Test 1
10) C
Test 2
36 = 9X
8) C
(32 )(32 )(32 ) = 9X
(9)(9)(9) = 9X
(9)3 = 9X
1) B (30 )(3-2)(32 ) =
1
(1)( )(9) = 1
9
1) C
X=3
5X + 2
5X
2
=
+
=
5X
5X
5X
1+
9) C
2
5X
8Y
Y
8Y(X - 1) + Y(X + 1)
X + 1 + X - 1 = (X + 1)(X - 1) (X - 1)(X + 1) =
8XY - 8Y + XY + Y
9XY - 7Y (X ≠ 1, -1)
=
(X + 1)(X - 1)
(X + 1)(X - 1)
(X ≠ 0)
2) A Y 8 ÷ Y2 = Y(8-2) = Y6
3) D
4) C
5) A
6) C
(3Q2 )3 = (3)3 (Q)(2x3) = 27Q6
P3 N-2 = (3-4) (-2-2) =
P
N
N2 P4
1
(N and P ≠ 0)
P-1N-4 or
N4 P
10) A
2) C
11) A X + 2Y = 5
(1/2 Y) + 2Y = 5
2(3)
5
1(3) 6 + 5 - 3
8
+
=
=
3 Y Y(3)
Y(3)
3Y
3Y
Y + 4Y = 10
5Y = 10
Y=2
3) D
2X
3X
=
X+2 X-2
(Y ≠ 0)
2X(X - 2)
3X(X + 2)
=
(X + 2)(X - 2) (X + 2)(X - 2)
(2X2 - 4X) - (3X2 + 6X)
=
(X + 2)(X - 2)
-X2 - 10X
(X ≠ 2, -2)
(X + 2)(X - 2)
12) B 4a = a + 6
3a = 6
a=2
11) D
3Y-1
= 81
3Y-1 = 34
Y-1=4
Y=5
A: 05 = 1
B: (-1)5 = -1
C: (?)5 = 16 (No whole number solution)
2 + 5
- Y-1 =
3Y
Y
13) B
14) D
D: 25 = 32
A straight line is 180°
180° - 20° = 160°
b + a = 160°
3 = 15
7
X
3X = 105
4) A
X2 + 2X = X2 + 2X = X + 2
X
X
X
X + 5 = 2X
X=5
12) A
( )( )( )
[(32 )-3]-1 = 3 2 -3 -1 = 36
13) D
-42 + 12 ÷ 4 - | 6 - 8 | =
(X ≠ 0)
5) B
X = 35
9 - 5 = 9(Y) - 5(X) =
4X 4 Y
4X(Y) 4Y(X)
-42 + 12 ÷ 4 - | -2 | =
+ 12 ÷ 4 - 2 =
-16 + 12 ÷ 4 - 2 =
-16 + 3 - 2 =
-15
9Y - 5X
4XY
35 cups of flour are needed
Parentheses
-42
(X, Y ≠ 0)
Exponents
Multiplication and Division
Addition and Subtraction
7) A A 2 B 4 + B3 A = AB3 (AB + 1)
A B3 is greatest common factor
15) A
counting on graph:
up
slope is
= 2
over
3
1 + P2 Q is left
+ A0 = 1 + 1 = 2
14) D
[0-(-2)]
= 2
[0-(-3)]
3
(Lesson 20 has a review of slope)
The area of a trapezoid is the average of the
2 bases times the height.
A = ÊË 3m + 5mˆ¯ m
2
(A ≠ 0)
A = ÊË 8mˆ¯ m = 4m2
2
using formula:
line passes through (0,0) and (-3,-2)
slope =
1
1
9) C (-2 + 4) -2 = (2)-2 = 2 =
4
2
A
A
7) B
†
18AB -12A2 = 18AB - 12A2 = 3B - 2A
6A
6A
6A
(X ≠ 0)
15) B
Test Solutions 1-2
8) D P2 Q + P4 Q 2 = P2 Q(1 + P2 Q)
6) D
Test 3
1) C
9) B
Test 4
3AABA-2 + 4AB - 6B =
3B + 4AB - 6B =
4AB - 3B
Ê 2ˆ
8) A Á ˜ 63Y12 =
Ë 3¯
Ê 2ˆ
Á ˜ 9 ⋅ 7Y 6Y 6 =
Ë 3¯
6,200 = 6.2 x 103
Ê 2ˆ 6
Á ˜ 3Y 7 = 2Y 6 7
Ë 3¯
1) B (2 3)(3 2 ) = 6 6
2) B
10) A
.268 = 2.68 x 10-1
2 -1
4X + 3XY Y + 8XY
Y1
4X + 3XY2 Y -2 + 8XY
3) C
.000073 x .0054 =
(7.3 x 10-5) x (5.4 x 10-3) =
4X + 3X + 8XY
(7.3 x 5.4)(10-5 x 10-3) = 39.42 x 10-8 =
7X + 8XY
2) C (2 3) + (3 3 ) = 5 3
9) A
3 9 16Y 4 =
()
3 3 Y2 = 9 Y2
4
4
(Y ≠ 0)
(3.942 x 101 ) x 10-8 = 3.942 x 10-7
4) A
3) A
32,000,000 ÷ 16,000 =
(3.2 x 107 ) ÷ (1.6 x 104 ) =
5 3 =5 3
3
3 3
10) A 3 50 - 2 18 =
3 25 ⋅ 2 - 2 9 ⋅2 =
(3.2 ÷ 1.6)(107 ÷ 104 ) = 2.0 x 103
15 2 - 6 2 = 9 2
11) B
5) B
2.0 x 10-2
Y= 1
9
=
(9.2 ÷ 2.0)(101 ÷ 10-2) = 4.6 x 103
-1
2ab-1 + 3a-1b - 4 b-1 =
a
+ 3b
a
3b
a
-
2a
b
-
4a
b
4) D
†
5
= 5 ÷ 8 = .625
8
5) B
7) B
Complementary
12) B
In the slope-intercept form Y = mX + b,
b indicates the Y intercept.
(See lesson 20 for review)
13) D
V = area of base times the height
3 (3 2Y + 3 3 ) =
†
=
13) A
3=1x3
†
6=1x2x3
(a and b ≠ 0)
8=1x2x2x2
6) C
=
V = (pb2 )6 or 6pb2
†
A(A2 ) + 3(6) = A 3 + 18
6(A2 ) A 2 (6)
6 A2
14) D P = 2(3b) + 2(a - b)
(A ≠ 0)
P = 6b + 2a - 2b
2
7) D
†
=
Y -1X-3
X3 Y - X3 Y + 2X3 Y = 2X3 Y
(X and Y ≠ 0)
15) C
102° An obtuse angle is greater
than 90° and less than 180°
2 5 + 5 2 =
5 5
2 2
()
14) A
6(Y - 4) = 3(Y - 4) = 3Y - 12
2
15) B
33 = 27 and 34 = 81, so four 3’s
()
2 5 2 +5 2 5 =
5 2
2 5
4 5 + 25 2
10
Test Solutions 3-4
P = 4b + 2a
†
8) C XXXY - YXXX +
75 5 = 25 ⋅ 3 ⋅ 5 =
5
5 5
5 15 = 15
5
†
+ 3
A2
11) C
3 ( 18Y + 3 3 ) =
3 6Y + 9
LCM = 1 x 2 x 2 x 2 x 3 = 24
A
6
2 32 = 2 16 ⋅ 2 =
2 ⋅4 2 = 8 2
(Y ≠ 0)
†
12) D
2a
b
= 6
1 = 9Y
(9.2 x 101 ) ÷ (2 x 10-2) =
6) D
- 3Y
Y
1 - 3Y = 6Y
(2.3 x 10-3)(4 x 104 )
=
2 x 10-2
(2.3 x 4.0)(10-3 x 104 )
1
Y
Test 5
9) D
1) B
2A + 6
2A + 6 ⋅ A - 1
2+ 6
A =
A A
A
3A + 9
=
12
3(A
1)
3A + 9 ⋅ A - 1
12
3+
A - 1 (A - 1) + A - 1
A - 1 3A + 9
(2A + 6)(A - 1)
2) C
A(3A + 9)
-5X2 + 20 =
-5(X2 - 4) =
-5(X - 2)(X + 2)
2(A + 3)(A - 1)
3A(A + 3)
=
9) B
3A
1) C
1 = 1
2 16
4
10002 3 = (3 1000 )2 = 102 = 100
Ê
Á
Ë
2
3 ˆ -2 Ê 1 4 ˆ 2
˜
= 16
= 2 =4
Ë 3 ¯
3
9
16 ¯
()
X2 + 9X + 20
X(2X2 - X - 3) =
X(X + 1)(2X - 3)
(X + 4)(X + 5) (X + 4)(X - 3)
⋅
X3 - 9X = X(X - 3)(X + 3) (X + 4)(X + 4) =
2
(X
+
4)(X
+
4)
(X + 4)(X - 3)
X + 8X + 16
⋅
(X + 4)(X - 3) (X + 4)(X + 4)
2
X + X - 12
Y 4 - 625 =
X(X + 3)
2X3 - X2 - 3X =
(3 64 )-2 = 64 -2 3 =
2(A - 1)
10) B
10) C
3) A
=
Test 6
2) B
( 2581)-1 2 =
81 9
=
25 5
(X + 5)
4) C
11) B
(Y2 - 25)(Y2 + 25) =
(Y - 5)(Y + 5)(Y2 + 25)
5) B
2X2 + 4X = 6 Æ
2
3˘
È
3 5 2 = Í 5 32 ˙ = (23 )2 = 82 = 64
32
( )
3) C
(
)
Ê 2 ˆ
Ë R1 3 ¯
-2
4) D
5) D
-3
3
Ê -1 ˆ
= 5
= -125
Ë 25 ¯
-1
Î
b
a
˚
a + b + c = 180 ° True for all triangles
c
2(X2 + 2X - 3) = 0
(X + 3)(X - 1) = 0
X+3=0
X-1=0
X = -3
6) A
†
-6X2 = -27X + 12 Æ
(2X - 1)(X - 4) = 0
2X - 1 = 0
X=
11) B
X=1
Any number raised to the zero power
equals one
†
=
2
2 3
Ê3 Rˆ
=R
= 1 R2 3
Ë 2 ¯
4
4
12) D
Equations with 3 variables produce a
3-dimensional figure
(Lesson 20)
13) C
Area of large rectangle
Æ
Area of shaded square
Æ
-3(2X2 - 9X + 4) = 0
†
X-4=0
1
2
12) C
If Y = mX + b, m is the slope and
a perpendicular line has a slope of -
X=4
or the negative reciprocal of m.
(See Lesson 21)
1
m
†
( )
Area of shaded triangle Æ
7) C
†
(2)4
2X(X + 2)
4 - 2X =
=
X+2 2
(2)(X + 2) 2(X + 2)
2
†
-(X2 + 2X - 4)
13) A
Æ
N+D=7
10N + 10D = 70
.05N + .10D = .50 Æ -5N - 10N = -50
5N
= 20
6) A
12
81 = (811 2 )
14
= 81
=3
†
N=4
(X ≠ 2)
X+2
2 x 2 = 4 ft2
1
( )(1)(2) = 1 ft2
2
30 - (4 + 1) = 25 ft2
†
2
8 - 2X − 4X = -2(X + 2X - 4) =
2(X + 2)
2(X + 2)
6 x 5 = 30 ft2
14) A
Sam missed 2 x 30% or 60%
(.60)(60) = 36 wrong answers.
60 - 36 = 24 correct answers for Sam.
15) C
93,000,000 = 9.3 x 107
(If you do not remember how to do these, don’t
worry. They are taught again in Lesson 28.)
†
2
8) A
7) A
2
3 - -2X + X
=
X+4 X-4
2
X - 16
14) B
2
3(X - 4)
-2X(X + 4)
+ X
=
(X + 4)(X - 4) (X - 4)(X + 4) X2 - 16
2
2
†
B
12
2
= (B12 3 ) = B
(22 + 1) - (22 - 2) = (4 + 1) - (4 - 2) = 5 - 2 = 3
†
7
2
3X - 12 + 2X + 8X + X = 3X + 11X - 12
2
2
X - 16
X - 16
3 12
15) D
(B2 + 1) + (B2 - 2) = 2B2 - 1
8) D
(3 27 )4 = 27 4 3 = (3)4 = 81
†
.3 = 3.0 x 10-1
9.3 ¥ 10 = (9.3 ÷ 3.0)(107 ÷ 10-1)
.3 ¥ 10 -1
= 3.1 x 108
Test Solutions 5-6
3 - 2X + X
=
X + 4 -X + 4
X2 - 16
Test 7
Test 8
1) B
1) B
Test 9
Test 10
1) A
X+2
-121 = 11i
2) C
3- 2
3) B
7+i
1) C
(X + 5)(X + 5) = X2 + 10X + 25
2) A
(2A + 4)(2A + 4) = 4 A2 + 8A + 8A + 16 =
2) B
3) A
2) A
3) C
4) C
5) D
-81 = 9i or 9 i
100 10 10
2- A
-16 = 4i ⋅ 7 = 4 7 i
7
7
7
7
5) D
Y ⋅ (4 + 3i) = 4Y + 3Yi = 4Y + 3Yi
(4 - 3i) (4 + 3i)
16 + 9
25
3) B
6) C
5Q ⋅ 2 - 7 = 10Q - 5Q 7 =
4-7
2+ 7 2- 7
4) A
-4 + -8 = 2i + 2i 2
(3 -6 )(5 -15 ) = (3 6 i)(5 15 i) = -15 90 =
†
8) B
(7i)(-3i) = (-21)(-1) = 21
9) C
(2 -4 )(5 -9) = (10)(2i)(3i) = (60)(-1) = -60
6) B
8) C
2X + 1 ⋅ i = 2Xi + i = -2Xi - i
i
i
-1
7) B
9) D
i ⋅ 1- 2 = i - 2 i = i - 2 i =
1- 2
-1
1+ 2 1- 2
8) C
Y3 (binomial theorem)
9) A
-3(3X)2 (1) = -27X2 (binomial theorem)
10) D
10) C
[(3i)(4i)]2
=
(-12)2
= 144
7) A
Coefficient will have 2 factors: (3 - 1)
6 ⋅ 5 = 15 Y exponent = 3 - 1 = 2
4 2
X exponent = 6 - 2 = 4 15X Y
1⋅ 2
8) C
Coefficient will have 3 factors (4 - 1)
2
B exponent = 4 -1 = 3
5 ⋅4⋅ 3 = 10 2A exponent = 5 - 3 = 2
1⋅ 2 ⋅ 3
10(2A)2 B 3 = 40A2 B 3
9) D
Coefficient will have 4 factors (5 - 1)
X3 + 6X2 + 12X + 8 (binomial theorem)
-6 ⋅ 9 + 3 3 = -54 - 18 3 =
81- (9)(3)
9-3 3 9+3 3
4/ ⋅3/ ⋅2
/ ⋅1
/ =1
1⋅
/ /2 ⋅/3 ⋅ 4/
(A - B)3 = A3 - 3A2 B + 3AB2 - B3
3 ⋅2+ Y = 6+ 3 Y
4 -Y
2- Y 2+ Y
10) D
( )4 = (1)(1)(161 ) = 161
(1)(X)0 - 1
2
10) D
11) A
( 21)3 = 81
11) A
The line has a negative slope.
By definition: 1 and 8 are also alternate
exterior angles
A2 + 4A2 = H2
† 12) D
11) A
Slope is -2, Y intercept is 2,
so equation is Y = -2X + 2.
† 14) A
By definition
X-Y=3 Æ X=Y+3
Substitution
Elimination
3(Y + 3) - Y = 13
3X - Y = 13
3Y + 9 - Y = 13
-(X - Y) = -3
2Y = 4
2X
= 10
Y=2
X=5
X = (2) + 3
X=5
(5, 2)
She must travel south. Latitude measures
distance north and south, starting with 0° at the
equator. Lines of longitude run through the north
and south poles.
12) D
15) B
6 triangles can be drawn, so area is
180º - 129º = 51º (Supplementary angles)
13) C
†
(5) - Y = 3
Y=2
(5, 2)
13) C †m–C =180º - 115º = 65º (Supplementary angles)
m–B + m–C = 51º + 65º = 116º
180º - 116º = 64º (180º in a triangle)
†
720º ÷ 6 = 120º
15) B
11 Æ 1¥ 21 + 1¥ 20 = 2 + 1 = 3
b2 + b2 = H2
2b2 = H2
b
H
2 b =H
b
(Two sides of a
45° - 45° - 90° triangle are congruent)
14) B
A kilogram is a little over 2 pounds
15) C
Area of 1st rectangle: XY ft2
4 x 180º = 720º
†
()
Area of triangle is 1 (4)(2 3 ) = 4 3
2
6 ⋅4 3 = 24 3 in2
14) C
14) B
12) C
†
(Two ways to solve same problem)
15) B
864 in2 1ft
1ft
864 ft2
⋅
⋅
=
= 6 ft2
1
12 in 12 in
144
Area of 2nd rectangle: (2X)(2Y) = 4XY ft2
Test Solutions 7-10
†
(2X)(?) = 6X2
Second dimension is 3X.
13) C
13) D
A 5 =H
†
†
12) A
†
A2 + (2A)2 = H2 (Pythagorean Theorem)
5 A2 = H2
11) C
†
Coefficient will have 1 factor (2-1)
2Y exponent = 2 - 1 = 1
7=7
X exponent = 7 - 1 = 6
1
7X6 (2Y)1 = 14X6 Y
†
†
(- 21) exponent is 4
X exponent is 4 - 4 or 0
2 i - i or i 2 - i
†
6+1=7
(X + 2)3 = X3 + 3X2 (2) + 3X(2)2 + (2)3 =
-54 - 18 3 = -3 - 3
54
3
( i ) = ({
i)( i)({
i)( i)({
i)( i) = ( -1)( -1)( -1) = -1
6) B
(2X - 2)(2X - 2) = 4X2 - 4X - 4X + 4 =
4X2 -8X + 4
5) D
32
81- -4 = 9 - 2i
(X + 4)(X + 4) = X2 + 8X + 16
10Q - 5 7Q
-3
7) B
7) A
5) D
4) A
-15 9 ⋅10 = -15 ⋅3 10 = -45 10
6) B
4) C
4 A2 + 16A + 16
Test 11
9) D
X2 - 10X
Test 12
= -3
10) C
X2 - 10X + 25 = -3 + 25
()
2) A
( 21)(-7) = ( -72 ), (- 72 )2 = 494
1) B
X - 5 = ± 22
1 (16) = 8, 82 = 64
2
X = 5 ± 22
10) C X2 + 4X
A = 4, B = 4, C = -10
-4 ± 16 - 4(4)(-10) -4 ± 176
=
=
2(4)
8
(X-5)2 = 22
1) D
4X2 + 4X - 10 = 0
= -8
2) B
X2 + 4X + 4 = -8 + 4
A, C and D are equations whose highest
exponent is equal to 2. The quadratic
equation works for equations of this nature
only.
-4 ± 4 11 = -1± 11
8
2
-B ± B2 - 4AC
2A
(X + 2)2 = -4
3) A
4) B
( )( )
2
()
1 3 = 3, 3
2 4
8 8
X + 2 = ± -4
X + 2 = ±2i
X = -2 ± 2i
= 9
64
81 = 9, 9 x 2 = 18, so 18X
is the middle term
11) D 42 + L2 = 82
5) C
L2 = 64 - 16
144 = 12, 12 x 2 = 24, so 24X
4
L2 = 48
†
L = 48 = 16 ⋅ 3 = 4 3
6) A
4) C
Standard form provides A, B and C with
the proper sign.
†
12) A
11) C
12) B
A rhombus and a parallelogram have
2 pairs of parallel sides. A regular
polygon may have any number
of sides.
13) C
SAS stands for side-angle-side
†
6) A
(For any 30º-60º-90º triangle, take 3
times the length of the short leg to find the
length of the long leg)
16 = 4 , 4 ¥ 2 = 8 , so 8 X
25 5 5
5
5
7) D (X - 7)(X - 7) = X2 - 14X + 49
All quadratic equations have constants A,
B and C which can be substituted into
the quadratic formula. Not every quadratic
equation can be factored.
5) A
8
L
3) C
X2 - 36 = 0
(X + 6)(X - 6) = 0
X = 6, -6
7) D
X2 + 3X + 3 = 0 A = 1, B = 3, C = 3
A
1 liter ª 1 qt
14) D
X2
=4
13) B
A = 5, B = 2, C = -1
Knowing that the angles of one triangle
are the same as the angles of another
triangle, proves similarity, not congruence.
A
D
14) C
X + 4 = ± 20
X + 4 = ± 4⋅ 5
C
The sum of the measures of the interior
angles of a triangle are 180º.
9) D
X + 4 = ±2 5
X = -4 ± 2 5
15) B
4X2 + 20X + 25 = 0
(2X + 5)(2X + 5) = 0
X = -5
2
15) A
B
F
1 yard is a little less than a meter
E
Test Solutions 11-12
-2 ± 2 6 = -1± 6
10
5
X2 + 8X + 16 = 20
†
5X2 + 2X - 1 = 0
-2 ± 4 - 4(5)(-1) -2 ± 24
=
=
2(5)
10
+ 8X + 16 = 4 + 16
(X + 4)2 = 20
B
-3 ± 9 - 4(1)(3) -3 ± -3 -3 ± 3 i
=
=
2(1)
2
2
8) B
8) B X2 + 8X +
25°
25°
Test 13
Test 14
1) B
1) C
250 - 200 = $50 Saved
WP x 250 = 50
WP = 50
250
2) A
3) D
5) C
2) A
(0)2 - 4(1)(-9) = 36
6) D
X2 + 5 = 2X Æ X2 - 2X + 5 = 0
3) C
7.83. - 7.25 = .58 raise
WP x 7.25 = .58
WP = .58 = .08 = 8%
7.25
4) D
P + .25P = 100,000
1.25P = 100,000
P = $80,000
(-2)2 - 4(1)(5) = 4 - 20 = -16
7) B
X2 + 9 = -6X Æ X2 + 6X + 9 = 0
(6)2 - 4(1)(9) = 36 - 36 = 0
8) A
X2 - 32 = -4X Æ X2 + 4X - 32 = 0
9) A
†
= -4 – 144 = -4 – 12 = -16 , 8 = -8,4
2
2
2 2
6) A
.45 x 75 = $33.75 off
75 - 33.75 = $41.25
X2 + 3X - 6 = 0
7) B
†
10) C
†
-3 ± 32 - 4(1)(-6) -3 ± 9 + 24
=
=
2(1)
2
±
-3
33
2
X=
†
X2 - 5X = -8 Æ X2 - 5X + 8 = 0
2
-(-5) ± (-5) - 4(1)(8)
=
2(1)
5 ± 25 - 32 = 5 ± -7 = 5 ± i 7
2
2
2
† 35
= 35 = .60 = 60%
23 + 35 58
9) D
2(16)
= 32 = .73 = 73%
12 + 32 44
†
10) C
†
X=
†
†
11) B
11) D
12) B
A and B are not true, C is true, but does
not prove triangles congruent
62,000 = 6.2 x 104
†
46.5 x 103 = 4.65 x 104
13) A
14) D
†
†
15) A
†
The figure has been translated or moved
over 2 and down 6
†
1 = 1
92 81
Area = (3.14)(3)2 = 28.26
unshaded area is 85%
28.26 x .85 = 24.02
14) C
F =5
243 9
F = 243⋅ 5 = 135 females
9
5) A
L =3
540 5
L = 540⋅ 3 = 324 who only looked
5
6) C
R=4
S 5
7) B
S2
= 64
CS2 76
S2
= 64
798 76
S2 = 798 ⋅ 64 = 672g
76
8) B
H2
= 2
H2 O 18
H2
= 2
720 18
H2 = 2 ⋅720 = 80g
18
(2 5)(5 12 ) = 10 60 =
10 4 ⋅15 = 20 15
Q
-R = 0
P
B = XYZ
A
Q
= R Q = RP
P
X(Y - Z) + D = 4
X(Y - Z) = 4 - D
X= 4-D
Y-Z
5) B
1 = 1 B=C
B C
6) B
X = S YZS = XT
YZ T
Y = XT or Y = TX
ZS
SZ
X-Z=Y+5
X - Z - 5 = Y or Y = X - Z - 5
9) D
A(B + C) - D = X
A(B + C) = X + D
B + C = X +D Æ B = X + D - C
A
A
-X + Y - 4 = A + B Æ X - Y + 4 = -A -B
X = -A - B + Y - 4
†
(X ≠ 0, -2)
R =4
100 5
R = 100⋅ 4 = 80 rainy days
5
12) A
Ex: (X + 3)(X + 2) = X2 + 5X + 6
14) B
13) B
6 + (-8) - 4 + 3 = 6 - 8 - 4 + 3 = -3
14) A
Circumference = (3.14)(10) = 31.4 ft.
or think “pi is a little more than 3.”
a ÷ b = a ⋅ B (invert and multiply)
A B A b
To prove:
a B
A ⋅ b = aB (B is the reciprocal of b )
b B Ab b
B
B b
O = 16
18
H2 O
O = 16
1440 18
K = 39, C = 12, N = 14 39 + 12 + 14 = 65
3
12
X = X1 3 = (X1 3 )
= X1 6
12) C
13) A
15) C
†
10) A
3 ⋅ 8 = 3 4 ⋅2 = 6 2 = 3 2
2 ⋅8
16
8
2 8 8
†
R = 2 ⋅21 = 6 red
7
H2 = 1440 ⋅16 = 1280g
18
11) B
†
†
9) C
11) D
10) B
R =2
21 7
RS = B
T
A
8) A
†
†
They are mirror images of each other
15) D
4) A
A
AB = XYZ
RSA = BT A = BT
RS
23
= 23 = .575ª58%
23 + 16 + 1 40
(round numbers ending in 5 to next
higher number)
12) A Y ⋅ X + 2 + 4Y ⋅ X =
X X+2 X +2 X
YX + 2Y + 4YX = 5XY + 2Y
X(X + 2)
X(X + 2)
13) B
R=2
P 7
A= D
BC
ABC = D
( 21)(-18) = -9,(-9)2 = 81
The easiest way to find the answer is to substitute
the given numbers in the second equation, and try
the resulting value of X in the first equation
X = (3) - 1 = 2
(2)2 - (2) + 2(2) = 6
4 - 2 + 4 = 6 Æ 6 = 6 (true)
You can also solve for y using substitution
15) D
Volume of cube: 10 x 10 x 10 = 1,000 cm3
Volume of tube: (3.14)(1)2 (10) = 31.4 cm3
1,000 - 31.4 = 968.6 cm3
Test Solutions 13-16
.75 = 7.5 x 10-1
(6.2 x 7.5)(104 x 10-1) =
3) D
7) C
†
†
Add 3 + 1 for total attempts of 4
.40 x 32 = $12.80 markup
32 + 12.80 = $44.80
8) D
†
2) B
4) D
38.95 x .05 = 1.95 tax
38.95 x .20 = 7.79 tip
38.95 + 1.95 + 7.79 = $48.69
-4 ± 42 - 4(1)(-32) -4 ± 16 +128
=
2(1)
2
3 + 5 = 8 total, not 15
3)
5) A
X=
1) D
2) C YZ = A
B
X
24 - 12 = $12 markup
WP x 12 = 12
WP = 12 = 1 = 100%
12
X2 + X = X + 9 Æ X2 - 9 = 0
Test 16
In this test, all unknowns are such that
denominators are not equal to zero
1) D
WP = 1 = .20 = 20%
5
4) C
Test 15
Test 17
Test 18
1) D
2) B
Two unit multipliers are needed for
square units
3) D
80 oz 1 lb
⋅
= 5 lbs
1 16 oz
4) A
6 yds 3 ft
⋅
= 18 ft
1 1yd
Test 19
Test 20
1) A
D = RT, B and D may be derived from this
1)
C
1) A
2) B
D = (60)(4) = 240 miles
2)
A
2) C
3)
A
3) C T = 270 = 90 minutes
3
4) D
360 cm
1m
1m
1m
⋅
⋅
⋅
=
1
100 cm 100 cm 100 cm
7) D
9 km .62 mi
⋅
= 5.58 mi
1
1km
7) D
8) C
RT = D (9)(4) = 36 mi or (6)(6) = 36 mi
DS
DS = DJ
DJ
10) A
† 9) A
†
10) B
10 qts .95 l
⋅
= 9.5 l
1
1qt
2 greens2 5 blue 5 blue 3 red 3 red
⋅
⋅
⋅
⋅
1
1green 1green 1blue 1blue
†
12) A
†
14) A
15) B
X ⋅ 6 - i = 6X - iX = 6X - iX
6 + i 6 - i 36 - (-1)
37
(6)(4) = 24 miles
8)
D
9)
C DA + DV = 130
RA TA + RV TV = 130
10)
2
11) B 4 - 1 = 3 factors 5 ⋅4 ⋅3 = 10
1⋅ 2 ⋅ 3
exponent of Y term is 4 - 1 = 3
exponent of 2X term is 5 - 3 = 2
10(2X)2Y 3 = 40X2Y 3
†
12) B m–2 = 180°- 132° = 48°
(supplementary angles)
m–7 = 48° (alternate exterior angles)
Ex:
14) C
Volume of a cylinder =
area of base times height =
2
B
3 - 82 = 9 - 64 = -55 = 55
A line only has one dimension: length
15) C
†
[
]
area of side = 2 p (2X)2 + (p4X)(H) =
2
8pX + 4pXH
4) B
Y = -2X, Y intercept is 0
has a negative slope
5) C
Y = 3X + 2, Y intercept is 2
moderate positive slope
6) C
(2) = 3(1) + b
2=3+b
-1 = b (y intercept)
7) A
The difference in Y divided by
the difference in X
8) D
5 - 1 = 4 = -2 slope
-1- 1 -2
(1) = (-2)(1) + b
1 = -2 + b
3 = b (Y intercept)
Y = -2X + 3
9) B
May also be written as 2X + Y - 3 = 0
10) A
D = (26)(2) = 52 miles
or 130 - 78 = 52 miles
1- (-2 ) = 3 slope
-1- 3
-4
(1) = -3 (-1) + b
4
1= 3 + b
4
1=b
Y = - 3X+ 1
4
4
4
( )
A
2¥2 =4
¸
Ô 2 is the greatest
2 ¥ 2 ¥ 2 ¥ 2 = 16˝
common factor
2 ¥ 3 ¥ 3 = 18 Ô˛
12)
B
(33 )(32 ) = (3 ⋅ 3 ⋅ 3)(3 ⋅3) = 35 = 243
13)
B
2X + 3 2X2 + 11X + 12
-2X2 - 3X
8X + 12
-8X - 12
11) A Quadratic formula
†
14)
15)
†
Surface Area = 2 times area of base plus
Y = 5X - 3, Y intercept is -3
has a steep positive slope
D = RT 130 = R(5)
R = 26 mph (their rates are the same)
11)
)
X + 4
B
12) D The discriminant is under a square root
sign, so a negative number will always
yield an imaginary result.
13) B
C
2
(p)(2X) (H) = p4X H
2
()
B
(26 )TA + (26 )( TA - 1) = 130 (TV = TA - 1)
52TA = 156
TA = 3
D = (26)(3) = 78 miles
(4 + 10)(4 - 10 ) = 16 - 10 = 6
2(2A ) B = 2AB units2
2
7)
4RS = 3RS + 6 (from above)
13) D
TR = 2TS = 2(3) = 6 hours
3) D
14) A
H
85 = H2
85 ª 9 miles
93,000,000 = 9.3 x 107
30,000 = 3.0 x 104
62 + 72 = H2
36 + 49 = H2
X + .15X + .05X = 68.15
1.20X = 68.15
X = $56.79
(9.3 ÷ 3.0)(107 ÷ 104 ) = 3.1 x 103 days
6
15) D
7
365 = 3.65 x 102
(3.1 ÷ 3.65)(103 ÷ 102) ª
.85 x 101 = 8.5 years
Test Solutions 17-20
13) C
†
D
†
†
= 450reds 2
11) A
6)
RS = 6 mph
56 oz 28 g
⋅
= 1,568 g
1
1oz
DR + DS = 39
RRTR + RS TS = 39
(5)(2TS ) + (3)( TS ) = 39 (TR = 2TS )
13TS = 39
TS = 3 hours
RS (4) = (RS + 2)(3)
9) C
†
B
DP = DH
RS TS = RJ TJ
8) C
5)
TP = 4 hours to park
2
6) C
A
(9)TP = (6)(6)
3
360
= .00036m
1,000,000
3 mi 5280 ft 5280 ft
⋅
⋅
= 83,635,200 ft2
1
1mi
1mi
4)
RPTP = RHTH
3
5) B
(4)T + (8)T = 24 (TA = TJ )
T = 2 hours
R = 400 = 228.57 ª 229mph
1.75
5) B
6) A
DA + DJ = 24
RA TA + RJTJ = 24
Test 21
Test 22
Test 23
Test 24
1) B
1) A
1) D
1) A
2) C
2) A
The equation of a parabola always has one
term squared
2) A
Coefficient of X term is positive, X is
squared term
3) D
Coefficient of Y term is negative, Y is
squared term
4) B
Coefficient of squared term has the
largest absolute value
5) C
If Y = (0)2 + 2, then Y = 2
6) A
Y intercept is 1, graph opens up and is
moderately narrow
7) C
X intercept is -2, graph opens to the left
and is wide
8) A
Y intercept is 1, graph opens down and is
wide
9) D
X intercept is 1, graph opens to the right
and is of average width
10) B
Y intercept is -1, graph opens up and is
of average width
11) B
The point naming the vertex is always the
middle of the three points naming an angle
2) D
3) A
3) B
2Y = 4X + 3 Æ Y = 2X + 3 (same slope)
2
4) C
4) A
4) D
Slope would be -4
5) C
Slope will be 2
(-2) = 2(2) + b
-2 = 4 + b
-6 = b Y = 2X - 6
5) D
6) B
7) A
Y = - 1X - 1
2
8) C
13) D
†
0 > 2(0) - 3
0 > -3 true
> sign indicates dotted line
9) B
10) C
0 < 2(0) - 3
0 < -3 false
< sign indicates dotted line
†
15) A
AB = Y
C
X
9) D
10) D
11) C
)
C = ABX
Y
12) D
By definition supplementary angles add to
180º. They may or may not be congruent.
†
13) A
3 -4 - 2 + 1- 2(3 - 6)2 =
14) C
15) D
51 = 3
T 7
3T = 51⋅ 7
T = 51⋅ 7 = 119
3
12) B
†
†192.5 cm .4 in 1 ft
⋅
⋅
= 6.4 ft (rounded)
1 cm 12 in
†1
14) A
B) (3)2 ÷ 3 + 5 =
9÷3+5=8
C) -(3)2 ÷ 3 + 5 =
D) (-3)2 ÷ 3 + 4 =
9÷3+4=7
2 + 6 - X-1 =
X 3X
2(3X) + 6(X) - 1(3X) =
X(3X) 3X(X) X(3X)
6X + 6X - 3X = 9X = 3
3X2
3X2 X
(X ≠ 0)
6 = 2TDav
3 = TDav
15) C
12) D
44 km
R1T1 = D1
R2 T2 = D2
R1T1 + R2 T2 = 44
D1 + D2 = 44
T1 + T2 = 6
T2 = 6 - T1
4 T1 + 9(6 - T1) = 44
4T1 + 54 - 9T1 = 44
-5T1 + 54 = 44
-5T1 = -10
T1 = 2
T2 = 6 - 2 = 4 hrs
13) C 2(X + 1) + 2(3X) = 2X + 2 + 6X = 8X + 2
14) C 2[3X(X + 1)] + 2[3X(X)] + 2[X(X + 1)] =
4TDav = 6TDav - 6
64 + 100 = H2
4 ⋅41 = H
DDav = DDan
RDav TDav = RDanTDan
4TDav = 6(TDav - 1)
82 + 102 = H2
2 41 = H
†
A) -32 ÷ 3 + 6 =
-9 ÷ 3 + 6 = 3
13) B
164 = H
†
8-1 3 = 113 = 31 = 1
8 2
8
Checking all four
-9 ÷ 3 + 5 = 2
CY = ABX
trapezoid
3 -6 + 1- 2(-3 =
†
3(6) + 1- 2(9) =
)
)
A = (-2, 1) B = (-1, 5)
( ) ( )
M = ÊË -2 + -1 , 1+ 5 ˆ¯ M = -3 , 3
2
2
2
Unequal coefficients, so an ellipse
2
2
Checking (0, 3) Æ 9(0) + 4(3) = 36
0 + 36 = 36 (true)
D = (-1, -2) C = (3, 2)
M = -1 + 3 , -2 + 2 M = (1, 0)
2
2
(
Substitute (0, 0)
18 + 1- 18 = 1
†
B = (-1, 5) D = (-1, -2)
( ) ( )
( )
M = ÊË -1 + -1 , 5 + -2 ˆ¯ M = -1, 3
2
2
2
(
)2
†
8) C
2(3X2 + 3X) + 2(3X2 ) + 2(X2 + X) =
(4)(3) = 12 miles
Area of original rectangle = XY
Area of new rectangle = (3X)(3Y) = 9Y
6X2 + 6X + 6X2 + 2X2 + 2X = 14X2 + 8X
15) A (X)(3X)(X + 1) = 3X2 (X + 1) = 3X3 + 3X2
Test Solutions 21-24
14) B
[-1- (-2 )]2 + (-2 - 1)2
(
Substitute (0, 0)
X4 - 1 factors to (X2 - 1)(X2 + 1), so
(X4 - 1) ÷(X2†- 1) = X2 + 1
7) A Equal coefficients, so a circle
center at (-3, -3)
D = 1+ 9 = 10
11) B
†
6) B
A = (-2, 1) D = (-1, -2)
D=
†
12) C
[3 - (-1)]2 + (2 - 5)2
D = 16 + 9 = 25 = 5
†
11) C
B = (-1, 5) C = (3, 2)
D=
9) C Same as above, but ≥ indicates solid line
†
(X - 3)2 + (Y + 5)2 = 42
5) A
6) B
7) A
10) B
(X - 3)2 + (Y + 5)2 = 16
A = (-2, 1) C = (3, 2)
D = 25 +1 = 26
( )
8) D
Completing the square
(X2 - 6X + 9) + (Y 2 + 10Y + 25 ) = -18 + 9 + 25
D = [3 - (-2)]2 + (2 - 1)2
Slope will be - 1
2
-2 = - 1 (2) + b
2
-2 = -1 + b
-1 = b
3) C
Ê X1 + X2 Y1 + Y2 ˆ
,
Ë 2
2 ¯
Test 25
Test 26
1) D
2) D
6) B
X=
3) A
4) C
5) B
1) D
-(-8 ) 8
= =4
2(1)
2
Y = (4)2 - 8(4) + 1
Y = 16 - 32 + 1 = -15
Test 27
2
2
7) C
Y = -3 (1)2 + 6(1)
X2 (X2 + 5) = 0
2) A
Y = -3 + 6 = 3
4) C
(-1)(-6) = 6
5) C
2
2
X - (-1) = 8
8) A
X2 = 9
X = ±3
-(-4 ) 4
= =4
1
2 1
2
Y = 1 (4)2 - 4(4)
2
Y = 8 - 16 = -8
X=
()
6) D
vertex = (4,- 8)
(
240 - 2X
2
9) D
X
X
240 - 2X
2
†
7) B
)
Area = X 240 - 2X =
2
X(120 - X) = -X2 + 120X
X = -120 = 60
2(-1)
= -3600 + 7200 = 3,600 ft
96 - 2X
X
X
X = 24’
96 - 2(24) = 48’
11) B
2
2
2
96X - 2X = -2X + 96X
X = -96 = 24
2(-2 )
3 -27 - 4 -8 = 3 -9 ⋅3 - 4 -4 ⋅ 2 =
†
†
9i 3 - 8i 2
†
12) A
†
13) A
†
X2 = 0
X=0
Y = (0)2 + 2
2i ⋅ (8 - 5i) = (2i)(8) - (2i)(5i) = 16i + 10
8 + 5i (8 - 5i)
64 + 25
89
12) A
14) †D
†
15) D
0 = -2(0) + b
0=b
(
)
6
3
2
1
[
†
) ( )
†
6) D
2
X
±4
* ª ±4.5 ± 20 *
0
2
X + Y = 34
X2 - Y 2 = 16
2X2
= 50
X2 - (0)2 = 16
X = ±4
Y
0
2
-16
X2 = 25
X = ±5
(5)2 + Y 2 = 34
25 + Y 2 = 34
Y2 = 9
Y = ±3
]
9) C
10) C
11) A
X
)
A = X 176 - 2X =
2
X
(area)
88X - X =
-X2 + 88X
X = -88 = 44
2(-1)
Y = - (44)2 + 88 (44 ) = -1936 + 3872 = 1936 ft2
12) A
A = D Æ AE = BCD Æ AE = B
BC E
CD
13) C
1,200,000,000 = 1.2 ¥ 10 9
3,000,000 = 3.0 ¥ 10 6
X2 + (X - 3)2 = 9
2X2 - 6X = 0
2X(X - 3) = 0 Æ X = 0, 3
Y = (3) -3
Y = (0) -3
Y=0
Y = -3
(3, 0)
(0, -3)
(
176-2X
2
2
(1.2 ¥ 3.0)(109 ¥ 10 6 ) = 3.6 ¥ 10 15
X2 + X2 - 6X + 9 - 9 = 0
299 - 254 = $45 markup
WP x 299 = 45
WP = 45 ª .15 = 15%
299
Work solution:
2
X - Y = 16
4) B circle and line
†
5) C
Make sketch:
2
substitute
2 in original equation, and solve for y:
†
( )
†3 2 + 2Y = 12 Æ Y = 3
(2, 3)
†
checking:†(2)(3) = 6
†
(true)
so solution is (2, 3)
†
23 = 23 ª .40 = 40%
23 + 35 58
1
2
3
6
8) A
X2 - 4Y + 4 = 0 Æ (X - 2)(X - 2) = 0 Æ X = 2
(-5 + 7) -3 + (-6)
,
Æ 2 , -9 Æ 1, -9
2
2
2 2
†2
(
Y
-1
-2
-3
-6
X = -5 = 5 i
invalid solution
14) B
D = RT
18 = (15 - 3)T
18 = T
12
1.5 = T
15) D
3 ¥ 1.6 = 4.8 km
2 ¥ 1.6 = 3.2 km
4.8 ¥ 3.2 = 15.36 km2
Test Solutions 25-27
13) B
-2 ± 22 - 4(2)(-5 )
=
2(2)
Slope = -2
Y = -2X
X
-6
-3
-2
-1
2
X = -5
Y = 2 Æ (0, 2)
plot points for
first equation:
XY = 6
substitute - 3 X + 6 in place of Y in
2
second equation, and solve:
(X) -3 X + 6 = 6 Æ - 2 -3 X2 + 6X - 6 = 0
2
3 2
†
14) D Using quadratic formula:
15) B
A) ellipse
B) difference of 2 squares form
C) regular hyperbola
D) parabola
10) D A) first and third quadrants
B) second and fourth quadrants
C) ellipse
D) difference of 2 squares form
11) C
9i -64 = 9i ⋅8i = 72 (i)()
i = -72
-2 ± 4 + 40 = -2 ± 4 ⋅11 =
4
4
-2 ± 2 11 = -1± 11
4
2
( )
Point (4, - 1) is in Quadrant IV
4
9) C A) first and third quadrants
B) difference of 2 squares
C) second and fourth quadrants
D) parabola
Area = X(96 - 2X)
Change second equation
to slope-intercept form:
3X + 2Y = 12
2Y = -3X + 12
Y = -3 X + 6
2
sketch both equations:
Point -4, 1 is in Quadrant II
4
8) A A) first and third quadrants
B) difference of 2 squares
C) second and fourth quadrants
D) parabola
Y (or area) = - (60 2 ) + 120(60 )
10) B
3) D
X2 - 1 = 8
vertex = (1, 3)
4
5X + X = 0
3) D
X = (-6 ) = -6 = 1
2 -3
-6
2
X2 + X4 + 4X2 + 4 - 4 = 0
1) C
2) A
vertex = (4, - 15 )
2
X + (X + 2) = 4
7) B
A) is obviously a hyperbola
B) X = 16 Æ XY = 16 (hyperbola)
Y
C) difference of 2 squares (hyperbola)
D) equation of an ellipse
Test 28
1) B
9) B
N + D = 11
.05N + .10D = .80
-5N - 5D = -55
5N + 10D = 80
5D = 25
D = 5
Test 29
F1 = 50%, F2 = 5%
F1 + F2 = 150lbs
.50F1 + .05F2 = .12(150)
1) A
-50F1 - 50F2 = -7500
50F1 + 5F2 = 1800
3) C
4) C
D+Q=7
-10D - 10Q = -70
.10D + .25Q = 1.15 10D + 25Q =115
15Q = 45
Q = 3
10) C
150 - 126.7 = 23.3 lbs
2) B
P + N = 25
.01P + .05N = .57
-P - N = -25
P + 5N = 57
4N = 32
N = 8
P + 8 = 25 Æ P = 17
11) D
·
DC + DB = 20
RC TC + RB TB = 20
TC = TB
4 T + 6T = 20
10T = 20
T = 2 hours
12 noon + 2 hours = 2:00 p.m.
3N + (N + 2) + 2 = 3(N + 4)
4N + 4 = 3N + 12
N = 8 Æ 8, 10 , 12
3) C
†
Each odd integer is 2 more than the
one before it
†
12) A
†
6) A
4N + 2(N + 1) = 4(N + 2)
4N + 2N + 2 = 4N + 8
6N + 2 = 4N + 8
2N = 6
N = 3 Æ 3, 4, 5
A) Parabola
B) Line (Y = -X + 9)
C) Hyperbola
D) Circle
†
†
10N + 10 (N + 2) = 10 + 10 (N + 4)
10N + 10N + 20 = 10N + 50
10N = 30
N = 3 Æ 3, 5, 7
7 is third number
†
13) C
14) C
†
†
A T + AE = 90 ml
.20AT + .08AE = .10(90)
-8A T - 8AE = -720
20A T + 8AE = 900
12A T
= 180 Æ A T = 15 ml
)
C - 5 = 1 (D - 5)
2
C - 5 = 1 D- 5
2
2
†
1
C = D+ 5
2
2
9) A
4) B
6 x 180º = 1080º
X
X
720 - 2X
Area = X(720 - 2X) = -2X2 + 720X
X = -720 = 180
2(-2)
X = 180’
720 - 2(180) = 360’
Dimensions: 180’
† x 360’
5) D
10) C
15) B
†
3
X +1
X ⋅ 2 = 3(X + 1) = 3X + 3
2
X +1
2X
2X
X+1 2
2 x 12 = 24 hours total
DD = (10 + W)2
DU = (10 - W)5
DD = 20 + 2W
DU = 50 - 5W
D = (10 + 4.3)2
D = 28.6 miles
11) B
13) B
14) C
R = 10C
C implies that the rate of the water is greater
than the rate of the boat, in which case the
boat would be traveling downstream, not
upstream.
A) Y = X - 4, slope is +1
B) not a line
C) not a line
D) Y = -X - 4, slope is -1
1¥3 = 3, 3 2 = 9
2
2 2
4
()
N + Q = 14
.05N + .25Q = 1.90
-5N - 5Q = -70
5N + 25Q =190
20Q = 120
Q = 6 Æ N + (6) = 14
N=8
8 - 3 = 5 nickels
Ï(10C) + 10 = 4C + 40
Ô
SubstitutingÌ
6C = 30
Ô
C=5
Ó
6) C
12 = (B - 2)T
12 = BT - 2T
20 + 2W = 50 - 5W
7W = 30
W = 4.3 mph (rounded )
(5) - 2 = 3
†
60 = (B + 2)T
)
S + 30 = 3 (D + 30 )
D - 5 = 4(S - 5)
5
D = 4S - 20 + 5
S + 30 = 3 D + 18
5
D = 4S - 15
5S + 150 = 3D + 90
5S + 60 = 3D
Ï5S + 60 = 3(4S - 15)
Ô5S + 60 = 12S - 45
Substituting ÔÌ
105 = 7S
Ô
ÔÓ
15 = S
R + 10 = 4(C + 10 )
42 = (B + 4)(3)
42 = 3B + 12
10 = B
8T
60 = BT + 2T
-12 = -BT + 2T
48 =
4T
12 = T one way
2 K + 32 + 4 = K + 2
3
3
2K + 32 + 12 = 3K + 6
38 = K
C + 3 = 5 (D + 3)
6
(
8) D
2(P + 2) = K + 2
2P + 4 = K + 2
( 31 K + 163 ) + 4 = K + 2
DU = (B - W )T
18 = (B - 4)T
15) A
9X2Y = 3X Y
Test Solutions 28-29
†
8) B
P - 8 = 1 (K - 8)
3
P = 1 K - 8 + 24
3
3 3
Ï2
P = 1 K + 16
3
3
Ô
Ô
Ô
SubstitutingÌ
Ô
Ô
ÔÓ
24 =
3=T
12) D
†
7) D
(
DD = (B + W )T
42 = (B + 4) T
42 = BT + 4 T
-18 = -BT + 4T
R - 3 = 6(S - 3)
Ï 1 D + 5 + 3 = 5 (D + 3)
Ô2
2
6
Ô3D + 15 + 18 = 5D + 15
SubstitutingÌ
Ô18 = 2D
†
Ô
ÔÓ9 = D
†
5) D
R + 2 = 7 (S + 2)
2
R+2 = 7S+7
2
R= 7 S+5
2
Ï 7 S + 5 - 3 = 6(S - 3)
Ô2
Ô
7 S + 2 = 6S - 18
ÔÔ
2
SubstitutingÌ
7S + 4 = 12S - 36
Ô
Ô
40 = 5S
Ô
ÔÓ
8=S
-45F2 = -5700 Æ F2 = 126.7 lbs
2) A
7) D
Test 30
1) C
Test 31
2) C
1) D
2) A
3) B
4) C
3) B
4) A
5) D
A} 4X + 2Y + 3Z = 10
B} 2X - 2Y + Z = -6
6X
+ 4Z = 4
B} 6X - 6Y + 3Z = -18
C} 2X + 6Y - 8Z = 44
8X
- 5Z = 26
30X + 20Z = 20
32X - 20Z = 104
62X
= 124 Æ X = 2
6(2) + 4Z = 4
12 + 4Z = 4
4Z = -8
Z = -2
6) D
(See solution for #5)
7) C
(See solution for #5)
8) A
A} 6X - 3Y + 12Z = 15
B} X + 3Y - 2Z = 8
7X
+ 10Z = 23
C} 3X + 3Y - Z = 9
B} -X - 3Y + 2Z = -8
2X
+ Z= 1
†
†
9) D
(See solution for #8)
10) B
(See solution for #8)
11) D
X = 3 Æ XY = 3
Y
(2)(7) - (4)(5) 14 - 20 -6
(2)(5) - (4)(3) = 10 - 12 = -2 = 3
6) C
3X - Y = -5
7) D
9
9) A
(3 27 )-1 = 3-1 = 31
14) A
CS2 = 12 + 32 + 32 = 76
2(-1) + Z = 1
-2 + Z = 1
Z=3
2
1
9
2
-4 1 -3 -4 1
0 -5 2 0 -5 (9)(1)(2) + (2)(-3)(0) + (1)(-4)(-5 ) - (0)(1)(1) - (-5 )(-3)(9) - (2)(-4 )(2) 18 + 0 + 20 - 0 - 135 + 16 -81
=
=1
= ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) =
2 2
1 2 2
4 - 24 - 15 - 4 - 30 - 12
-81
2 1 2 + 2 -3 4 + 1 3 -5 - 4 1 1 - -5 -3 2 - 2 3 2
3 1 -3 3 1
4 -5 2 4 -5
B} (-1) + 3Y - 2(3) = 8
-7 + 3Y = 8
3Y = 15
Y =5
11) C
F + 3 = 2(K + 3)
F + 3 = 2K + 6
(3K - 4) + 3 = 2K + 6
3K - 1 = 2K + 6
K=7
32 = (B + 6)2
32 = 2B + 12
20 = 2B
10 = B
14) D
T = 28%, S = 7%
T + S = 51 l
.28T + .07S = .14(51)
The difference in their ages is 10 years, so
when Kimberly is 15, Fritha will be 25.
F = 3(7) - 4 = 17
10 = (B - 6)(2.5)
10 = 2.5B - 15
25 = 2.5B
10 = B
Since there is only one unknown,
either equation may be used to
solve for B.
13) B
3(N) - [2 + (N + 2)] = -6 + (N + 4)
3N - 2 - N - 2 = -6 + N + 4
2N - 4 = -2 + N
N=2
2, 4, 6
2[(2 + X) - (-3 )] =
†
2(X + 5)2 =
2(2 + X + 3)2 =
2(X2 + 10X + 25 ) =
2
2X + 20X + 50
-7T - 7S = -357
28T + 7S = 714
21T
= 357
15) C
Volume of cone = 1 area of base x height
3
T = 17 l
V = 1 p (4)2 (6) ª 100 cm3 or ml
3
502 - 100 = 402 ml
Test Solutions 30-31
12) B
F - 2 = 3(K - 2)
F = 3K - 4
2
15) C
†
(-1, 2 )
2 9
1 2 9
3 -4 -3 3 -4
4 0
2 4 0
( )( )(2) + (9)(-3 )(4) + (1)(3)(0) - (4)(-4)(1) - (0)(-3 )(2) - (2)(3)(9) -16 - 108 + 0 + 16 - 0 - 54 -162
Y=
= (2 )(-4
=
=
=2
2 2
1 2 2
4 - 24 - 15 - 4 - 30 - 12
-81
2 1)(2) + (2)(-3 )(4) + (1)(3)(-5 ) - (4)(1)(1) - (-5 )(-3)(2) - (2)(3)(2)
3 1 -3 3 1
4 -5 2 4 -5
12 = C
Æ C = 12 ⋅1368 = 216 g
76 1368
76
†
3 -5
3 5
15 - (-15) 30
Y=
=
=
=2
3 -1
12 - (-3) 15
3 4
10) A The denominator remains the same for each variable. For the numerator of the Y variable, the first column is the same as the first column of the denominator.
The second column of the Y numerator matches the first column of the X numerator, and the last column of the numerator is the same for both X and Y.
Using #9, the determinants for Y look like this:
12) B First equation is a line and the second a circle. A line
cannot touch a circle in more than two places.
†
13) B
†
3X + 4Y = 5
-5 -1
5 4
-20 - (-5) -15
X=
=
=
= -1
3 -1
15
12 - (-3 )
3 4
8) D
A} 4(2) + 2Y + 3(-2) = 10
8 + 2Y - 6 = 10
2Y = 8
Y= 4
7X + 10Z = 23
-20X - 10Z = -10
-13X
= 13
X = -1
5) A