CBSE Class 10 Science Practice Paper 2026-27 Set 1 with Answer Key (PDF)

Q1 (1 mark · Life Processes)

(B) stomach

Gastric juice secreted by the gastric glands contains hydrochloric acid, which makes the contents of the stomach strongly acidic (about pH 1.5–2). Saliva in the mouth is nearly neutral, and the contents of the duodenum and ileum are made alkaline by bile and pancreatic juice.

Q2 (1 mark · Life Processes)

(B) lactic acid

During a hard sprint the muscles use oxygen faster than the blood can supply it. In this shortage of oxygen the pyruvate formed from glucose is broken down anaerobically in the muscle cells into lactic acid, and it is the build-up of lactic acid that causes cramps and the aching feeling. Ethanol and carbon dioxide are produced by yeast, not by human muscle.

Q3 (1 mark · Life Processes)

(B) pulmonary vein

The pulmonary vein is the one vein that carries oxygenated blood — it brings blood from the lungs back to the left atrium. The pulmonary artery is the one artery that carries de-oxygenated blood (heart to lungs). Both the superior vena cava and the renal vein return oxygen-poor blood to the heart.

Q4 (1 mark · Life Processes)

(B) glomerulus

The glomerulus is the cluster of capillaries sitting inside Bowman’s capsule. Blood enters it at high pressure and water, glucose, salts, urea and amino acids are filtered out into the capsule. The loop of Henle and the collecting duct are concerned with selective re-absorption, and the bladder only stores the urine that has already been formed.

Q5 (1 mark · Control and Coordination)

(a) Both A and R are true and R is the correct explanation of A.

The reflex arc is completed in the spinal cord, so the motor command to the arm muscles is issued while the sensory information is still travelling up to the brain. The hand therefore moves first and the feeling of pain is registered a moment later. This short-cut is exactly why reflex arcs evolved — they save time when a delay would cause injury.

Q6 (1 mark · Control and Coordination)

(B) collects on the shaded side and makes those cells longer

Auxin is made at the shoot tip and diffuses away from light. It therefore becomes more concentrated on the shaded side of the stem, where it makes the cells elongate more than the cells on the lit side. Because one side of the stem grows longer than the other, the shoot curves towards the light — this is positive phototropism.

Q7 (1 mark · Reproduction)

(C) condom

A condom simply prevents the sperm from reaching the egg — a physical barrier and nothing more. Oral pills act by changing hormone levels, the Copper-T is an intra-uterine device, and vasectomy is a surgical method in which the vas deferens is blocked.

Q8 (1 mark · Our Environment)

(C) 10 J

By the ten per cent law only about 10% of the energy at one trophic level passes to the next.

Grass 10 000 J → grasshopper 1000 J → frog 100 J → snake 10 J.

This steep loss at every step is also the reason food chains rarely have more than four or five links.

Q9 (1 mark · Our Environment)

(a) Both A and R are true and R is the correct explanation of A.

Decomposer bacteria and fungi carry enzymes that act on natural materials such as paper, cotton and food waste. Polythene is a man-made polymer for which they have no matching enzyme, so it is not decomposed and simply accumulates in the soil and in water bodies year after year.

Q10 (2 marks · Life Processes)

(i) The amount of oxygen dissolved in water is far smaller than the amount of oxygen present in the same volume of air, so a fish gets very little oxygen from each mouthful of water.

(ii) Water is much denser and moves more slowly than air, and diffusion through it is slow, so a large volume of water has to be pushed continuously over the gills to absorb enough oxygen.

Both reasons force the fish to raise its rate of breathing.

Marking: 1 mark for each correct reason.

Q11 (2 marks · Control and Coordination)

(i) The nervous system carries information as an electrical impulse along nerve fibres; the endocrine system carries it as a chemical (a hormone) dissolved in the blood.

(ii) A nerve impulse is extremely fast and its effect is short-lived and limited to the organ supplied by that nerve; a hormone travels slowly but its effect lasts much longer and can reach every part of the body through the blood.

Marking: 1 mark for each valid difference.

If you attempted the OR option

Promotes cell elongation — auxin. Auxin is made at the shoot tip and makes the cells just behind the tip grow longer, which is what produces the bending of a shoot towards light.

Inhibits growth — abscisic acid. Abscisic acid slows growth, closes the stomata when the plant is short of water, and causes wilting and the shedding of leaves.

Marking: 1 mark for each hormone with its function. Gibberellin, which also promotes stem elongation, is acceptable for the first part.

Q12 (2 marks · Reproduction)

Some plants in Farmer B’s field are more likely to survive.

Farmer B’s seed is the product of sexual reproduction between two different varieties, so his plants differ from one another — the population carries a good deal of variation. Among so many slightly different plants, some are likely to carry a feature that lets them resist the new fungus, and those plants survive and set seed.

Farmer A’s tubers are produced by asexual reproduction, so his plants are practically identical copies of the parent. If the fungus can attack one of them it can attack all of them, and the whole field may be lost.

Marking: 1 mark for identifying Farmer B, 1 mark for the reason based on variation from sexual reproduction.

Q13 (3 marks · Life Processes)

(a) Transpiration.

(b) As water evaporates from the mesophyll cells and escapes through the stomata, those cells draw water from the xylem next to them. This creates a suction called the transpiration pull. Since the column of water in the xylem is continuous from root to leaf, the pull is passed all the way down and lifts water up even to the top of a tall tree.

(c) The rate falls when the surrounding air is very humid (also acceptable: at night when the stomata close, in still air, or at a low temperature).

Marking: 1 mark for each part.

If you attempted the OR option

(a) Chlorophyll.

(b) 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (in the presence of sunlight and chlorophyll).

(c) The glucose that is not used at once is converted into starch and stored in the leaves, and later in other parts such as roots, stems and seeds.

Marking: 1 mark for each part; the equation must be balanced.

Q14 (3 marks · Our Environment)

(a) Biological magnification is the progressive increase in the concentration of a harmful, non-biodegradable substance (for example a pesticide) in the bodies of organisms at each successive higher level of a food chain.

(b) Such a substance is neither broken down nor excreted, so whatever an organism takes in stays in its body. Each consumer eats many organisms from the level below it and therefore receives the whole of the chemical they had stored. The amount keeps adding up at every step, so the concentration is highest in the organisms at the top of the chain — which is often ourselves.

(c) Any one workable step, for example: keep separate bins for biodegradable and non-biodegradable waste and send the second lot for recycling; replace single-use plastic water bottles with steel bottles and a refill station; ask the canteen to use paper or cloth bags instead of polythene.

Marking: 1 mark for each part.

Q15 (4 marks · Life Processes)

(a) Transpiration.

(b) In an ordinary broad leaf the stomata are far more numerous on the lower surface. In plant Q the jelly sealed only the upper surface, so nearly all its stomata stayed open and transpiration went on almost normally. In plant P the jelly sealed the lower surface, blocking most of the stomata, so very little water vapour escaped.

(c) So that the water vapour collected in the bag could have come only from the leaves and not from evaporation of water from the soil. This keeps the comparison a fair test.

(d) Hardly any droplets — or none at all. With jelly on both surfaces the stomata of both surfaces are blocked, so transpiration is almost completely stopped.

Marking: 1 mark for each part.

Q16 (5 marks · Life Processes)

A (i) Any three of the following:

• The alveoli are millions of tiny balloon-like sacs, which together give an enormous surface area for the exchange of gases.

• Their walls are extremely thin — only one cell thick — so gases diffuse across very quickly.

• Their walls carry a dense network of blood capillaries, so the exchanged gases are carried away at once and a steep diffusion gradient is maintained.

• The inner surface is moist, so the gases dissolve before diffusing.

A (ii) The left ventricle has to pump oxygenated blood into the aorta and drive it all the way round the body, against considerable resistance. Doing this needs a high pressure, which needs a thick, powerful muscular wall. The right ventricle only has to send blood the short distance to the lungs, and the delicate lung capillaries would be damaged by a high pressure, so a much thinner wall is enough.

Marking: 1 mark for each feature in (i); 2 marks for a correct explanation in (ii).

If you attempted the OR option

B (i) Mouth: the teeth chew the food and the salivary glands pour in saliva, whose enzyme salivary amylase starts breaking starch into sugar; mucus makes the food slippery and the tongue rolls it into a bolus.

Oesophagus: rhythmic waves of muscular contraction called peristalsis push the bolus down into the stomach.

Stomach: the muscular wall churns the food while the gastric glands pour in gastric juice — hydrochloric acid (which makes the medium acidic and kills germs), the enzyme pepsin (which digests proteins) and mucus (which protects the stomach lining). The food is released in small amounts into the small intestine through a sphincter.

Small intestine: bile from the liver and pancreatic juice from the pancreas (containing amylase, trypsin and lipase) act on the food, and finally the intestinal juice completes the digestion of carbohydrates, proteins and fats into glucose, amino acids, fatty acids and glycerol.

B (ii) Bile has two roles in the digestion of fats. First, it is alkaline and neutralises the acid coming from the stomach, giving the alkaline medium that the fat-digesting enzyme lipase needs. Second, the bile salts emulsify the large fat globules — they break them into very many tiny droplets, which greatly increases the surface area on which lipase can act, so fat is digested much faster. Bile itself contains no enzyme.

Marking: 3 marks for the account in (i) — 1 mark each for mouth, stomach and small intestine with the correct secretions; 2 marks for (ii).

Q17 (1 mark · Chemical Reactions and Equations)

(C) thermal decomposition

A single compound is broken down by heat into three simpler products. Because the energy needed is supplied as heat, it is a thermal decomposition (and, being a reaction that absorbs heat, it is endothermic).

Q18 (1 mark · Chemical Reactions and Equations)

(A) 2, 3, 3

The balanced equation is 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

Check: Al 2 = 2; S 3 = 3; O 12 = 12; H 6 = 6. Option (B) fails because it leaves 6 hydrogen atoms on the left and only 4 on the right.

Q19 (1 mark · Acids, Bases and Salts)

(A) 2

The lower the pH, the higher the concentration of hydrogen ions and the stronger the acid. A carbonate reacts with an acid to give carbon dioxide, so the most acidic solution (pH 2) gives the briskest effervescence. The solutions of pH 8 and 12 are basic and would give no fizzing at all.

Q20 (1 mark · Acids, Bases and Salts)

(B) baking soda

Baking soda is sodium hydrogencarbonate. On heating:

2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

The residue Y is sodium carbonate (washing soda), and the gas that turns lime water milky is carbon dioxide. This is exactly the reaction that makes a cake rise.

Q21 (1 mark · Metals and Non-metals)

(a) Both A and R are true and R is the correct explanation of A.

A displacement reaction happens only when the metal added is more reactive than the metal already in the salt. Copper is less reactive than zinc, so it cannot push zinc out of zinc sulphate, and nothing is seen to happen. (The reverse experiment does work — zinc added to copper sulphate turns the blue solution colourless and deposits red-brown copper.)

Q22 (1 mark · Metals and Non-metals)

(C) conduct electricity when molten or dissolved in water

In the solid state the ions are locked in position in the crystal lattice and cannot move, so no current flows. On melting or on dissolving in water the lattice breaks up, the ions become free to move towards the electrodes, and the substance conducts. Ionic compounds also have high melting points and dissolve in water rather than in petrol or kerosene.

Q23 (1 mark · Carbon and its Compounds)

(B) a –CH₂– group and 14 u

Successive members of a homologous series differ by one –CH₂– unit. Its mass is 12 + (2 × 1) = 14 u, which is why the molecular masses in a homologous series go up in steps of 14.

Q24 (1 mark · Carbon and its Compounds)

(B) calcium and magnesium ions in hard water form an insoluble scum with soap

Hard water contains dissolved calcium and magnesium salts. These react with the soap to form an insoluble, sticky curd called scum, which sticks to the cloth. The soap is used up making scum instead of lather, so cleaning is poor. Detergents work in hard water because their calcium and magnesium salts stay soluble.

Q25 (2 marks · Acids, Bases and Salts)

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Check: Zn 1 = 1, H 2 = 2, S 1 = 1, O 4 = 4 — the equation is balanced.

The gas evolved is hydrogen. It is collected and tested by bringing a burning splint near the mouth of the tube — the gas burns with a squeaky pop sound.

Marking: 1 mark for the correct balanced equation with state symbols, 1 mark for naming the gas.

Q26 (3 marks · Metals and Non-metals)

(a) Roasting — the sulphide ore is heated strongly in the presence of excess air, which converts it into the oxide.

(b) ZnO + C → Zn + CO

Check: Zn 1 = 1, O 1 = 1, C 1 = 1. Carbon is the reducing agent here.

(c) The metal obtained straight from reduction is impure — it still contains unreacted ore, bits of slag, carbon and traces of other metals. These impurities spoil the properties of the metal, so it is purified, most often by electrolytic refining.

Marking: 1 mark for each part.

If you attempted the OR option

(a) Corrosion is the slow eating away of the surface of a metal by the air, moisture and chemicals around it, which converts the surface into oxides, carbonates or sulphides.

(b) The metal is copper, and the green coating is basic copper carbonate.

(c) Any one method: painting the gate; oiling or greasing it; galvanising it (coating it with a layer of zinc); or electroplating it. Each works by keeping air and moisture away from the iron surface.

Marking: 1 mark for each part.

Q27 (3 marks · Acids, Bases and Salts)

(a) Bleaching powder, calcium oxychloride, CaOCl₂.

(b) Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Check: Ca 1 = 1, O 2 = 2, H 2 = 2, Cl 2 = 2 — balanced.

(c) Any one use: for disinfecting drinking water; for bleaching cotton and linen in the textile industry or wood pulp in paper factories; as an oxidising agent in chemical industries.

Marking: 1 mark for each part.

Q28 (4 marks · Acids, Bases and Salts)

(a) Lemon juice. It has the lowest pH (2), and the lower the pH the higher the concentration of hydrogen ions.

(b) A lemon drink leaves the mouth strongly acidic, well below pH 5.5, so the enamel starts to corrode. The baking-soda solution is basic (pH 9), so rinsing with it neutralises the extra acid and raises the pH of the mouth back above 5.5, and the enamel is no longer attacked.

(c) He could add quicklime, slaked lime or chalk — any basic substance. Soil of pH 5 is acidic; a base neutralises the excess acid and brings the pH back towards neutral, which suits the crop.

(d) Milk (6) < distilled water (7) < oven cleaner (13). The higher the pH, the more basic the liquid and the greater its hydroxide ion concentration.

Marking: 1 mark for each part.

Q29 (5 marks · Carbon and its Compounds)

A (i) The second member of the alkane series is ethane, C₂H₆.

Electron dot structure: each carbon atom has four valence electrons and each hydrogen one. The two carbon atoms share one pair of electrons with each other, forming the C–C single bond, and each carbon shares one pair with each of three hydrogen atoms. In this way every carbon completes its octet and every hydrogen its duplet, giving H₃C–CH₃ with each bond drawn as a shared pair of dots.

A (ii) Catenation — carbon atoms join to one another by strong covalent bonds to form long chains, branched chains and rings, with single, double or triple bonds. Tetravalency — carbon has a valency of four, so one carbon atom can bond with four other atoms of carbon, hydrogen, oxygen, nitrogen, sulphur or a halogen. Together these give an enormous number of possible structures. (The small size of the carbon atom, which makes its bonds unusually strong and its compounds stable, is also acceptable.)

A (iii) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O (+ heat and light)

Check: C 2 = 2, H 6 = 6, O 7 = 7 — balanced.

Marking: (i) 1 mark for the name, 1 for the structure; (ii) 1 mark for each reason; (iii) 1 mark for the balanced equation.

If you attempted the OR option

B (i) CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

Check: C 3 = 3, H 5 = 5, Na 1 = 1, O 5 = 5 — balanced. The gas released is carbon dioxide, seen as brisk effervescence; it turns lime water milky.

B (ii) Esterification is the reaction of a carboxylic acid with an alcohol, in the presence of a little concentrated sulphuric acid, to give a sweet-smelling ester and water.

CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O (conc. H₂SO₄)

Check: C 4 = 4, H 10 = 10, O 3 = 3 — balanced. The product is ethyl ethanoate.

B (iii) Add a pinch of sodium hydrogencarbonate to each liquid. Ethanoic acid gives brisk effervescence of carbon dioxide, which turns lime water milky; ethanol gives no effervescence at all. (Alternatively, ethanoic acid turns blue litmus red while ethanol has no effect on litmus.)

Marking: (i) 1 mark for the equation, 1 for the gas; (ii) 1 mark for the definition, 1 for the equation; (iii) 1 mark for the test with the observation.

Q30 (1 mark · Light)

(C) virtual, erect and magnified

This is the single position for which a concave mirror gives a virtual image. The reflected rays diverge and appear to come from a point behind the mirror, so the image is virtual, erect and larger than the object. It is why a concave mirror is used as a shaving mirror and by dentists.

Q31 (1 mark · Human Eye and the Colourful World)

(a) Both A and R are true and R is the correct explanation of A.

The molecules of the atmosphere are very small compared with the wavelength of visible light, and such small particles scatter the shorter (blue) wavelengths far more strongly than the longer (red) ones. Scattered blue light reaches our eyes from every direction in the sky, so the sky looks blue. An astronaut, who has almost no atmosphere above, sees a black sky instead.

Q32 (1 mark · Electricity)

(B) R/9

Resistance is proportional to length, so each of the three equal pieces has resistance R/3.

Three equal resistances in parallel give one-third of the value of one of them:

R_p = (R/3) ÷ 3 = R/9.

Q33 (2 marks · Electricity)

(a) For the parallel pair:

R_p = (6 × 3) / (6 + 3) = 18/9 = 2 Ω

This 2 Ω is in series with 4 Ω, so the total resistance is

R = 2 + 4 = 6 Ω

(b) I = V/R = 12/6 = 2 A

Marking: 1 mark for the total resistance, 1 mark for the current.

Q34 (2 marks · Electricity)

Power P = 1500 W = 1.5 kW. Time per day t = 45 min = 45/60 = 0.75 h.

Energy used in one day = P × t = 1.5 × 0.75 = 1.125 kWh

Energy used in 30 days = 1.125 × 30 = 33.75 kWh (33.75 units)

The voltage rating is not needed for this calculation.

Marking: 1 mark for the energy per day, 1 mark for the 30-day total with the correct unit.

Q35 (3 marks · Light)

Using the sign convention: u = –15 cm, f = –10 cm, h = +4 cm.

Mirror formula: 1/v + 1/u = 1/f

1/v = 1/f – 1/u = (–1/10) – (–1/15) = –1/10 + 1/15 = (–3 + 2)/30 = –1/30

v = –30 cm — the image is formed 30 cm in front of the mirror, on the same side as the object, so it is real.

Magnification m = –v/u = –(–30)/(–15) = –2

The negative sign shows the image is inverted, and its size is twice the object’s.

Height of image h′ = m × h = –2 × 4 = –8 cm, i.e. 8 cm tall and inverted.

Answer: real, inverted and magnified; 30 cm in front of the mirror; 8 cm tall.

Marking: 1 mark for v, 1 for m, 1 for the size and nature.

If you attempted the OR option

For a convex mirror: u = –30 cm, f = +20 cm.

1/v = 1/f – 1/u = 1/20 – (–1/30) = 1/20 + 1/30 = (3 + 2)/60 = 5/60 = 1/12

v = +12 cm — the image is 12 cm behind the mirror, so it is virtual and erect.

m = –v/u = –12/(–30) = +0.4

The magnification is positive (erect) and less than 1, so the image is diminished — it is 0.4 times the size of the object.

Marking: 1 mark for v, 1 for the nature, 1 for the magnification.

Q36 (3 marks · Human Eye and the Colourful World)

(a) Hypermetropia, also called long-sightedness or far-sightedness.

(b) A convex (converging) lens is used. In this defect the eyeball is too short, or the converging power of the eye lens is too low, so rays coming from a nearby object would be brought to a focus behind the retina. A convex lens converges the rays a little before they enter the eye, so the image is brought forward exactly onto the retina and near objects are seen clearly.

(c) Power P = 1/f, with f in metres.

f = 1/P = 1/2.5 = 0.4 m = +40 cm

Marking: 1 mark for each part.

Q37 (3 marks · Electricity)

In a parallel arrangement each resistor gets the full supply voltage of 24 V.

(a) I₁ = V/R₁ = 24/4 = 6 A;   I₂ = V/R₂ = 24/12 = 2 A

(b) Total current I = 6 + 2 = 8 A

Check: R_p = (4 × 12)/(4 + 12) = 48/16 = 3 Ω, and I = 24/3 = 8 A. ✓

(c) P = VI = 24 × 8 = 192 W

Check: P = V²/R_p = 576/3 = 192 W. ✓

Marking: 1 mark for the two branch currents, 1 for the total current, 1 for the power.

If you attempted the OR option

(a) 1/R_p = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1

R_p = 1 Ω

(b) I = V/R_p = 6/1 = 6 A

(c) Each resistor has the full 6 V across it, so the current in the 3 Ω resistor is

I₃ = 6/3 = 2 A

Check: 6/2 + 6/3 + 6/6 = 3 + 2 + 1 = 6 A, which matches the total. ✓

Marking: 1 mark for each part.

Q38 (4 marks · Light)

(a) Lens formula: 1/v – 1/u = 1/f, with f = +20 cm and u = –30 cm.

1/v = 1/f + 1/u = 1/20 + (–1/30) = (3 – 2)/60 = 1/60

v = +60 cm — the screen must be placed 60 cm from the lens, on the side away from the candle.

(b) m = v/u = 60/(–30) = –2, so h′ = –2 × 5 = –10 cm.

The image is 10 cm tall, real, inverted and magnified two times.

(c) With the candle 15 cm away it lies between the optical centre and the focus (15 cm < 20 cm). The rays leaving the lens then diverge and never actually meet; they only appear to come from a point on the candle’s own side. The image is therefore virtual, and a virtual image cannot be caught on a screen.

(d) 1/v = 1/20 + (–1/15) = (3 – 4)/60 = –1/60, so v = –60 cm — the image is 60 cm from the lens on the same side as the candle.

m = v/u = (–60)/(–15) = +4, so h′ = +4 × 5 = +20 cm.

The image is virtual, erect and magnified four times — 20 cm tall.

Marking: 1 mark for each part.

Q39 (5 marks · Electricity)

A (i) In a parallel circuit each appliance receives the full 220 V.

Kettle: I₁ = V/R = 220/40 = 5.5 A

Bulb: I₂ = V/R = 220/440 = 0.5 A

A (ii) Total current I = 5.5 + 0.5 = 6.0 A

Check: R_p = (40 × 440)/(40 + 440) = 17600/480 = 36.67 Ω, and 220/36.67 = 6.0 A. ✓

A (iii) The current actually drawn, 6.0 A, is greater than the fuse rating of 5 A. The fuse will blow — its thin wire melts and breaks the circuit, protecting the wiring and the appliances.

A (iv) Any one reason: in parallel every appliance gets the full 220 V and so works at its rated power; each appliance can be switched on and off on its own; and if one appliance fails the others keep working. In a series circuit the voltage would be shared, every appliance would under-perform, and a single break would stop everything.

Marking: (i) 1 mark for each current; (ii) 1 mark; (iii) 1 mark; (iv) 1 mark.

If you attempted the OR option

B (i) Joule’s law of heating: the heat produced in a resistor is directly proportional to the square of the current flowing through it, to the resistance of the resistor, and to the time for which the current flows — that is, H = I²Rt.

B (ii) In series the resistances add: R = 5 + 10 = 15 Ω.

Current I = V/R = 30/15 = 2 A — the same current flows through both resistors.

Time t = 2 minutes = 120 s.

Heat in the 5 Ω resistor: H₁ = I²R₁t = (2)² × 5 × 120 = 2400 J

Heat in the 10 Ω resistor: H₂ = I²R₂t = (2)² × 10 × 120 = 4800 J

Check: 2400 + 4800 = 7200 J, and VIt = 30 × 2 × 120 = 7200 J. ✓

B (iii) More heat is produced in the 10 Ω resistor. In a series circuit the current and the time are the same for both resistors, so by H = I²Rt the heat is directly proportional to the resistance — the larger resistance produces more heat.

Marking: (i) 1 mark; (ii) 1 mark for the current, 1 for each heat value; (iii) 1 mark.