88. (D) x > 4: It may be tempting to simplify this way:
The first step of this solution is wrong because you can’t divide by x
without knowing its sign. If x is negative, you would have to flip the sign.
Check both cases.
x x3 < 16x becomes: Take square root: Solve for x:
Positive x2 < 16 (don’t flip) |x| < 4 0 < x < 4
Negative x2 > 16 (flip) |x| > 4 x < −4
There are two ranges of solutions for x. The question asks for something
that is definitely NOT true. Check the answer choices for an answer that
does NOT include any values in these two ranges. All of the answer
choices include some of the values above, except for (D): if x must be
either between 0 and 4, or less than −4, it cannot be the case that x > 4.
However, it is possible, for certain values of x, for each of the other
answer choices to be true.
Alternatively, there is a way to do algebra with inequalities containing
variables. Instead of dividing by the variable, subtract it from both sides,
then factor.


In order for the product of three numbers to be negative, either all three
need to be negative, or exactly one of the three needs to be negative.
This cannot occur if x is greater than 4, since all three values would be
positive. Therefore, (D) cannot be true.
The correct answer is (D).
89. (D): Before diving into the statements, remind yourself of the divisibility
rules. One way to check for divisibility by 3 is to add the digits. If their
sum is divisible by 3, then the number itself is divisible by 3.
(1) SUFFICIENT: If m is divisible by 3, then the sum of the digits of m
must be divisible by 3 as well. Since m has the same digits as n, the sum
of the digits of n must also be a multiple of 3. Therefore, n itself is
divisible by 3.
(2) SUFFICIENT: This statement requires a bit more work because it’s not
immediately clear what would happen when you add m and n. Because
the question asks you to move digits around to diﬀerent places (the tens
place, the units place), rewrite the information using place-value
notation, where a = the tens digit of n and b = the units digit of n.


The statement indicates that the sum m + n is divisible by 3, so 11(a + b)
must also be divisible by 3. Eleven itself is not divisible by 3, so the sum
a + b must be divisible by 3. This sum represents the sum of the two
digits of the number m as well as the sum of the two digits of the
number n; if the sum of n’s two digits is divisible by 3, then n is also
divisible by 3.
The correct answer is (D).
90. (B): You can solve this problem algebraically or by Testing Cases. Both
methods are shown.
(1) INSUFFICIENT: If Printer A starts out with 100 sheets, then 20 sheets
are moved to Printer B, and Printer A ends up with 80 sheets. If B started
out with 75 sheets, then an increase of 20 sheets would be more than
25% of the starting number, and B would end up with 95 sheets. In this
case, B has more sheets than A a er the transfer.
If, instead, B started out with 40 sheets, then an increase of 20 sheets
would be more than 25% of the starting number, and B would end up
with 60 sheets. In this case, B has fewer sheets than A a er the transfer.
(2) SUFFICIENT: If Printer A starts out with 100 sheets and 20 are moved
to Printer B, then A ends up with 80 sheets. If B started out with 2,000
sheets, then an increase of 20 sheets would be less than 30% of the
starting number, and B would end up with many more sheets than A.


If B started out with 80 sheets, then a 20-sheet increase would still be
less than 30% of the starting number, and again B would end up with
more sheets than A.
How low can B go? If the 20 sheets represent a less-than-30% increase,
then use 30% as the limiting figure. If 20 represented exactly 30% of B’s
sheets, then B would have to start with
 sheets (ignore the fact
that this isn’t an integer). In this case, B would end up with
sheets, which is still greater than A’s 80 sheets. A er the transfer, B can’t
drop below A.
Algebraically, use a for the original number of sheets in Printer A and use
b for the original number of sheets in Printer B.
A er the transfer, A has a – 0.2a = 0.8a sheets. B has b + 0.2a sheets.
Rephrase the question: “Is b + 0.2a > 0.8a?” Or, “Is b > 0.6a?”
(1) INSUFFICIENT: Translate the statement into algebra.
Flip the statement around to compare it more easily to the rephrased
question: b < 0.8a. Is it true that b > 0.6a? Maybe.
(2) SUFFICIENT: Translate the statement into algebra.


Therefore,
 or
 .
The decimal equivalent of
 is
 , so b is indeed always greater than
0.6a.
The correct answer is (B).


Workout Set 10
91. If |x| ≠ |y|, xy ≠ 0,
 , and
 , then
 is
equal to which of the following?
(A)
(B)
(C)
(D)
(E)
92. If a certain culture of bacteria increases by a constant factor of x
every y minutes, how long will it take for the culture to increase to
10,000 times its original size?


(1)
In two minutes, the culture will increase to 100 times its
original size.
(2)
93. A cylinder of height h is
 full of water. When all of the water is
poured into an empty cylinder whose radius is 25% larger than that
of the original cylinder, the new cylinder is
 full. The height of the
new cylinder is what percent of h ?
  25%(A)
  50%(B)
  60%(C)
  80%(D)
100%(E)
94. If a and b are distinct positive integers, what is the units digit of
2a8b4a + b ?
b = 24 and a < 24(1)
The greatest common factor of a and b is 12.(2)
95. Employees of a certain company are each to receive a unique
seven-digit identification code consisting of the digits 0, 1, 2, 3, 4, 5,
and 6 such that no digit is used more than once in any given code.


In valid codes, the second digit in the code is exactly twice the first
digit. How many valid codes are there?
  42(A)
120(B)
210(C)
360(D)
840(E)
96. An n-sided die has sides labeled with the numbers 1 through n,
inclusive, and has an equal probability of landing on any side when
rolled. If the die is rolled twice, what is the probability of rolling 1
both times?
If the die is rolled twice, the probability that the two rolls are
diﬀerent is 80%.
(1)
If the die is rolled twice, the probability that the two rolls are
equal is
 .
(2)
97. If w, x, y, and z are positive integers and
 , what is
the proper order, increasing from le to right, of the following
quantities:
 ?


(A)
(B)
(C)
(D)
(E)
98. What is the value of |a + b| ?
(a + b + c + d) (a + b − c − d) = 16(1)
c + d = 3(2)
99. Set A consists of all the integers between 10 and 21, inclusive. Set B
consists of all the integers between 10 and 50, inclusive. If x is a
number chosen randomly from set A, y is a number chosen
randomly from set B, and y has no factor z such that 1 < z < y, what
is the probability that the product xy is divisible by 3 ?


(A)
(B)
(C)
(D)
(E)
100. At a birthday party, x children will be seated at two diﬀerent tables.
At the table with the birthday cake on it, exactly y children will be
seated, including the birthday girl, Sally. How many diﬀerent
groups of children may be seated at the birthday cake table?


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


Workout Set 10 Answer Key
91. D
92. D
93. D
94. B
95. D
96. A
97. B
98. C
99. B
 00. E


Workout Set 10 Solutions
91. (D)
 : If x = 2 and y = 3, then
 and
 .
The target number for testing the answer choices is
 .
(A)
(B)
 Eliminate.(C)
Correct.
(D)
 Greater
than 1. Eliminate.
(E)


This problem can be solved algebraically, but this path is not
recommended.
Now that the fractions have a common denominator of x, subtract one
from the other.
You are not alone if the algebraic solution was not obvious to you!
Algebraic “false starts” are common in this type of problem. In addition,
there are other, equally valid algebraic paths whose final forms would not
match any of the answers.


When you encounter a tough “pure algebra” problem that has variables in
the answer choices, picking numbers and testing the answer choices is
o en the best approach; it’s fast, easy, and correct.
The correct answer is (D).
92. (D): To understand the question stem, pick some numbers: the bacteria
culture begins with an initial quantity of I = 100 and increases by a factor
of x = 2 every y = 3 minutes. Construct a table to track the growth of the
bacteria.
Time (min.) Bacteria
3                100(2) = 100(2)1
6            100(2)(2) = 100(2)2
9       100(2)(2)(2) = 100(2)3
12 100(2)(2)(2)(2) = 100(2)4
… …
3n = t 100(2)n
n represents the number of growth periods, and
 where t is time
in minutes. For example, the fourth growth period in the table above


ended at 12 minutes, and
 .
From this example, construct a general formula for the quantity of
bacteria, F.
This question asks how long it will take for the bacteria to grow to 10,000
times their original amount. In other words, “What is t when F = 10,000I?”
The rephrased question is, “What is t when
 ?”
(1) SUFFICIENT: Note that the yth root of x is equivalent to x to the
power. This statement indicates that
 . If you plug this value
into the equation, you can solve for t (though stop the calculation at the
point that you can tell that you can solve for t).


(2) SUFFICIENT: The culture grows one-hundredfold in 2 minutes. In other
words, the sample grows by a factor of 102. Since exponential growth is
characterized by a constant factor of growth (i.e., by a factor of x every y
minutes), in another 2 minutes, the culture will grow by another factor of
102. Therefore, a er a total of 4 minutes, the culture will have grown by a
factor of 102 × 102 = 104, or 10,000.
The correct answer is (D).
93. (D) 80%: The answer choices contain percentages, and the problem never
oﬀers a real value for h. Choose a smart number.
Also note that the problem mentions the volume of the cylinder (V =
πr2h), so you’re going to need a radius. That radius needs to be increased
by 25% later in the problem, so pick a small integer that is easy to
increase by 25%, such as 4.
The problem also contains two fractions,
 and
 . Since one number is
already a multiple of 4, make the other number a multiple of 5.


Now, solve the problem using these numbers. Remember that the original
cylinder is only
 full.
The amount of water in the new cylinder has to be the same 60π.
Remember that it is only
 full.
The new height is 4. The original height was 5, so the new height is
 , or
80%, of the original height.
The correct answer is (D).
94. (B): This problem contains a common trap seen in many diﬀicult DS
questions. Answer choice (C) is a tempting short-cut answer as the


combined statements would provide the values of both a and b, which
could be plugged into the expression to answer the question.
Rephrase the question:
Remembering the units digit patterns for powers of 2 will help on this
problem:
The units digits for powers of 2 is a repeating pattern of [2, 4, 8, 6].
If you can determine the relationship of 3a + 5b to a multiple of 4 (i.e.,
where 23a + 5b is in the predictable four-term repeating pattern of units
digits), you will be able to answer the question. This question can be
rephrased as, “What is the remainder when 3a + 5b is divided by 4?”


(1) INSUFFICIENT: If b = 24, then 5b is a multiple of 4. However, a could be
any integer less than 24. Possible remainders when 3a is divided by 4 are
0, 1, 2, or 3.
(2) SUFFICIENT: If the greatest common factor of a and b is 12, then 12
must be a factor of both variables. That is, both a and b are multiples of 12
and thus also multiples of 4. As a result, 3a and 5b will be multiples of 4 as
well, so the remainder will be 0 when 3a + 5b is divided by 4.
The correct answer is (B).
95. (D) 360: Valid codes must have a second digit that is exactly twice the first
digit. There are three ways to do this with the available digits.
Scenario A: 12XXXXX
Scenario B: 24XXXXX
Scenario C: 36XXXXX
For each of these basic scenarios, there are 5! ways to shuﬀle the
remaining five numbers (represented by X’s above).
Thus, the total number of valid codes is 3 × 5! = 3 × 120 = 360.
The correct answer is (D).
96. (A): When an n-sided die is rolled once, the probability of rolling any
particular number (such as 1) is equal to
 . Therefore, the probability of


rolling 1 both times is equal to
 . Rephrase the
question as, “What is the value of n ?”
(1) SUFFICIENT: The probability that the two rolls are diﬀerent is 80%, or
.
When rolling an n-sided die twice, the first roll can have any value from 1
to n, inclusive. Regardless of the value of the first roll, the probability is
 that the second roll will have that same value. Therefore, the overall
probability of rolling the same value twice is
 , and the probability of
rolling two diﬀerent values is
 .
Therefore, according to this statement,
 . The value of n is
5, and this statement is suﬀicient.
(2) INSUFFICIENT: The probability that two rolls of an n-sided die will
have the same value is
 , regardless of the value of n. This statement
would be true for any value of n, so the value of n cannot be determined
and this statement is insuﬀicient.
The correct answer is (A).


97. (B)
 : It would require a lot of tricky
work to solve this algebraically, so test cases instead. Make sure to pick
values for the unknowns such that
 holds true. For
example, if w = 1, x = 2, y = 3, and z = 4, then
 is true.
Before plugging those values in for the quantities, check the answers. All
of them begin with 1 and
 , so don’t bother to test
 . In addition, check
the answer choices a er each term that you evaluate.

Now, place in ascending order:
 .
The third term is
 , so cross oﬀ answers (A), (C), and (D). The
fourth term is
 , so cross oﬀ answer (E).
The correct answer is (B).


98. (C): The question asks for the absolute value of a + b, so try to manipulate
the statements to isolate that combination of variables, a + b. Statement
(2) looks much easier, so start there.
(2) INSUFFICIENT: This provides information about c and d, and the
relationship between them, but no information about a or b.
(1) INSUFFICIENT: Manipulate the equation to group (a + b) and (c + d)
terms.
Note that this is of the form (x + y)(x − y), where x = (a + b) and y = (c + d).
This is the diﬀerence of squares special product, (x + y)(x − y) = x2 − y2. Use
this to transform this expression.
This is not enough to determine the value of (a + b)2.
(1) AND (2) SUFFICIENT: From statement (2), (a + b)2 = 16 + (c + d)2. From
statement (1), c + d = 3. You can substitute for c + d and solve. The
solution is shown below, but note that you don’t have to do that math:
The question asked for the absolute value of a + b, so the fact that the
equation has (a + b)2 doesn’t matter.


The correct answer is (C).
99. (B)
 : If y has no factor z such that 1 < z < y, then y must be prime.
Examine a few examples to see why this is true.
Because y is selected from set B, it is a prime number between 10 and 50,
inclusive. The only prime number that is divisible by 3 is 3, so y is
definitely not divisible by 3.
Thus, xy is only divisible by 3 if x itself is divisible by 3. Rephrase the
question: “What is the probability that a multiple of 3 will be chosen
randomly from set A?”
There are 21 − 10 + 1 = 12 terms in set A. Of these, 4 terms (12, 15, 18, and
21) are divisible by 3.
6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime.
15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime.
3 has NO factor between 1 and 3: 3 IS prime.
7 has NO factor between 1 and 7: 7 IS prime.


Thus, the probability that x is divisible by 3 is
 .
The correct answer is (B).
 00. (E)
 : This problem contains variables in the answer
choices, so Choose Smart Numbers.
There are 8 children at the party, and 5 will sit at the table with the cake.
Sally must sit at the birthday cake table, so pick 5 − 1 = 4 of the other 8 − 1
= 7 children to sit at that table with her. How many diﬀerent ways can you
choose 4 from a group of 7? Set up an anagram grid, where Y means “at
the cake table” and N means “at the other table.”
A B C D E F G
Y Y Y Y N N N
Now, you can calculate the number of possible groups:
Note that the answer choices are in factorial form; it may be the case that
you will find the unsimplified factorial form, rather than 35. Test each
answer choice by plugging in x = 8 and y = 5.


As an alternative to testing all five choices, you could use a hybrid
approach to determine the formula using variables.
The number of possible groups was
 , but remember that this
formula took Sally into account.
The 3 came from the diﬀerence between these numbers: 7 − 4 = (x − 1) − (y
− 1) = (x − y).
(A)
(B)
(C)
(D)
(E)


Substitute these variable expressions in place of the numbers.
The correct answer is (E).


Workout Set 11
101. A casino pays players with chips that are either turquoise or violet
colored. If each turquoise-colored chip is worth t dollars and each
violet-colored chip is worth v dollars, where t and v are integers,
what is the combined value of four turquoise-colored chips and two
violet-colored chips?
The combined value of six turquoise-colored chips and three
violet-colored chips is $42.
(1)
The combined value of five turquoise-colored chips and seven
violet-colored chips is $53.
(2)
102. Jonas and Amanda stand at opposite ends of a straight road and
start running toward each other at the same moment. Their rates
are randomly selected in advance so that Jonas runs at a constant
rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant
rate of 4, 5, 6, or 7 miles per hour. What is the probability that Jonas
has traveled farther than Amanda by the time they meet?


(A)
(B)
(C)
(D)
(E)
103. If p is a positive integer, is p2 divisible by 96 ?
p is a multiple of 8.(1)
p2 is a multiple of 12.(2)


104.
In the figure above, the trapezoid ABCD is inscribed in a circle.
Parallel sides AB and CD are 7 inches apart and 6 and 8 inches long,
respectively. What is the radius of the circle in inches?
4(A)
5(B)
7(C)
(D)
(E)
105. If 9y + 3b = 10(3b), then 2y is equal to which of the following?


b − 2(A)
b − 1(B)
b(C)
b + 1(D)
b + 2(E)
106. The radius of a circle is r yards. Is the area of the circle at least r
square yards? (1 yard = 3 feet)
The diameter of the circle is more than 2 feet.(1)
If the radius of the same circle is f feet, the area of the circle is
more than 2f square feet.
(2)
107. For how many values of x from 1 to 300, inclusive, is the sum of the
integers from 1 to x, inclusive, divisible by 3 ?
  30(A)
  99(B)
100(C)
199(D)
200(E)
108. If x and y are integers and |xy| = 8, what is the value of |x + y| ?
x and y are both divisible by 2.(1)


xy > 0(2)
109. Set X consists of exactly four distinct integers that are greater than
1. For each integer in the set, all of that integer’s unique prime
factors are also in the set. For example, if 20 is in set X, then 2 and 5
must also be in set X. How many of the distinct integers in set X are
prime?
The product of the integers in set X is divisible by 36.(1)
The product of the integers in set X is divisible by 60.(2)
110. If p = (22)(3x) and r = (22)(3y), where x and y are prime numbers and
x ≠ y, which of the following represents the least common multiple
of p and r ?
12xy(A)
6xy(B)
xy(C)
12(D)
6(E)


Workout Set 11 Answer Key
 01. D
 02. A
 03. C
 04. B
 05. E
 06. A
 07. E
 08. C
 09. B
 10. A


Workout Set 11 Solutions
 01. (D): The question asks for the value of 4t + 2v, where t and v represent the values
of the turquoise and violet chips, respectively. Note that the question asks for a
combination of variables, or a combo; it may not be necessary to be able to solve
for the individual values of t and v.
(1) SUFFICIENT: Translate and simplify the statement.
You can’t solve for t and v, so this statement might look suﬀicient—but remember
what the question is asking! You need to find the value of 4t + 2v. Multiply the
equation by 2: 4t + 2v = 28.
(2) SUFFICIENT: Translate the statement.
There isn’t a way to simplify this one, but notice all those prime numbers. Primes
tend to minimize the number of allowable scenarios, especially when the
question also specifies that the variables have to be integers. See what
combinations of integers would actually work here.
The 5t term could be 5, 10, 15, 20, and so on. This term can contribute only
numbers that end in 5 or 0. If that’s the case, the 7v term must have a units digit
of either 3 or 8. List out the possibilities for 7v, and try only the ones that end in 3
or 8.


7v: 7, 14, 21, 28, 35, 42, 49
Only one ends in 3 or 8! If 7v = 28, then v = 4 and the 5t term must equal 25, so t =
5. This one scenario works and it is the only possible scenario for this equation,
given that v and t must be integers.
The correct answer is (D).
 02. (A)
 : If Jonas and Amanda run at the same rate, they will meet each other
exactly in the middle. Jonas will only run farther than Amanda if Jonas’ rate is
greater than Amanda’s. In math terms, Distance = rate × time, and since Jonas
and Amanda run for the same time, their relative distances depend solely on
their relative rates. Rephrase the question as, “What is the probability that Jonas
ran faster than Amanda?”
There are four possible rates for Jonas (3, 4, 5, and 6) and four for Amanda (4, 5, 6,
and 7). In total, there are (4)(4) = 16 possible rate scenarios.
Of Jonas’ four possible rates, only two (5 and 6) are greater than some of
Amanda’s possible rates (4 and 5). List the three rate scenarios that result in a
faster speed (greater distance) for Jonas.
Since there are 16 possible combinations of rates, the probability that Jonas ran
farther than Amanda is
 .
The correct answer is (A).
Jonas ran 5 miles per hour (mph), and Amanda ran 4 mph.
Jonas ran 6 mph, and Amanda ran 4 mph.
Jonas ran 6 mph, and Amanda ran 5 mph.


03. (C): The prime factorization of 96 is (2)(2)(2)(2)(2)(3) = (25)(31). In order for p2 to be
divisible by 96, p2 would have to have the prime factors 2531 in its prime box. The
rephrased question is therefore, “Does p2 have at least five 2’s and one 3 in its
prime box?”
(1) INSUFFICIENT: If p is a multiple of 8 = (2)(2)(2), p has 23 in its prime box.
Therefore, p2 has (23)2 = 26 in its prime box, and thus has the required five 2’s.
However, it is uncertain whether p2 has at least one 3 in its prime box.
(2) INSUFFICIENT: If p2 is a multiple of 12 = (2)(2)(3), p2 has two 2’s and one 3 in its
prime box. It is uncertain whether p2 has at least five 2’s total as there may or may
not be three more 2’s in the prime box.
(1) AND (2) SUFFICIENT: Statement (1) indicates that p2 has 26 in its prime box
and statement (2) indicates that p2 has a 3 in its prime box. Therefore, it is certain
that p2 has at least five 2’s and one 3 in its prime box.
The correct answer is (C).
 04. (B) 5: Redraw the figure as closely to scale as possible (remember that the oﬀicial
scrap paper is graph paper!), labeling the known dimensions and the radius in
question.


In order for the trapezoid vertices to lie on the circle, the trapezoid must be
symmetrical about the dotted line, which passes through the center of the circle.
Draw this vertical and the radii to points B and C to create two right triangles,
allowing you to use the Pythagorean theorem.
In fact, you might play an educated hunch that the triangles are 3–4–5 common
right triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3
and 4 side. The unknown vertical sides are thus 4 and 3, which sum to 7 as they
must.
Algebraically, set up the following equations from the picture.
Setting the two equations for r2 equal.


Since (x + y)(x − y) = 7, (x − y) = 1.
Solve for x and y.
The radius of the circle is 5, because r2 = 32 + 42 = 25.
The correct answer is (B).
 05. (E)
 : First, simplify the given equation.
The correct answer is (E).
 06. (A): First, translate the question stem: Is πr2 ≥ r ? Simplify.


Note that the question is stated in yards, but the statements use feet, so convert
the question stem to feet.
(1) SUFFICIENT: D > 2 feet. Therefore, r > 1 foot. The value of
 is approximately
, or a little bit less than 1. Yes,
 .
(2) INSUFFICIENT: According to this statement, πf2 > 2f. Simplify.
In other words,
 or approximately
 . Therefore, f could be smaller
than
 feet (just a bit smaller than 1) but it could also be larger.
The correct answer is (A).
 07. (E) 200: Start at x = 1, then x = 2, and so on, looking for a pattern.


x Sum of the integers from 1 to x, inclusive Is the sum divisible by 3 ?x Sum of the integers from 1 to x, inclusive Is the sum divisible by 3 ?
1 1 No
2 3 Yes
3 6 Yes
4 10 No
5 15 Yes
6 21 Yes
7 28 No
It appears that the pattern is No, Yes, Yes, No, Yes, Yes, etc., repeating every three
terms. Therefore, 2 out of every 3 values of x yield a sum that is divisible by 3. So
200 of the 300 values of x will yield a sum that is divisible by 3. The answer is 200.
The correct answer is (E).
 08. (C): If the two variables must be integers and the absolute value of their product
is 8, then x and y have to represent either the factor pair (1, 8) or the factor pair (2,
4), in either order and with either sign (positive or negative). This narrows the
possible number of cases considerably.
(1) INSUFFICIENT: List out the possible cases allowed by this statement. If x and y
are both divisible by 2, then you’re dealing with the factor pair (2, 4). A er you try
your first case, ask yourself what diﬀerent case would be most likely to return a
diﬀerent answer to the question.
x y |xy| = 8 |x + y| = ?
2 4 ✓ 6


x y |xy| = 8 |x + y| = ?
−2 4 ✓ 2
In this case, just reversing the numbers (4, 2) won’t make a diﬀerence to the final
question, so don’t list that case second. Try one of the cases that allows a
negative value. Now that you have two diﬀerent values, you know the statement
is not suﬀicient.
(2) INSUFFICIENT: Follow the same process. This statement indicates that the two
values have the same sign: both positive or both negative.
x y |xy| = 8 |x + y| = ?
1 8 ✓ 9
2 4 ✓ 6
Again, list cases until you have two contradictory results (or until you’ve tried all
cases and realized that there is only one possible answer).
(1) AND (2) SUFFICIENT:
x y Both divisible by 2?
Same signs? |xy| = 8 |x + y| = ?
2 4 ✓
✓
✓ 6


x y Both divisible by 2?
Same signs? |xy| = 8 |x + y| = ?
−2 −4 ✓
✓
✓ 6
Only four possible cases are allowed: the variables must match the (2, 4) factor
pair and the signs have to be the same. The chart above shows two of the four
possible cases, but you could also reverse the order: (4, 2) and (−4, −2). Since
you’re just adding at the end, though, the order of the two variables doesn’t
matter. In all four cases, |x + y| = 6.
The correct answer is (C).
 09. (B): Set X contains exactly four distinct integers; distinct means the four integers
must be diﬀerent. Also, if a number is contained in the set, all of its unique prime
factors must also be in the set. If a prime number, such as 3, is in the set, no
additional numbers are required in the set: the only prime factor of 3 is 3 itself,
which is already included. However, if the composite number 8 is in the set, its
one unique prime factor, 2, must also be included.
To better understand the question, jot down some four-integer sets that fit the
description. Specifically, look for sets that have diﬀerent numbers of primes.
{2, 3, 5, 7} is a valid set that contains four primes.
{2, 4, 8, 16} is a valid set that contains only one prime, since all of the terms are
powers of 2.
(1) INSUFFICIENT: The question deals with primes and divisibility; start by
factoring 36.


The prime factorization of 36 is 36 = 2 × 2 × 3 × 3. For the product of the four
integers in the set to be divisible by 36, the four integers must include 2, 2, 3, and
3 as prime factors, in some combination. Therefore, based on the definition of
the set, 2 and 3 must both be included in the set. The remaining two numbers
must include at least one 2 and at least one 3 as prime factors between them.
However, there are several diﬀerent valid sets that fit this criterion.
Case 1: {2, 3, 6, 9} has a product divisible by 36, since it includes at least two 2’s
and at least two 3’s in its prime factors. Also, it is a valid set since it includes all of
the unique prime factors of 6 (2 and 3) and all of the unique prime factors of 9
(only 3). This set has two primes.
Case 2: The primes 2 and 3 must be included in the set; include a third prime as
well, such as 5. The final number in the set must include 2 and 3 as prime factors
in order for the product of the whole set to be divisible by 36. {2, 3, 5, 6} is one
example of a valid set that contains three primes and has a product divisible by
36. This set has three primes.
Because the set might have either two primes or three primes, the statement is
insuﬀicient.
(2) SUFFICIENT: Based on similar reasoning to that from statement (1), note that
2, 3, and 5 must be included in the set. The product of 2, 3, and 5 is 30, so the
product of the set will definitely be divisible by 30.
In order for the product to be divisible by 60, there must be at least one
additional multiple of 2 in the set. Because this number is a multiple of 2, and
because 2 itself is already included in the set, it is not a prime. Therefore, the only
primes in the set are 2, 3, and 5, and the answer to the question is 3.
The correct answer is (B).


10. (A) 12xy: p and r share three prime factors: 2, 2, and 3. Each also has an additional
prime factor that is not shared by the other. Draw overlapping circles in which to
place the shared and non-shared prime factors of p and r. To find the least
common multiple (LCM), multiply from le to right and include all the common
factors in the product.
The correct answer is (A).


Workout Set 12
111.
Which of the following equations represents a line parallel to line ℓ in the
figure above?
2y − 3x = 0(A)
2y + 3x = 0(B)
2y − 3x = 6(C)
3y + 2x = 6(D)
3y − 2x = 9(E)
112. If y = |x − 1| and y = 3x + 3, then x must be between which of the following
values?


2 and 3(A)
1 and 2(B)
0 and 1(C)
−1 and 0(D)
−2 and −1(E)
113. If a is a positive integer, is a2 a multiple of 8 ?
a3 is a multiple of 16.(1)
(a + 4)2 is a multiple of 8. (2)
114.
A road crew painted two black lines across a road, as shown in the figure
above, to mark the start and end of a one-mile stretch. Between the two
black lines, they will paint across the road a red line at each third of a
mile, a white line at each fi h of a mile, and a blue line at each eighth of a
mile. What is the smallest distance (in miles) between any of the painted
lines on this stretch of highway?


0(A)
(B)
(C)
(D)
(E)
115. Set S consists of n consecutive integers, where n > 1. What is the value of
n ?
The sum of the integers in set S is divisible by 7.(1)
The sum of the integers in set S is 14.(2)
116. A trapezoid is symmetrical about a vertical center line. If a circle is drawn
such that it is tangent to exactly three points on the trapezoid and is
enclosed entirely within the trapezoid, what is the diameter of the circle?
The parallel sides of the trapezoid are 10 inches apart.(1)
Of the parallel sides of the trapezoid, the shorter side is 15 inches
long.
(2)


117. Three boys are ages 4, 6, and 7, respectively. Three girls are ages 5, 8, and
9, respectively. If two of the boys and two of the girls are randomly
selected and the sum of the selected children’s ages is z, what is the
diﬀerence between the probability that z is even and the probability that
z is odd?
(A)
(B)
(C)
(D)
(E)
118. If two distinct integers x and y are randomly selected from the integers
between 1 and 10, inclusive, what is the probability that
  is a
positive odd integer?


(A)
(B)
(C)
(D)
(E)
119. Positive integers a and b are less than or equal to 9. If a and b are
assembled into the six-digit number ababab, which of the following must
be a factor of ababab ?
3(A)
4(B)
5(C)
6(D)
None of the above(E)
120. The two-digit positive integer s is the sum of the two-digit positive
integers m and n. Is the units digit of s less than the units digit of m ?
The units digit of s is less than the units digit of n.(1)


The tens digit of s is not equal to the sum of the tens digits of m and
n.
(2)


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


Workout Set 12 Solutions
 11. (B) 2y + 3x = 0: Parallel lines have the same slope. The slope of line ℓ is
. Work backwards from the
answers by rearranging into slope-intercept form:
 form,
where the slope is m. Since only one answer can be correct, only one
answer will have a slope of
 ; stop when you find the one with the
matching slope.
(A) 2y = 3x
(B) 2y = − 3x
Alternatively, use the slope to write the equation, manipulate, and look for
a match.
Only answer (B) is a match for the x and y portions.
The correct answer is (B).


12. (D) −1 and 0: Set the two equations for y equal and algebraically solve |x − 1|
= 3x + 3 for x. This requires two solutions: one for the case that x − 1 is
positive, the other for the case that x − 1 is negative.
x – 1 is positive: x – 1 is negative:
In the x − 1 is positive case, the answer is x = −2. In this case, x − 1 is not
actually positive, so this is a false case. Discard it. As a result, there is only
one solution:
 .
The correct answer is (D).
 13. (D): In order for a2 to be a multiple of 8, the prime factorization of a2 must
include all of the prime factors of 8. That is, to be a multiple of 8, a2 would
have to include 2, 2, and 2 among its prime factors.
When an integer is squared, all of its prime factors are duplicated.
Therefore, if a only includes a single 2 among its prime factors, a2 might
only include two 2’s. In order to guarantee that a2 includes at least three 2’s,
a must include at least two 2’s in its prime factors.
The question translates to: “Does the prime factorization of a include at
least two 2’s?” Or, “Is a divisible by 4?”


(1) SUFFICIENT: The positive integer a, when cubed, is a multiple of 16. Odd
integers, when cubed, will always yield an odd result, so a cannot be odd.
What positive even values could a have?
If a = 2, then a3 = 8, which is not a multiple of 16. Neither is 63 or 103.
However, if a = 4, then a3 = 64, which is a multiple of 16, as are 83 = (43)(23)
and 123 = (43)(33). Therefore, this statement indicates that a is a multiple of
4. When a multiple of 4 is squared, the result is always a multiple of 8, so
the answer is definitely Yes.
(2) SUFFICIENT: Start by multiplying out the expression in the statement.
(a + 4)2 is a multiple of 8.
a2 + 8a + 16 is a multiple of 8.
a2 + 8(a + 2) is a multiple of 8.
The second part of the expression, 8(a + 2), is a multiple of 8 for any value of
integer a. In order for the entire expression to be a multiple of 8, the first
part of the expression must be a multiple of 8 as well. So according to this
statement, a2 is a multiple of 8. The answer is a definite Yes, so this
statement is suﬀicient.
The correct answer is (D).
 14. (D)
 : When comparing fractional pieces of a whole, find a common
denominator. In this case, the one-mile stretch is divided into thirds, fi hs,
and eighths. The smallest common denominator of 3, 5, and 8 is 120. If the


one-mile highway is divided into 120 equal increments, where will the red,
white, and blue marks fall?
Red (thirds): 40, 80 (out of 120 increments)
White (fi hs): 24, 48, 72, 96 (out of 120 increments)
Blue (eighths): 15, 30, 45, 60, 75, 90, 105 (out of 120 increments)
The smallest distance between two marks is 75 − 72 = 3 or 48 − 45 = 3. This
equates to
 , or
 miles.
The correct answer is (D).
 15. (E): Both statements provide information about the sum of the set, so
rephrase with this in mind.
For n = odd, the median is the middle term, an integer. For n = even, the
median is the average of the two middle terms, a non-integer of the form
“integer + 0.5.” You can determine n if you can determine both the median
of set S and the sum of the integers in set S.
Next, glance at the statements. Statement (2) is actually a subset of
statement (1): if the sum is exactly 14, then the sum is divisible by 7. As a
result, the answer can’t be (A), because if statement (1) is suﬀicient, then
statement (2) would also have to be suﬀicient. The answer also can’t be (C),
because statement (2) doesn’t add new information to statement (1). Cross
oﬀ answers (A) and (C) and start with statement (2).
Sum of consecutive set = (Median)(Number of terms)
Sum of consecutive set = (Median)(n)


(2) INSUFFICIENT: Since n must be an integer, use divisibility rules to
narrow down possible median values.
Check some possible median and n values (remember that n > 1).
There are at least two possible values for n.
(1) INSUFFICIENT: The two possible cases for statement (2) also apply to
statement (1), since the numbers chosen for statement (2) must also work
in statement (1).
(1) AND (2) INSUFFICIENT: As noted, the final two cases tested for statement
(2) are also allowed by statement (1), so there are still at least two possible
values for n.
The correct answer is (E).
n = 2 and median = 7: Set S can’t have an integer median if there are only
two terms. Ignore.
n = 7 and median = 2: Set S is {−1, 0, 1, 2, 3, 4, 5}, which has a sum of 14.
Okay.
n = 4 and median = 3.5: Set S is {2, 3, 4, 5}, which has a sum of 14. Okay.
 16. (C): There are four basic ways this picture could look as there are four sides
of the trapezoid that could serve as the non-tangent side.


(1) INSUFFICIENT: If the circle is tangent to both of the parallel sides (Figure
A or B), then the diameter must be 10. If the circle is tangent to only one of
the parallel sides (Figure C or D), then the diameter is less than 10. Since
there are multiple possibilities for the diameter of the circle, statement (1)
does not contain enough information to answer the original question.
(2) INSUFFICIENT: Just knowing the length of the shorter parallel side is not
enough to determine which of the basic figures above describes the correct
situation. If Figure A or B represents the correct situation, the diameter of
the circle is determined solely by the distance between the parallel sides;
the diameter is independent of the length of the shorter parallel side. If
Figure C or D describes the correct situation, then the diameter would
depend on not only the 15-inch side but also the longer parallel side, which
has an unknown length.
(1) AND (2) SUFFICIENT: In Figures C and D, the diameter must be less than
10 but also greater than the length of the smaller parallel side. If the length
of the smaller parallel side is 15, then the diameter would have to be


greater than 15 for Figures C and D, but this is impossible (since the
diameter has to be less than 10). Therefore, Figure A or B represents the
correct situation, and the diameter of the circle must equal 10.
Alternatively, notice that Figure D could not represent the situation since
the circle would have to have a diameter larger than 15 inches in order to
be tangent to the short parallel side and the non-parallel sides of the
trapezoid. The parallel sides of the trapezoid are only 10 inches apart, so
the circle would be too large to be drawn entirely within the trapezoid as
required. Similar logic explains why Figure C is also impossible when
considering (1) and (2) together.
The correct answer is (C).
 17. (A)
 : List the possible cases. Wherever possible, avoid computing; use
Odd & Even principles to reduce your computation.
Boys Girls Sum z
4, 6 = Even 5, 8 = Odd E + O = O
4, 6 = Even   5, 9 = Even  E + E = E
4, 6 = Even 8, 9 = Odd  E + O = O
4, 7 = Odd 5, 8 = Odd  O + O = E
4, 7 = Odd 5, 9 = Even  O + E = O


Boys Girls Sum z
4, 7 = Odd 8, 9 = Odd  O + O = E
6, 7 = Odd 5, 8 = Odd  O + O = E
6, 7 = Odd 5, 9 = Even  O + E = O
6, 7 = Odd 8, 9 = Odd  O + O = E
Of the nine scenarios listed, five yield an even z and four yield an odd z.
The diﬀerence between the probability that z is even and the probability
that z is odd is therefore
 .
The correct answer is (A).
 18. (A)
 : This problem involves both probability and special quadratics. In
order to find the probability that
 is a positive odd integer, you
must find the number of values of x and y for which this is the case. The
problem is really asking you to count the values of x and y for which
 is a positive odd integer.
Because x2 — y2 is a special quadratic, the expression can be simplified to
.


This expression is an integer when (x — y)(x + y) is divisible by 4. Note that
x – y and x + y are always either both even or both odd. Whenever x – y and
x + y are both even, their product will be divisible by 4. In other words,
when x and y are either both even or both odd,
 is an
integer. But when x and y are diﬀerent in parity (one even and the other
odd),
 is not an integer.
The expression also needs to be positive. This will occur whenever x is
greater than y. Otherwise, x – y would not be positive and the entire
expression would be negative or zero.
Finally, when is
 an ODD integer? Test a few cases to
determine when this occurs. Based on the reasoning above, you only need
to test cases in which x is greater than y and in which x and y are either both
even or both odd.
x y
 Odd or even?
5 3
 Even
7 3
 Even


x y
 Odd or even?
9 3
 Even
3 1
 Even
7 1
 Even
It appears that whenever x and y are both odd,
 will be
even. Therefore, x and y cannot both be odd. Test some cases in which x
and y are both even:
x y
 Odd or even?
4 2
 Odd
6 2
 Even


x y
 Odd or even?
8 2
 Odd
10 2
 Even
6 4
 Odd
8 4
 Even
If x and y are both even, then whether the case is valid depends on whether
x and y diﬀer by a multiple of 4. Whenever x – y is a multiple of 4 in the table
above, so is x + y, and vice versa. Therefore, whenever x – y is a multiple of
4,
 is an even integer.
In summary:
x and y must both be even.
x must be greater than y.
x – y must NOT be a multiple of 4.


That leaves only a few cases to count:
x y
10 8
10 4
8 6
8 2
6 4
4 2
There are six valid cases. In total, there are (10)(9) = 90 ways to choose two
distinct values for x and y from the integers between 1 and 10, inclusive. So
the probability is
 .
The correct answer is (A).
 19. (A) 3: This problem is most eﬀiciently solved by Working Backwards from
the answers.
3: The sum of the digits of ababab is 3(a + b). This must be a multiple of
3. On the test, stop here and select answer (A). Below, you’ll find
explanations for the other answers.
(A)
4: An integer is divisible by 4 if its last two digits represent a two-digit
number that is itself divisible by 4. It is uncertain whether the two-digit
integer ab is divisible by 4.
(B)
5: An integer is divisible by 5 if the last digit is 0 or 5. It is uncertain
whether the positive integer b is 5.
(C)


The correct answer is (A).
6: An integer is divisible by 6 if it is even and divisible by 3. Answer (A)
established that ababab is divisible by 3, but it is uncertain whether the
last digit b is even, a requirement for ababab to be even. Note that you
can also use logic to eliminate this answer. If ababab were divisible by 6,
it would also have to be divisible by 3, but that would lead to two correct
answers!
(D)
 20. (D): Try out a few randomly chosen numbers to help you understand the
question.
If m = 10 and n = 10, then s = 20 and the units digits of m and s are
equal.
If m = 11 and n = 12, then s = 23 and the units digit of s is greater
than the units digit of m.
If m = 19 and n = 11, then s = 30 and the units digit of s is less than
the units digit of m.
What’s going on with all of these cases? Basically, if the units digits of m and
n are small enough, then the units digit of s will be equal to or greater than
the units digit of m (and of n).
On the other hand, if the units digits of m and n are large enough to cause
you to carry over a 1 to the tens digit, then the units digit of s will end up
being smaller than the units digit of m (and of n).
(1) SUFFICIENT: If the units digit of s is definitely less than the units digits of
one of the smaller numbers, then the carryover situation must apply, in


which case the units digits of both m and n must be larger than the units
digit of s.
(2) SUFFICIENT: There are only two ways in which the tens digit will not
equal the tens digits of the two smaller numbers:
Case 1: The tens digits of the two smaller numbers result in a number that
needs to carry over into the hundreds digit. This is impossible for this
problem because s is also a two-digit number.
Case 2: The units digits of the two smaller numbers result in a number that
carries over into the tens digit. This must be what is happening in this case.
If so, then the units digit of the larger number, s, must be smaller than the
units digits of the two smaller numbers, m and n.
The correct answer is (D).


Workout Set 13
121. If x and y are positive integers such that x2 – y2 = 48, how many
diﬀerent values of y are possible?
Two(A)
Three(B)
Four(C)
Five(D)
Six(E)
122. If the number 200! is written in the form p × 10q, where p and q are
integers, what is the maximum possible value of q ?
40(A)
48(B)
49(C)
55(D)
64(E)
123. If x and y are integers, and x ≠ 0, what is the value of xy ?
|x| = 2(1)


64x62x + y = 482x(2)
124. The three sides of a triangle have lengths p, q, and r, each an
integer. Is this triangle a right triangle?
The perimeter of the triangle is an odd integer.(1)
If the triangle’s area is doubled, the result is not an integer.(2)
125. If x is positive, what is the least possible value of
 ?
(A)
1(B)
2(C)
3(D)
4(E)
126. The average (arithmetic mean) cost of three computer models is
$900. If no two computers cost the same amount, does the most
expensive model cost more than $1,000 ?


The most expensive model costs 25% more than the model
with the median cost.
(1)
The most expensive model costs $210 more than the model
with the median cost.
(2)
127.
Which of the following equations represents a line perpendicular to
line k in the figure above?
3y + 2x = −12(A)
2y + x = 0(B)
2y − x = 0(C)
y + 2x = 12(D)
y − 2x = 12(E)
128. Is x > 0 ?
|2x − 12| < 10(1)


x2 − 10x ≥ −21(2)
129. K-numbers are positive integers with only 2’s as their digits. For
example, 2, 22, and 222 are K-numbers. The K-weight of a number n
is the minimum number of K-numbers that must be added together
to equal n.For example, the K-weight of 50 is 5, because 50 = 22 + 22
+ 2 + 2 + 2. What is the K-weight of 600 ?
10(A)
11(B)
12(C)
13(D)
14(E)
130. If the reciprocals of two consecutive positive integers are added
together, what is the sum in terms of the greater integer x ?


(A)
x2 − x(B)
2x − 1(C)
(D)
(E)


Workout Set 13 Answer Key
 21. B
 22. C
 23. B
 24. D
 25. C
 26. D
 27. D
 28. A
 29. A
 30. E


Workout Set 13 Solutions
 21. (B) Three: The choices themselves, as well as how many in the question text, suggest
that y could have more than one value. To begin, simplify the given information using
algebra and logical reasoning.
Since x and y are integers, both x – y and x + y are integers. The product of two integers
is 48. One of those integers is equal to x + y and the other is equal to x – y. Since x2 – y2
is positive and x and y are both positive, x must be greater than y. Therefore, x + y and
x – y are a pair of positive factors of 48, with x + y being the greater of the two factors.
List the pairs of factors of 48 and solve for the values of x and y.
Note one thing first. Since the factors are y above and y below x, the value of x equals
the average. Therefore, if for a particular case the average of (x – y) and (x + y) is not an
integer, that case is not valid. In order for the average of the two terms to be an integer,
the sum has to be even, so if the sum is odd, eliminate that possibility immediately.
x – y x + y x y
1 48 n/a (sum not even)
2 24 13 11
3 16 n/a (sum not even)
4 12   8   4
6   8   7   1


Of the five factor pairs, three yield integer values for x and y. Double-check that they fit
the equation.
The correct answer is (B).
 22. (C) 49: To maximize the value of q, you would need to constrain p to NOT be a multiple
of 10. In other words, q should count every factor of 10 in 200!, and p would be the
product of the remaining factors of 200!.
To count factors of 10 in 200!, you could start by counting the multiples of 10 between
1 and 200, inclusive. But this method would undercount the number of 10’s that are
factors of 200! because it would miss other pairs that can create 10, such as:
Therefore, this problem is really about counting the number of 2 × 5 factor pairs that
can be made from the factors of 200.
Of 2 and 5, there will be fewer 5’s overall, so count the number of 5’s found among the
factors of 200!.
Number of Multiples between 1 and 200, Inclusive
Multiples of 5 (= 51)
Multiples of 25 (= 52)
200! has 2 and 5 as factors, which multiply to 10, creating another factor of 10.
200! has 6 and 15 as factors, which multiply to 90, creating another factor of 10.


Number of Multiples between 1 and 200, Inclusive
Multiples of 125 (= 53) 125 is the only one: 1
The higher multiples contribute more than one factor of 5 to the total.
There are eight multiples of 25 in the range (namely, 25, 50, 75, 100, 125, 150, 175, and
200). Each of these is also a multiple of 5, so each has already been counted once, and
thus each of the eight multiples of 25 contributes one additional factor of 5. Finally,
125 contributes a total of three 5’s to the count—but two have already been counted,
leaving only one additional factor of 5 to include.
Therefore, the prime factorization for 200! includes 40 + 8 + 1 = 49 factors of 5.
The correct answer is (C).
 23. (B): Since x ≠ 0, x y does not equal 0 y or 0.
(1) INSUFFICIENT: This statement indicates that x is equal to 2 or −2. The statement
indicates nothing about y.
(2) SUFFICIENT: Simplify using exponent rules, noting the common factors of 6 and 8
on each side of the equation.
Since y = 0 and x ≠ 0 (as stated in the question stem), this information is suﬀicient to
conclude that x y = x0 = 1.


The correct answer is (B).
 24. (D): The question stem specifies that the triangle’s sides all have integer lengths.
(1) SUFFICIENT: Perimeter is calculated by summing the three side lengths. In order for
that sum to be odd, either all three numbers have to be odd or one of the three
numbers has to be odd.
If the triangle is a right triangle, then the Pythagorean theorem (a2 + b2 = c2) holds. If
all three side lengths are odd, this would mean odd2 + odd2 = odd2. This is impossible,
however; odd + odd = even.
If one side length is odd and the other two are even, then there are two possibilities:
Every right triangle possibility is impossible, so the triangle cannot be a right triangle.
(2) SUFFICIENT: A =
 bh, or 2A = bh. If, as this statement indicates, 2A is not an
integer, then A itself is not an integer. As a result, at least one of b and h is also not an
integer.
For right triangles, b and h are the two shorter sides of the triangle. The question stem
specifies that all three sides are integers, so if the triangle were a right triangle, then b
and h would both have to be integers. This is impossible, so the triangle cannot be a
right triangle.
The correct answer is (D).
O + O + O = odd
O + E + E = odd
odd2 + even2 = even2 (impossible: odd + even = odd)
even2 + even2 = odd2 (also impossible: even + even = even)


25. (C) 2: The problem specifies that x has to be positive, so try some small positive
integers to see how the expression
 works.
If x = 1, then the expression becomes
  . If x = 2, then the expression
becomes 1 + 1 = 2. Since x can equal 2, eliminate answers (D) and (E).
What if x is a fraction, such as
  ? In that case, the expression becomes
 … interesting. It’s even bigger than the integer options
above. How come?
The expressions
 and
 are reciprocals of each other. As long as x is positive, there
are two broad scenarios possible:
Answers (A) and (B), then, are impossible.
The correct answer is (C).
If one expression is less than 1, then the other has to be greater than 1.1.
If one expression equals 1, then the other also equals 1.2.
 26. (D): If the average of the cost of the three models is $900, then the sum is $2,700. Call
the three models a, b, and c, in order from least expensive to most.
(1) SUFFICIENT: The statement indicates that c = 1.25b, so a + b + 1.25b = 2,700. This
isn’t enough to solve for specific values, but it might be enough to tell whether c >
$1,000. Keep working.


By definition, a < b, so plug LTb (less than b) into the equation for a.
From here, estimate: 13 goes into 2,600 a total of 200 times and it goes into 100
approximately 7 times. Therefore, LTb = (207)(4) = 828.
If b is approximately 830, then 125% of b is definitely larger than 1,000. Therefore, the
most expensive model does cost more than $1,000.
(2) SUFFICIENT: Follow a similar path: c = b + 210. Therefore:
If b is a bit more than 830, then b + 210 is definitely over 1,000. Therefore, the most
expensive model does cost more than $1,000.
The correct answer is (D).
 27. (D) y + 2x = 12: The product of the slopes of two perpendicular lines is −1. The slope of
line k is:


Thus, the slope of a line perpendicular to k is −2.
Use the desired slope to create a slope-intercept equation and look for a match among
the answers.
Only answer (D) oﬀers a match for the le side of the equation.
Alternatively, put each choice in slope-intercept form: y = mx + b, where the slope is m.
Stop when you find the correct answer.
(A) 3y + 2x = −12
 Slope = −
 Not a match. Eliminate.
(B) 2y + x = 0
 Slope = −
 Not a match. Eliminate.
(C) 2y − x = 0
 Slope =
 Not a match. Eliminate.
(D) y + 2x = 12 y = −2x + 12 Slope = −2 Correct.
The correct answer is (D).
 28. (A):  The question stem asks a Yes/No question: Is x positive?
(1) SUFFICIENT: Manipulate the absolute value expression to represent this inequality
on a number line.


This can be interpreted as, “The distance between x and 6 is less than 5.” On a number
line, this is the region between 1 and 11.
All possible values for x are positive, so the answer to the question is a definite Yes.
(2) INSUFFICIENT: Manipulate the inequality to get 0 on one side, then factor the
resulting quadratic.
The factored quadratic on the le side will equal 0 when x = 3 and 7. These are the
boundary points. On a number line, check the regions on either side of these
boundary points to determine the valid region(s) for x.
Note that when 3 < x < 7, (x − 7)(x − 3) = (neg)(pos) = neg.
Since both positive and negative values are possible for x, the answer is Maybe.
The correct answer is (A).
 29. (A):


Since you are looking for the minimum number of K-numbers that sum to 600, a
practical place to start is with the largest K-number less than 600, or 222. There are
between two and three multiples of 222 in 600, so subtract out the two whole
multiples.
Now, the next largest K-number is 22. Again, subtract as many whole multiples as
possible.
The next largest K-number is 2.
Thus, 600 = 222(2) + 22(7) + 2(1), and the K-weight of 600 is 2 + 7 + 1 = 10.
The correct answer is (A).
 30. (E): If the greater of the two integers is x, then the two integers can be expressed as x −
1 and x. The sum of the reciprocals would therefore be:
Alternatively, choose smart numbers. For example, let the larger number x = 3. The
smaller number would therefore be 3 − 1 = 2. The sum of the reciprocals would be:


Plug x = 3 into the answers, and find the one that equals
 .
The correct answer is (E).
 Eliminate.(A)
x2 − x = An integer. Eliminate.(B)
2x − 1 = An integer. Eliminate.(C)
 Eliminate.(D)
 Correct.(E)


Workout Set 14
131. The sequence A is defined as follows: A1 = 1, and An = An − 1 + (−1)n +
1(n2) for all integer values n > 1. What is the value of A15 − A13 ?
  14(A)
  29(B)
169(C)
196(D)
421(E)
132. If d represents the hundredths digit and e represents the
thousandths digit in the decimal 0.4de, what is the value of this
decimal rounded to the nearest tenth?
d − e is a positive perfect square.(1)
(2)
133. If x is a positive integer greater than 1, and y is the smallest positive
integer that is evenly divisible by every integer between 1 and x,
inclusive, what is the value of x ?
y = 10x(1)


y = 60(2)
134. A chain is comprised of 10 identical links, each of which
independently has a 1% chance of breaking under a certain load. If
the failure of any individual link means the failure of the entire
chain, what is the probability that the chain will fail under the load?
(0.01)10(A)
10(0.01)10(B)
1 − (0.10)(0.99)10(C)
1 − (0.99)10(D)
1 − (0.99)(10 × 9)(E)
135. City A to City B City B to City C City C to City A
First Bus Departs 5:00 5:00 5:00
Length of Trip 25 minutes 35 minutes 60 minutes
Subsequent
Departures
Every 20
minutes
Every 30
minutes
Every 20
minutes
The table above shows the schedule for the bus routes between
three pairs of cities. If Ash plans to travel by bus from City A, to City
B, to City C, and back to City A, what is the shortest possible
duration, in minutes, of his entire trip?


120(A)
130(B)
140(C)
150(D)
160(E)
136. Are the positive integers x and y consecutive?
x2 − y2 = 2y + 1(1)
x2 − xy − x = 0(2)
137. What is the value of y − x2 − x  ?
y = −3x(1)
y = −4(x + 1)(2)
138. If f(x) = (x + 6)2 and g(x) = 9x, which of the following specifies all the
possible values of x for which f(g(x)) < g(f(x)) ?
–2 < x < 2(A)
x > 2(B)
0 < x < 3(C)
–4 < x < 2(D)
6 < x < 9(E)


139.
In the figure above, ABCD is a rectangle, and each of AP and CQ is
perpendicular to BD. If DP = PQ = QB = 1, what is the length of AB  ?
(A)
 (B)
 (C)
(D)
(E)


140.
In the rectangular coordinate system above, point A is not shown. If
the area of triangle OAB is at least 16, which of the following could
NOT be the coordinates of point A  ?
(1, 8)(A)
(4, 14)(B)
(5, −4)(C)
(8, −1)(D)
(14, 4)(E)


Workout Set 14 Answer Key
 31. B
 32. E
 33. A
 34. D
 35. C
 36. D
 37. C
 38. A
 39. C
 40. A


Workout Set 14 Solutions
 31. (B) 29: Generate the first several values of the sequence, using the given
relationship. Notice that a component such as (−1)n + 1 switches the sign of the
additive term back and forth. Use a table to keep your work organized.
n An = An − 1 + (−1)n + 1(n2)
1 1
2 1 + (−1)2 + 1(22) = 1 − 4 = −3
3 −3 + (−1)3 + 1(32) = −3 + 9 = 6
4 6 + (−1)4 + 1(42) = 6 − 16 = −10
5 −10 + (−1)5 + 1(52) = −10 + 25 = 15
6 15 + (−1)6 + 1(62) = −15 − 36 = −21
What is the pattern? The sign alternates between positive and negative. Ignoring
the sign of the terms (taking absolute values) might help to determine the full
pattern.
n An = An − 1 + (−1)n + 1(n2) |An| Change from Previous Term
1       1    1


n An = An − 1 + (−1)n + 1(n2) |An| Change from Previous Term
2    −3   3 +2
3     6  6 +3
4 −10 10 +4
5   15 15 +5
6 −21 21 +6
That is, the absolute value of each term equals the absolute value of the previous
term plus n. Following this pattern, |A15| is 15 greater than |A14|, and |A14| is 14
greater than |A13|. Thus, |A15| − |A13| = +15 + 14 = 29.
Since the odd-numbered terms in the original sequence are positive, the absolute
value of any odd-numbered term equals the term itself, so |A15| − |A13| = A15 − A13 =
29.
The correct answer is (B).
 32. (E): When rounding to the nearest tenth, use the hundredths digit, or d in this
case. If d ≥ 5, the decimal 0.4de is rounded up to 0.5. If d ≤ 4, the decimal 0.4de is
rounded down to 0.4. Thus, a rephrase of this question is, “Is d < 5 (or,
equivalently, is d ≥ 5)?”
(1) INSUFFICIENT: Since d − e is positive, d > e. Since d and e are digits (i.e., 0, 1, 2 .
. . 7, 8, 9), there is a maximum value for the diﬀerence: d − e ≤ 9. There are only
three perfect squares less than or equal to 9: d − e = 1, 4, or 9.


d e d − e = Perfect Square Is d < 5?d e d − e = Perfect Square Is d < 5?
9 0 9 − 0 = 9 ✓ No
1 0 1 − 0 = 1 ✓ Yes
It is possible for d to round up or down, so this statement isn’t suﬀicient.
(2) INSUFFICIENT: Since d and e are positive, you can square each side of the
inequality without worrying about flipping the sign: d > e4. Since d and e are digits
(i.e., 0, 1, 2 . . . 7, 8, 9), 9 ≥ d > e4, which means that e can only be 0 or 1. (Note that
24 = 16, which is too large.) Test cases again; whenever possible, reuse numbers
that you tried in the last statement, assuming those numbers are allowed by the
new statement.
d e
 Is d < 5?
9 0 3 > 0 ✓ No
1 0
 ✓ Yes
It is again possible for d to round up or down, so this statement isn’t suﬀicient.
(1) AND (2) INSUFFICIENT: Taking both statements together, e must be 0 or 1 and d
− e must equal 9, 4, or 1.
You’ve already tested the d = 9 and d = 1 scenarios for both of the statements
individually, so you don’t need to retest them here. Even together, the two
statements allow d to round up or down.


The correct answer is (E).
 33. (A): List a few values of x and y to gain understanding; doing so also prepares you
to most easily interpret the statements.
  x        y
2       2
3      6
4    12
5    60
6   60
7 420
(1) SUFFICIENT: Look in the table above for cases in which y = 10x. There is only
one: x = 6 and y = 60.
To be completely confident that this is the only valid case, approach the problem
theoretically. In order for y to equal 10x, 10x would have to be divisible by 1, 2, 3 . .
.   (x – 1), and x. However, x itself cannot have any common factors with x – 1, so 10
must be divisible by x – 1. So x – 1 equals 1, 2, 5, or 10, so x must be 2, 3, 6, or 11.
Therefore, x = 2 and x = 3 have already been eliminated as possibilities and x = 6
has already been confirmed to work. The only remaining case that needs to be
checked is x = 11. Since 110 is not divisible by every number smaller than 11 (e.g.,
110 is not divisible by 3 or 4), x = 11 is not a valid case. The only possible answer is
x = 6, so this statement is suﬀicient.
(2) INSUFFICIENT: Based on the table above, y = 60 when x = 5 or 6. There are two
possible answers, so the statement is insuﬀicient.
The correct answer is (A).


34. (D) 1 − (0.99)10: Qualitatively, many failure scenarios could occur.
Given the complexity of the failure scenarios, it is easier to look at the opposite
scenario.
For each of the links, the probability that it will not fail is 1 − 0.01 = 0.99. The
probability that all 10 will not fail is thus (0.99)10, since the probability that all 10
will not fail equals the product of the probabilities of the individual links not
failing.
Therefore, the probability that at least one link will fail equals 1 − (0.99)10.
The correct answer is (D).
None of the links will fail.
Exactly one of the links will fail.
Exactly two of the links will fail.
And so on.
Probability that at least one link will fail = 1 − probability that No links will fail
 35. (C) 140: Ash’s round trip consists of two parts: the time spent on the bus and the
time spent waiting for the next bus. No matter which buses he takes, the time
spent on the bus will always be equal: 25 + 35 + 60 minutes = 120 minutes. This
question is really about minimizing the time that Ash spends waiting. Note the
mismatch between the departure frequencies and the trip durations; it is unlikely
that the waiting time is 0, so eliminate (A).
Because buses leave each city every 20 or 30 minutes, and both of these time
periods divide evenly into one hour, the pattern of departures will always repeat
every hour. Therefore, it is only necessary to check three diﬀerent scenarios: Ash
will leave City A either on the hour, 20 minutes a er the hour, or 40 minutes a er
the hour.


Then, to minimize waiting, Ash should take the first available buses in City B and
City C. For each of the three possible times that Ash could leave City A, note the
next available buses for each leg of his trip.
Leave A Arrive in B Leave B Arrive in C Leave C    Arrive in A     Total Time
5:00 5:25 5:30 6:05 6:20 7:20 2 hr 20 min
5:20 5:45 6:00 6:30 6:40 7:40 2 hr 20 min
5:40 6:05 6:30 7:05 7:20 8:20 2 hr 40 min
Therefore, the shortest possible duration of the entire trip is 2(60) + 20 = 120 + 20 =
140 minutes.
The correct answer is (C).
 36. (D): Consecutive integers diﬀer by exactly 1. For x and y to be consecutive, either x
= y + 1 or x = y − 1. Rephrase the question as, “Does x equal either (y + 1) or (y −
1)?”
(1) SUFFICIENT: There’s only one term with x but a couple with y, so it is easiest to
solve for x.
Since x and y are both positive, you can drop the absolute value signs and
conclude that x = y + 1.


(2) SUFFICIENT: Factor out a common term.
Therefore, x = 0 or x − y − 1 = 0. The question stem indicates that x is positive, so x
cannot equal 0. As a result, it must be the case that x − y − 1 = 0, or x = y + 1. The
two integers are consecutive.
The correct answer is (D).
 37. (C):
(1) INSUFFICIENT: Solve for the expression in question by adding (−x2 − x) to both
sides.
The answer depends on the value of x, which is not given.
(2) INSUFFICIENT: Solve for the expression in question by adding (−x2 − x) to both
sides.
The answer depends on the value of x, which is not given.


(1) and (2) SUFFICIENT: Each statement provides a diﬀerent expression equal to y
− x2 − x. Set these two equal to each other and solve.
Once the squared terms drop out, the equation is solvable.
Stop when you know you can solve!
The correct answer is (C).
 38. (A) –2 < x < 2: This problem asks you to compare two “nested” functions: f(g(x))
and g(f(x)). To simplify a nested function, replace the variable term in the outer
function with the entire inner function.
To find the values of x for which f(g(x)) < g(f(x)), simplify the following inequality.
One possibility is to multiply out both quadratics and then simplify as much as
possible.


In turn, this simplifies to –2 < x < 2.
Another approach involves using the multiples of 9 in the initial inequality. Factor
out a 9 on the le side of the inequality.
Then, simplify:
The correct answer is (A).
 39. (C)
 : Start by labeling the three line segments with length 1. Label the
unknown length as x.


From here, there are two ways to approach the problem. One approach is to use
the Pythagorean theorem to label the lengths of the missing sides. Triangle ABP is
a right triangle, so the following is true:
Similarly, APD is a right triangle, and the following is true:
Finally, since BAD is a right triangle, the Pythagorean theorem can be used to
solve for the value of x.


The other solution uses the fact that triangles APD and BPA are similar. To prove
this, note that APD is a right angle. Therefore, angles PAD and ADP sum to 90
degrees. According to the diagram, angles PAD and BAP also sum to 90 degrees.
Therefore, ADP and BAP have the same degree measure. Since triangles APD and
BPA have two pairs of matching angles, the third pair of angles must match as
well, and the two triangles are similar. The sides line up as follows:
Triangle PAD Triangle PBA
Shortest Side side PD side AP
Middle Side side AP side BP
Longest Side side AD side AB
Since PD = 1, BP = 2, and AP = AP, you can set up a proportion to find the length of
side AP.


Finally, use the Pythagorean theorem to determine the length of AB.
Alternatively, since the diagram is drawn to scale, you can try to estimate. Side
AB appears to be longer than BP, but shorter than BD. Its length should be
between 2 and 3. Estimate the values of the answer choices.
Only answers (C) and (D) are close to the correct number.
The correct answer is (C).
 = Less than 2. Eliminate.(A)
 = Less than 2. Eliminate.(B)
  = between 2 and 3.(C)
 = between 2 and 3.(D)
 = More than 3. Eliminate.(E)
 40. (A) (1, 8): First, identify any answer choices that will be easy to eliminate before
you begin calculating the areas of triangles. Answers (B) and (E) are symmetrical;


the triangles they yield will have the same area. Therefore, neither can be the
correct answer. Eliminate answers (B) and (E).
The question is asking for triangles with area less than 16 square units, so first try
drawing a triangle with an area of exactly 16 square units. If the base of the
triangle is OB, it has a length
 . Use the area formula to find the height.
The height is also
 . Draw the height of the triangle perpendicular to OB
through the origin in either quadrant II or quadrant IV. The vertex of the height is
at either (–4, 4) or (4, –4).
Any triangle with the same height will have the same area. Therefore, if A lies on
the line y = x + 8 or on the line y = x – 8, the triangle will have an area of 16. If A lies
anywhere between those two lines, then the triangle will have an area less than
16. Of the answers, only (1, 8) lies between the two lines.
Alternatively, plot the triangles formed by the remaining answer choices on your
paper.


Rather than calculating the exact area of each triangle, notice that each one is
fairly close to a triangle whose area is easier to calculate.
In choice (A), the triangle has a slightly smaller area than the triangle with vertices
at (0, 8), (4, 4), and (0, 0). The latter triangle has an area of exactly 16 square units.
Therefore, triangle (A) has an area smaller than 16 square units, so answer (A) is
correct.
In choice (C), the triangle has a slightly greater area than the triangle with vertices
at (4, 4), (4, –4), and (0, 0). The latter triangle has an area of exactly 16 square
units. Therefore, the area of triangle (C) is slightly greater than 16.
In choice (D), the triangle has a slightly greater area than the triangle with vertices
at (4, 4), (8, 0), and (0, 0). The latter triangle has an area of exactly 16 square units.
Therefore, the area of triangle (D) is slightly greater than 16.
The correct answer is (A).


Workout Set 15
141. If
 , where a and b are integers and a
does not equal 1 or −2, which of the following could be the value of
b  ?
1I.
2II.
3III.
I only(A)
II only(B)
I and II only(C)
I and III only(D)
I, II, and III(E)
142. The currencies of three countries are called credits, units, and bells.
If 12 credits can be exchanged for a total of exactly 8 units and 5
bells, what is the smallest integer number of units that can be
exchanged for an integer number of bells, with no change le over?
15 credits can be exchanged for 4 units and 25 bells, with no
change le over.
(1)


25 bells can be exchanged for 8 units, with no change le over.(2)
143. What is the area of the quadrilateral bounded by the lines
  48(A)
  64(B)
  96(C)
100(D)
140(E)
144. Each digit 1 through 5 is used exactly once to create a five-digit
integer. If the 3 and the 4 cannot be adjacent digits in the integer,
how many five-digit integers are possible?
48(A)
66(B)
72(C)
78(D)
90(E)
145. For positive integers a, b, and c, b is 120% greater than a, and c is
25% less than the sum of a and b. Which of the following could be


the value of |a – c| ?
  6(A)
  7(B)
  8(C)
  9(D)
10(E)
146. A soccer competition consisted of two rounds. In each round, the 20
teams competing were divided randomly into 10 pairs, and each
pair played a single game. None of the games resulted in a tie, and
of the teams that lost in the first round, 7 also lost in the second
round. How many teams won both of their games?
3(A)
4(B)
5(C)
6(D)
7(E)
147. What is the units digit of the positive integer x  ?
, where y is a positive integer.(1)
, where z is a positive integer.(2)


148. If x and y are positive integers, what is the value of x + y ?
(x + y − 1)! < 100(1)
y = x2 − x + 1(2)
149.
In the figure above, two lines are tangent to a circle at points A and
B. What is x ?
The area of the circle is 81π.(1)
The length of arc ADB is 7π.(2)
150. For all n such that n is a positive integer, the terms of a certain
sequence B are given by the following rules:
Bn = Bn − 1 + 5 if n is odd and greater than 1
Bn = −Bn − 1 if n is even
B1 = 3


What is the sum of the first 65 terms in the sequence?
−5(A)
  0(B)
  3(C)
  5(D)
  8(E)


Workout Set 15 Answer Key
 41. C
 42. D
 43. C
 44. C
 45. B
 46. E
 47. C
 48. E
 49. C
 50. C


Workout Set 15 Solutions
 41. (C) I and II only: Factor the numerator:
Since the fraction equals 0, either ab + 6 = 0 or ab − 3 = 0. Thus, ab = −6 or 3.
Since a and b are integers, and ab = −6 or ab = 3, it must be the case that b
is a factor of either −6 or 3. Note that a ≠ 1 and a ≠ −2.
The correct answer is (C).
POSSIBLE: If b = 1, then a = −6 or a = 3, which are both allowed.I.
POSSIBLE: If b = 2, then a = −3 or a =
 . One of those (a = −3) is allowed.II.
IMPOSSIBLE: If b = 3, then a = −2 or a = 1, neither of which is allowed.III.
 42. (D): Use three variables to represent the unknown values in this problem.
Let c represent the value of one credit, u represent the value of one unit,
and b represent the value of one bell. The given information translates to
the following equation:
12c = 8u + 5b
The question asks about exchanging an integer number of units for an
integer number of bells. Therefore, your goal is to find the relationship
between the value of a unit and the value of a bell.


(1) SUFFICIENT: This statement can be rewritten as the following equation,
using the same variables as above.
  15c = 4u + 25b
Next, subtract one equation from the other and simplify.
Therefore, 8 units are worth the same amount as 25 bells. Because the
ratio 8 : 25 cannot be reduced further, 8 is the smallest number of units
that can be exchanged for an integer number of bells. For instance, 4 units
could only be exchanged for 12.5 bells, which is not an integer. The answer
to the question is 8.
Combine this with the equation from the question stem. Because the
goal is to find the relationship between b and u, focus on eliminating
the other variable, c. First, multiply the equations by 5 and 4,
respectively, so that the coeﬀicient of c is 60 in both.


(2) SUFFICIENT: If 8 units can be exchanged for 25 bells, then because 8
and 25 have no common factors, no smaller integer number of units can
be exchanged for bells, as described above. The answer to the question is
8, and this statement is suﬀicient.
The correct answer is (D).
 43. (C) 96: All of these line equations are of the form y = mx + b, where m is the
slope and b is the y-intercept. Two of these lines have a slope of
 and are
thus parallel to each other. The other two lines are parallel to one another
with a slope of
 . Two of the lines have a y-intercept of 6 while the
other two lines have a y-intercept of −6.
Sketch the lines.


In each quadrant, there is a triangle with the dimensions 6–8–10, a
multiple of the common 3–4–5 right triangle.
Alternatively, recognize that the quadrilateral is a rhombus (four equal
sides of length 10) and use the formula for the area of a rhombus:
, where D indicates the length of the diagonals.
The correct answer is (C).
 44. (C) 72: In a constrained combinatorics question such as this one, it is o en
easier to consider the “violating” cases instead of the “okay” cases.
# of permutations that obey the constraint = # of permutations total
− # of permutations that violate the constraint
At first glance, ignore the constraint that 3 and 4 cannot be adjacent to
determine the total number of five-digit integers possible:
# of permutations total = 5! = (5)(4)(3)(2)(1) = 120


Now, consider the basic ways that 3 and 4 might be adjacent to each other
in a five-digit number:
3 4 x x x
x 3 4 x x
x x 3 4 x
x x x 3 4
For each of these four base cases, there are two ways to order the 3 and 4
(3, 4 and 4, 3), as well as 3! ways the other digits (1, 2, and 5) can be
arranged in the x positions. Thus:
# of permutations that violate the constraint = 4 × 2 × 3! = (4)(2)(3)
(2)(1) = 48
Therefore, the number of permutations that do not violate the constraint
equals 120 − 48 = 72.
The correct answer is (C).
 45. (B) 7: Start by translating the text into math. Note that b is 120% greater
than a, not 120% of a.


There are three variables and only two equations, so you won’t be able to
solve in the traditional sense, but you may be able to determine
something about these integers. Substitute for b in the second equation.
Both a and c are integers, and 5 and 12 share no prime factors. So a must
be divisible by 5 and c must be divisible by 12. Try a = 5 and c = 12 to see
whether b is a positive integer in this case.
This is a valid case because b is a positive integer. In this case, the value of
|a – c| = |5 – 12| = |–7| = 7, which is one of the answer choices.
The correct answer is (B).
 46. (E) 7: Each team either won or lost the first game and either won or lost the
second game. Therefore, this problem can be approached using
overlapping sets. Create a Double-Set Matrix.


Won first game Lost first game Total
Won second game
Lost second game 7
Total 20
Initially, it appears that there is not enough information to determine how
many teams won both games. However, note that there were no ties and
each round consisted of exactly 10 games.
Since each of the 10 games yielded one winner and one loser, there were
exactly 10 teams that won in the first round and 10 teams that lost in the
first round. Similarly, there were 10 teams that won in the second round
and 10 teams that lost in the second round.
Based on this information, fill in the remainder of the matrix.
Won first game Lost first game Total
Won second game   7   3 10
Lost second game   3   7 10
Total 10 10 20
Seven teams won both of their games.
The correct answer is (E).
 47. (C):
(1) INSUFFICIENT: Manipulate the statement.


Thus, x is 1 greater than a multiple of 5. Since all multiples of 5 end in
either 0 or 5, x must end in either 1 or 6.
Alternatively, you could list numbers. Since y is a positive integer, 5y could
be 5, 10, 15, 20, etc. Thus, x could be 6, 11, 16, 21, etc. The units digit of x
could be 1 or 6.
(2) INSUFFICIENT: Manipulate the statement.
This indicates that x is odd, because 2z is an even number. Any odd single-
digit integer is a possible units digit for x: 1, 3, 5, 7, or 9.
(1) AND (2) SUFFICIENT: Statement (1) indicates that x must end in either 1
or 6. Because statement (2) indicates that x is odd, x must end in 1 and
cannot end in 6.
The correct answer is (C).
 48. (E): The question stem asks for the value of the combo x + y.


(1) INSUFFICIENT: (x + y − 1)! < 100. To see the possible values of (x + y − 1),
list the factorials of the first few integers.
Therefore, (x + y − 1) ≤ 4, and (x + y) ≤ 5.
(2) INSUFFICIENT: Add x to both sides of y = x2 − x + 1 to create (x + y) on
one side of the equation.
The exact value of (x + y) is unknown, as it depends on the value of x2,
which could be any positive integer.
(1) AND (2) INSUFFICIENT: (x + y) ≤ 5 and x + y = x2 + 1 combine to indicate
that x2 + 1 ≤ 5. There are two integer solutions: x = 1 or x = 2. Therefore:
The correct answer is (E).
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120 (too large)
If x = 1, then y = 12 − 1 + 1 = 1 and x + y = 1 + 1 = 2.
If x = 2, then y = 22 − 2 + 1 = 3 and x + y = 2 + 3 = 5.
 49. (C): The size of the angle depends on two things.


The size of the circle
 The distance of C from the circle (inversely related to arc
length ADB)
Larger circle → Larger x Farther C → smaller x
Larger arc ADB → smaller x
(1) INSUFFICIENT: The rubber band picture on the right indicates that for a
circle of fixed size, x can still vary with the length of arc ADB (i.e., x varies
with the placement of C).
(2) INSUFFICIENT: Draw some cases to prove that x can vary for a given arc
ADB.
For a circle with circumference 28π, the arc ADB is
 of the
circle, so x is 90°. For a circle with circumference 15π, the arc ADB is


, or nearly half of the circle, and the lines tangent to the
circle at A and B will join at a smaller angle x.
(1) AND (2) SUFFICIENT: If the area of the circle is πr2 = 81π, then r = 9. The
circumference of the circle is 2πr = 18. Thus, arc ADB is
  of
the circumference of the circle. There is only one way to draw the lines
tangent to the circle at A and B, so x must be one specific value.
The correct answer is (C).
 50. (C) 3: List the first few terms of the sequence according to the given rules.


Note that the pattern is a four-term repeat: 3, −3, 2, −2. Also note that the
sum of this repeating group is (3) + (−3) + (2) + (−2) = 0. This repeating
group will occur 16 times through term number 64. (Note: Because the
sum of the four terms is 0, you don’t actually have to figure out how many
times the pattern repeats.) Thus, the sum of the first 64 terms will be 0.
This leaves the 65th term, which will have the same value as B1: 3.
Therefore, the sum of the first 65 terms is 3.
The correct answer is (C).


Workout Set 16
151. Model Profit as a Percent of Model’s Total Sales
Model A 12%
Model B 16%
Model C 20%
A bicycle store earned a diﬀerent percent profit on each of the three
models of bicycles it sells, as shown in the table above. If the store
made a total profit of 16% on sales of these three models of
bicycles last month, and these three models represented the
entirety of the store’s profits, what percent of its total profits last
month came from sales of Model A bicycles?
The store earned 20% of its profits last month from sales of
Model B bicycles.
(1)
The store earned 50% of its profits last month from sales of
Model C bicycles.
(2)
152. A beaker contains 100 milligrams of a solution of salt and water that
is x% salt by weight such that x < 90. If the water evaporates at a


rate of y milligrams per hour, how many hours will it take for the
concentration of salt to reach (x + 10)%, in terms of x and y ?
(A)
(B)
(C)
(D)
(E)
153. Set A consists of five consecutive positive integers. Set B consists of
every integer that is in set A as well as the sum of every pair of two
distinct integers from set A, discarding any numbers that already
appear in set B. What is the greatest integer in set B  ?
The range of set B is 13.(1)
There are exactly 12 integers in set B.(2)


154. A survey included two questions, each of which could be answered
with either Yes or No. A total of 100 respondents took the survey,
and every respondent answered both questions. How many
respondents answered both questions with a Yes?
Of the respondents who answered Yes to the first question,
40% answered Yes to the second question.
(1)
Of the respondents who answered Yes to at least one question,
30% answered Yes to both questions.
(2)
155. If x is a positive two-digit integer, is x divisible by 17 ?
x is divisible by 7.(1)
x has exactly 3 unique prime factors.(2)
156. Green paint is made by mixing blue paint with yellow paint in a
ratio of 2 to x. Turquoise paint is made by mixing green paint with
blue paint in a ratio of y to 2. In terms of x and y, how many gallons
of yellow paint are required to make 10 gallons of turquoise paint?


(A)
(B)
(C)
(D)
(E)
157. Erin writes a list of every sixth integer in increasing order, starting
with the positive integer x. Harry writes a list of every ninth integer
in increasing order, starting with the positive integer y. Will any
integer appear on both Erin’s list and Harry’s list?
x is a multiple of y.(1)
x – y is a multiple of 3.(2)
158. The students in a certain college class took a midterm exam and a
final exam, both of which were scored out of a total of 100 points.


What percent of the students earned a higher score on the final
than on the midterm?
Of the students in the class, 28% scored at least 6 points higher
on the final than on the midterm.
(1)
Of the students who scored higher on the final than on the
midterm, 40% scored at least 6 points higher on the final.
(2)
159. A triathlon consists of three legs: a swim, a bike ride, and a run. A
race organizer analyzes the results of a certain triathlon and
observes that the ratio of swim distance to bike distance to run
distance was 1 to 50 to 12, and that the ratio of average swim time
to average bike time to average run time for all participants was 3
to 10 to 6. If the average speed of racers in each leg of the race does
not change, but the race organizer changes the distance of each leg
of the race so that the average swim time, average bike time, and
average run time are now equal to each other, what fraction of the
total distance of the race will be represented by the swim leg of the
triathlon?


(A)
(B)
(C)
(D)
(E)
160. An unpainted wall with a total area of 100 square feet needed to be
painted with three coats of paint. Nelson began the job by painting
an area of 60 square feet. Then, Jamaica painted a total area of 70
square feet. Finally, Bryan painted a total area of 80 square feet.
A er Bryan finished painting, an area of q square feet of the wall
was covered in three coats of paint. If the entire surface area of the
wall had received at least one coat of paint, which of the following
specifies all the possible values of q  ?
0 ≤ q ≤ 50(A)
0 ≤ q ≤ 60(B)
10 ≤ q ≤ 50(C)
10 ≤ q ≤ 55(D)
10 ≤ q ≤ 60(E)


Workout Set 16 Answer Key
 51. D
 52. D
 53. A
 54. E
 55. D
 56. A
 57. B
 58. C
 59. B
 60. D


Workout Set 16 Solutions
 51. (D): A store sold three diﬀerent models of bicycles last month. For each of these
models, the profit earned by the store represents a diﬀerent percentage of the sales
of that model. The question stem also states that the total profit was 16% of the
total sales.
One way to handle the question stem is to use variables A, B, and C to represent the
total sales of Model A, B, and C bicycles, respectively. Then, the profit from Model A
bicycles is 0.12A, the profit from Model B bicycles is 0.16B, and the profit from Model
C bicycles is 0.20C. Also, the total dollar amount of sales is A + B + C, so the following
is true and can be simplified:
You might also notice that 16% is exactly in the middle of the three percentages
provided in the question stem. The problem is eﬀectively a weighted average
problem: if more dollars worth of Model A bicycles were sold, then the average
profit would be pulled downwards, closer to 12%. If more dollars worth of Model C
bicycles were sold, then the average profit would be pulled upwards, closer to 20%.
Since the actual profit was in the middle, at 16%, the dollar value of the Model A
bicycles sold and the dollar value of the Model C bicycles sold must be equal.


The question asks about the share of profits that came from Model A bicycles. This
question can be rephrased as, “What is
 ?” which can then
be simplified as follows:
What is
 ?
What is
 ?
What is
 ?
The variables would cancel entirely if you could determine a known ratio of A to B.
(1) SUFFICIENT: 20% of the store’s profits came from Model B bicycles, which can be
translated and simplified as follows, remembering that A = C and the total profit
equals 0.16(A + B + C).
Substituting A = 2B into the question causes all unknowns to cancel out, yielding an
answer of
 , or 30%.


(2) SUFFICIENT: The store earned 50% of its profits from sales of Model C bicycles,
which can be translated and simplified as follows, remembering that A = C and the
total profit equals 0.16(A + B + C).
This is the same result that statement (1) produced. The answer is again 30%, as
shown above, so this statement is suﬀicient.
The correct answer is (D).
 52. (D)
 : The answer choices in this mixture problem include variables.
Therefore, it is possible to use Smart Numbers.
Suppose that x = 10 (initial concentration) and y = 2 (water evaporation rate in
milligrams [mg] per hour). Therefore, the beaker initially contains 10% of 100 mg =
10 mg of salt. The question asks how many hours it will take for the concentration of
salt to reach (x + 10)%, or 20%.
The beaker will always contain 10 mg of salt, since the salt does not evaporate. The
10 mg of salt will represent 20%, or
 , of the total weight when the total solution
weight has been reduced to 50 mg. To get to this point, 50 mg of water must
evaporate. This will take a total of
 = 25 hours.
Plug in x = 10 and y = 2 to see which answer choice equals 25 hours.


(A)
 Negative. Eliminate.
(B)
 = A non-integer close to 50. Eliminate.
(C)
 = 110. Eliminate.
(D)
 = 25. Correct.
(E)
 . Eliminate.
Alternatively, approach the problem with algebra. Initially, the beaker contains x mg
of salt in a total solution of 100 mg. A er h hours, hy mg of the solution will have
evaporated, so the solution will have a total weight of (100 – hy) mg, and the
amount of salt will still be x mg. So the new concentration, as a percentage, will be
. This percent should equal (x + 10)%. Set up an equation
and solve for h, the number of hours.


The correct answer is (D).
 53. (A): To better understand how the sets are constructed, write out an example.
Suppose that set A consists of the integers 1, 2, 3, 4, and 5. Then, set B will initially
include 1, 2, 3, 4, and 5, and more numbers will be added to it as follows, discarding
any numbers that already appear in set B.
1 + 2 = 3 discard
1 + 3 = 4 discard
1 + 4 = 5 discard
1 + 5 = 6 will be added to set B
2 + 3 = 5 discard
2 + 4 = 6 discard
2 + 5 = 7 will be added to set B
3 + 4 = 7 discard
3 + 5 = 8 will be added to set B
4 + 5 = 9 will be added to set B


Thus, if set A contains the integers from 1 to 5, set B consists of 1, 2, 3, 4, 5, 6, 7, 8,
and 9. The question asks for the greatest integer in set B.
(1) SUFFICIENT: The range of a set is the diﬀerence between the greatest and the
smallest numbers in that set. The smallest number in set B is the smallest number
from set A, since all of the additional numbers in set B are constructed by summing
numbers together and will therefore be greater. Call this number x.
The greatest number in set B is created by summing the two greatest numbers from
set A. Those numbers are x + 3 and x + 4, and their sum is 2x + 7.
Therefore, the range of set B is (2x + 7) – x. Set this equal to the given range of 13,
and solve for x:
The smallest number in set A is 6, and both sets can be listed with certainty. There is
only one answer to the question, so this statement is suﬀicient.
(2) INSUFFICIENT: There are exactly 12 integers in set B, and you will likely have to
test cases to determine how this could happen. Above, it was determined that if set
A contains the integers 1, 2, 3, 4, and 5, then set B contains 9 integers in total: the
integers from 1 to 9, inclusive. This is not enough integers: Find a case with fewer
duplicates to discard. Try larger numbers.
Case 1: If set A contains the integers 5, 6, 7, 8, and 9, then set B contains 5, 6, 7, 8, 9,
11, 12, 13, 14, 15, 16, and 17. This set B contains 12 integers, and the greatest integer
is 17.


Case 2: If set A contains the integers 100, 101, 102, 103, and 104, then set B contains
100, 101, 102, 103, 104, 201, 202, 203, 204, 205, 206, and 207. This set B again
contains 12 integers, but its greatest integer is 207.
The correct answer is (A).
 54. (E): This is an Overlapping Sets problem. Create a Double-Set Matrix and fill out the
known information.
Yes to Question 1 No to Question 1 Total
Yes to Question 2
No to Question 2
Total 100
The question asks for the number of respondents who gave two Yes answers. Try to
determine the number that belongs in the top le of the matrix.
(1) INSUFFICIENT: Let x represent the number of people who answered Yes to the
first question. The matrix can be filled out as follows:
Yes to Question 1 No to Question 1 Total
Yes to Question 2 0.4x
No to Question 2 0.6x
Total x 100 – x 100
If x = 5, then the answer is 0.4(5) = 2. If x = 90, then the answer is 0.4(90) = 36. More
than one value is possible for the answer, so this statement is not suﬀicient.
(2) INSUFFICIENT: Let y represent the number of people who answered No to both
questions. Then, the number of people who answered Yes to at least one question is


100 – y. Since 30% of these people answered Yes to both questions, the matrix can
be filled out as shown.
Yes to Question 1 No to Question 1 Total
Yes to Question 2 0.3(100 – y)
No to Question 2 y
Total 100
If y = 10, the answer is 0.3(90) = 27. If y = 80, the answer is 0.3(20) = 6. Since more
than one value is possible for the answer, this statement is insuﬀicient.
(1) AND (2) INSUFFICIENT: Using the information in both statements, you can
conclude that 0.3(100 – y) = 0.4x. This simplifies to 0.4x + 0.3y = 30, but both x and y
can still have multiple values.
Accordingly, the matrix can be filled out with numbers in several diﬀerent ways.
Yes to Question 1 No to Question 1 Total
Yes to Question 2 24 20 44
No to Question 2 36 20 56
Total 60 40 100
Yes to Question 1 No to Question 1 Total
Yes to Question 2 12 10 22
No to Question 2 18 60 78
Total 30 70 100
The correct answer is (E).


55. (D): The question stem includes a constraint that x is positive and has two digits.
Without this constraint, the answer would be (E), since knowing whether x is
divisible by 7 or has other prime factors does not help you determine whether x is
divisible by 17.
Use the constraint as you solve the problem. Since x is a positive two-digit integer, x
is between 10 and 99, inclusive. The question asks whether x is divisible by 17. Since
there are only a few integers in this range that are divisible by 17, jot them down for
later reference: 17, 34, 51, 68, and 85. The question can be rephrased as, “Is x equal
to 17, 34, 51, 68, or 85 ?”
(1) SUFFICIENT: Positive two-digit integer x is a multiple of 7. Since none of the
numbers listed earlier (17, 34, 51, 68, 85) are multiples of 7, x is definitely not one of
these numbers. Therefore, x is not a multiple of 17, and the answer is definitely No.
Alternatively, use logical reasoning to prove that the statement is suﬀicient without
doing too much arithmetic. Seven and 17 are both prime numbers, so the smallest
number that is divisible by both of them is their product, (7)(17) = 119. However, you
already know that x is smaller than 119, since x only has two digits. Therefore, x
cannot be a multiple of both 7 and 17, so the answer is definitely No.
(2) SUFFICIENT: None of the numbers listed earlier have exactly three unique prime
factors: 17 is prime, while 34, 51, 68, and 85 each have exactly two unique prime
factors. Therefore, x is not one of these numbers, so x is not a multiple of 17 and the
answer to the question is No.
It is also possible to show that x cannot be a multiple of 17 while doing minimal
arithmetic. If x has exactly three unique prime factors, then x will only be divisible by
17 if 17 is among those factors. However, the smallest number that has three unique
prime factors that include 17 is (2)(3)(17) = 102, which is too large to be the value of
two-digit integer x. Therefore, x can not be divisible by 17 and the answer is
definitely No.
The correct answer is (D).


56. (A)
 : Since the answer choices include variables, choose smart
numbers. It will be simpler to choose y first, since the value of y is more closely
related to the 10 gallons of turquoise paint. Choose y = 8, so that the sum of y and 2
in the ratio of the paints used to make turquoise can be 8 + 2 = 10. In this case, it
takes 8 gallons of green paint and 2 gallons of blue paint to make 10 gallons of
turquoise paint.
Next, choose a value for x. Because the total amount of green paint should be 8
gallons, choose x = 6, so that the of x and 2 in the ratio to make green paint is
2 + 6 = 8. In total, it will take 6 gallons of yellow paint to create the turquoise paint,
so the answer to the question is 6.
Plug x = 6 and y = 8 into the answer choices. The correct answer will simplify to 6.
(A)
 . Correct.
(B)
 Not an integer. Eliminate.
(C)
 Eliminate.
(D)
 Eliminate.
(E)
 = Not an integer. Eliminate.
The correct answer is (A).
 57. (B): Erin’s list starts with an unknown positive integer and then includes every sixth
integer. So Erin’s list could be any of the following, depending on the value of x:


x Erin’s Listx Erin’s List
1 1, 7, 13, 19, 25 . . .
2 2, 8, 14, 20, 26 . . .
3 3, 9, 15, 21, 27 . . .
etc. etc.
Similarly, Harry’s list could be any of the following:
y Harry’s List
1 1, 10, 19, 28, 37 . . .
2 2, 11, 20, 29, 38 . . .
3 3, 12, 21, 30, 39 . . .
etc. etc.
(1) INSUFFICIENT:
Case 1: If x = y = 1, then x is a multiple of y, and the number 1 appears on both lists.
Therefore, the answer to the question is Yes.
Case 2: If x = 3, all of the numbers on Erin’s list are multiples of 3. If y = 1, none of the
numbers on Harry’s list are multiples of 3. Therefore, x is a multiple of y, but no
numbers appear on both lists, so the answer to the question is No.
(2) SUFFICIENT: If x – y is a multiple of 3, then x – y = 3n, for some integer n. So x = 3n
+ y. The integers on the two lists are as follows:
Erin’s list: 3n + y, 3n + y + 6, 3n + y + 12, 3n + y + 18 . . .
Harry’s list: y, y + 9, y + 18, y + 27 . . .


If n is a multiple of 3, then 3n is a multiple of 9, so 3n + y appears on both lists. If n is
1 more than a multiple of 3, then 3n + 6 = 3(n + 2) = 3(multiple of 3) is a multiple of 9,
so y + 3n + 6 appears on both lists. Finally, if n is 2 more than a multiple of 3, then
3n + 12 = 3(n + 4) = 3(multiple of 3) is a multiple of 9 so y + 3n + 12 appears on both
lists. Regardless of the value of n, some number appears on both Erin’s and Harry’s
lists, so the answer to the question is Yes.
The correct answer is (B).
 58. (C): According to the question stem, each student in the class earned a certain score
on the midterm and a certain score on the final. A student’s final score could have
been higher than, equal to, or lower than his or her midterm score. To answer the
question, you’ll need to determine what percent of the students fell into the first of
these three categories.
(1) INSUFFICIENT: It is possible that all of the remaining students scored between 1
and 5 points higher on the final than on the midterm, which would make the answer
to the question 100%. It is also possible that all of the remaining students scored
higher on the midterm than on the final, which would make the answer to the
question 28%.
(2) INSUFFICIENT: This statement relates the percent of students who scored higher
on the final to the percent of students who scored at least 6 points higher on the
final. However, without any information about either of these percents, the
question cannot be answered definitively.
For instance, 50% of the students could have scored higher on the final, and 40% of
this 50% could have scored at least 6 points higher. Or, only 10% of the students
could have scored higher on the final, and 40% of this 10% could have scored at
least 6 points higher.
(1) AND (2) SUFFICIENT: According to statement (1), the following is true:
0.28(total students) = students who scored 6+ points higher on the final


According to statement (2), the following is true:
0.4(students who scored higher on the final) = students who scored 6+ points
higher on the final
Combine these two equations.
0.28(total students) = 0.4(students who scored higher on the final)
Simplify this equation to calculate the percent of students who scored higher on the
final.
Since an exact answer to the question can be calculated, the statements together
are suﬀicient.
The correct answer is (C).
 59. (B)
 : This problem provides two pieces of information. The first is the ratio of the
swim distance to the bike distance to the run distance: 1 to 50 to 12. The second
piece of information is the ratio of the average swim time to average bike time to
average run time: 3 to 10 to 6.
Use an unknown multiplier for each of these ratios.
Distance Time
Swim d 3t
Bike 50d 10t
Run 12d 6t


Now, use an RTD (rate/time/distance) chart to calculate the average speed for
competitors in each of the three legs, in terms of d and t.
Rate Time Distance
Swim
 3t d
Bike
 10t 50d
Run
 6t 12d
The question asks about a scenario in which the average swim time, bike time, and
run time are equal, but the rates stay the same. In this scenario, the following will be
the case:
Rate Time Distance
Swim
 t ?
Bike
 t ?
Run
 t ?
Use the RTD formula to fill in the missing column of this chart.
Rate Time Distance
Swim
 t
Bike
 t 5d
Run
 t 2d