AAMC MCAT test 5r a

Second, the author cites the critics’ claim that local and state competition does not contribute to national productivity lines 39-43.. C A: According to the passage, local leaders in th

MCAT PRACTICE TEST 5R

SOLUTIONS

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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 betweenchoices 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 isillustrated 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 fromReaction 1 that neither FeO nor Al2K2O4 is consumed in the reaction We conclude that this mixture is a catalyst for thereaction, and the role of a catalyst is to increase the reaction rate

4 D Since NH4+ is a cation with a +1 charge, and SO42– is an anion with a –2 charge, we would expect that a combination

of these ions would contain 2 NH4+ ions for each SO42– ion; that is, the compound would be (NH4)2SO4 The balancedacid–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 dipolemoment, we would expect dipole–dipole interactions between NH3 molecules In addition, NH3 is capable of hydrogenbonding (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 strongerbase 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

m t

F m

Since F is the same for both fragments but m is different, the fragments will have different KE values at each moment t.

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 range attractive force that balances the large repulsive force between the positive charges in the nucleus Once the nucleusget too large (that is, once the nucleus contains too many protons), we conclude that the short-range strong nuclear forcebecomes unable to hold the nucleus together, which is why such large nuclei do not occur naturally (they’ve spontaneousdecayed) The statement in choice C provides a reasonable and direct answer as to why the strong nuclear force becomesunable to hold large nuclei together

short-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 twonuclei together, there is no need to “overcome” an attractive force, eliminating choice D The answer must be A Sinceboth 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 • + OReaction 5: ClO • + O Cl • + O

2

ÆÆ

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 + OReaction 2: O + O ONet reaction: O O + O

2

ÆÆÆ

2

+

Now, to determine whether the overall reaction involves an increase or a decrease in free energy, we calculate DGrxn usingthe values for DGf given in Table 1:

kJ mol

kJ mol

Because DGrxn is positive, we can eliminate choices A and B Choice D is eliminated since it indicates a negative activationenergy (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:

J mol K

J mol K

18 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 drivethe 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 theatmosphere, 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 H2PO4– loses an

H+, it becomes HPO42– Therefore, HPO42– is the conjugate base of H2PO4–

22 C The equilibrium that would best account for an increase in pH would show the formation of OH– ions, so weeliminate choice A The reactions in choices B and D are incorrect, since, for example, neither is balanced electrically norstoichiometrically, 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(NH4)2HPO4 is a basic salt, so A is a better response than C

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 NaClNotice 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 isequivalent 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 MnO4– 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

27 B The circumference of the circular path is C = 2 pr = 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 atotal 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

29 D We apply Le Châtelier’s principle To increase the yield of the product, (CP)n, in Equation 1, we could increase theconcentration of the reactant (which would shift the equilibrium toward the product side) Therefore, we would expect thatincreasing 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 stoichiometriccoefficient of HPO42– is n Therefore, the concentration of (CP) n is 1/n times the concentration of HPO42–

32 C We use the Henderson–Hasselbalch equation:

[[

-

-= fi = =

Passage VI

33 A Because 14C undergoes beta decay, it will not emit an alpha particle or neutron in the decay process; this eliminateschoices 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

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

-1 2

2

1 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 electricalimpulses, 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 thedecaying object itself), thereby overestimating the radiation energy and rate of decay This would clearly produce aninaccurate 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

-

-hc J s)(3 10 m s)

450 m

J m4.5 m

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 inthese compounds, we would select choice C over choice D Furthermore, since Cl is higher in the periodic table than I, wewould 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 moleculeCd(NO3)2 to be neutral Since the chloride ion, Cl–, also has a –1 charge, we expect the combination of Cd and Cl to beCdCl2 and the balanced reaction between Cd(NO3)2 and NaCl to be Cd(NO3)2 + 2 NaCl Æ CdCl2 + 2 NaNO3

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 compoundswith 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 theaddition of HNO3, HNO3 must react with copper metal Copper metal (Cu0) must be oxidized during this reaction, andHNO3 must be reduced Of the remaining choices (A, C, and D), the only logical choice for the product of the reduction ofHNO3 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 NO3–) whileAgNO3 dissociates into only 2 ions (Ag+ + NO3–), we’d expect the boiling point of Zn(NO3)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 “newmetal [that] forms on the surface of the Cu strip.” Also, note that we can eliminate choices A, B, and C, since it is highlyunlikely 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

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, A sT4, 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 secondincreases by a factor of ( 2)4 =[( 2) ]2 2 =[ ]22 =4

48 A We use the equation q = mc DT, 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

-◊

D (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

V R

1 1

5 5

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:

1.04.6

54 B Because opposite charges attract, the negatively-charged particle will move toward the fixed positive charge Q; this

eliminates choices A and C

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 inTable 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 anuclear 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 haveidentical (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?) Theoxygen 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 solublethan toluene in H2O Choice B is false (since there are no hydrogen bonds between toluene and water to compare withthose 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 experiencesfriction 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 +

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

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 v 0y = 0, since the ball is thrown horizontally), we find that

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 atPoint B Therefore,

2 2

2

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

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 resultantstanding wave has displacement zero at the node positions (where the equal-amplitude traveling waves arrive exactly out ofphase 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 theanswer must be B (Note that we didn’t need to check Item III, but, if we did, since laser light is sharp and focused, itwould 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

Independent Questions

72 C By Archimedes’ Principle, the buoyant force exerted by the unknown liquid on the object is given by Bunknown =

B B

Vg Vg

125

710

8450

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,

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, theanswer 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.

divergingcorrective lens

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 = rH2OVg Thus, the ratio wsolid/B is

w B

Vg Vg

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.

B: As in choice A, this statement contradicts a main idea of the passage—that competition can have negative effects Asecondary 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).

80-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 decreasecompetition between jurisdictions (lines 70-83) If these efforts were followed by prosperity, the lesson would be tocooperate more, not less (and so compete less, not more) For both choices A and B, remember that the main idea of thepassage 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 theface of a regional economic downturn (lines 70-83) If this cooperation were followed by prosperity, less competitionwould 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 passagedraw 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 localjurisdictions (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 crossedstate 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 governmentsfailed (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 doesnot indicate that this can be extrapolated into a generalized comparison between the effectiveness of state and localleaders

D: This choice is both out of scope and too extreme The power of national governments to regulate competition is neverdiscussed Furthermore, the Monongahela case is an example of local leaders successfully regulating or limiting localcompetition (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 reasonwhy 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 inthe 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 ofschema) are activated in situations that are similar to events and experiences from our past In cases where scripts areactive, 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

restaurant scenario as an example of how this occurs.

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 ofthe 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 thatthe 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 afterinstantiation (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 existing schema (lines 12-15)], not its quantity

pre-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 fromthe 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).

D: Instantiation of multiple schemas is never discussed in the passage

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 knowledgestructures 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 memorystructures, but only the current text itself

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 quantummechanics, 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 theinsights of quantum mechanics (lines 40-45)

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), theauthor 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’sacceptance 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 noindication 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 thatthese 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 thetheory 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 humanbehavior 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 scientific theory (lines 32-40).

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 twodifferent 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 62) However, the increase in atmospheric CO2 would cause a decrease in CO2 in ocean waters (due to increases intemperatures) (lines 54-59) Decreases in dissolved CO2 would cause algae to decline, not to flourish (lines 59-60)

60-D: Yes An increase in atmospheric CO 2 would cause global warming (lines 46-51) Higher temperatures would cause less CO 2 to dissolve in ocean waters (lines 56-59) As dissolved CO 2 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 CO 2 theory posits that increases in atmospheric CO 2 would have caused decreases in levels of CO 2 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 wouldshow 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 thatthey 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 ofthe 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 theatmosphere 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 evidencesuggesting 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 CO 2 into the atmosphere by colliding with beds of carbonate rock on dry land or in shallow water (lines 20-23, 45-46, 49-50).