First, find the total number of outcomes for the triathlon. There are nine
competitors; three will win medals and six will not. Set up an anagram
grid where Y represents a medal, N no medal.
Competitor C1 C2 C3 C4 C5 C6 C7 C8 C9
Medal Y Y Y N N N N N N
Therefore, the number of ways three medals can be awarded is
.
Now, determine the number of instances when at least two brothers win
a medal. Practically speaking, this could happen when 1) exactly three
brothers win or 2) exactly two brothers win.
Start with all three triplets winning medals, where Y represents a medal.
Triplet A B C Non-triplet C1 C2 C3 C4 C5 C6
Medal Y Y Y Medal N N N N N N


The number of ways this could happen is
 . This makes
sense, as there is only one instance in which all three triplets would win
medals and all of the other competitors would not. (If you recognize this
immediately, no need to write out the math.)
Next, calculate the instances when exactly two of the triplets win
medals.
Triplet A B C Non-triplet C1 C2 C3 C4 C5 C6
Medal Y Y N Medal Y N N N N N
Since both triplets and non-triplets win medals in this scenario, consider
the possibilities for both sides of the grid. For the triplets, the number of
ways that two could win medals is
 .
For the non-triplet competitors, the number of ways that one could win
the remaining medal is
 .
Multiply these two numbers to get the total number of instances: 3 × 6 =
18.
The brothers win at least two medals in 18 + 1 = 19 cases. The total
number of cases is 84, so the probability is
 .


The correct answer is (B).
5. (B) 45%: To answer the question, you’ll need to identify the periods of
time during which diﬀerent numbers of hikers are on the trail. To
organize your work, start by drawing out a diagram showing the four
hours between 8:00 a.m. and noon.
Next, identify the time period during which each hiker was on the trail.
Floyd traveled the 5 miles at a rate of 3 miles per hour (mph), so he
spent
 = 1 hour and 40 minutes on the trail. Therefore, Floyd
was on the trail from 8:07 to 9:47.
Connie traveled the 5 miles at a rate of 2.5 mph, so she spent
 = 2 hours on the trail. Connie was on the trail from 8:30 to
10:30.
Finally, Karen traveled the 5 miles at a rate of 4 mph, so she spent
 = 1 hour 15 minutes on the trail. Karen was on the trail from
10:00 to 11:15.
Plot each of these time periods on your drawing.


There were no periods of time during which all three hikers were on the
trail. However, there were two periods during which exactly two of the
three hikers were on the trail. Floyd and Connie were both on the trail
from 8:30 to 9:47, and Connie and Karen were both on the trail from
10:00 to 10:30. This represents a total of 1 hour and 47 minutes, or
approximately 1.75 hours.
1.75 hours is more than a third of 4 hours, but less than half of 4 hours.
Therefore, the only reasonable answer is (B). It is also possible to
simplify the fraction more precisely.
The correct answer is (B).
6. (A) 14: Draw a diagram to illustrate the moment at which Car A and Car B
pass each other moving in opposite directions.


Try Working Backwards from the answers, starting with (B) or (D).
  B’s Distance
(miles)
B’s Rate
(mph)
=

=
A’s Rate
(mph)
= B’s Rate +
8
A’s Distance
(miles)
= R × T
= R × 0.75
Total Distance =
62?
(B) 12 48 56 42 54
The total distance is not 62, so answer (B) is incorrect. Furthermore, a
distance of 54 is too short, so the answer must be (A). If you aren’t sure,
confirm by checking answer (A).
  B’s Distance
(miles)
B’s Rate
(mph)
=

=
A’s Rate
(mph)
= B’s Rate +
8
A’s Distance
(miles)
= R × T
= R × 0.75
Total Distance =
62?
(A) 14 56 64 48 62
Alternatively, you could solve algebraically, using a rate–time–distance
(RTD) chart. Note that you must convert 15 minutes to 0.25 hours.
  Rate Time Distance


Rate Time Distance
Car A (r + 8) mph 0.75 hours (0.75) (r + 8) miles
Car B r mph 0.25 hours 0.25r miles
Total     62 miles
Set up and solve an equation for the total distance.
Therefore, in 15 minutes, Car B traveled a distance of 0.25r = (0.25)(56) =
14 miles.
The correct answer is (A).
7. (B) 24: Testing the choices would be a natural way to solve this problem,
since the question doesn’t ask you to solve for b in general, but rather
“for which of the following is x closest to zero?” However, numbers
between 220 and 242 are too great to plug and compute. Instead,
manipulate the terms with base 8 to see how they might balance with
2b.


The correct answer is (B).
8.
 Since there are variables in the
answer choices, choose a smart number to solve. If k = 2, then
,
which is less than 1. Now, test the answer choices and try to match the
target; stop if you can tell that an answer won’t equal the target.
(A) 2 Too high. Eliminate.
(B)
 Too high. Eliminate.
(C)
 Too high. Eliminate.
(D)
 Too high. Eliminate.
(E)
 Correct!
Alternatively, solve the problem algebraically. The expression given is of
the form
 , where
 and
 .


You need to either simplify or cancel the denominator, as none of the
answer choices have the starting denominator and most of the choices
have no denominator at all. First, try to eliminate the radical signs
entirely, leaving only a2 and b2 in the denominator. To do so, multiply by
a fraction that is a convenient form of 1.
Notice the “diﬀerence of two squares” special product created in the
denominator.
Substituting for a and b:
The correct answer is (E).
9. (D) 117%: An algebraic setup would be fairly ugly on this problem, but a
combination of two other techniques work quite nicely: choose smart
numbers and estimate.
There are no real values for the variables, so Choosing Smart Numbers is
a valid approach. Furthermore, the answer choices are far enough apart


that you can estimate if the Smart Numbers get a little messy.
Of the three variables, it is easiest to pick for x. If you were to pick for y
or z, you would have to do “reverse” calculations to find the other
variables. Since this is a percent problem, try 100 first.
e.g., Since x = 100, 90% of x is equal to 90. Therefore, you is equal
to 20% less than 90 = 90 − 18 = 72
This number is a little ugly, so round down to 70. Therefore, y = 70, so
130% of y is equal to 70 + 21 = 91.
The next step requires you to take 25% of that number, so 91 is going to
get messy. Estimate again, but this time round up a little bit (to oﬀset
the error you introduced when you rounded down earlier). Round to the
nearest number that is divisible by 4, which is 92. Next, find 25% more
than 92.
z = 92 + 23 = 115
If x = 100 and z ≈ 115, then the final calculation is
which is closest to 117%.
The correct answer is (D).
10. (B) 72: The values in set A and set B are unknown. Jot them down in
terms of variables. If x is a positive integer, then x, x + 1, and x + 2 are
consecutive integers, and 3x, 3(x + 1), and 3(x + 2) are three consecutive


multiples of 3. Similarly, set B contains 5y, 5(y + 1), 5(y + 2), 5(y + 3), and
5(y + 4), where y is a positive integer. Do not use the same variable for
both sets, since it is unclear whether or how they are related.
Sum the numbers in each set.
Set A: 3x + 3(x + 1) + 3(x + 2) = 3(3x + 3)
Set B: 5y + 5(y + 1) + 5(y + 2) + 5(y + 3) + 5(y + 4) = 5(5y + 10)
The sums of the sets are equal. Set them equal to each other and
simplify as much as possible.
3(3x + 3) = 5(5y + 10)
9(x + 1) = 25(y + 2)
Because x and y are positive integers, the right side of the equation is a
multiple of 25 and the le side is a multiple of 9. In other words, each
side of the equation must be a multiple of both 9 and 25. Because 9 and
25 have no common factors, x + 1 must be a multiple of 25 and y + 2
must be a multiple of 9. The least possible value for x is x = 24, and the
least number that could be a member of set A is 3x = 3(24) = 72.
Alternatively, the sum of the integers in set A can be calculated by
Working Backwards from the choices. Manually add them up or note
that for an evenly spaced set the sum is (# of terms) × (the median term).
The integers in set B are multiples of 5, so the sum of set B is a multiple
of 5. By extension, the (equivalent) sum of set A must be a multiple of 5.


Try the answer choices; since the problem asks for the least value, start
with answer (A).
(A) {69, 72, 75} sums to 3(72) = not a multiple of 5.
Eliminate.
(B) {72, 75, 78} sums to 3(75) = a multiple of 5.
Correct.
The problem asks for the least value that could be a member of set A, so
the correct answer is (B).


Workout Set 2
11. If a ≠ b, is
   ?
|a| > |b|(1)
a < b(2)
12. If m = 4n + 9, where n is a positive integer, what is the greatest
common factor of m and n ?
m = 9s, where s is a positive integer(1)
n = 4t, where t is a positive integer(2)
13. A museum sold 30 tickets on Saturday. Some of the tickets sold
were $10 general exhibit tickets and the rest were $70 special
exhibit tickets. How many general exhibit tickets did the museum
sell on Saturday?
The museum’s total revenue from ticket sales on Saturday was
greater than $1,570 and less than $1,670.
(1)
The museum sold more than 20, but fewer than 25, special
exhibit tickets on Saturday.
(2)


14. If x and y are integers, is xy = yx ?
x − y = 2(1)
xy = 8(2)
15. If ab3c4 > 0, is a3bc5 > 0 ?
b > 0(1)
c > 0(2)
16. What is the perimeter of isosceles triangle ABC ?
The length of side AB is 9.(1)
The length of side BC is 4.(2)
17. If x, y, and z are integers and 2x 5y z = 6.4 × 106, what is the value of
xy ?
z = 20(1)
x = 9(2)
18. Is x > y ?
y2 < z2 < x2(1)
z < x(2)


19. A school has a students and b teachers. If a < 150, b < 25, and
classes have a maximum of 15 students, can the a students be
distributed among the b teachers so that each class has the same
number of students? (Assume that any student can be taught by
any teacher.)
It is possible to divide the students evenly into groups of 2, 3, 5,
6, 9, 10, or 15.
(1)
The greatest common factor of a and b is 10.(2)
20. A set of five distinct positive integers has a median of 3 and a range
of 12. What is the mean of the set of integers?
The product of the integers in the set is a multiple of 14.(1)
The sum of the integers in the set is a multiple of 13.(2)


Workout Set 2 Answer Key
11. E
12. A
13. A
14. C
15. B
16. C
17. A
18. C
19. E
20. D


Workout Set 2 Solutions
11. (E): You can cross-multiply the inequality, as long as you consider both
cases.
This is a conditional rephrased question—importantly, one with
completely opposite questions as possibilities. Any statement that
doesn’t at least answer the question of whether a − b is positive or
negative is unlikely to be suﬀicient. You would also have to carry each
statement through both questions, so this path is not very eﬀicient. It’s
better to use the original inequality to test cases in order to eliminate
the incorrect answers.
(1) INSUFFICIENT: Test cases to determine whether this statement is
suﬀicient.
  a b
 Is
Case 1 2 1 ✓
 No
Case 2 2
 ✓
 Yes
If a − b is negative, the question becomes “Is 1 < ab(a − b)?”
(FLIPPED inequality sign)
If a − b is positive, the question becomes “Is 1 > ab(a − b)?”
(ORIGINAL inequality sign)


(2) INSUFFICIENT: Test cases again.
a b a < b Is
 ?
Case 1
 1 ✓
  Yes
Case 2
 ✓
 ? No
(1) AND (2) INSUFFICIENT: Whenever possible, reuse cases you’ve
already tested (you can only do this when the case makes both
statements true). The pairs tested for statement (2) are also valid for
statement (1).
a b Valid? Is
 ?
Case 1
 1 ✓
 ? Yes
Case 2
 ✓
 ? No
Once again, two valid sets of numbers return two diﬀerent answers, so
even together, the statements are insuﬀicient. The correct answer is (E).


12. (A): The greatest common factor (GCF) of any two numbers is given by
the product of the shared or overlapping primes in the two numbers.
Any such question can be rephrased accordingly: “What exactly are the
overlapping factors of m and n?”
(1) SUFFICIENT: This statement indicates that m must be a multiple of 9.
If m is a multiple of 9 and m = 4n + 9, then this equation is really saying.
multiple of 9 = 4n + 9
In order for this to be true, 4n must also be a multiple of 9. The number 4
does not contain any factors of 9, so n itself must be a multiple of 9.
If both m and n are multiples of 9, and m is exactly 9 units away from 4n,
then the greatest possible common factor is 9, so this statement is
suﬀicient.
(If you’re not sure about the logic of that last part, test out a couple of
real numbers. Remember that n must be a multiple of 9. If n = 9, then m
= 4(9) + 9 = 45. The GCF of the numbers 9 and 45 is 9. If n = 18, then m =
4(18) + 9 = 81. The GCF of the numbers 18 and 81 is still 9.)
Alternatively, methodically list possible values of s, m, and n. Ignore any
case for which the statement is not true; for the remaining cases,
determine the answer to the question.
s = pos int m = 9s
 n = pos int GCF of
m and n?


s = pos int m = 9s
 n = pos int GCF of
m and n?
1 9
 Zero ✗
2 18
 Fraction ✗
3 27
 Fraction ✗
4 36
 Fraction ✗
5 45
 ✓ 9
6 54
 Fraction ✗
7 63
 Fraction ✗
8 72
 Fraction ✗
9 81
 ✓ 9
10 90
 Fraction ✗


In this case, n is only a positive integer when (s − 1) is a multiple of 4.
Also, m and n are both multiples of 9. Here’s the proof:
Try the next number in the pattern, 9 + 4 = 13, to be confident that the
apparent GCF pattern will continue.
s = pos int m = 9s
 n = pos int GCF of
m and n?
5 (9)(5) = 45
 ✓ 9
9 (9)(9) = 81
 ✓ 9
13 (9)(13) = 117
 ✓ 9
The variables m and n always share an overlapping factor of 9, but there
is never any overlap between their remaining factors. The GCF is always
9.
(2) INSUFFICIENT: n = 4t, where t is a positive integer. In this case, n must
be a multiple of 4. This does not allow you to deduce anything
consistent about m, as statement (1) did.


t = pos int n = 4t m = 4n+ 9 GCF of
m and n ?
1 4 16 + 9 = 25 1
2 8 32 + 9 = 41 1
3 12 48 + 9 = 57 3
There is more than one possible GCF, so the statement is not suﬀicient.
The correct answer is (A).
13. (A):
(1) SUFFICIENT: Use the integer constraint to test possible cases. Let g
equal the number of general exhibit tickets sold. Because the special
tickets are so much more expensive, begin by choosing numbers for
which (70)(# of special tickets) is approximately $1,600.
If g = 7, total revenue = 10(7) + 70(23) = 70 + 1,610 = 1,680. Too high.
If g = 8, total revenue = 10(8) + 70(22) = 80 + 1,540 = 1,620. Ok.
If g = 9, total revenue = 10(9) + 70(21) = 90 + 1,470 = 1,560. Too low.
Only g = 8 gives a total revenue in the given range.
Don’t assume that having a range of values automatically means a
statement is insuﬀicient to answer a Value question. At times, a


constraint (such as an integer constraint) may limit the number of cases
to just one possibility.
Alternatively, you can use algebra, though that path is a bit long on this
problem. Let g equal the number of general exhibit tickets sold. Then
(30 − g) represents the number of special exhibit tickets sold. Set up and
solve the following inequality:
The only integer between 7.2 and 8.8 is 8, so g must be 8. The museum
sold 8 general exhibit tickets.
(2) INSUFFICIENT: This statement indicates that the museum sold 21, 22,
23, or 24 special exhibit tickets. Since the museum sold a total of 30
tickets, this means that it sold 9, 8, 7, or 6 general exhibit tickets.
The correct answer is (A).
14. (C):
(1) INSUFFICIENT: Test cases on this theory problem. If you can generate
both Yes and No cases, you will prove insuﬀiciency.
x y xy yx Does xy = yx ?


x y xy yx Does xy = yx ?
Case 1 2 0 20 = 1 02 = 0 No
Case 2 3 1 31 = 3 13 = 1 No
Case 3 4 2 42 = 16 24 = 16 Yes
(2) INSUFFICIENT: Continue to test cases.
x y xy yx Does xy = yx ?
Case 1 8 1 81 = 8 18 = 1 No
Case 2 4 2 42 = 16 24 = 16 Yes
(1) AND (2) SUFFICIENT: Test whether any cases work for both
statements. The case x = 4 and y = 2 was already used for both
statements and returned a Yes answer. Can any of the No cases work for
both statements?
They don’t. Now you have a choice. If you are good with numbers, you
can try to find another pair of integers that will work with both
equations. Since x and y have to be integers, there are only a limited
number of possibilities. The two numbers need to multiply to 4 and


have a diﬀerence of 2. In addition to 4 and 2, the pair −4 and −2 fit the
bill, as long as x = −2 and y = −4.
Alternatively, solve algebraically to determine the possible values for x
and y.
Therefore, x = −2 or x = 4.
You have already tested the case where x = 4 (and y = 2), so test the other
case.
x y xy yx Does xy = yx ?
−2 −4
 Yes
In either case, xy = yx. The correct answer is (C).
15. (B): Odd exponents do not “hide the sign” of the base. If ab3c4 > 0, then a
and b must have the same sign so that their product is positive. In that


case, a3b must also be positive. As a result, in order for the inequality
a3bc5 to be positive, c5 must be positive. The rephrased question is, “Is c
> 0?”
(1) INSUFFICIENT: No information about the sign of c.
(2) SUFFICIENT: Answers the rephrased question directly.
The correct answer is (B).
16. (C): The perimeter of a triangle is equal to the sum of the three sides.
(1) INSUFFICIENT: Knowing the length of one side of the triangle is not
enough to find the sum of all three sides.
(2) INSUFFICIENT: Knowing the length of one side of the triangle is not
enough to find the sum of all three sides.
(1) AND (2) SUFFICIENT: Triangle ABC is an isosceles triangle, which
means that two of the sides are equal in length. The statements provide
two of the side lengths, so the third side, AC, must equal one of the
given sides.
There is a hidden constraint in this problem: the triangle must be valid.
Recall that the sum of the lengths of any two sides of a triangle must be
greater than the length of the third side.
AB BC AC Perimeter Valid triangle?
9 4 4 9 + 4 + 4 = 17 No: 4 + 4 < 9


AB BC AC Perimeter Valid triangle?
9 4 9 9 + 4 + 9 = 22 Yes: 4 + 9 > 9
A “triangle” with three sides of 4, 4, and 9 is not really a triangle, as it
cannot be drawn with those dimensions.
Therefore, the actual sides of the triangle must be AB = 9, BC = 4, and
AC = 9. The perimeter is 22.
The correct answer is (C).
17. (A): Express both sides of the equation in terms of prime numbers.
The right side of the equation is composed of only 2’s and 5’s. The le
side of the equation has x number of 2’s and y number of 5’s along with
some factor z. This unknown factor z must be composed of only 2’s
and/or 5’s, or it must be 1 (i.e., with no prime factors).
If z = 1, then x = 11 and y = 5.


If z = 2?5?, where the exponents are not 0, then x and y will depend on
the value of those exponents.
The rephrased question is thus, “How many factors of 2 and 5 are in z?”
(1) SUFFICIENT: If z = 20 = 2251, then this answers the rephrased
question. Incidentally, this implies that 2x5y(2251 ) = 21155, so x = 9 and y
= 4, making xy = 36 (though you do not have to solve for x and y).
(2) INSUFFICIENT: x = 9, but the statement doesn’t indicate anything
about the value of y.
The correct answer is (A).
18. (C): This problem deals with both squares and inequalities. Remember
that negative numbers become positive when squared and that
squaring a number less than 1 results in a value that is even closer to 0.
(1) INSUFFICIENT:  Without any more information about z, this
statement only tells you that x2 is greater than y2. However, x could be
either greater than y or less than y, depending on whether x is positive
or negative.
Case 1: x = –100, y = 1. Since all squares are positive, x2 is much greater
than y2. However, x is less than y.
Case 2: x = 100, y = 1. In this case, x2 is greater than y2 and x is greater
than y.


(2) INSUFFICIENT: This statement does not provide any information
about y, so y could be greater than, less than, or equal to x.
(1) AND (2) SUFFICIENT: According to statement (2), x is greater than z.
According to statement (1), x2 is greater than z2. This implies that |x| is
greater than |z|.
There are only two situations in which x is greater than z and the
absolute value of x is greater than the absolute value of z. One
possibility is that x is positive and z is a lesser positive number. Another
possibility is that x is positive and z is a negative number with an
absolute value that is less than x. These scenarios are illustrated below.
There are no valid cases in which x is negative because in order to fit
statement (2), z would have to be even more negative than x, in which
case |z| would be greater than, not less than, |x|. Therefore, x is definitely
positive.
Statement (1) says that x2 > y2, which implies that |x| > |y|. Since x is
positive, |x| = x, so x > |y|. If x is greater than the absolute value of y, x
must also be greater than y itself, so the answer to the question is
definitely Yes.
The correct answer is (C).


19. (E): In order for the a students to be distributed evenly among the b
teachers, a must be divisible by b. This is a factor question in disguise:
“Is a divisible by b?”
(1) INSUFFICIENT: The given numbers represent factors of a. If the a
students can be divided evenly into a group of 2, then 2 is a factor of a.
Likewise, 3 and 5 must be factors of a.
If 2 and 3 are already factors of a, then 6 must also be a factor (since 2 × 3
= 6); you can ignore it. Likewise, you’ve already counted one factor of 3,
so you need only one more factor of 3 to make 9. You already have the
necessary factors to make both 10 and 15, so ignore those numbers as
well. The final list of factors is: 2, 3, 5, and 3.
As a result, a must equal (2)(3)(5)(3) = 90 or a multiple of 90. Since the
question stem indicated that a < 150, a must actually be 90. This
statement provides no information about the teachers, however, so it is
insuﬀicient.
(2) INSUFFICIENT: If the greatest common factor (GCF) of a and b is 10,
then a and b must both be multiples of 10. Since the question stem
indicated that b < 25, b must be 10 or 20. The value of a is any multiple of
10 up to 140, inclusive. This is not enough to determine whether a is
divisible by b. For example, if a = 50 and b = 10, then a is divisible by b. If
a = 50 and b = 20, then a is not divisible by b.
(1) AND (2) INSUFFICIENT: From statement (1), a = 90. From statement
(2), b = 10 or 20. If b = 10, then the 90 students can be divided up evenly
into classes of 9 students each. If b = 20, then the 90 students cannot be
divided up evenly.


The correct answer is (E).
20. (D): All five of the integers in the set are positive, and all of the integers
are distinct, or diﬀerent from each other. Jot down blanks to represent
the values in the set.
_ _ _ _ _
The median of the set is 3.
_ _ 3 _ _
The least two integers in the set must be less than the median, diﬀerent
from each other, and positive: the first two numbers must be 1 and 2.
1 2 3 _ _
Finally, the range of the set, or the diﬀerence between the greatest and
least integers, is 12. Therefore, the greatest integer in the set is 13.
1 2 3 _ 13
The only unknown value in the set is the second-greatest integer, which
must be between 3 and 13, exclusive. Finding this integer will give you
enough information to find the mean of the set.
(1) SUFFICIENT: For a number to be a multiple of 14, it must be a
multiple of both 2 and 7. The product of the four known values is already
a multiple of 2, since the set already contains the number 2. However,
there are no multiples of 7 among the known values. Therefore, the


missing fourth value must be a multiple of 7. The only multiple of 7
between 3 and 13 is 7 itself, so this is the missing integer and the mean
can now be calculated.
(2) SUFFICIENT: The four known integers sum to 19. Since the unknown
integer must be positive, the sum of the entire set must be greater than
19. The next multiples of 13 are 13(2) = 26 and 13(3) = 39, which would
imply that the missing integer is 26 – 19 = 7 or 39 – 19 = 20, respectively.
(For greater multiples of 13, the missing integer would have to be even
greater than 20.) However, the missing integer must be between 3 and
13, so 20 is too great. The missing integer must be 7, and the mean can
be calculated.
The correct answer is (D).


Workout Set 3
21. Let f(x) equal the sum of all of the integers from 1 to x, inclusive,
where x ≥ 1. If
, which of
the following is true of g ?
0 ≤ g < 5(A)
5 ≤ g < 10(B)
10 ≤ g < 15(C)
15 ≤ g < 20(D)
20 ≤ g < 25(E)
22. If x and y are positive integers, what is the remainder when xyis
divided by 10 ?
x = 26(1)
yx = 1(2)
23. For all positive integers n, the sequence An is defined by the
following relationship:


What is the sum of all the terms in the sequence from A1 through
A10, inclusive?
(A)
(B)
(C)
(D)
(E)
24. If x and y are positive integers and n = 5x + 7y + 3, what is the units
digit of n ?
y = 2x − 16(1)
y is divisible by 4.(2)


25. At a certain store, all notebooks have the same price and all pencils
have the same price. The price of four notebooks and three pencils
is more than $12 and less than $13. The price of two notebooks and
five pencils is more than $8 and less than $9. If a notebook costs x
more than a pencil, which of the following could be the value of x ?
$0.60(A)
$1.05(B)
$1.30(C)
$2.20(D)
$2.70(E)
26. A certain sequence is defined by the following rule: Sn = k(Sn − 1),
where k is a constant. If S1 = 2 and S13 = 72 , what is the value of S7  ?
  6(A)
12(B)
24(C)
36(D)
37(E)
27. If y =90 + 91 + 92 + … + 9n and integer n is greater than 2, what is the
remainder when y is divided by 5 ?
n is divisible by 3.(1)
n is odd.(2)


28. In a certain sequence, each term a er the first is twice the previous
term. If the first term of the sequence is 3, what is the sum of the
14th, 15th, and 16th terms in the sequence?
3(216)(A)
9(215)(B)
21(214)(C)
9(214)(D)
21(213)(E)
29. n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined
as tn = tn − 1 + n. If t0 = 3, is tn even?
n + 1 is divisible by 3.(1)
n − 1 is divisible by 4.(2)
30. To complete a 120-mile trip, a train first travels at a constant rate of
x miles per hour (mph) for 80 miles. Then it travels the remaining 40
miles at a constant rate of y mph. If the train had instead completed
the entire trip at a constant rate of z mph, the trip would have taken
1 hour longer. Which of the following is the value of z in terms of x
and y ?


(A)
(B)
(C)
(D)
(E)


Workout Set 3 Answer Key
21. B
22. A
23. C
24. B
25. D
26. B
27. B
28. E
29. B
30. C


Workout Set Solutions
21. (B) 5 ≤ g < 10: f(x) is the sum of all integers from 1 to x, inclusive.
Therefore, f(x) equals the average of the integers from 1 to x, which is
, multiplied by the number of integers from 1 to x, which is x.
Therefore, f(x) =
 .
The question asks for the sum of the values of
 ,
 , and so on
up to
 . Instead of summing these values by hand, look for a
pattern, since the problem only asks for a range and not a specific
answer.
One possibility is to use algebra.
  can be simplified as follows:


Using this formula, the first few terms of the sum are
 ,
,
 ,
 , and so on, until the
tenth term, which is
 . All of these terms are greater than
 and no
greater than 1. If all of the terms were 0.5, then the sum would be 10(0.5)
= 5. However, if all of the terms were 1, then the sum would be 10.
Therefore, the sum must be between 5 and 10.
This problem can also be solved without using a formula for f(x).
Manually calculate the first few values of f(x), and look for a pattern.
x f(x)
1 1
 = 1
2 3
3 6
4 10
5 15
6 21
7 28
etc. etc. etc.


The values appear to be decreasing and approaching
 . If each term in
the sum is greater than
 but no greater than 1, their sum will be
greater than 5 but no greater than 10.
The correct answer is (B).
22. (A): The question asks for the remainder when xy is divided by 10. The
remainder of any number when divided by 10 is equal to the units digit
of that number, so the question is really asking for the units digit of xy.
(1) SUFFICIENT: The tendency is to deem statement (1) insuﬀicient
because no information is given about the value of y. But 26 has a units
digit of 6, and remember that 6any positive integer has a units digit of 6 (the
pattern is a single-term repeat).
61 = 6
    62 = 36
    63 = 216
    etc.
Thus, 26 raised to ANY positive integer power will also have a units digit
of 6 and therefore a remainder of 6 when divided by 10.
(2) INSUFFICIENT: Given that yx = 1, there are a few possible scenarios:


x y yx = 1
0 anything nonzero (anything nonzero)0 = 1 ✓
anything 1 1anything = 1 ✓
even −1 −1even = 1 ✓
However, the question stem indicates that x and y are POSITIVE integers,
so eliminate the first and third scenarios.
The remaining scenario indicates that y = 1 and x = any positive integer.
Without more information about x, you cannot determine the remainder
when xy is divided by 10.
Since statement (1) indicates the value of x and statement (2) indicates
the value of y (y = 1), the temptation might be to combine the
information to arrive at an answer of (C). This is a common trap on
diﬀicult Data Suﬀiciency problems; in this case, you don’t need both
statements because statement (1) is suﬀicient by itself.
The correct answer is (A).
23. (C)
 : Set up a table to list the first few terms of the sequence
and also the cumulative sum.


n
 Cumulative Sumn
 Cumulative Sum
1
 0
2
3
4
As you build the table, compare the input column values (n) with the
output column values (cumulative sum), looking for a pattern. The
denominator of the cumulative sum is n! and the numerator is one less
than n!. Therefore:
Sum of terms through
Substitute to find the sum through A10.
Sum of terms through
The correct answer is (C).


24. (B): The units digit of n is determined solely by the units digit of the
expressions 5x and 7y + 3 because when two numbers are added
together, the units digit of the sum is determined solely by the units
digits of the added numbers.
Since x is a positive integer, and 5any positive integer always has a units digit
of 5, 5x always ends in a 5. However, the units digit of 7y + 3 is not certain,
as the units digit pattern for the powers of 7 is a four-term repeat: [7, 9,
3, 1].
The question can be rephrased as, “What is the units digit of 7y + 3?”
Note: Determining y would be one way of answering the question above,
but don’t rephrase to, “What is y?” Because the units digits of the
powers of 7 have a repeating pattern, you might get a single answer for
the units digit of 7y + 3 despite having multiple values for y.
(1) INSUFFICIENT: This statement indicates neither the value of y nor the
units digit of 7y + 3, as y depends on the value of x, which could be any
positive integer. For example, if x = 9, then y = 2, and 7y + 3 has a units
digit of 7. By contrast, if x = 10, then y = 4, and 7y + 3 has a units digit of 3.
(Note: The statement does indicate that y is an even number, so the
exponent y + 3 must be odd. As a result, there are only two possible
values for the units digit of the desired term: 7 or 3. That could be useful
to know if you later need to combine the two statements.)
(2) SUFFICIENT: Regardless of what multiple of 4 you pick, 7y + 3 will have
the same units digit.


y y + 3 Units Digit of 7y + 3y y + 3 Units Digit of 7y + 3
4 7 3
8 11 3
12 15 3
Ultimately, this means that n has a units digit of 5 + 3 = 8.
The correct answer is (B).
25. (D) $2.20: The answer choices in this problem are relatively simple
numbers, but they represent the diﬀerence between two prices, rather
than a single price. Therefore, Working Backwards may not be possible.
Instead, start by translating the information in the problem into math.
Let n be the price of one notebook, and p be the price of one pen. The
first piece of information in the problem translates to the following
inequality:
The second piece of information translates to this inequality:


The question asks for the diﬀerence between the cost of a notebook and
the cost of a pencil. In other words, what is the value of n – p ?
One option is to use the previous two inequalities to create a third
inequality, which includes the combo n – p. One way to accomplish this
is to multiply the second inequality by –1, taking care to flip the signs.
Then, reorder the result so that the inequality signs point in the
expected direction:
Now, sum the two inequalities. You can add inequalities as long as the
inequality signs are facing the same direction:
This simplifies to 1.5 < n – p < 2.5. The diﬀerence between the cost of a
notebook and the cost of a pencil must be between $1.50 and $2.50, so
the only answer choice that is in the correct range is answer (D).
It is also possible to “solve” the two inequalities for n and for p,
separately, then to use the resulting ranges to find the possible
diﬀerences between n and p.


The correct answer is (D).
26. (B) 12: To form each new term of the sequence, multiply the previous
term by k. If S1 = 2, then S2 = 2k, and S3 = 2k2, and Sn = 2kn − 1.
Thus, S13 = 2k12. Since the problem indicates that S13 = 72, set up an
equation to solve for k.
S7 is equal to 2k6. Plug the value of k into that expression to calculate
S7.
The correct answer is (B).
27. (B): Remember the units digit pattern for 9x, where x is an integer. The
units digit of 9x is 9 if x is odd, but the units digit is 1 if x is even: a
repeating pattern of [9, 1].
Now, consider the sums of the powers of 9 up to 9n:


n y = 90 + 91 + 92 + . . . + 9n Units Digit of yn y = 90 + 91 + 92 + . . . + 9n Units Digit of y
1 1 + 9 = 10 0
2 1 + 9 + 81 = 91 1
3 [1 + 9 + 81] + 729 = [units digit of 1] + units digit of 9 0
4 Units digit of 0 + units digit of 1 1
Odd Units digit: pairs of (1 + 9) 0
Even Units digit: pairs of (1 + 9) + 1 1
The alternating 1’s and 9’s in the units digits pair to a sum of 10, or a
units digit of 0. Thus, the units digit of the sum displays another two-
term repeating pattern. The units digit of y is 0 if n is odd, but 1 if n is
even.
The remainder when y is divided by 5 depends only on the units digit
and will be either 0 or 1 as well. The rephrased question is, “Is n odd or
even?”
(1) INSUFFICIENT: If n is a multiple of 3, it may be either odd or even.
(2) SUFFICIENT: If n is odd, the units digit of y is 0, and the remainder is 0
when y is divided by 5.


The correct answer is (B).
28. (E) 21(213): The sequence is 3, 6, 12, 24, 48, and so on. You could write
out the first 16 terms and add the 14th, 15th, and 16th together, but
such an approach would be prone to error and time-consuming.
Additionally, you don’t need to calculate the sum explicitly: the answer
choices all have some power of 2 as a factor, providing a hint at the best
solution method.
Write the sequence in terms of the powers of 2:
3, 6, 12, 24, 48, and so on
    3(20), 3(21), 3(22), 3(23), 3(24), and so on
So Sn, the nth term of the sequence, equals 3(2n − 1).
Thus, the sum of the 14th, 15th, and 16th terms equals 3(213) + 3(214) +
3(215).
All of the terms share the common factors 3 and 213, so factor those
terms out:


The correct answer is (E).
29. (B): Sequence problems are o en best approached by charting out the
first several terms of the given sequence. In this case, keep track of n, tn,
and whether tn is even or odd.
n tn Is tn even or odd?
0 3 Odd
1 3 + 1 = 4 Even
2 4 + 2 = 6 Even
3 6 + 3 = 9 Odd
4 9 + 4 = 13 Odd
5 13 + 5 = 18 Even
6 18 + 6 = 24 Even
7 24 + 7 = 31 Odd
8 31 + 8 = 39 Odd
Notice that beginning with n = 1, a four-term repeating cycle of [even,
even, odd, odd] emerges for tn. Thus, a statement will be suﬀicient only


if it indicates how n relates to a multiple of 4 (i.e., n = a multiple of 4 ±
known constant).
(1) INSUFFICIENT: This statement does not indicate how n relates to a
multiple of 4. If n + 1 is a multiple of 3, then n + 1 could be 3, 6, 9, 12, 15,
etc. This means that n could be 2, 5, 8, 11, 14, etc. From the chart, if n = 2
or n = 5, then tn is even. However, if n = 8 or n = 11, then tn is odd.
(2) SUFFICIENT: This statement indicates exactly how n relates to a
multiple of 4. If n − 1 is a multiple of 4, then n − 1 could be 4, 8, 12, 16, 20,
etc., and n could be 5, 9, 13, 17, 21, etc. From the chart (and the
continuation of the four-term pattern), tn must be even.
The correct answer is (B).
30. (C)
 : The answer choices are complex and include
two diﬀerent variables. If you choose to use Smart Numbers, select your
numbers cautiously to simplify the arithmetic.
For instance, if x is equal to y, the problem may be much simpler to
solve, although some of the answer choices may then yield the same
result. Choose x = y = 40, so that the train completes the entire trip at 40
mph, taking
  = 3 hours.
If the train had instead traveled at z mph, it would have taken one more
hour, or 4 hours in total. Therefore, z =
 = 30 mph.


The correct answer will yield a value of z = 30 mph when x = 40 and y =
40 are plugged in.
(A)
 is a fraction that is significantly less
than 1. Eliminate.
(B)
 , which
is too low. Eliminate.
(C)
Correct!
(D)
 is much greater than 30.
Eliminate.
(E)
  is not an integer, so it cannot equal 30. Eliminate.
Alternatively, solve algebraically by creating an RTD chart.
Rate Time Distance
x mph
  hours 80 miles


Rate Time Distance
y mph
  hours 40 miles
(Original speed)
  hours 120 miles
z mph
 hours 120 miles
Use the last row of the chart to create an equation, which can then be
solved for z.
The correct answer is (C).


Workout Set 4
31. If
 , and x and y are integers such that x < y, what is
the value of y − x ?
−1(A)
  1(B)
  3(C)
  5(D)
  7(E)
32. If Sn = 5n + 94 and K = (S80 + S82 + S84) − (S81 + S83 + S85), what is the
value of K ?
−282(A)
  −84(B)
  −30(C)
  −15(D)
    −3(E)
33. Is xy + xy < xy ?


(1)
(2)
34. If y ≠ x, then
 is the equivalent of
which of the following?
(x − 1)2y(A)
(x + 1)2(B)
x2 + x + 1(C)
(x2 + x + 1)y(D)
(x2 + x + 1)(x − y)(E)
35. In the sequence gn defined for all positive integer values of n, g1 = g2
= 1 and, for n ≥ 3, gn = gn − 1 + 2n − 3. If the function
 equals the
sum of the terms g1, g2 . . . gi, what is


g3(A)
g8(B)
(C)
(D)
(E)
36. If 3k + 3k = (39)39
 − 3k, then what is the value of k ?
(A)
(B)
242(C)
310(D)
311 − 1(E)
37. If x and y are positive integers, what is the value of xy ?


x! = 6y!(1)
(2)
38. If k is an integer and
 is divisible by 6k, what is the maximum
possible value of k ?
3(A)
4(B)
5(C)
6(D)
7(E)
39. In a certain sequence, the term Sn is given by the formula Sn = (n +
1)! for all integers n ≥ 1. Which of the following is equivalent to the
diﬀerence between S100 and S99 ?
101!(A)
100!(B)
992(98!)(C)
1002(99!)(D)
(100!)2(E)


40. If
 and
 , what is the value of
(A)
(B)
(C)
(D)
(E)


Workout Set 4 Answer Key
31. B
32. D
33. E
34. C
35. A
36. E
37. C
38. D
39. D
40. C


Workout Set 4 Solutions
31. (B) 1: Simplify the right side of the equation.
This still needs to be manipulated in order to match the le side of the
equation, since the bases need to match their exponents. The number 33
is okay, but 28 is not. Try turning it into a base of 4 instead.
That’s it: x = 3 and y = 4, since the problem states that y > x. So y − x = 4 −
3 = 1.
The correct answer is (B).
32. (D) −15: Note that the expression can be rewritten as follows by pairing
up consecutive terms in the sequence.
In each of these pairs, the 94 term cancels out.


Therefore, the value of K is as follows.
The correct answer is (D).
33. (E): First, rephrase the question stem by subtracting xy from both sides:
“Is xy < 0?” The question is whether xy is negative, or “Do x and y have
opposite signs?”
Be careful! Do not rephrase as follows:
Is xy + xy < xy ?
Is 2xy < xy ?
Is
 ? Incorrect! Dividing by variables is the mistake:
What if xy = 0? What if xy < 0 ?
Is 2 < 1? Incorrect as a result of the mistake in the previous step.


Not only does this rephrase make the statements moot (2 is definitely
not less than 1, no matter what the statements say), but it also ignores
some special cases. If xy = 0, then dividing by xy yields an undefined
value. If xy < 0, you should have flipped the sign of the inequality.
Instead, use the correct rephrasing: “Do x and y have opposite signs?”
(1) INSUFFICIENT: If
 , then x2 and y must have opposite signs.
Since x2 must be positive, y must be negative. However, x could be either
positive or negative.
(2) INSUFFICIENT: If x3y3 < (xy)2, then you can divide both sides by (xy)2
(since that quantity is positive). The simplified inequality is xy < 1, which
is not suﬀicient to answer the question.
(1) AND (2) INSUFFICIENT: Each statement indicates that xy could be
either positive or negative. The statements are equally insuﬀicient and
neither provides any additional information to the other.
The correct answer is (E).
34. (C) x2 + x + 1: First, distribute the numerator.
None of the answer choices are fractions, so the x − y in the denominator
must be canceled by a x − y in the numerator. Group the numerator
terms with x and −y in mind.


Alternatively, you could choose smart numbers. If x = 2 and y = 3, then.
Plug the selected values into the choices. The choice that equals 7 is the
correct answer.
The correct answer is (C).
(x − 1)2y = (2 − 1)2(3) = 3 Eliminate.(A)
(x + 1)2 = (2 + 1)2 = 9 Eliminate.(B)
x2 + x + 1 = 4 + 2 + 1 = 7 Correct!(C)
(x2 + x + 1)y = (4 + 2 + 1)(3) = Too great. Eliminate.(D)
(x2 + x + 1)(x − y) = (4 + 2 + 1)(2 − 3) = Negative. Eliminate.(E)
35. (A) g3: Begin by listing some values of gn to get a sense for how gn
progresses.


For n ≥ 3, gn = 2n − 2 .
Now, look for a pattern in the sums defined by
 .
Each value is double the previous value:
 .
Therefore:
Now, all you need to do is scan the answer choices to find an expression
that equals 2. You have already discovered that g3 = 2, so you can select
g3 as the answer.
The correct answer is (A).
36. (E) 311 − 1: The common term in this problem is the recurring base of 3.
Group like terms (i.e., all the terms with k on the le side, all the other


powers of 3 on the right side), then simplify each power of 3 using
exponent rules.
The correct answer is (E).
37. (C): Note the constraints that x and y are positive integers.
(1) INSUFFICIENT: If y = 5 and x = 6, then this statement is true: 6! = 6 × 5!.
Are these the only possible positive integer values for x and y ?
The constant 6 is equal to 3!, so there’s one more: when y = 1 and x = 3,
the statement is also true: 3! = 6 × 1!.
(2) INSUFFICIENT: You can multiply both sides by y to eliminate that
variable entirely; in other words, this statement doesn’t indicate
anything about y.


(1) AND (2) SUFFICIENT: Continue to simplify the equation from
statement (2).
The quadratic produces two solutions, but only x = 6 is valid, since x
must be a positive integer. From statement (1), if x = 6, then y = 5. This is
suﬀicient to calculate a value for xy.
The correct answer is (C).
38. (D) 6: The question asks for the greatest value of k such that
 is
divisible by 6k. Since 6 = 3 × 2, the greatest value of k will equal the
number of 3 × 2 pairs among the prime factors of
 .


To count the number of times 3 appears as a factor of
 , rewrite the
expression, pulling out any factor(s) of 3 from each term.
   (33)(32)(31)(30)(29)(28)(27)(26)(25)(24)(23)
   = (3 × 11)(32)(31)(3 × 10)(29)(28)(33)(26)(25)(3 × 8)(23)
There are six factors of 3 in
 .
To count the number of times 2 appears as a factor of
 , rewrite the
expression, pulling out any factor(s) of 2 from each term.
(33)(32)(31)(30)(29)(28)(27)(26)(25)(24)(23)
= (33)(25)(31)(2 × 15)(29)(22 × 7)(27)(2 × 13)(25)(23 ×3)(23)
There are twelve factors of 2 in
 .
Since there are twelve 2’s but only six 3’s, there are only six 3 × 2 pairs
among the prime factors of
 . In general, focus on the greatest prime
in the divisor (in this case, the six maximum possible factors of 3 in 6k),
as it will be the limiting factor.
The correct answer is (D).


39. (D) (1002)(99!): S100 = 101! and S99 = 100!.
Factor the diﬀerence.
The correct answer is (D).
40. (C)
 Not only does
 , but you can also factor
 out of p.
Thus,
 , and
 . Watch out! This is one of the
incorrect answers.


The question asks for
 , which is
 .
The correct answer is (C).


Workout Set 5
41. If x and y are positive integers, what is the value of
 ?
x2 = 2xy − y2(1)
2xy = 8(2)
42. What is the remainder when (47)(49) is divided by 8 ?
1(A)
3(B)
4(C)
5(D)
7(E)
43. If a = 4x2 + 4xy and b = 4y2 + 4xy, which of the following is equivalent
to x + y ?


(A)
(B)
(C)
(D)
(E)
44. What is the value of
 ?
a + b = 7(1)
ab = 12(2)
45. What is the value of
 ?


48(A)
(B)
(C)
1(D)
(E)
46. If x is the square of an integer, is y the square of an integer?
xy is the square of an integer.(1)
 is the square of an integer.(2)
47. If x and y are positive integers such that x > y and
, which of the following is equivalent
to 2b ?


(A)
(B)
(C)
(D)
(E)
48. If (x + y)2 = 16 and x2 − y2 = 16, what is the value of 2x2 ?
  8(A)
16(B)
18(C)
32(D)
50(E)
49. Which of the following is equal to 5,9952 – 4,7962 ?
1,1992(A)
2,6872(B)
3,4252(C)
3,5972(D)
4,8202(E)


50. If
 , what is the value of xy ?
y > x(1)
x < 0(2)


Workout Set 5 Answer Key
41. A
42. E
43. E
44. B
45. D
46. B
47. B
48. D
49. D
50. A


Workout Set 5 Solutions
41. (A):
(1) SUFFICIENT: Move all terms to one side.
Factor this “square of a diﬀerence” special product and simplify.
Since x = y, the value of
 is 1.
(2) INSUFFICIENT: If 2xy = 8, then xy = 4. Test diﬀerent values for x and y.
x (positive) y (positive) xy = 4
2 2 2 × 2 = 4 ✓ 1


x (positive) y (positive) xy = 4
1 4 1 × 4 = 4 ✓
Alternatively, manipulate the equation xy = 4 to get
 on one side.
The expression
 does not equal a constant, so
 could take on many
diﬀerent values.
The correct answer is (A).
42. (E) 7: One way to solve would be to multiply (47)(49), then either divide
the result by 8 or repeatedly subtract known multiples of 8 from the
result until you are le with a remainder less than 8.
An alternative is to rewrite the given product as an equivalent yet easier-
to-manipulate product. Note that 47 and 49 are equidistant from 48, a
multiple of 8. Write each of the original factors as terms in the form (a + b)
or (a − b).


(47)(49) = (48 + 1)(48 − 1)
This form is the diﬀerence of two squares special product, (a + b)(a − b) =
a2 − b2; continue to simplify.
(48 + 1)(48 − 1) = (482 − 12)
Forty-eight is a multiple of 8, and therefore so is 482. Thus, (482 − 12) is 1
less than a multiple of 8. All such numbers (e.g., 7, 15, 23, 31) have a
remainder of 7 when divided by 8.
The correct answer is (E).
43. (E)
 : Add a and b to get a + b = 4x2 + 8xy + 4y2 = 4(x2 + 2xy + y2).
The right side is the square of a sum, so factor and solve.
Note that you could safely take the square root of both sides because any
square is non-negative.


Alternatively, choose smart numbers. For example, if x = 2 and y = 3, then
the final answer x + y = 5.
For values to plug into the choices, first compute a and b.
Next, test each answer choice; the one that equals 5 is the correct answer.
The correct answer is (E).
 Eliminate.(A)
(B)
(C)
(D)
 Correct!(E)
44. (B): Manipulate the question expression, noting the special products in
the exponents.


The rephrased question is, “What is the value of ab?” Knowing the values
of a and b individually would be suﬀicient, of course, but the individual
values are not required as long as you can determine ab.
(1) INSUFFICIENT: It is impossible to manipulate a + b = 7 to get ab, nor
can you solve for a and b individually.
(2) SUFFICIENT: This statement answers the rephrased question directly.
The correct answer is (B).
45. (D) 1:
Notice the special product of the form (a + b)(a − b) = a2 − b2 under the
square root symbol.
The correct answer is (D).
46. (B): Jot down the information provided in the question stem. If x is the
square of an integer, then x = i2, where i is an unknown integer. Is y a


perfect square?
(1) INSUFFICIENT: x is a perfect square, and when multiplied by the
unknown value y, it turns into a (possibly diﬀerent) perfect square.
Case 1: This situation can occur if y is also a perfect square. For example,
if x = 22 = 4, and y = 32 = 9, then xy = 36 = 62. Since y is the square of an
integer, the answer is Yes.
Case 2: If you only consider integer values for y, it seems as though y
must be a square. For instance, if x = 22, y cannot equal 2, 3, 5, 6, or any
non-square, since 2(22), 3(22), 5(22), etc., are not squares.
However, the question does not restrict y to integer values. To find a non-
integer value of y that fits the known information, start by choosing
square values for both x and xy. Suppose that x = 22 = 4 and xy = 32 = 9. In
this case, y =
 , which is not a perfect square. Therefore, the answer is
No.
(2) SUFFICIENT:
  is the square of an integer, so
 = j2, for some
unknown integer j. From the question stem, x = i2, for some unknown
integer i. Simplify.


Therefore, y must be the square of an integer. The answer is definitely
Yes, and this statement is suﬀicient.
The correct answer is (B).
47. (B)
 : To solve this problem, isolate b in the equation.
Note that it is okay to divide by
 , since x > y, which implies
that
 .
The question asks for 2b, but the result does not match any of the answer
choices. Most of the choices are not fractions, so try to cancel the
denominator. Since x − y is a well-disguised diﬀerence of two squares,
factor the numerator and denominator.


Cancel
 in the numerator and denominator to get
.
The correct answer is (B).
48. (D) 32: Each of the expressions given is equal to 16, so set them equal to
each other. Note that you have the square of a sum and the diﬀerence of
two squares special products. Put them both in distributed form, then
simplify.
Since 2y(x + y) = 0, it must be true that either 2y or (x + y) is equal to 0.
However, (x + y) cannot equal 0, since the problem indicates that (x + y)2 =
16 . So it must be that 2y = 0, and therefore y = 0.
Plug in 0 for y in x2 – y2 = 16 to get x2 = 16. Thus, 2x2 = 2(16) = 32.
Alternatively, you can solve this problem by Working Backwards from the
answer choices. Try (B) or (D) first.
(D) If 32 = 2x2, then x = ±4. Use this to solve for y in the second equation.


Do x = ±4 and y = 0 also work in the first equation? Yes!
The correct answer is (D).
49. (D) 3,5972: There are several ways to approach this problem without
extensive calculation.
Units Digit
One option is to note that four of the five answer choices have diﬀerent
units digits. The units digit of 5,9952 must equal the units digit of 52,
which is 5. Similarly, the units digit of 4,7962 must equal the units digit of
62, which is 6.
Subtracting a number with a units digit of 6 from a number with a units
digit of 5 results in a units digit of 9.
However, this does not mean that answer (A) is correct! Since the answer
choices are squared, the correct answer must have a units digit of 9 a er
being squared. Therefore, the correct answer is either (B) or (D), since 72
has a units digit of 9. See the estimation approach below for one way to
diﬀerentiate between choices (B) and (D).
Estimation
It is also possible to estimate the answer: 5,995 is approximately equal to
6,000, and 4,796 is approximately equal to 5,000.


6,0002 – 5,0002 = 36,000,000 – 25,000,000 = 11,000,000
Since 3,0002 = 9,000,000, and 4,0002 = 16,000,000, the correct answer
must be between 3,0002 and 4,0002. Eliminate any answer other than
answers (C) and (D). Combined with the units digit approach above, this
yields the correct answer, choice (D).
Factoring
The number 5,995 is a multiple of 5 just under 6,000, which equals
5(1,200).  The number 4,796 is a multiple of 4 just under 4,800, which
equals 4(1,200). This “coincidence” suggests factoring.
5,995 = 6,000 – 5 = 5(1,200 – 1) = 5(1,199)
4,796 = 4,800 – 4 = 4(1,200 – 1) = 4(1,199)
The shared factor of 1,199 means that simplification is possible.
The correct answer is (D).


50. (A): Begin by simplifying the equation given in the question.
At this stage, you might be tempted to divide both sides of the equation
by xy, in order to arrive at the simpler equation y2 + x2 = 2xy. However,
that would be a mistake—never divide both sides of an equation by an
unknown (in this case, xy), unless you are certain that the unknown
cannot equal zero. (Division by zero is undefined, and can lead to
nonsensical results.) So rather than divide by xy, subtract 2x2y2 from both
sides to group all terms on one side of the equals sign.
This last line implies that either xy = 0 or y – x = 0. In other words, either
xy = 0 or y = x.
(1) SUFFICIENT: If y > x, then it is impossible that y = x. Therefore, xy = 0.
(2) INSUFFICIENT: If x < 0, then it is possible that xy = 0 (i.e., if y = 0) or that
y = x (i.e., y is negative, too). If y = x = any negative number, then there are
infinitely many solutions for xy.


The correct answer is (A).


Workout Set 6
51. Is it possible for a wooden block in the shape of a rectangular solid
measuring l by w by h centimeter to pass through a square hole
with sides of length 4 centimeter?
The volume of the wooden block is 16 cm3.(1)
L > w > h > 1(2)
52. Three of the four vertices of a rectangle in the xy-coordinate plane
are (–5, 1), (–4, 4), and (8, 0). What is the fourth vertex?
(–4.5, 2.5)(A)
(–4, 5)(B)
(6, –2)(C)
(7, –3)(D)
(10, 1)(E)
53. In the coordinate plane, circle C has radius r and its center is at
point (x, y). Is at least 50% of the area of circle C contained in a
single quadrant of the coordinate plane?
x > y > 0(1)


x > r(2)
54. A group of men and women gathered to compete in a marathon.
Each competitor was weighed before the competition; the average
weight of the women was 120 pounds, and the average weight of
the men was greater than 120 pounds. If the average weight of the
entire group was w pounds, what percent of the competitors were
women?
The average weight of the men was 150 pounds.(1)
The diﬀerence between w and the average weight of the
women was twice the diﬀerence between w and the average
weight of the men.
(2)
55. What is the average (arithmetic mean) of these numbers: 12; 13; 14;
510; 520; 530; 1,115; 1,120; and 1,125 ?
419(A)
551(B)
601(C)
620(D)
721(E)
56. A set of 5 numbers has an average (arithmetic mean) of 50. The
greatest element in the set is 5 greater than 3 times the least


element in the set. If the median of the set equals the mean, what is
the greatest possible value in the set?
85(A)
87(B)
88(C)
92(D)
93(E)
57.
In the triangle above, DE is parallel to AC. What is the length of DE ?
AC = 14(1)
BE = EC(2)
58. 5, 2, 4, m, 9, 5
For the list of numbers above, what is the median?
The median is an integer.(1)


m = 8(2)
59. Set M contains seven consecutive integers, and set N contains three
values chosen from set M. Is the standard deviation of set N greater
than the standard deviation of set M ?
Set N contains the median of set M.(1)
The range of set M and set N are equal.(2)
60. Four diﬀerent children have jelly beans: Aaron has 5, Bianca has 7,
Callie has 8, and Dante has 11. How many jelly beans must Dante
give to Aaron to ensure that each child has within 1 jelly bean of all
the other children?
2(A)
3(B)
4(C)
5(D)
6(E)


Workout Set 6 Answer Key
51. C
52. D
53. B
54. B
55. B
56. D
57. C
58. D
59. B
60. B


Workout Set 6 Solutions
51. (C): Start by considering the diﬀerent types of blocks that could or could
not fit through the hole.
If all dimensions of the block are less than 4 centimeters (cm), the block
can definitely fit through the hole.
If exactly one of the three dimensions is greater than 4 cm, the block can
still definitely fit through the hole. To do so, rotate the block so that the
two shorter dimensions are aligned with the sides of the hole.


However, if at least two of the dimensions of the block are greater than 4
cm, further investigation is required. It may be possible to fit the block
through the hole by rotating it to take advantage of the longest
dimension of the hole, which is the diagonal of
 ,
but if the dimensions of the block are too great, this may be impossible.
(1) INSUFFICIENT: The block can have any dimensions such that Lwh =
16. A block with dimensions 1, 1, and 16 can fit through the hole as
shown in the second diagram above. However, a block with dimensions
8, 8, and 0.25 cannot fit through the hole. Since the answer can be Yes or
No, this statement is insuﬀicient.
(2) INSUFFICIENT: A block with dimensions 2, 3, and 4 can fit through the
hole, but a block with dimensions 100, 101, and 102 cannot.
(1) and (2) SUFFICIENT: Lwh = 16 and h is greater than 1, so Lw is less
than 16. Therefore, L and w cannot both be greater than 4. Since h is the
least dimension, it must also be less than 4. Therefore, the block has at
least two dimensions that are less than 4, and it can fit through the hole
as shown in the second diagram.
The correct answer is (C).
52. (D) (7, −3): Your GMAT scratch pad has a grid; use it to plot the diagram to
scale.


“Eyeball” solution: Complete the rectangle with the dashed lines shown.
The fourth point must be located approximately where the bigger dot is
drawn. Answers (A), (B), and (E) must be incorrect. The closest answer
choice is the point (7, −3). Alternatively, you could plot the remaining
two answer choice points to see which one “works” with the three given
points.
Alternatively, compute the location of the fourth point, using the fact
that the short sides have the same slope. The known short side connects
the points (−5, 1) and (−4, 4). In other words, the bottom le corner is 1
to the le and 3 down from the top le corner. The unknown bottom
right corner should therefore be 1 to the le and 3 down from the top
right corner, or x = 8 − 1 = 7 and y = 0 − 3 = −3, corresponding to the point
(7, −3).
The correct answer is (D).
53. (B): The question asks whether at least half of the circle is contained in
one quadrant. However, it does not specify a specific quadrant. If at
least half of the circle is contained in any one of the four quadrants, the
answer to the question will be Yes.
Note that if circle C lies in exactly one or two quadrants, then the answer
to the question must be Yes, because at least half of the circle must be in


one of those quadrants.
(1) INSUFFICIENT: If x = 2, y = 1, and r = 0.5, then the answer to the
question is Yes, since the entire circle is contained in the first quadrant.
However, if x = 2, y = 1, and r is much greater than 2, then approximately
a quarter of the circle is contained in each quadrant and the answer to
the question is No.
(2) SUFFICIENT: Since the x coordinate is greater than the radius of the
circle, no part of the circle can be in the second or third quadrants.


Therefore, at least half of the circle must be in the first quadrant or at
least half of the circle must be in the fourth quadrant. The answer to the
question is definitely Yes, and this statement is suﬀicient.
The correct answer is (B).
54. (B): The question stem indicates that the average weight of the women
was 120 and the average weight of the men was greater than 120. In
order to determine what percent of the competitors were women, you
would need to know more about the weight of the men and you’d also
need to know something about the relative number of women vs. men.
If there were not an equal number of each, then this is a weighted
average question.
(1) INSUFFICIENT: Statement (1) provides the average weight of the men
but does not indicate whether there was an equal number of men and
women.
(2) SUFFICIENT: This does not provide the weight of the men or even the
value of w, but it does indicate where the weighted average falls
between the two values.


If the weighted average is twice as far from the women’s end of the line,
then the men are responsible for
 of the total weight and the women
are responsible for
 of the total weight. Therefore,
 , or
approximately
 of the competitors were women. (Note: You do
not need to calculate this figure. You can stop whenever you understand
that this figure can be calculated.)
The correct answer is (B).
55. (B) 551: The simple average formula
 applies to this problem.


However, the chance of computational error is high on a problem with
this many terms of such great value.
Try grouping the similar terms.
Group A: 12, 13, 14 (equidistant terms with an average of 13, the
middle term)
Group B: 510, 520, 530 (equidistant terms with an average of 520, the
middle term)
Group C: 1,115; 1,120; 1,125 (equidistant terms with an average of
1,120, the middle term)
Since each group of terms consists of three values (and are therefore
equally weighted in the set of nine terms), the average of all nine
original terms is the average of the respective averages of Groups A, B,
and C.
Average =
You can make that division easier by breaking 1,653 into lesser numbers
that are divisible by 3: 1,653 = 1,500 + 150 + 3. Divide each separately by
3 to get 500 + 50 + 1 = 551.


The correct answer is (B).
56. (D) 92: Two techniques will help you eﬀiciently interpret the information
given in the question. First, draw a number line with five dots
representing the five numbers in the set. Second, label these numbers A,
B, C, D, and E, with the understanding that A ≤ B ≤ C ≤ D ≤ E.
The problem indicates that:
A + B + C + D + E = 250 (The set of five numbers has an average of 50.)
E = 5 + 3A (The greatest element is five greater than three times the
least element in the set.)
C = 50 (The median of the set equals the mean.)
You’re asked to maximize E. Arrange the dots on the number line such
that you obey the constraints, yet also note where you have some
flexibility.
Point D can be anywhere on the line from point C to point E. Given the
information from the question stem, you can maximize E by minimizing
D. Therefore, make D = C = 50.


Similarly, point B can be anywhere on the line from point A to point C.
Maximize E by minimizing B, so make B = A.
Therefore, E = 5 + 3A = 5 + 3(29) = 5 + 87 = 92.
The correct answer is (D).
57. (C):
(1) INSUFFICIENT: Many elements in this triangle could vary; you don’t
even know the placement of B relative to AC, so the triangle itself might
stretch. Even for a fixed triangle, DE could slide up or down, so various
lengths are possible for DE, as shown here:


(2) INSUFFICIENT: You don’t know the lengths of any sides of the
triangle. The side that most aﬀects the length of DE is AC, so stretch that
side. Stretching the triangle out to the right stretches DE.
(1) AND (2) SUFFICIENT: AC must be 14 and DE must be parallel to AC
and halfway between AC and B, in order to maintain BE = EC. Even
though vertex B is free to move, DE will always be the average of the
width of the triangle at AC (14) and the width at B (0). Thus, DE must be
7, no matter how the picture shi s.


The correct answer is (C).
58. (D): To find the median of a set of numbers, line them up in order of
value. The question of interest is, “Where is m relative to the other
values?” This set has six values, an even number of terms, so the median
is the average of the two middle terms.
These are the three scenarios shown above.
If m ≤ 4, then the two middle terms are 4 and 5, and the median is
4.5.
If 4 < m < 5, then the two middle terms are m and 5, and the median
is
If m ≥ 5, then the two middle terms are 5 and 5, and the median is 5.
(1) SUFFICIENT: For the case where 4 < m < 5, the median is

Thus, if the median is an integer, it must be 5.
(2) SUFFICIENT: If m = 8, then m ≥ 5, and the median is 5.
The correct answer is (D).


59. (B): Standard deviation is a measure of the “spread” of a group of
numbers.
(1) INSUFFICIENT: If set M contains the numbers {1, 2, 3, 4, 5, 6, 7}, then 4
is the median. Set N could be {3, 4, 5}, which has a lesser standard
deviation than set M because the two sets have the same average but N
is not as spread out as M.
Alternatively, N could be {1, 4, 7}, which has a greater standard deviation
than set M because the two sets still have the same average, but N is
now more spread out than M. (It doesn’t have additional values that are
closer to the average.)
(2) SUFFICIENT: If set M contains the numbers {1, 2, 3, 4, 5, 6, 7}, and set
N has the same range, then N must contain 1 and 7. If set N contains {1,
4, 7}, then N has a greater standard deviation than M. If set N contains {1,
2, 7}, the standard deviation actually increases even more (because the
numbers are no longer evenly distributed}. No matter what combination
you try, the standard deviation of N has to be greater than the standard
deviation of M.
The correct answer is (B).
60. (B) 3: Conceptually, the transfer of jelly beans from Dante to Aaron
reduces the range of the number of jelly beans held by individual
children. The constraint is that the final distribution represents a range
of just 1 jelly bean, a condition Bianca and Callie already satisfy.
Draw a picture (a number line) to visualize the scenario.


Aaron and Dante must end up with a number of jelly beans that is either
7 or 8. If either Aaron or Dante has a number of jelly beans other than 7
or 8, he will diﬀer too much from either Bianca’s or Callie’s number. You
can count out the necessary change on the number line above or you
can write out the algebra.
The solution to both equations is x = 3: Add 3 to Aaron to get 8, and
subtract 3 from Dante to get 8. The resulting number of jelly beans is A =
8, B = 7, C = 8, and D = 8.
The correct answer is (B).


Workout Set 7
61. Is rectangle R a square?
At least one side of rectangle R has an integer length.(1)
The diagonals of rectangle R have integer lengths.(2)
62. A group of friends charters a boat for $540 and each person
contributes equally to the cost. They determine that if they can get
three more of their friends to join them, every person in the group
will pay $9 less. If they find three more friends to join them, what is
the total number of people renting the boat?
  6(A)
  9(B)
15(C)
18(D)
21(E)
63. In Smithtown, the ratio of right-handed people to le -handed
people is 3 to 1 and the ratio of men to women is 3 to 2. If the
number of right-handed men is maximized, then what percent of all
the people in Smithtown are le -handed women?


50%(A)
40%(B)
25%(C)
20%(D)
10%(E)
64. The sum of the interior angle measures for any n-sided polygon
equals 180(n − 2). If Polygon A has interior angle measures that
correspond to a set of consecutive integers, and if the median angle
measure for Polygon A is 140°, what is the least angle measure in
the polygon?
130°(A)
135°(B)
136°(C)
138°(D)
140°(E)
65. When the positive integer x is divided by the positive integer y, the
quotient is 2 and the remainder is z. When x is divided by the
positive integer a, the quotient is 3 and the remainder is b. Is z > b ?
The ratio of y to a is less than 3 to 2.(1)
The ratio of y to a is greater than 2 to 3.(2)


66. If a and b are odd integers, a ∆ b represents the product of all odd
integers between a and b,inclusive. If y is the least prime factor of (3
∆ 47) + 2, which of the following must be true?
y > 50(A)
30 ≤ y ≤ 50(B)
10 ≤ y < 30(C)
3 ≤ y < 10(D)
y = 2(E)
67. Set S is the set of all prime integers between 0 and 20. If three
numbers are chosen randomly from set S, and no number is chosen
more than once, what is the probability that the sum of all three
numbers is odd?
(A)
(B)
(C)
(D)
(E)


68. Sets A and B each consist of three terms selected from the first five
prime integers. No term appears more than once within a set, but
any integer may be a term in both sets. If the average of the terms
in set A is 4 and the product of the terms in set B is divisible by 22,
how many terms are shared by both sets?
The product of the terms in set B is not divisible by 5.(1)
The product of the terms in set B is divisible by 14.(2)
69.
In the figure above, SAND and SURF are squares, and O is the center
of the circle. If Q is the sum of the areas of squares SAND and SURF
and C is the area of the circle, then the fraction
 is


less than
(A)
between
 and
(B)
between
 and
(C)
between
 and 1(D)
greater than 1(E)
70. Set S consists of n consecutive positive integers. If n > 3, what is the
value of n ?
The number of multiples of 2 contained in set S is equal to the
number of multiples of 3 contained in set S.
(1)
n is odd.(2)


Workout Set 7 Answer Key
61. C
62. C
63. C
64. C
65. A
66. A
67. D
68. D
69. C
70. E


Workout Set 7 Solutions
61. (C): A rectangle is a square if and only if its length equals its width.
(1) INSUFFICIENT: If the length is 1 and the width is 2, the rectangle is
not a square. If the length and width are both 2, the rectangle is a
square.
(2) INSUFFICIENT: Since R is a rectangle, both of its diagonals have the
same length. It is only necessary to examine one of the diagonals. Since
the diagonal and two of the sides form a right triangle, the length of the
diagonal is always equal to
 .
The diagonal of the rectangle below has an integer length, but the
rectangle is not a square:


However, in the rectangle below, the diagonal has an integer length and
the rectangle is a square. Note that it is not necessary to calculate the
exact length of the sides of the square, since this is a Data Suﬀiciency
problem. However,
  in this case.
The answer could be Yes or No, so this statement is insuﬀicient.
(1) AND (2) SUFFICIENT: According to the statements, the rectangle has
at least one integer side, and it also has integer diagonals.
In a square, the legs and hypotenuse must be in the ratio
. To be a square, a rectangle must have a diagonal that is
  times as
long as its sides.
However,
  is an irrational number, so if x is an integer, then
  is
not an integer (and vice versa). It is impossible for the rectangle
described by these statements to be a square, and the answer to the
question is definitely No. The statements together are suﬀicient.
The correct answer is (C).


62. (C) 15: Call the initial number of friends in the group f and the initial cost
c. The problem allows you to write two equations.
The question asks for f + 3. These equations can be solved algebraically,
but the math is going to result in having to solve a quadratic. The
numbers in the answer choices are fairly small; try plugging them into
the problem to find the correct answer. Start with (B) or (D).
  f + 3 f Orig Cost
 New Cost
 Are the costs $9 apart?
(B) 9 6
 No
The two costs are much too far apart. More people need to join the
group in order to bring the costs closer together. Eliminate (A) and (B).
Try (D) next.
f + 3 f Orig Cost
 New Cost
 Are the costs $9 apart?
(D) 18 15
 No
This time, the two costs are only $6 apart. They’re too close. The answer
must be between 9 and 18; therefore, the answer must be (C).


Check the math if you’re not sure, but do practice this technique enough
that you know when you can actually tell what the answer must be
(without doing the math).
The correct answer is (C).
63. (C) 25%: Use a double-set matrix to solve this problem.
Right-Handed Le -Handed Total
Men 3x
Women 2x
Total 3y y 4y = 5x
There is a hidden constraint: the number of people must be an integer.
Thus, both x and y are integers. Moreover, the total number of people
must be a multiple of 4 and of 5 in order for the given ratios to be
possible. From this constraint, there are two ways to solve.
1. Algebraic Solution
Since the question specifies that the number of right-handed men be as
great as possible, assume that all the men are right-handed; of course,
that means that none of the men are le -handed. Because each column
in a double-set matrix must total, you can also fill in the number of le -
handed women (the group you want).


Right-Handed Le -Handed Total
Men 3x 0 3x
Women y 2x
Total 3y y 4y = 5x
Thus, le -handed women represent
 of the total
population.
2. Smart Number Solution
Since the total number of people in Smithtown must be a multiple of 20,
set the total to 20 and determine the subtotals of men, women, le -
handed, and right-handed based on the ratios given in the problem.
Right-Handed Le -Handed Total
Men 12
Women 8
Total 15 5 20


To maximize the number of right-handed men, assign all the men to the
“right-handed men” cell, and fill in the remaining cells.
  Right-Handed Le -Handed Total
Men 12 0 12
Women 3 5 8
Total 15 5 20
Therefore, le -handed women represent
 of the
population.
The correct answer is (C).
64. (C) 136°: If the median angle measure is 140° and the interior angle
measures correspond to a set of consecutive integers, then the average
angle measure must equal 140°. Since the sum of the angles must equal
180(n − 2), the average angle must equal
 .


Therefore, the polygon has nine sides and nine interior angles, and the
measures of these angles are equal to a set of consecutive integers
centered at 140. The set of nine consecutive integers must therefore be:
{136, 137, 138, 139, 140, 141, 142, 143, 144}
The least angle measure is 136°.
The correct answer is (C).
65. (A): If
  has a quotient of 2 and a remainder of z, then x is z more than
2y. Mathematically, x = 2y + z. Therefore, z = x − 2y.
If
 has a quotient of 3 and a remainder of b, then x is b more than 3a.
Mathematically, x = 3a + b. Therefore, b = x − 3a.
The question asks whether z > b, and the statements give information
about
 . Simplify the question by replacing z and b with their
equivalents and solving for the combination
 .


Note that the inequality sign flipped in the last step because of the
division by −2. The variable a is a positive integer, so no additional flip is
required for that manipulation. Rephrase the question as, “Is
?”
(1) SUFFICIENT: This directly answers the rephrased question: “Yes,
.” Therefore, z > b.
(2) INSUFFICIENT: This indicates only that
 . The answer might
be Yes if
 . However, the answer might be No if
 is
greater than
 .
The correct answer is (A).
66. (A) y > 50: The function (3 ∆ 47) equals the product (3)(5)(7) … (43)(45)
(47). This product is a very large odd number, as it is the product of only
odd numbers and thus does not have 2 as a factor. Therefore, (3 ∆ 47) + 2
= Odd + Even = Odd, and (3 ∆ 47) + 2 does not have 2 as a factor either.
Every odd prime number between 3 and 47 inclusive is a factor of (3 ∆
47), since each of these primes is a component of the product. For
example, (3 ∆ 47) is divisible by 3, since dividing by 3 yields an integer
minus the product (5)(7)(9) … (43)(45)(47).
Now consider the sum (3 ∆ 47) + k, where k is an integer. The sum will
only be divisible by 3 if k is also divisible by 3. In other words, when you


divide (3 ∆ 47) + k by 3, you are evaluating
 . Because
 is an integer,
 must also be an integer to yield an integer
sum.
In this problem, k = 2, which is not divisible by any of the odd primes
between 3 and 47. Since (3 ∆ 47) IS divisible, but 2 is NOT divisible, the
sum (3 ∆ 47) + 2 is NOT divisible by any of the odd primes between 3 and
47.
So (3 ∆ 47) + 2 is not divisible by any prime number less than or equal to
47. The least prime factor of (3 ∆ 47) + 2 must be greater than 47. Thus,
the minimum possible prime factor is 53, since that is the least prime
greater than 47.
The correct answer is (A).
67. (D): If set S is the set of all prime integers between 0 and 20, then S = {2,
3, 5, 7, 11, 13, 17, 19}.
There are seven odd terms and one even term in set S. If the even term is
among those selected, the sum will be even (E + O + O = E). The sum will
be odd if all three terms selected are odd (O + O + O = O).
The probability of selecting three odd terms is
 .
The correct answer is (D).


68. (D): The first five prime integers are 2, 3, 5, 7, and 11. These are the only
terms that can appear in sets A and B. There are some other restrictions
on the sets:
Set A: The average of the terms in set A is 4, so the sum of the terms is (4)
(3) = 12. There is only one way for three of the first five primes to sum to
12: 2 + 3 + 7. Set A is {2, 3, 7}.
Set B: The product of the terms in set B is divisible by 22, so 2 and 11 are
terms in set B. Set B is {2, 11, x}, where x can be 3, 5, or 7, but not 2 or 11
(no duplicates).
Sets A and B share at least one term: the 2. If x is either 3 or 7, the sets
will share two terms. If x is 5, the sets will only share one term.
The rephrased question is, “Is x = 5?” A definite Yes or No answer leads to
a definite value answer for the number of shared terms (i.e., Yes = one
shared term, No = two shared terms).
(1) SUFFICIENT: If the product of the terms in set B is not divisible by 5, x
≠ 5 and the answer to the rephrased question is a definite No.
(2) SUFFICIENT: If the product of the terms in set B is divisible by 14,
then 2 and 7 are terms in B. Therefore, x = 7 and the answer to the
rephrased question is a definite No.
The correct answer is (D).


69. (C) between
 and
 : The area of a square is equal to the length of
one side squared. The area of a circle is equal to πr2. The question asks
for the fraction
 .
Inscribed angle USA cuts oﬀ a diameter (UA) of the circle, so angle USA
must be a right angle. Therefore, the triangle is a right triangle with
hypotenuse UA.
The Pythagorean theorem indicates that US2 + SA2 = UA2.
US2 also represents the area of the smaller square. SA2 also represents
the area of the larger square. The sum of the two equals the quantity Q
mentioned in the question stem, so Q = UA2.
UA is a diameter of the circle, so the quantity Q = (2r)2 = 4r2. The fraction
is:
The value of π is approximately 3.14, which falls between 3 and 3.5.
Therefore:


The correct answer is (C).
70. (E): The value of n must be 4 or greater. Start with statement (2), as it is
much easier than statement (1).
(2) INSUFFICIENT: If n is odd, it could equal 3, 5, 7, or so on.
(1) INSUFFICIENT: Test cases to figure out what is possible.
n Set S # Multiples 2 = # Multiples 3?
4 3, 4, 5, 6 Yes: 2 multiples of each
5 3, 4, 5, 6, 7 Yes: 2 multiples of each
(1) AND (2) INSUFFICIENT: The case of n = 5 was already proven in the
last step. Is there a way to have n = another odd number? Keep testing.
n Set S # Multiples 2 = # Multiples 3?
5 3, 4, 5, 6, 7 Yes: 2 multiples of each
7 3, 4, 5, 6, 7, 8, 9 Yes: 3 multiples of each


Since there are still at least two possible values for n, none of the
information is suﬀicient to answer the question.
The correct answer is (E).


Workout Set 8
71.
Energy usage (units) 11 10 8 7
Number of days   4   5 n 3
The table above shows daily energy usage for an oﬀice building and
the number of days that amount of energy was used. If the average
(arithmetic mean) daily energy usage was greater than the median
daily energy usage, what is the least possible value for n ?
2(A)
3(B)
4(C)
5(D)
6(E)
72. A painting crew painted 80 houses. They painted the first y houses
at a rate of x houses per week. Then, more painters arrived and
everyone worked together to paint the remaining houses at a rate
of 1.25x houses per week. How many weeks did it take to paint all
80 houses, in terms of x and y  ?


(A)
(B)
(C)
(D)
(E)
73. a = x + y and b = x − y. If a2 = b2, what is the value of y ?
(1)
(2)
74. A herd of 33 sheep is sheltered in a barn with 7 stalls, each of which
is labeled with a unique letter from A to G, inclusive. Is there at least
one sheep in every stall?
The ratio of the number of sheep in stall C to the number of
sheep in stall E is 2 to 3.
(1)


The ratio of the number of sheep in stall E to the number of
sheep in stall F is 5 to 2.
(2)
75. If (s × 10q) − (t × 10r) = 10r, where q, r, s, and t are positive integers
and q > r, then what is the units digit of t ?
0(A)
1(B)
5(C)
7(D)
9(E)
76. Is
 ?
|y − 3| ≤ 1(1)
y × |y| > 0(2)
77. What is the greatest prime factor of 21054 − 21352 + 214 ?
  2(A)
  3(B)
  7(C)
11(D)
13(E)


78. A decimal is called a “shrinking number” if its value is between 0
and 1 and each digit to the right of the decimal is not less than the
digit to its immediate right. For instance, 0.86553221 is a shrinking
number. If x is a shrinking number, which of the following must be
true?
 is a shrinking number.I.
 is a shrinking number.II.
 is a shrinking number. III.
I only(A)
II only(B)
I and II only(C)
III only(D)
I, II, and III(E)
79. Each of the seven paintings in an art gallery has a diﬀerent price. Is
it possible to purchase at least three paintings for no more than
$1,800 in total?
The median price of the seven paintings is $550.(1)
It is possible to purchase four paintings at the gallery for a total
of $2,300.
(2)


80. An (x, y) coordinate pair is to be chosen at random from the xy-
plane. What is the probability that y ≥ |x| ?
(A)
(B)
(C)
(D)
(E)


Workout Set 8 Answer Key
71. E
72. B
73. B
74. C
75. E
76. A
77. C
78. B
79. D
80. E


Workout Set 8 Solutions
71. (E) 6: Here’s what you know about the arithmetic mean:
In contrast, the median depends on n, but not in a linear way. To find the
median, you order the terms and pick the middle one, so try various n
values (i.e., vary the number of 8’s in the list). This implies the eventual
need to plug n values into the mean formula above, so draw a picture to
help eliminate some answers first.
The “low” and “high” grouping is a fast way to find the relationship
between median and mean for this set.


If the number of “lows” and “highs” are equal, the median is the average
of the middle terms. That is, if n = 6, then the median = 9.
When n = 6, the mean must be greater than 9. Why? Pairs of 8 and 10
terms average to 9, but there is one “extra” 8. Pairs of 7 and 11 terms
average to 9, but there is one “extra” 11. These “extra” terms diﬀer from
9 by −1 and +2 respectively, for a total diﬀerence of +1. That positive
diﬀerence implies that (mean > 9), or (mean > median).
To prove that n = 5 is too low, you could take a more conventional
approach. If n = 5, the number of terms is 3 + 5 + 5 + 4 = 17. The median is
the ninth term in this ordered set: 7, 7, 7, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10,
11, 11, 11, 11. Thus, the median is 10, while the mean is closer to 9.
That is, if n = 5, then median > mean.
If n = 6, then mean > median.
The correct answer is (E).


72. (B)
 : This is a combined work problem, so use the work
formula: rate × time = work. The work and rates are given, but you need
to calculate time, so manipulate the formula:
 . This
problem also has variables in the answer choices, so it is eﬀicient to
choose smart numbers.
There are 80 houses, y houses are painted at a rate of x houses per week,
and the rate increases to 1.25x houses per week for the remaining 80 − y
houses. To make the math easier, choose values such that x and 1.25x
are integers (i.e., x is a multiple of 4) and y and 80 − y are divisible by x
and 1.25x, respectively.
Variable Value Units
Total Houses 80 80 Houses
Houses Painted at Slower Rate y 20 Houses
Houses Painted at Faster Rate 80 − y 60 Houses
Initial Rate x 4 Houses/Week
Increased Rate 1.25x 5 Houses/Week
The total painting time is:


20 houses painted at a rate of 4 houses/week = 5 weeks
60 houses painted at a rate of 5 houses/week = 12 weeks
Total time for 80 houses = 5 + 12 = 17 weeks
The correct answer is (B).
 Eliminate.(A)
 Correct!(B)
 Not an integer.
Eliminate.
(C)
 Not an integer.
Eliminate.
(D)
 Eliminate.(E)
73. (B): If a2 = b2, then (x + y)2 = (x − y)2. Distribute both sides using the
square of a sum and square of a diﬀerence special products, then
simplify.


There are three basic scenarios.
x y xy = 0
0 Any nonzero ✓
Any nonzero 0 ✓
0 0 ✓
(1) INSUFFICIENT: This statement indicates that x and y must be non-
negative for their square roots to be real values. The statement also
eliminates the last scenario, in which x = y = 0. But y could still be 0 or
any positive value.
x y xy = 0
0 Any positive ✓ ✓
Any positive 0 ✓ ✓
0 0 ✓ ✗
(2) SUFFICIENT: This statement indicates that x and y must be non-
negative for their square roots to be real values. This statement


eliminates the last scenario. If x and y were both 0,
 would
equal 0. It also eliminates the first scenario. If
 , then
. Therefore,
 . Thus, you can conclude that y = 0.
x y xy = 0
0 Any positive ✓ ✗
Any positive 0 ✓ ✓
0 0 ✓ ✗
The correct answer is (B).
74. (C):
(1) INSUFFICIENT: Set up a table and assign sheep to stalls.
Stall A B C D E F G
# of Sheep 2x 3x
Since a fractional sheep is not possible in this problem, x must be a
positive integer. Suppose x = 2, so there are 4 sheep in C and 6 sheep in
E. With 23 sheep remaining, it is possible for each of the other stalls to


hold at least 1 sheep (a Yes answer). However, the 23 other sheep might
all be in stall B, leaving stalls A, D, F, and G empty (a No answer).
(2) INSUFFICIENT: Set up a table and assign sheep to stalls.
Stall A B C D E F G
# of Sheep 5y 2y
If y = 1, there are 5 sheep in E and 2 sheep in F. With 26 sheep remaining,
it is possible for each of the other stalls to hold at least 1 sheep (a Yes
answer). However, 13 sheep might be in both stalls A and B, leaving
stalls C, D, and G empty (a No answer).
(1) AND (2) SUFFICIENT: Set up a table and assign sheep to stalls.
Stall A B C D E F G
# of Sheep 2x 3x = 5y 2y
Since a fractional sheep is not possible in this problem, x and y must
both be positive integers that satisfy the equation 3x = 5y. The only
possibility is x = 5 and y = 3, since higher multiples would require more
than 33 sheep total. Thus, 31 sheep are allocated among these stalls as
follows:


Stall A B C D E F G
# of Sheep 10 15 6
Thirty-one of the 33 sheep are in three of the 7 pens. With only 2 sheep
unaccounted for, there is no way to place at least 1 sheep in each of the
remaining four pens (a definite No answer).
The correct answer is (C).
75. (E) 9: Group the 10r terms together.
Now, solve for y.
Since q > r, the exponent on 10q − r is positive. Since s is a positive
integer, s × 10q − r is a multiple of 10 and therefore ends in 0. Any multiple
of 10 minus 1 yields an integer with a units digit of 9.


The correct answer is (E).
76. (A): The complicated expression in the question stem leads to a
disguised Positive/Negative problem. In general,
 . Think
about this relationship with a real example.
In both cases (positive or negative 3), the end result is 3. Thus, in
general,
 will always result in a positive value, or |x|. Rephrase the
original question using the absolute value symbol in place of the
“square root of the square” symbols; then try to make the right-hand
side look more like the le .
Is |y − 4| = 4 − y ? becomes Is |y − 4| = −(y − 4) ?
Since the absolute value of y − 4 must be positive or zero, you can
rephrase the question further.
Is −(y − 4) ≥ 0 ? becomes Is (y − 4) ≤ 0 ? and then Is y ≤ 4 ?
(1) SUFFICIENT: The absolute value |y − 3| can be interpreted as the
distance between y and 3 on a number line. Thus, y is no more than 1
unit away from 3 on the number line, so 2 ≤ y ≤ 4. Thus, y ≤ 4.


(2) INSUFFICIENT: If y × |y| > 0, then y × |y| is positive. This means y and
|y| must have the same sign. The term |y| is non-negative, so y must be
positive. However, knowing that y is positive is not enough to indicate
whether y ≤ 4.
The correct answer is (A).
77. (C) 7: A base of 2 is common to each term, and 10 is the smallest
exponent appearing with that base. Factor 210 out from all of the terms
in the expression.
Clearly, 2 is a prime factor, but is it the greatest? Examine (54 − 2352 + 24)
to determine whether it has a larger prime factor. The expression is of
the form x2 − 2xy + y2, where:
Write the expression in factored form.
The prime factors of 21 are 3 and 7, so the largest prime factor of the
original expression is 7.


Alternatively, if you did not see the quadratic template in (54 − 2352 + 24),
you could also perform the computation and factor the result.
The correct answer is (C).
78. (B) II only: According to the text, a “shrinking number” is a decimal
between 0 and 1 in which each digit to the right of the decimal is no
smaller than the digit to its immediate right.
That is, in a shrinking number, each digit is either larger than the one to
its right, or the same as the one to its right. As you go through the digits
from le to right, the digits will either stay the same or get smaller. For
instance, 0.5331 is a shrinking number, but 0.37654 is not a shrinking
number.
This Roman Numeral problem asks which of three options must be true.
Test cases and eliminate any options that are not necessarily true.
One shrinking number is x = 0.1000 . . . This is a shrinking number
because the digits do not get larger a er the decimal point. Since this


number is relatively easy to do math with, start by using x = 0.1 to test
cases.
I. As a fraction,
 , so
. However, this is not a
shrinking number, since 9 is greater than 0. So statement I is not
necessarily true. Eliminate choices (A), (C), and (E).
II.
 , which is a shrinking number.
Statement II could be true.
In fact, statement II must always be true. Adding 9 and then dividing by
10 is equivalent to inserting the digit 9 to the immediate right of the
decimal. Since 9 is the largest possible digit, it will never be lower than
the digit to its immediate right, so the decimal will still be a shrinking
number.
III.
 , which is not a shrinking number, since the digit
two places to the right of the decimal is greater than the digit
immediately to the right of the decimal. Statement III is not necessarily
true. Eliminate choice (D).
The correct answer is (B).


79. (D): A gallery has seven paintings with diﬀerent prices. Is it possible to
purchase at least three of the paintings for no more than $1,800? This is
possible only if the three least expensive paintings at the gallery have a
total price of $1,800 or less. The question can be rephrased as, “Is the
total price of the three least expensive paintings $1,800 or less?”
(1) SUFFICIENT: The median price of the seven paintings is $550; since
all of the paintings have diﬀerent prices, the three least expensive
paintings have prices below the median. Therefore, the total price of the
three least expensive paintings is less than $550 + $550 + $550 = $1,650.
The answer is definitely Yes, so this statement is suﬀicient.
(2) SUFFICIENT: This statement implies that the four least expensive
paintings have a total price of $2,300 or less. Jot down some possible
cases.
Case 1: $1, $2, $3, and $2,000. The three least expensive paintings can be
purchased for $1,800, and the answer to the question is Yes.
Case 2: To find a No answer, try to maximize the lowest three prices out
of the four. To do so, make the highest price as low as possible, or only
slightly more than one-fourth of the total. Since one-fourth of $2,300 is
$575, a good case to test is $573, $574, $576, and $577. However, the
sum of the lowest three prices is $1,723, which is still well below $1,800,
and the answer is still Yes.
Even when maximizing the lowest of the three prices, the three least
expensive paintings can still be purchased for no more than $1,800, and
the answer to the question is definitely Yes, so this statement is
suﬀicient.


The correct answer is (D).
80. (E)
 : Graph the equation y = |x|.
The inequality y > |x| represents everything above the line (on either side
of the y-axis)—that is, the shaded region. Since the equation y = |x| forms
a 45-degree angle from the x-axis, there are 90 degrees above the line
(on both sides of the y-axis). This represents one-fourth of the xy-plane.
Therefore, if a random pair of (x, y) coordinates is chosen from the
plane, the probability is
 that the point will fit the criterion y ≥ |x|.
The correct answer is (E).


Workout Set 9
81.
As shown in the figure, square ABCD is inscribed in a circle with
circumference
 . What is the area of the shaded region in the
diagram above?
2x(A)
(B)
(C)
(D)
(E)


82. If 3a + 3a − 2 = (90)(3b), what is b in terms of a ?
a − 4(A)
a − 2(B)
a + 4(C)
3a + 2(D)
3a + 4(E)
83. If xy ≠ 0, what is the value of
 ?
y = 4 − x(1)
x(x − 6y) = −9y2(2)
84. When one new integer is added to an existing list of six integers,
does the median of the list change?
The mean of the original six numbers is 50.(1)
At least two of the numbers in the original list were 50.(2)
85. Let abc and dcb represent three-digit positive integers. If abc + dcb
= 598, then which of the following must be equivalent to a ?


d − 1(A)
d(B)
3 − d(C)
4 − d(D)
5 − d(E)
86. For nonzero integers a, b, c and d, is
 negative?
ad + bc = 0(1)
abcd = −4(2)
87. Set A consists of 8 distinct prime numbers. If x is equal to the range
of set A and y is equal to the median of set A, is the product xy
even?
The smallest integer in the set is 5.(1)
The largest integer in the set is 101.(2)
88. If x3 < 16x, which of the following CANNOT be true?
|x| > 4(A)
x > −4(B)
x < −4(C)
x > 4(D)
x < 4(E)


89. Is the two-digit positive integer n divisible by 3 ?
If the digits of n are reversed to produce the two-digit integer
m, then m is divisible by 3.
(1)
If the digits of n are reversed to produce the two-digit integer
m, then m + n is divisible by 3.
(2)
90. Of all the sheets of paper in the tray of Printer A, 20% are removed
and transferred to the tray of Printer B. A er the transfer, are there
more sheets of paper in the tray of Printer B than in the tray of
Printer A ?
The transfer increases the number of sheets of paper in the tray
of Printer B by more than 25%.
(1)
The transfer increases the number of sheets of paper in the tray
of Printer B by less than 30%.
(2)


Workout Set 9 Answer Key
81. B
82. A
83. B
84. E
85. D
86. D
87. A
88. D
89. D
90. B


Workout Set 9 Solutions
81. (B) πx − 2x: The circumference of the circle is equal to
so
 .
The area of the circle is equal to
 .
The area of a square is side2. The diagonal of this square is the diameter
of the circle, which is equal to
 . The diagonal of a square is
always
 , so the side of the square is equal to
 .
Therefore, side
 and the area of square ABCD is:
The shaded area is the area of the circle minus the area of the square =
πx − 2x.


The correct answer is (B).
82. (A) a − 4: Because the problem never provides real values for a or b, you
can choose smart numbers to solve. Choose something for a that will
make both exponents on the le side positive. If a = 3, then:
Therefore, b = −1. Plug a = 3 into the answers and look for −1.
Alternatively, you can solve algebraically. Some manipulation is
required to determine b in terms of a, but spend a minute thinking
strategically about what manipulations to do. It would help to isolate
terms with an a by factoring 3a out on the le side. This will let you
compare 3a to 3b, and thus compare a to b, once you clean up the
constant terms that are le over. Also, 90 is a multiple of 9, which is a
3 – 4 = −1 Correct.(A)
3 – 1 = 1 Eliminate.(B)
3 + 4 = Too big. Eliminate.(C)
3(3) + 2 = Too big. Eliminate.(D)
3(3) + 4 = Too big. Eliminate.(E)


power of 3. Ninety is also the sum of 81 and 9, both powers of 3
themselves. It is likely that some constant terms will cancel, as shown
below.
Therefore, b = a − 4.
The correct answer is (A).
83. (B): The constraint in the question stem indicates that neither x nor y
equals zero.
(1) INSUFFICIENT: You can prove insuﬀiciency by Testing Cases. If y = 1,
then x = 3, in which case
 is 3. If y = 2, then x = 2, in which case
 is 1.
(2) SUFFICIENT: Testing Cases might not be such a great idea for this
statement because there are multiple instances of each variable. Begin
by distributing the le -hand side of the equation.


Now, you’ve got a quadratic, so solve that way.
The correct answer is (B).
84. (E): You can test cases to try to prove or disprove the statements.
(1) INSUFFICIENT: If the original set is 50, 50, 50, 50, 50, 50, the mean and
median were 50 originally. The new term may be any value and the
median will not change.
If, on the other hand, the original set is 0, 0, 0, 100, 100, 100, the mean
and median were 50 originally. If the new term is 75, the median
increases to 75. More generally for this set, if the new term is not equal
to the original median of 50, the median will change.
(2) INSUFFICIENT: If the original set is 50, 50, 50, 50, 50, 50, the median
was 50 originally. The new term may be any value and the median will
not change.
If, on the other hand, the original set is 0, 0, 0, 50, 50, 50, the median was
25 originally. If the new term is 40, the median increases to 40. More


generally for this set, if the new term is not equal to the original median
of 25, the median will change.
(1) AND (2) INSUFFICIENT: The mean must be 50 and at least two terms
in the original set must be 50.
If the original set is 50, 50, 50, 50, 50, 50, the mean and median were 50
originally. The new term may be any value and the median will not
change.
If the original set is 0, 0, 0, 50, 50, 200, the mean was 50 and the median
was 25 originally. If the new term is greater than 25, the median
increases. If the new term is less than 25, the median decreases.
The correct answer is (E).
85. (D) 4 − d: Set up the addition and extract several equations by summing
each digit place individually.
Note that both the ones digit and tens digit come from the sum c + b.
Since the result is diﬀerent in the ones digit (8) and the tens digit (9), a 1
must be carried from the ones to the tens digit. Thus, c + b ≠ 8; instead, c
+ b = 18.
If c + b = 18, then both c and b must be 9 (the largest digit). Place an 8 in
the ones digit of the sum and carry a 1 to the tens place. The sum in the


tens digit is thus 1 + b + c = 1 + 18 = 19. Next, place a 9 in the tens digit of
the sum and carry a 1 to the hundreds place.
In the hundreds place, the sum is 1 + a + d = 5.
Now, solve for a.
The correct answer is (D).
86. (D): If an even number (0, 2, or 4) of the integers a, b, c, and d is negative,
each pair of negatives will cancel, because (−1)(−1) = +1 and
. This would yield a positive result for
 .
Thus, a way to rephrase the question is, “Among the integers a, b, c, and
d, are an odd number (one or three) of them negative?”
(1) SUFFICIENT: This statement can be rephrased as ad = −bc.
a d b c ad = −bc Odd number of negatives?
+ + − + ✓ Yes
+ − + + ✓ Yes


a d b c ad = −bc Odd number of negatives?
− − − + ✓ Yes
+ − − − ✓ Yes
Though the table doesn’t list all possibilities, it lists enough to realize
that, in order for the signs of ad and bc to be opposite one another,
either one or three of the four integers must be negative.
(2) SUFFICIENT: You might recognize that the (−1)(−1) = +1 property
implies that abcd is only negative when there are non-paired negatives
among the integers. That is, an odd number (one or three) of the
integers a, b, c, and d must be negative. If not, you could list a few cases
to see the pattern.
a b c d abcd = negative Odd number of negatives?
+ + − + ✓ Yes
+ − + + ✓ Yes
− + − − ✓ Yes
− − + − ✓ Yes


The correct answer is (D).
87. (A): The product xy will be even if x is even, y is even, or both are even.
The prime numbers include 2, 3, 5, 7, 11, 13, 17, 19, etc. The smallest
possible term in set A is 2, which is the only even prime.
x = the range of set A = largest term − smallest term in set A. If the
smallest term in set A is 2, then x = odd − even = odd. If the smallest term
in set A is odd (i.e., not 2), then x = odd − odd = even.
The median of set A is the average of the two middle terms, since the
number of terms in the set is even. Thus,
. However, y could be
either even (e.g., when the middle terms are 11 and 13) or odd (e.g.,
when the middle terms are 7 and 11).
A useful rephrase of this question is, “Is either x or y even?”
(1) SUFFICIENT: If the smallest prime in the set is 5, x = even, and
therefore xy is even.
(2) INSUFFICIENT: If the largest integer in the set is 101, the range of the
set can be odd or even (e.g., 101 − 3 = 98 or 101 − 2 = 99). The median of
the set can also be odd or even, as discussed. Therefore, xy can be either
odd or even.
The correct answer is (A).