Palacký University Medicine Entrance Exam Sample Test 1, with Answers (2027)

Answer: D

Mitochondria (and chloroplasts) keep a small circular genome and bacterial-type 70S ribosomes, which is the main evidence for the endosymbiotic theory. The Golgi apparatus, lysosomes and rough ER belong to the cell's own membrane system and carry no DNA. The nucleolus does contain DNA, but it is part of the linear nuclear chromosomes, and it assembles the subunits of the cytoplasm's 80S ribosomes rather than containing 70S ones.

Answer: C

Meiosis I separates homologous chromosomes, not sister chromatids, so each daughter cell receives one chromosome of each pair (23) and each is still made of two chromatids joined at the centromere. The chromatids only separate in anaphase II, which gives 23 single-chromatid chromosomes at the end of meiosis II. 46 chromosomes per cell would be the result of mitosis, not of a reduction division.

Answer: E

The genetic code is read in triplets, so 903 ÷ 3 = 301 codons. The stop codon codes for no amino acid, which leaves 300. The start codon AUG does code for an amino acid (methionine), so it is not subtracted as well; 299 removes it wrongly, and 301 forgets that the stop codon codes for nothing.

Answer: B

The man's mother was OO, so he must be heterozygous, IᴮIᴼ. The woman gives Iᴬ or Iᴮ and the man Iᴮ or Iᴼ, so the four equally likely children are IᴬIᴮ (AB), IᴬIᴼ (A), IᴮIᴮ (B) and IᴮIᴼ (B). Only IᴬIᴼ is group A, a probability of 1/4. 1/2 is the probability of group B, and 0 would be right only if the father were IᴮIᴮ.

Answer: A

The unaffected father passes a normal X to his daughters, so the affected brother's X came from their mother, who must be a carrier (XᴴXʰ). The woman therefore has a 1/2 chance of being a carrier. If she is, the chance of a child being a son who receives Xʰ is 1/2 × 1/2 = 1/4; overall 1/2 × 1/4 = 1/8. 1/4 assumes she is certainly a carrier, and 1/2 is the chance for a son of a known carrier, not for any child.

Answer: D

Mature mRNA has already been spliced, so the complementary DNA (cDNA) made from it contains only exons; bacteria have no spliceosomes and could not remove introns from a genomic copy. The promoter is not in the mRNA at all, so cDNA lacks it rather than supplying it. Cutting the plasmid is the job of a restriction enzyme, sealing the backbone is DNA ligase, and the resistance marker is already part of the plasmid vector.

Answer: E

The appendicular skeleton is the limbs together with the girdles that attach them, and the clavicle is part of the pectoral (shoulder) girdle. The sternum and hyoid bone are axial. The sacrum and the atlas are vertebrae, so they are axial too, even though the sacrum articulates with the hip bones of the pelvic girdle.

Answer: A

At the start of ventricular systole the pressure in the ventricles rises above that in the atria, and the tricuspid and mitral valves snap shut, giving the first sound. The second heart sound is the closure of the aortic and pulmonary (semilunar) valves at the end of systole. Valves opening make no audible sound, and atrial contraction is normally silent.

Answer: C

Fetal red cells enter the mother's circulation mainly at delivery; the injected anti-D binds them and they are cleared before her B cells can mount a primary response and form memory cells. Without memory cells there is no rapid IgG response to attack an Rh-positive fetus next time. The injection is passive, not active, immunity and lasts only weeks. It cannot help once she has already made her own anti-D, and it does not alter her genotype or the placenta.

Answer: B

Inside the epithelial cells the fatty acids and monoglycerides are rebuilt into triglycerides and packed with protein into chylomicrons. These are too large to enter the blood capillaries, so they pass into the lacteal and travel in the lymph to the thoracic duct. Micelles exist only in the gut lumen, carrying the products to the cell surface, and the bile duct carries bile towards the intestine, not absorbed food away from it.

Answer: E

Vitamin K is needed for the liver to produce active prothrombin and factors VII, IX and X, so a deficiency prolongs clotting time. Vitamin C deficiency (scurvy) also causes bleeding gums, which makes it tempting, but that is due to weak collagen in vessel walls, not to missing clotting factors. Vitamin A, B12 and D deficiencies cause night blindness, anaemia and rickets respectively.

Answer: A

Glucose is always freely filtered at the glomerulus, and in health all of it is reabsorbed by co-transport carriers in the proximal convoluted tubule. When blood glucose is very high, the filtered load exceeds the carriers' maximum rate (the renal threshold), so the excess stays in the urine. Filtration of glucose does not depend on insulin, the collecting duct does not secrete it, and ADH controls water, not glucose, reabsorption.

Answer: D

Aldosterone makes the distal tubule and collecting duct reabsorb Na⁺ in exchange for K⁺, which is secreted into the urine. Water follows the retained sodium, so blood volume and blood pressure rise while plasma potassium falls. Low blood pressure with high potassium is the picture of too little aldosterone, sodium-rich urine is the opposite of its effect, and plasma calcium is controlled by parathyroid hormone.

Answer: B

In the knee-jerk reflex the sensory neurone from the muscle spindle synapses straight onto the motor neurone in the spinal cord, so there is a single synapse in the central nervous system and no relay neurone. The arc still has two neurones, not one. The reflex is a spinal reflex that does not need the cortex, and the term says nothing about the transmitter or about the side of the body.

Answer: C

Human chorionic gonadotrophin from the trophoblast acts like LH and keeps the corpus luteum secreting progesterone until the placenta takes over; it is also the hormone a pregnancy test detects. In a normal cycle progesterone suppresses the pituitary's gonadotrophins, so the pituitary cannot rescue the corpus luteum, and FSH acts on follicles anyway. Oxytocin acts at birth and in milk ejection, and the corpus luteum cannot sustain itself.

Answer: A

The inner cell mass forms the embryo itself, so its cells can become any body tissue, but not the trophoblast-derived parts of the placenta: they are pluripotent. The zygote is totipotent, since it also gives rise to the placenta. Haematopoietic stem cells are multipotent, making only blood cells, mature red cells have no nucleus and cannot divide, and the trophoblast forms only placental tissue.

Answer: B

Type 2 diabetes, strongly linked to obesity and inactivity, begins with insulin resistance: liver, muscle and fat respond poorly to insulin, which the pancreas at first secretes in normal or even raised amounts. A raised fasting glucose and glucose in the urine occur in both types, so they do not distinguish them. Autoimmune destruction of beta cells and an absolute lack of insulin needing injections from the start describe type 1.

Answer: C

Efficiency = energy in the higher level ÷ energy in the level below × 100 = 1 200 ÷ 15 000 × 100 = 8%. 0.08 is the fraction before multiplying by 100, and 12.5 is the ratio the wrong way round (15 000 ÷ 1 200). An efficiency close to 10% is typical, because most energy is lost in respiration, excretion and uneaten material.

Answer: B

Neutrons = mass number − atomic number = 32 − 16 = 16. The atom has 16 electrons and the 2− charge means it has gained two, giving 18. 14 electrons would be a 2+ ion, and 32 is the mass number, not the neutron count; the charge changes only the electrons, never the neutrons.

Answer: D

NH₃ has three bonding pairs and one lone pair around nitrogen; the lone pair repels more strongly than bonding pairs and squeezes the tetrahedral 109.5° to about 107° (trigonal pyramidal). CH₄ has four bonding pairs and keeps 109.5°, H₂O has two lone pairs and closes to about 104.5°, BF₃ is trigonal planar at 120° and CO₂ is linear at 180°.

Answer: D

Use p₁V₁/T₁ = p₂V₂/T₂ with kelvin temperatures: 127 °C = 400 K and 527 °C = 800 K. V₂ = 4.0 × (150/50) × (800/400) = 4.0 × 3 × 2 = 24.0 dm³. 8.0 dm³ allows only for the temperature, 2.7 dm³ inverts the pressure ratio, and 49.8 dm³ uses Celsius temperatures, which an ideal gas law never accepts.

Answer: C

Mᵣ(CO₂) = 44, so 4.4 g is 0.10 mol. The equation is 1 : 1, so 0.10 mol CaCO₃ is needed, and Mᵣ(CaCO₃) = 40 + 12 + 48 = 100, giving 10.0 g. 5.6 g is the mass of calcium oxide formed, and 4.4 g assumes mass is conserved between one reactant and one product alone.

Answer: E

Moles of H₂SO₄ = 0.150 × 0.0200 = 0.00300 mol, and each mole neutralises 2 mol NaOH, so n(NaOH) = 0.00600 mol. Concentration = 0.00600 ÷ 0.0250 = 0.240 mol dm⁻³. 0.120 ignores the 1 : 2 ratio, and 0.188 and 0.375 put the two volumes the wrong way round.

Answer: C

n(H⁺) = 0.10 × 0.050 = 0.0050 mol and n(OH⁻) = 0.080 × 0.050 = 0.0040 mol, leaving 0.0010 mol H⁺ in excess. The total volume is 100 cm³, so [H⁺] = 0.0010 ÷ 0.100 = 0.010 mol dm⁻³ and pH = 2.0. 1.7 divides by 50 cm³ instead of the mixed volume, 1.0 is the pH of the acid before mixing, and 7.0 would need equal amounts of acid and alkali.

Answer: A

ΔH = ΣΔfH(products) − ΣΔfH(reactants) = [−394 + 2(−286)] − [−75 + 0] = −966 + 75 = −891 kJ mol⁻¹; oxygen is an element, so its ΔfH is zero. −605 counts only one mole of water, −1041 adds the methane value instead of subtracting it, and +891 has the sign reversed although combustion is exothermic.

Answer: C

Only a change of temperature changes Kc. The forward reaction is exothermic, so lowering the temperature shifts the equilibrium to the right and increases Kc and the yield. Raising pressure and removing ammonia also increase the yield, but Kc stays the same; a catalyst changes neither; raising the temperature changes Kc but lowers the yield.

Answer: B

A conjugate pair differs by exactly one proton. HCO₃⁻ donates a proton and becomes CO₃²⁻, so they are a pair (the other pair in the equation, H₂O and H₃O⁺, is not offered). HCO₃⁻ and H₂O are the two reactants, an acid and a base but not conjugate to each other, and the other combinations do not differ by a single H⁺.

Answer: A

A buffer needs a weak acid and its conjugate base together in significant amounts. In A, 0.0025 mol NaOH neutralises half of the 0.0050 mol ethanoic acid, leaving equal amounts of CH₃COOH and CH₃COO⁻. In B and D the weak acid or weak base is neutralised completely, leaving only a salt, and C and E contain a strong acid, whose anion Cl⁻ cannot take up added H⁺.

Answer: E

Manganese falls from +7 to +2, so each MnO₄⁻ gains 5 electrons: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Each Fe²⁺ loses only one electron, so 5 Fe²⁺ are needed per MnO₄⁻. 8 is the number of H⁺ ions in the half-equation, not the number of electrons transferred.

Answer: C

Oxidising power falls down Group 17, so a halogen displaces only the halide ions below it: Cl₂ + 2KBr → 2KCl + Br₂, turning the solution orange. Iodine is the weakest oxidising agent offered and cannot displace bromide or chloride, and bromine cannot displace chloride or fluoride, which lie above it.

Answer: B

The four are 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. Only 2-chlorobutane has a carbon bonded to four different groups (H, Cl, CH₃ and C₂H₅), so only it has two enantiomers. In 1-chloro-2-methylpropane the branching carbon carries two identical CH₃ groups, so it is not chiral, which rules out two.

Answer: E

FeBr₃ polarises Br₂ to generate the electrophile Br⁺, which attacks the electron-rich ring; a hydrogen is then lost so the stable delocalised ring is restored, which makes it substitution. Addition, as in ethene with bromine, would destroy the aromatic system and does not happen here. Free-radical substitution needs UV light and is typical of alkanes, and nucleophiles are repelled by the ring's π electrons.

Answer: D

2-Methylpropan-2-ol is a tertiary alcohol: the carbon carrying the OH has no hydrogen atom to lose, so it resists oxidation and the orange dichromate stays orange. Ethanol and propan-1-ol are primary and are oxidised to aldehydes and then carboxylic acids; propan-2-ol and butan-2-ol are secondary and are oxidised to ketones.

Answer: A

A diamine and a dicarboxylic acid each have two reactive groups, so they join end to end by amide bonds, losing a molecule of water each time: this pair gives nylon-6,6. Ethene, propene and chloroethene form addition polymers with no amide links. Ethanol and ethanoic acid have only one reactive group, so they would end any chain rather than extend it.

Answer: B

A sugar reduces Benedict's or Fehling's reagent through a free anomeric carbon that lets the ring open to a carbonyl form. In sucrose the bond links the anomeric carbons of both glucose and fructose, so neither is free. Sucrose is a disaccharide that does contain glucose and is very soluble, and lactose has a β(1→4) bond yet is reducing, so that cannot be the reason.

Answer: A

A competitive inhibitor resembles the substrate and binds reversibly to the active site, so more substrate is needed to reach half the maximum rate: the apparent Km rises. At very high substrate concentration the substrate outcompetes the inhibitor, so the same Vmax is still reached. A lower Vmax with unchanged Km is the mark of a non-competitive inhibitor, which binds elsewhere on the enzyme.

Answer: B

Work is force times distance, so 1 J = 1 N m. A newton is kg m s^-2 (from F = ma), so 1 J = kg m s^-2 × m = kg m^2 s^-2. The other options are other derived units: kg m s^-2 is the newton itself, kg m^2 s^-3 is the watt (a joule per second) and kg m^-1 s^-2 is the pascal (a newton per square metre).

Answer: A

1 litre = 1 dm^3 = 10^-3 m^3, so 5.0 L = 5.0 × 10^-3 m^3. One minute is 60 s, so the flow rate is 5.0 × 10^-3 / 60 = 8.3 × 10^-5 m^3/s. 8.3 × 10^-2 is 5.0/60, the rate in litres per second, with the litre never converted. 3.0 × 10^-1 multiplies by 60 instead of dividing. 8.3 × 10^-8 takes a litre as 10^-6 m^3, which is the cubic centimetre.

Answer: D

When quantities are multiplied, their percentage uncertainties add. Length: 0.1/20.0 = 0.5%. Width: 0.2/10.0 = 2.0%. Area = 200 cm^2 with an uncertainty of 0.5% + 2.0% = 2.5% (that is, ± 5 cm^2). 0.3% adds the absolute uncertainties (0.1 + 0.2) and then treats the sum as a percentage, 2.0% keeps only the larger contribution, and 1.25% averages the two percentages instead of adding them.

Answer: C

From rest, s = (1/2)at^2 = 0.5 × 2.0 × 5.0^2 = 0.5 × 2.0 × 25 = 25 m. 50 m forgets the factor of 1/2. 10 m is at = 10, which is the final speed in m/s, not a distance. 5 m is (1/2)at, with the time not squared.

Answer: D

The vertical motion starts from zero vertical speed: 20 = (1/2) × 10 × t^2, so t^2 = 4 and t = 2.0 s. The vertical speed on impact is gt = 10 × 2.0 = 20 m/s, and the horizontal speed stays 15 m/s. The speed is sqrt(15^2 + 20^2) = sqrt(225 + 400) = sqrt(625) = 25 m/s. Check by energy: v^2 = 15^2 + 2 × 10 × 20 = 625, v = 25 m/s. 35 m/s adds the components as if they pointed the same way; 20 m/s and 15 m/s are only the vertical or only the horizontal component.

Answer: B

Work against gravity = mgh = 60 × 10 × 3.0 = 1800 J. Power = work / time = 1800 / 4.0 = 450 W. 1800 W is the work in joules, never divided by the time. 7200 W multiplies by the time instead of dividing. 45 W leaves out g, using mass instead of weight (60 × 3.0 / 4.0).

Answer: A

The normal reaction equals the weight, mg = 4.0 × 10 = 40 N, so the friction force is 0.20 × 40 = 8.0 N, opposing the motion. Net force = 20 - 8.0 = 12 N, and a = F/m = 12 / 4.0 = 3.0 m/s^2. 5 m/s^2 ignores friction (20/4.0). 2 m/s^2 is the friction force alone divided by the mass. 7 m/s^2 adds friction to the pull (28/4.0), as if friction helped the motion.

Answer: C

By Archimedes' principle the upthrust equals the weight of fluid displaced: ρVg = 1000 × 2.0 × 10^-3 × 10 = 20 N. The displaced water has a mass of 2.0 kg (two litres). 2 N leaves out g, giving the mass in kg rather than a weight in newtons; 0.2 N and 200 N are slips of a power of ten in handling 10^-3.

Answer: B

For an incompressible fluid the volume flow rate Av is the same everywhere along the tube (continuity equation), so v is inversely proportional to the cross-sectional area. Area is πr^2: halving the radius divides the area by 2^2 = 4, so the speed rises by a factor of 4. 'It doubles' scales with the radius instead of the area. A factor of 16 is the r^4 of Poiseuille's law, which governs resistance to flow, not speed. 'It halves' has the relationship the wrong way round: fluid speeds up in a constriction.

Answer: A

Q = mcΔT = 0.50 × 4200 × (80 - 20) = 0.50 × 4200 × 60 = 126 000 J = 126 kJ. 168 kJ uses the final temperature, 80, as the temperature change. 42 kJ uses the starting temperature, 20. 12.6 kJ is a power-of-ten slip in converting joules to kilojoules.

Answer: D

At constant volume p/T is constant, with T in kelvin: 27 °C = 300 K and 127 °C = 400 K. p2 = 120 × 400/300 = 160 kPa. 564 kPa uses Celsius temperatures (120 × 127/27), which is wrong because pressure is proportional to absolute temperature. 90 kPa inverts the ratio, and 120 kPa assumes the pressure cannot change, which would only hold if the volume could expand.

Answer: C

λ = v/f = 1500 / (3.0 × 10^6) = 5.0 × 10^-4 m = 0.5 mm. 2 mm comes from dividing the other way round (f/v = 2000) and misreading the result. 5 mm and 0.05 mm are power-of-ten slips, usually from taking 1 MHz as 10^5 or 10^7 Hz, or from converting metres to millimetres wrongly.

Answer: B

E = (1/2)CV^2 = 0.5 × 20 × 10^-6 × 12^2 = 0.5 × 20 × 10^-6 × 144 = 1.44 × 10^-3 J = 1.44 mJ. Check with E = (1/2)QV: Q = CV = 240 µC, and 0.5 × 240 × 10^-6 × 12 = 1.44 mJ. 2.88 mJ forgets the 1/2. 0.24 mJ is CV, the charge (2.4 × 10^-4 C) read as an energy. 0.12 mJ is (1/2)CV, with V not squared.

Answer: A

E = Pt = 2000 W × 360 s = 720 000 J = 720 kJ (equivalently 2.0 kW × 0.1 h = 0.2 kWh). 12 kJ uses the time in minutes without converting to seconds. 0.2 kJ is the kilowatt-hour figure with the wrong unit attached. 43 MJ treats 6.0 minutes as 6.0 hours.

Answer: D

The parallel pair has resistance (6.0 × 3.0)/(6.0 + 3.0) = 18/9 = 2.0 Ω. Total resistance = 2.0 + 4.0 = 6.0 Ω, so the battery current is 12/6.0 = 2.0 A. The potential difference across the 4.0 Ω resistor is 2.0 × 4.0 = 8.0 V, leaving 4.0 V across the parallel pair (8.0 + 4.0 = 12 V). 4.0 V is the voltage across the parallel pair, not the 4.0 Ω resistor. 3.7 V puts all three resistors in series (12 × 4/13). 6.0 V assumes the battery voltage splits equally.

Answer: C

For an ideal transformer Vs/Vp = Ns/Np, so Vs = 240 × 100/2000 = 240/20 = 12 V. This is a step-down transformer, so the output must be lower than 240 V. 4800 V uses the turns ratio upside down (a step-up). 120 V and 24 V divide by 2 and 10 rather than by the actual ratio of 20.

Answer: B

Using 1/f = 1/u + 1/v (real is positive): 1/v = 1/10 - 1/15 = 3/30 - 2/30 = 1/30, so v = +30 cm, a real image on the far side of the lens. Magnification = v/u = 30/15 = 2, and a real image formed by a single lens is inverted. 6.0 cm comes from adding 1/10 and 1/15 instead of subtracting. A virtual, upright image needs the object inside the focal length, which 15 cm is not. A real image the same size as the object needs the object at 2f = 20 cm, not 15 cm.

Answer: C

24 hours is 24/6.0 = 4 half-lives, and each halves what is left: (1/2)^4 = 1/16. 1/4 takes the number of half-lives (4) as the denominator instead of 2^4. 1/8 counts only three half-lives. 1/24 has no basis: decay is exponential, not a fraction per hour.

Answer: C

84 = 2^2 × 3 × 7 and 120 = 2^3 × 3 × 5. The LCM takes the highest power of each prime: 2^3 × 3 × 5 × 7 = 840. Check: HCF × LCM = product, and the HCF is 2^2 × 3 = 12, so LCM = 10080/12 = 840. 12 is the highest common factor. 420 takes 2^2 instead of 2^3, and 420 is not a multiple of 120. 10080 is the product, a common multiple but not the lowest.

Answer: D

72 = 2^3 × 3^2. A divisor is 2^a × 3^b with a from 0 to 3 (4 choices) and b from 0 to 2 (3 choices), so there are 4 × 3 = 12. Listed: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. 6 multiplies the exponents (3 × 2) without adding 1 to each. 10 leaves out 1 and 72 themselves.

Answer: A

Only the last digit matters for division by 5. Powers of 7 end in 7, 9, 3, 1 and then repeat with period 4 (7, 49, 343, 2401, ...). Since 100 = 4 × 25, 7^100 ends in 1, like 7^4, and leaves remainder 1 on division by 5. Equivalently, 7 ≡ 2 (mod 5) and 2^4 = 16 ≡ 1, so 7^100 ≡ (2^4)^25 ≡ 1. 2, 4 and 3 are the remainders of 7^1, 7^2 and 7^3, which come from misplacing 100 in the cycle.

Answer: B

Daily amount = 15 × 18 = 270 mg; divided into three equal doses, each is 270/3 = 90 mg. 270 mg is the whole day's amount given at once. 810 mg multiplies by 3 instead of dividing. 5 mg divides 15 by 3 and never uses the child's mass.

Answer: B

The multipliers are 1.20 and 0.80, and 1.20 × 0.80 = 0.96, so the final price is 96% of the original: a 4% decrease. For example, 100 becomes 120 and then 120 - 24 = 96. 'No change' assumes the two percentages cancel, but the 20% reduction is taken from a larger amount. The decrease is 4%, not 2%, and it cannot be an increase.

Answer: C

Solute: 0.30 × 200 = 60 mL plus 0.10 × 300 = 30 mL, a total of 90 mL in 500 mL of mixture. 90/500 = 0.18 = 18%. 20% is the plain average of 30% and 10%, which ignores that there is more of the weaker solution. 22% weights the percentages the wrong way round (0.10 × 200 + 0.30 × 300 = 110 mL). 40% adds the two percentages.

Answer: A

From the second equation x = y + 1. Substituting: 2(y + 1) + 3y = 12, so 5y + 2 = 12, y = 2 and x = 3. Check: 2(3) + 3(2) = 12 and 3 - 2 = 1. 2 is the value of y, 5 is x + y and 6 is xy, all answers to a different question.

Answer: A

Squaring both sides: x + 7 = x^2 - 10x + 25, so x^2 - 11x + 18 = 0, which factorises as (x - 2)(x - 9) = 0. Squaring can introduce false roots, so check each in the original: x = 9 gives sqrt(16) = 4 and 9 - 5 = 4, which works. x = 2 gives sqrt(9) = 3 but 2 - 5 = -3, and a square root is never negative, so x = 2 is rejected. 'x = 2 and x = 9' keeps the extraneous root; 'x = 2 only' keeps the wrong one.

Answer: D

8^(2/3) = (cube root of 8)^2 = 2^2 = 4. 4^(-1/2) = 1/sqrt(4) = 1/2. The product is 4 × 1/2 = 2. 4 is the first factor alone, 1/2 the second alone, and 16 multiplies 4 by 4, ignoring both the negative sign and the root in 4^(-1/2).

Answer: C

2 hours = 120 minutes = 6 doubling periods, so the count is 500 × 2^6 = 500 × 64 = 32 000. 16 000 counts only five doublings, and 64 000 counts seven (an off-by-one in the number of periods). 3000 adds 500 six times, treating doubling as linear growth.

Answer: B

From the 3rd to the 7th term is 4 steps, so 4d = 23 - 11 = 12 and d = 3. The first term is a = 11 - 2(3) = 5 and the 10th term is 5 + 9(3) = 32. Sum = (10/2)(5 + 32) = 5 × 37 = 185. 155 takes the first term as 11 - 3(3) = 2, counting three steps back from the 3rd term instead of two. 200 takes the 10th term as a + 10d = 35. 320 is 10 times the 10th term, not the sum.

Answer: D

log_2 32 = 5 because 2^5 = 32. log_3 (1/9) = -2 because 3^-2 = 1/9. The sum is 5 + (-2) = 3. 7 drops the minus sign on the second logarithm. -10 multiplies the two values, and 2.5 divides them; neither is what a sum of logarithms means.

Answer: A

Combining the logarithms: log_3 [x(x - 6)] = 3, so x(x - 6) = 3^3 = 27, x^2 - 6x - 27 = 0 and (x - 9)(x + 3) = 0. A logarithm needs a positive argument, so x > 6, and x = -3 is rejected. Check x = 9: log_3 9 + log_3 3 = 2 + 1 = 3. Answers that keep x = -3 ignore the domain; x = 27 sets x alone equal to 3^3 and forgets the second logarithm.

Answer: B

cos^2 θ = 1 - sin^2 θ = 1 - 9/25 = 16/25, so cos θ = ±4/5. In the second quadrant cosine is negative, so cos θ = -4/5 and tan θ = sin θ / cos θ = (3/5)/(-4/5) = -3/4. 3/4 forgets that tangent is negative between 90° and 180°. 4/3 and -4/3 are cos/sin, the cotangent, rather than sin/cos.

Answer: C

The radius is half the diameter, 5 cm, so the area is πr^2 = π × 5^2 = 25π cm^2 (about 78.5 cm^2). 10π is the circumference, πd, which is a length, not an area. 100π uses the diameter as the radius. 20π is 2πd, mixing the circumference formula with the diameter.

Answer: D

The gradient of the given line is (10 - 2)/(7 - 1) = 8/6 = 4/3. Perpendicular gradients multiply to -1, so the perpendicular gradient is -1/(4/3) = -3/4. 4/3 is the original line's gradient (a parallel line). -4/3 changes the sign but does not take the reciprocal, and 3/4 takes the reciprocal but does not change the sign.

Answer: B

Order does not matter in a committee, so the count is 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. 336 is the number of ordered selections (for example a chair, a secretary and a treasurer), which counts each committee 3! = 6 times. 512 = 8^3 allows the same person to be chosen more than once. 24 is simply 8 × 3.

Answer: C

LEVEL has 5 letters: L twice, E twice and V once. Arrangements = 5!/(2! × 2!) = 120/4 = 30. 120 treats all five letters as different. 60 allows for only one of the repeated letters (120/2). 15 divides by a third 2!, as if V were repeated too.