Use these selected concepts to connect scientific principles with calculations, spatial interpretation, and passage analysis. Each focuses on a precise distinction, an original resolved example, and a specific mistake to avoid.
Cell transport and physiological feedback
1. Selective permeability and molecular properties
The membrane's hydrophobic lipid core admits small, uncharged, nonpolar molecules by simple diffusion, while ions such as sodium, potassium, calcium, and chloride are blocked by charge and require specialized transport pathways. This concerns lipid-core passage only, not transport direction.
Worked example: Comparing carbon dioxide with chloride ions, a student predicts carbon dioxide crosses the lipid core readily, while chloride needs a specific membrane pathway; this comparison alone indicates nothing about either substance's movement direction.
Mistake to avoid: Assuming a sodium ion crosses the lipid core easily just because it is small.
Reference: [4]
2. Diffusion gradients and dynamic equilibrium
Each solute diffuses down its own concentration gradient, independent of other solutes' gradients. For a permeable, uncharged solute, equal concentrations produce dynamic equilibrium: molecules keep moving randomly, but net transfer between the two sides stops.
Worked example: A permeable neutral solute sits at 10 units outside and 4 inside, so net movement occurs inward. Once both sides reach 7 units, molecules keep crossing both ways, yet the inside count no longer changes.
Mistake to avoid: Predicting glucose's diffusion direction from sodium's gradient instead of glucose's own concentration difference.
Reference: [4]
3. Facilitated diffusion and carrier saturation
Facilitated diffusion moves polar molecules and ions through channel or carrier proteins down the existing gradient, without cellular energy expenditure. Because any membrane contains a finite number of carriers, transport saturates once all are bound; a steeper gradient cannot raise the carrier system's maximum rate.
Worked example: A carrier system moves a solute at 15 units per minute when every carrier is occupied. Adding more solute steepens the gradient, but the rate stays near 15 because no free carriers remain.
Mistake to avoid: Labeling protein-mediated transport as active transport, forgetting facilitated diffusion follows the gradient without ATP spending.
Reference: [4]
4. Negative feedback and control-system roles
A sensor monitors a variable and reports it to a control center, which compares it with the normal range around a set point. If deviation is large enough, effectors are activated to reverse it; homeostasis maintains a fluctuating range, not a fixed value.
Worked example: When body warming exceeds the normal range, the brain's temperature regulation center activates sweating; sweat glands carry out the response, so they function as effectors promoting evaporative heat loss toward the set point's range.
Mistake to avoid: Calling sweat glands the sensor because they respond, when they actually act as effectors.
Reference: [6]
5. Positive feedback and its terminating event
Positive feedback intensifies a change in the body's physiological condition rather than reversing it, so the system moves farther from its normal range. Such a reinforcing loop is normal in the body only when there is a definite end point that stops the cycle.
Worked example: A diagram shows vessel injury initiating clotting-substance release, where each clotting step stimulates release of more clotting substances and accelerates clotting until the damaged area is sealed off; the definite endpoint is the sealed damaged area, not completed vessel-wall repair.
Mistake to avoid: Describing the reinforcing clotting cascade as a reversal of the initiating change instead of an intensification with a definite endpoint.
Reference: [6]
Genetics, evolution, and ecosystems
6. Allele segregation and offspring probabilities
Paired alleles segregate equally into gametes, so offspring have an equal likelihood of inheriting either factor. The physical basis is the first division of meiosis, in which homologous chromosomes with their different gene versions are segregated into daughter nuclei, making Punnett-square prediction accurate.
Worked example: In a Bb × bb cross, the heterozygote produces B and b gametes equally, so half the offspring are Bb and half bb; assuming complete dominance, the recessive phenotype therefore appears with probability one-half.
Mistake to avoid: Applying the 3:1 phenotypic ratio to any heterozygote cross without confirming complete dominance, which does not always hold.
Reference: [5]
7. Independent assortment in multigene crosses
When genes assort independently, alleles at each locus sort into gametes without influencing one another. To find a combined outcome's probability, determine each locus's probability separately and multiply using the product rule; this calculation is valid only when independent assortment actually applies.
Worked example: For AaBb crossed with aaBb at loci on separate chromosomes, the probability of aa is one-half and of bb is one-quarter; multiplying gives one-eighth aabb offspring via the product rule.
Mistake to avoid: Adding probabilities for independent outcomes that must occur together, instead of multiplying each locus's individual probability.
Reference: [5]
8. Linkage and recombinant allele combinations
Genes physically close on the same chromosome tend to be inherited as a pair, violating independent assortment. Recombination during meiosis combines maternal and paternal alleles onto one chromosome, and as distance between genes increases, crossovers become more likely, so distant loci can behave approximately independently.
Worked example: With homologous chromosomes carrying AB and ab, a crossover between the two loci produces Ab and aB gametes alongside the parental types, showing that recombination combines maternal and paternal alleles on one chromosome.
Mistake to avoid: Treating all same-chromosome genes as completely linked, ignoring that sufficient recombination makes distant loci behave approximately independently.
Reference: [5]
9. Heritable variation and natural selection
Natural selection acts only on heritable variation, because alleles pass from parents to offspring, while acquired traits generally are not inherited. Environmental conditions still influence phenotypes and can favor particular inherited alleles, but an environmentally induced difference alone does not demonstrate evolutionary change.
Worked example: In a hypothetical mouse population, individuals inheriting thicker fur survive cold winters and leave more offspring, so the fur allele rises in frequency; mice that grew heavier merely from abundant food show no inherited change.
Mistake to avoid: Assuming a phenotype change caused only by diet or climate is a genetic adaptation passed to offspring.
Reference: [7]
10. Genetic drift, bottlenecks, and founder effects
Genetic drift is change in allele frequencies caused by chance rather than advantage, and small populations are especially susceptible because random events can remove a large share of the gene pool (s7). A bottleneck samples random survivors of a catastrophe, while a founder effect samples the individuals that establish a new population.
Worked example: A hypothetical storm randomly kills most of an island lizard population, and the few survivors happen to carry a rare tail-color allele, which then becomes common; separately, five birds colonizing an empty island carry only their own allele mix.
Mistake to avoid: Treating every allele-frequency increase as adaptive selection when random survival in a small population can produce the same shift by chance.
Reference: [7]
11. Mutation versus gene flow
Mutation is a change to an organism's DNA and a major source of new variation; its effects may be beneficial, harmful, or neutral. Gene flow instead moves existing alleles between populations through migrating individuals or gametes, contributing genetic material when migrants reproduce.
Worked example: Wind-blown pollen introduces an existing flower-color allele into a plant population, illustrating gene flow; a newly altered DNA sequence within one plant's lineage, possibly beneficial, harmful, or neutral, is a mutation.
Mistake to avoid: Labeling an allele merely new to a population as a fresh mutation without evidence the DNA sequence changed.
Reference: [7]
12. Food webs, trophic levels, and energy loss
A food chain follows one linear feeding path, while a food web holistically models interconnected feeding relationships. Energy is lost as heat between trophic levels by the second law of thermodynamics, so after several transfers too little energy may remain for additional consumers.
Worked example: In a hypothetical stream, producers hold 1,000 energy units and herbivores receive 150, a measured 15 percent transfer; a fish eating both insect larvae and snails creates interconnected paths only a food web captures.
Mistake to avoid: Assuming a universal ten-percent trophic transfer rule instead of calculating energy loss from the ecosystem's stated measurements.
Reference: [8]
13. Ecosystem resistance versus resilience
Resistance measures how well an ecosystem stays near equilibrium during a disturbance, while resilience measures how quickly it returns to equilibrium afterward. A system can show one property strongly and the other weakly, and severe disturbance may eliminate resilience entirely.
Worked example: A hypothetical coral reef barely changes during a mild warming event, showing resistance, while a kelp forest loses canopy but recovers within two seasons, showing resilience; severe disturbance could still prevent either from recovering.
Mistake to avoid: Treating a large initial change in species composition as proof of slow recovery afterward.
Reference: [8]
General chemistry: quantities, acids, and gases
14. Balanced coefficients as mole ratios
Coefficients in a balanced equation state relative numbers of particles, so they serve equally as molecule ratios, dozen ratios, and mole ratios. This lets you convert the molar amount of one substance into the molar amount of any other substance in the reaction, provided the reaction proceeds as written and other reactants are available.
Worked example: For N2 + 3 H2 → 2 NH3, 1.5 mol H2 corresponds through the 3:2 ratio to 1.0 mol NH3, assuming the reaction occurs as written with sufficient nitrogen and complete conversion of the hydrogen supplied.
Mistake to avoid: Reading coefficients as gram ratios and multiplying a given mass directly by the coefficient ratio without converting to moles first.
Reference: [9]
15. Mass-to-mass stoichiometry
When a problem gives a mass and asks for a mass, three steps are needed: convert the starting mass to moles using molar mass, apply the stoichiometric factor from the balanced equation, and convert product moles back to mass. The mole ratio remains the essential link between the substances.
Worked example: For CH4 + 2 O2 → CO2 + 2 H2O with molar masses 16.0 and 44.0 g/mol, 32 g of methane corresponds to 88 g of carbon dioxide, assuming complete combustion with sufficient oxygen available.
Mistake to avoid: Skipping the mole conversion when the two coefficients happen to be equal, treating the masses as directly interchangeable.
Reference: [9]
16. Ionization constants and acid-base strength
The ionization constant Ka or Kb measures the extent to which an acid or base ionizes in water, so a larger constant means a stronger acid or base. Strength describes ionization extent, not how much solute is dissolved, so a dilute strong acid differs from a concentrated weak acid.
Worked example: At the same temperature in water, a base with Kb = 1.8 × 10−5 ionizes more fully than one with Kb = 5.6 × 10−10, so equal-concentration solutions of the first base contain more hydroxide ions, though concentration differences can reverse observed hydroxide amounts.
Mistake to avoid: Describing a more concentrated solution of a weak acid as a stronger acid, confusing solution concentration with ionization extent.
Reference: [10]
17. Conjugate acid-base strength
For any conjugate acid-base pair in water at a given temperature, Ka times Kb equals Kw, because adding the pair's two ionization equations yields water's autoionization. This creates an inverse relationship: the stronger one member is, the weaker its conjugate partner.
Worked example: At 25 °C, a weak acid with Ka = 2.0 × 10⁻⁵ has a conjugate base with Kb = Kw / Ka = 1.0 × 10⁻¹⁴ / 2.0 × 10⁻⁵ = 5.0 × 10⁻¹⁰, a genuinely weak base.
Mistake to avoid: Multiplying the Ka of one acid by the Kb of an unrelated base and calling the product Kw.
Reference: [10]
18. Boyle's law and inverse pressure-volume changes
For a fixed amount of ideal gas at constant temperature, pressure and volume are inversely proportional, so the product PV stays constant and P1V1 = P2V2. Use absolute pressure and consistent units, assuming ideal behavior and no phase change.
Worked example: A theoretical gas sample occupies 6.0 L at 0.50 atm; compressed at unchanged temperature to 1.5 atm, its volume becomes V2 = 0.50 × 6.0 / 1.5 = 2.0 L.
Mistake to avoid: Assuming volume rises in the same proportion when pressure increases, or applying Boyle's law while temperature changes.
Reference: [19]
19. Charles's law requires absolute temperature
For a fixed amount of gas at constant pressure, volume is directly proportional to kelvin temperature, giving V1/T1 = V2/T2. Temperatures must be converted to kelvin for any gas law calculation, because Celsius ratios give wrong answers.
Worked example: An ideal gas occupies 4.00 L at 250 K under constant pressure; heated to 375 K with no condensation, its volume rises to 4.00 × 375 / 250 = 6.00 L.
Mistake to avoid: Dividing volumes by Celsius readings, such as using 27 °C as 27 instead of converting it to 300 K.
Reference: [19]
General chemistry: thermodynamics and gas calculations
20. Free-energy sign and reaction direction
At constant temperature and pressure, ΔG's sign predicts spontaneity for the process as written: negative means the forward change is spontaneous, positive means it is nonspontaneous in that direction, and zero means equilibrium. Reversing a process flips the sign while keeping the magnitude.
Worked example: A hypothetical forward process has ΔG = +4 kJ at constant temperature and pressure, so it is nonspontaneous as written. The reverse process therefore has ΔG = −4 kJ and proceeds spontaneously under the same conditions.
Mistake to avoid: Reversing a reaction's direction while keeping the original ΔG sign instead of changing its sign.
Reference: [11]
21. Calculating ΔG with consistent units
Gibbs free energy change is ΔG = ΔH − TΔS. Enthalpy is usually tabulated in kilojoules while entropy appears in joules per kelvin, so convert one before subtracting. Temperature must be absolute kelvin, which is always positive, and matching units make the subtraction meaningful.
Worked example: A process has ΔH = 48 kJ and ΔS = 120 J/K at 350 K. Converting gives TΔS = 42 kJ, so ΔG = 48 − 42 = +6 kJ, meaning the process is nonspontaneous under these specified conditions.
Mistake to avoid: Subtracting a TΔS term in joules from a ΔH value in kilojoules, inflating the entropy term a thousandfold.
Reference: [11]
22. The ideal gas law connects amount with gas conditions
The ideal gas law PV = nRT links pressure, volume, molar amount, and kelvin temperature. The gas constant's form must match the units used for the other quantities. Ideal behavior is only a reasonable approximation at relatively low pressure and high temperature.
Worked example: A gas sample at 1.00 atm and 300 K occupies 4.92 L; using R = 0.08206 L atm mol⁻¹ K⁻¹, n = PV/RT = 4.92/(0.08206 × 300) ≈ 0.200 mol, under low-pressure conditions where ideal behavior is a reasonable approximation.
Mistake to avoid: Inserting a pressure in kilopascals into PV = nRT while using the atmosphere-based value of R.
Reference: [19]
23. Standard free energy and equilibrium constants
For the same reaction and temperature, ΔG° = −RT ln K, with T in kelvin and consistent energy units: K above one gives negative ΔG°, K below one positive. Q is the numerical value of the reaction's mass action expression, and at equilibrium Q = K, so actual ΔG is zero.
Worked example: At 298 K, a hypothetical dimerization with K = 0.01 gives ΔG° = −8.314 × 298 × ln(0.01), roughly +11.4 kJ/mol. Once the mixture reaches equilibrium, Q equals K and the actual ΔG becomes exactly zero.
Mistake to avoid: Equating actual ΔG with ΔG° at equilibrium, or comparing individual concentrations without examining the reaction's mass action expression.
Reference: [11]
Organic chemistry: structure, mechanisms, and analysis
24. Bond rotation and butane conformations
Rotation about butane's central C2–C3 bond interconverts conformations without changing atom connectivity. Viewed along C2–C3, the anti conformation places the methyl groups 180° apart, while the gauche conformation places them 60° apart. This concept does not establish a complete conformational energy ranking.
Worked example: Starting from anti butane at 180°, two successive 60° rotations of one carbon in the decreasing-angle direction give 120°, then 60°; the resulting gauche arrangement is still butane, its connectivity unchanged.
Mistake to avoid: Measuring the dihedral angle around an unrelated bond, or treating a rotated conformation as a different constitutional isomer.
25. Curved arrows track electron movement
On the DAT, a full-headed curved arrow represents movement of an electron pair, while a half-headed fishhook arrow represents movement of a single electron. Arrows track electrons, not atoms; predicting a complete product requires the starting structure plus all relevant arrows.
Worked example: One full-headed curved arrow drawn from a bond shows two electrons moving together; two separate fishhook arrows from that same bond show each electron moving independently, as in homolytic cleavage.
Mistake to avoid: Reading a curved arrow as showing atom movement, or treating a fishhook arrow as a two-electron arrow.
Reference: [3]
26. Electronegativity and acidity across a period
Across a period, the bonded nonmetal's electronegativity increases, which raises the polarity of the H–A bond, so acidity of the binary hydrogen compounds increases; this concerns removal of one proton from these specified compounds, not substituted organic molecules.
Worked example: Comparing phosphine (PH3) with hydrogen sulfide (H2S) across period 3, sulfur's higher electronegativity polarizes the H–S bond more, so the rule predicts H2S is the stronger acid of the pair.
Mistake to avoid: Applying this electronegativity-only trend to substituted organic acids where resonance and other structural effects govern acidity.
Reference: [10]
27. Bond strength and acidity down a group
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals increases down a group as the H–A bond strength decreases, as in group 16: H2O < H2S < H2Se < H2Te. In water, hydronium is the strongest acid that can exist, so distinguishing strong acids requires a solvent less basic than water.
Worked example: Comparing H2S with H2Te in a solvent less basic than water, the weaker H–Te bond makes H2Te the stronger acid, following the group 16 order of increasing acid strength.
Mistake to avoid: Claiming aqueous pH readings distinguish equally concentrated, fully ionized strong acids despite water's leveling effect.
Reference: [10]
28. Carbonyl addition changes bonding and geometry
A nucleophile adding to an aldehyde or ketone bonds to the electrophilic carbonyl carbon as the C=O π-electron pair shifts toward oxygen. In the anionic-nucleophile pathway, the carbon rehybridizes from sp2 to sp3, producing a tetrahedral alkoxide intermediate; protonation is a separate step.
Worked example: In a schematic addition to acetaldehyde, CH3− attacks the carbonyl carbon, the π electrons move onto oxygen, and the tetrahedral alkoxide CH3–CH(O−)–CH3 forms; a later protonation step would give the alcohol.
Mistake to avoid: Drawing the new C–Nu bond while keeping the C=O double bond, giving carbon five bonds.
Reference: [18]
29. Substitution and elimination depend on combined conditions
Predicting substitution or elimination requires the substrate together with nucleophile basicity, base steric bulk, and solvent; substrate class alone does not settle the mechanism. These are trends for the stated alkyl halides rather than rigid universal answers.
Worked example: The same secondary, nonallylic alkyl halide undergoes SN2 with a weakly basic nucleophile in a polar aprotic solvent, but replacing those conditions with a strong base favors E2 instead.
Mistake to avoid: Selecting SN1 for every secondary substrate without checking the nucleophile, base, and solvent conditions.
Reference: [20]
30. Infrared interpretation combines band position and shape
IR evidence supports functional-group patterns, not a unique complete structure. A terminal alkyne shows a sharp, fairly intense C–H stretch near 3300 cm−1 plus a 2100–2260 cm−1 triple-bond band, whereas an alcohol's O–H band near 3400–3650 cm−1 is usually broad; N–H bands overlap this region.
Worked example: A hypothetical spectrum shows a sharp band near 3300 cm−1 and a band at 2150 cm−1; together they support a terminal alkyne pattern, though the data do not specify one unique molecule.
Mistake to avoid: Calling every sharp absorption near 3300 cm−1 an alcohol O–H stretch while ignoring band shape.
Reference: [21]
31. Aldehyde–ketone reactivity combines steric and electronic effects
Aldehydes are generally more reactive than ketones in nucleophilic addition when substrates and conditions are otherwise comparable: a nucleophile approaches an aldehyde's carbonyl more readily because only one substituent crowds it, and only one alkyl group inductively stabilizes the carbonyl carbon's partial positive charge.
Worked example: Comparing ethanal with acetone under comparable conditions, predict easier nucleophilic addition to ethanal, because hydrogen is one carbonyl substituent and only one methyl group donates inductively, unlike acetone's two methyl groups.
Mistake to avoid: Explaining the reactivity trend only by counting substituents while ignoring their inductive electron donation.
Reference: [18]
32. Conformational strain and equilibrium populations
Not all staggered conformations share one energy: in butane viewed along C2–C3, gauche lies 3.8 kJ/mol above anti because its methyl groups experience steric strain even without eclipsing interactions. At room temperature all conformers interconvert in equilibrium, and a larger percentage occupies the more stable arrangement.
Worked example: Two explanations claim anti and gauche butane are equally stable because both are staggered, or that gauche is higher because its methyl groups crowd each other; the second is correct, and interconverting conformers still persist.
Mistake to avoid: Equating absence of eclipsing interactions with absence of strain, or treating the anti conformation as permanently fixed.
Reference: [23]
Perceptual ability: spatial interpretation
33. Aperture entry uses one fixed orientation
Under the ADA aperture rules, the object may be turned in any direction before entry begins and may start on a side not shown, but once passage starts it must continue straight through without turning or twisting.
Worked example: A T-shaped block can begin entering an opening on a side not shown, which the rules permit, but if finishing the passage would require a mid-course twist, that opening is invalid.
Mistake to avoid: Imagining a rotation partway through the opening to make an incompatible aperture seem passable.
Reference: [13]
34. Aperture matching requires shape and scale
An aperture must equal the appropriate external outline of the object in both shape and drawn scale. An opening with the correct silhouette but visibly smaller overall must be rejected, since the official rules guarantee size differences are large enough to judge by eye.
Worked example: A T-shaped block is shown alongside two openings with identical T silhouettes; one opening's stem and arms are visibly narrower throughout, so that undersized opening is rejected despite the matching shape.
Mistake to avoid: Mentally shrinking the object to squeeze it through an opening that is visibly too small.
Reference: [13]
35. Hidden geometry follows the aperture assumptions
Aperture items assume there are no irregularities in any hidden portion of the object, and symmetric indentations continue symmetrically. These are item-specific assumptions rather than properties of real objects, so a matching outline must not be rejected for imagined hidden bumps or asymmetry.
Worked example: A block bears a symmetric U-shaped groove on its visible face; the rules keep such indentations symmetric on the hidden portion, so an invented hidden ledge cannot justify rejecting an otherwise matching outline.
Mistake to avoid: Adding unseen protrusions or hidden asymmetry that the aperture rules explicitly exclude from every item.
Reference: [13]
36. View recognition uses parallel projection
Top, front, and end views are projections along parallel lines of vision, without perspective. Parallel edges stay parallel and equal lengths keep their drawn size regardless of depth, so farther edges are never drawn smaller. Respect the fixed labeled positions of each view.
Worked example: A rectangular solid five units wide, two deep, and three high yields front, top, and end outlines of five by three, five by two, and two by three, with no edge drawn smaller for depth.
Mistake to avoid: Drawing farther edges progressively smaller as in a perspective sketch, contradicting the parallel lines of vision.
Reference: [13]
37. Dotted lines represent hidden features
A dotted line marks a feature hidden from that view's direction. One dotted line alone does not reveal the feature; the full set of views is needed, since the front view may rule out a hole that the top view's outline suggests.
Worked example: A slot in a block's top face appears as solid outlines in the top view, where its opening is visible, but as dotted lines in the front view, where the slot's floor is hidden.
Mistake to avoid: Treating every dotted line as a hole or as a visible surface boundary in that view.
Reference: [13]
38. A missing view must agree with both given views
In view recognition items, two views are shown and the missing view may be top, front, or end. A proposed view must agree with the shape information in both supplied projections, and the views are drawn without perspective, using parallel viewing directions. Two supplied projections do not uniquely determine every possible solid.
Worked example: A face-aligned cuboid has a front view 6 wide by 2 high and a top view 6 wide by 4 deep; its end view is 4 by 2, so a 6-wide end option is wrong.
Mistake to avoid: Accepting an option because it agrees with the front view while ignoring contradictory information in the top view.
Reference: [13]
39. Unfolding restores hole positions
Paper folding items require mentally reversing each depicted fold, in reverse order, to locate every punched hole on the original square. The paper is never turned or twisted during folding, so the original orientation is preserved, and each hole must be mirrored across every crease it crosses when unfolded.
Worked example: A unit square folded along the lines x equals 0.5 and y equals 0.5 into its lower-left quarter receives a punch at the point (0.2, 0.3), which unfolds to four holes at (0.2, 0.3), (0.8, 0.3), (0.2, 0.7), and (0.8, 0.7).
Mistake to avoid: Rotating the paper between folds, or stopping after reversing only one fold and leaving remaining creases unmirrored.
Reference: [13]
40. Layer count determines replicated holes
A punch penetrates every paper layer overlapping at that location, producing one hole per pierced thickness that separates into distinct positions when the paper is unfolded. Candidates should count the layers actually present at the punch point, because some locations may not contain every folded layer of the sheet.
Worked example: A punch made through four fully overlapping layers of a folded sheet, positioned away from every crease and edge, yields four separate holes on the original square once all folds are reversed.
Mistake to avoid: Doubling the hole count for every fold without verifying which layers the punch actually crosses at its location.
Reference: [13]
41. Painted faces depend on exposure and ground contact
Cube groups are assembled first and painted afterward, so any cemented face receives no paint, and the bottom resting on the ground stays unpainted. Count painted faces by viewing the complete stack from all sides, since faces hidden in the drawing may still be exposed.
Worked example: In a four-cube L resting on the ground, the corner cube touches two neighbors and the ground, leaving three painted faces, while each endpoint cube touching one neighbor shows four.
Mistake to avoid: Assuming every downward-facing face is unpainted, when an elevated cube's exposed bottom face actually receives paint.
Reference: [13]
42. Hidden supporting cubes belong in the inventory
Because the only hidden cubes are those required to support other cubes, add exactly the cubes needed beneath visible cubes to hold them at their drawn heights, and avoid inventing extra hidden cubes the geometry does not require.
Worked example: A visible cube sitting one level above the ground with its support obscured requires exactly one supporting cube beneath it, in addition to the visible neighbors already shown in the drawing.
Mistake to avoid: Counting only visible cubes, or inventing an unnecessary hidden row that no visible cube's position requires.
Reference: [13]
Reading scientific passages
43. Main ideas versus supporting details
A main-idea summary restates the passage's central explanation in your own words, checking comprehension. In informational texts, repeated statements or sentences that pull ideas together often signal the main idea, while vivid individual details merely illustrate that broader message.
Worked example: An invented passage explains three water-quality measurement methods and illustrates one with a single pond reading; its main idea concerns the methods, so a summary built on that lone pond result misses the central message.
Mistake to avoid: Selecting the vivid pond result as the passage's overall message instead of judging the whole passage.
Reference: [14]
44. Purpose follows the passage's function
Determine a passage's purpose by examining what its language and reasoning accomplish across the whole text, whether explaining, evaluating, or arguing. For argumentative texts, examine the claim, reasoning, and supporting evidence; a single adjective does not by itself determine purpose.
Worked example: An invented passage compares two instruments, explains each one's limitations, and recommends one for a defined application; its purpose includes evaluation and recommendation, so labeling it purely descriptive ignores the weighing the passage performs.
Mistake to avoid: Labeling the passage purely descriptive while ignoring its recommendation, or inferring argument from one adjective.
Reference: [14]
45. Claims, evidence, and reasoning
When analyzing an argument, label the author's conclusion as the claim, identify the observations offered in its support as evidence, and describe the logic that links them as reasoning. Sorting these roles clarifies the argument's structure, though it does not by itself confirm that the argument is valid or well supported.
Worked example: A lab report states that enzyme X degrades starch faster at 37°C; the cited measurements show faster glucose release at that temperature, and the author reasons that higher kinetic energy accelerates catalysis.
Mistake to avoid: Treating the author's final conclusion as if it were a directly measured observation rather than an interpretation drawn from the data.
Reference: [14]
46. Paraphrase preserves meaning; summary selects main ideas
A paraphrase restates a passage in fresh wording while keeping its full meaning, including any limits on population, setting, or certainty. A summary compresses only the main ideas. Neither may add conclusions the passage does not support, and both require crediting the original source.
Worked example: A passage reports that a whitening gel reduced surface staining in a two-week trial on extracted teeth; a faithful paraphrase keeps the trial length and extracted-teeth setting, while a summary notes only the main finding with those limits intact.
Mistake to avoid: Dropping a limiting condition, such as the study population or duration, while keeping the passage's general topic.
Reference: [14]
47. Inferences are limited by available evidence
A sound interpretation follows from the text's own evidence and reasoning: examine the claim, reasoning, and each piece of supporting evidence, and locate specific textual evidence for your analysis. A reported result in some examined cases supports that result in those cases; it does not by itself establish an unstated universal conclusion, though if a passage supplies a basis for broader inference, evaluate that reasoning and its stated limits.
Worked example: A fictional passage reports reduced demineralization in eighteen of twenty examined teeth and supplies no basis for a universal conclusion. It supports the reported result in those eighteen teeth, not in all twenty or every tooth elsewhere.
Mistake to avoid: Adding a universal conclusion about every tooth that the passage's evidence and reasoning do not support.
Reference: [14]
48. Alternative explanations test causal conclusions
A proposed cause is stronger when the author has considered whether another explanation could fit the same observations. Noting a plausible alternative weakens the claim that the author's interpretation is uniquely established, although it does not by itself prove that the alternative explanation is correct.
Worked example: A report links frequent snacking to more enamel erosion among children who also drink more acidic beverages; because both habits differ together, the observations alone cannot isolate snacking as the cause of the erosion.
Mistake to avoid: Treating simultaneous differences as proof that one selected factor caused the outcome while ignoring other co-occurring variables.
Reference: [14]
49. Tone needs evidence from wording
Tone is inferred from the author's specific word choices, emphasis, and presentation rather than guessed from topic alone. Point to particular diction or framing that supports your reading, and weigh each expression in its context, since a single phrase may not characterize the whole passage. Qualifying language often signals cautious rather than extreme attitudes.
Worked example: A report concludes that early results 'warrant further study before routine use.' The hedge 'before routine use' establishes caution or reservation about the intervention, so an answer describing the author as reserved fits; the wording alone does not establish support for it.
Mistake to avoid: Choosing an extreme attitude while overlooking hedges such as 'preliminary' or 'may' that qualify the author's actual stance.
Reference: [14]
50. Paragraph roles reveal argument structure
A paragraph's role is the work it performs for surrounding ideas, such as presenting evidence, conceding a limitation, or qualifying earlier claims. Judge that function from relationships between paragraphs, without assuming a uniform structure, and answer with the job performed rather than the topic.
Worked example: An invented passage reports strong trial outcomes, then the next paragraph notes that participants were unusually healthy volunteers. That paragraph's role is limiting how far the favorable results generalize.
Mistake to avoid: Summarizing a paragraph's topic, such as 'it discusses sample size,' instead of naming the work it performs.
Reference: [14]
Quantitative reasoning
51. Equivalent operations in linear equations
Equality survives only when the same valid operation is applied to both sides, and distribution must reach every term the multiplier affects. Multiplying or dividing an entire side means each term on that side, not just the first; division additionally requires a nonzero divisor. Checking the answer by substitution confirms the result.
Worked example: Solving 5(x − 3) + 2 = 17 requires distributing 5 to both x and −3, giving 5x − 15 + 2 = 17, so 5x = 30 and x = 6. Substituting back yields 17, confirming the solution.
Mistake to avoid: Multiplying only the first term inside parentheses by the outside factor, leaving later terms unchanged and breaking the equation's balance.
Reference: [15]
52. One solution, no solution, or infinitely many
Once a linear equation is simplified, the remaining statement determines the solution set. A nonzero coefficient on the variable gives exactly one solution; variables canceling to a false numeric claim, such as 5 = 8, gives no solution; canceling to a true identity gives infinitely many solutions. These conclusions fit unrestricted real-variable linear equations.
Worked example: Simplifying 4x + 9 = 4x − 2 leaves 9 = −2, a false statement, so no value of x works. Changing −2 to 9 would instead produce the identity 9 = 9 and infinitely many solutions.
Mistake to avoid: Concluding x = 0 after variable terms cancel, when the leftover constant statement actually shows no solution or infinitely many.
Reference: [15]
53. Rearranging formulas for a requested variable
To rearrange a formula, apply the same operations to both sides until the requested variable stands alone with coefficient 1 and all other variables and constants sit on the opposite side. Valid operations may be applied in varying orders, and any divisor must be nonzero.
Worked example: From A = bh/2, isolating h gives h = 2A/b; dividing by b first or doubling first both work. With A = 24 square units and b = 8 units, h = 6 units.
Mistake to avoid: Dropping the factor of 2 when rearranging the triangle area formula for height.
Reference: [15]
54. Building a linear model from a word problem
When a scenario charges a one-time fixed amount plus a constant rate per unit, write the total cost as the fixed amount plus the rate times the count of units. Set that expression equal to the stated total, then solve the resulting linear equation for the count, checking that the answer satisfies the original wording.
Worked example: A printing order costs 12 units for setup plus 3 per booklet. A total of 57 gives 12 + 3n = 57, so n = 15 booklets.
Mistake to avoid: Treating the one-time setup charge as a per-booklet rate and multiplying it by the number of booklets ordered.
Reference: [15]
55. Conditional probability restricts the denominator
A conditional probability updates an event's likelihood once another outcome is known. The denominator becomes the outcomes consistent with the given condition, not the full sample. Count-based ratios apply when the remaining eligible outcomes are equally likely within the condition's restricted group.
Worked example: One of 20 samples is chosen uniformly at random; 8 carry label A and 3 of those contain marker X. Given label A, those eight remain equally likely: P(X given A) = 3/8, not 3/20.
Mistake to avoid: Keeping the whole-sample denominator of 20 after conditioning on the subgroup of 8 labeled samples.
Reference: [16]
56. Joint probabilities account for dependence
For a two-stage experiment, the probability that both stages occur equals the first stage's probability times the second stage's probability given the first result. When stages depend on each other, the second probability must reflect the changed conditions, such as a reduced pool after removal without replacement.
Worked example: A bag holds 5 red and 3 blue tokens. Drawing two tokens without replacement, P(both red) equals 5/8 times 4/7, which simplifies to 5/14.
Mistake to avoid: Using 5/8 for the second draw after a red token has been removed.
Reference: [16]
57. Complement probabilities simplify at-least-one events
When a question asks for the probability of at least one occurrence, direct enumeration requires summing many mutually exclusive paths. The complement—no occurrences at all—is usually a single path whose probability follows from repeated multiplication. Subtracting that complement from one yields the desired probability, a technique the birthday problem illustrates clearly.
Worked example: Two independent trials each succeed with probability 1/4. The probability of at least one success is one minus the probability that both trials fail: 1 − (3/4)² = 1 − 9/16 = 7/16.
Mistake to avoid: Subtracting the probability of exactly one success from one instead of subtracting the probability that every trial fails.
Reference: [16]
58. Unordered outcomes combine exclusive paths
Some questions count an outcome regardless of the order in which its parts appear. Each possible ordering forms a separate path through a tree diagram, and multiplying along a path gives that path's probability. Because the orderings cannot happen together, the Addition Rule lets you add the path probabilities directly.
Worked example: A box holds 2 red and 3 blue tokens drawn uniformly without replacement. One of each color occurs as red-then-blue with probability 2/5 × 3/4, or blue-then-red with probability 3/5 × 2/4; these exclusive paths sum to 3/5.
Mistake to avoid: Counting only one drawing order when the question accepts either order, which understates the probability of the combined outcome.
Reference: [16]
59. Frequencies versus proportions
Raw frequencies are hard to interpret without knowing the total, since twelve responses mean different things for twenty-five versus a thousand participants. Dividing each category's frequency by its group's total yields a proportion, expressible as a fraction, decimal, or percentage. Proportions normalize the counts, allowing fair comparison between groups of different sizes.
Worked example: Two clinics report positive samples using the same category definition. Clinic A records 18 positive results among 60 tests, a proportion of 30%, while Clinic B records 20 positive results among 100 tests, or 20%; Clinic A's rate is higher despite its smaller count.
Mistake to avoid: Comparing raw positive counts across groups with different totals, as if both denominators were equal.
Reference: [17]
60. Histograms versus bars for labeled measurements
Histograms group numerical observations into adjoining intervals, with touching bars whose heights show frequencies. Bar charts for labeled data instead display one bar per label, such as a team or school, with the vertical axis labeled in the measurement's units rather than frequency.
Worked example: Measurements 2, 4, 7, and 8 fill equal-width bins [0,5) and [5,10) with two observations each, so bars of height two touch at boundary 5; a value of 5 belongs only to the second bin.
Mistake to avoid: Reading a histogram bar's height as the measured value rather than the count of observations in that interval.
Reference: [17]
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