I85PREP
⌘ K
Zhongkao - independent subject foundations study materials
Zhongkao - independent subject foundations study resources

Zhongkao - independent subject foundations practice questions

One question at a time. Build your Zhongkao - independent subject foundations understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Lower-secondary number, algebra, geometry and data

数据 8，10，12 的平均数是：

  1. 9
  2. 10 Correct answer.
  3. 11
  4. 12

Explanation

平均数等于总和除以个数，(8 + 10 + 12) ÷ 3 = 30 ÷ 3 = 10。

Question 2Lower-secondary number, algebra, geometry and data

学校本月回收纸张 25 千克，上月回收 20 千克。本月比上月增长了：

  1. 5%
  2. 20%
  3. 25% Correct answer.
  4. 45%

Explanation

增长量是 25 - 20 = 5 千克，以原来的 20 千克为基数，5 ÷ 20 = 25%。

Question 3Lower-secondary number, algebra, geometry and data

若 x + y = 10，x - y = 2，则 x 和 y 分别为：

  1. 5 和 5
  2. 6 和 4 Correct answer.
  3. 7 和 3
  4. 8 和 2

Explanation

两式相加得 2x = 12，所以 x = 6；代入 x + y = 10 得 y = 4。

Question 4Lower-secondary number, algebra, geometry and data

计算 15^2 - 5^2 的值，最简便的方法得到：

  1. 100
  2. 150
  3. 200 Correct answer.
  4. 250

Explanation

用平方差公式，15^2 - 5^2 = (15 - 5)(15 + 5) = 10 × 20 = 200。

Question 5Lower-secondary number, algebra, geometry and data

函数 y = 2x - 3 中，当 x = 4 时，y 等于：

  1. 3
  2. 4
  3. 5 Correct answer.
  4. 8

Explanation

把 x = 4 代入，y = 2 × 4 - 3 = 8 - 3 = 5。

Question 6Lower-secondary number, algebra, geometry and data

直角三角形两条直角边分别为 6 cm 和 8 cm，斜边长为：

  1. 9 cm
  2. 10 cm Correct answer.
  3. 12 cm
  4. 14 cm

Explanation

根据勾股定理，斜边平方为 6^2 + 8^2 = 36 + 64 = 100，所以斜边为 10 cm。

Question 7Lower-secondary number, algebra, geometry and data

方程 x^2 - 5x + 6 = 0 的两个根是：

  1. 1 和 6
  2. 2 和 3 Correct answer.
  3. -2 和 -3
  4. 0 和 6

Explanation

因式分解得 x^2 - 5x + 6 = (x - 2)(x - 3)，所以根为 2 和 3。

Question 8Lower-secondary number, algebra, geometry and data

一个长方形的周长为 30 cm，面积为 54 cm^2。若长大于宽，则长和宽分别是：

  1. 6 cm 和 9 cm
  2. 9 cm 和 6 cm Correct answer.
  3. 10 cm 和 5 cm
  4. 12 cm 和 3 cm

Explanation

设长宽为 a、b，则 a + b = 15，ab = 54。满足条件的是 9 和 6，且长大于宽，所以长 9 cm、宽 6 cm。

Question 9Lower-secondary number, algebra, geometry and data

半径 OA 垂直于圆 O 在点 A 处的切线 l。若直线 l 与另一条过 A 的直线夹角为 35°，则该过 A 直线与 OA 的夹角是：

  1. 35°
  2. 55° Correct answer.
  3. 65°
  4. 90°

Explanation

半径 OA 与切线 l 垂直，夹角为 90°。另一条直线与切线夹角为 35°，则它与 OA 的锐角为 90° - 35° = 55°。

Question 10Lower-secondary number, algebra, geometry and data

袋中有 2 个红球和 3 个蓝球，不放回连续取 2 个，两个都是红球的概率是：

  1. 1/10 Correct answer.
  2. 1/5
  3. 2/5
  4. 4/25

Explanation

第一次取红球概率为 2/5；不放回后还剩 1 个红球和 4 个球，第二次取红球概率为 1/4。相乘得 2/5 × 1/4 = 1/10。

Question 11Lower-secondary number, algebra, geometry and data

直线 y = x + 1 与 y = -x + 5 的交点坐标是：

  1. (1, 2)
  2. (2, 3) Correct answer.
  3. (3, 2)
  4. (4, 1)

Explanation

令 x + 1 = -x + 5，得 2x = 4，x = 2；代入 y = x + 1 得 y = 3，所以交点为 (2, 3)。

Question 12English grammar, contextual language and reading evidence

The plural form of “child” is:

  1. childs
  2. children Correct answer.
  3. childes
  4. childrens

Explanation

“Child” has the irregular plural form “children.” It does not take the regular -s ending.

Question 13English grammar, contextual language and reading evidence

Choose the correct sentence:

  1. I has finished my homework.
  2. I have finished my homework. Correct answer.
  3. I finishing my homework.
  4. I am finish my homework.

Explanation

The subject “I” takes “have,” and “finished” is the past participle used in the present perfect tense.

Question 14English grammar, contextual language and reading evidence

Choose the correct form: She reads ______ than before.

  1. careful
  2. more careful
  3. more carefully Correct answer.
  4. most carefully

Explanation

“Reads” is a verb, so the adverb “carefully” is needed. The comparison with “than before” requires “more carefully.”

Question 15English grammar, contextual language and reading evidence

Read the sentence: “Tom missed the early bus, so he ran all the way to school.” What can we infer?

  1. Tom wanted to be late.
  2. Tom was trying not to be late. Correct answer.
  3. Tom did not go to school.
  4. Tom drove a car to school.

Explanation

Missing the bus creates a risk of being late, and running to school shows Tom tried to arrive on time.

Question 16English grammar, contextual language and reading evidence

The passive form of “Workers built the bridge last year” is:

  1. The bridge built workers last year.
  2. The bridge was built last year. Correct answer.
  3. The bridge is building last year.
  4. Workers were built last year.

Explanation

The object “the bridge” becomes the subject, and past passive uses “was/were + past participle”: “was built.”

Question 17English grammar, contextual language and reading evidence

In “Although the problem was difficult, Li Hua solved it,” the word “Although” introduces:

  1. a contrast Correct answer.
  2. a place
  3. a list
  4. a command

Explanation

“Although” shows contrast between the difficulty of the problem and Li Hua's success in solving it.

Question 18English grammar, contextual language and reading evidence

Choose the best word: The girl ______ helped me is my classmate.

  1. which
  2. who Correct answer.
  3. where
  4. when

Explanation

The relative pronoun “who” refers to a person and acts as the subject of “helped me.”

Question 19English grammar, contextual language and reading evidence

Choose the best word: ______ the heavy rain, the volunteers arrived on time.

  1. Because
  2. Despite Correct answer.
  3. Unless
  4. Since

Explanation

“Despite” is followed by a noun phrase and shows contrast: the heavy rain did not prevent the volunteers from arriving on time.

Question 20Mechanics, fields, thermal behavior, waves and modern models

两个 6 Ω 电阻并联后再与一个 2 Ω 电阻串联，接在 10 V 电源上。电路总电流是：

  1. 1 A
  2. 2 A Correct answer.
  3. 3 A
  4. 5 A

Explanation

两个 6 Ω 并联的等效电阻为 3 Ω，再与 2 Ω 串联，总电阻为 5 Ω。总电流 I = 10 ÷ 5 = 2 A。

Question 21Mechanics, fields, thermal behavior, waves and modern models

某装置输入能量 500 J，其中有用输出能量 350 J。该装置效率为：

  1. 35%
  2. 50%
  3. 70% Correct answer.
  4. 85%

Explanation

效率 = 有用输出能量 ÷ 输入能量 × 100% = 350 ÷ 500 × 100% = 70%。

Question 22Mechanics, fields, thermal behavior, waves and modern models

An electrical resistor has a resistance $R = 40\ \Omega$ and carries a direct current $I = 0.25\text{ A}$. According to Ohm's law, what is the voltage $U$ across its terminals?

  1. $160\text{ V}$
  2. $2.5\text{ V}$
  3. $10\text{ V}$ Correct answer.
  4. $100\text{ V}$

Explanation

According to Ohm's law, $U = R \times I$. Substituting the given values: $U = 40\ \Omega \times 0.25\text{ A} = 10\text{ V}$.

Question 23Mechanics, fields, thermal behavior, waves and modern models

Two resistors $R_1 = 30\ \Omega$ and $R_2 = 60\ \Omega$ are connected in parallel (en dérivation). What is their equivalent resistance $R_{\text{eq}}$?

  1. $90\ \Omega$
  2. $45\ \Omega$
  3. $15\ \Omega$
  4. $20\ \Omega$ Correct answer.

Explanation

For two resistors in parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega$. Note that the equivalent resistance in parallel is always strictly less than the smallest individual resistance ($20 < 30$).

Question 24Mechanics, fields, thermal behavior, waves and modern models

An incandescent lamp rated $100\text{ W}$ operates for 5 hours every day during a 30-day month. If the electricity tariff is $50\text{ FCFA}$ per kilowatt-hour ($\text{kWh}$), what is the total monthly operating cost of this lamp?

  1. $750\text{ FCFA}$ Correct answer.
  2. $1,500\text{ FCFA}$
  3. $7,500\text{ FCFA}$
  4. $250\text{ FCFA}$

Explanation

Total operating hours $= 5\text{ h/day} \times 30\text{ days} = 150\text{ h}$. Power in kilowatts $= 100\text{ W} = 0.1\text{ kW}$. Electrical energy consumed $E = P \times t = 0.1\text{ kW} \times 150\text{ h} = 15\text{ kWh}$. Total cost $= 15\text{ kWh} \times 50\text{ FCFA/kWh} = 750\text{ FCFA}$.

Question 25Mechanics, fields, thermal behavior, waves and modern models

According to Joule's law ($Q = R \cdot I^2 \cdot t$), if the current $I$ flowing through a resistor of constant resistance $R$ is doubled for the same duration $t$, by what factor is the thermal energy released multiplied?

  1. Multiplied by 2
  2. Multiplied by 4 Correct answer.
  3. Multiplied by 8
  4. Remains unchanged

Explanation

Thermal energy dissipation due to the Joule effect is directly proportional to the square of the current intensity: $Q \propto I^2$. If the current becomes $I' = 2I$, then $Q' = R(2I)^2 t = R(4I^2)t = 4Q$. Thus, the energy is multiplied by 4.

Question 26Mechanics, fields, thermal behavior, waves and modern models

A converging thin lens has a focal length $f = 20\text{ cm}$. What is its optical vergence (vergence $C$) in diopters ($\delta$)?

  1. $+0.05\ \delta$
  2. $+20\ \delta$
  3. $+5\ \delta$ Correct answer.
  4. $-5\ \delta$

Explanation

Vergence is defined by $C = \frac{1}{f}$, where the focal length $f$ must be expressed in meters ($\text{m}$). Converting $f = 20\text{ cm} = 0.20\text{ m}$. Thus $C = \frac{1}{0.20\text{ m}} = +5\ \delta$ (diopters). Converging lenses always have positive vergence.

Question 27Mechanics, fields, thermal behavior, waves and modern models

An object $AB$ is placed perpendicularly to the optical axis of a converging lens at a distance $OA > 2f$ (greater than twice the focal length). What are the characteristics of the image $A'B'$ formed on the screen?

  1. Virtual, upright, and larger than the object
  2. Real, upright, and larger than the object
  3. Virtual, inverted, and same size as the object
  4. Real, inverted, and smaller than the object Correct answer.

Explanation

When an object is placed beyond twice the focal length ($OA > 2f$) of a converging lens, the refracted light rays converge behind the lens between $F'$ and $2F'$, creating an image that can be projected on a screen (real), upside down (inverted), and diminished in height (smaller than the object).

Question 28Mechanics, fields, thermal behavior, waves and modern models

An astronaut has a mass $m = 70\text{ kg}$ on Earth. What are his mass $m$ and his weight $P$ on the Moon, where gravitational acceleration is $g_{\text{moon}} = 1.6\text{ N/kg}$?

  1. $m = 11.2\text{ kg}$ and $P = 112\text{ N}$
  2. $m = 70\text{ kg}$ and $P = 112\text{ N}$ Correct answer.
  3. $m = 70\text{ kg}$ and $P = 700\text{ N}$
  4. $m = 43.75\text{ kg}$ and $P = 70\text{ N}$

Explanation

Mass is an intrinsic measure of the quantity of matter and is invariable regardless of location: $m = 70\text{ kg}$. Weight is the gravitational force $P = m \times g$. On the Moon: $P = 70\text{ kg} \times 1.6\text{ N/kg} = 112\text{ N}$.

Question 29Mechanics, fields, thermal behavior, waves and modern models

Under what condition is a solid body in static equilibrium when subjected to the action of two mechanical forces $\vec{F}_1$ and $\vec{F}_2$?

  1. The two forces are perpendicular with equal magnitudes
  2. The two forces have the same direction and same line of action
  3. The two forces have the same line of action (même droite d'action), the same magnitude ($F_1 = F_2$), and opposite directions (sens opposés: $\vec{F}_1 + \vec{F}_2 = \vec{0}$) Correct answer.
  4. The two forces have different lines of action but equal magnitudes

Explanation

For a solid subject to two forces to remain in static equilibrium, the vector sum must be zero ($\vec{F}_1 + \vec{F}_2 = \vec{0} \implies \vec{F}_1 = -\vec{F}_2$), which requires both forces to share the exact same line of action, possess identical intensities ($F_1 = F_2$), and act in opposite directions.

Question 30Mechanics, fields, thermal behavior, waves and modern models

A solid metal object with volume $V = 2 \times 10^{-3}\text{ m}^3$ is completely submerged in pure water of density $\rho = 1000\text{ kg/m}^3$. Taking $g = 10\text{ N/kg}$, what is the Archimedes buoyant force ($F_A$) exerted by the water on this object?

  1. $200\text{ N}$
  2. $2\text{ N}$
  3. $500\text{ N}$
  4. $20\text{ N}$ Correct answer.

Explanation

Archimedes' principle states that the buoyant force equals the weight of the displaced fluid: $F_A = \rho_{\text{fluid}} \times V_{\text{displaced}} \times g$. Substituting: $F_A = 1000\text{ kg/m}^3 \times (2 \times 10^{-3}\text{ m}^3) \times 10\text{ N/kg} = 2 \times 10 = 20\text{ N}$.

Question 31Mechanics, fields, thermal behavior, waves and modern models

A worker lifts a bag of cement of mass $m = 50\text{ kg}$ vertically at constant speed to a height $h = 4\text{ m}$ in $5\text{ seconds}$. Taking $g = 10\text{ N/kg}$, calculate the mechanical work $W$ performed and the mechanical power $P$ developed.

  1. $W = 2000\text{ J}$ and $P = 400\text{ W}$ Correct answer.
  2. $W = 200\text{ J}$ and $P = 40\text{ W}$
  3. $W = 2000\text{ J}$ and $P = 10000\text{ W}$
  4. $W = 500\text{ J}$ and $P = 100\text{ W}$

Explanation

The lifting force equals the weight $F = P_{\text{weight}} = m \times g = 50 \times 10 = 500\text{ N}$. Work done is $W = F \times h = 500\text{ N} \times 4\text{ m} = 2000\text{ J}$. Power developed is the work divided by time: $P = \frac{W}{t} = \frac{2000\text{ J}}{5\text{ s}} = 400\text{ W}$.

Question 32Mechanics, fields, thermal behavior, waves and modern models

What is the correct sequential order of the 4 strokes in a four-stroke internal combustion engine (moteur thermique à 4 temps)?

  1. 1. Admission (Intake) $\rightarrow$ 2. Compression $\rightarrow$ 3. Explosion-Détente (Power) $\rightarrow$ 4. Échappement (Exhaust) Correct answer.
  2. 1. Compression $\rightarrow$ 2. Admission $\rightarrow$ 3. Échappement $\rightarrow$ 4. Détente
  3. 1. Explosion-Détente $\rightarrow$ 2. Échappement $\rightarrow$ 3. Admission $\rightarrow$ 4. Compression
  4. 1. Admission $\rightarrow$ 2. Échappement $\rightarrow$ 3. Compression $\rightarrow$ 4. Explosion

Explanation

A four-stroke engine operates in the continuous cycle: 1. Admission (intake of air-fuel mixture), 2. Compression (piston compresses mixture), 3. Explosion-Détente (spark ignition drives piston down in the only driving/motor stroke), and 4. Échappement (exhaust valve opens to expel burned gases).

Question 33Chemical structure, reactions and organic models

反应 2Mg + O2 -> 2MgO 中，若有 4 mol Mg 和 1 mol O2 完全反应，最多生成 MgO 的物质的量是：

  1. 1 mol
  2. 2 mol Correct answer.
  3. 3 mol
  4. 4 mol

Explanation

1 mol O2 只能与 2 mol Mg 反应，生成 2 mol MgO。虽然有 4 mol Mg，但氧气不足，是限制反应物。

Question 34Chemical structure, reactions and organic models

把 10 g 食盐完全溶于 90 g 水中，所得溶液的溶质质量分数是：

  1. 9%
  2. 10% Correct answer.
  3. 11.1%
  4. 90%

Explanation

溶液质量为 10 + 90 = 100 g，溶质质量分数 = 10 ÷ 100 × 100% = 10%。

Question 35Chemical structure, reactions and organic models

What is the stoichiometric balanced equation for the complete combustion of butane gas ($\text{C}_4\text{H}_{10}$) in dioxygen?

  1. $\text{C}_4\text{H}_{10} + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}$
  2. $2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}$ Correct answer.
  3. $\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}$
  4. $2\text{C}_4\text{H}_{10} + 9\text{O}_2 \rightarrow 8\text{CO} + 10\text{H}_2\text{O}$

Explanation

In complete combustion of a hydrocarbon, carbon forms $\text{CO}_2$ and hydrogen forms $\text{H}_2\text{O}$. For butane: 8 carbons produce $8\text{CO}_2$ (16 oxygen atoms) and 20 hydrogens produce $10\text{H}_2\text{O}$ (10 oxygen atoms). Total oxygen atoms needed $= 16 + 10 = 26$, which corresponds to $13\text{O}_2$. Thus, $2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}$.

Question 36Chemical structure, reactions and organic models

Given molar atomic masses $M(\text{H}) = 1\text{ g/mol}$, $M(\text{S}) = 32\text{ g/mol}$, and $M(\text{O}) = 16\text{ g/mol}$, what is the molar mass of sulfuric acid ($\text{H}_2\text{SO}_4$) and how many moles are contained in $19.6\text{ g}$ of pure $\text{H}_2\text{SO}_4$?

  1. $M = 98\text{ g/mol}$ and $n = 5.0\text{ mol}$
  2. $M = 49\text{ g/mol}$ and $n = 0.40\text{ mol}$
  3. $M = 98\text{ g/mol}$ and $n = 0.20\text{ mol}$ Correct answer.
  4. $M = 82\text{ g/mol}$ and $n = 0.24\text{ mol}$

Explanation

Molar mass $M(\text{H}_2\text{SO}_4) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98\text{ g/mol}$. Number of moles $n = \frac{m}{M} = \frac{19.6\text{ g}}{98\text{ g/mol}} = 0.20\text{ mol}$.

Question 37Chemical structure, reactions and organic models

An aqueous solution has a $\text{pH} = 3$ at $25^\circ\text{C}$. What happens to its acidity and $\text{pH}$ value when it is diluted by adding pure distilled water?

  1. The acidity increases and the $\text{pH}$ decreases towards 0
  2. The $\text{pH}$ remains strictly unchanged at 3
  3. The $\text{pH}$ instantly jumps to 14
  4. The acidity decreases and the $\text{pH}$ increases towards 7 (without ever exceeding 7) Correct answer.

Explanation

Diluting an acidic solution decreases the concentration of hydronium ions $[\text{H}_3\text{O}^+]$, which reduces its acidity. Consequently, the $\text{pH}$ value rises closer to the neutral $\text{pH}$ of pure water ($7.0$) but can never exceed 7 through pure water dilution alone.

Question 38Chemical structure, reactions and organic models

Which reagent is used in the laboratory to identify chloride ions ($\text{Cl}^-$) in an aqueous solution, and what is the positive visual result?

  1. Barium chloride solution ($\text{BaCl}_2$), which forms a yellow precipitate
  2. Silver nitrate solution ($\text{AgNO}_3$), which forms a white precipitate that blackens under light ($\text{AgCl}$) Correct answer.
  3. Sodium hydroxide solution ($\text{NaOH}$), which forms a blue precipitate
  4. Limewater (eau de chaux), which turns milky

Explanation

Chloride ions ($\text{Cl}^-$) react with silver ions ($\text{Ag}^+$) from silver nitrate ($\text{AgNO}_3$) to form silver chloride ($\text{AgCl}$), a characteristic white curdy precipitate that turns purple/black upon exposure to sunlight.

Question 39Chemical structure, reactions and organic models

In the dry reduction of black copper(II) oxide by carbon powder ($2\text{CuO} + \text{C} \rightarrow 2\text{Cu} + \text{CO}_2$), which substance acts as the reducing agent (réducteur)?

  1. Copper(II) oxide ($\text{CuO}$), because it loses oxygen
  2. Metallic copper ($\text{Cu}$)
  3. Carbon ($\text{C}$), because it gains oxygen and is oxidized to $\text{CO}_2$ Correct answer.
  4. Carbon dioxide ($\text{CO}_2$)

Explanation

A reducing agent (réducteur) is the species that captures oxygen (or donates electrons) and undergoes oxidation. Carbon ($\text{C}$) captures oxygen from $\text{CuO}$ to become $\text{CO}_2$, thereby reducing $\text{CuO}$ to metallic copper.

Question 40Chemical structure, reactions and organic models

During the electrolysis of acidified water in a voltameter, what are the gas volumes collected at the negative electrode (cathode) and positive electrode (anode)?

  1. Cathode: 1 volume of dioxygen ($\text{O}_2$); Anode: 2 volumes of dihydrogen ($\text{H}_2$)
  2. Cathode: 1 volume of chlorine ($\text{Cl}_2$); Anode: 1 volume of dihydrogen ($\text{H}_2$)
  3. Cathode: 2 volumes of carbon monoxide ($\text{CO}$); Anode: 2 volumes of dioxygen ($\text{O}_2$)
  4. Cathode: 2 volumes of dihydrogen ($\text{H}_2$); Anode: 1 volume of dioxygen ($\text{O}_2$) Correct answer.

Explanation

The decomposition of water follows $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$. Dihydrogen gas is released at the cathode (negative electrode) with twice the volume ($2V$) of the dioxygen gas released at the anode (positive electrode, $1V$).

Question 41Biological organization, inheritance, regulation and ecosystems

In the human nervous system, what is the anatomical pathway of a spinal reflex arc (arc réflexe médullaire) in response to a painful skin stimulus?

  1. Brain cortex $\rightarrow$ Spinal cord $\rightarrow$ Sensory nerve $\rightarrow$ Skin
  2. Effector muscle $\rightarrow$ Motor neuron $\rightarrow$ Spinal cord $\rightarrow$ Receptor
  3. Receptor (skin) $\rightarrow$ Sensory / Afferent neuron (via dorsal root) $\rightarrow$ Spinal cord (integrating center) $\rightarrow$ Motor / Efferent neuron (via ventral root) $\rightarrow$ Effector muscle Correct answer.
  4. Receptor $\rightarrow$ Ventral root $\rightarrow$ Brain $\rightarrow$ Dorsal root $\rightarrow$ Effector

Explanation

A spinal reflex is an involuntary, rapid response. The stimulus excites sensory receptors in the skin, which transmit nerve impulses along sensory (afferent) fibers entering the spinal cord via the dorsal root. The spinal cord processes the signal and immediately directs impulses through motor (efferent) neurons exiting via the ventral root to the effector muscle.

Question 42Biological organization, inheritance, regulation and ecosystems

How is a nerve impulse transmitted across the synaptic cleft (fente synaptique) between two neurons?

  1. By direct metallic electrical conductivity through copper ions
  2. By the physical contraction of myofibrils bridging the two cells
  3. By the continuous mechanical circulation of red blood cells
  4. By the exocytosis and diffusion of chemical neurotransmitters (e.g., acetylcholine) released from the presynaptic vesicle into postsynaptic receptors Correct answer.

Explanation

When an action potential reaches the presynaptic axon terminal, it triggers the influx of calcium and the exocytosis of neurotransmitters (such as acetylcholine or dopamine) into the synaptic cleft. These molecules diffuse across the gap and bind to specific receptors on the postsynaptic membrane to propagate the signal.

Question 43Biological organization, inheritance, regulation and ecosystems

In the human digestive system, which enzyme found in saliva breaks down cooked starch (amidon cuit) into maltose at neutral $\text{pH}$ and $37^\circ\text{C}$?

  1. Salivary amylase (ptyaline) Correct answer.
  2. Gastric pepsin (pepsine)
  3. Pancreatic lipase (lipase)
  4. Intestinal lactase (lactase)

Explanation

Salivary amylase (also historically known as ptyalin) is secreted by salivary glands in the mouth. It catalyzes the hydrolysis of complex cooked starch polysaccharides into maltose disaccharides at body temperature ($37^\circ\text{C}$) and near-neutral $\text{pH}$.

Question 44Biological organization, inheritance, regulation and ecosystems

Through which anatomical structure and circulatory pathway are water-soluble nutrients (glucose, amino acids, minerals, water) absorbed from the small intestine into the general circulation?

  1. Intestinal villi directly into central lymphatic chyliferous vessels to the brain
  2. Intestinal villi into blood capillaries, which drain through the hepatic portal vein (veine porte hépatique) to the liver Correct answer.
  3. Stomach wall directly into the renal artery to the kidneys
  4. Esophagus mucosal glands into the pulmonary veins

Explanation

The microvilli of the intestinal epithelium provide an immense surface area for absorption. Water-soluble nutrients (glucose, amino acids, vitamins B and C, mineral salts) pass into blood capillaries that converge into the hepatic portal vein, which delivers them directly to the liver for metabolic processing.

Question 45Biological organization, inheritance, regulation and ecosystems

During pulmonary gas exchange across the alveolar-capillary membrane, how do oxygen ($\text{O}_2$) and carbon dioxide ($\text{CO}_2$) diffuse?

  1. Both $\text{O}_2$ and $\text{CO}_2$ diffuse simultaneously from blood into alveolar air
  2. $\text{CO}_2$ diffuses into blood and $\text{O}_2$ is exhaled out of the body
  3. Gas exchange only occurs by active cellular ATP pumping against concentration gradients
  4. $\text{O}_2$ diffuses from alveolar air (high partial pressure) into capillary blood, while $\text{CO}_2$ diffuses from capillary blood into alveolar air to be exhaled Correct answer.

Explanation

Gas exchange occurs by passive diffusion down partial pressure gradients across the ultra-thin alveolar wall: oxygen diffuses from high concentration in alveolar air into blood where it binds to hemoglobin (forming oxyhemoglobin), while carbon dioxide diffuses from high concentration in deoxygenated venous blood into the alveolus to be expired.

Question 46Biological organization, inheritance, regulation and ecosystems

What is the functional sequence of urine formation in the microscopic nephron of the human kidney?

  1. 1. Glomerular filtration (filtration glomérulaire) $\rightarrow$ 2. Tubular reabsorption (réabsorption tubulaire) $\rightarrow$ 3. Tubular secretion (sécrétion tubulaire) Correct answer.
  2. 1. Tubular secretion $\rightarrow$ 2. Glomerular filtration $\rightarrow$ 3. Storage in ureter
  3. 1. Tubular reabsorption $\rightarrow$ 2. Excretion $\rightarrow$ 3. Blood coagulation
  4. 1. Digestion $\rightarrow$ 2. Bile secretion $\rightarrow$ 3. Urine condensation

Explanation

Nephrons form urine in three sequential physiological stages: 1. Glomerular filtration across Bowman's capsule (producing primitive urine free of proteins and cells), 2. Tubular reabsorption along the proximal/distal tubules and loop of Henle (recovering glucose, amino acids, water, and essential ions), and 3. Tubular secretion (actively clearing metabolic wastes like creatinine and protons) to form final definitive urine.

Question 47Biological organization, inheritance, regulation and ecosystems

Which specific cells of the human immune system are preferentially targeted, infected, and destroyed by the Human Immunodeficiency Virus (HIV / VIH)?

  1. Red blood cells (erythrocytes)
  2. Blood platelets (thrombocytes)
  3. Bone osteocytes
  4. Helper T-lymphocytes (T4 / CD4+ lymphocytes) Correct answer.

Explanation

HIV surface glycoprotein gp120 binds specifically to the CD4 receptor found on helper T-lymphocytes (T4 / $ ext{CD4}^+$). By hijacking their replication machinery and destroying them, HIV collapses the immune coordination, leading to Acquired Immunodeficiency Syndrome (AIDS / SIDA) and opportunistic infections.

Question 48Scientific questions, evidence and quantitative reasoning

In a GCE A-Level physics practical, a student records the diameter of a wire as 0.52 mm using a micrometer screw gauge with a zero error of +0.03 mm. What is the corrected diameter?

  1. 0.52 mm
  2. 0.58 mm
  3. 0.49 mm Correct answer.
  4. 0.55 mm

Explanation

A positive zero error means the instrument reads a value when the jaws are closed, so the error must be subtracted from every reading. Here 0.52 − 0.03 = 0.49 mm. Adding the error would double-count the fault rather than remove it.

Question 49Scientific questions, evidence and quantitative reasoning

In an experiment to verify Ohm's law, a student plots a graph of current I against potential difference V. What does the gradient of this graph represent?

  1. The reciprocal of the resistance Correct answer.
  2. The resistance of the resistor
  3. The power dissipated
  4. The charge passed per second

Explanation

Ohm's law gives I = V/R, so plotting I on the vertical axis against V gives a straight line through the origin with gradient 1/R. The resistance itself would be the gradient only if V were plotted against I.

Question 50Scientific questions, evidence and quantitative reasoning

A student takes six readings of the time for 20 oscillations of a simple pendulum. Which quantity is the most appropriate to report as the period?

  1. The mean of the six times divided by 20 Correct answer.
  2. The mean of the six times multiplied by 20
  3. The difference between the largest and smallest times
  4. The largest of the six times divided by 20

Explanation

The period is the time for one oscillation, so the mean time for 20 oscillations is divided by 20. Averaging repeated timings reduces the effect of random reaction-time errors.