Sat - MC Grawhill part 78 potx

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760 MCGRAW-HILL’S SAT
3. C There are 180° in a triangle. Set up equations
for the two triangles in the ﬁgure.
a + b + 52 = 180
Subtract 52: a + b = 128
c + d + 52 = 180
Subtract 52: c + d = 128
a + b + c + d =
Substitute: 128 + 128 = 256
(Chapter 10, Lesson 2: Triangles)
4. B f(x) = x
2
− 4
Set f(x) equal to 32: x
2
− 4 = 32
Add 4: x
2
= 36
Take positive square root: x = 6
(Chapter 11, Lesson 2: Functions)
5. C The ratio of the nuts is a part-to-part-to-part-
to-part ratio. Adding these numbers gives the total
number of parts: 2 + 4 + 5 + 7 = 18. Since four of these
parts are almonds, the fraction of the mixture that is
almonds is 4/18, or 2/9.
(Chapter 7, Lesson 4: Ratios and Proportions)
6. D If 20 students scored an average of 75 points,
then the sum of their scores is 20 × 75 = 1,500 total
points.
If 12 of those students scored an average of 83 points,

then the sum of their scores is 12 × 83 = 996 points.
Therefore, the remaining 8 students scored 1,500 −
996 = 504 points altogether, so their average score is
504 ÷ 8 = 63 points.
(Chapter 9, Lesson 2: Mean/Median/Mode Problems)
7. A The sides of square EFGH all have length
. A diagonal of this square can be found with the
Pythagorean theorem: (8 )
2
+ (8 )
2
= EG
—
2
.
Simplify: 128 + 128 = EG
—
2
256 = EG
—
2
Take square root: 16 = EG
—
(Or, more simply, you can remember that the length of
the diagonal of a 45°-45°-90° triangle is the length of
the side times . So the diagonal is .)
By the same reasoning, since the sides of square
ABCD all have length 14 : AD
—
= 14 ×=28.

2
2
2
82 2 16×=
2
2
2
82
Notice that AC
—
= AE
—
+ EG
—
+ CG
—
; therefore,
28 = AE
—
+ 16 + CG
—
, so AE
—
+ CG
—
= 12. By the same rea-
soning, BF
—
+ DH
—

= 12, so AE
—
+ BF
—
+ CG
—
+ DH
—
= 24.
(Chapter 10, Lesson 3: The Pythagorean Theorem)
8. D Although you were probably taught to add the
“rightmost” digits ﬁrst, here the “leftmost” digits pro-
vide more information about the number, so it’s best
to start there.
RS
+
SR
TR4
The largest possible 3-digit number that can be
formed by adding two 2-digit numbers is 99 + 99 =
198. Therefore, T must be 1.
RS
+
SR
1R4
Therefore, there must be a “carry” of 1 from the ad-
dition of R + S in the 10s column. Looking at the units
column tells us that S + R yields a units digit of 4, so
S + R = 14. The addition in the 10s column tells us that
R + S + 1 = R + 10. (The “+10” is needed for the carry

into the 100s column.)
R + S + 1 = R + 10
Substitute R + S = 14: 14 + 1 = R + 10
Subtract 10: 5 = R
So 2R + T = 2(5) + 1 = 11.
(Chapter 9, Lesson 3: Numerical Reasoning Problems)
9. 1 n =
If it helps, you can think of this as f(n) =
.
Find the value of (f(4))
2
Plug in 4 for n: f(4) =
Plug in 1 for f(4): ((f(4))
2
= (1)
2
= 1
(Chapter 9, Lesson 1: New Symbol or Term Problems)
4
16
16
16
1
2
==
n
2
16
n
2

16
14
DC
AB
EF
GH
2
66
66
8
8
8
8
14 2
14 214 2 82 82
82
82
14. 3 Draw a line with points P, Q, R, and S on the
line in that order. You are given that PS
—
= 2PR
—
and
that PS
—
= 4PQ
—
, so choose values for those lengths, like
PS
—

= 12, PR
—
= 6, and PQ
—
= 3.
This means that QS
—
= 9, so QS
—
/PQ
—
= 9/3 = 3.
(Chapter 6, Lesson 2: Analyzing Problems)
CHAPTER 16 / PRACTICE TEST 3 761
10. 750 25% of $600 is $150. Therefore, the club
earned $150 more in 2007 than it did in 2006, or $600 +
$150 = $750. Remember, also, that increasing any
quantity by 25% is the same as multiplying that quan-
tity by 1.25.
(Chapter 7, Lesson 5: Percents)
11. 3 Set up equations: x + y = 4
x − y = 2
Add straight down: 2x = 6
Divide by 2: x = 3
Plug in 3 for x:3 + y = 4
Subtract 3: y = 1
Final product: (x)(y) = (3)(1) = 3
(Chapter 8, Lesson 2: Systems)
12. 32 Let LM = x, and let LO = y. Since x is twice the
length of y, x = 2y.

x + x + y + y = P
Substitute for x: 2y + 2y + y + y = P
Combine terms: 6y = P
Plug in 48 for P: 6y = 48
Divide by 6: y = 8
Solve for x: x = 2y = 2(8) = 16
To ﬁnd the area of the shaded region, you might notice
that if PM is the base of the shaded triangle, then LO is
the height, so area =
1
⁄
2
(base)(height) =
1
⁄
2
(8)(8) = 32.
If you don’t notice this, you can ﬁnd the shaded area
by ﬁnding the area of the rectangle and subtracting
the areas of the two unshaded triangles.
Area of rectangle = (length)(width)
Area of rectangle = (x)(y) = (16)(8) = 128
Area of triangle PLO =
1
⁄
2
(base)(height)
Area of triangle PLO =
1
⁄

2
(8)(8) = 32
Area of triangle MNO =
1
⁄
2
(base)(height)
Area of triangle MNO =
1
⁄
2
(16)(8) = 64
Area of triangle OPM = 128 − 64 − 32 = 32
(Chapter 10, Lesson 5: Areas and Perimeters)
13. 9 64
3
= 4
x
Substitute 4
3
for 64: (4
3
)
3
= 4
x
Simplify: 4
9
= 4
x

Equate the exponents: x = 9
(Chapter 8, Lesson 3: Working with Exponentials)
LP
16
8
88
8 M
ON
33 6
PQR S
y
x
(4, 6)
(–1, 6)
(m, n)
O
15. 1.5 Since the graph is a parabola, it has a vertical
axis of symmetry through the vertex. The points (−1, 6)
and (4, 6) have the same y-coordinate, so each one is the
reﬂection of the other over the axis of symmetry. This
axis, therefore, must be halfway between the two
points. Since the average of −1 and 4 is (−1 + 4)/2 = 1.5,
the axis of symmetry must be the line x = 1.5, and there-
fore m = 1.5.
(Chapter 11, Lesson 2: Functions)
16. 18 Since these numbers are “evenly spaced,”
their mean (average) is equal to their median (middle
number). The average is easy to calculate: 110/5 = 22.
Therefore, the middle number is 22, so the numbers
are 18, 20, 22, 24, and 26.

Alternatively, you can set up an equation to ﬁnd the
sum of ﬁve consecutive unknown even integers,
where x is the least of these:
x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 110
Combine like terms: 5x + 20 = 110
Subtract 20: 5x = 90
Divide by 5: x = 18
So the ﬁve integers are 18, 20, 22, 24, and 26.
(Chapter 9, Lesson 2: Mean/Median/Mode Problems)
17. 20 Use the percent change formula:
(Chapter 7, Lesson 5: Percents)
24 000 20 000
20 000
100 20
,,
,
%%
−
×
()
=
Final Original
Original
−
×
()
100%
762 MCGRAW-HILL’S SAT
18. 25 Let b = the number of black marbles, w = the
number of white marbles, and r = the number of red

marbles in the jar. If you are four times as likely to
choose a black marble as a white one, then b = 4w.
If you are ﬁve times as likely to choose a red marble
as a black one, then r = 5b. To ﬁnd the least possible
number of marbles in the jar, imagine you have only
one white marble. This would mean you have 4(1) = 4
black marbles and 5(4) = 20 red marbles, for a total of
1 + 4 + 20 = 25 marbles.
In general, you can represent the total number of
marbles as total = b + w + r
Since r = 5b: total = b + w + 5b
Since b = 4w: total = 4w + w + 5(4w)
Simplify: total = 4w + w + 20w
Simplify: total = 25w
In other words, the number of marbles in the jar must
be a multiple of 25. The smallest positive multiple of
25 is, of course, 25.
(Chapter 9, Lesson 6: Probability Problems)
Section 6
1. A If the ﬁght did not ensue, John must have in-
tervened to stop it. intervene = get in the way of some-
thing; coalesce = fuse together; intermingle = mix
together; exacerbate = make worse
2. D The defendant hoped the testimony would
corroborate (support) his alibi, which would clear him
of blame. convoke = call together; synthesize = gener-
ate; absolve = free of blame; impeach = accuse
3. E Being ensnarled (tied up) in trafﬁc is an un-
pleasant experience that Rachel would have an aver-
sion to or dislike for. antipathy = feeling against;

penchant = liking; predilection = liking; proclivity = ten-
dency to do something; aversion = feeling of dislike;
insufferable = intolerable
4. A If the practices are no longer considered state
of the art, they must now be considered outdated or
unsophisticated. The physicians are incredulous (not
able to believe) that such barbaric acts were once sup-
ported or condoned. primitive = old, unsophisticated;
sanctioned = approved; ingenious = incredible, bril-
liant; boorish = rude, censured = publicly condemned;
innovative = new; endorsed = supported; foolhardy =
recklessly bold; condemned = criticized
5. B The Prime Minister had vetoed the law in the
past many times, so he didn’t want it to pass. What
would come as a great surprise? The Prime Minister’s
suddenly supporting the law. articulated = expressed
clearly; championed = defended; denounced = spoke
out against; initiated = began; abbreviated = shortened
6. C Lines 3–4 state that the tradition is that a man
never lifts his hand against a woman. Furthermore, if
a man offends a woman, she is entitled to give him a
sound thrashing (line 6). Therefore, a man who dis-
respected a woman would face censure.
7. E Saying that it is not an unusual thing for a
squaw to administer a sound thrashing to a warrior
husband (lines 5–7) is like saying that it is not unusual
for her to give him a beating, or dispense it.
8. C Lines 5–6 say that merely receiving palliative
care . . . provides no hope of a cure. Therefore, pallia-
tive care only reduces the discomfort of the symptoms,

without curing the disease, as something analgesic
does.
9. A Lines 8–11 ask, How can a doctor know if a pa-
tient has the mental capacity to decide for herself that
the time has come to stop ﬁghting the disease? This
question indicates that there may be some difﬁculty
in determining a patient’s state of mind.
10. B The ﬁrst sentence of the passage says there
was great optimism about earthquake prediction. Each
paragraph discusses potential precursors, or predic-
tors, of earthquakes.
11. E Lines 8–10 say that because foreshocks look
just like any other earthquake, they are not in them-
selves very useful in prediction.
12. D Support for choice II can be found in line 19,
which says that groundwater has become cloudy prior
to an earthquake. Choice III is supported in lines
16–18, which say that before a large earthquake,
marked changes have been reported in the level or ﬂow
of wells and springs. Nothing is said about density
changes in the groundwater.
13. A The passage says (lines 8–10) that since fore-
shocks look just like any other earthquakes, they are not
in themselves very useful in prediction but later (lines
39–42) mentions that because the Haicheng earth-
quake had hundreds of foreshocks, it was easier than
average . . . to predict, thereby suggesting that fore-
shocks are, in fact, useful in predicting earthquakes.
14. A This paragraph describes a particular appli-
cation of the theory of earthquake prediction, de-

scribed in the previous paragraphs, which led to
scientists’ predicting a large earthquake and saving
many lives. Although this is said to have prov[ed] that
. . . earthquake prediction is possible (lines 38–39), it
was not a scientiﬁc experiment, as there was no con-
trol group.
CHAPTER 16 / PRACTICE TEST 3 763
15. C Lines 49–50 mention that seismologists
missed predicting the Tangshan earthquake and that
over 250,000 people died. This was far worse than the
Haicheng earthquake, which was successfully pre-
dicted, so that many lives were saved.
16. D The word “evacuation” in line 46 is placed in
quotations to indicate that it is not being used in the
traditional sense. The task of evacuating a population
from a natural disaster does not typically involve
showing movies, so doing so is unconventional.
17. C Lines 7–8 say that one of the missionaries who
met the ship took us under his wing.
18. E Saying that he could hardly believe that we
were really restored to him is like saying he couldn’t be-
lieve that we were returned to him.
19. B The narrator states that she could use tools as
well as [her] brothers did (lines 20–21), that her ﬁrst
childhood friendship was with a male ship-builder
next door, and that she was eager and able to work
with the ship-builders around her. Thus, she conveys
a clear sense that she considers herself the equal of
the males in her life.
20. D The author was emancipated from her con-

ﬁning clothing so that she could work with tools, such
as her hatchet, in the shipyard.
21. C The big movements of the day refer to the
changes in culture and civilization (line 43).
22. A Choice II is supported by lines 38–40, which
say that we had around us the ﬁne ﬂower of New Eng-
land civilization, as opposed to Michigan, which the
author characterizes as the wilderness (line 45). The
passage does not suggest that New England had ﬁner
gardens or humbler citizens than Michigan had.
23. D The author describes the move to Michigan as
a complete upheaval (lines 37–38), and an unwelcome
move from the ﬁne ﬂower of New England civilization
(lines 39–40), thereby suggesting that she resents the
move. She conveys no sign of bewilderment, fear, or
awe in this passage, since she describes the move with
insight and equanimity.
24. A The passage says that the sisters were so
pained by (the lumber wagon’s) appearance that we re-
fused to ride in it (lines 55–56) and that they wanted
to look as if we had no association with it (lines 57–58).
Section 7
1. D 4x + 5 = 20
Add 3: 4x + 8 = 23
(Chapter 8, Lesson 1: Solving Equations)
2. C First ﬁnd out how many cups are in 3 pints.
Set up a ratio:
Cross-multiply: x = 6 cups
Set up a ratio to solve for servings:
Cross-multiply:

1
⁄
3
x = 6
Divide by
1
⁄
3
: x = 18
(Chapter 7, Lesson 4: Ratios and Proportions)
3. A Since the angle shown is a right angle, the arc
represents
1
⁄
4
of the circumference.
length of arc =
1
⁄
4
(2πr)
Substitute 4 for r: length of arc =
1
⁄
4
(2π(4))
Simplify: length of arc = 2π
(Chapter 10, Lesson 8: Circles)
4. C This question tests your understanding of 30°-
60°-90° triangles. The hypotenuse, which corresponds

to 2x, is 14. This means that the base is x = 7. The
height is therefore x = 7.
(Chapter 10, Lesson 5: Areas and Perimeters)
(Chapter 10, Lesson 3: The Pythagorean Theorem)
5. A Given that ∇x = 3x − 3, ﬁnd ∇7.
∇7 = 3x − 3
Plug in 7 for x: 3(7) − 3 = 18
Find ∇3: ∇3 = 3x − 3
Plug in 3 for x: 3(3) − 3 = 6
Be careful not to pick answer choice (B) ∇3, because
∇3 = 3(3) − 3 = 6, not 3. Answer choice (A) ∇2 is cor-
rect, because ∇2 = 3(2) − 3 = 3.
(Chapter 9, Lesson 1: New Symbol or Term Problems)
∇
∇
==
7
3
18
6
3
33
1 serving
cups
servings
6cups
1
3
=
x

1pint
2cu
p
s
3pints
cu
p
s
=
x
x
3 7
7
3
2x 14
x
30°30°
60° 60°
764 MCGRAW-HILL’S SAT
6. B A little common sense should tell you that they
will not need a full hour to clean the pool, because
Stephanie can clean it in an hour all by herself, but
Mark is helping. Therefore, you should eliminate
choices (C), (D), and (E) right away. You might also
notice that it can’t take less than 30 minutes, because
that is how long it would take if they both cleaned
one pool per hour (so that the two working together
could clean it in half the time), but Mark is slower, so
they can’t clean it quite that fast. This eliminates
choice (A) and leaves (B) as the only possibility.

But you should know how to solve this problem if it
were not a multiple-choice question, as well:
Stephanie’s rate for cleaning the pool is one pool per
hour. Mark’s rate for cleaning the pool is one pool ÷
1.5 hours =
2
⁄
3
pools per hour. Combined, they can
clean 1 +
2
⁄
3
=
5
⁄
3
pools per hour. Set up a rate equation
using this rate to determine how much time it would
take to clean one pool:
1 pool = (
5
⁄
3
pools per hour)(time)
Divide by
5
⁄
3
:

3
⁄
5
hours to clean the pool
Multiply by 60:
3
⁄
5
(60) = 36 minutes
(Chapter 9, Lesson 4: Rate Problems)
7. A Change each expression to a base-10 exponential:
(A) = ((10
2
)
3
)
4
= 10
24
(B) = ((10
2
)
5
)((10
2
)
6
) = (10
10
)(10

12
) = 10
22
(C) = ((10
4
)
4
) = 10
16
(D) = (((10
2
)
2
)((10
2
)
2
))
2
= ((10
4
)(10
4
))
2
= (10
8
)
2
= 10

16
(E) = (10
6
)
3
= 10
18
(Chapter 8, Lesson 3: Working with Exponentials)
8. B Consider the points (0, 2) and (3, 0) on line l.
When these points are reﬂected over the x-axis, (0, 2)
transforms to (0, −2) and (3, 0) stays at (3, 0) because
it is on the x-axis. You can then use the slope formula
to ﬁnd the slope of line m:
It’s helpful to notice that whenever a line is reﬂected
over the x-axis (or the y-axis, for that matter—try it),
its slope becomes the opposite of the original slope.
(Chapter 10, Lesson 4: Coordinate Geometry)
9. C (a + b)
2
= (a + b)(a + b)
FOIL: a
2
+ ab + ab + b
2
Combine like terms: a
2
+ 2ab + b
2
Plug in 5 for ab: a
2

+ 2(5) + b
2
Simplify: a
2
+ b
2
+ 10
Plug in 4 for a
2
+ b
2
:4 + 10 = 14
(Chapter 8, Lesson 5: Factoring)
yy
xx
21
21
02
30
2
3
−
−
=
−−
()
−
=
10. D The total area of the patio to be constructed is
24 × 12 = 288 ft

2
. The slab shown in the ﬁgure has an
area of 8 ft
2
. Therefore, to ﬁll the patio you will need
288 ÷ 8 = 36 slabs.
(Chapter 10, Lesson 5: Areas and Perimeters)
11. D The prize money ratio can also be written as
7x:2x:1x. Because the total prize money is $12,000,
7x + 2x + 1x = 12,000
Combine like terms: 10x = 12,000
Divide by 10: x = 1,200
The ﬁrst place prize is 7x = 7(1,200) = $8,400.
(Chapter 7, Lesson 4: Ratios and Proportions)
12. E Always read the problem carefully and notice
what it’s asking for. Don’t assume that you must solve
for x and y here. Finding the value of 6x − 2y is much
simpler than solving the entire system:
2x + 3y = 7
4x − 5y = 12
Add straight down: 6x − 2y = 19
(Chapter 8, Lesson 2: Systems)
13. D Think carefully about the given information
and what it implies, then try to ﬁnd counterexamples
to disprove the given statements. For instance, try to
disprove statement I by showing that s can be even.
Imagine s = 2:
s + 1 = 2r
Substitute 6 for s: 6 + 1 = 2r
Combine like terms: 7 = 2r

Divide by 2: 3.5 = r (nope)
This doesn’t work because r must be an integer. Why
didn’t it work? Because 2r must be even, but if s is
even, then s + 1 must be odd and cannot equal an even
number, so s must always be odd and statement I is
true. (Eliminate choice (B).)
Statement II can be disproven with r = 1:
s + 1 = 2r
Substitute 1 for r: s + 1 = 2(1)
Subtract 1: s = 1 (okay)
Since 1 is an integer, we’ve proven that r is not nec-
essarily even, so II is false. (Eliminate choices (C)
and (E).)
Since we still have two choices remaining, we have to
check ugly old statement III. Try the values we used
before. If r = 1 and s = 1, then , which
is an integer. But is it always an integer? Plugging in
more examples can’t prove that it will ALWAYS be an
integer, because we can never test all possible solu-
tions. We can prove it easily with algebra, though.
Since s + 1 = 2r:
Divide by r:
Distribute:
s
rr
+=
1
2
s
r

+
=
1
2
s
rr
+=+=
11
1
1
1
2
CHAPTER 16 / PRACTICE TEST 3 765
Since 2 is an integer, statement III is necessarily true.
(Chapter 9, Lesson 3: Numerical Reasoning Problems)
(Chapter 6, Lesson 7: Thinking Logically)
14. C Find all the possible products of the values on
two chips: (1)(2) = 2; (1)(3) = 3; (1)(4) = 4; (1)(5) = 5;
(1)(6) = 6; (2)(3) = 6; (2)(4) = 8; (2)(5) = 10; (2)(6) = 12;
(3)(4) = 12; (3)(5) = 15; (3)(6) = 18; (4)(5) = 20; (4)(6) =
24; (5)(6) = 30. There are 15 different combinations of
chips. Of these, only the last 2 yield products that are
greater than 20. So the probability is 2/15.
(Chapter 9, Lesson 6: Probability Problems)
15. D In this problem, only the signs of the terms
matter. By following the rule of the sequence, you
should see that the ﬁrst six terms of the sequence are
+, −, −, +, −. −, . . . The pattern {+, −, −} repeats forever.
In the ﬁrst 100 terms, the pattern repeats 100 ÷ 3 =
33

1
⁄
3
times. Because each repetition contains two neg-
ative numbers, in 33 full repetitions there are 33 × 2 =
66 negative numbers. The 100th term is the ﬁrst term
of the next pattern, which is positive, so the total
number of negative terms is 66.
(Chapter 11, Lesson 1: Sequences)
16. B Draw the ﬁve triangles. The simplest way to
solve this problem is to compare the choices one pair
at a time. For instance, it should be clear just by in-
spection that RB > RA and SB > SA, so we can elimi-
nate A. Similarly, it should be clear that RB > RC and
SB > SC, so we can eliminate C. Likewise, since RB > RD
and SB > SD, we can eliminate D. Finally, we compare
B with E. Since RB and RE are each a diagonal of one
of the square faces, they must be equal. But SB is
clearly longer than SE, because SB is the hypotenuse
of triangle SEB, while SE is one of the legs.
(Chapter 10, Lesson 7: Volumes and 3-D Geometry)
(Chapter 6, Lesson 7: Thinking Logically)
Section 8
1. C If the review suggested that the décor of the
restaurant was insipid (tasteless), but that the cuisine
came close to compensating for it, the review must
have been part positive and part negative, that is, am-
bivalent. indefatigable = untiring; banal = lacking orig-
inality; ambivalent = characterized by conﬂicting
feelings; sublime = supreme, impressive; piquant =

spicy; tepid = lukewarm
2. C The sentence suggests that Dr. Thompson
should have characterized the results as unusual, but
didn’t. meticulous = concerned with detail; belligerent =
prone to fighting; anomalous = deviating from the
norm; convergent = coming together; warranted = ap-
propriate to the situation
3. B They would hope that bad news did not predict
further bad news. amalgam = a combination of diverse
elements; harbinge = omen; arbiter = judge; talisman =
an object with magical power
4. C To bring slaves out of bondage is to free or un-
fetter them. encumber = burden; forgo = relinquish
5. D A writer who can produce both decorative po-
etry and a keenly analytical mystery novel is a versatile
writer; that is, she is able to write in divergent styles.
ﬂamboyant = ornate; immutability = permanence, un-
changeability; austere = plain; ﬂorid = ornate; grand-
iloquent = characterized by pompous language
6. B The word because indicates that the sentence
shows a cause-and-effect relationship. There are sev-
eral ways to complete this sentence logically, but the
only one among the choices is (B), because multifari-
ous (widely varied) mechanisms would logically
“stymie” (impede) scientists who are trying to inves-
tigate them. efﬁcacious = capable of producing a de-
sired effect; bilked = cheated; conspicuous = obvious;
thwarted = prevented; hampered = hindered; lucid =
clear; proscribed = forbidden
7. B If the cultural assumption that there are many

alien civilizations . . . stems in no small way from . . .
the “Drake Equation,” then this equation has had
quite an inﬂuence on public opinion.
8. E The ﬁrst two paragraphs discuss how the
Drake Equation has led to the belief that there are
many alien civilizations in the universe. The third
paragraph discusses the author’s contrasting view
that there is indeed probably much simple life in the
universe but very little if any other complex life.
9. B The sentence states that a planet could go from
an abiotic state to a civilization in 100 million years
thereby implying that a civilization must, by deﬁni-
tion, not be abiotic. Choice (B) is the only choice that
necessarily cannot apply to a civilization.
10. A The author states his thesis in lines 38–39: per-
haps life is common, but complex life is not, and goes
on to explain this thesis, stating in lines 61–67 that re-
search shows that while attaining the stage of animal
life is one thing, maintaining that level is quite another.
. . . Complex life is subject to an unending succession
of planetary disasters, creating what are known as
mass-extinction events.
11. A The phrase the evolutionary grade we call
animals refers to the level of life form produced by
evolution.
766 MCGRAW-HILL’S SAT
12. C Statement (A) is supported in lines 48–50,
statement (B) is supported in lines 74–76, statement
(D) is supported in lines 38–39, and statement (E) is
supported in lines 51–55.

13. C The sample size of one refers to the uniqueness
of Earth history (line 78).
14. A The ﬁrst quotation in lines 101–103 is de-
scribed as a rejoinder, or an opposing response, to the
author’s thoughts. The author then responds with his
own quotation.
15. C The author says that he does not conclude that
there are no other cats (Rare Cat Hypothesis), only that
there are no other cats exactly like Wookie in order to
convey the idea that one should not draw conclusions
based on one occurrence.
16. B The author says that life is opportunistic to
summarize the next statement that the biosphere has
taken advantage of the myriad of strange idiosyncrasies
that our planet has to offer.
17. D The passage says that these creatures might
naively assume that these qualities, very different from
Earth’s, are the only ones that can breed complexity,
that is, that all life evolved the same way.
18. A The author of Passage 1 believes that complex
life, once evolved, faces numerous dangers that push
it toward extinction. The author would point this fact
out in response to the statement in lines 134–135 of
Passage 2.
19. D The author of Passage 1 says in line 26, In my
view, life in the form of microbes or their equivalents is
very common in the universe, perhaps more common
than even Drake and Sagan envisioned. The author of
Passage 2 says in line 139, My bet is that many other
worlds, with their own peculiar characteristics and his-

tories, co-evolve their own biospheres. Both authors
seem to agree that there is a lot of undiscovered life
out there in the universe.
Section 9
1. B When you list items in a sentence, the items
should have the same grammatical form. If the ﬁrst
item is in the gerund, they should all be in the gerund.
Because the sentence says Eating an english mufﬁn
and sitting down, drink coffee should instead be drink-
ing coffee.
(Chapter 15, Lesson 3: Parallelism)
2. D As written, the sentence suggests that Mark’s
attempt was pretending to be hurt rather than Mark
himself. Answer choice (D) corrects this error.
(Chapter 15, Lesson 7: Dangling and Misplaced
Participles)
3. C The verb are is the improper tense. It should
be be as in answer choice (C).
(Chapter 15, Lesson 9: Tricky Tenses)
4. C When you list items in a sentence, the items
should have the same grammatical form. If the ﬁrst
term is in the noun form, then they all should be in
the noun form. Because the sentence says his temper,
impatience, how easily he can be irritated should in-
stead be irritability.
(Chapter 15, Lesson 3: Parallelism)
5. B Before she gave the gracious speech, she won
the match. The verb winning should instead be in the
past perfect form, having won.
(Chapter 15, Lesson 9: Tricky Tenses)

6. C The sentence begins by describing something
that was the most inﬂuential science treatise of the
20th century. The pronoun to follow the comma
should describe this treatise. Choice (C) corrects the
error in the most logical and concise fashion.
(Chapter 15, Lesson 7: Dangling and Misplaced
Participles)
7. B The subject neither is singular and therefore
were should instead be was.
(Chapter 15, Lesson 1: Subject-Verb Disagreement)
CHAPTER 16 / PRACTICE TEST 3 767
8. A This sentence is correct as written.
9. C Because the waves were described to have
been covering the stores (gerund form), the verb swept
should also be in the gerund form—sweeping.
(Chapter 15, Lesson 3: Parallelism)
10. B The problem with this question is that it is not
a complete sentence as written. Answer choice (B)
corrects that ﬂaw in the most logical fashion.
(Chapter 15, Lesson 15: Coordinating Ideas)
11. D The problem with this sentence is that it sug-
gests that the father was 7 years old when he took his
son to the game. That is not likely. Answer choice (D)
best corrects the problem.
(Chapter 15, Lesson 7: Dangling and Misplaced
Participles)
12. A This sentence is correct as written.
13. E The pronoun them refers to a plural subject.
However, anyone is singular. Answer choice (E)
clears up this pronoun-antecedent disagreement in

the most concise and logical way.
(Chapter 15, Lesson 5: Pronoun-Antecedent
Disagreement)
14. A Although the original phrasing is not the most
concise option, it is the only one that logically coor-
dinates the ideas in the sentence.
PRACTICE TEST 4
768
CHAPTER 16 / PRACTICE TEST 4 769
ANSWER SHEET
Last Name:________________________________ First Name:____________________________________________
Date:_____________________________________ Testing Location:_______________________________________
Directions for Test
• Remove these answer sheets from the book and use them to record your answers to this test.
• This test will require 3 hours and 20 minutes to complete. Take this test in one sitting.
• The time allotment for each section is written clearly at the beginning of each section. This test contains
six 25-minute sections, two 20-minute sections, and one 10-minute section.
• This test is 25 minutes shorter than the actual SAT, which will include a 25-minute “experimental” section that
does not count toward your score. That section has been omitted from this test.
• You may take one short break during the test, of no more than 10 minutes in length.
• You may only work on one section at any given time.
• You must stop ALL work on a section when time is called.
• If you ﬁnish a section before the time has elapsed, check your work on that section. You may NOT work on any
other section.
• Do not waste time on questions that seem too difﬁcult for you.
• Use the test book for scratchwork, but you will receive credit only for answers that are marked on the answer
sheets.
• You will receive one point for every correct answer.
• You will receive no points for an omitted question.
• For each wrong answer on any multiple-choice question, your score will be reduced by

1
⁄
4
point.
• For each wrong answer on any “numerical grid-in” question, you will receive no deduction.
When you take the real SAT, you will be asked to ﬁll in your personal information in grids as shown below.
YOUR NAME2
DATE OF BIRTH
4
TEST
CENTER
7
Last Name
(First 4 Letters.)
First
Init.
Mid.
Init.
−
′
− −
A
B
C
D
E
F
G
H
I