Sat math Essentials

.03/pound
–
M AT H P R E T E S T
–
2 2

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A n s w e r s
1.
b.
Substitute 
1
8
for 
w
. To raise 
1
8
to the exponent
2
3
, square 
1
8
and then take the cube root.
1
8
2
6
1
4
, and the cube root of
6
1
4
1
4
.
2.
d.
Samantha is two years older than half of
Michele’s age. Since Michele is 12, Samantha
is (12 
2) 
2 
8. Ben is three times as old
as Samantha, so Ben is 24.
3.
e.
Factor the expression 
x
2
– 8
x
12 and set
each factor equal to 0:
x
2
– 8x 
12 
(
x
– 2)(
x
– 6)
x
– 2 
0, so 
x
2
x
– 6 
0, so 
x
6
4.
d.
Add up the individual distances to get the
total amount that Mia ran; 0.60 
0.75 
1.4
2.75 km. Convert this into a fraction by
adding the whole number, 2, to the fraction
1
7
0
5
0
2
2
5
5
3
4
. The answer is 2
3
4
km.
5.
c.
Since lines 
EF
and 
CD
are perpendicular, tri-
angles 
ILJ
and 
JMK
are right triangles.
Angles 
GIL
and 
JKD
are alternating angles,
since lines 
AB
and 
CD
are parallel and cut by
transversal 
GH
. Therefore, angles 
GIL
and
JKD
are congruent—they both measure 140
degrees. Angles 
JKD
and 
JKM
form a line. A
line has 180 degrees, so the measure of angle
JKM
180 – 140 
40 degrees. There are
also 180 degrees in a triangle. Right angle
JMK
, 90 degrees, angle 
JKM
, 40 degrees, and
angle 
x
form a triangle. Angle 
x
is equal to
180 – (90 
40) 
180 – 130 
50 degrees.
6.
c.
The area of a circle is equal to 
π
r
2
, where r is
the radius of the circle. If the radius,
r
, is
doubled (2
r
), the area of the circle increases
by a factor of four, from 
π
r
2
to 
π
(2
r
)
2
4
π
r
2
.
Multiply the area of the old circle by four to
find the new area of the circle:
6.25
π
in
2
4 
25
π
in
2
.
7.
a.
The distance formula is equal to 
((
x
2
– 
x
1
)
2
(
y
2
– 
y
1
)
2
)
. Substituting the
endpoints (–4,1) and (1,13), we find that
((–4 –
1)
2
(
1 – 13)
2
)
((–5)
2
(–12
)
2
)
25 
1
44
169
13, the length of David’s line.
8.
b.
A term with a negative exponent in the
numerator of a fraction can be rewritten
with a positive exponent in the denominator,
and a term with a negative exponent in the
denominator of a fraction can be rewritten
with a positive exponent in the numerator.
(
a
b
–
–
3
2
) 
(
a
b
3
2
). When (
a
b
3
2
) is multiplied by (
a
b
3
2
),
the numerators and denominators cancel
each other out and you are left with the frac-
tion 
1
1
, or 1.
9
.
e.
Since triangle 
ABC
is equilateral, every angle
in the triangle measures 60 degrees. Angles
ACB
and 
DCE
are vertical angles. Vertical
angles are congruent, so angle 
DCE
also
measures 60 degrees. Angle 
D
is a right
angle, so 
CDE
is a right triangle. Given the
measure of a side adjacent to angle 
DCE
, use
the cosine of 60 degrees to find the length of
side 
CE
. The cosine is equal to 
(
(
a
h
d
y
j
p
ac
o
e
t
n
en
t
u
si
s
d
e
e
)
)
,
and the cosine of 60 degrees is equal to 
1
2
;
1
x
2
1
2
, so 
x
24.
10.
d.
First, find 25% of
y
; 16 
0.25 
4. 10% of
x
is equal to 4. Therefore, 0.1
x
4. Divide
both sides by 0.1 to find that 
x
40.
11.
e.
The area of a triangle is equal to (
1
2
)
bh
, where
b
is the base of the triangle and 
h
is the height
of the triangle. The area of triangle 
BDC
is 48
square units and its height is 8 units.
48 
1
2
b
(8) 
48 
4
b
b
12
The base of the triangle,
BC
, is 12. Side 
BC
is
equal to side 
AD
, the diameter of the circle.
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M AT H P R E T E S T
–
2 3

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The radius of the circle is equal to 6, half its
diameter. The area of a circle is equal to 
π
r
2
,
so the area of the circle is equal to 36
π
square
units.
12.
d.
The sides of a square and the diagonal of a
square form an isosceles right triangle. The
length of the diagonal is 
2
times the
length of a side. The diagonal of the square
is 16 
2
cm, therefore, one side of the
square measures 16 cm. The area of a square
is equal to the length of one side squared:
(16 cm)
2
256 cm
2
.
13.
a.
If both sides of the inequality 
m
2
> 
n
2
are mul-
tiplied by 2, the result is the original inequal-
ity,
m
> 
n
.
m
2
is not greater than 
n
2
when 
m
is
a positive number such as 1 and 
n
is a nega-
tive number such as –2.
mn
is not greater than
zero when 
m
is positive and 
n
is negative. The
absolute value of
m
is not greater than the
absolute value of
n
when 
m
is 1 and 
n
is –2.
The product 
mn
is not greater than the prod-
uct –
mn
when 
m
is positive and 
n
is negative.
14.
c.
There are 60 minutes in an hour and 120
minutes in two hours. If 4 liters are poured
every 3 minutes, then 4 liters are poured 40
times (120 
3); 40 
4 
160. The tank,
which holds 2,000 liters of water, is filled with
160 liters;
2
1
,0
6
0
0
0
1
8
00
. 8% of the tank is full.
15.
d.
The curved portion of the shape is 
1
4
π
d
,
which is 4
π
. The linear portions are both the
radius, so the solution is simply 4
π
16.
16.
4
Multiply the number of appetizers, entrées,
and desserts offered at each restaurant. The
Scarlet Inn offers (2)(5)(4) 
40 meal com-
binations, and the Montgomery Garden
offers (3)(4)(3) 
36 meal combinations.
The Scarlet Inn offers four more meal
combinations.
17.
35
Angles 
OBC
and 
OCB
are congruent, so both
are equal to 55 degrees. The third angle in the
triangle, angle 
O
, is equal to 180 – (55 
55)
180 – 110 
70 degrees. Angle 
O
is a cen-
tral angle; therefore, arc 
BC
is also equal to 70
degrees. Angle 
A
is an inscribed angle. The
measure of an inscribed angle is equal to half
the measure of its intercepted arc. The meas-
ure of angle 
A
70 
2 
35 degrees.
18.
4
The function 
f
(
a
) 
(4
a
2
(
a
2
–
12
1
a
6
)
9)
is undefined
when its denominator is equal to zero;
a
2
– 16
is equal to zero when 
a
4 and when 
a
–4.
The only positive value for which the func-
tion is undefined is 4.
19.
46
Over 12 hours, Kiki climbs (1,452 – 900) 
552 feet. On average, Kiki climbs (552 
12)
46 feet per hour.
20.
55
The total weight of the first three dogs is
equal to 75 
3 
225 pounds. The weight of
the fourth dog,
d
, plus 225, divided by 4, is
equal to the average weight of the four dogs,
70 pounds:
d
4
225
70
d
225 
280
d
55 pounds
21.
260
The weight Kerry can lift now, 312 pounds, is
20% more, or 1.2 times more, than the
weight,
w
, he could lift in January:
1.2
w
312
w
260 pounds
22.
485
2,200(0.07) equals $154; 1,400(0.04) equals
$56; 3,100(0.08) equals $248; 900(0.03)
equals $27. Therefore, $154 
$56 
$248 
$27 
$485.
23.
19
Let 
x
,
x
1, and 
x
2 represent the consec-
utive integers. The sum of these integers is 60:
x
x
1 
x
2 
60, 3
x
3 
60, 3
x
57,
x
19. The integers are 19, 20, and 21, the
smallest of which is 19.