2012 PSAT/NMSQT Practice Test Explanations - Section 2

7/21/2019 2012 PSAT/NMSQT Practice Test Explanations - Section 2

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2012 PSAT/NMSQT Answer Explanations 2012 The College Board. All Rights Reserved

Mathematics: Section 2

Mathematics Question 1

Choice (C) is correct. If 2x+ 4 = 8, then 2x= (2x+ 4)4 = 84 = 4, and sox= 2. Therefore, 6x+ 4 = (6)(2) + 4 = 12 + 4 = 16.

Choice (A) is not correct. The value ofx is 2, but the value of 6x+ 4 is (6)(2) + 4 = 16.

Choice (B) is not correct. The value of 2x+ 4is 8, but the value of 6x+ 4 is 16.

Choice (D) is not correct. Since 2x+ 4 = 8, it follows that 3(2x+ 4) = 6x + 12 = 24. Therefore, thevalue of 6x+ 4 is 248 = 16, not 20.Choice (E) is not correct. Since 2x+ 4 = 8, it follows that 3(2x+ 4) = 6x + 12 = 24, but the value of6x+ 4 is 248 = 16.


Choice (E) is correct. Since the veterinarian treated 2 mice, 3 cats, 6 dogs and no other animals,the total number of animals the veterinarian treated was 2 + 3 + 6 = 11. Therefore, the ratio of the

number of cats treated to the total number of animals treated by the veterinarian was 3 to 11.

Choice (A) is not correct. Had the veterinarian treated a total of 12 animals, the ratio of the

number of cats treated to the total number of animals treated by the veterinarian would have been

3 to 12, or 1 to 4. However, the total number of animals the veterinarian treated was 2 + 3 + 6 =

11, and so the ratio of the number of cats treated to the total number of animals treated was 3 to

11.

Choice (B) is not correct. The veterinarian treated 3 cats and a total of 2 + 3 + 6 = 11 animals.Therefore, the ratio of the number of cats treated to the total number of animals treated by the

veterinarian was 3 to 11, not 1 to 6.

Choice (C) is not correct. The veterinarian treated 3 cats and a total of 2 + 3 + 6 = 11 animals.Therefore, the ratio of the number of cats treated to the total number of animals treated by theveterinarian was 3 to 11, not 1 to 13.

Choice (D) is not correct. The ratio of the number of cats treated by the veterinarian to the totalnumber of other animals treated was 3 to 8, but the ratio of the number of cats treated to the totalnumber of animals, including cats, treated was 3 to 11.


Choice (E) is correct. Since the measure of an exterior angle of a triangle is equal to the sum ofthe measures of the interior angles at the other two vertices, it follows that 110 = 40 + x. Solvingforxgivesx= 70.

Choice (A) is not correct. If the value ofx were 30, then 40 +x would be equal to 70. However, 40+xis equal to 110, not 70, so the value of xcannot be 30.

Choice (B) is not correct. If the value ofx were 40, then 40 +x would be equal to 80. However, 40+xis equal to 110, not 80, so the value of xcannot be 40.


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Choice (C) is not correct. If the value ofx were 50, then 40 +x would be equal to 90. However, 40+xis equal to 110, not 90, so the value of xcannot be 50.

Choice (D) is not correct. If the value ofx were 60, then 40 +x would be equal to 100. However,40 +xis equal to 110, not 100, so the value of xcannot be 60.


Choice (D) is correct. For the 12 books sold, the three different sale prices represented were$2.50, $4.50 and $5.00. Alex sold 4 books for $2.50 each and Dana sold 0 books for $2.50, for atotal of 4 + 0 = 4 books sold at $2.50 each. Alex sold 2 books for $4.50 each and Dana sold 3books for $4.50 each, for a total of 2 + 3 = 5 books sold at $4.50 each. Alex sold 1 book for $5.00and Dana sold 2 books for $5.00 each, for a total of 1 + 2 = 3 books sold at $5.00 each.Therefore, the most frequently occurring sale price for the 12 books was $4.50.

Choice (A) is not correct. There were 4 books sold for $2.50 each. However, there were 5 bookssold for $4.50 each. Therefore, the most frequently occurring sale price for the 12 books was not$2.50.

Choice (B) is not correct. There were no books sold for the price of $4.00 each. Therefore, the

most frequently occurring sale price for the 12 books was not $4.00.

Choice (C) is not correct. There were no books sold for the price of $4.25 each. Therefore, themost frequently occurring sale price for the 12 books was not $4.25.

Choice (E) is not correct. There were 3 books sold for the price of $5.00 each. However, therewere 5 books sold for $4.50 each. Therefore, the most frequently occurring sale price for the 12books sold was not $5.00.


Choice (C) is correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) =48 feet. Therefore, 16 yards is 48 16 = 32 feet longer than 16 feet.

Choice (A) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3)= 48 feet. However, 16 yards is 48 16 = 32, not 48, feet longer than 16 feet.

Choice (B) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3)= 48 feet. Thus 16 yards is 488 = 40 feet longer than 8 feet. However, 16 yards is 48 16 = 32feet longer than 16 feet.

Choice (D) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3)= 48 feet. Thus 16 yards is 4824 = 24 feet longer than 24 feet. However, 16 yards is 48 16 =32 feet longer than 16 feet.

Choice (E) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3)

= 48 feet. Thus 16 yards is 4832 = 16 feet longer than 32 feet. However, 16 yards is 48 16 =32 feet longer than 16 feet.


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Choice (E) is correct. For both rows of the table, r2is equal to s: 2

2= 4 and 32=9. Hence r2= ks,

where k= 1, for both rows of the table. Therefore, r2could be directly proportional to s.

Choice (A) is not correct. It could be true that s is proportional to r, but it is r2, not ,r that

could be proportional to s.

Choice (B) is not correct. If rwere directly proportional to s, then it would be true that r= ksfor

some constant k. But 2 =1

2(4) and 3 =

1

3(9), so there is no constant kfor which r= ks.

Therefore, rcannot be directly proportional to s.

Choice (C) is not correct. If r+ 1 were directly proportional to s, then it would be true that r+ 1 =

ks, for some constant k. But 2 + 1 =3

4(4) and 3 + 1 =

4

9(9), so there is no constant kfor which r

+ 1 = ks. Therefore, r+ 1 cannot be directly proportional to s.

Choice (D) is not correct. If 2rwere directly proportional to s, then it would be true that 2r= ks, for

some constant k. But 2(2) = 1(4) and 2(3) =2

3(9), so there is no constant kfor which 2r= ks.

Therefore, 2rcannot be directly proportional to s.


Choice (D) is correct. Since 20 percent of the surveyed customers answered black, it follows

that the probability is20 1

100 5that a customer chosen at random from those surveyed will be

one who answered black.

Choice (A) is not correct. The probability cannot be 120

that a customer chosen at random from

those surveyed will be one who answered black, because the graph indicates that 20 percent,which is 1 in 5 (not 5 percent, which is 1 in 20), of the customers surveyed answered black.

Choice (B) is not correct. The probability cannot be1

18that a customer chosen at random from

those surveyed will be one who answered black, because the graph indicates that 20 percent,which is 1 in 5 (not 1 in 18), of the customers surveyed answered black.

Choice (C) is not correct. The probability cannot be1



Choice (E) is not correct.The probability cannot be1




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Choice (A) is correct. Since point Plies on line segment ,RU the sum of the measures of

RPS and SPUis 180. Similarly, since point Plies on segment ,QT the sum of the

measures of QPS and SPT is 180. The measure of SPUis 70, so the measure of

RPS is 180 70 = 110. Since RPS has measure 110 and RPQ has measure 40,the measure of SPT is 180 (110 + 40) = 30.

Choice (B) is not correct. The measure of RPQ is 40, so the measure of TPUis also 40,

since vertical angles are of equal measure. The measure of SPUis 70, so the measure of

RPS is 180 70 = 110. If the measure of SPT were 35, then the measure of RPUwould be 110 + 35 + 40 = 185. However, RPUis a straight angle, which has measure

180, not 185. Therefore, the measure of SPT cannot be 35.

Choice (C) is not correct. The measure of RPQ is 40, as is the measure of ,TPU but the

measure of SPT is 30.

Choice (D) is not correct. The measure of RPQ is 40, so the measure of TPUis also 40,




Choice (E) is not correct. The measure of RPQ is 40, so the measure of TPUis also 40,





Choice (D) is correct. The function fis defined by f(x) = 2x, so f(5t) = 2(5t) = 10t.

Choice (A) is not correct. The function fis defined by f(x) = 2x, so f(t) = 2t, but f(5t) = 2(5t) = 10t.

Choice (B) is not correct. The function fis defined by f(x) = 2x, so f(2.5t) = 2(2.5t) = 5t, but f(5t) =2(5t) = 10t.

Choice (C) is not correct. The function fis defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not (2 + 5)t=7t.

Choice (E) is not correct. The function fis defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not 52t= 25t.


Choice (B) is correct. Adding the respective sides of the equations ax + by =14 and axby = 4gives ax + by + axby=14 + 4, which simplifies to 2ax= 18. It follows that ax= 9.

Choice (A) is not correct. Adding the respective sides of the given equations shows that the valueof axmust be 9, not 5.


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Choice (C) is not correct. Adding the respective sides of the given equations shows that the valueof ax, not by, must be 9.

Choice (D) is not correct. Subtracting the respective sides of the given equations shows that thevalue of 2by, not by, must be 10.

Choice (E) is not correct. The value of (2ax)(2by), not ax2by2, must be 180.


Choice (A) is correct. Since the perimeter of the equilateral triangle is 18, it follows that the each

side of the triangle is18

63

units long. From the figure, the length of a radius of the circle is

equal to the length of a side of the equilateral triangle, 6 units. The length of a diameter of thecircle is double the length of a radius, and therefore, it is (2)(6) = 12.

Choice (B) is not correct. If the length of a diameter of the circle were 10, the length of a radiuswould be 5, from which it would follow that the equilateral triangle has perimeter (3)(5) = 15. But

the perimeter of the triangle is 18, not 15, and so the length of a diameter of the circle cannot be10.

Choice (C) is not correct. If the length of a diameter of the circle were 9, the length of a radiuswould be 4.5, from which it would follow that the equilateral triangle has perimeter (3)(4.5) = 13.5.But the perimeter of the triangle is 18, not 13.5, and so the length of a diameter of the circlecannot be 9.

Choice (D) is not correct. If the length of a diameter of the circle were 8, the length of a radiuswould be 4, from which it would follow that the equilateral triangle has perimeter (3)(4) = 12. Butthe perimeter of the triangle is 18, not 12, and so the length of a diameter of the circle cannot be8.

Choice (E) is not correct. The length of a radius of the circle is 6; however, the question asks forthe length of a diameter of the circle, which is 12.


Choice (E) is correct. The car travels at a rate of (2r + 4) miles per hour. The number of miles thecar can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2.Therefore, in terms of r,the car can travel (2)(2r +4) = 4r + 8 miles in 2 hours.

Choice (A) is not correct. The number of miles the car can travel in 2 hours can be found bymultiplying the rate by the number of hours, which is 2. Therefore, in 2 hours,the car can travel

(2)(2r +4) = 4r + 8 miles, not2 4

2

r= r+ 2miles.

Choice (B) is not correct. The number of miles the car can travel in 2 hours can be found bymultiplying the rate by the number of hours, which is 2. Therefore, in 2 hours,the car can travel

(2)(2r +4) = 4r + 8 miles, not2

2

r+ 4 = r+ 4 miles.

Choice (C) is not correct. The number of miles the car can travel in 2 hours can be found bymultiplying the rate by the number of hours, which is 2. Therefore, in 2 hours,the car can travel(2)(2r +4) = 4r + 8 miles, not (2)(2r) + 4 = 4r+ 4 miles.


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Choice (D) is not correct. The number of miles the car can travel in 2 hours can be found bymultiplying the rate by the number of hours, which is 2. Therefore, in 2 hours,the car can travel(2)(2r +4) = 4r + 8 miles, not (2)(2r) + (2 + 4) = 4r+ 6 miles.


Choice (B) is correct. The shaded region can be divided into two regions: a rectangle withvertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4).The rectangle has length 40 = 4 and width 30 = 3, so its area is (4)(3) = 12. The right trianglehas one leg of length 30 = 3 and one leg of length 10 4 = 6. Thus the area of the triangle is

(3)(6)9.

2Therefore, the area of the shaded region is 12 + 9 = 21.

Choice (A) is not correct. The shaded region is contained within a rectangle with vertices (0, 0),(0, 10), (3, 10) and (3, 0). This rectangle has length 10 and width 3, so its area is 30. However,the area of the shaded region is 30 9 = 21, which is the area of the rectangle minus the area ofthe unshaded triangular region inside of the rectangle.

Choice (C) is not correct. The area of the shaded region is the sum of the area of a rectangle ofarea 12 and a right triangle of area 9. Thus the area of the shaded region is 12 + 9 = 21, not 9 + 9= 18.

Choice (D) is not correct. The shaded region can be divided into two regions: a rectangle withvertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4).The rectangle has area 12. However, the total area of the shaded region is the area of therectangle plus the area of the triangle. The area of the triangle is 9, so the area of the shadedregion is 12 + 9 = 21.

Choice (E) is not correct. The shaded region can be divided into two regions: a rectangle withvertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4).The triangle has area 9. However, the total area of the shaded region is the area of the triangleplus the area of the rectangle. The area of the rectangle is 12, so the area of the shaded region is9 + 12 = 21.


Choice (E) is correct. Since renting video games from The Game Garage costs $3 per game, plus

an additional $60 for an annual membership, the cost to rent nvideo games in a year from The

Game Garage, in dollars, is 3n+ 60. Since renting video games from The Video Vendor costs $6

per game with no annual membership fee, the cost to rent nvideo games in a year from The

Video Vendor, in dollars, is 6n. Hence it costs less to rent ngames in a year from The Game

Garage than from The Video Vendor if and only if 3n+ 60 < 6n. This inequality is equivalent to 60

< 3n, which is equivalent to n> 20. Therefore, n> 20 gives all values of n for which it costs less

to rent ngames in a year from The Game Garage than from The Video Vendor.

Choice (A) is not correct. If n< 10, then it costs more, not less, to rent ngames in a year fromThe Game Garage than from The Video Vendor. For example, if n= 5, it costs $75 to rent n= 5games in a year from The Game Garage and $30 from The Video Vendor. Therefore, n< 10cannot give all values of n for which it costs less to rent ngames in a year from The GameGarage than from The Video Vendor.

Choice (B) is not correct. If n= 10, then it costs more, not less, to rent ngames in a year fromThe Game Garage than from The Video Vendor: it costs $90 to rent n= 10 games in a year from


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The Game Garage and $60 from The Video Vendor. Therefore, n= 10 cannot give all values of nfor which it costs less to rent ngames in a year from The Game Garage than from The VideoVendor.

Choice (C) is not correct. If 10 < n< 20, then it costs more, not less, to rent ngames in a yearfrom The Game Garage than from The Video Vendor. For example, if n= 15, it costs $105 to rentn= 15 games in a year from The Game Garage and $90 from The Video Vendor. Therefore, 10 46.

Choice (C) is not correct. The sum of the five different positive integers is 100, and the smallest ofthe integers is 10. If 64 were one of the integers, then the sum of the integers other than 10 and64 would be 100 (10 + 64) = 26. However, if the sum of three positive integers is 26, at leastone of them must be less than 10, and then 10 would not be the smallest of the five integers.Therefore, 64 cannot be the maximum value of one of the four integers other than 10.


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Choice (D) is not correct. The sum of the five different positive integers is 100, and the smallest ofthe integers is 10. If 84 were one of the integers, then the sum of the integers other than 10 and84 would be 100 (10 + 84) = 6. However, if the sum of three positive integers is 6, all of themmust be less than 10, and then 10 would not be the smallest of the five integers. Therefore, 84cannot be the maximum value of one of the four integers other than 10.

Choice (E) is not correct. The sum of the five different positive integers is 100, and the smallest ofthe integers is 10. If 90 were one of the integers, then the sum of the integers other than 10 and90 would be 100 (10 + 90) = 0. However, the sum of three positive integers cannot be 0. Therefore, 90 cannot be the maximum value of one of the four integers other than 10.


Choice (A) is correct. Since 2x= 3y= 4z, it follows that y=2

3xand z=

1.

2x Thusx+ y+ z=x+

2

3x+

1.

2x It follows that 6(x+ y+ z) = 6x+

26

3x +

16

2x = 6x+ 4x+ 3x= 13x, and sox+

y+ z=13

.6

x Therefore, the value ofx+ y+ zis13

6

times the value ofx.

Choice (B) is not correct. Since 6(x+ 2y+ z) = 6x+ 8x+ 3x= 17x, the value ofx+ 2y+ zis17

6

times the value ofx, but the value ofx+ y+ zis13

6times the value ofx.

Choice (C) is not correct. The value ofx+ y+ zis13

2times the value of ,

3

xbut the value ofx+

y+ zis13


Choice (D) is not correct. Since 2(x+ 2y+ z) =6 8 3

3

x x x= 17 ,

3

xthe value ofx+ 2y+ z

is17

2times the value of ,

3

xbut the value ofx+ y+ zis

13


Choice (E) is not correct. It can be determined from the information given that the value ofx+ y+

zis13



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Choice (D) is correct. When the point in the xy-plane with coordinates (a, b) is reflected about theline y=x, the image is the point with coordinates (b, a); that is, reflecting a point about the line y=xhas effect of switching its coordinates. Hence when A(2, 5) and B (3, 2) are reflected about y=x, their images are the points with coordinates (5, 2) and(2, 3), respectively. The slope of the

line through the points with coordinates (5, 2) and(2, 3) is given by the quotient 3 ( 2) ,2 ( 5)

which is equal to5.

7

Choice (A) is not correct. IfA(2, 5) and B (3, 2) are reflected about thex-axis, the points towhichAand Bare reflected have coordinates (2, 5) and(3, 2), respectively, and the slope of

the line through these two points is7.

5However, whenA(2, 5) and B (3, 2) are reflected

about y=x, their images are the points with coordinates (5, 2) and(2, 3), respectively, and the

slope of the line through these points is5.

7

Choice (B) is not correct. IfA(2, 5) and B (3, 2) are rotated 90 degrees clockwise, the points towhichAand Bare rotated have coordinates (5, 2) and(2, 3), respectively, and the slope of the

line through these two points is5.

7However, whenA(2, 5) and B (3, 2) are reflected about y

=x, their images are the points with coordinates (5, 2) and(2, 3), respectively, and the slope of

the line through these points is5.

7

Choice (C) is not correct. WhenA(2, 5) and B (3, 2) are reflected about the line y=x, theirimages are the points with coordinates (5, 2) and (2, 3), respectively. The slope of the line

through (5, 2) and(2, 3) is given by the quotient3 ( 2)

,2 ( 5) which is equal to5,7 not

1.7

Choice (E) is not correct. The line throughA(2, 5) and B (3, 2) has slope7,

5but whenA(2,

5) and B (3, 2) are reflected about y=x, the slope of the line through their images is5.

7


Choice (A) is correct. Dividing both sides of 5 5y x by 5x yields5

1.5

y

xSince

5 ( 5 )( 5 ), it follows that5

5.5

Thus5 5

1,5

y y

xxand so

5 1 5 1 1(1) .

5 5 5 5

y y

x x


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Choice (B) is not correct. If5

5

y

xwere equal to

1,

5then cross-multiplying would give 5y= 5x.

But 5 5,y x not 5x. Therefore,5

5

y

xis not equal to

1.

5

Choice (C) is not correct. If5

5

y

xwere equal to 1, then it would follow that 5 5 ,y x which is

equivalent to 5.y x But 5 5 ,x y not y. Therefore,5

5

y

xis not equal to 1.

Choice (D) is not correct. If5

5

y

xwere equal to 5, then multiplying both expressions by

5

5

x

would give y= 5x, which is equivalent to 5y= 25x. But 5 5,y x not 25x. Therefore,5

5

y

xis

not equal to 5.

Choice (E) is not correct. If5

5

y

xwere equal to 5 5, then multiplying both expressions by

5

5

x

would give y= 25x, which is equivalent to 5y= 125x. But 5 5,y x not 125x. Therefore,5

5

y

x

is not equal to 5 5.


Choice (B) is correct. Since the upper line graph shows the sum of the number of soft drinks andthe number of containers of popcorn sold, and the lower line graph shows only the number of soft

drinks sold, the number of popcorn containers sold each month can be found by subtracting thevalues of the lower line graph from the corresponding values of the upper line graph. Thus, inmonth 2, there were 2,750 1,750 = 1,000 containers of popcorn sold. Similarly, in month 3,there were 3,250 2,250 = 1,000; in month 4, there were 3,500 1,500 = 2,000; in month 5,there were 3,750 1,500 = 2,250; in month 6, there were 3,250 2,000 = 1,250; and in month 7,there were 3,750 2,250 = 1,500. Thus, from month 2 to month 3, there was no increase ordecrease in the number of containers sold. From month 3 to month 4, there was an increase of1,000. From month 4 to month 5, there was an increase of 250. From month 5 to month 6, therewas a decrease of 1,000. From month 6 to month 7, there was an increase of 250. Therefore,month 4 showed the greatest increase in the number of containers of popcorn sold from theprevious month.

Choice (A) is not correct. From month 2 to month 3, there was no increase in the number of

containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000.Therefore, month 3 did not show the greatest increase in the number of containers of popcornsold from the previous month.

Choice (C) is not correct. From month 4 to month 5, there was an increase of 250 in the numberof containers of popcorn sold. However, from month 3 to month 4, there was an increase of1,000. Therefore, month 5 did not show the greatest increase in the number of containers ofpopcorn sold from the previous month.


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Choice (D) is not correct. From month 5 to month 6, there was a decrease of 1,000 in the numberof containers of popcorn sold. However, from month 3 to month 4, there was an increase of1,000. Therefore, month 6 did not show the greatest increase in the number of containers ofpopcorn sold from the previous month.

Choice (E) is not correct. From month 6 to month 7, there was an increase of 250 in the numberof containers of popcorn sold. However, from month 3 to month 4, there was an increase of1,000. Therefore, month 7 did not show the greatest increase in the number of containers ofpopcorn sold from the previous month.


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2012 PSAT/NMSQT Practice Test Explanations - Section 2

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Transcript of 2012 PSAT/NMSQT Practice Test Explanations - Section 2