Mathematics: Section 2
Mathematics Question 1
Choice (A) is correct. If 2
a
k
1 9, then 2
a
k
9 1 8, and so
a
k
8
2
4.
a
a
1 would equal (2)(5) + 1 = 11.
were equal to 5, then 2
k
k
a
1 is equal to 9, not 11. Therefore,
cannot equal 5.
k
Choice (B) is not correct. If
However, 2
a
k
a
a
1 would equal (2)(6) + 1 = 13.
were equal to 6, then 2
k
k
a
1 is equal to 9, not 13. Therefore,
cannot equal 6.
k
Choice (C) is not correct. If
However, 2
a
k
a
a
1 would equal (2)(7) + 1 = 15.
were equal to 7, then 2
k
k
a
1 is equal to 9, not 15. Therefore,
cannot equal 7.
k
Choice (D) is not correct. If
However, 2
a
k
a
a
1 would equal (2)(8) + 1 = 17.
were equal to 8, then 2
k
k
a
1 is equal to 9, not 17. Therefore,
cannot equal 8.
k
Choice (E) is not correct. If
However, 2
a
k
Mathematics Question 2
Choice (E) is correct. In rectangle OBCD, point B (0, 5) is 5 units directly above point O (0, 0).
Hence point C must likewise be 5 units directly above point D (10, 0). Therefore, the coordinates
of point C must be (10, 5).
Choice (A) is not correct. The point with coordinates (0, 10) is 5 units directly above point B (0, 5).
But point C is 5 units directly above point D (10, 0), not point B, and has coordinates (10, 5).
Choice (B) is not correct. The point with coordinates (5, 0) is 5 units to the left of point D (10, 0).
But point C is 5 units directly above, not to the left of, point D and has coordinates (10, 5).
Choice (C) is not correct. Point C is 5 units directly above point D (10, 0) and thus has
coordinates (10, 5). The point with coordinates (5, 5) is 5 units to the left of point C.
Choice (D) is not correct. Point C is 5 units directly above point D (10, 0) and thus has
coordinates (10, 5). The point with coordinates (5, 10) is 5 units to the left of and 5 units above
point C.
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Mathematics Question 3
Choice (E) is correct. Of the 4 cans of paint, only one contains green paint, and so the remaining
3 do not contain green paint. Therefore, the probability that the can chosen by the painter does
not contain green paint is
3
.
4
Choice (A) is not correct. The probability that the can chosen by the painter contains green paint
1
. However, the question asks for the probability that the can chosen does not contain green
4
1
1 3
paint, which is 1
, not .
4
4 4
is
Choice (B) is not correct. The ratio of the number of cans that contain green paint to the number
that do not contain green paint is 1 to 3. However, the probability that the can chosen by the
painter does not contain green paint is
3
1
, not .
4
3
Choice (C) is not correct. Of the 4 cans of paint, 3, not 2, do not contain green paint. Therefore,
the probability that the can chosen by the painter does not contain green paint is
3
1
, not .
4
2
Choice (D) is not correct. If 1 of a total of 3 cans of paint contained green paint, then the
probability that the can chosen by the painter does not contain green paint would be
2
.
3
However, 1 of a total of 4 cans contain green paint, so the probability that the can chosen by the
painter does not contain green paint is
3
2
, not .
4
3
Mathematics Question 4
Choice (C) is correct. Considering each of the given possibilities separately, substitute the values
2
given in I, II and III for y in the equation 49 = (7y) .
2
2
I When −1 is substituted for y in the equation, the result is 49 = [(7)(−1)] = (−7) = 49, so y could
equal −1.
2
2
II When 1 is substituted for y in the equation, the result is 49 = [(7)(1)] = (7) = 49, so y could
equal 1.
2
2
III When 7 is substituted for y in the equation, the result is 49 = [(7)(7)] = 49 Since 49 is not
2
equal to 49 , y cannot equal 7.
Therefore, of the three values, only the values in I and II could be values of y that satisfy the
equation.
Choice (A) is not correct. The y value in II satisfies the equation, but so does the y value in I.
Choice (B) is not correct. The y value in III does not satisfy the equation.
Choice (D) is not correct. The y value in I satisfies the equation, but the y value in III does not.
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Choice (E) is not correct. The y value in II satisfies the equation, but the y value in III does not.
Mathematics Question 5
Choice (B) is correct. Square ACEG has one side of length x. It follows that AC = CE = EG = GA
= x. Triangles ABC, CDE, EFG, and GHA are all equilateral, and each has one side of length x,
so AB = BC = CD = DE = EF = FG = GH = HA = x. Thus each of the 12 segments in the figure
has length x. Therefore, the sum of the lengths of all the segments in the figure is 12x.
Choice (A) is not correct. The perimeter of the figure consists of 8 segments, each of length x,
and so the perimeter of the figure is 8x. However, the question asks for the sum of the lengths of
all the segments in the figure, which is 12x.
Choice (C) is not correct. Each of the 4 triangles in the figure has perimeter 3x, and square ACEG
has perimeter 4x, so the sum of the perimeters of all the triangles and the square is (4)(3x) + 4x =
12x + 4x = 16x. However, the question asks for the sum of the lengths of all the segments in the
figure, which is 12x.
2
2
2
2
Choice (D) is not correct. Square ACEG has area x , so 3 times the area of ACEG is 3x .
However, the question asks for the sum of the lengths of all the segments in the figure, which is
12x.
Choice (E) is not correct. Square ACEG has area x , so 5 times the area of ACEG is 5x .
However, the question asks for the sum of the lengths of all the segments in the figure, which is
12x.
Mathematics Question 6
Choice (D) is correct. By definition, a point (x, y) in the xy-plane lies on the graph of f if and only if
the value of f(x) is equal to y. From the graph, the point (b, k) lies on the graph of f, and so f(b) =
k. Likewise, the point (j, k) lies on the graph of f, and so f(j) = k. Therefore, f(b) = f(j).
Choice (A) is not correct. From the graph, the point (a, h) lies on the graph of f, and so f(a) = h.
Also, the point (b, k) lies on the graph of f, and so f(b) = k. Since h ≠ k, it follows that f(a) ≠ f(b).
Choice (B) is not correct. From the graph, the point (a, h) lies on the graph of f, and so f(a) = h.
Also, the point (c, r) lies on the graph of f, and so f(c) = r. Since h ≠ r, it follows that f(a) ≠ f(c).
Choice (C) is not correct. From the graph, the point (b, k) lies on the graph of f, and so f(b) = k.
Also, the point (c, r) lies on the graph of f, and so f(c) = r. Since k ≠ r, it follows that f(b) ≠ f(c).
Choice (E) is not correct. From the graph, the point (c, r) lies on the graph of f, and so f(c) = r.
Also, the point (j, k) lies on the graph of f, and so f(j) = k. Since r ≠ k, it follows that f(c) ≠ f(j).
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Mathematics Question 7
Choice (D) is correct. Line p forms a 180° angle with vertex at the point of intersection of the three
lines , p and m. This angle is made up of three nonoverlapping angles one of measure 40°, one
unmarked and one of measure x°. Since
m, the unmarked angle measures 90°. It follows
that 40 + 90 + x = 180. Solving this equation for x gives x = 50.
Choice (A) is not correct. Line p forms a 180° angle with vertex at the point of intersection of the
three lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,
90° and x°. If the value of x were 20, then 40 + 90 + 20 would equal 180. However, 40 + 90 + 20
= 150, not 180. Therefore, the value of x cannot be 20.
Choice (B) is not correct. Line p forms a 180° angle with vertex at the point of intersection of the
three lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,
90° and x°. If the value of x were 30, then 40 + 90 + 30 would equal 180. However, 40 + 90 + 30
= 160, not 180. Therefore, the value of x cannot be 30.
Choice (C) is not correct. Line p forms a 180° angle with vertex at the point of intersection of the
three lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,
90° and x°. If the value of x were 40, then 40 + 90 + 40 would equal 180. However, 40 + 90 + 40
= 170, not 180. Therefore, the value of x cannot be 40.
Choice (E) is not correct. Line p forms a 180° angle with vertex at the point of intersection of the
three lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,
90° and x°. If the value of x were 60, then 40 + 90 + 60 would equal 180. However, 40 + 90 + 60
= 190, not 180. Therefore, the value of x cannot be 60.
Mathematics Question 8
Choice (C) is correct. By the commutative property of multiplication, wx = xw, so wx + yw − 2w is
equal to xw + yw − 2w. By the commutative property of addition, xw + yw − 2w is equal to yw −
2w + xw. By the distributive property of multiplication over addition, yw − 2w + xw is equal to (y −
2 + x)w. Finally, by the commutative property of multiplication, (y − 2 + x)w is equal to w(y − 2 +
x). Therefore, for all values of x, y, and w, the expression wx + yw − 2w is equal to w(y − 2 + x).
Choice (A) is not correct. If x = y = 1 and w = 2, then wx + yw − 2w = (2)(1) + (1)(2) − (2)(2) = 0,
but w(x + y) − 2 = (2)(1 + 1) − 2 = 2. Therefore, it is not true that the expression wx + yw − 2w is
equal to w(x + y) − 2 for all values of x, y, and w.
Choice (B) is not correct. If x = y = 1 and w = 0, then wx + yw − 2w = (0)(1) + (1)(0) − (2)(0) = 0,
but 2(x + y) − 2w = (2)(1 + 1) − (2)(0) = 4. Therefore, it is not true that the expression wx + yw −
2w is equal to 2(x + y) − 2w for all values of x, y, and w.
Choice (D) is not correct. If x = y = 1 and w = 0, then wx + yw − 2w = (0)(1) + (1)(0) − (2)(0) = 0,
but y(x + w) − 2 = (1)(1 + 0) − 2 = −1. Therefore, it is not true that the expression wx + yw − 2w is
equal to y(x + w) − 2 for all values of x, y, and w.
Choice (E) is not correct. If x = y = 2 and w = 3, then wx + yw − 2w = (3)(2) + (2)(3) − (2)(3) = 6,
but 2w − w(x + x) = (2)(3) − (3)(2 + 2) = −6. Therefore, it is not true that the expression wx + yw −
2w is equal to 2w − w(x + x) for all values of x, y, and w.
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Mathematics Question 9
Choice (A) is correct. If x = –2, then f ( x )
f ( 2)
1
2 2
1
. Division by 0 is not defined, so
0
f(−2) is undefined.
Choice (B) is not correct. The function f is defined at x = −1, since
1
f ( x ) f ( 1)
1
1. However, the question asks for a value of x for which the
1
1 2
function is not defined.
Choice (C) is not correct. The function f is defined at x = 0, since
f ( x ) f (0)
1
1
0 2
2
2
. However, the question asks for a value of x for which the
2
function is not defined.
Choice (D) is not correct. The function f is defined at x = 1, since
f ( x ) f (1)
1
1
1 2
3
3
. However, the question asks for a value of x for which the
3
function is not defined.
Choice (E) is not correct. The function f is defined at x = 2, since f ( x )
f (2)
1
2 2
1
.
2
However, the question asks for a value of x for which the function is not defined.
Mathematics Question 10
Choice (A) is correct. Let r be the number of red apples Ms. Jones bought, and let g be the
number of green apples she bought. Since each red apple cost $0.25, Ms. Jones spent r($0.25)
on red apples; since each green apple cost $0.35, she spent g($0.35) on green apples. Since Ms.
Jones spent a total of $7.65 on these apples, it follows that r($0.25) + g($0.35) = $7.65. Since she
bought twice as many red apples as green apples, it follows that r =2g, and so 2g($0.25) +
g($0.35) =$7.65. Simplifying the previous equation gives g($0.85) = $7.65, and so g = 9.
Therefore, Ms. Jones bought 9 green apples.
Alternatively, for each green apple Ms. Jones bought, she bought 2 red apples, and so she spent
$0.35 + $0.25 + $0.25 = $0.85. Since Ms. Jones spent a total of $7.65 on these apples, and
$7.65 divided by $0.85 is 9, she must have bought 9 green apples.
Choice (B) is not correct. If Ms. Jones had bought 13 green apples, she would have bought 2 ×
13 = 26 red apples. Thus she would have spent a total of 13($0.35) + 26($0.25) = $11.05 on
these apples. But Ms. Jones spent only $7.65 on these apples. Therefore, she did not buy 13
green apples.
Choice (C) is not correct. Ms. Jones bought 18 red apples, but the question asks how many
green apples she bought, which is 9.
Choice (D) is not correct. If Ms. Jones had bought 22 green apples, she would have bought 2 ×
22 = 44 red apples. Thus she would have spent a total of 22($0.35) + 44($0.25) = $18.70 on
these apples. But Ms. Jones spent only $7.65 on these apples. Therefore, she did not buy 22
green apples.
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Choice (E) is not correct. Ms. Jones bought 27 apples in all, but the question asks how many
green apples she bought, which is 9.
Mathematics Question 11
Choice (D) is correct. Let d be the positive integer that is added to each term to obtain the next
term in the sequence. Then z = 10 + d and 18 = z + d. Substituting 10 + d for z in 18 = z + d gives
18 = 10 + 2d. Hence d = 4. Thus y + 4 = 10, which gives y = 2; and then x + 4 = 6, from which x =
2. Therefore, the first five terms of the sequence are 2, 6, 10, 14, and 18, and the sum of these
terms is 50.
Choice (A) is not correct. From the conditions given, z is halfway between 10 and 18, that is, z =
14. Thus each term is the sequence after x is 4 more than the preceding term. It follows that the
first five terms of the sequence are 2, 6, 10, 14, and 18. Their sum is 50, not 44.
Choice (B) is not correct. From the conditions given, z is halfway between 10 and 18, that is, z =
14. Thus each term is the sequence after x is 4 more than the preceding term. It follows that the
first five terms of the sequence are 2, 6, 10, 14, and 18. Their sum is 50, not 46.
Choice (C) is not correct. From the conditions given, z is halfway between 10 and 18, that is, z =
14. Thus each term is the sequence after x is 4 more than the preceding term. It follows that the
first five terms of the sequence are 2, 6, 10, 14, and 18. Their sum is 50, not 48.
Choice (E) is not correct. From the conditions given, z is halfway between 10 and 18, that is, z =
14. Thus each term is the sequence after x is 4 more than the preceding term. It follows that the
first five terms of the sequence are 2, 6, 10, 14, and 18. Their sum is 50, not 52.
Mathematics Question 12
Choice (B) is correct. In the xy-plane, the points with coordinates (3, 2), (13, 2) and (a, 2) all lie on
the line y = 2. Since the points (3, 2) and (13, 2) are on a circle and the point (a, 2) is the center of
the circle, the order of the three points from left to right on the line is (3, 2), (a, 2), and (13, 2);
hence the distance from (3, 2) to (a, 2) is a − 3, and the distance from (a, 2) to (13, 2) is 13 − a.
Since (a, 2) is the center of the circle and the other two points are on the circle, the distance from
(3, 2) to (a, 2) is the same as the distance from (a, 2) to (13, 2), so a − 3 = 13 − a. This equation
simplifies to 2a = 16, and so a = 8.
Choice (A) is not correct. Since (a, 2) is the center of the circle and the other two points are on
the circle, the distance from (3, 2) to (a, 2), which is a − 3, is the same as the distance from (a, 2)
to (13, 2), which is 13 − a. If the value of a were 5, then 5 − 3 = 2 would equal 13 − 5 = 8.
However, 2 does not equal 8. Therefore, the value of a cannot be 5.
Choice (C) is not correct. Since (a, 2) is the center of the circle and the other two points are on
the circle, the distance from (3, 2) to (a, 2), which is a − 3, is the same as the distance from (a, 2)
to (13, 2), which is 13 − a. If the value of a were 10, then 10 − 3 = 7 would equal 13 − 10 = 3.
However, 7 does not equal 3. Therefore, the value of a cannot be 10.
Choice (D) is not correct. In the xy-plane, the points with coordinates (3, 2), (13, 2) and (a, 2) all
lie on the line y = 2. Since the points (3, 2) and (13, 2) are on a circle and the point (a, 2) is the
center of the circle, the order of the three points from left to right on the line is (3, 2), (a, 2) and
(13, 2). If the value of a were 16, then the order of the three points from left to right on the line
would be (3, 2), (13, 2) and (16, 2), and the point (a, 2) could not be the center of the circle.
Therefore, the value of a cannot be 16.
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Choice (E) is not correct. In the xy-plane, the points with coordinates (3, 2), (13, 2) and (a, 2) all
lie on the line y = 2. Since the points (3, 2) and (13, 2) are on a circle and the point (a, 2) is the
center of the circle, the order of the three points from left to right on the line is (3, 2), (a, 2) and
(13, 2). If the value of a were 18, then the order of the three points from left to right on the line
would be (3, 2), (13, 2) and (18, 2), and the point (a, 2) could not be the center of the circle.
Therefore, the value of a cannot be 18.
Mathematics Question 13
Choice (D) is correct. Each year, the manufacturer distributed 20, 30, and 30 percent of its fabric
to retailers A, B, and C, respectively, so each year the manufacturer distributed 100 − 20 − 30 −
30 = 20 percent if its fabric to retailer D.
In 2000 the manufacturer distributed a total of 1,500,000 yards of fabric to these retailers, so 20
percent — or (0.20)(1,500,000) = 300,000 yards — was distributed to retailer D.
In 2005 the manufacturer distributed a total of 2,800,000 yards of fabric to these retailers, so 20
percent — or (0.20)(2,800,000) = 560,000 yards — was distributed to retailer D. Therefore, the
increase in the number of yards of fabric distributed to retailer D from 2000 to 2005 was 560,000
− 300,000 = 260,000.
Choice (A) is not correct. The increase in the number of yards of fabric distributed to retailer D
from 2000 to 2005 was (0.2)(1,300,000) = 260,000, not (0.2)(100,000) = 20,000.
Choice (B) is not correct. The increase in the number of yards of fabric distributed to retailer D
from 2000 to 2005 was (0.2)(1,300,000) = 260,000, not (0.02)(1,300,000) = 26,000.
Choice (C) is not correct. The increase in the number of yards of fabric distributed to retailer D
from 2000 to 2005 was (0.2)(1,300,000) = 260,000, not (0.03)(1,300,000) = 39,000.
Choice (E) is not correct. The increase in the number of yards of fabric distributed to each of
retailers B and C from 2000 to 2005 was (0.3)(1,300,000) = 390,000. However, the question asks
for the increase in the number of yards of fabric distributed to retailer D from 2000 to 2005, which
was (0.2)(1,300,000) = 260,000.
Mathematics Question 14
14
is in lowest terms, since 14 and 3 have no prime factors in
3
14 a
common. Then, since a and b are integers and
, it follows that a = 14k and b = 3k for
3 b
Choice (A) is correct. The fraction
some nonzero integer k. Hence a + b = 17k for some nonzero integer k; in other words, a + b
must be a multiple of 17. Of the given choices, only 68 = 17 × 4 is a multiple of 17, which
corresponds to
14
3
a
b
14 4
3 4
56
.
12
14
is in lowest terms, since 14 and 3 have no prime
3
14 a
factors in common. Then, since a and b are integers and
, it follows that a = 14k and b =
3 b
Choice (B) is not correct. The fraction
3k for some nonzero integer k. Hence a + b = 17k for some nonzero integer k; in other words, a +
2013 PSAT/NMSQT Answer Explanations
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b must be a multiple of 17. Therefore, since 66 is not a multiple of 17, the value of a + b cannot
be 66.
14
is in lowest terms, since 14 and 3 have no prime
3
14 a
factors in common. Then, since a and b are integers and
, it follows that a = 14k and b =
3 b
Choice (C) is not correct. The fraction
3k for some nonzero integer k. Hence a + b = 17k for some nonzero integer k; in other words, a +
b must be a multiple of 17. Therefore, since 63 is not a multiple of 17, the value of a + b cannot
be 63.
14
is in lowest terms, since 14 and 3 have no prime
3
14 a
factors in common. Then, since a and b are integers and
, it follows that a = 14k and b =
3 b
Choice (D) is not correct. The fraction
3k for some nonzero integer k. Hence a + b = 17k for some nonzero integer k; in other words, a +
b must be a multiple of 17. Therefore, since 60 is not a multiple of 17, the value of a + b cannot
be 60.
14
is in lowest terms, since 14 and 3 have no prime
3
14 a
factors in common. Then, since a and b are integers and
, it follows that a = 14k and b =
3 b
Choice (E) is not correct. The fraction
3k for some nonzero integer k. Hence a + b = 17k for some nonzero integer k; in other words, a +
b must be a multiple of 17. Therefore, since 58 is not a multiple of 17, the value of a + b cannot
be 58.
Mathematics Question 15
Choice (C) is correct. Each match in the tournament took place between a single pair of
participants. Since each participant played 2 matches against each of the other participants, the
total number of matches played is twice the number of all possible pairs formed from the 6
participants. Let the participants be called A, B, C, D, E and F. The possible pairs of participants
can be listed A-B, A-C, A-D, A-E, A-F, B-C, B-D, B-E, B-F, C-D, C-E, C-F, D-E, D-F and E-F.
There are 15 distinct pairs of participants, and so there were 2 × 15 = 30 matches played during
the tournament.
Another way to find the number of possible distinct pairs that can be formed from the 6
participants is to note that to form a pair involves first choosing one member of the pair and then
the second member. Any of the 6 participants can be chosen as the first member of the pair; then
any of the other 5 participants can be chosen as the second member. This gives a total of 6 × 5 =
30 choices. However, this double counts all of the pairs, as each pair is counted a second time
with the order of selection reversed. (For example, if the participants are called A, B, C, D, E and
F, then the choice A, then B gives the same pair as the choice B, then A.) Therefore, the total
number of distinct pairs of participants is
6 5
2
15, and so the total number of matches played
during the tournament is 2 × 15 = 30.
Alternatively, the number of distinct pairs of participants taken from the 6 participants in the
tournament is the number of combinations of 2 participants from the field of 6 participants. This
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number is
6
2
6 5
2
15, and so the total number of matches played during the tournament is
2 × 15 = 30.
Choice (A) is not correct. The total number of matches played during the tournament is twice the
number of distinct pairs chosen from the 6 participants, which is 2 × 15 = 30, not twice the
number of participants (2 × 6 = 12).
Choice (B) is not correct. The number distinct pairs of participants that can be chosen from the 6
participants in the tournament is 15. However, each participant played 2 matches, not 1, against
each of the other participants. Therefore, the total number of matches played during the
tournament is 2 × 15 = 30.
Choice (D) is not correct. The total number of ordered pairs of participants, including those that
contain the same participant twice, is 36. However, this is not the same as the number of the total
matches played in the tournament, because one plays matches against the other players only,
not against oneself.
Choice (E) is not correct. The total number of matches played during the tournament is 30, not
48.
Mathematics Question 16
1
1
1
1
and
must be negative and −x,
and
must
3
x
x
x
x2
1
1
be positive. Thus, of the choices given, only
or
could have the least value. If −1 < x < 0,
x
x3
1 1
1
3
then −1 < x < x < 0, and so
has the
1. Therefore, of the five given expressions,
3
x
x
x3
Choice (E) is correct. If −1 < x < 0, then
least value.
1
1
1
1
and
must be negative and −x,
and
3
x
x
x
x2
1
1
must be positive. Thus, of the choices given, only
or
could have the least value. In
x
x3
Choice (A) is not correct. If −1 < x < 0, then
particular, −x cannot have the least value.
1
1
1
1
and
must be negative and −x,
and
x
x
x3
x2
1
1
must be positive. Thus, of the choices given, only
or
could have the least value. If −1 < x
x
x3
1 1
1
3
< 0, then −1 < x < x < 0, and so
1. Therefore, of the five given expressions,
, not
3
x
x
x3
1
, has the least value.
x
Choice (B) is not correct. If −1 < x < 0, then
2013 PSAT/NMSQT Answer Explanations
© 2013 The College Board. All Rights Reserved
1
1
1
1
and
must be negative and −x,
and
x
x
x3
x2
1
1
must be positive. Thus, of the choices given, only
or
could have the least value. In
x
x3
1
particular,
cannot have the least value.
x
Choice (C) is not correct. If −1 < x < 0, then
1
1
1
1
and
must be negative and −x,
and
3
x
x
x
x2
1
1
must be positive. Thus, of the choices given, only
or
could have the least value. In
x
x3
1
Choice (D) is not correct. If −1 < x < 0, then
particular,
x2
cannot have the least value.
Mathematics Question 17
Choice (B) is correct. The lengths of the legs of right triangle RST are 6 and 8, so the hypotenuse
has length 62 82
36 64
100 10. Thus the perimeter of RST is 6 + 8 + 10 = 24.
Triangle XYZ is isosceles and has the same perimeter as triangle RST, so if the lengths of the
sides of XYZ are a, a and b, then 2a + b = 24.
Since a and b are integers and 2a and 24 are even, it follows that b must be even. If b were 12,
then a would be 6, but b < 2a by the triangle inequality. Thus the greatest possible value for b is
10, in which case a = 7.
The greatest possible value of a occurs when b is as small as possible. In this case, b = 2 and a =
11. Therefore, the greatest possible length for one of the sides of triangle XYZ is 11.
Choice (A) is not correct. Triangle XYZ is an isosceles triangle with sides of integer length and
perimeter 24. If the sides of triangle XYZ are a, a and b, the greatest possible value for b is 10,
but the greatest possible value for a is 11. Therefore, the greatest possible length for one of the
sides of triangle XYZ is 11, not 10.
Choice (C) is not correct. Triangle XYZ is an isosceles triangle with sides of integer length and
perimeter 24. If one of the sides of triangle XYZ were of length 14, then the sum of the lengths of
the other two sides of the triangle would be 24 − 14 = 10. However, by the triangle inequality, 14
would have to be less than 10, which is not true. Therefore, 14 is not a possible length for a side
of triangle XYZ and, thus, cannot be the greatest possible length for one of the sides.
Choice (D) is not correct. Triangle XYZ is an isosceles triangle with sides of integer length and
perimeter 24. If one of the sides of triangle XYZ were of length 16, then the sum of the lengths of
the other two sides of the triangle would be 24 − 16 = 8. However, by the triangle inequality, 16
would have to be less than 8, which is not true. Therefore, 16 is not a possible length for a side of
triangle XYZ and, thus, cannot be the greatest possible length for one of the sides.
Choice (E) is not correct. The greatest possible sum of the lengths of two sides of triangle XYZ is
11 + 11 = 22, but the greatest possible length for one of the sides of this triangle is 11.
Mathematics Question 18
2013 PSAT/NMSQT Answer Explanations
© 2013 The College Board. All Rights Reserved
Choice (E) is correct. Each of the three statements must be true.
Statement I It is given that x < y. Hence x + 1 < y + 1. Since y +1 < y +2 for any value of y, it
follows that x + 1 < y + 1 < y + 2.
Statement II It is given that x < y. Hence 2x < 2y. Since x <0, it follows that 3x < 2x. Therefore, 3x
< 2x < 2y.
Statement III Since x < y < 0, the number x is further from 0 than is the number y. Thus |x| > |y|. It
follows that
and so
x
y
x
y
1, and so
x
y
1. Since x and y are both negative, the quotient
x
x
. Therefore,
y
y
x
y
x
is positive,
y
1.
Choice (A) is not correct. Statement I must be true, but so must statements II and III.
Choice (B) is not correct. Statement II must be true, but so must statements I and III.
Choice (C) is not correct. Statement III must be true, but so must statements I and II.
Choice (D) is not correct. Statements I and II must be true, but so must statement III.
Mathematics Question 19
Choice (C) is correct. If there are k male employees at the company, then, since there are n more
male employees than female employees, there are k − n female employees at the company.
Thus the total number of employees at the company is k + (k − n) = 2k − n. Therefore, the fraction
of the employees who are male is
k
2k
n
.
Choice (A) is not correct. There are 2k − n, not 2k + n, total employees at the company, and the
fraction of employees at the company who are male is
k
2k
n
, not
1
2k
n
.
Choice (B) is not correct. If there were k male employees and n female employees at the
company, then the fraction of employees at the company who are male would be
k
k
n
.
However, n is not the number of female employees, but the number by which the male
employees outnumber the female employees, and the fraction of employees at the company who
are male is
k
.
2k n
Choice (D) is not correct. There are 2k − n, not 2k + n, total employees at the company, and the
fraction of employees at the company who are male is
k
2k
n
, not
n
2k
n
.
Choice (E) is not correct. There are 2k − n, not 2k, total employees at the company, and the
fraction of employees at the company who are male is
2013 PSAT/NMSQT Answer Explanations
k
2k
n
, not
k
n
.
2k
© 2013 The College Board. All Rights Reserved
Mathematics Question 20
Choice (A) is correct. The surface area of the cylinder, not including the bases, is given as both
2
2πrh and 70π. Setting 2πrh = 70π yields rh = 35. The volume V of a cylinder is given by πr h,
which can be rewritten as πrrh = (rh)(πr). Since rh = 35, it follows that the volume of the right
circular cylinder in terms of r is 35πr.
Choice (B) is not correct. The surface area of the cylinder, not including the bases, is given as
both 2πrh and 70π. Setting 2πrh = 70π yields rh = 35. The volume V of a cylinder is given by
2
2
πr h. If the volume of the cylinder were 70πr, then 70πr would equal πr h, and rh would equal 70.
However, rh = 35, not 70. Therefore, the volume of the cylinder in terms of r cannot be 70πr.
2
Choice (C) is not correct. The volume V of a right circular cylinder is given by πr h. If h =
1
, and
2
r = 70, the surface area of the cylinder, not including the bases, would be
2 rh
2 (70)
1
2
70 , and the volume of the cylinder would be
r2
1
2
1 2
r .
2
However, if h = 1 and r = 35, the surface area of the cylinder, not including the bases, would still
2
2
2
be 2πrh = 2π(35)(1) = 70π, but the volume of the cylinder would be πr h = πr (1) = πr , which is
not equal to
1 2
1 2
r . Therefore, the volume of the cylinder in terms of r need not be
r .
2
2
2
Choice (D) is not correct. The volume V of a right circular cylinder is given by πr h. If h = 35 and r
= 1, the surface area of the cylinder, not including the bases, would be 2πrh = 2π(1)(35) = 70π,
2
2
2
and the volume of the cylinder would be πr h = πr (35) = 35πr . However, if h = 1 and r = 35, the
surface area of the cylinder, not including the bases, would still be 2πrh = 2π(35)(1) = 70π, but
2
2
2
2
the volume of the cylinder would be πr h = πr (1) = πr , which is not equal to 35πr . Therefore,
2
the volume of the cylinder in terms of r need not be 35πr .
Choice (E) is not correct. The volume V of a right circular cylinder is given by
r 2h. If h = 70 and
1
, the surface area of the cylinder, not including the bases, would be
2
1
2 rh 2
(70) 70 , and the volume of the cylinder would be πr2(70) = 70πr2. However, if
2
r=
h = 1 and r = 35, the surface area of the cylinder, not including the bases, would still be 2πrh =
2
2
2
2π(35)(1) = 70π, but the volume of the cylinder would be πr h = πr (1) = πr , which is not equal to
2
2
70πr . Therefore, the volume of the cylinder in terms of r need not be 70πr .
2013 PSAT/NMSQT Answer Explanations
© 2013 The College Board. All Rights Reserved
2013 PSAT/NMSQT Answer Explanations
© 2013 The College Board. All Rights Reserved