Sat math Essentials

6.
d.
Triangles 
ABC
and 
BED
have two pairs of
congruent angles. Therefore, the third pair of
angles must be congruent, which makes these
triangles similar. If the area of the smaller
triangle,
BED
, is equal to 
b
2
h
, then the area of
the larger triangle,
ABC
, is equal to 
(5
b
)
2
(5
h
)
or
25(
b
2
h
). The area of triangle 
ABC
is 25 times
larger than the area of triangle 
BED
. Multiply
the area of triangle 
BED
by 25: 25(5
a
2
+ 10)
= 125
a
2
+ 250.
7.
b.
The positive factors of 180 (the positive num-
bers that divide evenly into 180) are 1, 2, 3, 4,
5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90,
and 180. Of these numbers, 8 (6, 12, 18, 30,
36, 60, 90, and 180) are multiples of 6.
8.
c.
A positive number minus a negative number
will not only always be a positive number,
but will also be a positive number greater
than the first operand.
gh
will always be neg-
ative when one multiplicand is positive and
the other is negative.
g
+ 
h
will be positive
when the absolute value of
g
is greater than
the absolute value of
h
, but 
g
+ 
h
will be neg-
ative when the absolute value of
g
is less than
the absolute value of
h
. |
h
| – |
g
| will be posi-
tive when |
h
| is greater than 
g
, but |
h
| – |
g
| will
be negative when |
h
| is less than 
g
.
h
g
will be
positive when 
g
is an even, whole number, but
negative when 
g
is an odd, whole number.
9.
23
If
x
is the width of the room, then 3 + 2
x
is the
length of the room. The perimeter is equal to
x
+ 
x
+ (3 + 2
x
) + (3 + 2
x
) = 66; 6
x
+ 6 = 66;
6
x
= 60;
x
= 10. The length of the room is
equal to 2
x
+ 3, 2(10) + 3 = 23 feet.
10.
11
The labeled angle formed by lines 
M
and 
K
and the supplement of the labeled angle
formed by lines 
L
and 
N
are alternating
angles. Therefore, they are congruent. The
angle labeled (10
a
+ 5) and its supplement,
which is equal to (8
b
+ 1), total 180 degrees:
(10
a
+ 5) + (8
b
+ 1) = 180. If
b
= 8, then:
(10
a
+ 5) + (8(8) + 1) = 180
10
a
+ 70 = 180
10
a
= 110
a
= 11
11.
2
The first expression, 6
x
+ 9
y
– 15, is –3 times
the second expression, –2
x
– 3
y
+ 5 (multiply
each term in the second expression by –3 and
you’d get the first expression). Therefore, the
value of the first expression, –6, is –3 times
the value of the second expression. So, you
can find the value of the second expression by
dividing the value of the first expression by
–3:
–
–
6
3
= 2. The value of –2
x
– 3
y
+ 5 (2) is just
–
3
1
times the value of 6
x
+ 9
y
– 15 (–6) since
–2
x
– 3
y
+ 5 itself is –
1
3
times 6
x
+ 9
y
– 15.
12.
90
Triangle 
DBC
and triangle 
DEF
are isosceles
right triangles, which means the measures of
BDC
and 
EDF
both equal 45°; 180 –
(m
BDC
+ m
EDF
) = m
Z
; 180 – 90 =
m
Z
; m
Z
= 90°.
13.
7
First, use the distance formula to form an
equation that can be solved for 
m
:
Distance = 
(
x
2
– 
x
1
)
2
+ (
y
2
– 
y
1
)
2
10 = 
(4 – (–
2))
2
+
((–1) –
m
)
2
10 = 
(6)
2
+
(–1 – 
m
)
2
10 = 
36 + 
m
2
+ 2
m
+ 1
10 = 
m
2
+ 2
m
+ 37
100 = 
m
2
+ 2
m
+ 37
m
2
+ 2
m
– 63 = 0
Now, factor 
m
2
+ 2
m
– 63:
(
m
+ 9)(
m
– 7) = 0
m
= 7,
m
= –9. The positive value of
m
is 7.
14.
27
Substitute 3 for 
a
:
= 9. To solve for 
z
, raise
both sides of the equation to the power 
3
2
:
= ,
z
= 
9
3
= 3
3
= 27.
9
3
2
z
2
3
3
2
z
2
3
–
P R A C T I C E T E S T 1
–
1 9 3

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15.
24
If the height of the prism is 
h
, then the length
of the prism is four times that, 4
h
. The length
is one-third of the width, so the width is three
times the length: 12
h
. The volume of the
prism is equal to its length multiplied by its
width multiplied by its height:
(
h
)(4
h
)(12
h
) = 384
48
h
3
= 384
h
3
= 8
h
= 2
The height of the prism is 2 in, the length of
the prism is (2 in)(4) = 8 in, and the width of
the prism is (8 in)(3) = 24 in.
16.
3
Solve 2
a
2
+ 
b
= 10 for 
b
:
b
= 10 – 2
a
2
. Substi-
tute (10 – 2
a
2
) for 
b
in the second equation
and solve for 
a
:
–
10 –
4
2
a
2
+ 3
a
= 11
–10 + 2
a
2
+ 12
a
= 44
2
a
2
+ 12
a
– 54 = 0
(2
a
– 6)(
a
+ 9) = 0
2
a
– 6 = 0,
a
= 3
a
+ 9 = 0,
a
= –9
The positive value of
a
is 3.
17.
4.20
If one pound of almonds costs $1.00, then 4
pounds of almonds costs 4($1.00) = $4.00. If
Stephanie pays a 5% tax, then she pays
($4.00)(0.05) = $0.20 in tax. Her total bill is
$4.00 + $0.20 = $4.20.
18.
5
The circumference of a circle = 2
π
r
and the
area of a circle = 
π
r
2
. If the ratio of the num-
ber of linear units in the circumference to
the number of square units in the area is 2:5,
then five times the circumference is equal to
twice the area:
5(2
π
r
) = 2(
π
r
2
)
10
π
r
= 2
π
r
2
10
r
= 2
r
2
5
r
= 
r
2
r
= 5
The radius of the circle is equal to 5.
Section 3 Answers
1.
b.
Two numbers are in the ratio 4:5 if the second
number is 
5
4
times the value of the first number;
1
4
is 
5
4
times the value of
1
5
.
2.
a.
Substitute –3 for 
x
:
–2(–3)
2
+ 3(–3) – 7 = –2(9) – 9 – 7 = –18 – 16
= –34
3.
a.
First, convert the equation to slope-intercept
form:
y
= 
mx
+ 
b
. Divide both sides of the equa-
tion by –3:
–
–
3
3
y
= 
12
–
x
3
– 3
y
= –4
x
+ 1
The slope of a line written in this form is equal
to the coefficient of the 
x
term. The coefficient
of the 
x
term is –4, so the slope of the line is –4.
4.
d.
The equation of a parabola with its turning
point 
c
units to the right of the 
y
-axis is written
as 
y
= (
x
– 
c
)
2
. The equation of a parabola with
its turning point 
d
units below the 
x
-axis is writ-
ten as 
y
= 
x
2
– 
d
. The parabola shown has its
turning point three units to the right of the 
y
-
axis and two units below the 
x
-axis, so its equa-
tion is 
y
= (
x
– 3)
2
– 2. Alternatively, you can
plug the coordinates of the vertex of the
parabola, (3,–2), into each equation. The only
equation that holds true is choice 
d
:
y
= (
x
– 3)
2
– 2, –2 = (3 – 3)
2
– 2, –2 = 0
2
– 2, –2 = –2.
5.
c.
1
5
6
= 0.3125 and 
2
9
0
= 0.45;
3
8
= 0.375, which is
between 0.34 and 0.40, and between 0.3125
and 0.45.
6.
d.
20% of $85 = (0.20)($85) = $17. While on sale,
the coat is sold for $85 – $17 = $68; 10% of $68
= (0.10)($68) = $6.80. After the sale, the coat is
sold for $68 + $6.80 = $74.80.
7.
e.
Set the expression 4
x
2
– 2
x
+ 3 equal to 3 and
solve for 
x
:
4
x
2
– 2
x
+ 3 = 3
4
x
2
– 2
x
+ 3 – 3 = 3 – 3 
4
x
2
– 2
x
= 0
4
x
(
x
– 
1
2
) = 0
x
= 0,
x
= 
1
2
–
P R A C T I C E T E S T 1
–
1 9 4

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8.
a.
There are three numbers on the wheel that are
less than four (1, 2, 3), but only one of those
numbers (3) is greater than two. The probabil-
ity of Jenna spinning a number that is both less
than 4 and greater than 2 is 
1
8
.
9.
e.
The volume of a cylinder is equal to 
π
r
2
h
. The
volume of the cylinder is 160
π
and its radius is 4.
Therefore, the height of the cylinder is equal to:
160
π
= 
π
(4)
2
h
160 = 16
h
h
= 10
The length of an edge of the cube is equal to half
the height of the cylinder. The edge of the cube
is 5 units. The surface area of a cube is equal to
6
e
2
, where 
e
is the length of an edge of the cube.
The surface area of the cube = 6(5)
2
= 6(25) =
150 square units.
10.
c.
m
#
n
is a function definition. The problem is
saying “
m
#
n
” is the same as “
m
2
– 
n
”. If
m
#
n
is
m
2
– 
n
, then 
n
#
m
is 
n
2
– 
m
. So, to find 
m
#(
n
#
m
),
replace (
n
#
m
) with the value of (
n
#
m
), which is
n
2
– 
m
:
m
#(
n
2
– 
m
).
Now, use the function definition again.
The function definition says “take the value
before the # symbol, square it, and subtract the
value after the # symbol”:
m
squared is 
m
2
,
minus the second term, (
n
2
– 
m
), is equal to 
m
2
– (
n
2
– 
m
) = 
m
2
– 
n
2
+ 
m
.