The BioMedical Admissions Test (BMAT) was withdrawn in 2024. Use these selected concepts to review foundational distinctions, interpret original examples and recognise specific errors. Each group moves from basic relationships towards applications. Compare each example with its assumptions before applying the same reasoning to a different problem.
Reasoning and evaluating arguments
1. Separate premises from the conclusion
Identify which statement the argument is trying to establish and which statements are offered in its support. The supporting statements are the premises and the supported statement is the conclusion. In an inspection argument, the batch record and the item's batch membership work together as premises, while the claim that the item passed inspection is the conclusion they support.
Worked example: Example: a laboratory log states that every sample in consignment 12 was tested, a register shows this blood vial belongs to consignment 12, so the technician concludes the vial was tested; the two records are premises and the tested claim is the conclusion.
Mistake to avoid: Avoid counting unrelated background details, such as the vial's storage temperature, as premises supporting the conclusion.
Reference: [3]
2. Distinguish validity from soundness
Validity concerns only whether the conclusion follows from the premises, while soundness additionally requires every premise to be true. Assess these separately: a structurally perfect argument can rest on a false premise, and an argument with a true conclusion can still be invalid. Check premise truth independently of the argument's form before accepting its conclusion.
Worked example: Example: suppose a hospital policy states that all ward four patients have broken limbs; a patient is on ward four; the conclusion that she has a broken limb follows validly, yet the false policy premise makes the argument unsound.
Mistake to avoid: Avoid accepting a premise as true simply because the argument leading from it is structurally valid.
Reference: [3]
3. Recognise deductive and inductive reasoning
Deductive reasoning aims at a conclusion that its premises require, whereas inductive reasoning extends an observed pattern towards a broader claim. An inductive argument is judged strong or weak according to the evidence offered, and even a strong sample-based argument leaves room for exceptions, so its general conclusion remains supported rather than guaranteed.
Worked example: Example: three randomly chosen syringe packets from a shipment are found underweight; this pattern reasonably supports concern about the shipment's packaging but does not prove every packet is underweight, because the remaining packets were never individually checked.
Mistake to avoid: Avoid presenting a sample-based generalisation as a necessary conclusion that the sample alone cannot guarantee.
Reference: [3]
4. Apply a conditional when its antecedent holds
When a conditional states that if P then Q, and the antecedent P is known to hold, the consequent Q follows. This inference depends on the conditional holding without exception; if other conditions could also produce Q, the link from Q back to P fails, so the antecedent must never be inferred from the consequent.
Worked example: Example: a storeroom rule says that every container stamped with a red tag must be quarantined; a container carries the red tag; the rule therefore requires its quarantine. The reverse fails, since a container could be quarantined for damage without carrying any red tag.
Mistake to avoid: Avoid reversing the rule by inferring the red tag from quarantine alone, since another condition could require it.
Reference: [3]
5. Reject an antecedent when its consequence is absent
Modus tollens lets you deny the antecedent when its consequence fails: from 'if P then Q' and 'not Q', infer 'not P'. The rule makes Q necessary whenever P occurs, so Q's absence rules P out. The absence of Q must be established, not merely unreported, because another process could still produce Q.
Worked example: Every accepted application receives an acknowledgement letter. Priya's application received no acknowledgement letter, and the admissions office confirmed none was sent, so her application was not accepted under that rule; a lost letter would undermine this reasoning.
Mistake to avoid: Inferring that no receipt was issued merely because acceptance was not reported, when another office could have sent one.
Reference: [3]
6. Join compatible conditional statements
The chain rule combines compatible conditionals: if P implies Q and Q implies R, then P implies R. The intermediate statement must keep the same meaning in both premises, and chaining implications establishes no instances and no converses, so the existence of any amber object remains unproven.
Worked example: Every amber-labelled vial in the protocol is fragile, and every fragile vial requires protective packaging; therefore every amber-labelled vial requires protective packaging. This establishes a rule only, not that any amber-labelled vial exists in the laboratory's current stock.
Mistake to avoid: Treating the chained conclusion as proof that an amber object exists, when the premises state only relationships between categories.
Reference: [3]
7. Detect a hasty generalisation
A hasty generalisation draws a broad conclusion from inadequate or biased evidence. No universal sample-size threshold guarantees a justified generalisation, so consider how observations were selected as well as how many were made. Two volunteers' preferences support a claim only about those volunteers, not about every participant.
Worked example: A student surveys two friends leaving an evening revision class and reports that all course participants prefer evening sessions. The established preference belongs only to those two friends, and a convenient sample of attendees cannot represent absent participants.
Mistake to avoid: Assuming that a convenient sample, such as whoever is nearby, automatically represents the whole population being generalised about.
Reference: [5]
8. Identify circular support
A circular argument merely restates its conclusion as if it were a reason, so it provides no independent support. For example, saying an appointment system is efficient because it operates efficiently repeats the same claim in different words. Reliable, verifiable evidence would be needed instead, and its sufficiency would still require evaluation.
Worked example: A student argues that a clinic's booking system works well because it functions effectively; the reason restates the conclusion, so no independent evidence about waiting times or scheduling outcomes has actually been offered.
Mistake to avoid: Treating a rephrased conclusion as supporting evidence instead of checking whether an independent reason has been given.
9. Match evidence to the scope of a claim
A conclusion is only as strong as the evidence that actually addresses it. A fact can be verified and still support a narrower claim than the one being evaluated, so reliability alone does not establish relevance or sufficiency. When assessing an argument, check whether the presented facts, rather than opinions or unsupported inferences, cover the specific conclusion the author draws.
Worked example: A council report confirms that a new cycle lane opened in March. That date supports the claim about when the lane opened, but it does not by itself prove that cycling increased; separate usage data would be needed for that conclusion.
Mistake to avoid: Treating one verified fact as proof of every claim made about the same subject, ignoring relevance and sufficiency.
Reference: [4]
10. Test an interpretation against alternatives
An observation usually admits more than one explanation, so a sound evaluation considers alternatives before settling on one. If an author jumps to a preferred conclusion without examining other possibilities, the reasoning shows bias. Alternative explanations remain possibilities to be weighed against evidence; they are not established facts until evidence supports them over the alternatives.
Worked example: A student submits an essay late. The teacher first assumes carelessness, but a note from the student's family about a house move offers another explanation. Considering both possibilities before judging is fairer than accepting the first interpretation.
Mistake to avoid: Accepting a preferred explanation immediately without asking whether other plausible interpretations could account for the same observation.
Reference: [4]
Quantitative problem solving and mathematics
11. Convert centimetres to metres before substitution
Formulas often require lengths in a specific unit, so convert the measurement before substituting it. A reciprocal length expressed per metre must use the length in metres: divide one by the converted value, not by the centimetre figure. Skipping the conversion produces a numerical answer that is correct in appearance but carries the wrong unit and wrong size.
Worked example: A physics problem gives a lens focal length of 40 cm. Converting first, 40 cm = 0.40 m, so the reciprocal is 1 / 0.40 = 2.5 m⁻¹. The conversion happens before substitution, so the unit works out correctly.
Mistake to avoid: Computing 1 divided by the centimetre value and labelling the result per metre, which silently keeps centimetres in the calculation.
Reference: [16]
12. Solve a linear equation while preserving equality
Solving a linear equation means simplifying each side first, then applying the same operation to both sides so equality is preserved. Multiplication and division must reach every term, and division by a nonzero divisor only. After finding the variable's value, substitute it into the original equation to confirm that both sides genuinely balance.
Worked example: Solve 3(x − 1) + 4 = 13. Expanding every term gives 3x − 3 + 4 = 13, so 3x + 1 = 13, then 3x = 12 and x = 4. Checking: 3(4 − 1) + 4 = 13, so both sides balance.
Mistake to avoid: Distributing a multiplier to only the first term inside brackets, or dividing only part of one side while leaving the rest unchanged.
Reference: [6]
13. Translate a situation into an equation
Define the unknown clearly and translate the stated relationship into an equation before doing any arithmetic. Identify the fixed quantity and the per-unit rate, then check that the equation mirrors the wording. After solving, interpret the answer in its original context rather than leaving a bare number.
Worked example: A photographer charges a $40 session fee plus $12 per print; with a $100 budget, 40 + 12p = 100 gives p = 5 prints.
Mistake to avoid: Writing 12p = 100 by dropping the fixed session fee, then reporting about eight prints instead of the correct five.
Reference: [6]
14. Classify an equation's solution set
When variable terms cancel completely, inspect the remaining numerical statement before naming the solution set. A true statement means every real value of the variable satisfies the equation, giving infinitely many solutions, while a false statement means no value works. Cancellation is only valid when it follows legitimate algebraic operations on both sides.
Worked example: Simplifying 4(x + 3) = 4x + 12 leaves 12 = 12, true for every real x, whereas 4(x + 3) = 4x + 9 leaves 12 = 9, which has no solution.
Mistake to avoid: Concluding x = 0 simply because the variable terms cancelled, without checking whether the remaining statement is true or false.
Reference: [6]
15. Rearrange a formula for the required quantity
Isolate the requested variable algebraically before substituting any numbers, keeping every other quantity on the opposite side. Perform the same operation on both sides, and confirm any denominator you introduce is nonzero. Retain the units throughout so the final answer is identified as the required quantity rather than an ambiguous number.
Worked example: From P = 2l + 2w, subtracting 2w and dividing by 2 gives l = (P − 2w)/2; a 30 cm perimeter with width 6 cm yields l = 9 cm.
Mistake to avoid: Dividing only the 2w term by 2, or subtracting w instead of 2w, which produces an incorrect isolated expression.
Reference: [6]
16. Recover a triangle's height from its area
The triangle area formula A = bh/2 links area, base and height. To isolate the height, multiply both sides by two and then divide by the base, giving h = 2A/b. This rearrangement works only when the given area and base are the ones used in the formula, the units are compatible, and the base is nonzero.
Worked example: A triangular sign panel has area 24 cm² and base 8 cm. Applying h = 2A/b gives h = 48/8 = 6 cm. Checking by substitution, (8 × 6)/2 = 24 cm², so the recovered height is consistent with the original area formula.
Mistake to avoid: Dividing the area by the base directly, forgetting to double the area first, yields half the true height.
Reference: [6]
17. Use the conditioned population as the denominator
A conditional probability updates the chance of an event assuming another event occurred. For equally likely cases, restrict the sample space to the group satisfying the condition and use that group's size as the denominator. The conditioning event must have positive probability, and the count ratio assumes uniform selection within the conditioned group.
Worked example: A clinic roster lists 20 nurses on day shift, 9 of whom hold a phlebotomy certificate. Selecting a nurse uniformly at random, the probability the nurse holds the certificate given she is on day shift is 9/20, using only the day-shift group as the denominator.
Mistake to avoid: Using the whole roster as the denominator ignores that the condition has already narrowed the population.
Reference: [7]
18. Multiply probabilities along a sequence
In a multistage experiment, the probability of reaching a final outcome equals the product of the first-stage probability and each later probability conditional on the earlier results. When sampling without replacement, each draw is uniform among the items remaining, so the counts shrink at every dependent stage before multiplying along the path.
Worked example: A dispenser holds 6 red and 3 blue test tubes, drawn uniformly without replacement. The probability of two red tubes is (6/9) × (5/8) = 30/72 = 5/12, because only five red tubes remain among eight after the first red draw.
Mistake to avoid: Reusing the original counts for the second draw treats dependent stages as independent and overstates the probability.
Reference: [7]
19. Add mutually exclusive routes to an outcome
When an outcome can arise through several distinct sequences, compute each sequence's probability with the multiplication rule and add the results, provided the sequences cannot occur together. If either order of two draws qualifies, both orderings must be counted; adding only one route understates the probability.
Worked example: A bag contains 3 black and 2 white counters. Drawing two counters uniformly at random without replacement, P(one of each colour) equals (3/5)(2/4) + (2/5)(3/4) = 3/5, because black-then-white and white-then-black are separate, nonoverlapping routes.
Mistake to avoid: Counting only the black-then-white ordering when the question asks for one counter of each colour in any order.
Reference: [7]
20. Check all branches leaving a probability node
At every node of a tree diagram, the branches must cover all mutually exclusive possibilities for the next stage, so their probabilities should sum to one. Verify this using the collection as it stands at that node; comparing conditional branch probabilities against the original, pre-draw collection signals an error.
Worked example: After one green counter is drawn uniformly at random from a box holding 4 green, 2 orange and 2 purple counters, the next-draw branches are 3/7, 2/7 and 2/7, which sum to one and use the seven counters remaining at that node.
Mistake to avoid: Checking the second-draw branch probabilities against the original eight counters instead of the seven remaining after the first draw.
Reference: [7]
21. Identify the condition imposed by selected records
Records chosen because they share an outcome describe probabilities conditional on that outcome, so their features characterise only the selected group. A hypothetical archive containing only projects finished on time shows which features completed projects had, not which features predict completion across all projects. Reasoning from such records ignores the excluded projects and can mislead.
Worked example: A hypothetical coaching log lists only runners who finished a season uninjured. Their common stretching routine describes that surviving group; it cannot establish whether stretching predicts injury-free seasons overall, because injured runners who stopped appear nowhere in the records.
Mistake to avoid: Concluding that features common in selected records predict the outcome for everyone, while overlooking the excluded cases entirely.
Reference: [7]
Biology: membranes, enzymes and inheritance
22. Interpret selective membrane permeability
A plasma membrane is selectively permeable, allowing some substances to pass while blocking others. Uncharged molecules such as oxygen and carbon dioxide cross the lipid core by simple diffusion, whereas ions such as sodium, potassium, calcium, and chloride cannot, because their charge prevents direct passage and they require special means of penetration.
Worked example: In a hypothetical cell, oxygen molecules pass directly through the lipid bilayer by simple diffusion, while sodium ions of similar size cannot cross the lipid core because their charge blocks passage, so they would need a special penetration pathway.
Mistake to avoid: Judging membrane permeability by particle size alone while ignoring that ionic charge prevents direct passage through the lipid core.
Reference: [8]
23. Distinguish net diffusion from molecular motion
Diffusion drives net movement of a substance down its own concentration gradient, but reaching equilibrium halts only the net transfer, not molecular motion. Once a permeable solute is evenly distributed, particles continue crossing in both directions at matching rates, so describing equilibrium particles as stationary mistakes the absence of net flow for absent motion.
Worked example: A hypothetical membrane separates two solutions of one permeable dye. After the dye concentrations equalise, dye molecules keep crossing the membrane in each direction at equal rates, so molecular motion persists while net transfer between the two sides stays zero.
Mistake to avoid: Claiming that molecules stop moving once diffusion equilibrium is reached, confusing the absence of net transfer with absent motion.
Reference: [8]
24. Compare diffusion surfaces and distances
Diffusion is favoured by greater available surface area and hindered by a thicker membrane or longer travel distance. When comparing two exchange surfaces, other relevant factors such as gradient, temperature and molecule mass must be held constant; the comparison is qualitative and yields no numerical rate, so a larger area alone does not guarantee faster diffusion.
Worked example: Two hypothetical exchange sacs share the same concentration gradient, temperature and diffusing molecule, but one wall is thinner. The thinner wall favours faster diffusion, showing that membrane thickness, not a difference in area, explains the expected difference in exchange speed.
Mistake to avoid: Choosing the surface with larger area as automatically faster without checking whether thickness or distance also differs between the cases.
Reference: [8]
25. Predict osmosis from membrane permissions
Osmosis is the net movement of water through a semipermeable membrane, driven by the water concentration gradient, which is inversely related to solute concentration. Because the membrane blocks the solute, water alone can move, travelling towards the side with higher osmolarity. The movement can be balanced by hydrostatic pressure, and this membrane model says nothing about giving fluids to patients.
Worked example: In a beaker divided by a membrane permeable to water but not to dissolved sugar, the side containing the stronger sugar solution has less free water, so water tends to cross towards it until hydrostatic pressure balances the osmotic pull.
Mistake to avoid: A common error is predicting that the dissolved sugar will also diffuse across the membrane, which would wrongly cancel the osmotic water movement.
Reference: [8]
26. Distinguish plant turgor from animal cell swelling
Under hypotonic conditions water enters both plant and animal cells, but a rigid cell wall changes the outcome. A plant cell swells only until the membrane presses against its wall, building turgor pressure that prevents lysis. An animal cell lacks this wall, so excessive swelling can stretch the plasma membrane beyond its capacity and cause lysis.
Worked example: A plant cell placed in a hypotonic solution takes in water and becomes turgid as its wall resists further expansion, whereas an animal cell in the same solution may lyse if swelling continues beyond what its membrane can tolerate.
Mistake to avoid: Students often claim that any water entry bursts an animal cell immediately, ignoring that lysis requires swelling beyond the membrane's expansion limit.
Reference: [8]
27. Identify what an enzyme changes
Enzymes catalyse reactions by lowering the activation energy needed to reach the transition state, often by holding substrates in a favourable orientation or environment. They do not alter the reactants' or products' free energy, so the overall free-energy difference, and therefore whether a reaction is exergonic or endergonic, stays the same.
Worked example: Suppose an uncatalysed barrier of 70 arbitrary energy units drops to 35 when an enzyme is present, while the energy difference between reactants and products remains unchanged; catalysis is faster, but an endergonic reaction does not become spontaneous.
Mistake to avoid: A frequent error is treating a lowered activation energy as making an energetically unfavourable reaction favourable, confusing the barrier with the overall change.
Reference: [9]
28. Explain specificity through the active site
An active site's unique combination of amino acid residues creates a specific chemical environment, so binding depends on shape and on attractions between functional groups and the substrate. Binding is dynamic: induced fit involves a mild shift in the enzyme's structure as enzyme and substrate come together, rather than a rigid lock-and-key fit.
Worked example: Imagine a molecule whose shape matches an active site but whose charge pattern repels the residues lining it; shape similarity alone would not establish effective binding, because the site's chemical environment must also favour the substrate, with induced fit adjusting both partners.
Mistake to avoid: Candidates sometimes describe specificity as a purely rigid geometric lock-and-key match, overlooking charge interactions and the dynamic shape change of induced fit.
Reference: [9]
29. Interpret temperature and pH effects on enzymes
Raising temperature generally speeds reactions, but beyond an enzyme's favourable range it disrupts bonds in the active site, and high temperatures eventually denature the protein. Similarly, active-site residues work best within a certain pH range, and extreme pH impairs substrate binding or denatures the enzyme. Favourable ranges are enzyme-specific, so no universal optimum applies, and a rate change alone does not reveal its cause.
Worked example: A hypothetical enzyme catalyses fastest near 37°C and pH 7; after a student heats it well above that range, activity drops sharply because the structure denatures, not because substrate has run out.
Mistake to avoid: Concluding from a slowed reaction that all substrate was consumed, when denaturation or a pH shift could equally explain the reduced rate.
Reference: [9]
30. Separate genotype from dominant phenotype
Under complete dominance at a diploid single locus, heterozygotes and homozygous dominant individuals are phenotypically identical, so a dominant appearance cannot distinguish AA from Aa; only offspring with two recessive copies express the recessive trait. Because the law of dominance does not always hold, other inheritance patterns need not produce this relationship.
Worked example: In a stipulated diploid plant model where allele A gives tall stems under complete dominance, a tall specimen could be either AA or Aa, so its visible appearance alone cannot resolve which genotype it carries.
Mistake to avoid: Labelling every dominant-appearing organism homozygous, forgetting that heterozygotes share the identical phenotype under complete dominance.
Reference: [10]
31. Predict a monohybrid cross
Mendel's law of segregation gives each Aa parent an equal likelihood of passing A or a, so Aa × Aa yields genotype proportions 1:2:1 and, with complete dominance, phenotype proportions 3:1. These are probabilistic expectations best approached with large samples; observed counts can deviate, so an expected value is not a guaranteed outcome.
Worked example: Two heterozygous pea plants are crossed for a stipulated non-human seed trait; among 80 offspring, one quarter, 20, are expected to show the recessive form, though the actual count may reasonably differ.
Mistake to avoid: Insisting an exact 3:1 split must appear in a small sample, treating the expected recessive count as certain rather than probabilistic.
Reference: [10]
32. Recognise the limits of a single-gene model
Mendel's monohybrid ratios assume each characteristic is controlled by one gene, but single observable characteristics are almost always influenced by multiple genes acting together. Human eye colour, for instance, involves several genes. Consequently, a 3:1 diagram need not describe every trait, and knowing one gene's genotype may not determine the outcome.
Worked example: Suppose a stipulated plant trait depends on three genes; identifying a dominant allele at one locus leaves the other two unknown, so the phenotype cannot be predicted from that single gene alone.
Mistake to avoid: Applying a 3:1 monohybrid ratio to a characteristic that several genes jointly control, then blaming segregation errors when observations disagree.
Reference: [10]
Chemistry: quantities, acidity and molecular properties
33. Read coefficients as amount ratios
Balanced-equation coefficients give relative numbers of particles or moles, not masses. Because different substances have different molar masses, a coefficient ratio such as 2:3 cannot be read as a gram ratio; convert any given mass to moles first, then apply the coefficient ratio to find the molar amount of the other species.
Worked example: A stipulated balanced equation 2A + B → 3C is used with 4 mol A and sufficient B; assuming complete conversion, the 2:3 coefficient ratio gives 6 mol C, showing coefficients compare amounts, not grams.
Mistake to avoid: Treating the coefficient ratio 2:3 as a gram ratio, concluding 4 g of A yields 6 g of C without converting to moles.
Reference: [11]
34. Convert a substance's mass to moles
To find the amount of a pure substance, divide its mass by its molar mass, keeping units compatible. This conversion is the first step in most reacting-mass problems. Do not multiply mass by molar mass, and remember that equal masses of different substances generally contain different numbers of moles because their molar masses differ.
Worked example: A stipulated substance has molar mass 50 g/mol; a 15 g sample therefore contains 15 ÷ 50 = 0.30 mol, since dividing the sample's mass by its molar mass gives the amount in moles.
Mistake to avoid: Multiplying the sample's mass by its molar mass, or assuming that equal masses of two different substances always contain equal mole amounts.
Reference: [11]
35. Convert moles to molecule numbers
Once a substance's amount is known in moles, multiply by Avogadro's number, 6.022 × 10²³ particles per mole, to obtain the number of molecules. Check what particle type the question requests, since atoms or ions need different factors. Never apply Avogadro's number directly to a mass; convert grams to moles first.
Worked example: For a stipulated molecular substance, 0.20 mol contains 0.20 × 6.022 × 10²³ ≈ 1.2 × 10²³ molecules, rounded to two significant figures, because multiplying moles by Avogadro's number gives the molecule count.
Mistake to avoid: Multiplying a sample's mass in grams directly by Avogadro's number instead of first converting the mass to moles using molar mass.
Reference: [11]
36. Connect reaction masses through moles
A reacting-mass problem links two substances through a mass-to-moles conversion, the balanced equation's coefficient ratio, and a moles-to-mass conversion using the second molar mass. Coefficients never multiply masses directly. Actual yields and limiting-reactant questions need extra information beyond the balanced equation and the stated masses.
Worked example: For stipulated A → 2B with molar masses 40 and 20 g/mol, 12 g A is 0.30 mol; the 1:2 ratio gives 0.60 mol B, which is 12 g assuming complete conversion to B.
Mistake to avoid: Multiplying the given mass directly by the coefficient ratio, skipping both molar-mass conversions and so ignoring that coefficients relate moles, not grams.
Reference: [11]
37. Distinguish acid strength from concentration
Acid strength describes how completely an acid ionises in water, whereas concentration describes how much acid is dissolved in a given volume of solution. A dilute solution of a fully ionising acid can contain fewer hydronium ions per litre than a concentrated solution of a partially ionising acid, so the two properties must be assessed separately rather than treated as interchangeable terms.
Worked example: Suppose solution P contains a 0.010 M acid that ionises completely, while solution Q contains a 0.10 M acid that ionises only slightly. Solution P is the stronger acid even though solution Q is ten times more concentrated.
Mistake to avoid: Describing a 0.10 M partially ionising acid as stronger than a 0.010 M fully ionising acid simply because its concentration is higher.
Reference: [13]
38. Interpret the logarithmic pH relationship
The pH scale is logarithmic with base ten: pH equals the negative logarithm of hydronium ion molarity. Each one-unit rise in pH therefore corresponds to a tenfold decrease in hydronium ion concentration, so a lower pH value indicates a higher hydronium concentration when solutions are compared this way.
Worked example: Two solutions are stipulated: one has hydronium concentration 0.0010 M, giving pH 3.00, and the other has 0.00010 M, giving pH 4.00. The second concentration is one-tenth the first, so its pH is one unit higher.
Mistake to avoid: Reading pH 4 as meaning more hydronium ions than pH 3, forgetting that the logarithmic scale runs inversely to concentration.
39. Calculate the fraction of a weak acid ionised
Percent ionisation compares the equilibrium hydronium concentration with the initial acid concentration: divide the equilibrium hydronium concentration by the stated initial acid concentration and multiply by 100. The denominator is the starting acid concentration, not the remaining un-ionised amount, and the resulting percentage typically falls as the initial concentration rises.
Worked example: If a 0.10 M monoprotic acid solution reaches an equilibrium hydronium concentration of 0.0020 M, assuming other hydronium sources are negligible, the percent ionisation is (0.0020 ÷ 0.10) × 100, which equals 2.0%.
Mistake to avoid: Dividing the equilibrium hydronium concentration by the un-ionised acid remaining at equilibrium instead of by the initial acid concentration.
Reference: [13]
40. Identify the direction of a polar bond
When two bonded atoms have different electronegativities, the shared electrons are pulled towards the more electronegative atom, giving it a partial negative charge and the other atom a partial positive charge. The bond dipole is drawn as an arrow along the bond pointing from the less electronegative atom towards the more electronegative one.
Worked example: In a hypothetical polar covalent A–B bond where B is stipulated to be more electronegative than A, the bond dipole arrow points from A towards B, and B carries a partial negative charge while A carries a partial positive charge.
Mistake to avoid: Treating the partial negative charge on the more electronegative atom as if it were a complete formal ionic charge.
Reference: [12]
41. Distinguish bond polarity from molecular polarity
Bond polarity describes charge separation within a single bond, while molecular polarity describes the molecule's net dipole after all bond dipoles are added as vectors. When equal dipoles point in opposite directions, they cancel and the molecule is nonpolar despite polar bonds. Molecular structure must therefore be considered; deciding from one polar bond alone can mislead.
Worked example: A stipulated linear molecule A–B–A has two identical polar bonds whose dipoles are equal in magnitude and point in opposite directions; the vectors cancel, so the molecule is predicted to be nonpolar overall even though each individual bond is polar.
Mistake to avoid: Judging a molecule polar merely because it contains a polar bond, without checking whether the bond dipoles cancel in the molecular geometry.
Reference: [12]
42. Use polarity as a solvent comparison
Like dissolves like is a qualitative guide: polar solvents generally favour polar substances and nonpolar solvents generally favour nonpolar substances. It predicts which solvent is the better initial choice but does not calculate how much dissolves. The substances' polarities must be supplied, and the tendency should not be treated as proof of unlimited solubility.
Worked example: Given a stipulated nonpolar solute and two otherwise comparable liquid solvents, one nonpolar and one strongly polar, the nonpolar solvent is the better initial prediction for dissolving it, while the actual amount dissolved still requires experimental data.
Mistake to avoid: Assuming the polarity rule guarantees complete dissolution, treating a qualitative tendency as a numerical solubility value.
Reference: [12]
Physics: forces, circuits and refraction
43. Choose a system before classifying forces
Whether a force is internal or external depends on the chosen system boundary, so the boundary must be defined first. Forces between objects inside the system are internal and cancel in the net force, while forces from outside are external. Redrawing the boundary can reclassify a force, so interactions within the system must not be counted again as external.
Worked example: For a horse and cart analysed as one system, the push and pull between horse and cart is internal and cancels in the net force, whereas a rope pulled by a person standing outside the system counts as an external force.
Mistake to avoid: Counting the force between trolley and passenger as external after already including both objects inside the chosen system boundary.
Reference: [14]
44. Calculate acceleration from net force
Newton's second law uses the vector sum of all external forces, not the largest single force. For constant mass along one axis, subtract opposing forces to find the net force, then divide by mass. The resulting acceleration points along the net force, which need not match the object's current direction of motion.
Worked example: A 4 kg box on a level, frictionless surface experiences 15 N rightward and 3 N leftward; the 12 N net rightward force gives 3 m/s² rightward acceleration, even if the box is currently sliding leftward and slowing.
Mistake to avoid: Using only the largest force instead of the vector sum, or assuming acceleration must point along the velocity.
Reference: [14]
45. Distinguish mass from gravitational weight
Mass measures the quantity of matter and stays the same when an intact object is moved between gravitational fields, while weight is the gravitational force on it, calculated as mg and measured in newtons. Because g varies with location, weight changes even though mass does not, so gravitational weight should never be reported in kilograms.
Worked example: A sealed 5.0 kg instrument container is moved from a laboratory where stipulated g is 9.8 N/kg to a lunar base where stipulated g is 1.6 N/kg; its mass stays 5.0 kg while its weight falls from 49 N to 8.0 N.
Mistake to avoid: Reporting an object's lunar weight as '3.2 kg' instead of converting the gravitational force into newtons using the stipulated field strength.
Reference: [14]
46. Recognise the free-fall assumption
Free fall is a modelling assumption in which gravity alone acts on the falling object, so its acceleration equals g and the net force equals the weight. Real falling objects experience upward air resistance, which reduces the net force below the weight, so gravity-only acceleration applies only when drag is negligible or explicitly excluded.
Worked example: A stipulated skydiver weighs 700 N while drag acts upward with 250 N; the resultant 450 N downward force shows drag is not negligible, so treating the descent as ideal free fall with acceleration g would overstate the acceleration.
Mistake to avoid: Applying g as the acceleration to a falling object while its stated air resistance remains a substantial fraction of the weight.
Reference: [14]
47. Check force units through Newton's second law
The newton is defined so that one newton accelerates one kilogram at one metre per second squared, giving N = kg·m/s². Substituting SI units into F = ma therefore yields a force in newtons, providing a dimensional check; however, matching units alone does not confirm the physical situation is modelled correctly.
Worked example: A stipulated 0.50 kg trolley accelerates at 6.0 m/s²; multiplying gives 3.0 kg·m/s², which converts to 3.0 N, confirming the unit consistency of the force calculation before the answer is reported.
Mistake to avoid: Stating the force from F = ma in kilograms, or confusing metres per second, a velocity unit, with acceleration.
Reference: [14]
48. Analyse resistors in series
In a series circuit the same current passes through every resistor because there is no branching, the equivalent resistance is the sum of individual resistances, and each resistor takes only its share of the source voltage, with the drops summing to the source voltage under the stated idealising assumptions.
Worked example: A stipulated 24 V source drives 3 Ω and 5 Ω resistors in series; the 8 Ω total gives 3 A through both, with drops of 9 V and 15 V that sum to 24 V.
Mistake to avoid: Assigning the full source voltage across every series resistor instead of dividing it in proportion to each resistance.
Reference: [15]
49. Analyse parallel branches
Resistors wired in parallel each experience the full source voltage, so each branch current follows Ohm's law independently. Conservation of charge means the source current equals the sum of the branch currents, and the equivalent resistance comes out smaller than the smallest branch. This holds only with negligible wire resistance, positive ohmic resistors and a maintained source voltage.
Worked example: A 12 V source feeds parallel 4 Ω and 12 Ω branches: the 4 Ω branch carries 3 A and the 12 Ω branch carries 1 A, giving 4 A total and an equivalent resistance of 3 Ω, smaller than either branch.
Mistake to avoid: Assuming two parallel branches carry equal currents simply because they share the same voltage across their ends.
Reference: [15]
50. Reduce a mixed circuit in stages
When a network decomposes into series and parallel groups, collapse each group to its equivalent and repeat until one resistance remains. Series resistances add, while parallel reciprocals combine. This staged method applies only to networks built from series and parallel connections, because some arrangements of loops and junctions require other analysis rules instead.
Worked example: A 12 V battery drives 6 Ω and 3 Ω in parallel, giving 2 Ω, in series with 4 Ω, so the total is 6 Ω and the battery current is 2 A, with 8 V across the 4 Ω resistor.
Mistake to avoid: Adding every resistance together as if the whole network were one series chain without checking the connections.
Reference: [15]
51. Calculate power using a component's own voltage
Power formulas must combine quantities belonging to the same component: P = IV uses that resistor's current and its own voltage drop, and P = I²R or P = V²/R follow from Ohm's law. Substituting the source voltage is valid only when the whole voltage appears across that resistor, in a steady DC circuit with an ohmic resistor.
Worked example: A 12 V source drives a series circuit in which 4 V appears across an 8 Ω resistor. That resistor carries 0.5 A and dissipates 4 V times 0.5 A, namely 2 W, not the 18 W that the full 12 V would wrongly suggest.
Mistake to avoid: Dividing the full battery voltage by the resistor to compute power when only part of that voltage appears across it.
Reference: [15]
52. Explain a voltage loss in connecting wires
When connecting wires have non-negligible resistance, they act like series resistors, so the drop across them grows in proportion to the current drawn. Devices then receive less than the source voltage, and the shortfall widens whenever a larger current flows through the same path. The connecting resistance must be stipulated constant for this comparison.
Worked example: With a stipulated 0.25 Ω connecting resistance, a 3 A load drops 0.75 V across the wires, leaving the load 0.75 V below the source; if the current rises to 6 A, the wire drop doubles to 1.5 V under the same source voltage.
Mistake to avoid: Treating connecting wires as always delivering the full source voltage even when a stipulated wire resistance carries a large current.
Reference: [15]
53. Measure refraction direction from the normal
When light travels from air into a material whose refractive index is higher, the transmitted ray bends towards the normal at the entry surface, and on leaving back into air it bends away from the normal. Angles must be measured from the normal, not from the surface, for this comparison to hold.
Worked example: A ray is stipulated to strike a lens, whose refractive index exceeds that of the surrounding air, obliquely and to be transmitted at the entry surface; the ray inside the lens then lies closer to the perpendicular than the incident ray did.
Mistake to avoid: Measuring the refraction angle from the surface rather than the normal, which reverses the comparison of bending direction.
Reference: [16]
54. Relate voltage to energy transferred per charge
Voltage measures energy transferred per unit charge, so the electrical potential energy associated with charge q moved through a potential difference V is qV, with q in coulombs and V in volts. Multiplying voltage by current instead gives power, which is only needed when a question asks about energy transferred per unit time.
Worked example: A charge of 3 C is moved through a stated potential difference of 4 V, so the energy transferred is 3 × 4 = 12 J; no current or time enters this magnitude calculation.
Mistake to avoid: Multiplying the voltage by the current to find energy when the question supplies a charge, which yields power-related quantities rather than the energy asked for.
Reference: [15]
Writing a reasoned response
55. Represent a proposition accurately
Before judging an argument, restate the proposition in your own words and check its exact scope, keeping qualifiers such as some, all and only intact. Attacking a stronger or weaker version than the writer actually claimed, for example converting some into only, misrepresents the position and makes the evaluation invalid.
Worked example: A passage states that some revision techniques suit most learners; a response treating this as claiming every technique suits every learner attacks a stronger proposition than the one actually written.
Mistake to avoid: Paraphrasing a claim about some cases as a claim about all cases, then refuting the exaggerated version instead of the proposition actually stated.
Reference: [4]
56. State a focused position
A thesis should advance an arguable position that the essay's reasons and evidence can actually sustain. A sentence that merely announces a topic gives the response nothing to defend, while an overly broad claim invites objections the developed paragraphs cannot answer, so narrow the thesis whenever the available support falls short.
Worked example: In an essay on hospital waiting times, a writer argues that publishing average wait data would help patients choose services, supporting this with two studies and a patient survey rather than promising to fix all delays.
Mistake to avoid: Writing a thesis that promises to cover every cause of a problem when the body paragraphs develop only one reason, leaving the position unsupported.
Reference: [5]
57. Connect a paragraph's point, evidence and explanation
A body paragraph works only when its topic sentence states a reason, the evidence bears directly on that reason, and the explanation says why the evidence matters. A statistic supports only the conclusion it actually measures; the writer must interpret it and must not stretch it beyond what the sampled records show.
Worked example: Suppose stipulated records show that 12 of 30 sampled notices omit a displayed revision date. The paragraph may conclude that 40% of those sampled notices lack a displayed revision date, but not that version control is absent or that the content is outdated.
Mistake to avoid: Claiming that missing revision dates prove notices lack version control, when identifiers or other records could still provide it.
Reference: [4]
58. Answer a realistic objection fairly
A strong counterargument paragraph names an objection a reasonable opponent would actually raise, concedes any part that is true, and answers the remainder with reasoning and evidence. Conceding one point need not abandon the thesis; the response shows fairness while the core position stands on its remaining support.
Worked example: An objection states that shorter notices lose necessary detail. The writer concedes that risk, proposes a concise summary with access to fuller detail as a tentative response, and states that reader testing would be needed to establish whether it resolves the concern.
Mistake to avoid: Presenting an untested proposal as a proven solution instead of identifying the evidence still needed to support it.
Reference: [5]
59. Distinguish logical support from emotional appeal
Logical support uses reasoning and credible evidence to establish a claim, while emotional appeal can show why an issue matters. A vivid anecdote illustrates a possible consequence but cannot establish how often it occurs; a prevalence claim needs observations of readers or reading events whose scope and denominator match the conclusion.
Worked example: A fictional reader's frustration after misreading a warning notice illustrates a possible consequence of unclear wording, but showing how often misreading occurs would require observing sampled readers attempting the notice and counting misunderstandings per reading event.
Mistake to avoid: Treating a survey of notice wording as a count of reader misunderstandings, although it measures text features rather than reader behaviour.
Reference: [5]
60. Reach a qualified reconciliation
A conclusion may reconcile competing concerns by keeping the valid element of each side, but it must stay within the support presented. Weighing evidence unequally is legitimate when the evidence itself is unequal, and a reconciling proposal remains a proposal until evidence shows it works.
Worked example: After weighing clarity against full disclosure, the writer proposes one-sentence uncertainty statements in main notices with an accessible appendix of detail, while acknowledging that no stipulated evidence yet tests whether readers actually use the appendix, so the proposal stays tentative.
Mistake to avoid: Presenting the compromise as proven simply because it accommodates both concerns, without any evidence that it works.
Sources
- Cambridge Assessment Admissions Testing: Website Closure and Cessation of Test Delivery
- UCL Medical School Selection Procedure
- OpenStax Contemporary Mathematics: Logical Arguments
- OpenStax Writing Guide: Thinking Critically About a Text
- OpenStax Writing Guide: Creating a Position Argument
- OpenStax Contemporary Mathematics: Linear Equations in One Variable with Applications
- OpenStax Contemporary Mathematics: Conditional Probability and the Multiplication Rule
- OpenStax Biology 2e: Passive Transport
- OpenStax Biology 2e: Enzymes
- OpenStax Biology 2e: Laws of Inheritance
- OpenStax Chemistry 2e: Reaction Stoichiometry
- OpenStax Chemistry 2e: Molecular Structure and Polarity
- OpenStax Chemistry 2e: Relative Strengths of Acids and Bases
- OpenStax College Physics 2e: Newton’s Second Law
- OpenStax College Physics 2e: Resistors in Series and Parallel
- OpenStax College Physics 2e: Image Formation by Lenses
- Cambridge University Press & Assessment: BMAT Content Specification, August 2023 to July 2024 (archived original publisher document; MyMathsCloud mirror)