2. Knowing that 5 is a prime factor of a
indicates that 5 is a factor of ab, but does not indicate whether 3 is a factor.
(2) INSUFFICIENT: Statement (2) indicates that 63 is a factor of b2, which
means that 3, 3, and 7 are prime factors of b2 (3 × 3 × 7 = 63). Knowing that
3 is a prime factor of b indicates that 3 is a factor of ab, but does not
indicate whether 5 is a factor.
(1) AND (2) SUFFICIENT: According to the combined statements, a is
divisible by 5 and b is divisible by 3. This is suﬀicient information to answer
the rephrased question.
The correct answer is (C).
Notice that the solution never indicated exactly what consecutive integers
a and b are—this isn’t necessary. Also notice a potential trap in this
problem—assuming that a < b because of the way the question is phrased.
This assumption would not be fatal in this case, but if you realize that a and
b can come in either order, then it’s easier to create an accurate list of
diﬀerent scenarios for a and b.


Try-It #8-4
A carnival card game gives the contestant a one in three probability
of successfully choosing the correct card and thereby winning the
game. If a contestant plays the game repeatedly, what is the
minimum number of times that he must play the game so that the
probability that he never loses is less than 1% ?
This problem primarily tests Probability theory. In addition, Exponents are
needed to represent the impact of playing multiple games on the
probability of the outcomes. The probabilities are given in fractions, yet the
question is asked in terms of percents, so Fraction, Decimal, and Percent
(FDP) connections are relevant. Finally, the question is phrased in terms of
an inequality, so Inequalities come into play.
“The probability that he never loses” can be rephrased as “the probability
that he always wins.” This probability can be expressed as
 , where
 is the chance of winning on a single play and n is the number of times
the contestant plays.
Track scenarios in a chart:
Number of Plays P(All Wins) Approx. Equivalent
1
 0.33 = 33%


Number of Plays P(All Wins) Approx. Equivalent
2
 0.11 = 11%
3
 = 0.04 = 4%
4
 = 1.25%
5
  = 0.4% < 1%
You might also have noticed that only the denominator in the probabilities
mattered, as the numerator was always 1. To have a probability of less than
1%, the fractional probability must be “1 over something greater than 100.”
In order for 3n >100, n must be at least 5.
The correct answer is 5.


Problem Set
Solve the following problems and identify the topics being tested.
1. The average (arithmetic mean) of a list of six numbers is equal to 0.
What is the positive diﬀerence between the number of positive
numbers in the list and the number of negative numbers in the list?
Each of the positive numbers in the list equals 10.(1)
Each of the negative numbers in the list equals −5.(2)
2. Simplify:
3. If a, b, and c are positive, is a > b ?
(1)
b + c < a(2)
4. If c is randomly chosen from the integers 20 to 99, inclusive, what is
the probability that c3 − c is divisible by 12 ?


5. If x and y are positive integers greater than 1 such that x − y and
are both even integers, which of the following numbers must be
non-prime integers?
xI.
x + yII.
III.
I only(A)
II only(B)
III only(C)
I and II only(D)
I, II, and III(E)
6. A number cube with faces numbered 1 through 6 has an equal
chance of landing on any face when rolled. If the number cube is
rolled twice, what is the probability that the sum of the two rolls is
a prime number?
7. The function {x} is defined as the area of a square with diagonal of
length x. If x > 0 and {x2} = x2, what is the value of x  ?


1(A)
(B)
(C)
2(D)
4(E)
8. A circular microchip with a radius of 2.5 centimeters is
manufactured by following a circular diagram. The scale of the
diagram is such that a measurement of 1 centimeter on the
diagram corresponds to a measurement of 0.5 millimeters on the
microchip. What is the area of the diagram, in square centimeters?
(1 centimeter = 10 millimeters)
9. Three consecutive integers are selected from the integers 1 to 50,
inclusive. What is the sum of the remainders that result when each
of the three integers is divided by x  ?
When the greatest of the consecutive integers is divided by
x, the remainder is 0.
(1)
When the least of the consecutive integers is divided by
x, the remainder is 1.
(2)


10. If x, y, and z are all distinct positive integers, and the percent
increase from x to y is equal to the percent increase from y to z,
what is x  ?
y is prime.(1)
z = 9(2)


Solutions
1. (E): Since the average of the six numbers in the list is 0, the sum of the
six numbers is 0. There could be positive numbers and negative
numbers in the set. Zero is not mentioned, but this does not rule it out.
In order for the sum of the numbers in the set to be 0, either all the
terms are 0 or there are some positives and some negatives.
(1) INSUFFICIENT: Statement (1) indicates that the list contains at least
one positive number and that each positive term is 10. The list could be
{−2, −2, −2, −2, −2, 10}, and the positive diﬀerence between the number
of positive terms and the number of negative terms would be 4.
Alternatively, the list could be {−20, −20, 10, 10, 10, 10}, and the positive
diﬀerence would be 2.
(2) INSUFFICIENT: Statement (2) indicates that the list contains at least
one negative number and that each negative term is −5. The set could be
{−5, 1, 1, 1, 1, 1}, and the positive diﬀerence between the number of
positive terms and the number of negative terms would be 4. The set
could be {−5, −5, −5, 5, 5, 5}, and the positive diﬀerence would be 0.
(1) AND (2) INSUFFICIENT: The statements together suggest that the set
has twice as many −5 terms as 10 terms, in order to maintain a sum of 0.
If every term is negative or positive, then the set would have to be {−5,
−5, −5, −5, 10, 10}, and the positive diﬀerence would be 2. However, zero


terms are possible, so the set could be {−5, −5, 0, 0, 0, 10}, and the
positive diﬀerence would be 1.
2. 30: You could simplify the numerator arithmetically (multiply out the
terms and then add). Alternatively, factor a 22 out of the numerator.
Then, distribute the denominator (which becomes the diﬀerence of
squares).
3. (D): Statement (2) appears to be a bit easier to work with so begin there.
(2) SUFFICIENT: If a, b, and c are all positive, then a > b + positive.
Therefore, a must be greater than b. You could also prove this fact by
Testing Cases.
Case b + c < a Is this case possible according to (2)?
a > b b + positive < number greater than b Possible
a = b b + c < b Impossible, since c is positive
a < b b + c < number less than b Impossible, since c is positive


Only the a > b case is possible, so the answer is a definite Yes.
(1) SUFFICIENT: Statement (1) can be cross-multiplied without flipping
the inequality sign, since the denominators are positive.
a(a + c) > b(b + c)
 a2 + ac > b2 + bc
 a2 + ac − b2 − bc > 0
 a2 − b2 + ac − bc > 0 Group similar terms to simplify.
 [a2 − b2] + [ac − bc] > 0 Note the Quadratic Template.
 (a − b)(a + b) + c(a − b) > 0
 (a − b)[(a + b) + c] > 0 Factor out a − b.
Therefore, a + b + c is positive, because all three additive terms are
positive. So (a − b)(positive) > 0. By Number Properties sign rules, (a − b)
must also be positive in order for the product to be greater than 0. Thus,
a > b.
This algebra is very tough; it is hard to see where to begin or what series
of manipulations will be productive. If you did not see this, you could try
Testing Cases to see which are allowed by statement (1). Note: LT = less
than and GT = greater than. GTb means “a number greater than b.”
Case
 Is this case possible according to (1)?
a > b
 Possible. The le is greater than the right.


Case
 Is this case possible according to (1)?
a = b
 Impossible; the two expressions are equal.
a < b
 Impossible; the le side is actually less than the right
side, not greater than as (1) requires.
The correct answer is (D).
4.
 : The words divisible and probability are used, so this question is
about Divisibility & Primes and Probability.
Probability is
 . There are 99 − 20 + 1 = 80
possible values for c, so the unknown is how many of these c values
yield a c3 − c that is divisible by 12.
The prime factorization of 12 is 2 × 2 × 3. There are at least two ways to
think about this: numbers are divisible by 12 if they are divisible by 3
and by 2 twice, or if they are multiples of both 4 and 3.
The expression involving c can be factored.
These are consecutive integers. It may help to put them in increasing
order: (c − 1)c(c + 1). Thus, this question has a lot to do with Consecutive
Integers, and not only because the integers 20 to 99 themselves are
consecutive.


In any set of three consecutive integers, a multiple of 3 will be included.
Thus, (c − 1)c(c + 1) is always divisible by 3 for any integer c. This takes
care of part of the 12. So the question becomes: “How many of the
possible (c − 1)c(c + 1) values are divisible by 4?” Since the prime factors
of 4 are 2’s, it makes sense to think in terms of Odds and Evens.
For example,  (c − 1)c(c + 1) could be (E)(O)(E), which is definitely
divisible by 4, because the two evens would each provide at least one
separate factor of 2. Thus, c3 − c is divisible by 12 whenever c is odd,
which are the cases c = 21, 23, 25 … 95, 97, 99. That’s
 possibilities.
Alternatively, (c − 1)c(c + 1) could be (O)(E)(O), which will only be
divisible by 4 when the even term itself is a multiple of 4. Thus, c3 − c is
also divisible by 12 whenever c is a multiple of 4, which are the cases c =
20, 24, 28, . . . , 92, 96. That’s
possibilities.
The probability is thus
5. (D) I and II only: x cannot equal y, as that would make
 ≠ even.
So either x > y or y > x.
x and y are both positive and
 is an integer, so x > y.
Since x − y is even, either x and y are both even or they are both odd.
Since
 = an even integer, x = y × (even integer).


Therefore, x is an even integer, as is y.
The correct answer is (D).
TRUE. x and y are both positive even integers and x > y. No even
number greater than 2 is prime, so x can’t be prime.
I.
TRUE. x and y are each positive even integers and x > y. Thus, x +
y is even and the least possible value of x + y = 4 + 2 = 6. All even
numbers greater than or equal to 6 are non-prime.
II.
FALSE. It could be that x = 4 and y = 2, so
 , which is not
prime, but is also not an integer. In fact, if
 an even integer,
then
III.
6.
 : First, think about the prime numbers less than 12, the maximum
sum of the numbers. These primes are 2, 3, 5, 7, 11.
The probability of rolling 2, 3, 5, 7, or 11 is equal to the number of ways
to roll any of these sums divided by the total number of possible
outcomes. The total number of possible outcomes is 6 × 6 = 36.
Sum of 2 can happen 1 way: 1 + 1
Sum of 3 can happen 2 ways: 1 + 2, 2 + 1
Sum of 5 can happen 4 ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1
Sum of 7 can happen 6 ways: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1


Sum of 11 can happen 2 ways: 5 + 6, 6 + 5
That’s a total of 1 + 2 + 4 + 6 + 2 = 15 ways to roll a prime sum.
Thus, the probability is
 .
7. (B)
 : The problem defines the function in words; you’ll need to
translate into math.
“The function {x} is defined as the area of a square with diagonal of
length x.”
{x} = s2, where s is the side of the square.
If the diagonal equals x, then the side of the square equals
 .
The question stem also indicates that {x2} = x2. In other words, applying
the defined function to x2 will result in an answer of x2. First, find {x2}.
Therefore:


Now, set that equal to x2 and solve.
(Note: It is acceptable to divide by x2 because the question stem
indicates that x ≠ 0.) Therefore, x is equal to
  or
 . Since the
question stem indicates that x > 0, only
 is a valid solution.
Alternatively, you could work backwards from the answers. Start with
(B) or (D).
(B) If
 , then x2 = 2. Next, the question specifies that that
 becomes {2}, so what does this function
return?
“The function {x} is defined as the area of a square with diagonal
of length x.”
The function {2} is defined as the area of a square with diagonal length
2. The side length of this square is
 , so the area of this square is
. Therefore, {2} = 2, which matches the question
stem specification that {x2} = x2.


The correct answer is (B).
8. 2,500π: Microchip radius = (2.5 cm)(10 mm/cm) = 25 mm
9. (C): When consecutive integers are divided by the same number, the
remainders follow a repeating pattern. For instance, when consecutive
integers are divided by 4, the remainders form a repeating pattern of [1,
2, 3, 0], with every fourth integer being divisible by 4.
(1) INSUFFICIENT: This statement indicates that the greatest of the three
integers is divisible by x. However, the sum of the remainders depends
on the value of x. For instance, if x = 10, the three remainders would be 8,
9, and 0, respectively, and their sum would be 17. If x = 5, the three
remainders would be 3, 4, and 0, with a sum of 7.


(2) INSUFFICIENT: Statement (2) indicates that the least of the three
integers is one greater than a multiple of x. However, the sum of the
remainders again depends on the value of x. If x = 10, the three
remainders would be 1, 2, and 3, with a sum of 6. If x = 2, the three
remainders would be 1, 0, and 1, with a sum of 2.
(1) AND (2) SUFFICIENT: Call the three consecutive integers y, y + 1, and
y + 2. The remainder when y is divided by x is 1 and the remainder when
y + 2 is divided by x is 0. Write out the known remainders and write out
enough other terms to see where the pattern repeats. In this case, the
term y – 1 also has a remainder of 0, so the pattern repeats every three
terms.
y − 1 y y + 1 y + 2
0 1 ? 0
The only way for the remainders to have a consistent repeating pattern
is for the missing remainder to equal 2 (in which case x must equal 3).
Therefore, the sum of the three remainders is 1 + 2 + 0 = 3.
10. (A): The question asks for the value of x. Translate the equation given in
the question stem.


The problem stem specifies that x is an integer, so the right-hand side of
the equation must also be an integer. Therefore, z must be a factor of y2.
The question asks for the value of x.
(1) SUFFICIENT: If y is prime, then there are two possible scenarios.
Case 1: y = z. This isn’t allowed, though, because the question stem
indicates that the variables represent three diﬀerent positive integers.
Case 2: y2 = z, in which case x must equal 1. This is the only possible
case, so statement (1) is suﬀicient.
(2) INSUFFICIENT: If z = 9, then y2 is a multiple of 9, so z must be a
multiple of 3. It’s possible that z = 9, y = 3, and x = 1. It’s also possible
that z = 9, y = 36, and x = 4.
The correct answer is (A).


PART THREE
Practice


CHAPTER 9
Workout Sets


In This Chapter...
Workout Set 1
Workout Set 2
Workout Set 3
Workout Set 4
Workout Set 5
Workout Set 6
Workout Set 7
Workout Set 8
Workout Set 9
Workout Set 10
Workout Set 11
Workout Set 12
Workout Set 13
Workout Set 14
Workout Set 15
Workout Set 16


CHAPTER 9 Workout Sets
Workout Set 1
1.
Figure not drawn to scale.
The circle with center O has a circumference of
 . If AC is a
diameter of the circle, what is the length of BC ?


(A)
3(B)
(C)
9(D)
(E)
2. A batch of widgets costs p + 15 dollars for a company to produce
and each batch sells for p(9 − p) dollars. For which of the following
values of p, does the company make a profit?
3(A)
4(B)
5(C)
6(D)
7(E)
3. If K is the sum of the reciprocals of the consecutive integers from 41
to 60 inclusive, which of the following is less than K ?
I.


II.
III.
None(A)
I only(B)
II only(C)
I and II only(D)
I, II, and III(E)
4. Triplets Adam, Bruce, and Charlie enter a triathlon. There are nine
competitors in the triathlon. If every competitor has an equal
chance of winning, and three medals will be awarded, what is the
probability that at least two of the triplets will win a medal?


(A)
(B)
(C)
(D)
(E)
5. Floyd started at one end of a trail at 8:07 a.m. and hiked the entire 5
miles to the other end of the trail at an average rate of 3 miles per
hour. Connie started at the same end of the trail at 8:30 a.m. and
hiked the entire length of the trail at an average rate of 2.5 miles per
hour. Finally, Karen started her hike in the same place at 10:00 a.m.
and hiked the entire length of the trail at an average rate of 4 miles
per hour. During approximately what percent of the 4-hour period
from 8:00 a.m. to noon were at least two of the three hikers on the
trail?


34%(A)
45%(B)
55%(C)
59%(D)
68%(E)
6. Half an hour a er Car A started traveling from Newtown to
Oldtown, a distance of 62 miles, Car B started traveling along the
same road from Oldtown to Newtown. The cars met each other on
the road 15 minutes a er Car B started its trip. If Car A traveled at a
constant rate that was 8 miles per hour greater than Car B’s
constant rate, how many miles had Car B driven when they met?
14(A)
12(B)
10(C)
  9(D)
  8(E)
7. If x = 2b − (88 + 86), for which of the following b values is x closest to
zero?
20(A)
24(B)
25(C)
30(D)
42(E)


8. If k > 1, which of the following must be equal to
 ?
2(A)
(B)
(C)
(D)
(E)
9. If y is 20% less than 90% of x and z is 25% more than 130% of y,
then z is what percent of x ?
  72%(A)
  92.5%(B)
108.5%(C)
117%(D)
135%(E)
10. Set A consists of three consecutive positive multiples of 3, and set B
consists of five consecutive positive multiples of 5. If the sum of the


integers in set A is equal to the sum of the integers in set B, what is
the least number that could be a member of set A ?
69(A)
72(B)
75(C)
78(D)
81(E)


Workout Set 1 Answer Key
1. D
2. B
3. D
4. B
5. B
6. A
7. B
8. E
9. D
10. B


Workout Set 1 Solutions
1. (D) 9: If AC is a diameter of the circle, then triangle ABC is a right triangle,
with angle ABC = 90 degrees. The shortest side of a triangle is across
from its least angle and the longest side of a triangle is across from its
greatest angle. Therefore, AC > BC > AB.
The circumference of the circle =
 , so
. Thus, AC ≈ 10.2 and BC < 10.2.
Answer (E) is too great and answer (D), while less than 10.2, is
questionable, because side BC is opposite a 60-degree angle. The value
should be less than 9.
Since angle ACB is 30 degrees, angle CAB is 60 degrees.
The sides in a 30–60–90 triangle have the ratio
 , so use
the ratio to compute the desired side, BC.
Find the length of BC.


Therefore, BC has length 9.
The correct answer is (D).
2. (B) 4: You can work backwards from the answers. Start with (B) or (D).
Eliminate (D) and try (B) next.
The company makes a profit when p = 4.
Alternatively, profit equals revenue minus cost. The company’s profit is:
(D) p = 6
Cost: 6 + 15 = 21
Revenue: 6(9 − 6) = 18
Profit: 18 − 21 = −3              Loss, not profit!
(B) p = 4
Cost: 4 + 15 = 19
Revenue: 4(9 − 4) = 20
Profit: 20 − 19 = 1


Profit will be zero if p = 5 or p = 3, which eliminates answers (A) and (C).
For p > 5, both (p − 5) and (p − 3) are positive. In that case, the profit is
negative (i.e., the company loses money). The profit is only positive if (p
− 5) and (p − 3) have opposite signs, which occurs when 3 < p < 5.
The correct answer is (B).
3. (D) I and II only: The sum
 has 20
fractional terms. It is impossible to compute this by hand in two
minutes. Instead, look at the maximum and minimum possible values
for the sum.
Maximum: The greatest fraction in the sum is
  . K is definitely less
than 20 ×
 , or
  , which is less than
  .
Minimum: The least fraction in the sum is
  . K is definitely greater
than 20 ×
 =
  .
Therefore,
 < K <
  .
 <
I.

II.


The correct answer is (D).

III.
4. (B)
 : With nine competitors and only three medals awarded, only
of the competitors will win overall. Although a simplification, it is
reasonable for each competitor to see his or her chance of winning a
medal as
 , or to expect to win
 of a medal (pretending for a
moment that medals can be shared).
The question asks for the probability that at least two of the triplets will
win a medal. In other words, you want
 to
 of the triplets to win
medals or for each triplet to win
 to
 of a medal. Since
 and
are both greater than
 , you are looking for the probability that the
triplets will win medals at a rate greater than that expected for
competitors overall. In other words, this would be an unusual outcome.
Thus, the probability should be less than
 . Eliminate (D) and (E). You
could then guess from among the remaining answers with a 1 in 3
chance of guessing correctly.
To solve, use the probability formula and combinatorics.