Medical University of Warsaw Competency Test Practice Paper 1 with Answers

Answer: B

A virus is nucleic acid in a protein capsid, sometimes inside an envelope, with no metabolism and no ribosomes of its own, so it can be copied only by taking over a host cell's machinery. A virus particle carries either DNA or RNA, not both. Ribosomes belong to cells: 70S ones to prokaryotes. Penicillin blocks the synthesis of the bacterial cell wall; viruses have no cell wall, which is why antibiotics do not treat viral infections.

Answer: C

Most bacteria are enclosed by a cell wall of peptidoglycan (murein); animal cells, human ones included, have no cell wall at all. Both kinds of cell have ribosomes (70S in bacteria, 80S in the human cytoplasm) and a phospholipid plasma membrane. Mitochondria are the other way round: human cells have them and bacteria, being prokaryotes, have no membrane-bound organelles.

Answer: A

Cholesterol molecules sit between the phospholipids. At body temperature they restrain the movement of the fatty-acid tails, so the membrane is less fluid; at low temperature they stop the tails packing tightly, so it does not become rigid: a fluidity buffer. Ion channels, hormone receptors and glucose carriers are all membrane proteins, not lipids.

Answer: A

DNA is replicated in S phase, so between G1 and the end of G2 the amount doubles, from 6 to 12 pg. Mitosis then shares the sister chromatids equally between two nuclei, each receiving 6 pg, the same as the G1 cell: mitosis keeps the amount of DNA constant from one generation of cells to the next. Halving it again, to 3 pg, is what meiosis II does to gametes, not mitosis.

Answer: C

A 0.3% solution has a lower solute concentration (a higher water potential) than the cytoplasm of a red cell, so it is hypotonic: water moves into the cells by osmosis. Red blood cells have no cell wall to resist the pressure, so they swell and can burst (haemolysis). In a hypertonic solution they would shrink (crenation), and in 0.9% there would be no net movement. Osmosis is passive: no pump is involved.

Answer: A

To evaporate, water molecules must break the hydrogen bonds that hold them to their neighbours, which takes a lot of energy: water's latent heat of vaporisation is high. That energy comes from the skin, which cools. Water is polar, not non-polar, and its specific heat capacity is high, not low. Evaporation breaks hydrogen bonds between molecules; the covalent O–H bonds inside each molecule stay intact.

Answer: A

A triglyceride forms by condensation: each of glycerol's three –OH groups reacts with the –COOH group of a fatty acid, forming an ester bond and releasing one water molecule, so three ester bonds and three waters in all. Peptide bonds join amino acids. Water is used, not released, when a triglyceride is hydrolysed. Glycerol with two fatty acids and a phosphate group is a phospholipid.

Answer: A

Adenine pairs with thymine and guanine with cytosine, so A = T and G = C (Chargaff's rule). If A is 30%, T is also 30%, together 60%. The remaining 40% is shared equally between G and C: 20% each. 40% is G and C together, and 70% is everything except adenine.

Answer: A

RNA polymerase builds the mRNA complementary and antiparallel to the template strand, with uracil in place of thymine: T→A, A→U, C→G, G→C, giving 5′-AUG CCU-3′. The anticodon is complementary to the codon: AUG pairs with 3′-UAC-5′ on the tRNA, the one that carries methionine. B keeps thymine, which RNA does not contain; C copies the template instead of pairing with it; D repeats the codon as its own anticodon.

Answer: A

A competitive inhibitor resembles the substrate and competes with it for the active site. The more substrate there is, the more often the substrate wins, so a high substrate concentration restores the maximum rate. A non-competitive inhibitor binds elsewhere (an allosteric site) and changes the enzyme's shape, so extra substrate cannot undo it; nor can it undo an irreversible inhibitor, which permanently blocks the enzyme molecules it has bound.

Answer: C

Glycolysis and the Krebs cycle each make only 2 ATP per glucose directly; their main product is reduced NAD and FAD. These carriers pass electrons to the electron transport chain on the inner mitochondrial membrane (the cristae), which pumps protons into the intermembrane space; the protons flow back through ATP synthase, making roughly 26–28 of the 30–32 ATP per glucose. The link reaction happens in the matrix and makes no ATP.

Answer: A

The light-dependent reactions on the thylakoid membranes split water, release oxygen and make ATP and NADPH. The Calvin cycle runs in the stroma, where RuBisCO fixes carbon dioxide, and it spends that ATP and NADPH to reduce the product to triose phosphate. Oxygen is a waste product of the light-dependent stage, not something the Calvin cycle uses, and water is split on the thylakoids.

Answer: C

UAA is one of the three stop codons (with UAG and UGA), so translation ends early and the protein is truncated: a nonsense mutation. A silent mutation would give another codon for the same amino acid, and a missense mutation a codon for a different amino acid. A frameshift needs an insertion or deletion of bases not in a multiple of three; a substitution keeps the reading frame.

Answer: C

Genes on different chromosomes assort independently. The AaBb parent makes four kinds of gamete in equal numbers (AB, Ab, aB, ab); the aabb parent makes only ab. So the offspring are AaBb, Aabb, aaBb and aabb in a 1 : 1 : 1 : 1 ratio, and aabb, recessive for both traits, is 1/4. This is a test cross. 1/16 and 9/16 belong to the dihybrid cross AaBb × AaBb.

Answer: C

The mother is XᴴXʰ and the father XᴴY. A son receives his Y chromosome from his father and one of his mother's two X chromosomes, each with probability 1/2, so he is XʰY, affected, with probability 1/2. The 1/4 is the probability that any child of theirs is an affected son, before the child's sex is known; the question already says it is a son.

Answer: A

Each PCR cycle has three steps. At about 95 °C the DNA is denatured: the hydrogen bonds between the bases break and the strands separate. At about 50–65 °C the primers anneal, and at about 72 °C a heat-stable polymerase (Taq) extends them. Ligase is not part of PCR; it joins DNA fragments in cloning.

Answer: C

Affected people are homozygous recessive, so q² = 1/400 and q = 1/20 = 0.05. Then p = 1 − 0.05 = 0.95, and the carriers are 2pq = 2 × 0.95 × 0.05 = 0.095, or 9.5%: about one person in ten. 0.25% is q², the affected people themselves; 5% is q, the allele frequency, not a genotype frequency; 90.25% is p², the homozygous dominant people.

Answer: C

Sunlight, mostly visible and short-wave, passes through the atmosphere and warms the surface, which radiates the energy as long-wave infrared. Carbon dioxide, methane and water vapour absorb that infrared and re-emit it in all directions, part of it back towards the surface. Absorbing ultraviolet is the ozone layer's job, and the ozone hole is a separate problem from the greenhouse effect.

Answer: C

Count the transfers: producers → primary consumers (5,000 kJ) → secondary consumers (500 kJ) → tertiary consumers (50 kJ). Three steps, each keeping about a tenth. The rest is lost mostly as heat from respiration, and in material not eaten or not digested, which is why food chains rarely have more than four or five links. 5 kJ is one step too many, a quaternary consumer.

Answer: A

The two groups were separated geographically, so gene flow stopped; mutation, selection and drift then changed each gene pool independently until the groups were reproductively isolated: two species. That is allopatric speciation. Sympatric speciation happens without geographical separation; convergent evolution is unrelated species coming to look alike; artificial selection is breeding by humans.

Answer: A

Pepsin is secreted by the stomach wall as pepsinogen, activated by hydrochloric acid, and breaks proteins into polypeptides at about pH 2. Trypsin, from the pancreas, also digests proteins, but in the small intestine. Lipase digests lipids, and salivary amylase digests starch.

Answer: B

The pulmonary veins return oxygenated blood from the lungs to the left atrium; they are the only veins in the adult that carry oxygenated blood. The pulmonary artery does the opposite, taking deoxygenated blood from the right ventricle to the lungs. The venae cavae return deoxygenated blood from the body to the right atrium, and the aorta carries oxygenated blood away from the left ventricle to the body.

Answer: B

A vaccine presents the pathogen's antigens without causing the disease. The primary immune response makes plasma cells, which produce antibodies, and long-lived memory cells. On a later infection the memory cells start a secondary response that is quicker and larger, usually stopping the pathogen before symptoms. Injected antibodies give passive immunity, which lasts only weeks to months; a vaccine is not an antibiotic and does not alter the body's DNA.

Answer: B

The rate of diffusion rises with the surface area and with the concentration (partial pressure) difference, and falls as the diffusion distance grows (Fick's law). A larger alveolar surface therefore speeds it up. A thicker barrier lengthens the distance, a smaller gradient slows it, and slower blood flow lets the blood fill with oxygen and the gradient shrink.

Answer: A

The action potential depolarises the terminal and opens voltage-gated Ca²⁺ channels; calcium flows in and makes the vesicles fuse with the presynaptic membrane, releasing acetylcholine by exocytosis. Potassium leaving the cell repolarises the membrane, and sodium entry is what depolarises the postsynaptic cell after acetylcholine binds. Acetylcholinesterase breaks acetylcholine down in the cleft afterwards, ending the signal.

Answer: A

Sweating loses water, so the blood becomes more concentrated; osmoreceptors in the hypothalamus detect it and the posterior pituitary releases more ADH. ADH inserts aquaporins into the walls of the collecting ducts, more water is reabsorbed into the blood, and the urine is small in volume and concentrated. B is what happens with little ADH, after drinking a lot. Glucose is normally all reabsorbed in the proximal tubule, with or without ADH.

Answer: A

Water evaporates from the leaf cells and diffuses out through the stomata (transpiration). This creates tension that pulls water up the xylem, because water molecules cling to each other by hydrogen bonds (cohesion) and to the vessel walls (adhesion): the cohesion-tension theory. Companion cells belong to the phloem, which carries sugars (translocation), and root pressure is too small to lift water to the top of a tall tree.

Answer: B

Mosses are seedless plants with alternation of generations in which the haploid gametophyte dominates: it is the leafy green plant, and it makes gametes by mitosis. After fertilisation the diploid sporophyte grows as a stalk and capsule on top of the gametophyte and depends on it; it makes haploid spores by meiosis. In ferns and seed plants the diploid sporophyte is the dominant generation instead.

Answer: B

Every chordate has, at least in the embryo, a notochord, a dorsal hollow nerve cord, pharyngeal slits and a post-anal tail. A vertebral column belongs only to vertebrates, one subphylum of the chordates (tunicates and lancelets have none). A ventral solid nerve cord and an exoskeleton are features of arthropods and other invertebrates.

Answer: C

Fungi are eukaryotic heterotrophs. Their walls contain chitin, and they digest food outside their bodies with secreted enzymes, then absorb the products through their hyphae: saprotrophic nutrition, which makes many of them decomposers. Cellulose walls and photosynthesis are features of plants, and fungi do not engulf food.

Answer: C

The molar mass of H₂O is 2 × 1 + 16 = 18 g/mol, so 9.0 g is 0.50 mol of molecules, 3.0 × 10²³ molecules. Each molecule has three atoms, so there are 1.5 mol of atoms: 1.5 × 6.0 × 10²³ = 9.0 × 10²³. A counts the molecules instead of the atoms; B takes 9 g as one mole; D treats 9 g as a mole of molecules and then triples it.

Answer: C

M(CH₄) = 12 + 4 = 16 g/mol, so 8.0 g is 0.50 mol. The equation gives one mole of CO₂ per mole of CH₄, so 0.50 mol of CO₂ forms, and M(CO₂) = 12 + 32 = 44 g/mol: 0.50 × 44 = 22 g. 44 g is the mass from a whole mole of methane, and 11 g halves the answer a second time.

Answer: B

Take 100 g and divide each mass by its atomic mass: C 40.0 / 12 = 3.33 mol, H 6.7 / 1 = 6.7 mol, O 53.3 / 16 = 3.33 mol. Dividing by the smallest, 3.33, gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (the formula of methanal, and the empirical formula of glucose and ethanoic acid too). C₂H₄O would need 54.5% carbon.

Answer: A

The atomic number gives the protons: 17. The mass number 37 is protons plus neutrons, so there are 37 − 17 = 20 neutrons. A neutral atom has 17 electrons, and the 1− charge means one extra: 18. B is the neutral atom, C takes the mass number for the neutrons, and D swaps protons and electrons.

Answer: B

The iron atom is [Ar] 4s² 3d⁶ (18 + 2 + 6 = 26 electrons). When a transition metal forms a cation, the 4s electrons are lost first, because once the 3d orbitals are occupied the 4s electrons are the outermost. Removing two gives Fe²⁺ = [Ar] 3d⁶. A removes 3d electrons instead, C is the neutral atom, and [Ar] 3d⁵ is Fe³⁺.

Answer: B

The highest shell occupied is n = 3, so the element is in period 3. Its last electrons enter a p sub-shell, so it is in the p-block. It has 2 + 4 = 6 valence electrons, which places it in group 16 (the oxygen group): it is sulfur, 1s² 2s² 2p⁶ 3s² 3p⁴, atomic number 16.

Answer: C

Each element down the group has one more occupied shell, so the atoms get larger and the outer electron is further from the nucleus and better shielded. It is lost more easily: ionisation energy and electronegativity fall and reactivity rises, so potassium reacts with water far more vigorously than lithium. The metallic bonding weakens as the atoms grow, so the melting point falls too.

Answer: B

Nitrogen has four electron pairs around it, three bonding and one lone pair, arranged roughly tetrahedrally (sp³). The lone pair repels the bonding pairs more strongly than they repel each other, squeezing the angle from 109.5° to about 107°, and with only three atoms attached the shape is a trigonal pyramid. 109.5° is methane, with four bonding pairs; 104.5° is water, with two lone pairs.

Answer: D

A hydrogen bond needs a hydrogen atom bonded directly to N, O or F, attracted to a lone pair on an N, O or F of another molecule. Methanol has an O–H group, so its molecules hydrogen-bond to each other. Methoxymethane has an oxygen but all its hydrogens are on carbon, so it can accept a hydrogen bond (from water) but not form one with itself. HCl is polar but chlorine is not one of the three, and methane has only London forces.

Answer: B

At constant temperature Boyle's law applies: p₁V₁ = p₂V₂, so V₂ = (100 × 6.0) / 300 = 2.0 dm³. Tripling the pressure divides the volume by three. 18 dm³ multiplies instead of dividing.

Answer: B

Oxygen gas is diatomic, M(O₂) = 32 g/mol, so 8.0 g is 0.25 mol, and 0.25 × 22.4 = 5.6 dm³. 11.2 dm³ comes from using 16 g/mol, the mass of an oxygen atom, instead of the O₂ molecule.

Answer: B

By Hess's law, burning carbon straight to CO₂ releases the same energy as burning it to CO and then the CO to CO₂. So ΔH(C → CO) + (−283) = −394, and ΔH(C → CO) = −394 + 283 = −111 kJ/mol: reverse the second equation and add it to the first. −677 adds the two values instead, and +111 gets the sign wrong; forming CO from carbon is exothermic.

Answer: C

A reaction is spontaneous when ΔG = ΔH − TΔS is negative. With ΔH positive and ΔS positive, the −TΔS term is negative and grows with temperature, so ΔG becomes negative only once TΔS exceeds ΔH, above T = ΔH / ΔS. Negative ΔH with positive ΔS is spontaneous at every temperature, positive ΔH with negative ΔS at none, and negative ΔH with negative ΔS only at low temperature.

Answer: A

A catalyst offers a different route from reactants to products whose energy barrier is lower, so at the same temperature a larger fraction of collisions have enough energy to react. It changes neither ΔH, which depends only on the reactants and products, nor the position of equilibrium, since it speeds the forward and reverse reactions alike. Raising the kinetic energy of the particles is what heating does.

Answer: A

The half-life of a first-order reaction does not depend on the concentration, so 60 minutes is three half-lives: 0.80 → 0.40 → 0.20 → 0.10 mol/dm³. The concentration falls by half in each 20 minutes, not by a fixed amount: 0.27 divides by three as if the decrease were linear, and 0.20 stops after two half-lives.

Answer: B

There are four moles of gas on the left and two on the right, so raising the pressure shifts the equilibrium towards the side with fewer gas molecules, the ammonia (Le Chatelier's principle). The forward reaction is exothermic, so raising the temperature shifts the equilibrium back towards nitrogen and hydrogen. A catalyst only brings the system to equilibrium faster, and removing a reactant shifts it to the left.

Answer: C

Kc = [HI]² / ([H₂][I₂]) = 0.80² / (0.10 × 0.10) = 0.64 / 0.010 = 64. The coefficient 2 becomes a square in the expression. 80 forgets to square [HI] (0.80 / 0.010), 8.0 divides by 0.10 only once, and 0.016 is the expression upside down.

Answer: C

If s mol/dm³ dissolves, [A⁺] = [B⁻] = s, so Ksp = s² = 1.0 × 10⁻¹⁰ and s = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol/dm³. A takes Ksp itself as the solubility, B halves it, and D doubles the right answer, as if each formula unit released two of each ion.

Answer: C

Ostwald's law gives Ka = α²c / (1 − α); for small α this is Ka ≈ α²c, so α ≈ √(Ka / c). Ka is constant at constant temperature, so dividing c by 100 multiplies α by √100 = 10. Dilution increases the fraction of the acid that dissociates, though the concentration of H⁺ still falls, by about 10-fold, as c falls 100-fold and α rises 10-fold.

Answer: C

Sodium hydroxide is a strong base, fully dissociated, so [OH⁻] = 0.010 = 10⁻² mol/dm³ and pOH = 2. At 25 °C pH + pOH = 14, so pH = 12. Equivalently, [H⁺] = Kw / [OH⁻] = 10⁻¹⁴ / 10⁻² = 10⁻¹² mol/dm³. 2 is the pOH reported as the pH, which would make the solution acidic.

Answer: A

A conjugate pair differs by exactly one proton. HCO₃⁻ donates a proton (it is the acid) and becomes CO₃²⁻, its conjugate base; H₂O accepts the proton (the base) and becomes H₃O⁺, its conjugate acid. So the two pairs are HCO₃⁻/CO₃²⁻ and H₃O⁺/H₂O. Each of the other options pairs a species from one couple with one from the other.

Answer: C

Sodium ethanoate is the salt of a strong base and a weak acid. The ethanoate ion is a weak base and hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so the solution is alkaline. Ammonium chloride, from a weak base and a strong acid, gives an acidic solution (NH₄⁺ donates protons). Sodium chloride and potassium nitrate, from a strong acid and a strong base, are neutral.

Answer: D

The oxidation numbers in a neutral compound add up to zero. Potassium is +1 and each oxygen −2, so +1 + x + 4(−2) = 0 and x = +7, which is what the (VII) in the name says. +2 is manganese in Mn²⁺, +4 in MnO₂ and +6 in manganate(VI), MnO₄²⁻.

Answer: C

Manganese goes from +7 to +2, gaining 5 electrons; each iron goes from +2 to +3, losing 1. The electrons gained must equal those lost, so five Fe²⁺ react with each MnO₄⁻: MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O. Check the charges: 1− + 10+ + 8+ = 17+ on the left, 2+ + 15+ = 17+ on the right. 8 is the number of H⁺ ions.

Answer: A

EMF = E°(cathode) − E°(anode) = +0.34 − (−0.76) = 1.10 V. Zinc has the more negative potential, so it is oxidised (Zn → Zn²⁺ + 2e⁻): it is the anode and, in a galvanic cell, the negative electrode, from which electrons flow through the wire to the copper, where Cu²⁺ is reduced. 0.42 V subtracts the magnitudes and ignores the minus sign.

Answer: A

At the cathode water is reduced more easily than Na⁺, so hydrogen forms: 2 H₂O + 2e⁻ → H₂ + 2 OH⁻, and the solution around it becomes sodium hydroxide. At the anode, in concentrated chloride, chloride ions are oxidised: 2 Cl⁻ → Cl₂ + 2e⁻. Sodium metal forms only from molten sodium chloride, and oxygen at the anode only when the chloride is very dilute.

Answer: B

Cis–trans isomerism needs restricted rotation about a C=C double bond and two different groups on each carbon of the double bond. In but-2-ene each double-bond carbon carries H and CH₃, so the two methyl groups can be on the same side (cis) or opposite sides (trans). In the other three, one carbon of the double bond carries two identical groups (two H atoms, or two CH₃ groups), so no isomers are possible.

Answer: A

Aldehydes are easily oxidised and ketones are not. Propanal reduces Tollens' reagent, depositing a silver mirror, while propanone does not react; Fehling's solution distinguishes them the same way. 2,4-DNPH gives an orange precipitate with any carbonyl compound, aldehyde or ketone. Neither compound has a C=C bond to decolourise bromine water or an O–H group to react with sodium.

Answer: A

A carboxylic acid and an alcohol react, with an acid catalyst, to give an ester and water (esterification, a condensation). The ester is named alcohol part first: ethyl from ethanol, ethanoate from ethanoic acid, so CH₃COOCH₂CH₃, ethyl ethanoate. Methyl ethanoate would need methanol, and ethyl methanoate methanoic acid. Ethanal comes from oxidising ethanol, not from this reaction.

Answer: B

The primary structure is the amino acid sequence, joined by peptide bonds. The secondary structure is the regular local folding of the backbone into α-helices and β-sheets, held by hydrogen bonds between C=O and N–H groups of peptide bonds. The overall folding held by R-group interactions (ionic, hydrophobic, hydrogen and disulfide bonds) is the tertiary structure, and the combination of several chains, as in haemoglobin, the quaternary.

Answer: B

Hospital Q's rate fell by 25% with no programme at all, so some factor shared by both hospitals (the season, a fading outbreak, a change in testing) probably lowered infections at both. That makes it likely that much of Hospital P's fall had nothing to do with its programme. The programme may still have helped, perhaps the extra five points, so 'caused the whole fall' and 'does not work' both go beyond the data, and nothing is said about the patients at either hospital.

Answer: A

The evidence is a correlation, and the conclusion predicts that changing breakfast will change scores, which holds only if breakfast causes the difference. If breakfast-eaters simply come from homes with more support or sleep, giving everyone breakfast would not close the gap. The argument does not need most pupils to skip breakfast (some doing so is enough for scores to rise), and cost and the value of tests are outside it.

Answer: C

Colds normally resolve on their own in about a week to ten days, so all 40 patients would very probably have recovered without the remedy. Without a comparison group that did not take it, recovery tells us nothing about the remedy. Cost is irrelevant to whether it works, forty is not 'too many', and the argument makes no claim about which virus causes colds.

Answer: B

A city-wide cut in the speed limit at the same moment is a strong alternative explanation for fewer accidents, so the fall can no longer be credited to the lanes. How safe cyclists feel is not evidence about accidents, and cost says nothing about safety. A rise in a similar city without lanes would, if anything, strengthen the council's case.

Answer: B

Marta is a nurse on Ward 3, and all of them have completed resuscitation training, so she has. The 'some' who also have paediatric training may or may not include Marta or any Ward 3 nurse: they could all work elsewhere, so A and D may be true but need not be. C reverses the first statement: all Ward 3 nurses are trained does not mean all trained staff are Ward 3 nurses.

Answer: B

The rise is 300 − 240 = 60 patients, and a percentage change is measured against the starting figure: 60 / 240 = 1/4 = 25%. 20% divides by the new figure, 300, instead of the old; 60% mistakes the number of patients for a percentage; 125% is April as a percentage of March, not the increase.

Answer: B

The daily amount is 15 × 24 = 360 mg (15 × 24 = 15 × 20 + 15 × 4 = 300 + 60). Split into three equal doses, each is 360 / 3 = 120 mg. 360 mg is the whole day's amount given as one dose, 1,080 mg multiplies by three instead of dividing, and 40 mg divides the day's 360 mg by three twice.

Answer: B

Divide nurses by doctors: 2023, 120 / 40 = 3.0; 2024, 150 / 45 = 10/3 ≈ 3.33; 2025, 160 / 50 = 3.2. The ratio was highest in 2024. 2025 has the most nurses, but also the most doctors; a ratio has to be worked out, not read from the largest number.

Answer: B

From 08:40 to 11:10 is 2 hours 30 minutes, 2.5 hours. Average speed = distance / time = 300 / 2.5 = 120 km/h. The other options are the usual slips: about 130 km/h reads 2 h 30 min as 2.3 hours, and 150 and 100 km/h count only 2 or 3 whole hours.

Answer: C

35% of 800 is 280 walkers and a quarter of 800 is 200 cyclists, together 480. The rest, 800 − 480 = 320, take the bus; as a share, 100% − 35% − 25% = 40%, and 40% of 800 is 320. 280 is the walkers and 520 is everyone except the cyclists.

Answer: A

Carla is fifth, so the other four fill places 1 to 4. Ewa comes before Anna, who comes before Dev, so Dev is at least third; Ben follows Dev directly, so Dev cannot be fourth. Dev is therefore third and Ben fourth, leaving Ewa first and Anna second: Ewa, Anna, Dev, Ben, Carla.

Answer: A

Three pharmacists working five days each give 15 pharmacist-days a week. Seven days with at least two each need 14 of them, which leaves exactly one more: one day has three on duty and the other six have two. In symbols, if x days have three, 3x + 2(7 − x) = 15, so x = 1. Any other number would need more or fewer than 15 pharmacist-days.

Answer: B

Work backwards: if it is full at 6 pm and doubles every hour, it was half full at 5 pm and a quarter full at 4 pm. 4.30 pm treats the growth as if it were steady, adding the same amount each hour, and 5 pm is when it was half full.

Answer: A

Add every cost, and two bus fares where the trip needs them: P = 12 + 2 + 2 = €16; Q = €18 with no fare; R = 10 + 3 + 2 + 2 = €17. P is cheapest. R has the cheapest test, but the booking fee puts it a euro above P. Q's faster results are true but irrelevant to the cost, and are there to distract.

Answer: B

30 − 5 = 25 students study at least one of the two. Adding 18 and 15 gives 33, which counts the students who study both twice, so 33 − 25 = 8 study both. Check: 10 biology only + 7 chemistry only + 8 both + 5 neither = 30.

Answer: C

Two rules work at once. The dot moves one corner clockwise each step (top left, top right, bottom right, bottom left), so in box 5 it is back at the top left. The number of lines goes up by one each step, 1, 2, 3, 4, so box 5 has 5. Only C has both. A moves the dot one corner too far, B forgets to add a line, and D keeps the dot where it was in box 4.

Answer: D

In A, B and C the number of dots equals the number of sides: 3 in the triangle, 4 in the square, 5 in the pentagon. D is a hexagon, six sides, with only five dots, so it breaks the rule. The shading (A and C grey, B and D white) and the sizes are distractors: they split the figures two and two, or vary at random, so they cannot single one out.

Answer: B

Every row and every column contains each shape once (circle, square, triangle) and each shade once (black, white, grey). Row 3 already has a square and a triangle, so the missing figure is a circle; it already has grey and black, so the circle is white. The third column agrees: it has a grey triangle and a black square, so it needs a white circle. A and D repeat a shade already in the row and column; C repeats the triangle.

Answer: A

From Figure 1 to Figure 2 the arrow turns 90° clockwise (up to right) and its colour is reversed (black to white). Applied to Figure 3, a white arrow pointing down, a 90° clockwise turn points it left and reversing the colour makes it black: option A. B turns it anticlockwise, C forgets to change the colour, and D turns it 180°.

Answer: C

The hand turns clockwise by a growing angle: 45° (up to up-right), then 90° (to down-right), then 135° (to left). The next turn is 180°, from pointing left to pointing right. The background alternates white, grey, white, grey, so dial 5 is white: option C. A keeps the background grey; B adds 90° again; D adds only 45°.