Sat 2015 Practice Test #1 Answer Explanations

Choice B is correct.
Since the right-hand side of the equation is 
_ 
U
_ 
U
N
1 +
( 
) ( 
)
3
times the expression 
N 
, multiplying both 
_ 
U
(
1 + 
) 
ơ
sides of the equation by the reciprocal of this expression results 
_ 
U
1 + 
ơ
( 
)
N 
in 
N 
P
= 
3
. 
_ 
U
_ 
U
1 +
( 
) ( 
) 
Choice A is incorrect and is the result of multiplying both sides of the 
_ 
U
_ 
U
N
1 +
( 
) ( 
)
equation by the rational expression 
N
rather than 
_ 
U
(
1 + 
) 
ơ
_ 
U
(
1 + 
)
N
ơ
by the reciprocal of this expression
N
. Choices C 
_ 
U
_ 
U
1 +
( 
) ( 
) 
and D are incorrect and are likely the result of errors while trying to 
solve for 
3
. 
QUESTION 8 
_ 
_ 
_
1
Choice C is correct.
Since 
D
E
= 2, it follows that 
E
D
= 2. Multiplying both
_ 
E
_
1 
_
4
E
sides of the equation by 4 gives 4
= 4 
, or 
= 2.
(
D
) 
(
2
) 
D
Choice A is incorrect because if 4 
D
E
WKHQ
D
E
ZRXOGEHXQGHƮQHG
Choice B is incorrect because if 4 
D
E
= 1, then 
D
E
= 4. Choice D is
incorrect because if 4 
D
E
= 4, then 
D
E
= 1.
QUESTION 9 
Choice B is correct.
Adding 
[
and 19 to both sides of 2
y
ơ[ ơ
gives 
[
= 2
y
+ 19. Then, substituting 2
y
+ 19 for 
[LQ[
+ 4
y
ơJLYHV
y
+ 19) + 4
y
ơ7KLVODVWHTXDWLRQLVHTXLYDOHQWWR
y
ơ
6ROYLQJ
y
ơJLYHV
y
ơ)LQDOO\VXEVWLWXWLQJơIRU
y
in 
2
y
ơ[ ơJLYHVơơ[ ơRU[ 7KHUHIRUHWKHVROXWLRQ
[
, 
y
) 
WRWKHJLYHQV\VWHPRIHTXDWLRQVLVơ
Choices A, C, and D are incorrect because when the given values of 
[
and 
y
are substituted in 2
y
ơ[ ơWKHYDOXHRIWKHOHIWVLGHRIWKH
HTXDWLRQGRHVQRWHTXDOơ
QUESTION 10 
Choice A is correct.
Since 
g
is an even function, 
g 
(
ơ 
g
(4) = 8. 
Alternatively: First find the value of 
D
, and then find 
g 
ơ
Since 
g
(4) = 8, substituting 4 for 
[
and 8 for 
g
(
[
) gives 
8 = 
D
(4)
2
+ 24 = 16
D
+ 24. Solving this last equation gives 
D ơ
Thus 
g
(
[
)
ơ[
2
+ 24, from which it follows that 
g 
(
ơ ơơ
2
+ 24; 
g 
(
ơ ơDQG
g 
(
ơ

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__ 
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_
Choices B, C, and D are incorrect because 
g
is a function and there can 
only be one value of 
(
g 
ơ
QUESTION 11 
Choice D is correct.
To determine the price per pound of beef 
when it was equal to the price per pound of chicken, determine 
the value of 
[
(the number of weeks after July 1) when the two 
prices were equal. The prices were equal when 
E
= 
F
; that is, when 
[
[
. This last equation is equivalent to 
[
, and so 
[
= 
= 4. Then to determine 
E
, the price per
pound of beef, substitute 4 for 
[
in 
E [
, which gives 
E GROODUVSHUSRXQG
Choice A is incorrect. It results from substituting the value 1, not 4, for 
[
in 
E [
. Choice B is incorrect. It results from substituting 
the value 2, not 4, for 
[
in 
E [
. Choice C is incorrect. It 
UHVXOWVIURPVXEVWLWXWLQJWKHYDOXHQRWIRU[
in 
F [
. 
QUESTION 12 
Choice D is correct.
In the 
[\
-plane, all lines that pass through 
the origin are of the form 
y
= 
P[
, where 
P
is the slope of the line. 
_
1
Therefore, the equation of this line is 
y
= 
[
, or 
[
= 7
y
. A point
7 
with coordinates (
D
, 
E
) will lie on the line if and only if 
D
= 7
E2IWKH
JLYHQFKRLFHVRQO\FKRLFH'VDWLVƮHVWKLVFRQGLWLRQ
&KRLFH$LVLQFRUUHFWEHFDXVHWKHOLQHGHWHUPLQHGE\WKHRULJLQ
DQGLVWKHYHUWLFDOOLQHZLWKHTXDWLRQ[ WKDWLVWKH
y
-axis. 
_
The slope of the 
y
D[LVLVXQGHƮQHGQRW
1
7
7KHUHIRUHWKHSRLQW
_
does not lie on the line that passes the origin and has slope 17.
Choices B and C are incorrect because neither of the ordered pairs has 
_
a 
y
-coordinate that is 17 the value of the corresponding 
[
-coordinate.
QUESTION 13 
1
Choice B is correct.
To rewrite
—
, multiply 
[
1
+ 2 + 
1
—
—
[
(
[
+ 2)(
[
(
[
+ 2)(
[
by
—
. This results in the expression 
, which
(
[
+ 2)(
[
(
[
+ 2) + (
[
is equivalent to the expression in choice B. 
Choices A, C, and D are incorrect and could be the result of common 
algebraic errors that arise while manipulating a complex fraction. 
48(67,21
_
Choice A is correct.
2QHDSSURDFKLVWRH[SUHVV
8
[
so that the
2
y 
numerator and denominator are expressed with the same base. Since 
2 and 8 are both powers of 2, substituting 2
for 8 in the numerator 

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_ 
_ 
_ 
_ 
_ 
(2
)
[
of 8
[
gives 
, which can be rewritten as 2
[
. Since the numerator
2
y 
2
y 
2
y 
_
and denominator of 2
[
have a common base, this expression can be
2
y 
rewritten as 2
[ơ\
,WLVJLYHQWKDW[ơ
y
= 12, so one can substitute 12 
_
for 
WKHH[SRQHQW[ơ
y
, given that the expression 8
[
is equal to 2
12
.
2
y 
Choice B is incorrect. The expression 8
[
can be rewritten as 22 
y 
[
, or
2
y 
2
[ơ\
. If the value of 2
[ơ\
is 4
4
, which can be rewritten as 28, then 
2
[ơ\
= 2
8 
, 
ZKLFKUHVXOWVLQ[ơ
y
= 8, not 12. Choice C is incorrect. If 
_
the value of 8
[
is 8
2
, then 2
[ơ
 y
= 8
2
ZKLFKUHVXOWVLQ[ơ
y
= 6, not 12.
2
y 
_
Choice D is incorrect because the value of 82 
[
y
can be determined. 
QUESTION 15 
Choice D is correct.
2QHFDQƮQGWKHSRVVLEOHYDOXHVRID
and 
E
in (
D[
+ 2) (
E[
+ 7) by using the given equation 
D
+ 
E
= 8 and 
ƮQGLQJDQRWKHUHTXDWLRQWKDWUHODWHVWKHYDULDEOHVD
and 
E
. Since 
(
D[
+ 2)(
E[
+ 7) = 15
[
2
+ 
F[
+ 14, one can expand the left side of the 
equation to obtain 
DE[
2
+ 7
D[
+ 2
E[
+ 14 = 15
[
2
+ 
F[
+ 14. Since 
DE
is the 
FRHƱFLHQWRI[
2
RQWKHOHIWVLGHRIWKHHTXDWLRQDQGLVWKHFRHƱFLHQW
of 
[
2
on the right side of the equation, it must be true that 
DE
= 15. Since 
D
+ 
E
= 8, it follows that 
E ơD
. Thus, 
DE
= 15 can be rewritten as 
D
ơD
) = 15, which in turn can be rewritten as 
D
2
ơD )DFWRULQJ
gives (
Dơ
Dơ 7KXVHLWKHUD DQGE
= 5, or 
D
= 5 and 
E ,I
D DQGE
= 5, then (
D[
+ 2)(
E[ [
+ 2)(5
[
+ 7) = 15
[
2
[
+ 14. 
Thus, one of the possible values of 
FLV,ID
= 5 and 
E WKHQ
(
D[
+ 2)(
E[
+ 7) = (5
[[
+ 7) = 15
[
2
+ 41
[
+ 14. Thus, another possible 
value for 
F
is 41. Therefore, the two possible values for 
FDUHDQG
&KRLFH$LVLQFRUUHFWWKHQXPEHUVDQGDUHSRVVLEOHYDOXHVIRU
D
and 
E
, but not possible values for 
F
. Choice B is incorrect; if 
D
= 5 
and 
E WKHQDQGDUHWKHFRHƱFLHQWVRI[
when the expression 
(5
[[
+ 7) is expanded as 15
[
2
[
+ 6
[
+ 14. However, when 
WKHFRHƱFLHQWVRI[DUHDQGWKHYDOXHRIF
is 41 and not 6 and 
&KRLFH&LVLQFRUUHFWLID DQGE WKHQDQGDUHWKH
FRHƱFLHQWVRI[ZKHQWKHH[SUHVVLRQ[
+ 2)(5
[
+ 7) is expanded as 
15
[
2
+ 21
[[+RZHYHUZKHQWKHFRHƱFLHQWVRI[DUHDQG
21, the value of 
FLVDQGQRWDQG
QUESTION 16 
The correct answer is 2.
To solve for 
t
, factor the left side of 
t 
2
ơ
giving (
t
ơ
t
7KHUHIRUHHLWKHU
t
ơ RU
t
,I
t
ơ WKHQ
t
= 2, and if 
t
WKHQ
t
ơ6LQFHLWLVJLYHQWKDW
t
!WKHYDOXHRI
t
must be 2. 
Another way to solve for 
t
is to add 4 to both sides of 
t 
2
ơ JLYLQJ
t
2
= 4. 
Then, taking the square root of the left and the right side of the equation 
_ 
gives 
t
= ±
Ƥ
4 = ± 2. Since it is given that 
t
!WKHYDOXHRI
t
must be 2. 

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̝_̝
6$73UDFWLFH7HVW
QUESTION 17 
The correct answer is 1600.
It is given that 
∠
$(%
and 
∠
&'%
have 
the same measure. Since 
∠
$%(
and 
∠
&%'
are vertical angles, they 
have the same measure. Therefore, triangle 
($%
is similar to triangle 
'&%
because the triangles have two pairs of congruent corresponding 
angles (angle-angle criterion for similarity of triangles). Since 
the triangles are similar, the corresponding sides are in the same 
_
_ 
%'
proportion; thus 
CD 
= 
(%
6XEVWLWXWLQJWKHJLYHQYDOXHVRIIRU
CD
,
[
_
_ 
%'
_
_ 
for 
%'DQGIRU(%
in 
CD 
= 
(%
gives 
 = 
. Therefore,
[
[
(
[
= 
—
QUESTION 18