Simulazione Test San Raffaele Medicina 1 con soluzioni

Risposta: D

To deny that every train left on time, it is enough for a single train not to have done so, so the negation of "all are" is "at least one is not". "No train left on time" and "all left late" are far stronger claims that could both be false while the original is false too. "Most left late" says too much, and "some left on time" is compatible with the original statement rather than contradicting it.

Risposta: A

Insomnia is a lack of sleep, and famine is a lack of food. Fatigue is a result of exercise, not an absence of it; a fever is too high a temperature, not a missing one; a glutton has too much appetite; and a library contains books. Only famine and food share the relationship of absence.

Risposta: C

Let the son be s, so Marta is 3s. In 12 years: 3s + 12 = 2(s + 12), so 3s + 12 = 2s + 24 and s = 12. Marta is 3 × 12 = 36 (check: in 12 years she is 48 and he is 24). 48 is her age in 12 years' time, and 24 is twice the son's present age, not Marta's.

Risposta: E

30 − 5 = 25 students study at least one language. Adding the two groups gives 18 + 14 = 32, which counts the students who study both twice, so both = 32 − 25 = 7. Answering 2 (32 − 30) forgets the 5 who study neither.

Risposta: B

"Only if" introduces a necessary condition: booking a scan requires a referral letter, so a patient without one cannot book (the contrapositive). The rule does not say a letter is sufficient, so the first and last options reverse it. The option about patients who may not book also reverses it, and nothing in the rule tells us whether some letter-holders are refused, so that option is not guaranteed either.

Risposta: D

The ring road offers a different explanation for the fall: with half as much traffic, fewer pedestrians would be hit whatever the speed limit, so the data no longer point to the limit. The neighbouring town's rise supports the council. Residents' opinions and the ages of the injured have no bearing on what caused the fall, and a similar fall for cyclists fits the speed-limit explanation as well as any other.

Risposta: A

With S in position 3, the pair P then R must take positions 1–2 or 4–5. If P and R take 4–5, then T and Q fill positions 1–2 with T first, which is not allowed. So P is 1 and R is 2, leaving T and Q for 4 and 5; T comes before Q, so the order is P, R, S, T, Q and Q is fifth. This is the only arrangement that meets all three conditions.

Risposta: C

The chess club members who are afraid of water cannot be swimmers, because no swimmer is afraid of water, so at least some chess club members cannot swim. Nothing is said about chess club members who are not afraid of water: they might swim and might row, so "no chess member rows" and "no chess member swims" go too far, and an overlap between the clubs is possible but not certain. "Everyone who can swim rows" reverses the first premise.

Risposta: E

Run only: 45 − 15 − 12 + 5 = 23. Cycle only: 40 − 15 − 10 + 5 = 20. Swim only: 35 − 12 − 10 + 5 = 18. Exactly one: 23 + 20 + 18 = 61. Forgetting to add back the 5 who do all three (they are removed twice) gives 46, and 83 is the number who do at least one activity (88) less the 5 who do all three, which still includes those who do exactly two.

Risposta: B

The argument moves from "Lucia owns no typing machine" to "Lucia could not have typed the letter". That step holds only if she could not have used someone else's machine, at a library, a friend's house or work. The newspaper's policy and Lucia's motives play no part in the reasoning, and the argument needs no claim about who else could have written it or whether the typist owned the machine used.

Risposta: D

Together the pumps fill 1/2 of the tank per hour, and A fills 1/6, so B fills 1/2 − 1/6 = 1/3 per hour. After 1 hour together the tank is half full; B fills the remaining 1/2 in (1/2) ÷ (1/3) = 1.5 hours. The total is 1 + 1.5 = 2.5 hours. 2 hours assumes A never breaks down, and 4 hours adds B's full 3 hours to the first hour.

Risposta: A

The two rules chain together: contaminated → positive → retested. Since the patient was not retested, the result cannot have been positive, and so the sample cannot have been contaminated. That is a definite conclusion, so "nothing can be concluded" is wrong, and a positive result or contamination would each have led to a retest. The last option denies the condition rather than the result: a clean sample could still give a positive result for another reason.

Risposta: C

The passage gives two reasons for low winter intake (weak sun, few food sources), which is why a supplement is recommended; the cautious conclusion follows directly. It does not say everyone is deficient, that the skin makes none outside summer, or that supplements beat sunlight. It says nothing at all about southern countries.

Risposta: E

Without restriction there are 7C3 = (7 × 6 × 5) ÷ (3 × 2 × 1) = 35 committees. Those with no nurse are all-doctor committees: 4C3 = 4. So 35 − 4 = 31 include at least one nurse. 18 counts only committees with exactly one nurse (3 × 6), and 45 (3 × 6C2) counts committees with two or three nurses more than once.

Risposta: B

"Not X unless Y" means X only if Y: the trial goes ahead only if funding and approval are both confirmed. So missing either one, including the ethics approval, stops the trial. The statement gives necessary conditions, not sufficient ones, so the first option reverses it. The remaining three all reason backwards from the trial not going ahead, which could happen for some other reason even with both confirmed.

Risposta: C

Count the failures instead: 12 failed biology, 15 failed chemistry and 10 failed physics, 37 failures in all. Even if no student failed more than one subject, at most 37 students failed something, so at least 40 − 37 = 3 passed everything, and 3 is achievable with the failures kept separate. 0 ignores this, 13 is the minimum overlap of biology and chemistry alone, and 25 is the maximum, not the minimum.

Risposta: A

The age difference shows that the study cannot isolate the effect of stretching, but both age and stretching could contribute. Jumping from "age might explain it" to "stretching does nothing" is the flaw. 500 runners is not a tiny sample, the argument says nothing about severity or about how injuries were reported, and it claims only that stretching does not help, not that it does harm.

Risposta: D

The free days are Mon, Tue, Thu, Fri and Sat. Anatomy and Physiology two days apart must be Tue–Thu or Thu–Sat (Mon–Wed and Wed–Fri use Wednesday). Biochemistry then Genetics needs two consecutive free days: Mon–Tue, Thu–Fri or Fri–Sat. Thu–Fri leaves no room for Anatomy and Physiology; Fri–Sat forces Anatomy–Physiology onto Tue–Thu and Chemistry onto Monday, before Genetics, which is not allowed. So Biochemistry is Monday, Genetics Tuesday, Anatomy Thursday, Physiology Saturday and Chemistry Friday.

Risposta: C

The passage follows Semmelweis from the comparison of the two clinics, through the handwashing order and its result, to the resistance that lasted until Pasteur and Lister supplied a theory: that is the whole arc. Crowding is explicitly ruled out (the second clinic was more crowded), so the first option is wrong. The passage never compares the skill of midwives and doctors, nor says Pasteur and Lister deserve the credit. Slow publication is one reason among several, so 'only because' is too strong.

Risposta: A

The second paragraph says the midwives of the second clinic did not dissect bodies, and the third says Semmelweis traced the fever to particles carried from the dissecting room, so the midwives were less likely to carry that material. The second clinic was the more crowded, which rules out 'fewer women at a time'. Nothing in the passage says the midwives washed in chlorinated lime earlier, suspected hands themselves, or were less skilled.

Risposta: D

The colleague's death is the clue that led Semmelweis to his conclusion about cadaverous particles: it settled what he had been trying to find out. 'Firmly expressed' is the sense used of people ('a decisive manager'), not of clues. The passage does not present the clue as the most recent, as surprising, or as widely reported.

Risposta: B

'It would be pleasing to report that his colleagues were persuaded' shows regret, while the last sentence offers a reason the resistance is understandable: evidence that something works rarely wins acceptance when no one can say how it works. The author never says the rejection was right, so 'fully justified' goes too far. Jealousy is not mentioned, the author treats the doctors' doubts as a real difficulty rather than indifference to evidence, and slow publication is listed as one factor, not as the one that would have changed the outcome.

Risposta: E

The paragraph's own summary is its last sentence: the railways created a need for agreement, while the form of the agreement was settled by people arguing. The first option says the opposite ('the railway did not decide the answer'), and the third generalises the very view the author is qualifying. The United States is an example of a different choice, not a better one, and pride is only one of three things people argued over.

Risposta: B

The first paragraph says 'As long as travel was slow, nobody minded', and gives the day-long coach journey on which ten minutes 'hardly' mattered. Every town set its own clocks by the sun, so it did not keep London time, and Bristol is an example of a general pattern, not the only case. The passage says nothing about the accuracy of clocks or about coach timetables.

Risposta: D

The passage says noon 'arrives a little later at each step westward'. Read the other way, it arrives a little earlier at each step eastward, so a town east of London reaches noon before London does. 'A little after' reverses the direction. Since Bristol's noon is after London's, the eastern town's noon is well before Bristol's, not after it. Railway stations play no part before railway time.

Risposta: A

The author's claim is that people, not the railways, settled the form of standard time. If the railway companies' choice prevailed everywhere regardless of what anyone else wanted, the railways did decide the answer, which undercuts the claim. Years of national debate and towns holding out both fit the claim that people argued the matter out. Fewer collisions show why an agreed time was needed, which the author already accepts, and the order in which telegraph offices and newspapers changed says nothing about who chose the form.

Risposta: C

The final sentence, introduced by 'in other words', draws the conclusion: the vault's value may lie in insuring individual collections against ordinary misfortune. The first option is the popular picture the text argues against. The text says the centre could no longer use its collection, not that the collection was destroyed, and it says nothing about Arctic seed banks being the safest or about who deposited seeds first.

Risposta: E

The conclusion rests on the trips having doubled. If this year's count was taken in ideal weather and last year's over a mixed month, the apparent doubling may reflect the conditions of the count rather than a real rise, and the per-trip risk may not have fallen. Where injured cyclists were riding bears on the councillor's blame of the lanes, not on the per-trip rate, and injuries on other roads do not change these figures. The 28 threshold (14 x 2) is true and supports the writer, and seriousness is a different matter from the risk of being injured at all.

Risposta: B

Small pack: €12.00 / 30 = €0.40 a tablet. Large pack: €18.00 / 50 = €0.36 a tablet. The saving is €0.04 on €0.40, and 0.04 / 0.40 = 10%. 'About 11%' divides by the wrong base (0.04 / 0.36), 4% mistakes four cents for a percentage, and 50% compares the pack prices (€6 on €12) without allowing for the different numbers of tablets.

Risposta: D

Total before: 20 x 62 = 1,240. Total after: 21 x 63 = 1,323. The new mark is 1,323 - 1,240 = 83. The new student has to cover their own 63 and also add 1 mark for each of the 20 earlier students: 63 + 20 = 83. Answers of 63 or 64 ignore the effect on the other twenty marks.

Risposta: C

Original rate: 1,200 ml / 480 min = 2.5 ml a minute. In the first 140 minutes, 140 x 2.5 = 350 ml is delivered, leaving 850 ml. At the doubled rate of 5 ml a minute, that takes 850 / 5 = 170 minutes, or 2 h 50 min. Total: 2 h 20 min + 2 h 50 min = 5 h 10 min. The other times come from applying the doubled rate to the wrong amount or the wrong stretch of time: only the 850 ml left after 140 minutes runs at 5 ml a minute.

Risposta: E

Let there be n nurses and 4n patients at first. Afterwards 5(n + 1) = 4n + 10, so 5n + 5 = 4n + 10 and n = 5. There were 5 nurses and 20 patients; now there are 6 nurses and 30 patients, and 6 : 30 = 1 : 5. 20 is the number before the admissions, and 25 or 24 come from applying one of the ratios to the wrong number of nurses.

Risposta: A

A: 240 / 1,200 = 20%. B: 225 / 900 = 25%. C fell by 80 / 1,600 = 5%. D: 210 / 750 = 28%. D has the largest percentage rise. A has the largest rise in numbers (240), which is the trap: a percentage rise is measured against each clinic's own starting figure.

Risposta: C

A: 15% of 1,440 = 216. B: 20% of 1,125 = 225. C: 20% of 1,520 = 304. D: 30% of 960 = 288. C has the most, 304. D has the highest percentage, which tempts many to choose it, but it saw far fewer patients than C.

Risposta: B

2024 total: 1,200 + 900 + 1,600 + 750 = 4,450. 2025 total: 1,440 + 1,125 + 1,520 + 960 = 5,045. The rise is 595, and 595 / 4,450 is about 0.134, or 13%. 17% is the simple average of the four clinics' own changes ((20 + 25 - 5 + 28) / 4), which wrongly gives a small clinic the same weight as a large one.

Risposta: D

Each clinic needs at least its patients / 200 doctors, rounded up. A: 1,440 / 200 = 7.2, so 8 doctors, 2 more than 6. B: 1,125 / 200 = 5.625, so 6, 1 more. C: 1,520 / 200 = 7.6, so 8, which it already has. D: 960 / 200 = 4.8, so 5, 2 more. Total 2 + 1 + 0 + 2 = 5. Pooling the clinics (5,045 / 200 = 25.2, so 26 against 22 doctors) gives 4, as does rounding A's 7.2 down to 7, but doctors cannot be shared and 7 doctors would leave A above 200 each.

Risposta: B

The carboxyl group of one amino acid reacts with the amino group of the next, releasing a molecule of water and forming a peptide (amide) bond, -CO-NH-. Glycosidic bonds join monosaccharides, ester bonds join fatty acids to glycerol, and phosphodiester bonds join nucleotides. Hydrogen bonds hold the secondary structure together but are not covalent and do not form the chain itself.

Risposta: D

Every cell must make proteins, so prokaryotes and eukaryotes both have ribosomes (70S in prokaryotes, 80S in the eukaryotic cytoplasm). The nucleus, mitochondria, Golgi apparatus and endoplasmic reticulum are membrane-bound organelles, which prokaryotes lack: their DNA lies free in the cytoplasm in a region called the nucleoid.

Risposta: A

The woman received her father's only X, which carries the allele, so she is a carrier, X^B X^b; the man is X^B Y. The four equally likely children are X^B X^B, X^B X^b, X^B Y and X^b Y, and only X^b Y is a colour-blind boy, so the probability is 1/4. The tempting 1/2 is the probability that a son is colour-blind, which answers a different question: it is conditional on the child already being a boy.

Risposta: E

Pyruvate from glycolysis is carried into the mitochondrion, converted to acetyl-CoA in the link reaction, and oxidised in the Krebs cycle, all in the matrix. Glycolysis is the stage that happens in the cytoplasm. The inner membrane holds the electron transport chain and ATP synthase, and the intermembrane space is where protons accumulate during oxidative phosphorylation.

Risposta: C

ADH, released by the posterior pituitary, makes the cells of the collecting duct (and the late distal tubule) insert aquaporins into their membranes, so more water leaves the filtrate by osmosis into the concentrated medulla. The glomerulus and Bowman's capsule filter the blood rather than reabsorb from it. The proximal tubule and the descending limb do reabsorb water, but they do so whatever the ADH level.

Risposta: C

Restriction enzymes (restriction endonucleases) recognise a short, specific base sequence and cut both strands of the DNA there, often leaving sticky ends. Joining fragments is the job of DNA ligase, copying RNA into DNA is reverse transcriptase, unwinding the helix is helicase, and extending a strand in PCR is a heat-stable DNA polymerase.

Risposta: A

The atomic number gives 26 protons. Neutrons = mass number - atomic number = 56 - 26 = 30. A neutral atom has 26 electrons, and a 3+ charge means 3 have been lost, leaving 23. Adding electrons for a positive charge gives 29, taking 56 as the neutron count confuses it with the mass number, and the number of protons never changes when an ion forms.

Risposta: E

Each C=O bond in CO₂ is polar, but the molecule is linear (O=C=O), so the two equal bond dipoles point in opposite directions and cancel. H₂O is bent and NH₃ is trigonal pyramidal, so their dipoles do not cancel. HCl has a single polar bond, and in CH₃Cl the C-Cl bond is not balanced by the three C-H bonds, so both are polar molecules.

Risposta: B

Moles needed = concentration × volume = 0.2 mol/L × 0.250 L = 0.05 mol. Mass = 0.05 mol × 40 g/mol = 2.0 g. The answer 8.0 g is the mass for a whole litre (0.2 × 40), forgetting that only 250 mL is made; 0.5 g and 20 g are each out by a factor from mishandling the 250 mL to 0.250 L conversion or the molar mass.

Risposta: D

In KMnO₄, K is +1 and four O are -8, so Mn is +7; in MnCl₂, with two Cl at -1, Mn is +2. Its oxidation number falls, so manganese is reduced, from +7 to +2. Chlorine does change, from -1 in HCl to 0 in Cl₂, but a rise is oxidation, so that is the species oxidised. Hydrogen and potassium stay at +1 throughout, and +6 is the oxidation number of Mn in the manganate ion, not in KMnO₄.

Risposta: C

There are 4 moles of gas on the left and 2 on the right, so by Le Chatelier's principle a higher pressure shifts the equilibrium towards the side with fewer gas moles, the ammonia side. Raising the temperature favours the endothermic, backward reaction, and lowering the pressure favours the side with more gas moles. Removing nitrogen shifts the equilibrium to the left. A catalyst speeds up both directions equally, so equilibrium is reached sooner but its position does not change.

Risposta: A

The carbon skeleton can be a chain of five (pentane), a chain of four with a methyl branch on the second carbon (2-methylbutane), or a chain of three with two methyl groups on the middle carbon (2,2-dimethylpropane). Every other drawing is one of these three turned or renumbered: a methyl on carbon 3 of butane is still 2-methylbutane, and an ethyl branch on a four-carbon chain just makes a longer chain. Two is the answer for C₄H₁₀, not C₅H₁₂.

Risposta: D

Write every number as a power of 2: 4^3 = (2^2)^3 = 2^6 and 8^3 = (2^3)^3 = 2^9. The expression is 2^5 × 2^6 / 2^9 = 2^(5 + 6 - 9) = 2^2 = 4. Writing 8^3 as 2^6 instead of 2^9 gives 32, and similar slips with the indices give the other powers of 2 offered.

Risposta: B

Factorise: x^2 - x - 6 = (x - 3)(x + 2), which is zero at x = 3 and x = -2. The parabola opens upwards, so it is at or below zero between its roots: -2 ≤ x ≤ 3. x ≤ -2 or x ≥ 3 describes where the expression is positive (or zero), -2 < x < 3 wrongly drops the endpoints where it equals zero, and the options with -3 and 2 come from a sign error in the factors.

Risposta: E

Apply f first, then g: g(f(x)) = (2x - 3)^2 + 1 = 4x^2 - 12x + 9 + 1 = 4x^2 - 12x + 10. Check with x = 2: f(2) = 1 and g(1) = 2, and 16 - 24 + 10 = 2. The answer 2x^2 - 1 is f(g(x)), the other order; 4x^2 + 10 forgets the middle term of the square, and the cubic is the product f(x) × g(x).

Risposta: C

Complete the square: (x - 3)^2 - 9 + (y + 4)^2 - 16 = 0, so (x - 3)^2 + (y + 4)^2 = 25. The centre is (3, -4) and the radius is sqrt(25) = 5. Taking the signs of the coefficients directly gives (-3, 4), 25 is the radius squared, and (6, -8) uses the coefficients without halving them.

Risposta: A

By the cosine rule, cos θ = (5^2 + 8^2 - 7^2) / (2 × 5 × 8) = (25 + 64 - 49) / 80 = 40/80 = 1/2, so θ = 60°. It cannot be 90°, because 5^2 + 8^2 = 89 is not 49. A slip in the sign of 49 gives cos θ = -1/2 and 120°, which would also make 7 cm the longest side, and it is not.

Risposta: D

There are 6 × 6 = 36 equally likely ordered outcomes. A total of 8 comes from (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2): 5 outcomes, so the probability is 5/36. Counting (2, 6) and (6, 2) as one outcome would undercount, and 1/6 = 6/36 is the probability of a total of 7, the most likely sum.

Risposta: E

From rest, s = (1/2)at^2 = (1/2) × 2 × 5^2 = 25 m. Equivalently, the final speed is 10 m/s, the average speed is 5 m/s, and 5 m/s × 5 s = 25 m. The answer 50 m uses the final speed as if it applied all the way, and 10 m is the final speed mistaken for a distance.

Risposta: B

Take upwards as positive with the origin at the cliff edge. The sea is at -25 m, so -25 = 20t - 5t^2, which gives t^2 - 4t - 5 = 0 and (t - 5)(t + 1) = 0. The physical root is t = 5 s. Check: after 5 s the displacement is 100 - 125 = -25 m. 2 s is the time to the top of the flight and 4 s the time to return to the cliff edge, before the extra 25 m fall.

Risposta: C

The resultant force is 20 N - 8 N = 12 N in the direction of the pull, and a = F/m = 12/4 = 3 m/s^2. Ignoring friction gives 20/4 = 5 m/s^2, adding the friction to the pull gives 28/4 = 7 m/s^2, and 8/4 = 2 m/s^2 uses the friction force alone.

Risposta: A

On the frictionless slope all the potential energy becomes kinetic energy: mgh = 2 × 10 × 5 = 100 J at the bottom. On the rough surface friction does work F × d to remove it, so 4 × d = 100 and d = 25 m. 50 m and 12.5 m come from doubling or halving that energy, for instance by mishandling the 1/2 in (1/2)mv^2, and 2.5 m divides mgh by the weight rather than the friction force; the slope's length does not matter, only its height.

Risposta: D

Hydrostatic pressure is p = ρgh = 1000 × 10 × 5 = 50 000 Pa = 5 × 10^4 Pa. The answer 1.5 × 10^5 Pa adds atmospheric pressure (about 1 × 10^5 Pa), which the question excludes, and the other powers of ten come from slips in the arithmetic.

Risposta: E

The parallel pair has resistance (6 × 3)/(6 + 3) = 2 Ω, so the total is 2 + 2 = 4 Ω and the battery current is 12/4 = 3 A. The parallel pair takes a potential difference of 3 A × 2 Ω = 6 V, so the current in the 3 Ω resistor is 6/3 = 2 A (and 1 A in the 6 Ω, which adds back to 3 A). 3 A is the total current, and 4 A comes from putting the full 12 V across the 3 Ω resistor.