MCAT
®Practice Test 5R
Solutions
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5R
Solutions
MCAT P
RACTICET
EST5R
S
OLUTIONSEdited, produced, typeset, and illustrated by Steven A. Leduc
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MCAT P
RACTICET
EST5R
S
OLUTIONSC
ONTENTS:
Physical Sciences ... 3
Verbal Reasoning ... 14
PHYSICAL SCIENCES
Passage I
1. B. The value of DH∞ for Reaction 1 is given as –92 kJ. Since DH∞ is negative, the reaction is exothermic, by definition. 2. C. According to the data in Table 1, the equilibrium concentration of NH3(g) (% by volume) increases as the pressure increases. This eliminates choices A and B, which show the yield decreasing as the pressure increases. To decide between choices C and D, we first notice that when the pressure increases from 1 atm to 100 atm—which is an increase of 99 atm— the % NH3(g) by volume at equilibrium jumps significantly, from 15.3 to 80.6. However, when the pressure increases from 100 atm to 200 atm—an increase of 100 atm—the % NH3(g) by volume only increases from 80.6 to 85.8. This indicates that the % yield increases sharply with pressure initially, but then increases less abruptly at high pressures. This behavior is illustrated by the graph in choice C.
3. A. The passage states that Reaction 1 is carried out “in the presence of” FeO/Al2K2O4. Furthermore, we notice from Reaction 1 that neither FeO nor Al2K2O4 is consumed in the reaction. We conclude that this mixture is a catalyst for the reaction, and the role of a catalyst is to increase the reaction rate.
4. D. Since NH4+ is a cation with a +1 charge, and SO 4
2– is an anion with a –2 charge, we would expect that a combination
of these ions would contain 2 NH4+ ions for each SO 4
2– ion; that is, the compound would be (NH
4)2SO4. The balanced
acid–base reaction would be 2 NH3 + H2SO4Æ (NH4)2SO4.
5. D. Choice A is eliminated since the product of Reaction 1, NH3, is not an ion. Because NH3 has a permanent dipole moment, we would expect dipole–dipole interactions between NH3 molecules. In addition, NH3 is capable of hydrogen bonding (the partial positive charge on an H atom of one molecule of NH3 attracted to the partial negative charge on the N atom of another NH3 molecule).
6. A. Of the ions listed, Mg2+ and H+ are cations, and being electron deficient, are acids (eliminating choices C and D).
Since nitrogen is less electronegative than oxygen and since N3– has a greater negative charge than –OH, N3– is a stronger
base than –OH.
Passage II
7. D. According to the passage, “there is a large electrical repulsion between these two fragments that causes them to . . .
gain kinetic energy.”
8. A. According to Newton’s Third law, the magnitude of the force by Fragment 1 on Fragment 2 is equal to the magnitude
of the force by Fragment 2 on Fragment 1, so choice A must be correct. Since the forces the fragments feel have the same magnitude, but the fragments have different masses, the accelerations of the fragments must be different; since a = F/m, the fragment with the greater mass experiences a smaller acceleration, eliminating choice B. Because both fragments start from rest and have different accelerations, the speeds of the fragments at any moment t will be different (because v = at); this eliminates choice C. To eliminate choice D, notice that
KE mv m at m t F mt F m = 1 = =
( )
= 2 2 1 2 2 1 2 2 2 2 2 ( )9. C. First, eliminate choice B. If all of the heavier elements were stable, then there would be naturally-occurring elements
that have more protons in their nucleus than uranium does. As for choices A and D, even if we grant that they are true, they simply beg the question and do not provide an answer. Why have all of the heavier elements radioactively decayed? Why is it that heavier elements can be made only in nuclear reactors? According to the passage, the strong nuclear force is a short-range attractive force that balances the large repulsive force between the positive charges in the nucleus. Once the nucleus get too large (that is, once the nucleus contains too many protons), we conclude that the short-range strong nuclear force becomes unable to hold the nucleus together, which is why such large nuclei do not occur naturally (they’ve spontaneous decayed). The statement in choice C provides a reasonable and direct answer as to why the strong nuclear force becomes unable to hold large nuclei together.
10. D. According to Coulomb’s law, the force between the two charged fragments, +Q and +Q, is given by the equation
F = kQQ/r2 = kQ2/r2, where r is their separation. Since F is inversely proportional to r2, the graph of F vs. r must decrease
nonlinearly, which is shown in the graph in choice D.
11. A. Neither choice B nor C is applicable here, since there is no mention of electrons in the fusion (or fission) of nuclei.
And according to the passage, the strong nuclear force is an attractive force between the charges in a nucleus, so to fuse two nuclei together, there is no need to “overcome” an attractive force, eliminating choice D. The answer must be A. Since both nuclei are positively charged, energy must be provided to force them together and overcome their electrical repulsion.
12. C. Since each fragment (of charge +Q) experiences a decreasing force as it moves away from the other fragment
(because, according to Coulomb’s law, F = kQQ/r2 = kQ2/r2, and r is increasing), each fragment will experience a
decreasing acceleration (a = F/m = kQ2/mr2, where m is the mass of the fragment).
Passage III
13. D. The passage states that CFCs “can undergo photolysis in the upper atomsphere and subsequently assist in the
decomposition of ozone. . . .” Therefore, we can conclude that if a compound were inert in the upper atmosphere, then it would not “significantly assist” in the depletion of ozone.
14. A. The chlorine atom in Reaction 4 is a radical, Cl•. By definition, a radical is an atom or molecule fragment with one
or more unpaired electrons.
15. B. If we combine Reactions 4 and 5, crossing out the Cl• and ClO• radicals, the net reaction is given in choice B:
Reaction 4: Cl • + O ClO • + O Reaction 5: ClO • + O Cl • + O 3 2 2 Æ Æ Net reaction: O + O 3 Æ 2 O2
+
16. C. First, we find the net reaction of Reactions 1 and 2:
Reaction 1: O O + O Reaction 2: O + O O Net reaction: O O + O 2 2 3 2 3 Æ Æ Æ 2
+
Now, to determine whether the overall reaction involves an increase or a decrease in free energy, we calculate DGrxn using the values for DGf given in Table 1:
D D D D D D G n G n G G G G rxn f, products f, reactants f, O f, O f, O kJ mol kJ mol kJ mol 3 2 = ◊ - ◊ = ◊ + ◊ - ◊ = + - ◊ >
Â
Â
( ) ( ) ( . . ) ( ) 1 1 2 230 1 163 4 2 0 0Because DGrxn is positive, we can eliminate choices A and B. Choice D is eliminated since it indicates a negative activation energy (since it shows the activated complex at a lower energy level than the reactants). The answer must be C.
17. D. We calculate DS for the reaction 2 O3Æ 3 O2 using the values for S given in Table 1:
DS n S n S S S rxn products reactants O O J mol K J mol K J mol K 2 3 = ◊ - ◊ = ◊ - ◊ = ◊ - ◊ =
Â
Â
◊ ◊ ◊ ( ) ( ) ( . ) ( . ) . 3 2 3 205 0 2 238 8 137 418. C. In Reactions 3–5, we see that the Cl• generated by the cleavage of a CFC (Reaction 3) causes the decomposition of
O3 (Reaction 4). However, the Cl• is regenerated in Reaction 5, so that only a catalytic amount of CFC is needed to drive the formation of O2.
Passage IV
19. A. Apply Le Châtelier’s principle to Equation 2. “Excessively moist soil conditions” describe conditions where the
amount of the H2O(l) is increased. Since H2O(l) is a reactant in Equation 2, we would thus expect that an increase in H2O(l) would shift the equilibrium toward the product side, causing a greater degree of ionization and releasing more OH–(aq). 20. D. Because the statements in choices A, B, and C are all false (since N2 accounts for more than 75% by volume of the atmosphere, the N2 molecule is nonpolar, and N2 is not a noble gas), the correct response must be D.
21. D. The species that results when an acid loses an H+ is called the conjugate base of that acid. When H 2PO4 – loses an H+, it becomes HPO 4 2–. Therefore, HPO 4
2– is the conjugate base of H 2PO4
–.
22. C. The equilibrium that would best account for an increase in pH would show the formation of OH– ions, so we
eliminate choice A. The reactions in choices B and D are incorrect, since, for example, neither is balanced electrically nor stoichiometrically, so the best answer here is C.
23. A. Pure liquids are omitted from equilibrium expressions since their concentrations remain essentially constant (not that
they’re zero), so we first eliminate choices B and D. Since Equation 2 shows the formation of OH–, we conclude that
Independent Questions
24. B. First, to balance the given equation, we need only place a coefficient of 2 in front of both the HCl and the NaCl:
Na2CO3 + 2 HCl Æ CO2 + H2O + 2 NaCl
Notice that this affects neither the Na2CO3 nor the CO2. Now, if we treat CO2 as an ideal gas, then 11.2 L of CO2 at STP is equivalent to 1/2 mole of CO2 (since, by Avogadro’s law, one mole of any ideal gas at STP occupies a volume of 22.4 L). According to the reaction above, we would need 1/2 mole of Na2CO3 to produce 1/2 mole of CO2. Since a 2 M solution of Na2CO3 contains 2 moles of Na2CO3 per liter, we would need 1/4 L = 250 mL of this solution to obtain 1/2 mole of Na2CO3.
25. C. The molecule MnO4– is not an exception to the rule that the oxidation number of oxygen is –2. So, if we let x denote
the oxidation number of Mn in MnO4–, then x + 4(–2) = –1, so x = +7. Now, the oxidation number of Mn for the cation
Mn2+ is clearly +2, so the oxidation number of Mn in MnO 4
– differs from its oxidation number in Mn2+ by (+7) – (+2) = 5. 26. B. First, if the waves strike the shore every 3.0 seconds, then the period of the waves is T = 3.0 seconds. Next, if the
horizontal distance between adjacent crests and troughs is 1.0 m, then the wavelength is twice as much, l = 2.0 m. We now use the equation v = lf. Since f = 1/T, we have
v T = =l 2 0 = 3 0 0 67 . . . m s m s
27. B. The circumference of the circular path is C = 2pr = 2p(4 cm) = 8p cm. Since the particle complete 4 revolutions
(or cycles) per second—that is what “moving on a circular path . . . with a frequency of 4 Hz” means—the particle travels a total distance of 4 ¥ (8p cm) = 32p cm in one second. Therefore, to travel half this distance, 16p cm, would require half the time: that is, 1/2 second.
Passage V
28. C. The buffered solution at the beginning of Experiment 1 contains 16 mmol of CDP in 1 L of aqueous solution. Since
10 mL = 1/100 L, we conclude that 10 mL of the solution contains (16/100) mmol of CDP. To find the mass of CDP in 10 mL of this solution, we multiply (16/100) mmol of CDP by its molecular mass:
m=(10016 ¥10-3 mol CDP)¥ 403 g ª(10016 ¥10-3 ¥(400 ¥ ¥ -3 ¥ -3 ¥ -2
mol CDP ) g) = (16 4) 10 g = 64 10 g = 6.4 10 g
29. D. We apply Le Châtelier’s principle. To increase the yield of the product, (CP)n, in Equation 1, we could increase the concentration of the reactant (which would shift the equilibrium toward the product side). Therefore, we would expect that increasing the amount of the reactant, CDP, would increase the yield of the product, (CP)n.
30. A. The reactant, CDP, and one of the products, namely HPO42–, both have a stoichiometric coefficient of n in the
balanced reaction (Equation 1 in the passage). Therefore, in the expression for the equilibrium constant for this reaction, both [CDP] and [HPO42–] must appear with an exponent of n. This eliminates choices B and C. The expression in A is a
better choice than the one in D since the other product of Reaction 1 is (CP)n, not simply CP.
31. C. In Equation 1, which is balanced, the stoichiometric coefficient of the polymer, (CP)n, is 1, and the stoichiometric coefficient of HPO42– is n. Therefore, the concentration of (CP)
n is 1/n times the concentration of HPO4 2–. 32. C. We use the Henderson–Hasselbalch equation:
pH p conjugate base]
[weak acid] a
= K +log[
Since the solution is buffered at pH 8.7, and pKa = 6.7, we have pH – pKa = 2, so
log[ [ [ [ HPO ] H PO ] HPO ] H PO ] 4 4 2 2 2 2 10 100 -= fi = =
Passage VI
33. A. Because 14C undergoes beta decay, it will not emit an alpha particle or neutron in the decay process; this eliminates
choices B and C. Since the radioactive decay process is 146
7 14
1 0
C Æ N + -e-, we see that 14C undergoes b– decay and emits
an electron, -01e .
-34. D. According to the passage, the half-life of 14C is approximately 6000 years. Therefore, a time period of 18,000 years
represents approximately 3 half-lives. If the object currently contains 1000 atoms of 14C, then 1 half-life ago, it contained
2000 atoms of 14C; 2 half-lives ago, it contained 4000 atoms of 14C; and 3 half-lives ago, it contained 8000 atoms of 14C. 35. C. The passage gives the mass of a beta particle as 9 ¥ 10–31 kg. So, if its speed is 3 ¥ 107 m/s, its kinetic energy is
KE= mv = ¥ ¥ = ¥ ¥ ª ¥ = ¥ -1 2 2 1 2 31 7 1 2 2 31 14 17 16 9 10 3 10 9 3 10 10 4 0 10 ( )( ( )( ) . kg m s) J 40 10 J J 2
36. A. As stated in the passage, a scintillator is a substance that produces light when it absorbs the energy accompanying
radioactive decay. The scintillator is attached to a photomultiplier that collects this light and converts it into electrical impulses, which are then counted. These pulses then serve to measure the rate at which decay occurs. If the scintillator were to be non-transparent to the light it emits, then it would reabsorb some of that light, which the photomultiplier would then turn into electrical pulses and add to the count (that is, in addition to the pulses that are actually due directly to the decaying object itself), thereby overestimating the radiation energy and rate of decay. This would clearly produce an inaccurate reading. To prevent (or at least to minimize) this reabsorption of light, the scintillator should therefore be (nearly) transparent to the light it emits.
37. A. The energy of a photon of frequency f is given by the equation E = hf, where h is Planck’s constant. Since f = c/l,
we can rewrite the equation for photon energy as E = hc/l. For the photon described in this question, then, we have
E= = ¥ ◊ ¥ ¥ ª ¥ ◊ ¥ ª ¥ -hc J s)(3 10 m s) 450 m J m 4.5 m J 8 l ( . . 6 6 10 10 20 10 10 4 4 10 34 9 26 7 19 Passage VII
38. C. We first eliminate choices A and B; statement A is false (after all, Cl2 is a gas at room temperature, whereas I2 is not) and although statement B is true, the relative boiling points of Cl2 and I2 are irrelevant to the determination of the oxidation state of copper in the compounds it forms in reactions with these substances. To decide between choices C and D, we look at Table 1, and notice that Cu and Cl2 form CuCl2, while Cu and I2 form CuI. In CuCl2, copper is in a +2 oxidation state, while in CuI, copper is in only a +1 oxidation state. Since Cu “gives up” two electrons to chlorine but only one to iodine in these compounds, we would select choice C over choice D. Furthermore, since Cl is higher in the periodic table than I, we would expect that each Cl atom in Cl2 would have a stronger attraction for electrons than each I atom in I2.
39. C. The nitrate ion, NO3–, has a –1 charge, so a cadmium cation would have a +2 charge in order for the molecule
Cd(NO3)2 to be neutral. Since the chloride ion, Cl–, also has a –1 charge, we expect the combination of Cd and Cl to be
40. B. Looking to the passage for a clue about the product that would most likely form between Cd and S, we notice in
Table 1 that Zn and S form the compound ZnS. Since Cd is in the same family as Zn, it is reasonable to expect that Cd and S would form the compound CdS. Now sulfur, like oxygen, is most commonly in a –2 oxidation state in its compounds with other atoms. If the oxidation state of S in CdS is –2, then the oxidation state of Cd must be +2.
41. A. First, eliminate choice B (where is the source of carbon to form CO2?). Since evolution of gas occurs only with the addition of HNO3, HNO3 must react with copper metal. Copper metal (Cu0) must be oxidized during this reaction, and
HNO3 must be reduced. Of the remaining choices (A, C, and D), the only logical choice for the product of the reduction of HNO3 is NO.
42. B. Because boiling-point elevation is a colligative property, the solution whose solute dissociates into the greater
number of ions will be the one with the higher boiling point. Since Zn(NO3)2 dissociates into 3 ions (Zn2+ + 2 NO 3
–) while
AgNO3 dissociates into only 2 ions (Ag+ + NO 3
–), we’d expect the boiling point of Zn(NO
3)2(aq) to be higher than that of
AgNO3(aq).
43. D. The AgNO3(aq) solution contains Ag+ ions; as Cu atoms are oxidized, Ag+ ions are reduced to Ag, which is the “new
metal [that] forms on the surface of the Cu strip.” Also, note that we can eliminate choices A, B, and C, since it is highly unlikely that the cation Ag+ would be oxidized—or that Ag or Cu would be reduced—in this situation.
Passage VIII
44. C. The temperature, T, is 673 K in Trial 5. According to the data for Trial 5 given in Table 1, the current I is 2 A, and
the voltage across the wire is 28 V. Therefore, the power dissipated by the wire is P = IV = (2 A)(28 V) = 56 W. [Alternatively, since the resistance R is approximately 14 W, the power dissipated is P = I2R = (2 A)2(14 W) = 56 W.] 45. D. In the first paragraph of the passage, the mass of the wire is given to be m = 4 ¥ 10–3 kg. Since the volume of the
wire is given in the question to be V = 5 ¥ 10–7 m3, the density of the wire is r = = ¥ ¥ = ¥ = ¥ = -m V 4 10 5 10 0 8 10 8 10 8 000 3 7 4 3 kg m3 kg m kg m kg m 3 3 3 . ,
46. A. According to the data in Table 1, R increases as T increases (as we can see by reading the values of R as the
temperature increases from Trial 1 through Trial 5). This eliminates the graphs in B and D, which show the resistance R either constant or decreasing with temperature. Since the only choices left are the graphs in choices A and C, the question becomes, “Does R increase linearly with T?” Comparing Trials 2 and 3, then 3 and 4, then 4 and 5, we see that R increases by a steady 2.6 W for every 100 K increase in temperature. Therefore, R does increase linearly with T, and the answer is A.
47. D. From the expression given in the last paragraph of the passage, AsT4, we see that the energy radiated from the
heated wire each second is proportional to T 4. So, if T increases by a factor of 2, the energy radiated each second
increases by a factor of ( 2)4 =[( 2) ]2 2 =[ ]22 =4.
48. A. We use the equation q = mcDT, where m is the mass of the sample being heated and c is the specific heat of the
sample (in this case, it is the iron wire). Since the mass of the wire is m = 4 ¥ 10–3 kg, the specific heat of iron is c = 460
J/kg·K, and DT = 573 K – 373 K = 200 K, we have
q=mc TD = ¥(4 10-3 kg)(460 kg KJ◊ )(200 K) = 368 J [Note: The question uses the term “heat capacity” where it should use the term “specific heat.”]
49. D. According to the data in Table 1 for Trial 1, the resistance of the wire was R1 = 4 W when T = 293 K. When T = 673 K (Trial 5), the resistance rose to R5 = 13.9 W ª 14 W. Therefore, if the voltage remained constant at 28 V, the current decreased, from I V R I V R 1 1 5 5 28 7 28 2 = = V= = = = 4 A to V 14 A W W Independent Questions
50. B. Because the activity decreased to 60/240 = 1/4 its initial value, this means that 2 half-lives elapsed, since (1/2)2 is
equal to 1/4. If a time period of 2 half-lives is equal to 24 minutes, then one half-life must be 12 minutes.
51. C. Since the molecules in the gas phase of a substance are much more disordered than in the solid phase, the phase
change from solid to gas (sublimation) represents an increase in the entropy, S. That is, DS > 0.
52. C. The Doppler Effect implies that when the source of a sound moves away from the observer, the perceived frequency
is lower than the emitted frequency.
53. A. As the diagram below shows, water at –0.1∞C and 1.0 torr is vapor, and as the pressure is increased at constant
temperature, the vapor will become a solid and then a liquid:
pressure (torr) 1.0 4.6 temperature (∞C) –3.0 0.01 solid liquid vapor –0.1
off the diagram at P = 200 atm
54. B. Because opposite charges attract, the negatively-charged particle will move toward the fixed positive charge Q; this
eliminates choices A and C.
–q +Q
r
The negatively-charged particle experiences a force (F = kQq/r2) as it approaches +Q, so it will undergo an acceleration
(a = F/m = kQq/mr2, where m is the mass of the –q particle). Since the –q charge is accelerating, its speed of approach
Passage IX
55. D. As stated in the first paragraph of the passage, the sample XT-n contains n% Ti, where n = 0, 1, 3, or 5. Looking at
the data in Table 1, we notice that the solubility of XT-n increases as n increases, so choices A and B are eliminated. Since the entry in each row is greater for toluene than for THF, we conclude that the XTs are more soluble in toluene than in THF.
56. D. Even if there were any indication in the passage that the XTs are even capable of hydrogen bonding, the formation
of hydrogen bonds would not decrease the weight of a sample, so we eliminate choice A. Next, according to the data in Table 2, the masses (and therefore the weights) of the samples decrease by 20% when heated from 20∞C to 700∞C; the loss of some electrons, even if they escaped from the heating chamber, could not account for this much of a decrease in mass, so we eliminate choice B. As for choice C, the removal of protons from nuclei requires extreme conditions (like those in a nuclear reactor); it is highly unlikely that simply heating the compound to 700∞C would cause a nuclear reaction. Choice D provides the most reasonable explanation for the loss of mass by the samples as they are heated.
57. B. The transition metal titanium (Ti, atomic number 22) is in the 3d “block” of the Periodic Table, which means its
valence electrons are in 3d orbitals. Titanium doesn’t contain electrons in 4p or 5f orbitals, so choices C and D are eliminated, and titanium’s 2s electrons (choice A) are not in the valence shell, so they’re unavailable to form bonds.
58. C. Of the elements listed in the choices, only zirconium (Zr, atomic number 40) is in the same family (group) as
titanium (Ti, atomic number 22). The elements in each family of the Periodic Table have similar properties and have identical (or very similar) outer configurations.
59. B. Since oxygen is an element in Period 2, it has only s and p orbitals and can form no more than four hybrid orbitals,
so choice C is eliminated immediately. Choice D should be eliminated immediately as well (s2p2 hybridization?). The
oxygen atom in THF is bonded to two carbon atoms, so there must be four equivalent hybrid orbitals on the oxygen atom, formed by sp3 hybridization; two contain lone pairs and two contain a single electron each, which will form the s bonds
with the carbons.
60. A. Since THF can participate in hydrogen bonding with H2O, but toluene cannot, we’d expect THF to be more soluble than toluene in H2O. Choice B is false (since there are no hydrogen bonds between toluene and water to compare with those between THF and water), and while the statements in choices C and D are true, they don’t answer the question.
Passage X
61. C. When the toboggan begins its slide from Point A, it has gravitational potential energy (relative to Point B), which is
converted to kinetic energy as the toboggan slides down the hill. Since the passage states that the toboggan experiences friction as it slides, some of the potential energy is also converted to heat (thermal energy). Therefore, the energy conversion is best described by choice C: potential to kinetic and thermal.
62. B. According to the passage, the toboggan is opposed by a constant 60 N frictional force when it’s sliding down the hill.
Since the toboggan feels this force for the entire length, l, along the hill, the work done by sliding friction on the toboggan is equal to –(60 N)(l), so the energy lost to friction is (60 N)(l).
63. A. We apply Conservation of Total Momentum to this completely inelastic collision. Since the toboggan and rider
(T&R) are stationary before the collision, their momentum before the collision is zero, so the total momentum before the collision is simply MS&RvS&R, the momentum of the sled and rider (S&R). The momentum after the collision is (MS&R +
MT&R)v¢. Therefore,
M v M M v v M v
M M
S& R S& R S& R T& R
S& R S& R S& R T& R kg kg) (10 m s) kg kg) kg kg) m s 4.55 m s = + ¢ fi ¢ = + = + + + ◊ + = ª ( ) ( ( ( 3 47 3 47 6 54 50 11
64. A. The passage states that, to the right of Point B, the sled and rider are opposed by a 50 N frictional force, so Ffrict = 50 N. Since Ffrict = mN, and N = MS&Rg = (3 kg + 47 kg)(10 m/s2) = 500 N, we have
m = F = = = N frict N 500 N 50 1 10 0 1.
65. B. The snowball will land when it has fallen a vertical distance of y = 20 m. Let’s first figure out how long this will
take. Using Big Five # 3 (with v0y = 0, since the ball is thrown horizontally), we find that
y=12gt2 fi 20 m =12(10 ms2)t2 fi t=2 s
Because the snowball’s horizontal speed is vx = 25 m/s, the snowball would, in this time, travel a horizontal distance of
x = vxt = (25 m/s)(2 s) = 50 m
Since this equals the horizontal distance from Point A to Point B, the snowball will land at Point B.
snowball
Point A
y = 20 m
x = 50 m Point B
t = 2 sec
66. D. We use Conservation of Mechanical Energy. The passage states that the sled and rider slide free of friction from
Point A to Point B, so the gravitational potential energy of the sled and rider at Point A is transformed into kinetic energy at Point B. Therefore,
PEÆKE fi Mgh=12Mv2 fi v= 2gh fi vµ h
If the sled and rider start at a point on the hill that is 10 m lower than Point A, then the sled and rider’s initial height above Point B is being reduced from 20 m to 10 m, a decrease by a factor of 2. Since v is proportional to h , if h is reduced by a factor of 2, then v will be reduced by factor of 2 .
Passage XI
67. D. As the diagram below shows, the distance between adjacent nodes (labeled N) on a standing wave is always equal to half the wavelength, l/2:
1 2l
l
68. A. The traveling waves whose superposition generates a standing wave travel in opposite directions, eliminating
choices B and C. The oppositely-directed traveling waves must have equal amplitudes (choice A) since the resultant standing wave has displacement zero at the node positions (where the equal-amplitude traveling waves arrive exactly out of phase with each other and thus completely cancel).
69. B. That the light emitted by the laser is coherent is stated in the first sentence of the passage, so Item II is true, and we
can eliminate choices A and C. If the laser has only one mode of oscillation, then it produces only one wavelength of light (since, according to the passage, “laser cavities have mode numbers that are related to the allowed cavity wavelength[s].” Because the laser produces light of only one wavelength, the light is monochromatic (“one color”), so Item I is true, and the answer must be B. (Note that we didn’t need to check Item III, but, if we did, since laser light is sharp and focused, it would not be characterized as diffuse.)
70. B. If we substitute lm = c/fm into the equation mlm = 2L (both of which are given in the passage), we find that
m f L m Lf ◊ = fi = = ¥ ¥ = ¥ c c m)( Hz) m s m m 2 2 2 9 10 3 10 2 10 1 3 14 8 6 (
71. D. Because l = c/f, we have lbeat = c/fbeat. Therefore, since fbeat = fm+1 – fm, we have
lbeat beat m m c c = = -+ f f 1 f Independent Questions
72. C. By Archimedes’ Principle, the buoyant force exerted by the unknown liquid on the object is given by Bunknown =
runknownVg, and the buoyant force exerted by benzene is given by Bbenzene = rbenzeneVg. Thus, the ratio Bunknown/Bbenzene is
B B Vg Vg unknown benzene unknown benzene unknown benzene = r = r rr
Since we’re given that Bunknown = 12 N and Bbenzene = 5 N, the ratio Bunknown/Bbenzene is equal to 12/5. Therefore,
r runknownbenzene r r r r r r r unknown benzene H O2 H O2 H O2 H O2 H O2 =12 fi = = ÊË ˆ¯ = = = ª 5 12 5 12 5 7 10 84 50 168 100 1 68. 1 7. This tells us that the specific gravity of the unknown liquid is approximately 1.7.
73. B. We apply Conservation of Total Momentum to this perfectly inelastic collision. Since the block is stationary before
the collision, its momentum before the collision is zero, so the total momentum before the collision is simply mobjvobj, the momentum of the sliding object. The momentum after the collision is (mobj + mblock)v¢. Therefore,
m v m m v v m
m m v v v
obj obj obj block
obj obj block
obj obj obj
kg 1 kg kg = (8 m s) 2 m s = + ¢ fi ¢ = + = + = = ( ) 1 3 1 4 1 4
74. B. The equilibrium is Ca(OH)2(s) Ca2+(aq) + 2 OH–(aq). Increasing the acidity of the solution has the effect of
reducing the concentration of OH–(aq). By Le Châtelier’s principle, removing a product causes a shift toward the products;
therefore, additional Ca(OH)2(s) will dissolve. Since the value of Ksp can be changed only by a change in temperature, the answer must be B.
75. A. For nearsighted individuals, light from distant objects is focused in front of the retina, which occurs if the focal
length of the lens of the eye is smaller than the distance to the retina. A divergent corrective lens is required to cause the light from distant objects to diverge slightly before entering the lens of the eye, so that it may be focused on the retina.
retina retina
diverging corrective lens
nearsighted eye nearsightedness
corrected
76. A. The principal quantum number, n, has nothing to with the number of valence electrons of an atom (choice C) or with
the mass number (choice D). And choice B is eliminated since, for example, the 3s, 3p, and 3d orbitals all have different shapes but the same principal quantum number (namely, n = 3), so the principal quantum number cannot be a “measure” of the approximate shape of an electron cloud. The best response here must be A (and the higher the value of n, the larger the radial size of the electron cloud).
77. B. The weight of the unknown solid is given by the equation wsolid = rsolidVg, and, by Archimedes’ Principle, the
buoyant force exerted by the water on the solid is given by B = r
H2OVg. Thus, the ratio wsolid/B is
w B Vg Vg solid solid H O solid H O 2 2 = r = r rr
Because the solid weighs 31.6 N but has an apparent weight of only 19.8 N when submerged in water, the buoyant force on the object must be 31.6 – 19.8 = 11.8 N. Since wsolid = 31.6 N and B = 11.8 N, the ratio wsolid/B is also equal to 31.6/11.8. Therefore, r rsolidH O2 = 31 6ª 11 8 2 68 . . . This tells us that the specific gravity of the solid is approximately 2.68.
VERBAL REASONING
Passage I
78. D
A: A main theme of the passage is that competition between state and local governments can be destructive. No positive benefits are mentioned. Furthermore, competition between private firms is never discussed; subnational governments compete with each other to attract private firms to their area.
B: As in choice A, this statement contradicts a main idea of the passage—that competition can have negative effects. A secondary problem with this choice is that efficiency is never mentioned in the passage.
C: The passage makes only two references to the national level. First, the author indicates that national economic development is not characterized by competition (lines 4-6). Second, the author cites the critics’ claim that local and state competition does not contribute to national productivity (lines 39-43). Neither reference supports a link between subnational competition and incentives for national development policy.
D: Yes. The author cites the critics’ claim that cooperation would be more likely than competition to result in increased national (“overall”) productivity (lines 39-44). The passage also describes a case in which increased cooperation between jurisdictions in the Monongahela River Valley may better address those jurisdictions’ economic problems than would competition (lines 72-83).
79. B
A: The primary purpose of the author is to discuss competition and cooperation on a subnational level. No role for national officials is mentioned.
B: Yes. The author describes one successful attempt at cooperation, an effort organized by local leaders (lines 80-83).
C: Academic researchers are never discussed in the passage. D: The author does not discuss any role for media representatives.
80. C
A: According to the passage, local leaders in the Monongahela region worked to increase cooperation and decrease competition between jurisdictions (lines 70-83). If these efforts were followed by prosperity, the lesson would be to cooperate more, not less (and so compete less, not more). For both choices A and B, remember that the main idea of the passage is that cooperation may be more advantageous than competition at subnational levels.
B: According to the final paragraph, local jurisdictions had competed in the past, but now are managing to cooperate in the face of a regional economic downturn (lines 70-83). If this cooperation were followed by prosperity, less competition would be called for.
C: Yes. If local leaders’ cooperative efforts (lines 80-83) were followed by prosperity, it is likely that continued cooperation would lead to continued prosperity for the region.
D: The author does not say that local governments in the Valley region ever received federal grants, nor does the passage draw any link between federal moneys and either prosperity or cooperation.
81. C
Note: To be the credited response, the answer choice must both present an assumption that can reasonably be attributed to
the author and be inconsistent with the new information given in the question.
A: This choice is too extreme; the author does not make this assumption. The author indicates that cooperation is difficult (lines 44-55), not that it is impossible. In fact, the passage presents a specific example of cooperation between local jurisdictions (lines 77-83).
B: This choice is out of scope (“state law”) and too extreme (“only if”); the author does not make this assumption. State
law is never mentioned, only pacts, policies, and agreements within and between states and localities. Furthermore, an
example of cooperation is given with no suggestion that state law played any role (lines 70-83).
C: Yes. In the final paragraph, the author states that “Economic conditions may ultimately serve as the catalyst for greater cooperation….” (lines 70-71). This statement is followed by the one example of a successful attempt at cooperation given in the passage; this cooperation occurred when the region was experiencing economic difficulties (lines 73-80). Thus, the author does make the assumption that cooperation is most likely in times of economic stress. The study cited in the question gives cases in which cooperation followed a period of economic
D: This is not an assumption made by the author. In fact, the final paragraph presents a case in which local leaders crossed state lines to foster cooperation between jurisdictions (lines 80-83).
82. D
Item I: The passage indicates that a difference between national and state or local economic activity is that state and local economic development is competitive and that state and local governments are “awash in competition” (lines 4-9). Thus we can infer that national governments do not exhibit competitive economic behavior.
Item II: Yes. This is directly stated in lines 4-9. Item III: Yes. This is directly stated in lines 4-9. 83. A
A: Yes. In lines 44-46, the passage mentions the failure of Great Lakes states’ governors to keep communities within those states from pirating or stealing economic developments from each other. This example is introduced with the more general statement that cooperation is elusive (line 44).
B: The experience of the Great Lakes states is one of a failed attempt to restrict competition. Furthermore, the author does not indicate that many other state governments have made similar attempts, and so the word “usually” is too extreme. C: The passage makes no such comparison between state and local governments. The fact that some state governments
failed (lines 44-45) does not in and of itself indicate that local governments are or would have been more successful. Finally, while local leaders organized the successful Monongahela attempt at cooperation (lines 80-83), the author does not indicate that this can be extrapolated into a generalized comparison between the effectiveness of state and local leaders.
D: This choice is both out of scope and too extreme. The power of national governments to regulate competition is never discussed. Furthermore, the Monongahela case is an example of local leaders successfully regulating or limiting local competition (lines 72-83).
Passage II
84. B
A: The author explains in lines 59-64 that false memories cannot always be distinguished from correct or true memories.
B: Yes. This is the main idea of the passage. In paragraphs 2, 3 and 4 the author explains how schemas, once instantiated, help us to understand new information and act upon it. In the final paragraph, the passage discusses how there may be glitches or flaws in that process that also could affect our memory and comprehension of events.
C: While the author does state that certain things may be forgotten (lines 60-64), the passage does not indicate that this would necessarily be true for important things. For that reason, this choice is too extreme. Furthermore, while imperfect recall is one aspect of schema theory, it is not presented as the most important aspect of it, or as the reason why schema theory is itself important. Compare this choice to answer choice B.
D: Schemas are activated and instantiated when we are confronted with situations similar to those we have experienced in the past (lines 8-15). Thus memory is activated in these cases by familiar, not unfamiliar situations.
85. D
A: This choice is inconsistent with the main idea of the passage, which is in part that schemas and scripts (a special type of schema) are activated in situations that are similar to events and experiences from our past. In cases where scripts are active, then, encounters with certain events are not new learning experiences, but are affected by our past experiences. B: In lines 60-64, the passage states that we may remember information that was never part of the original event.
C: As in choices A and B, this statement contradicts the passage. The very nature of a script is that once developed, it can be used to help us understand and take action in new instances (paragraphs 3 and 4). Compare this choice with choice D. Choice C is the opposite of the credited response.
D: Yes. Scripts are a special form of schema (lines 22-24), and scripts not only help us understand new instances of familiar situations, but also guide our behavior in those contexts (lines 22-24). The author presents the
86. A
Item I: Yes. Slot filling occurs when a script created by previous experiences helps us to understand and fill in gaps in other, similar experiences (lines 35-52). In this choice, the child has already had one year of school, presumably with significant similarities to the second year.
Item II: A child first learning to ride a bicycle would not necessarily have had other similar experiences in the past. Item III: As for Item II, a child first learning the alphabet would not necessarily have had previous experience with
similar situations that could have created a schema or that would lead to slot filling within that schema.
87. D
A: The passage only discusses inferences, correct or incorrect, that are made within schemas (lines 35-52, 60-64). B: The author does not discuss partial activation of scripts nor suggest in any way that partial activation cannot occur.
This choice may sound familiar, as the author does discuss instantiation of a script based on partial information (lines 15-19). Always be sure to go back to the passage and reread carefully.
C: While it is likely that scripts are instantiated subconsciously (without our direct knowledge), the author never states it. Most importantly, the author does not make a connection between the way in which schemas are instantiated (lines 12-21, 35-46) and recall error as described in the question and in the passage (lines 60-64).
D: Yes. The author states that inferencing may lead to recall errors and that some information may be forgotten (lines 56-64). If different readers recalled the same text differently, this would provide evidence that recall errors do in fact occur.
88. B
A: The passage does not suggest that scripts are instantiated through a deliberate or conscious act. Notice the wording of the passage. For example, in lines 14-15, the author states that a “schema is thus instantiated by the new information,” not that the person instantiates it through a deliberate act of will.
B: Yes. The author explains that scripts, when activated, affect both how we process new information and how we behave based on that information (lines 10-12, 22-24). Schemas and scripts are based on memory or prior knowledge (lines 3-4, for example).
C: While scripts provide general information about particular circumstances (lines 32-34), slot filling provides specific, not generalized information within that script (lines 35-52). Notice the word “however” in line 35. It is at that point that the author shifts from discussing scripts in general to describing the particular function of slot filling.
D: Activation of the script influences processing of new information (lines 10-12). Inferencing occurs after activation and instantiation, when slot filling occurs (lines 35-52). Thus inferencing depends on the availability of specific pieces of information from the past, not on processing the new information itself.
89. D
A: Instantiation depends on the availability of an appropriate old, pre-existing schema, not a new schema.
B: Alteration of a schema may occur when slot filling and inferencing occurs (lines 56-64). This happens during or after instantiation (lines 35-46). It does not determine whether or not instantiation has occurred.
C: As described by the author, instantiation depends on the quality of the information [whether or not it matches a pre-existing schema (lines 12-15)], not its quantity.
D: Yes. In lines 12-15, the author explains that instantiation occurs when new information is judged to be “similar enough to the content of the schema.”
90. C
A: Careful reading is never suggested as a factor in recall errors, certainly not as a cause of error.
B: According to the passage, errors occur when slots are inappropriately filled in a current situation with information from the schema (lines 59-64), not when the schema (prior knowledge) is incomplete.
C: Yes. A person may fill in slots in a current situation [inferencing (lines 39, 49)] with information from the schema. While the situations are similar (as they must be for the schema to be activated and instantiated), they may not match exactly, and those slots may be filled with information that does not match the current reality. The person may then later remember these things as if they actually happened in that more recent situation (lines 35-45, 53-64).
91. A
A: Yes. An understanding of schema theory would most likely lead the teacher to teach reading in a way that created and/or utilized pre-existing knowledge structures. The development of background knowledge would help create such structures.
B: Nothing in the passage suggests that improved pronunciation would either utilize or develop “organized knowledge structures in memory” (lines 3-4).
C: While memory is involved in this choice, rote memorization would be unlikely to provide the kind of generic knowledge structures that could be applied to new situations (lines 3-7). Compare this choice to answer choice A. D: The credited response must involve the development or use of schemas or “organized knowledge structures in memory”
(lines 3-4). Nothing in the passage indicates that an explanation of an unfamiliar word would help create a schema. Given that the word is unfamiliar, it is also unlikely that the explanation would draw on or utilize a schema.
92. B
Item I: Inferences occur when a person fills in empty spaces in the current text with details from an instantiated script. A script is not instantiated unless it is significantly similar to, and so most likely appropriate for, the current text or situation (lines 12-15). The passage does not discuss the activation of inappropriate or wrong texts.
Item II: Yes. Inferencing occurs when empty slots in a text are filled in with details from a pre-existing schema (lines 35-45). When information came only from the schema and did not in fact exist in the text being read, a person may still later incorrectly remember them as part of that text (lines 59-64), essentially
“rewriting” the text in their memory.
Item III: Inferences are made when the reader takes information from a pre-existing schema and inserts it into gaps in a text (lines 35-45). Skimming a text to acquire specific facts does not involve this use of pre-existing memory structures, but only the current text itself.
93. B
A: The passage does not indicate that the formation or use of knowledge structures requires effort or purpose, as in concentrated study.
B: Yes. The author indicates that schemas are formed through previous experiences (lines 3-7), and that schemas provide a context that helps us to comprehend and utilize new information (lines 10-12, 22-24).
C: Schemas help us to comprehend new situations when not all of the facts are present, as shown by the example of recognition of a face when only a few details are available (lines 15-21).
D: As in choice A, the passage does not indicate that activation and instantiation of schemas requires any conscious act of will.
Passage III
94. B
A: The author uses quantum mechanics as an example of a new [“in our own time” (lines 40-43)] scientific research tradition that has not managed to overcome most people’s common-sense view of “the world as being populated by substantial objects, with fixed and precise properties” (lines 40-45).
B: Yes. The main idea of the third paragraph is that some world views persist despite the appearance of scientific traditions that contradict those views. The author offers quantum physics as an example of a scientific theory that has not managed to shake most people’s common-sense view of “the world as being populated by substantial objects, with fixed and precise properties” (lines 40-45).
C: This is the right answer to the wrong question. It is people’s belief in indetermination (lines 62-68), not quantum mechanics, that is incompatible with the idea that we live in a universe governed by natural laws (lines 68-72). D: See answer choice A. According to the passage, most people refuse to change their world view to accommodate the
95. C
A: This choice takes words out of the context of the passage. Our social, political, and moral beliefs contradict the idea that all physical changes are subject to the same unchanging natural laws (lines 62-72).
B: This choice also takes words and ideas from the passage out of context. Our social, political, and moral beliefs are
themselves a broader system of cultural attitudes. Such systems sometimes conflict with scientific theories and
traditions (lines 32-40, 62-72).
C: Yes. According to the passage, the scientific tradition that holds that all physical changes are subject to natural law has been accepted since the 17th century (lines 55-59). However, most of our social, political, and moral beliefs are inconsistent with the idea that these laws could apply to human beings and perhaps higher animals (lines 68-72).
D: While the author does argue that Darwinism and Marxism have been accepted by “reflective people” (lines 26-31), the author draws no direct connection between those research traditions and the social, political, and moral beliefs
discussed in lines 64-72.
96. B
A: Newton’s ideas were “eventually” (lines 21-22) accepted after a “process of readjustment” (line 26). Thus the words “readily” and “quickly” are too extreme.
B: Yes. This choice corresponds to the author’s description of the process of readjustment that led to acceptance of Newton’s ideas (lines 14-26).
C: The passage indicates that most people eventually modified their world view to bring it into line with Newton’s ideas (lines 14-26). There is no indication that the implications of Newton’s theories were ignored, or that most people’s acceptance of those theories was feigned or only superficial.
D: The Newtonian view of reality was eventually accepted, not rejected (lines 21-26).
97. D
A: The author refers to this claim in order to make the argument that some long-standing scientific traditions (in this case, since the 17th century) have still not been accepted by most people (lines 49-68), and that the strength of old world views may not necessarily fade over time. This choice is not consistent with the main ideas of the last two paragraphs. B: The author writes that natural laws may be either statistical or nonstatistical (lines 58-59). The passage gives no
indication that the laws governing human actions must be statistical in nature, nor does the application of physical laws to humans indicate that some physical changes can only be explained by statistical laws.
C: The passage states that in the 17th century, physical laws, not theories, were thought to apply to all physical objects or changes (lines 55-59). Secondly, it is the author, not (as far as we know) people in the 17th century who assume that these laws apply equally to human beings (lines 62-64).
D: Yes. The view that all physical changes are completely determined would, according to the author, apply to human beings as well (lines 62-64).
98. A
Item I: Yes. The new information in the question describes people changing their views of reality in order to accommodate new scientific findings.
Item II: See the explanation for Item I. The question gives an example of adaptability and acceptance of changing world views. This choice indicates that some new scientific ideas may never find acceptance in the face of
contradictory world views.
Item III: See the explanations for Items I and II. This choice describes resistance to, not acceptance of new scientific ideas (lines 62-72).
99. C
A: Remember the main idea of the passage. The author argues that behaviorism is not widely accepted due to the strength of the old world view in which inner mental states exist (lines 32-37, 45-47). The passage never indicates that the theory of behaviorism is itself weak.
B: This is the right answer to the wrong question. The author’s discussion of the application of natural law to human behavior comes later in the passage, in a different context (lines 55-68).
C: Yes. According to the author, the fact that people still believe in inner mental states indicates that the scientific tradition of behaviorism has not been able to supplant or transform an old world view that contradicts this
D: While behaviorism may be a relatively new (very new is too extreme) tradition, the author does not discuss it or the contradictory belief in inner states in order to make that claim. Always keep the main idea of the paragraph in mind when answering “support” or “in order to” questions.
Passage IV
100. D
A: The passage never suggests that CO2 is poisonous to fish. According to the CO2 theory, it could have been a decrease in the dissolved CO2 level in the ocean waters (and an increase in atmospheric CO2) that lead to extinctions of marine life by inhibiting the growth of algae, the base of the food chain (lines 56-62).
B: Dust is never mentioned in the context of the CO2 theory. This choice illegitimately combines aspects of the two different theories presented in the passage (lines 8-11 for the first, 56-62 for the second).
C: This choice contradicts the passage. Fish may in fact eat algae, or other creatures that themselves eat algae (lines 60-62). However, the increase in atmospheric CO2 would cause a decrease in CO2 in ocean waters (due to increases in temperatures) (lines 54-59). Decreases in dissolved CO2 would cause algae to decline, not to flourish (lines 59-60).
D: Yes. An increase in atmospheric CO2 would cause global warming (lines 46-51). Higher temperatures would cause less CO2 to dissolve in ocean waters (lines 56-59). As dissolved CO2 levels fell, so would the population of algae, which sit at the base of the oceanic food chain (lines 59-61). Disruption of the food chain could then have led to the extinction of a variety of marine species (lines 60-62), including species of fish.
101. A
Note: Notice the word “would” in the question. The credited response must be something that these measurements would,
not just could show, prove or indicate.
A: Yes. The CO2 theory posits that increases in atmospheric CO2 would have caused decreases in levels of CO2 dissolved in ocean waters (lines 56-59). Ice cores would indicate if such a change did in fact occur.
B: The new information in the question describes the possibility of measurement, not what those measurements would show. Compare this choice to answer choice A. This choice is too extreme.
C: The new information in the question states only that measurements could be made, not what those measurements would be. Even if it were shown that CO2 levels had fallen in the oceans at the K–T boundary, that would not be sufficient to
prove that the cause was an asteroid strike. This choice is also too extreme.
D: See the explanation for choice C. Even if decreased levels of oceanic CO2 levels were found (and we don’t know that they would be), that would be insufficient evidence to either disprove the dust scenario or prove the CO2 theory.
102. D
Note: Notice that the passage discusses only asteroid strikes on dry land (lines 8-11, 22) or in shallow ocean beds (line 22).
Avoid a trap by going back to the passage and re-reading carefully.
A: Such warming would occur if a 10 km asteroid hit in a bed of carbonate rock (lines 48-51), not in the deepest part of the ocean.
B: A large enough asteroid could cause mass extinctions if it hit a layer of carbonate rock [on dry land or in ocean shallows (lines 20-27)]. There is no indication in the passage that a hit in deep ocean would release CO2 into the atmosphere and set off a chain of events leading to large-scale extinctions.
C: Nowhere in the passage are extinctions limited to aquatic species discussed. Neither does the author present evidence suggesting that an asteroid hit in deep water would cause any extinction of marine species.
D: Yes. The author only discusses asteroid impacts which might send up clouds of smoke and dust (lines 8-11) (and so presumably occur on dry land), or which release CO2 into the atmosphere by colliding with beds of carbonate rock on dry land or in shallow water (lines 20-23, 45-46, 49-50).
103. A
A: Yes. The CO2 hypothesis is based on the claim that a major asteroid impact on a bed of carbonate rock would release CO2 into the air, causing global warming and a decrease in oceanic CO2 levels. If a major asteroid hit on carbonate rock were not followed by climatic change, it would significantly undermine the causal claims at the heart of the CO2 theory.
B: In the first paragraph, the author explains that high iridium levels at the K–T boundary inspired the idea that an asteroid struck the Earth at that time. The iridium evidence is consistent with both theories presented in the passage, and would support, not challenge the CO2 theory.
C: The CO2 theory argues that increased levels of atmospheric CO2 would lead to global warming (lines 28-32, 44-53). Further correlation between CO2 levels and global warming would strengthen, not weaken the CO2 theory.
D: The CO2 theory does not rest on the claim that only asteroid impact could lead to increases in global CO2. In fact, in the context of presenting the CO2 theory, the author describes other factors that increase atmospheric CO2 levels, such as burning of fossil fuels (lines 32-38).
104. C
A: The possibility of a loss of carbonate rock is not addressed by the passage. Even if we were to speculate that an asteroid hit that released CO2 from carbonate rocks would entail a reduction in the rocks’ mass (and this speculation would not be sufficiently supported by the passage), there is no connection to be made to an ash-spewing volcano.
B: The passage indicates just the opposite. Dust and smoke filling the atmosphere would block sunlight from reaching the Earth (lines 10-11). It is reasonable to assume that massive amounts of ash would have a similar blocking effect.
C: Yes. The author states that the dust and smoke from an asteroid impact could have temporarily cooled the Earth by blocking out sunlight (lines 8-17). It is reasonable to assume that massive amounts of ash in the atmosphere would have a similar effect.
D: This is the right answer to the wrong question. The passage makes no connection between dust, smoke, or ash in the atmosphere (the first theory presented, in the first paragraph) and increases in CO2 (the second theory described, in the rest of the passage).
105. D
A: This choice is too extreme. According to the passage, if the CO2 theory is valid, then high levels of CO2 may have lead to large-scale extinctions, including that of the dinosaurs, by causing global warming. However, the author does not suggest that all forms of life were or would have been destroyed. Finally, the wording of this answer choice (“survive in high levels of CO2”) appears to indicate that it is the CO2 itself that kills, not its indirect effects through global warming.
B: According to the passage, large amounts of dust and smoke in the atmosphere would lead to a rapid decrease in temperature and a temporary “winter” (lines 8-11). Beware of opposite answer choices.
C: The passage makes no such comparison. The only reference in the passage to a species that is directly sensitive to CO2 levels is to a marine species, algae (lines 57-60).
D: Yes. The author describes two theories in the passage, both of which explain the extinctions as the result of the impact of a large asteroid or comet. The author appears to prefer the CO2 theory as shown by the problems raised regarding the first theory (lines 13-17) and by the last line of the passage, “Thus, global warming could have led to the extinction of the dinosaurs.”
106. A
A: Yes. The author states that low-level global warming of 2–4 ∞C due to burning of fossil fuels may lead to melting of the polar ice caps and flooding (lines 38-43). We can infer that this flooding would be caused by an increase in sea level due to the melting ice. Sea levels would have risen much more dramatically at the K–T boundary if the temperature had increased 5–20 ∞C, lasting for 10,000 years (lines 46-53). Thus, if the main theory presented in the passage, the CO2 theory, is correct, we would find increased sea levels at the K–T boundary.
B: Excess iridium is mentioned in the context of the “dust and smoke” theory (lines 5-11), a theory which according to the author has serious weaknesses (lines 13-17). However, the presence of excess iridium would also be consistent with the CO2 theory, which also claims that the primary cause of extinction was an asteroid strike (lines 18-20). Thus, if the main hypothesis presented in the passage (the CO2 theory) is correct, excess iridium could well be present.
C: The main theory presented in the passage, the CO2 theory, argues that major extinctions did occur at the K–T boundary (lines 60-66).
D: As in choices B and C, this choice contradicts the passage. The CO2 theory is predicated on the claim that atmospheric CO2 levels rose (lines 23-27), not that they (and the composition of the atmosphere) remained constant.
107. B
A: The author believes that asteroids may have indirectly, not directly, caused the dinosaurs’ extinction. Global warming would have affected the entire planet; there would have been no need for the dinosaurs to be located near the impact site.
B: Yes. In both of the scenarios described, a consequence of the asteroid or comet impact would be slowing of plant growth (lines 11-13, 60-65). In the context of discussing the CO2 theory (the author’s preferred explanation), the author explains that reduction in plant growth disrupts the food chain, and so indicates that starvation was likely a major cause of extinction.
C: The author may well believe that dust filled the air after an impact, just that the dust itself was not sufficient to cause mass extinction (lines 13-20)
D: The passage states just the opposite. Decreased levels of CO2 in the oceans could have lead to a decrease in the algae population, and so disruption of the entire marine food chain (lines 56-62).
Passage V
Note: This passage is particularly confusing because the author’s own position on the issue of whether or not Picasso was in
fact a “cerebral structurer” as a young child, and whether or not he did in fact “begin again” as an adult artist is never made clear (see lines 8-12, 35-41). Note as well that the questions never ask for the author’s own views on those points. Don’t get distracted by things that are left unclear in the passage if they are not necessary for answering the questions.
108. D
A: The author argues that Picasso was not a genius or “frankly precocious” as a child (lines 41-45), nor was his work effortless (lines 22-26, 32-35). His achievements arose from the transformation or metamorphosis that came later (lines 43-45, 52-54).
B: While we know from the passage that Picasso and Braque invented cubism (line 10), the passage does not indicate that Picasso rejected the art movements of his time (he could have built upon them, for example), or that such rejection would itself be a cause of his artistic accomplishments. Be careful not to use outside knowledge.
C: Only one example of revision is given (lines 61-64). We do not know from the passage that Picasso had a practice of making many revisions, or that revision was a reason for his level of artistic achievement.
D: Yes. The primary purpose of the passage is to assess Picasso’s work along two related lines. First, what is the relationship between his mature work and his artistic creation as a child. Second, what is the relationship between his work as a child and as an adult, and a childlike or “unreflective” and spontaneous approach to art. According to Apollinaire and to the author, Picasso’s “spectacular progress” came when he transformed himself into an “unreflective virtuoso, who relies on nature” (lines 46-54). Picasso’s later apparent spontaneity is contrasted with his academic, literal, and precise work as a child around the age of nine (lines 24-29). 109. C
A: This choice is too literal. The passage never suggests that Picasso [an example of an artist who relies on nature (lines 50-52)] painted naturalistic scenes. In fact, the “fragmented and distorted” works from his childhood, the “harbingers [forerunners] of his later genius” (lines 3-7) indicate that he did not paint naturalistic scenes or objects. His unrealistic portrait of Gertrude Stein provides additional evidence along these lines (lines 59-64).
B: It is unlikely that the “unreflective virtuoso who relies on nature” would consciously refer to or employ defined aesthetic principles, even elementary ones. This is more likely true of an academic painter or “cerebral structurer.”
C: Yes. The author centers the passage on a basic contrast between childhood or childlike art on one hand, and academic, literal, and precise art on the other. The “cerebral structurer” represents the latter, while the “unreflective virtuoso” represents the former category. An adult artist who draws like a child works not from rules and tradition, but with intuitive spontaneity and adventurousness (lines 22-29, 59-64).
