PRACTICE TESTS Math Section2 2013 Answers

7/27/2019 PRACTICE TESTS Math Section2 2013 Answers

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2013 PSAT/NMSQT Answer Explanations © 2013 The College Board. All Rights Reserved

Mathematics: Section 2

Mathematics Question 1

Choice (A) is correct. If 2 1 9,

a

k then 2 9 1 8,

a

k and so

8

4.2

a

k

Choice (B) is not correct. If a

k were equal to 5, then 2 1

a

k would equal (2)(5) + 1 = 11.

However, 2 1a

k is equal to 9, not 11. Therefore,

a

k cannot equal 5.

Choice (C) is not correct. If a


a

k would equal (2)(6) + 1 = 13.

However, 2 1

a


a

k cannot equal 6.

Choice (D) is not correct. If a


a

k would equal (2)(7) + 1 = 15.

However, 2 1a


a

k cannot equal 7.

Choice (E) is not correct. If a


a

k would equal (2)(8) + 1 = 17.

However, 2 1a


a

k cannot equal 8.


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 coordinatesof 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 hascoordinates (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 hascoordinates (10, 5). The point with coordinates (5, 10) is 5 units to the left of and 5 units abovepoint C .





Choice (E) is correct. Of the 4 cans of paint, only one contains green paint, and so the remaining3 do not contain green paint. Therefore, the probability that the can chosen by the painter does

not contain green paint is3.

4

Choice (A) is not correct. The probability that the can chosen by the painter contains green paint

is1.

4However, the question asks for the probability that the can chosen does not contain green

paint, which is1 3

1 ,4 4

not1.

4

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 is3,

4not

1.

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 is3,

4not

1.

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 be2.

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 is3,

4not

2.

3


Choice (C) is correct. Considering each of the given possibilities separately, substitute the valuesgiven in I, II and III for y in the equation 49 = (7y )

2.

I When −1 is substituted for y in the equation, the result is 49 = [(7)(−1)]2 = (−7)2 = 49, so y couldequal −1.

II When 1 is substituted for y in the equation, the result is 49 = [(7)(1)]2 = (7)2 = 49, so y couldequal 1.

III When 7 is substituted for y in the equation, the result is 49 = [(7)(7)]2

= 492

Since 49 is notequal to 492, y cannot equal 7.

Therefore, of the three values, only the values in I and II could be values of y that satisfy theequation.

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.




Choice (E) is not correct. The y value in II satisfies the equation, but the y value in III does not.


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 figurehas length x . Therefore, the sum of the lengths of all the segments in the figure is 12 x .

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 8 x . However, the question asks for the sum of the lengths of all the segments in the figure, which is 12 x .

Choice (C) is not correct. Each of the 4 triangles in the figure has perimeter 3 x , and square ACEGhas perimeter 4 x , so the sum of the perimeters of all the triangles and the square is (4)(3 x ) + 4 x =12 x + 4 x = 16 x . However, the question asks for the sum of the lengths of all the segments in thefigure, which is 12 x .

Choice (D) is not correct. Square ACEG has area x 2, so 3 times the area of ACEG is 3 x

2.

However, the question asks for the sum of the lengths of all the segments in the figure, which is12 x . Choice (E) is not correct. Square ACEG has area x

2, so 5 times the area of ACEG is 5 x 2.

However, the question asks for the sum of the lengths of all the segments in the figure, which is12 x .


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 ).





Choice (D) is correct. Line p forms a 180° angle with vertex at the point of intersection of the threelines , 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 followsthat 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 thethree 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 thethree lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,


Choice (C) is not correct. Line p forms a 180° angle with vertex at the point of intersection of thethree lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,


Choice (E) is not correct. Line p forms a 180° angle with vertex at the point of intersection of thethree lines , p and m. This angle is made up of three nonoverlapping angles of measures 40°,



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 isequal 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 isequal 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 .





Choice (A) is correct. If x = –2, then1 1

( ) ( 2) .02 2

f x f Division by 0 is not defined, so

f (−2) is undefined.

Choice (B) is not correct. The function f is defined at x = −1, since1 1

( ) ( 1) 1.11 2

f x f However, the question asks for a value of x for which the

function is not defined.

Choice (C) is not correct. The function f is defined at x = 0, since

1 1 2( ) (0) .

20 2 2f x f However, the question asks for a value of x for which the


Choice (D) is not correct. The function f is defined at x = 1, since

1 1 3( ) (1) .31 2 3

f x f However, the question asks for a value of x for which the


Choice (E) is not correct. The function f is defined at x = 2, since1 1

( ) (2) .22 2

f x f

However, the question asks for a value of x for which the function is not defined.


Choice (A) is correct. Let r be the number of red apples Ms. Jones bought, and let g be thenumber 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 shebought 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 onthese apples. But Ms. Jones spent only $7.65 on these apples. Therefore, she did not buy 13green apples.

Choice (C) is not correct. Ms. Jones bought 18 red apples, but the question asks how manygreen 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 onthese apples. But Ms. Jones spent only $7.65 on these apples. Therefore, she did not buy 22green apples.




Choice (E) is not correct. Ms. Jones bought 27 apples in all, but the question asks how manygreen apples she bought, which is 9.


Choice (D) is correct. Let d be the positive integer that is added to each term to obtain the nextterm in the sequence. Then z = 10 + d and 18 = z + d . Substituting 10 + d for z in 18 = z + d gives18 = 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 theseterms 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 thefirst 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 thefirst 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 thefirst 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 thefirst five terms of the sequence are 2, 6, 10, 14, and 18. Their sum is 50, not 52.


Choice (B) is correct. In the xy -plane, the points with coordinates (3, 2), (13, 2) and (a, 2) all lie onthe 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 equationsimplifies 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 onthe 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 onthe 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) alllie on the line y = 2. Since the points (3, 2) and (13, 2) are on a circle and the point ( a, 2) is thecenter 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 linewould 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.




Choice (E) is not correct. In the xy -plane, the points with coordinates (3, 2), (13, 2) and (a, 2) alllie on the line y = 2. Since the points (3, 2) and (13, 2) are on a circle and the point ( a, 2) is thecenter 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 linewould 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.


Choice (D) is correct. Each year, the manufacturer distributed 20, 30, and 30 percent of its fabricto 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 20percent — 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 20percent — or (0.20)(2,800,000) = 560,000 yards — was distributed to retailer D. Therefore, theincrease 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 Dfrom 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 Dfrom 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 Dfrom 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 asksfor 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.


Choice (A) is correct. The fraction14

3is in lowest terms, since 14 and 3 have no prime factors in

common. Then, since a and b are integers and14

,3

a

bit follows that a = 14k and b = 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. Of the given choices, only 68 = 17 × 4 is a multiple of 17, which

corresponds to14 14 4 56

.3 3 4 12

a

b

Choice (B) is not correct. The fraction14

3is in lowest terms, since 14 and 3 have no prime

factors in common. Then, since a and b are integers and14

,3

a

bit follows that a = 14k and b =

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 66 is not a multiple of 17, the value of a + b cannotbe 66.

Choice (C) is not correct. The fraction14


factors in common. Then, since a and b are integers and

14

,3

a

b it follows that a = 14k and b =

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 cannotbe 63.

Choice (D) is not correct. The fraction14



,3

a



Choice (E) is not correct. The fraction14



,3

a




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, thetotal number of matches played is twice the number of all possible pairs formed from the 6participants. Let the participants be called A, B, C , D, E and F . The possible pairs of participantscan 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 duringthe tournament.

Another way to find the number of possible distinct pairs that can be formed from the 6participants is to note that to form a pair involves first choosing one member of the pair and thenthe second member. Any of the 6 participants can be chosen as the first member of the pair; thenany 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 timewith 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 is6 5

15,2

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 thetournament is the number of combinations of 2 participants from the field of 6 participants. This




number is6 6 5

15,22

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 thenumber of distinct pairs chosen from the 6 participants, which is 2 × 15 = 30, not twice thenumber of participants (2 × 6 = 12).

Choice (B) is not correct. The number distinct pairs of participants that can be chosen from the 6participants in the tournament is 15. However, each participant played 2 matches, not 1, againsteach of the other participants. Therefore, the total number of matches played during thetournament is 2 × 15 = 30.

Choice (D) is not correct. The total number of ordered pairs of participants, including those thatcontain the same participant twice, is 36. However, this is not the same as the number of the totalmatches 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.


Choice (E) is correct. If −1 < x < 0, then1

x and

3

1

x must be negative and − x ,

1

x and

2

1

x must

be positive. Thus, of the choices given, only1

x or

3

1

x could have the least value. If −1 < x < 0,

then −1 < x < x 3

< 0, and so3

1 11.

x x Therefore, of the five given expressions,

3

1

x has the

least value.

Choice (A) is not correct. If −1 < x < 0, then1

x and

3

1


1

x and

2

1

x

must be positive. Thus, of the choices given, only1

x or

3

1

x could have the least value. In

particular, − x cannot have the least value.

Choice (B) is not correct. If −1 < x < 0, then1

x and

3

1


1

x and

2

1

x


x or

3

1

x could have the least value. If −1 < x

< 0, then −1 < x < x 3 < 0, and so

3

1 11.

x x Therefore, of the five given expressions,

3

1,

x not

1,

x has the least value.




Choice (C) is not correct. If −1 < x < 0, then1

x and

3

1


1

x and

2

1

x


x or

3

1


particular,

1

x cannot have the least value.

Choice (D) is not correct. If −1 < x < 0, then1

x and

3

1


1

x and

2

1

x


x or

3

1


particular,2

1

x cannot have the least value.


Choice (B) is correct. The lengths of the legs of right triangle RST are 6 and 8, so the hypotenuse

has length 2 26 8 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 thesides 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 is10, 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 andperimeter 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 thesides of triangle XYZ is 11, not 10.

Choice (C) is not correct. Triangle XYZ is an isosceles triangle with sides of integer length andperimeter 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, 14would have to be less than 10, which is not true. Therefore, 14 is not a possible length for a sideof 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 andperimeter 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, 16would 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 is11 + 11 = 22, but the greatest possible length for one of the sides of this triangle is 11.





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 , itfollows that x + 1 < y + 1 < y + 2.

Statement II It is given that x < y . Hence 2 x < 2y . Since x <0, it follows that 3 x < 2 x . Therefore, 3 x < 2 x < 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 1, x

y and so 1.

x

y Since x and y are both negative, the quotient

x

y is positive,

and so . x x

y y Therefore, 1.

x x

y y

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.


Choice (C) is correct. If there are k male employees at the company, then, since there are n moremale 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 .2

k

k 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 ,2

k

k nnot

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 maleemployees outnumber the female employees, and the fraction of employees at the company who

are male is .2

k

k n

Choice (D) is not correct. There are 2k − n, not 2k + n, total employees at the company, and the


k

k nnot .

2

n

k n

Choice (E) is not correct. There are 2k − n, not 2k , total employees at the company, and the


k

k nnot .

2

k n

k





Choice (A) is correct. The surface area of the cylinder, not including the bases, is given as both2π rh and 70π . Setting 2π rh = 70π yields rh = 35. The volume V of a cylinder is given by π r

2h,

which can be rewritten as π rrh = (rh)(π r ). Since rh = 35, it follows that the volume of the rightcircular cylinder in terms of r is 35π r .

Choice (B) is not correct. The surface area of the cylinder, not including the bases, is given asboth 2π rh and 70π . Setting 2π rh = 70π yields rh = 35. The volume V of a cylinder is given byπ r

2h. If the volume of the cylinder were 70π r , then 70π r would equal π r

2h, and rh would equal 70.

However, rh = 35, not 70. Therefore, the volume of the cylinder in terms of r cannot be 70π r .

Choice (C) is not correct. The volume V of a right circular cylinder is given by π r 2h. If h =

1,

2and

r = 70, the surface area of the cylinder, not including the bases, would be

12 2 (70) 70 ,

2rh and the volume of the cylinder would be 2 21 1

.2 2

r r

However, if h = 1 and r = 35, the surface area of the cylinder, not including the bases, would stillbe 2π rh = 2π (35)(1) = 70π , but the volume of the cylinder would be π r

2h = π r

2(1) = π r 2, which is

not equal to 21.

2r Therefore, the volume of the cylinder in terms of r need not be 21

.2

r

Choice (D) is not correct. The volume V of a right circular cylinder is given by π r 2h. If h = 35 and r

= 1, the surface area of the cylinder, not including the bases, would be 2π rh = 2π (1)(35) = 70π ,and the volume of the cylinder would be π r

2h = π r

2(35) = 35π r 2. However, if h = 1 and r = 35, the

surface area of the cylinder, not including the bases, would still be 2π rh = 2π (35)(1) = 70π , butthe volume of the cylinder would be π r

2h = π r

2(1) = π r 2, which is not equal to 35π r

2. Therefore,the volume of the cylinder in terms of r need not be 35π r

2.

Choice (E) is not correct. The volume V of a right circular cylinder is given by 2 .r h If h = 70 and

r =1,2

the surface area of the cylinder, not including the bases, would be

12 2 (70) 70 ,

2rh and the volume of the cylinder would be π r

2(70) = 70π r 2. However, if

h = 1 and r = 35, the surface area of the cylinder, not including the bases, would still be 2π rh =2π (35)(1) = 70π , but the volume of the cylinder would be π r

2h = π r

2(1) = π r

2, which is not equal to

70π r 2. Therefore, the volume of the cylinder in terms of r need not be 70π r

2.

PRACTICE TESTS Math Section2 2013 Answers

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