Post on 03-Jun-2018
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SAT MATH 1 & 2 SUBJECT TEST2
3. When the figure below is spun around its vertical axis, the volume
of the solid formed will be
(A) 9 (B) 36 (C) 72 (D) 144 (E) 288
4. Iff(x) =86
6
2
2
x
xx
,f(2) =
(A) 0 (B) 5.75 (C) 6.25 (D) 24.5 (E) Undefined
5. A high school musical production sells student tickets for $5 each
and adult tickets for $8 each. If the ratio of adult to student tickets pur-
chases is 3:1, what is the average income per ticket sold?
(A) $5.50 (B) $5.75 (C) $6.50 (D) $7.25 (E) $14.50
6. Due to poor economic conditions, a company had to lay off 20% of
its workforce. When the economy improved, it was able to restore the
number of employees to its original number. By what percent was the
depleted workforce increased in order to return to the original number
of employees?
(A) 20 (B) 25 (C) 80 (D) 120 (E) 125
7. A value zis multiplied by1
3,
1
2is subtracted from the result, and
the square root of the end result is 4. What was the original number?
(A) 1
2 (B) 15
6 (C) 115
2 (D) 16 (E)
149
2
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SAT MATH 1 & 2 SUBJECT TEST4
15. If the binary operation a# b= ab __b, then (2 # 4) (4 # 2) =
(A) 32 (B) __2 2 (C) 0 (D) __2 2 (E) 32
16. Of the 45 countries in Europe, 7 get 100% of their natural gas from
Russia, and 6 get 50% of their natural gas from Russia. If 25% of all the
natural gas imported into Europe comes from Russia, what is the average
percent of imported natural gas from Russia for the remaining countries
in Europe?
(A) 3.9% (B) 20% (C) 25% (D) 75% (E) 78.1%
17. A(3,9) and B(9,1) are the endpoints of the diameter of a circle.
The equation for this circle is
(A) (x 3)2+ (y 4)2= 61 (B) (x 7)2+ (y+ 4)2= 269
(C) (x+ 7)2+ (y 4)2= 61 (D) (x+ 3)2+ (y+ 4)2= 169
(E) (x+ 3)2+ (y 4)2= 25
18. Isosceles trapezoidACDEwith ||AC DE is shown below. Eis the
midpoint ofAB, and BD= DCandBC = DE.
The ratio of the area of triangle BDCto trapezoidACDEis
(A) 1:2 (B) 1:3 (C) 1:4 (D) 1:5 (E) 1:6
19. If2 3
3 24 1 3 4
x y=
10 11
5z
, thenx+y z=
(A) 21 (B) 15 (C) 0 (D) 15 (E) 21
20. Iff(x) = 5x+ 3 andg(x) =x2 1, thenf(g(2)) =
(A) 3 (B) 13 (C) 18 (D) 39 (E) 168
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 5
21. Chords AB and CD of circle O intersect at point E. If CE= 3,
ED= 12, andAEis 5 units longer than EB,AB=
(A) 4 (B) 9 (C) 11 (D) 13 (E) 18
22. Which is the equation of the line perpendicular to 4x 5y= 17 that
passes through the point (5,2)?
(A) 4x 5y= 10 (B) 5x+ 4y= 33 (C) 4x+ 5y= 30
(D) 5x 4y= 17 (E) y x4
5
2
15=-
+
23. A stone is thrown vertically into the air from the edge of a build-
ing with height 12 meters. The height of the stone is given by the for-
mula h= 4.9t2+ 34.3t+ 12. What is the maximum height, in meters,
of the stone?
(A) 3.5 (B) 12 (C) 72.025 (D) 114.9 (E) 468.2
24. In ABC,AB= 40, the measure of angle B= 50, and BC= 80. The
area of ABCto the nearest integer is
(A) 613 (B) 1024 (C) 1226 (D) 2240 (E) 2252
25. If 42
a b+= , and aand bare non-negative integers, which of the
following cannot be a value of ab?
(A) 0 (B) 7 (C) 14 (D) 15 (E) 16
26. The perpendicular bisector of the segment with endpoints (3,5) and
(1,3) passes through
(A) (5,2) (B) (5,3) (C) (5,4)
(D) (5,5) (E) (5,6)
27. The difference between the product of the roots and the sum of the
roots of the quadratic equation 6x2 12x+ 19 = 0 is
(A)
7
6 (B)
31
6 (C)
7
12 (D)
31
12 (E)7
6
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SAT MATH 1 & 2 SUBJECT TEST6
28. In right triangleABC, ||D E B C, CD= 1.5, and BE= 2.0.
The sine of angle is equal to
(A)1
2 (B)
3
4 (C)
2
2 (D)
3
2 (E)
3
5
29. QUESTis a pentagon. The measure of angle Q= 3x 20, the mea-
sure of angle U= 2x+ 50, the measure of angle E=x+ 30, the measure
of angle S= 5x 90, and the measure of angle T= x+ 90. Which two
angles have equal measures?(A) Eand S (B) Qand U (C) Uand T
(D) Tand E (E) Uand E
30. The vertices of triangle PQRare P(3,2), Q(1,4), and R(7,0). The
altitude drawn from Qintersects the line PRat the point
(A) (1,2) (B) (2,1) (C) ( 1,2)
(D) (3,2) (E) (7,0)
31. If qis a positive integer > 1 such that2 4
31
nn
q
= , n=
(A) 1 (B) 1 (C) 1, 4___3 (D) 1, 4__
3 (E)
1 47
6
i
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 7
32. The measure of arcABin circle Ois 108.
3
a b c+ +=
(A) 18 (B) 27 (C) 36 (D) 45 (E) 54
33. Alex observed that the angle of elevation to the top of 800-foot
Mount Colin was 23. To the nearest foot, how much closer to the base
of Mount Colin must Alex move so that his angle of elevation is doubled?
(A) 200 (B) 400 (C) 489 (D) 1112 (E) 1600
34. Iff(x) =2
2
6
6 8
x x
x x
, solvef(x) = 3.
(A) {5, 1} (B) {2, 7.5} (C) 1 3 7 1 3 7,2 2
(D)17 73 17 73
,6 6
(E)
35. InQRS,Xis on QR and Yis on QS , so that ||X Y R Sand1
4
QX
XR = .
The ratio of the area of QXYto the area of trapezoidXYSRis
(A) 1:4 (B) 1:15 (C) 1:16 (D) 1:24 (E) 1:25
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SAT MATH 1 & 2 SUBJECT TEST8
36. In quadrilateral KLMN, KL= LM, KN=MN, and diagonals KM
and NL intersect at P. If KP= PM, then which of the following state-
ments is true?
I. NP= PL.
II. KLMNis a rhombus.
III. The area of KLMNis1
2(KM)(NL).
(A) I only (B) II only (C) III only
(D) II and III only (E) I and III only
37. If 7x+ 9y= 86 and 4x 3y= 19,x+ 4y=
(A)18
3119
(B)1
223
(C)18
3119
(D) 35 (E) 105
38. The solution set to 10x2+ 11x 6 0 is
(A) 0.4 x 1.5 (B) 1.5 x 0.4 (C)x 0.4 orx 1.5
(D)x 1.5 orx 0.4 (E) 1.5 x 0.4
39. In simplest form,
12
3
11
3
x
x
is equivalent to
(A) 2x 7______x 2
(B) 7 2x______x 2
(C) 2x 5______x 2
(D) 2x+ 7______x 2
(E) 1
40. In right triangle QRS, QRis perpendicular to RS, QR= 12, and RS
= 12 3 . The area of the circle that circumscribes triangle QRSis
(A) 108 (B) 144 (C) 288 (D) 576 (E) 1728
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 9
41. The solution set for the equation |2x 1| |x+ 2| = 5 is
(A) {2} (B) {3} (C) {8} (D) {3, 7} (E) {2, 8}
42. Given the graph off(x) below, letg(x) =f(x 2) + 1. For what set of
values of willg(x) = 0?
(A) {2, 2, 4} (B) {0, 4, 6} (C) {4, 2}
(D) {1, 5} (E)
43. SquareABCDhas sides with length 20. Each of the smaller figuresis formed by connecting midpoints of the next larger figure.
What is the area of EFGH?
(A)25
64
(B)25
16
(C)25
4
(D) 25 (E) 100
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SAT MATH 1 & 2 SUBJECT TEST10
44. Which of the statements about the graphs off(x) = x2x 6_________x 3
and
g(x) =x+ 2 are true?
I. f(x) +g(x) = 2x+ 4
II. They intersect at one point
III. They are the same except for one point
(A) I only (B) II only (C) III only
(D) I and III only (E) II and III only
45. A 25-foot ladder leans against a building. As the bottom of the lad-
der at pointAslides away from the building, the top of the ladder, B,
slides from a height of 24 feet above the ground to a height of 16 feet.
How many feet did the bottom of the ladder slide?
(A) 7 (B) 8 (C) 9 (D) 12.2 (E) 19.2
46. In parallelogram ABCD, W is the midpoint of AD , and X is the
midpoint of BC . CW and DX are drawn and intersect at point E.
What is the ratio of the area of DEWto the area ofABCD?
(A) 1:2 (B) 1:4 (C) 1:6 (D) 1:8 (E) 1:16
47. Point O(3,2) is the center of a circle with radius 4. OA is paral-
lel to thex-axis and mAOB= 120 degrees. To the nearest tenth, the
y-coordinate of point Bis
(A) 3.5 (B) 4.0 (C) 5.5 (D) 6.6 (E) 6.9
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 11
48. The vertices of ABC have coordinates A(7,3), B(1,0), and
C(2,8). The coordinates of the center of the circumscribed circle are
(A)1 2
3 , 33 3
(B)5 5
2 , 36 6
(C)1
1 , 42
(D) 14,12
(E)1 1
4 , 52 2
49. Each side of the base of a square pyramid is increased in length by
25%, and the height of the pyramid is decreased byx%, so that the vol-
ume of the pyramid is unchanged.x=(A) 20 (B) 25 (C) 36 (D) 50 (E) 64
50. If2 1
( )2
xf x
x
, thenf(f(x)) =
(A)2
2
4 4 1
4 4
x x
x x
(B)3 4
4 3
x
x
(C)4 3
3 4
x
x
(D) 34 3
xx +
(E) 3 14 3xx
+
+
Level 1 Practice Test Solutions
1. (A) The percent increase is computed aspopulation
original populationD
. The
greatest percent change will occur when the numerator is large and the
denominator is small.
2010 2025 Pct
Africa 1,033,043 1,400,184 367,141 35.5
Asia 4,166,741 4,772,523 605,782 14.5
Latin Americaand theCaribbean
588,649 669,533 80,884 13.7
North America 351,659 397,522 45,863 13
Oceania 35,838 42,507 6,669 18.6
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SAT MATH 1 & 2 SUBJECT TEST12
2. (B) Let Bhave coordinates (x,y). The formula for finding the mid-
point of a line segment is to average the x-coordinates and average
the y-coordinates. This gives the equations10
22
x+= and
126
2
y+= .
Multiply each equation by 2. 10 + x= 4 gives x= 6 and 12 + y= 12
givesy= 0. Point Bmust have coordinates (6,0).
3. (E) The solid formed will be a hemisphere with radius 6. The vol-
ume of a sphere is given by the formula V = 4__3
r3. The volume of the
hemisphere will be half that number. With r= 6, the volume of the
hemisphere is 2__3(6)3 = 288.
4. (A)f(2) =
2
2
(2) 2 6 0
(2) 6(2) 8 24
= 0.
5. (D) Because the tickets are sold in the ratio of 3:1 you can work with
just 4 tickets. The three adult tickets raise $24 while the student ticketraises $5. The 4 tickets bring in $29 or an average of $7.25 each.
6. (B) It may help to think of the business as having 100 employees.
After the layoffs, the workforce is 80 employees. To bring the workforce
back to 100, 20 people must be hired. 20 out of the current level of 80
is a 25% increase.
7. (E) The equation described by the sentence is 1 1
3 2z = 4. Square
both sides of the equation to get 1 1
163 2z . Add
1
2to both sides of the
equation to get1 1
163 2
z = . Multiply by 3 for the answer:1
492
z = .
8. (C) At most 5 means 5 or less. There is one way to get a 2 (1 on
each die). There are two ways to get a 3 (1 and 2 or 2 and 1), three ways
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 13
to get a 4 (1,3 or 2,2, or 3,1), and four ways to get a 5 (1,4 or 2,3 or 3,2
or 4,1). This is a total of 10 outcomes out of the possible 36 outcomes
when two dice are rolled.
9. (A) Solve the equation by subtracting3
2 from both sides of the equation
and then multiplying by5
8 to getx=15
2- . Half this amount is
15
1- .
10. (B)2
2
2 8 6 3
4 20 5
x x x
x x
=
( 4)( 2) 3(2 )
( 2)( 2) 5(4 )
x x x
x x x
=
( 4) (2 )( 2) (4 )
x
x
x
x
35
. 2 xandx 2 are negatives of one another, so they
reduce to be 1. The same is true forx 4 and 4 x. The two factors of
1 multiply to 1 so the answer is 3/5.
11. (D) (a+ b)2= a2+ 2ab+ b2, so (5 + 6i)2= 52+ 2(5)(6i) + (6i)2= 25
+ 60i+ 36i2= 25 + 60i 36, or 11 + 60i.
12. (C) The mean of the 6 numbers is 37, so the sum of the six numbers
is 222. The four given numbers sum to 135, sox+ 2x= 222 135 = 87
andx= 29.
13. (D) 32 = 8 4 = 23 4, so 32 2 43 3= . x x63 2= and
y y y y y y83 8 31
3
82
3
22 23
= = = =^ h . x y32 6 83 = 2x y y42 232 .
14. (D) The radius of the inscribed circle is 3. The area of the circle is
32, or 9. The area of the square is 62= 36. Subtract the area of the
circle from the area of the square to get 36 9.
15. (B) 2 #4 = 24 __4 = 16 2 = 14. 4 #2 = 42 __2 = 16 __2 . (2 #
4) (4 #2) = 14 (16
__
2 ) = 2 + __
2 .
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 15
21. (D) When chords intersect inside a circle, the products of the seg-
ments formed by the chords are equal. That is, (AE)(EB) = (CE)(ED).
Solve for EB: (EB+ 5)(EB) = (3)(12), so (EB)2+ 5EB 36 = 0 or (EB 4)
(EB+ 9) = 0 and EB= 4.AE= EB+ 5 = 9 andABhas a length of 13.
22. (B) The slope of the line 4x 5y= 17 is5
4 . The slope of the perpen-
dicular line is4
5- . The equation of the line is y x b
4
5=-
+ . Substituting
5 forxand 2 forygives b24
55=
-+^ h so that b=
4
33 . y x4
5
4
33=-
+
becomes 4y = 5x + 33 or 5x + 4y = 33. It is faster to know that a
line perpendicular toAx+ By= Chas the equation BxAy= D. You
could then have started the problem with 5x+ 4y= Dand determined
that D= 33.
23. (C) The time the stone reaches its maximum height can be com-
puted using the axis of symmetry for the equation. Maximum height is
reached at t= 34.3/(2)(4.9) = 3.5 seconds. Substitute this number
into the equation for the height of the stone 4.9(3.5)2+ 34.3(3.5) +
12 = 72.025 meters.
24. (C) The altitude to side BC in ABCcan be calculated using trigo-
nometry. Drop the height fromAto BC and call the foot of the altitude
D. Then sin(50)40
AD= so thatAD= 40 sin(50). The area of the triangle
is1
( )( )2
BC AD =1
2(80)(40) sin(50).
25. (C) a+ b= 8 so the values for (a,b) could be (0,8), (1,7), (2,6),
(3,5), (4,4), (5,3), (6,2), (7,1), and (8,0). The only product not available
from the choices listed is 14.
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SAT MATH 1 & 2 SUBJECT TEST16
26. (C) The segment with endpoints (3,5) and (1,3) has its midpoint
(1,1) and its slope is 2. The slope of the perpendicular line is 1/2 and
the equation of the perpendicular bisector isy 1 = 1/2 (x 1). Of the
points available as choices, only (5,4) lies on this line.
27. (A) The product of the roots is19
6
c
a = . The sum of the roots is
12
6
b
a
= . The difference between these numbers is
6
7 .
28. (B) sin( )
AD
AE . With ||D E B C,
AD DC
AE EB= . Therefore,
.
.sin
EB
DC
2 0
1 5
4
3i = = =^ h .
29. (C) The sum of the measures of the angles is 12x+ 60. The sum
of the interior angles of a pentagon is equal to 3(180) = 540. (Five
sides implies three non-overlapping triangles.) Solve forx= 40. mU=
mT= 130.
30. (B) The slope of PR is 1/5, so the slope of the altitude to PR is 5.
The only point among the choices that satisfies that the slope to point Q
will be 2/3 is (2,1).
31. (D) The exponent must be equal to 0. 3n2 n 4 = 0 becomes
(3n 4) (n+ 1) = 0 so n= 4/3, or 1.
32. (C) Inscribed Chas a measure of 54. Draw OC . Aand ACO
are congruent, as areBandBCO. mA+ mB= mACO+ mBCO
= 54. Therefore,3
cba ++=
54 54
3
+
= 36.
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SAT MATH 1 & 2 SUBJECT TEST18
38. (B)y= 10x2+ 11x 6 is a parabola that opens up and crosses the
x-axis at x= 1.5 and x= 0.4. The parabola will be below the x-axis
between these values, so y 0, or 10x2+ 11x 6 0, is satisfied by
1.5 x 0.4.
39. (A) Because 3 x= (x 3),
x
x
3
11
3
12
=
3
11
3
12
+
x
x . Multiply the
numerator and denominator by the common denominatorx 3 to get
3
3
3
11
3
12
+
x
x
x
x
=2
72
13
1)3(2
+
=
+
x
x
x
x.
40. (B) Triangle QRSmust be a 30-60-90 triangle because the longer
leg is 3 times longer than the shorter leg. The hypotenuse of the right
triangle has length 24. The circumscribed circle about a right triangle
must have its center at the midpoint of the hypotenuse, so the radius of
the circle is 12, and the area of the circle is 144.
41. (E) The graph off(x) = |2x 1| |x+ 2| intersects the graph ofy=
5 atx= 2 and atx= 8.
42. (D) Because the graph is translated right 2 and up 1, you need to
find those points for whichf(x) = 1. These are atx= 3 and 3. The graph
ofg(x) will cross thex-axis 2 points to the right of wheref(x) = 1, so
x= 1 and 5.
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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 19
43. (D) Joining the midpoints of the sides of a quadrilateral forms paral-
lelograms whose lengths are 1/2 the length of the diagonal of the larger
square. The lengths of the sides of the squares, in reduced order, are
AB= 20, 10 2 , 10, 5 2 , 5 = EF. The area of EFGHis 25.
44. (C)f(x) =6
3
x x
x
=( 2)( 3)
3
x x
x
=x + 2. With the exception
of the point (3,5), which is removed from the graph of f(x), the two
graphs are exactly the same.
45. (D) When the 25-foot ladder originally was at a height of 24 feet,
the foot of the ladder was 7 feet from the base of the building. This can
be determined using the Pythagorean Theorem: 242+ 72= 252. When
the top of the ladder falls to a height of 16 feet, the distance from the
foot of the ladder to the base of the building must satisfy the equation
162+ b2= 252, so that b2= 252 162= 369, so b= 19.2 ft. The ladder
slipped 12.2 feet further from the building.
46. (D) Draw WX. WXCDis a parallelogram with one-half the area of
ABCD because WD=XCand ||WD XC . Therefore, Eis the midpoint
of the diagonals, and the distance from E to WD is half the distance
from Bto WD . If his the distance from Bto WD , the area ofDEW is
1 1 1
( )2 2 4
WD h WD h
=1 1 1
( )4 2 8
AD h AD h
=1
8areaABCD.
47. (C) Extend radius AO through Oto intersect the circle at D. The
altitude from Bto the diameter forms a right triangle with mDOB=
60. The length of the altitude is the opposite leg of the right triangle, so
its length is 4sin(60) = 3.46. Add this to 2, and they-coordinate of B, to
the nearest tenth, is 5.5.
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SAT MATH 1 & 2 SUBJECT TEST20
48. (B) The circumcenter is the intersection of the perpendicular bisec-
tors of the sides of the triangle. The midpoint of AC is (4.5,5.5), and
the slope of AC is 1. The equation of the perpendicular bisector to AC
isy= x+ 1. The midpoint of AB is (4,1.5), and the slope of AB is
0.5. The equation of the perpendicular bisector to AB isy= 2x+ 9.5.
Solve this system of equations to get the point5 5
2 , 36 6
.
49. (C) If his the height of the original pyramid and His the height of
the new pyramid, then the volumes of the two pyramids are related by
the equation (1/3)hs2= (1/3)(x)h(1.25s)2, wheresrepresents the length
of a side of the base of the original pyramid. Solve forxto getx= .64.
The height of the new pyramid must be 64% of the original pyramid, so
the height of the original pyramid was reduced by 36%.
50. (B)2
2
12
12
122
))((+
+
+
=
x
x
x
x
xff ) = 2
2
22
12
12
122
+
+
+
+
+
x
x
x
x
x
x
=
)2(212
)2(1122
xx
xx
=4212
224
xx
xx
=34
43
x
x
.