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5.5 Problem Solving Answer Explanations
s(1 + r) dollars to spend next year.  is amount is
to be
1
2
the amount he spends this year.  e task
is to fi nd
s
I
, where I and s satisfy the
condition s(1 + r) =
1
2
(I – s). Solve for I as follows:
s(1 + r) =
1
2
(I – s)
2s(1 + r) = I – s
2s + 2rs = I – s
3s + 2rs = I
s(3 + 2r) = I
 en,
s
I
=
s
sr32+
()

=
1
23r +
.
 e correct answer is E.
164. If m
–1
= –
1
3
, then m
–2
is equal to
(A)
–9
(B)
–3
(C) –
1
9
(D)
1
9
(E) 9
Arithmetic Negative exponents
Using rules of exponents, m
–2
= m
–1
∙

2
= (m
–1
)
2
,
and since m
–1
= –
1
3
, m
–2
= =
1
9
.
 e correct answer is D.
165. Lois has x dollars more than Jim has, and together
they have a total of y dollars. Which of the following
represents the number of dollars that Jim has?
(A)
y − x
2

(B) y −
x
2
(C)
y

2
− x
(D) 2y – x
(E) y – 2x
Algebra Simplifying algebraic expressions
Let J be the number of dollars that Jim has.
 en, the amount that Lois has can be expressed
as J + x dollars. If Lois and Jim together have a
total of y dollars, then:
y = J + ( J + x) total dollars =
Jim’s dollars + Lois’s dollars
Solve this for J to determine the number of dollars
that Jim has:
y = 2 J + x
y − x = 2 J

y − x
2
= J
 e correct answer is A.
166. During a certain season, a team won 80 percent of
its first 100 games and 50 percent of its remaining
games. If the team won 70 percent of its games for
the entire season, what was the total number of
games that the team played?
(A) 180
(B) 170
(C) 156
(D) 150
(E) 105

Arithmetic; Algebra Percents; Applied problems
Let G equal the number of games played by the
team this season.  e given information can be
expressed as (0.80)(100) + 0.50(G − 100) = 0.70G,
that is, 80 percent of the first 100 games plus
50 percent of the remaining games equals
70 percent of the total number of games played.
 is equation can be solved for G to determine
the answer to the problem:
(0.80)(100) + 0.50(G − 100) = 0.70G
80 + 0.50G − 50 = 0.70G simplify and distribute
30 = 0.20G simplify and subtract
0.05G from both sides
150 = G multiply by 5
 e correct answer is D.
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167. Of 30 applicants for a job, 14 had at least 4 years’
experience, 18 had degrees, and 3 had less than
4 years’ experience and did not have a degree.
How many of the applicants had at least 4 years’
experience and a degree?
(A) 14
(B) 13
(C) 9

(D) 7
(E) 5
Arithmetic Operations on rational numbers
 e problem classified the job applicants into two
categories: whether they had more or less than
4 years’ experience, and whether they had a
degree.  e given information can be summarized
in the following table:
At least 4 years’
experience
Less than 4 years’
experience Total
Degree 18
No degree 3
Tota l 14 30
 us, according to the given information,
30 – 14 = 16 applicants had less than 4 years’
experience.  en, of those applicants with
less than 4 years’ experience, it is given that
3 applicants did not have a degree, so 16 – 3 =
13 applicants had less than 4 years’ experience
and had a degree.  erefore, out of the given 18
applicants that had degrees, 13 applicants had less
than 4 years’ experience, so 18 – 13 = 5 applicants
had at least 4 years’ experience with a degree.
 ese results are shown in the following table.
At least 4 years’
experience
Less than 4 years’
experience Total

Degree 5 13 18
No degree 3
Tota l 14 16 30
 e correct answer is E.
168.
(A) –1
(B)
1
3
(C)
2
3
(D) 2
(E) 3
Algebra First-degree equations
Work the problem to solve the equation for x.

x + 1 = 2x − 2 multiply through by x
3 = x solve for x by adding 2 to
and subtracting x from
both sides
 e correct answer is E.
169. Last year, for every 100 million vehicles that traveled
on a certain highway, 96 vehicles were involved in
accidents. If 3 billion vehicles traveled on the highway
last year, how many of those vehicles were involved in
accidents? (1 billion = 1,000,000,000)
(A) 288
(B) 320
(C) 2,880

(D) 3,200
(E) 28,800
Arithmetic Operations on rational numbers
According to the given information, 96 out of
every 100 million vehicles were in an accident
last year.  us, of the 3 billion vehicles on the
highway last year, the number of vehicles involved
in accidents was:
 e correct answer is C.
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5.5 Problem Solving Answer Explanations
170. Thirty percent of the members of a swim club have
passed the lifesaving test. Among the members
who have not passed the test, 12 have taken the
preparatory course and 30 have not taken the course.
How many members are there in the swim club?
(A) 60
(B) 80
(C) 100
(D) 120
(E) 140
Algebra Applied problems
If 30 percent of the club members have passed
the test, then 70 percent have not. Among the
members who have not passed the test, 12 have
taken the course and 30 have not, for a total of
12 + 30 = 42 members who have not passed the
test. Letting x represent the total number of
members in the swim club, this information can

be expressed as 0.70x = 42, and so x = 60.
 e correct answer is A.
171. What is the difference between the sixth and the fi fth
terms of the sequence 2, 4, 7, … whose nth term is
n + 2
n – 1
?
(A) 2
(B) 3
(C) 6
(D) 16
(E) 17
Algebra Simplifying algebraic expressions
According to the given formula, the sixth term
of the sequence is 6 + 2
6 – 1
= 6 + 2
5
and the fi fth
term is 5 + 2
5 – 1
= 5 + 2
4
.  en,
(6 + 2
5
) – (5 + 2
4
) = (6 – 5) + (2
5

– 2
4
)
= 1 + 2
4
(2 – 1)
= 1 + 2
4
= 1 + 16
= 17
 e correct answer is E.
172. If (x – 1)
2
= 400, which of the following could be the
value of x – 5 ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26
Algebra Second-degree equations
Work the problem by taking the square root of
both sides and solving for x.
(x − 1)
2
= 400
x − 1 = ± 20
x − 1 = −20, or x − 1 = 20
x = −19, or x = 21
 us, x − 5 = −24 or 16.

 e correct answer is C.
173. Which of the following describes all values of x for
which 1 – x
2
≥ 0 ?
(A) x
≥ 1
(B) x ≤ – 1
(C) 0 ≤ x ≤ 1
(D) x ≤ – 1 or x
≥ 1
(E) – 1 ≤ x ≤ 1
Algebra Inequalities
 e expression 1 – x
2
can be factored as
(1 – x)(1 + x).  e product is positive when both
factors are positive (this happens if 1 ≥ x and
x ≥ –1, or equivalently if –1
≤ x ≤ 1) or both
factors are negative (this happens if 1
≤ x and
x
≤ 1, which cannot happen), and therefore the
solution is –1
≤ x ≤ 1.
 e correct answer is E.
174. The probability is
1
2

that a certain coin will turn up
heads on any given toss. If the coin is to be tossed
three times, what is the probability that on at least
one of the tosses the coin will turn up tails?
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(A)
1
8
(B)
1
2
(C)
3
4
(D)
7
8
(E)
15
16
Arithmetic Probability
Another way of stating that a coin toss will turn
up tails at least once is to say that it will not turn
up heads every time.  e probability that on at

least one of the tosses the coin will not turn up
heads is 1 minus the probability that the coin
will turn up heads on all three tosses. Each toss
is an independent event, and so the probability
of getting heads all three times is
1
2
1
8
3
⎛
⎝
⎜
⎞
⎠
⎟
= .

 us, the probability of not getting heads all
three times (that is, getting tails at least once) is
1
1
8
7
8
−=.
 e correct answer is D.
175. Of the final grades received by the students in a
certain math course,
1

5
are A’s,
1
4
are B’s,
1
2
are C’s,
and the remaining 10 grades are D’s. What is the
number of students in the course?
(A) 80
(B) 1 10
(C) 160
(D) 200
(E) 400
Algebra Applied problems
Let x be the number of students in the course.
en or or
19
20
o
1
5
1
4
1
2
4
20
5

20
10
20
++
++xxx
ff the students received grades of A, B, or C.
 is means the 10 remaining grades represent
1
20
of the students in the course.
 us,
1
20
x = 10, and x = 200.
 e correct answer is D.
176. As x increases from 165 to 166, which of the following
must increase?
I. 2x – 5
II. 1 –
1
x
III.
1
x
2
–
x
(A) I only
(B) III only
(C) I and II

(D) I and III
(E) II and III
Algebra Simplifying algebraic expressions
Investigate each of the functions to determine if
they increase from x = 165 to x = 166.
I. Graphically, this represents a line with
positive slope.  erefore, the function
increases between any two values of x.
A direct computation can also be used:
[2(166) – 5] – [2(165) – 5] = 2(166 – 165) =
2, which is positive, and thus the function
increases from x = 165 to x = 166.
II. Between any two positive values of x,
1
x

decreases, and hence both –
1
x
and 1 –
1
x
increase. A direct computation can also be
used:


=


1

165
1
166
166 165
165 166
1
165 166
− =
−
()()
=
()()
,

which is positive, and thus the function
increases from x = 165 to x = 166.
III. For x = 165, the denominator is 165
2
– 165 =
(165)(165 – 1) = (165)(164), and for
x = 166, the denominator is 166
2
– 166 =
(166)(166 – 1) = (166)(165).  erefore,
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5.5 Problem Solving Answer Explanations
166
2
– 166 > 165

2
– 165, and hence

,
which shows that
decreases from
x = 165 to x = 166.
 e correct answer is C.
177. A rectangular box is 10 inches wide, 10 inches long,
and 5 inches high. What is the greatest possible
(straight-line) distance, in inches, between any two
points on the box?
(A) 15
(B) 20
(C) 25
(D) 10
(E) 10
Geometry Pythagorean theorem
 e greatest possible distance between any two
points in a rectangular solid is the space diagonal
(AD) of the rectangular solid as shown below.
A
B
C
D
10
10
5
To compute the length of AD, the Pythagorean
theorem must be used twice as follows:

For ΔABC:
For ΔACD:
 e correct answer is A.
Club
Number of
Students
Chess 40
Drama 30
Math 25
178. The table above shows the number of students in
three clubs at McAuliffe School. Although no student is
in all three clubs, 10 students are in both Chess and
Drama, 5 students are in both Chess and Math, and
6 students are in both Drama and Math. How many
different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
Arithmetic Interpretation of graphs and tables
A good way to solve this problem is to create a
Venn diagram. To determine how many students
to put in each section, begin by putting the given
shared-student data in the overlapping sections.
Put 0 in the intersection of all three clubs, 10 in the
Chess and Drama intersection, 5 in the Chess and
Math intersection, and 6 in the Drama and Math
intersection, as shown in the Venn diagram below.
25 14

14
10
0
56
DramaChess
Math
Subtracting the shared students from the totals in
each club that are listed in the table establishes the
members who belong only to that club.  rough
this process, it can be determined that the Chess
club has 25 such members (40 – 10 – 5 = 25), the
Drama club has 14 such members (30 – 10 – 6 =
14), and the Math club has 14 such members
(25 – 5 – 6 = 14). Putting the number of unshared
club members into the Venn diagram and then
adding up all the sections of the diagram gives
25 + 14 + 14 + 10 + 5 + 6 = 74 students.
 e correct answer is C.
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179. The ratio of two quantities is 3 to 4. If each of the
quantities is increased by 5, what is the ratio of these
two new quantities?
(A)
3

4
(B)
8
9
(C)
18
19
(D)
23
24
(E) It cannot be determined from the information
given.
Algebra Applied problems
Both 3 to 4 and 6 to 8 are examples of two
quantities in the ratio 3 to 4. Increasing both
numbers in each of these examples by 5 gives
8 to 9 and 11 to 13. Since
8
9
≠
11
13
, the ratio of
the two new quantities cannot be determined
from the information given.
 e correct answer is E.
180. If the average (arithmetic mean) of x and y is 60 and
the average (arithmetic mean) of y and z is 80, what is
the value of z – x ?
(A) 70

(B) 40
(C) 20
(D) 10
(E) It cannot be determined from the information
given.
Arithmetic; Algebra Statistics; Applied
problems
 e given information gives the following two
equations:
xy+
=
2
60

since the average of x and y is 60 and
yz+
=
2
80

since the average of y and z is 80.
 e two equations can be rewritten as x + y = 120
and y + z = 160. Subtracting the fi rst equation
from the second equation gives (y + z) – (x + y) =
160 – 120, or z – x = 40.
 e correct answer is B.
181. If
1
2
of the air in a tank is removed with each stroke

of a vacuum pump, what fraction of the original
amount of air has been removed after 4 strokes?
(A)
15
16
(B)
7
8
(C)
1
4
(D)
1
8
(E)
1
16
Arithmetic Operations on rational numbers
With each stroke’s removal of
1
2
of the tank’s
air, the amount of air being removed from the
tank on that stroke is equal to the amount of air
remaining in the tank after that stroke. With the
first stroke of the pump,
1
2
of the air is removed;
with the second stroke,

of the air is
removed, leaving
1
4
of the air. With the third
stroke,
of the air is removed, leaving
1
8
of the air, and with the fourth stroke,

of the air is removed.  erefore, with four strokes,
of the
air has been removed.
 e correct answer is A.
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5.5 Problem Solving Answer Explanations
182. If the two-digit integers M and N are positive and have
the same digits, but in reverse order, which of the
following CANNOT be the sum of M and N ?
(A) 18 1
(B) 165
(C) 121
(D) 99
(E) 44
Algebra Applied problems
It is given that M and N have the same digits in
reverse order. Let M = 10t + u and N = 10u + t,
where t and u are two digits.  en, M + N =

(10t + u) + (10u + t ) = 11t + 11u = 11(t + u).  is
means that any sum of the two integers M and
N must also be a multiple of 11. Of the answer
choices, only 181 is not a multiple of 11 and thus
cannot be the sum of M and N.
 e correct answer is A.
183. Car X and Car Y traveled the same 80-mile route. If
Car X took 2 hours and Car Y traveled at an average
speed that was 50 percent faster than the average
speed of Car X, how many hours did it take Car Y to
travel the route?
(A)
2
3
(B) 1
(C) 1
1
3
(D) 1
3
5
(E) 3
Arithmetic Operations on rational numbers
Substituting the given information in the formula
rate =
distance
time
, it can be determined that Car X
traveled at a rate of
80 miles

2 hours
, or 40 miles per
hour.  us, Car Y traveled at 1.50(40) = 60 miles
per hour. At this speed, Car Y would travel the
80-mile route in hours.
80
60
4
3
1
1
3
==
 e correct answer is C.
184. If the average (arithmetic mean) of the four numbers
K, 2K + 3, 3K – 5, and 5K + 1 is 63, what is the value
of K ?
(A) 11
(B) 15
3
4
(C) 22
(D) 23
(E) 25
3
10
Arithmetic Statistics
Using the formula
sum of n values
n

= average,
the given information can be expressed in the
following equation and solved for K.
KK K K+++−++
=
()()()233551
4
63
KKKK+ + + +−+
=
235351
4
63
K −
=
11 1
4
63
K11 −−=1 252
=11 253K
= 23K
 e correct answer is D.
185. If p is an even integer and q is an odd integer, which of
the following must be an odd integer?
(A)
p
q
(B) pq
(C) 2p + q
(D) 2(p + q)

(E)
3p
q
Arithmetic Properties of numbers
Since it is given that p is even and q is odd, use
these properties to test the outcome of each answer
choice to determine which one must be odd.
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A
even
odd
= even must be even
B (even)(odd) = even must be even
C 2(even) + odd
= even + odd = odd must be odd
D 2(even + odd)
= 2(odd) = even must be even
E
3(even)
odd
=
even
odd
= even must be even

 e correct answer is C.
186. Drum X is
1
2
full of oil and Drum Y, which has twice
the capacity of Drum X, is
2
3
full of oil. If all of the oil
in Drum X is poured into Drum Y, then Drum Y will be
filled to what fraction of its capacity?
(A)
3
4
(B)
5
6
(C)
11
12
(D)
7
6
(E)
11
6
Algebra Applied problems
Let y represent the capacity of Drum Y. Since Y
has twice the capacity of Drum X, Drum X has
half the capacity of Drum Y, and thus the

capacity of Drum X can be expressed as
1
2
y.
Since Drum X is half full, the amount of oil in
Drum X is equal to
According
to the given information, the initial amount of
oil in Drum Y is
2
3
y. When the oil in Drum X is
poured into Drum Y, Drum Y thus contains
 e correct answer is C.
187. If x > 0,
is what percent of x ?
(A) 6%
(B) 25%
(C) 37%
(D) 60%
(E) 75%
Algebra; Arithmetic Simplifying algebraic
expressions; Percents
Because the question asks for a percent, use a
common denominator of 100 to combine the two
terms.
, which
is 6 percent of x.
 e correct answer is A.
188. If the operation is defi ned for all a and b by the

equation a b =
a
2
b
3
, then 2
(3 –1) =
(A) 4
(B) 2
(C) −
4
3
(D) – 2
(E) – 4
Arithmetic Operations on rational numbers
Because 3 –1 is within the parentheses, its value
is computed fi rst:
3
–1 =
=

=

= –3
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5.5 Problem Solving Answer Explanations
 en, 2 (3 –1) = 2 –3, which has the
following value:
2

(3 –1) = 2 –3
=
=
= 4(–1)
= –4
 e correct answer is E.
189. The inside dimensions of a rectangular wooden box
are 6 inches by 8 inches by 10 inches. A cylindrical
canister is to be placed inside the box so that it stands
upright when the closed box rests on one of its six
faces. Of all such canisters that could be used, what is
the radius, in inches, of the one that has maximum
volume?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
Geometry Volume
 e largest cylinder that can fit in a rectangular
box will have the same height as the box and a
diameter equal to the smaller dimension of the
top of the box. By definition, the diameter of the
canister is twice its radius. One possible canister
placement in the box is illustrated below.
10 inches
8 inches
6 inches
However, since the box can rest on any of three
diff erently sized faces, it is necessary to consider

the volume of each possibility.  e formula for
calculating volume is volume = π(radius)
2
(height),
or v = π r
2
h; the possible volumes for the canister
are those shown in the following table:
Dimensions
of the box top rhv
6 by 8 3 10
90π
6 by 10 3 8
72π
8 by 10 4 6
96π
 us, the radius, in inches, of the canister having
the maximum volume is 4.
 e correct answer is B.
190. What is the units digit of (13)
4
(17)
2
(29)
3
?
(A) 9
(B) 7
(C) 5
(D) 3

(E) 1
Arithmetic Operations on rational numbers
 e units digit of 13
4
is 1, since 3 × 3 × 3 × 3 = 81;
the units digit of 17
2
is 9, since 7 × 7 = 49; and
the units digit of 29
3
is 9, since 9 × 9 × 9 = 729.
 erefore, the units digit of (13)
4
(17)
2
(29)
3
is 1,
since 1 × 9 × 9 = 81.
 e correct answer is E.
4th Street
3rd Street
2nd Street
1st Street
A
venue C
Avenue B
Avenue A
X
Y

191. Pat will walk from Intersection X to Intersection Y
along a route that is confined to the square grid of
four streets and three avenues shown in the map
above. How many routes from X to Y can Pat take
that have the minimum possible length?
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(A) 6
(B) 8
(C) 10
(D) 14
(E) 16
Arithmetic Elementary combinatorics
In order to walk from Intersection X to
Intersection Y by one of the routes of minimum
possible length, Pat must travel only upward or
rightward between the intersections on the map.
Let U represent upward movements and R
represent rightward movements. It takes 3 upward
and 2 rightward movements to complete the route.
 e following 10 routes are possible:
U U U R R U R R U U
U U R U R R R U U U
U U R R U R U U U R
U R U U R R U U R U

U R U R U R U R U U
 e correct answer is C.
192. The ratio, by volume, of soap to alcohol to water in
a certain solution is 2:50:100. The solution will be
altered so that the ratio of soap to alcohol is doubled
while the ratio of soap to water is halved. If the altered
solution will contain 100 cubic centimeters of alcohol,
how many cubic centimeters of water will it contain?
(A) 50
(B) 200
(C) 400
(D) 625
(E) 800
Arithmetic Operations on rational numbers
From


we get
,

and from


we get

.

 erefore, the amount of soap in the altered
solution can be found by solving
,

which gives
, and now
the amount of water in the altered solution can be
found by solving
, which gives
.
 e correct answer is E.
193. If 75 percent of a class answered the first question
on a certain test correctly, 55 percent answered the
second question on the test correctly, and 20 percent
answered neither of the questions correctly, what
percent answered both correctly?
(A) 10%
(B) 20%
(C) 30%
(D) 50%
(E) 65%
Arithmetic Percents
For questions of this type, it is convenient to draw
a Venn diagram to represent the conditions in the
problem. For example, the given information can
be depicted:
20%
Q1
75%
Q2
55%
In the diagram it can be seen that the 80% of the
class answering a question correctly is represented
by the two circles. Let x represent the percent of

the class that answered both questions correctly,
that is, the shaded region above. Since the sum of
the circles minus their overlap equals 80% of the
class, the information given in the problem can
then be expressed as 75% + 55% – x = 80%.  is
equation can be solved for x as follows:
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5.5 Problem Solving Answer Explanations
75% + 55% – x = 80%
130% – x = 80%
–x = –50%
x = 50%
 e correct answer is D.
O
A
y = x
y
x
194. In the rectangular coordinate system above, the line
y = x is the perpendicular bisector of segment AB (not
shown), and the x-axis is the perpendicular bisector of
segment BC (not shown). If the coordinates of point A
are (2,3), what are the coordinates of point C ?
(A) (–3,–2)
(B) (–3,2)
(C) (2,–3)
(D) (3,–2)
(E) (2,3)
Geometry Simple coordinate geometry

Since the line y = x is the perpendicular bisector
of AB
, B is the reflection of A through this line.
In any reflection through the line y = x, the
x -coordinate and the y-coordinate of a point
become interchanged.  us, if the coordinates of
A are (2,3), the coordinates of B are (3,2).
O
A
y = x
y
x
3
2
1
1 2 3
B
Since the x-axis is the perpendicular bisector of
BC
, C is the reflection of B through the x-axis. In
any reflection through the x-axis, the x-coordinate
remains the same, and the sign of the y-coordinate
changes.
O
A
y = x
y
x
3
2

1
–1
–2
1 2
B
C
Since the coordinates of B are (3,2), the
coordinates of C are therefore (3,–2).
 e correct answer is D.
195. A store currently charges the same price for each
towel that it sells. If the current price of each towel
were to be increased by $1, 10 fewer of the towels
could be bought for $120, excluding sales tax. What is
the current price of each towel?
(A) $ 1
(B) $ 2
(C) $ 3
(D) $ 4
(E) $12
Algebra Applied problems
Let p be the current price per towel, and let n be
the number of towels that can be bought for $120.
 en the information in the problem can be
expressed in the following equations:
(i) pn = 120
(ii) (p + 1)(n – 10) = 120 or equivalently
(iii) pn + n – 10p – 10 = 120.
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 en replace pn in (iii) with 120 to get:
120 + n – 10p – 10 = 120
n – 10p – 10 = 0
n – 10(p + 1) = 0
n = 10(p + 1)
np = 10p(p + 1)
120 = 10p(p + 1)
12 = p(p + 1)
0 = p
2
+ p – 12
0 = (p + 4)(p – 3)
p = 3
 e correct answer is C.
Number of Solid-Colored Marbles in Three Jars
Jar
Number o
f
red marbles
Number of
green marbles
Total number
of red and
green marbles
P x y 80
Q y z 120

R x z 160
196. In the table above, what is the number of green
marbles in Jar R ?
(A) 70
(B) 80
(C) 90
(D) 100
(E) 110
Arithmetic; Algebra Interpretation of tables;
Applied problems
First, set up an equation to find the total number
of marbles in the three jars as follows:
x + y + y + z + x + z = 80 + 120 + 160
2x + 2y + 2z = 360 combine the like terms
x + y + z = 180 divide both sides by 2
 en, since it can be seen from the table that the
number of green marbles in Jar R is z, solve for z
to answer the problem. To do this most efficiently,
use the information from the table for Jar P, which
is that x + y = 80.
x + y + z = 180
80 + z = 180 substitute 80 for x + y
z = 100
 e correct answer is D.
197. A point on the edge of a fan blade that is rotating in a
plane is 10 centimeters from the center of the fan.
What is the distance traveled, in centimeters, by this
point in 15 seconds when the fan runs at the rate of
300 revolutions per minute?
(A) 750π

(B) 1,500π
(C) 1,875π
(D) 3,000π
(E) 7,500π
Geometry Circles; Circumference;
Applied problems
Since 15 seconds is
15
60
=
1
4
minute, if the fan
blade is rotating at the rate of 300 revolutions
per minute, then in
1
4
minute it rotates
300
4
= 75 revolutions.  e dist
ance the point on
the edge of the fan blade rotates in 75 revolutions
is 75 times the circumference of a circle with
radius 10 centimeters.  e circumference C of a
circle with radius r is C = 2πr.  us the distance
the point travels is 75[2π(10)] = 1,500π.
 e correct answer is B.
198. If n = 4p, where p is a prime number greater than 2,
how many different positive even divisors does n have,

including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
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5.5 Problem Solving Answer Explanations
Arithmetic Properties of numbers
Since p is a prime greater than 2, p must be odd.
 erefore, the possible even divisors of n = 4p are
2, 4, 2p, and 4p. Alternatively, choose such a
prime, for example p = 3, and determine the
number of positive even divisors that n = 4p =
12 has.
 e correct answer is C.
I. 72, 73, 74, 75, 76
II. 74, 74, 74, 74, 74
III. 62, 74, 74, 74, 89
199. The data sets I, II, and III above are ordered from
greatest standard deviation to least standard deviation
in which of the following?
(A) I, II, III
(B) I, III, II
(C) II, III, I
(D) III, I, II
(E) III, II, I
Arithmetic Statistics
 e data set with the least standard deviation will

be the data set with elements most closely
clustered around the mean of the data set, and the
data set with the greatest standard deviation will
be the data set with elements that are spread out
farthest from the mean of the data set.
Because set I is symmetric about 74 (73 is 1 less
than 74 and 75 is 1 more than 74; 72 is 2 less
than 74 and 76 is 2 more than 74), the mean of
set I is 74. Because every number in set II is 74,
the mean of set II is 74.  e mean of set III is
62
5
+
()
+
==
374 89
373
5
74 6. .

e element
s of set II do not deviate at all from
74, so set II has the least standard deviation.  e
most that any element of set I diff ers from 74 is 2,
but there are elements of set III that diff er from
74.6 by 12.6 and 14.4.  erefore, set III has a
greater standard deviation than set I, which has a
greater standard deviation than set II.
 e correct answer is D.

200. Of the 50 researchers in a workgroup, 40 percent will
be assigned to Team A and the remaining 60 percent
to Team B. However, 70 percent of the researchers
prefer Team A and 30 percent prefer Team B. What is
the lowest possible number of researchers who will
NOT be assigned to the team they prefer?
(A) 15
(B) 17
(C) 20
(D) 25
(E) 30
Arithmetic Percents
 e number of researchers assigned to Team A
will be (0.40)(50) = 20, and so 30 will be assigned
to Team B.  e number of researchers who prefer
Team A is (0.70)(50) = 35, and the rest, 15, prefer
Team B.
If all 15 who prefer Team B are assigned to
Team B, which is to have 30 researchers, then
15 who prefer Team A will need to be assigned
to Team B. Alternatively, since there are only
20 spots on Team A, 35 – 20 = 15 who prefer
Team A but will have to go to Team B instead.
 e correct answer is A.
201. If m is the average (arithmetic mean) of the fi rst
10 positive multiples of 5 and if M is the median
of the fi rst 10 positive multiples of 5, what is the
value of M – m ?
(A) –5
(B) 0

(C) 5
(D) 25
(E) 27.5
Arithmetic Statistics
 e fi rst 10 positive multiples of 5 are 5, 10, 15,
20, 25, 30, 35, 40, 45, and 50. From this, the
average (arithmetic mean) of the 10 multiples,
that is,
sum ofvalues
number ofvalues
, can be calculated:
m =
+++++++++
==
5101520253035404550
10
275
10
27 5
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Since there is an even number of multiples, the
median, M, is the average of the middle two
numbers, 25 and 30:
M =

25 + 30
2
= 27.5.
 erefore, the median minus the average is:
M – m = 27.5 – 27.5 = 0.
 is problem can also be solved as follows. Since
the values can be grouped in pairs (i.e., 5 and 50,
10 and 45, 15 and 40, etc.), each of which is
symmetric with respect to the median, it follows
that the average and median are equal.
 e correct answer is B.
202. If m > 0 and x is m percent of y, then, in terms of m, y
is what percent of x ?
(A) 100m
(B)
1
100m
(C)
1
m
(D)
10
m
(E)
10,000
m
Arithmetic Percents
 e information that x is m percent of y can be
expressed as x
m

y=
100
and solved for y as follows:

x
m
y=
100
100
m
xy=
 en, to convert the fraction
100
m
to an equivalent
percent, multiply by 100, thus obtaining the value
10,000
m
.
 e correct answer is E.
203. What is the 25th digit to the right of the decimal point
in the decimal form of
6
11
?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Arithmetic Properties of numbers
 e fraction in its decimal form is
6
11
= 0.545454.… Every odd-numbered digit to
the right of the decimal point is 5, so the 25th
digit must be 5.
 e correct answer is C.
204. John and Mary were each paid x dollars in advance to
do a certain job together. John worked on the job for
10 hours and Mary worked 2 hours less than John. If
Mary gave John y dollars of her payment so that they
would have received the same hourly wage, what was
the dollar amount, in terms of y, that John was paid in
advance?
(A) 4y
(B) 5y
(C) 6y
(D) 8y
(E) 9y
Algebra Applied problems
Let w be the amount of Mary and John’s same
hourly wage. To set their hourly pay equal, John,
who worked 10 hours, needs to be paid 10w, and
Mary, who worked 8 hours, needs to be paid 8w.
Since Mary gave John y dollars, Mary now has
x – y dollars and John now has x + y dollars.
 eir pay can thus be expressed as follows:
x – y = 8w Mary’s pay
x + y = 10w John’s pay

Subtract the first equation from the second and
solve for w.
2y = 2w
y = w
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5.5 Problem Solving Answer Explanations
Substitute y for w in the second equation, and
solve for x, the amount each was paid in advance.
x + y = 10 y
x = 9 y
 e correct answer is E.
y
x
P (4,0)
O
205. In the rectangular coordinate system above, if point R
(not shown) lies on the positive y-axis and the area of
triangle ORP is 12, what is the y-coordinate of point R ?
(A) 3
(B) 6
(C) 9
(D) 12
(E) 24
Geometry Simple coordinate geometry; Area
Since O and P of triangle ORP are already drawn
and R has to be on the positive y-axis, the triangle
is a right triangle with its base length the distance
from the origin O (0,0) to P (4,0), which is 4.
Since the area of a triangle =

(base)(height)
2
,
the information about the area and base can be
expressed as follows and solved for the height of
triangle OPR:
12 =
(4)(height)
2
12 = 2(height) simplify the right side
6 = height solve for the height
On the y-axis, the x-coordinate is 0 and the
y-coordinate is the distance above the axis that
the point is located. In this case, the y-coordinate
is the height of the triangle.
 e correct answer is B.
206. Car A is 20 miles behind Car B, which is traveling in
the same direction along the same route as Car A.
Car A is traveling at a constant speed of 58 miles
per hour and Car B is traveling at a constant speed
of 50 miles per hour. How many hours will it take for
Car A to overtake and drive 8 miles ahead of Car B ?
(A) 1.5
(B) 2.0
(C) 2.5
(D) 3.0
(E) 3.5
Arithmetic Operations on rational numbers
Understand that Car A first has to travel 20 miles
to catch up to Car B and then has to travel an

additional 8 miles ahead of Car B, for a total of
28 extra miles to travel relative to Car B. It can
be stated that Car A is traveling 58 – 50 = 8 miles
per hour faster than Car B. Solving the
distance = (rate)(time) formula for time yields
distance
rate
= time.
By substitution into this formula, it will take
Car A
28 miles
8 miles per hour
= 3.5 hours to overtake
and drive 8 miles ahead of Car B.
 e correct answer is E.
207. For the past n days, the average (arithmetic mean)
daily production at a company was 50 units. If today’s
production of 90 units raises the average to 55 units
per day, what is the value of n ?
(A) 30
(B) 18
(C) 10
(D) 9
(E) 7
Arithmetic; Algebra Statistics; Applied
problems; Simultaneous equations
Let x be the total production of the past n days.
Using the formula average =
sum of values
number of values

,
the information in the problem can be expressed
in the following two equations:
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50 =
x
n
daily average of 50 units
over the past n days
55
90
1
=
+
+
x
n
increased daily average
when including today’s
90 units
Solving the first equation for x gives x = 50n.
 en substituting 50n for x in the second
equation gives the following that can be solved
for n:

55
50 90
1
=
+
+
n
n
55(n + 1) = 50n + 90 multiply both sides
by (n + 1)
55n + 55 = 50n + 90 distribute the 55
5n = 35 subtract 50n and 55 from
both sides
n = 7 divide both sides by 5
 e correct answer is E.
x
x
+
−
⎛
⎝
⎜
⎞
⎠
⎟
1
1
2
208. If x ≠ 0 and x ≠ 1, and if x is replaced by

1
x
everywhere
in the expression above, then the resulting expression
is equivalent to
(A)
x
x
+
−
⎛
⎝
⎜
⎞
⎠
⎟
1
1
2
(B)
x
x
−
+
⎛
⎝
⎜
⎞
⎠
⎟

1
1
2
(C)
x
x
2
2
1
1
+
−
(D)
x
x
2
2
1
1
−
+
(E)
−
−
+
⎛
⎝
⎜
⎞
⎠

⎟
x
x
1
1
2
Algebra Simplifying algebraic expressions
Substitute
1
x
for x in the expression and simplify.

1
1
1
1
2
x
x
+
−
⎛
⎝
⎜
⎜
⎜
⎞
⎠
⎟
⎟

⎟
Multiply the numerator and denominator inside
the parentheses by x to eliminate the compound
fractions.
x
x
x
x
1
1
1
1
2
+
⎛
⎝
⎞
⎠
−
⎛
⎝
⎞
⎠
⎛
⎝
⎜
⎜
⎜
⎞
⎠

⎟
⎟
⎟
Distribute the x’s.
1
1
2
+
−
⎛
⎝
⎜
⎞
⎠
⎟
x
x
Since this is not one of the answer choices,
it is necessary to simplify further. With the
knowledge that 1 + x = x + 1 and 1 – x = –(x – 1),
it can be stated that
because the negative, when squared, is positive.
 e correct answer is A.
y°
z°
x°
209. In the figure above, if z = 50, then x + y =
(A) 230
(B) 250
(C) 260

(D) 270
(E) 290
Geometry Angles; Measures of angles
Refer to the figure below.
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5.5 Problem Solving Answer Explanations
A
E
BD C
y°
z°
x°
Triangle ABC is a right triangle, and segment
AB is parallel to segment ED since they are
both perpendicular to the same segment (BC
).
 erefore, m∠DEC = m∠BAC = z° = 50°. So,
since ∠DEC and ∠AED form a straight line
at E, y + 50 = 180, or y = 130.
 e measure of an exterior angle of a triangle is
the sum of the measures of the nonadjacent
interior angles.  us,
m∠x = m∠z + 90°, or
m∠x = 50° + 90° = 140°
 us, x + y = 140 + 130 = 270.
 e correct answer is D.
O
1
y

x

1
210. In the coordinate system above, which of the following
is the equation of line
C ?
(A) 2x – 3y = 6
(B) 2x + 3y = 6
(C) 3x + 2y = 6
(D) 2x – 3y = –6
(E) 3x – 2y = –6
Geometry Simple coordinate geometry
 e line is shown going through the points (0,2)
and (3,0).  e slope of the line can be found with
the formula slope =
change in
change in

y
x
yy
xx
=
−
−
21
21
,

for two points (x

1
,y
1
) and (x
2
,y
2
).  us, the slope
of this line equals
. Using the formula
for a line of y = mx + b, where m is the slope and
b is the y-intercept (in this case, 2), an equation for
this line is
y
x=− +
2
3
2
. Since this equation must
be compared to the available answer choices, the
following further steps should be taken:
y
x=− +
2
3
2
3y = –2x + 6 multiply both sides by 3
2x + 3y = 6 add 2x to both sides
 is problem can also be solved as follows. From
the graph, when x = 0, y is positive; when y = 0,

x is positive.  is eliminates all but B and C. Of
these, B is the only line containing (0,2). Still
another way is to use (0,2) to eliminate A, C, and
E, and then use (3,0) to eliminate D.
 e correct answer is B.
211. If a two-digit positive integer has its digits reversed,
the resulting integer differs from the original by 27.
By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Algebra Applied problems
Let the one two-digit integer be represented by
10t + s, where s and t are digits, and let the other
integer with the reversed digits be represented
by 10s + t.  e information that the diff erence
between the integers is 27 can be expressed in
the following equation, which can be solved for
the answer.
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(1 0 s + t ) − (10t + s) = 27
10s + t − 10t − s = 27 distribute the negative

9s − 9t = 27 combine like terms
s − t = 3 divide both sides by 9
 us, it is seen that the two digits s and t diff er
by 3.
 e correct answer is A.
O
y
x
C
212. The circle with center C shown above is tangent to
both axes. If the distance from O to C is equal to k,
what is the radius of the circle, in terms of k ?
(A) k
(B)

k
2
(C)

k
3
(D)
k
2
(E)
k
3
Geometry Circles; Simple coordinate
geometry
In a circle, all distances from the circle to the

center are the same and called the radius, r.
O
y
x
C
r
k
r
Since the horizontal distance from C to the y-axis
is also a radius, the base of the triangle drawn will
be r as well.  is creates a right triangle, and so
the Pythagorean theorem (or a
2
+ b
2
= c
2
) applies.
r
2
+ r
2
= k
2
substitute values into


Pythagorean theorem;
2r
2

= k
2
combine like terms

r
k
2
2
2
=
divide both sides by 2

r
k
=
2
2
take the square root of
both sides

r
k
=
2
simplify the square root
 e correct answer is B.
213. In an electric circuit, two resistors with resistances x
and y are connected in parallel. In this case, if r is the
combined resistance of these two resistors, then the
reciprocal of r is equal to the sum of the reciprocals of

x and y. What is r in terms of x and y ?
(A) xy
(B) x + y
(C)
1
x + y
(D)
xy
x + y
(E)
x + y
xy
Algebra Applied problems
Note that two numbers are reciprocals of each
other if and only if their product is 1.  us the
reciprocals of r, x, and y are
11 1
rx y
,, and
,
respectively. So, according to the problem,
111
rxy
=+.
To solve this equation for r, begin by
creating a common denominator on the right side
by multiplying the first fraction by
y
y
and the

second fraction by
x
x
:
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5.5 Problem Solving Answer Explanations
111
rxy
=+
1
r
y
xy
x
xy
=+
1
r
xy
xy
=
+
combine the fractions on the
right side
r
xy
xy
=
+

invert the fractions on both sides
 e correct answer is D.
214. Xavier, Yvonne, and Zelda each try independently to
solve a problem. If their individual probabilities for
success are
1
4
,
1
2
, and
5
8
, respectively, what is the
probability that Xavier and Yvonne, but not Zelda, will
solve the problem?
(A)
11
8
(B)
7
8
(C)
9
64
(D)
5
64
(E)
3

64
Arithmetic Probability
Since the individuals’ probabilities are
independent, they can be multiplied to figure out
the combined probability.  e probability of
Xavier’s success is given as
1
4
, and the probability
of Yvonne’s success is given as
1
2
. Since the
probability of Zelda’s success is given as
5
8
, then
the probability of her NOT solving the problem
is
 us, the combined probability is
 e correct answer is E.
215. If
1
x
1
x + 1
1
x + 4
– =
, then x could be

(A) 0
(B) –1
(C) –2
(D) –3
(E) –4
Algebra Second-degree equations
Solve the equation for x. Begin by multiplying all
the terms by x(x + 1)(x + 4) to eliminate the
denominators.

(x + 1)(x + 4) – x(x + 4) = x(x + 1)
(x + 4)(x + 1 – x) = x(x + 1) factor the (x + 4) out
front on the left side
(x + 4)(1) = x(x + 1) simplify
x + 4 = x
2
+ x distribute the x on
the right side
4 = x
2
subtract x from both
sides
±2 = x take the square root
of both sides
Both –2 and 2 are square roots of 4 since (–2)
2
= 4
and (2)
2
= 4.  us, x could be –2.

 is problem can also be solved as follows.
Rewrite the left side as ,
then set equal to the right side to get
1
1
1
4
xx
x
+
()
=
+
. Next, cross multiply:
(1)(x + 4) = x(x + 1)(1).  erefore, x + 4 = x
2
+ x,
or x
2
= 4, so x = ± 2.
 e correct answer is C.
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216.
1

2
1
4
1
16
32 1
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
=
−− −
(A)
1
2
48

⎛
⎝
⎜
⎞
⎠
⎟
−
(B)
1
2
11
⎛
⎝
⎜
⎞
⎠
⎟
−
(C)
1
2
6
⎛
⎝
⎜
⎞
⎠
⎟
−
(D)

1
8
11
⎛
⎝
⎜
⎞
⎠
⎟
−
(E)
1
8
6
⎛
⎝
⎜
⎞
⎠
⎟
−
Arithmetic Operations on rational numbers
It is clear from the answer choices that all three
factors need to be written with a common
denominator, and they thus become
1
2
1
2
1

4
1
2
1
2
33
22
2
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠

=
⎛
⎝
⎞
⎠
−−
−
−
−−
−
−
−
−
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞

⎠
⎛
⎝
⎞
⎠
4
14
1
4
3
1
16
1
2
1
2
1
2
1
So,
44
1
16
1
2
1
2
1
2
1

2
21
344
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
=
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
=
⎛
⎝
−−
−−−
⎞⎞

⎠
=
⎛
⎝
⎞
⎠
−−− −344 11
1
2

.
 e correct answer is B.
217. In a certain game, a large container is filled with red,
yellow, green, and blue beads worth, respectively, 7,
5, 3, and 2 points each. A number of beads are then
removed from the container. If the product of the point
values of the removed beads is 147,000, how many
red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0
Arithmetic Properties of numbers
From this, the red beads represent factors of 7 in
the total point value of 147,000. Since 147,000 =
147(1,000), and 1,000 = 10
3
, then 147 is all that
needs to be factored to determine the factors of 7.

Factoring 147 yields 147 = (3)(49) = (3)(7
2
).  is
means there are 2 factors of 7, or 2 red beads.
 e correct answer is D.
218.
If , then
2
1
2
1
+
==
y
y
(A) – 2
(B) −
1
2
(C)
1
2
(D) 2
(E) 3
Algebra First-degree equations
Solve for y.
2
1
2
1

+
=
y
1
2
2+=
y
multiply both sides by
1
2
+
y

2
1
y
=
subtract 1 from each side
y = 2 solve for y
 e correct answer is D.
219. If a, b, and c are consecutive positive integers and
a < b < c, which of the following must be true?
I. c – a = 2
II. abc is an even integer.
III.
a + b + c
3
is an integer.
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259

5.5 Problem Solving Answer Explanations
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
Arithmetic Properties of numbers
Since a, b, and c are consecutive positive integers
and a < b < c, then b = a + 1 and c = a + 2.
I. c - a = (a + 2) - a = 2 MUST be true
II. (odd)(even)(odd) = even MUST be true
(even)(odd)(even) = even MUST be true
III.
abc a a a++
=
++
()
++
()
3
12
3
33
a
ab=
+
=+=
3
1
b is an integer MUST be true

 e correct answer is E.
220. A part-time employee whose hourly wage was
increased by 25 percent decided to reduce the
number of hours worked per week so that the
employee’s total weekly income would remain
unchanged. By what percent should the number of
hours worked be reduced?
(A) 12.5%
(B) 20%
(C) 25%
(D) 50%
(E) 75%
Algebra Applied problems
Let w represent the original hourly wage.
Letting h be the original number of hours the
employee worked per week, the original weekly
income can be expressed as wh. Given a 25%
increase in hourly wage, the employee’s new wage
is thus 1.25w. Letting H be the reduced number
of hours, the problem can then be expressed as:
1.25wH = wh (new wage)(new hours) =
(original wage)(original hours)
By dividing both sides by w, this equation can be
solved for H:
1.25H = h
H = 0.8h
Since the new hours should be 0.8 = 80% of the
original hours, the number of hours worked
should be reduced by 20 percent.
 e correct answer is B.

221. Of the 200 students at College T majoring in one or
more of the sciences, 130 are majoring in chemistry
and 150 are majoring in biology. If at least 30 of the
students are not majoring in either chemistry or
biology, then the number of students majoring in both
chemistry and biology could be any number from
(A) 20 to 50
(B) 40 to 70
(C) 50 to 130
(D) 110 to 130
(E) 110 to 150
Arithmetic Operations on rational numbers
A Venn diagram will help with this problem.
 ere are two extremes that need to be
considered: (1) having the least number of
students majoring in both chemistry and biology
and (2) having the greatest number of students
majoring in both chemistry and biology.
(1) If at least 30 science majors are not majoring
in either chemistry or biology, then at most
200 – 30 = 170 students can be majoring in
either or both. Since there are 130 + 150 =
280 biology and chemistry majors (some of
whom are individual students majoring in both
areas), then there are at least 280 – 170 = 110
majoring in both.  e diagram following shows
this relationship.
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®
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20 110 40
Chemistry
Biology
170 TOTAL STUDENTS
FOR CHEMISTRY AND BIOLOGY MAJORS
(2)  e maximum number of students who can
be majoring in both chemistry and biology is 130,
since 130 is the number given as majoring in
chemistry, the smaller of the two subject areas.
Logically, there cannot be more double majors
than there are majors in the smaller field.  e
diagram below shows this relationship in terms of
the given numbers of majors in each subject area.
0 130 20
Chemistry
Biology
Additionally, from this diagram it can be seen
that the total number of students who are
majoring in chemistry, or in biology, or in both
is 130 + 20 = 150.  us, there are 200 – 150 =
50 students who are neither chemistry nor biology
majors.  is number is not in conflict with the
condition that 30 is the minimum number of
nonchemistry and nonbiology majors.
 us, the number of students majoring in both
chemistry and biology could be any number from

a minimum of 110 to a maximum of 130.
 e correct answer is D.
222. If 5 –
6
x
= x, then x has how many possible values?
(A) None
(B) One
(C) Two
(D) A fi nite number greater than two
(E) An infi nite number
Algebra Second-degree equations
Solve the equation to determine how many values
are possible for x.
5 –
6
x
= x
5x – 6 = x
2
0 = x
2
– 5x + 6
0 = (x – 3)(x – 2)
x = 3 or 2
 e correct answer is C.
223. Seed mixture X is 40 percent ryegrass and 60 percent
bluegrass by weight; seed mixture Y is 25 percent
ryegrass and 75 percent fescue. If a mixture of X and
Y contains 30 percent ryegrass, what percent of the

weight of the mixture is X ?
(A) 10%
(B) 33
1
3
%
(C) 40%
(D) 50%
(E) 66
2
3
%
Algebra Applied problems
Let X be the amount of seed mixture X in the
final mixture, and let Y be the amount of seed
mixture Y in the final mixture.  e final
mixture of X and Y needs to contain 30 percent
ryegrass seed, so any other kinds of grass seed are
irrelevant to the solution to this problem.  e
information about the ryegrass percentages for X,
Y, and the final mixture can be expressed in the
following equation and solved for X.
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261
5.5 Problem Solving Answer Explanations
0.40X + 0.25Y = 0.30(X + Y)
0.40X + 0.25Y = 0.30X + 0.30Y distribute the
0.30 on the
right side
0.10X = 0.05Y subtract 0.30X

and 0.25Y from
both sides
X = 0.5Y divide both
sides by 0.10
Using this, the percent of the weight of the
combined mixture (X + Y) that is X is
X
XY
Y
YY
Y
Y+
=
+
=== =
05
05
05
15
05
15
0333 33
1
3
.
.
.
.
.
.

.%
 e correct answer is B.
224. If n is a positive integer, then n(n + 1)(n + 2) is
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
Arithmetic Properties of numbers
 e numbers n, n + 1, and n + 2 are consecutive
integers.  erefore, either their product is
(odd)(even)(odd) = even, or their product is
(even)(odd)(even) = even. In either case, the
product of n(n + 1)(n + 2) is even.  us, each
of answer choices A, B, and C is false.
A statement is false if a counterexample can be
shown. Test the statement using an even multiple
of 3 as the value of n in the equation. When
n = 6, n(n + 1)(n + 2) = 6(7)(8) = 336. Since in this
counterexample n is even but 336 is still divisible
by 3, answer choice D is shown to be false.
When n is even (meaning divisible by 2), n + 2
is also even (and also divisible by 2). So
n(n + 1)(n + 2) is always divisible by 4.
 e correct answer is E.
225. A straight pipe 1 yard in length was marked off in
fourths and also in thirds. If the pipe was then cut into
separate pieces at each of these markings, which of
the following gives all the different lengths of the
pieces, in fractions of a yard?

(A)
1
6
and
1
4
only
(B)
1
4
and
1
3
only
(C)
1
6
,
1
4
, and
1
3
(D)
1
12
,
1
6
, and

1
4
(E)
1
12
,
1
6
, and
1
3
Arithmetic Operations on rational numbers
BC DE F
0
11
1
4
1
3
1
2
2
3
3
4
A
 e number line above illustrates the markings
on the pipe. Since the pipe is cut at the five
markings, six pieces of pipe are produced.
 e length of each piece, as a fraction of a yard,

is given in the following table.
Pipe piece Length
A
B
C
D
E
F
 e correct answer is D.
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226. If
= 5 × 10
7
, then m – k =
(A) 9
(B) 8
(C) 7
(D) 6
(E) 5
Arithmetic Operations on rational numbers
 e left side is easier to work with when the
expressions are rewritten so that integers are
involved:

= 5 × 10
7


= 5 × 10
7

× = 5 × 10
7
5 × = 5 × 10
7

= 10
7
10
m – 4 – (k – 2)
= 10
7
m – 4 – (k – 2) = 7
m – k – 2 = 7
m – k = 9
 e correct answer is A.
227. If x + y = a and x – y = b, then 2xy =
(A)
(B)
(C)
(D)
(E)
Algebra Simplifying algebraic expressions
Begin by adding the two given equations to

establish a value for x. Adding x + y = a and
x – y = b gives 2x = a + b and thus x =
.
 en, substitute this value of x into the fi rst
equation and solve for y:
Finally, solve the equation, substituting the values
now established for x and y:
 is problem can also be solved as follows: Since
the squares of x + y and x – y, when expanded,
each include the expression x
2
+ y
2
along with a
multiple of xy, we can obtain a multiple of xy by
subtracting these squares:
a
2
– b
2
= (x + y)
2
– (x – y)
2
= x
2
+ 2xy + y
2
– (x
2

– 2xy + y
2
)
= 4xy
= 2(2xy)

= 2xy
 e correct answer is A.
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263
5.5 Problem Solving Answer Explanations
p, r, s, t, u
228. An arithmetic sequence is a sequence in which
each term after the first is equal to the sum of the
preceding term and a constant. If the list of letters
shown above is an arithmetic sequence, which of the
following must also be an arithmetic sequence?
I. 2p, 2r, 2s, 2t, 2u
II. p – 3, r – 3, s – 3, t – 3, u – 3
III. p
2
, r
2
, s
2
, t
2
, u
2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
Algebra Concepts of sets; Functions
It follows from the definition of arithmetic
sequence given in the first sentence that there is a
constant c such that r – p = s – r = t – s = u – t = c.
To test a sequence to determine whether it is
arithmetic, calculate the diff erence of each pair
of consecutive terms in that sequence to see if a
constant diff erence is found.
I. 2r – 2p = 2(r – p) = 2c
2s – 2r = 2(s – r) = 2c
2t – 2s = 2(t – s) = 2c
2u – 2t = 2(u – t) = 2c MUST be
arithmetic
II. (r – 3) – (p – 3) = r – p = c MUST be
arithmetic
Since all values are just three less than the
original, the same common diff erence applies.
III. r
2
− p
2
= (r − p)(r + p) = c(r + p)
s
2
− r

2
= (s − r)(s + r) = c(s + r) NEED NOT
be arithmetic
Since p, r, s, t, and u are an arithmetic sequence,
r + p ≠ s + r, because p ≠ s unless c = 0.
 e correct answer is D.
229. Right triangle PQR is to be constructed in the xy-plane
so that the right angle is at P and PR is parallel to the
x-axis. The x- and y-coordinates of P, Q, and R are to
be integers that satisfy the inequalities –4 ≤ x ≤ 5 and
6 ≤ y ≤ 16. How many different triangles with these
properties could be constructed?
(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100
Geometry; Arithmetic Simple coordinate
geometry; Elementary combinatorics
In the xy-plane, right triangle PQR is located in
the rectangular region determined by −4 ≤ x ≤ 5
and 6 ≤ y ≤ 16 (see following illustration).
y
x
P
6
5–4
16
Since the coordinates of points P, Q, and R are
integers, there are 10 possible x values and 11

possible y values, so point P can be any one of
10(11) = 110 points in the rectangular area.
Since PR

has to be horizontal, R has the same
y value as P and can have 9 other x values. PQ

has to be vertical, so Q has the same x value as
P and can have 10 other y values.  is gives
110(9)(10) = 9,900 possible triangles.
 e correct answer is C.
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