2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

(a) A reflex action is a rapid, automatic (or involuntary) response to a stimulus, which does not require conscious thought by the brain.

(b) The pathway starts when a receptor in the skin detects the thermal stimulus (heat). This generates an electrical impulse which travels along a sensory neurone to the central nervous system (spinal cord). Inside the spinal cord, the impulse is transmitted across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse out of the spinal cord to the effector, which is a muscle (e.g., biceps), causing it to contract and pull the hand away from the hot object.

(c) When an electrical impulse reaches the end of the presynaptic neurone, it causes vesicles containing neurotransmitters to fuse with the cell membrane, releasing the chemical neurotransmitters into the synaptic cleft. These neurotransmitter molecules diffuse across the gap and bind with specific receptor proteins on the postsynaptic membrane. This binding triggers a new electrical impulse in the postsynaptic neurone.

(a) [Max 2 marks]
- rapid / fast response [1]
- automatic / involuntary / does not involve conscious decision-making [1]

(b) [Max 4 marks]
- receptor detects heat and generates impulse [1]
- impulse travels along sensory neurone to spinal cord / CNS [1]
- impulse passes across relay neurone [1]
- impulse passes along motor neurone to muscle / effector, causing contraction [1]

(c) [Max 4 marks]
- impulse triggers release of neurotransmitters from vesicles [1]
- neurotransmitters diffuse [1]
- across the synaptic cleft / gap [1]
- neurotransmitter binds to specific receptor molecules on postsynaptic membrane [1]

(a) Water in the soil enters root hair cells by osmosis, moving down a water potential gradient. It then moves through the cortex of the root to the xylem vessel. Water is pulled up the xylem vessels in a continuous column to the leaves because of the transpiration pull. This pull is created by the evaporation of water vapour from the surfaces of mesophyll cells in the leaves, which then diffuses out through the open stomata. Cohesion between water molecules keeps the column unbroken.

(b) (i) An increase in temperature increases the rate of transpiration because water molecules gain more kinetic energy, leading to a faster rate of evaporation from the mesophyll cell walls and faster diffusion of water vapour out through the stomata.

(ii) An increase in wind speed increases the rate of transpiration because the moving air sweeps away water vapour accumulating around the leaf surface, which maintains a steep concentration gradient of water vapour between the inside and outside of the leaf.

(a) [Max 4 marks]
- water enters root hair cells by osmosis [1]
- moves down water potential gradient / from high to low water potential [1]
- evaporation of water from mesophyll cell surfaces (creates transpiration pull / tension) [1]
- water molecules form a continuous column due to cohesion [1]

(b) (i) [Max 3 marks]
- rate of transpiration increases [1]
- water molecules gain more kinetic energy [1]
- leads to faster evaporation / faster diffusion out of the stomata [1]

(b) (ii) [Max 3 marks]
- rate of transpiration increases [1]
- wind blows away water vapour near the leaf surface [1]
- maintains/steepens concentration gradient (for diffusion) [1]

(a) Mechanical digestion is the physical breakdown of food into smaller pieces without chemically changing the molecules, whereas chemical digestion is the chemical breakdown of large, insoluble food molecules into small, soluble molecules using enzymes.

(b) (i) Substrate: starch. Product: maltose (or glucose).
(ii) Site of production: gastric glands (stomach wall). Site of action: stomach.
(iii) Fatty acids and glycerol.

(c) Bile is an alkaline mixture produced in the liver and stored in the gallbladder. It emulsifies fats, meaning it physically breaks large fat globules into millions of tiny fat droplets. This increases the surface area of the fat exposed to lipase molecules, speeding up chemical digestion. Bile also contains hydrogencarbonate ions, which neutralize stomach acid and provide the optimum alkaline pH for lipase action in the small intestine.

(a) [Max 2 marks]
- mechanical digestion is physical breakdown of food into smaller pieces / without chemical change [1]
- chemical digestion is breakdown of large, insoluble molecules into small, soluble molecules (by enzymes) [1]

(b) [Max 5 marks]
(i) substrate: starch [1]; product: maltose / glucose [1]
(ii) site of production: stomach (wall) [1]; site of action: stomach [1]
(iii) products: fatty acids and glycerol [1]

(c) [Max 3 marks]
- bile emulsifies fats / breaks large drops into small droplets [1]
- increases surface area [1]
- for lipase action / chemical digestion [1]
- neutralizes stomach acid / provides optimum alkaline pH for pancreatic enzymes [1]

(a) Chemical digestion is the process where large, insoluble nutrient molecules (such as starch, proteins, or lipids) are chemically broken down by enzymes into small, soluble molecules (such as glucose, amino acids, fatty acids, and glycerol). This is essential because only small, soluble molecules are capable of passing through the selectively permeable membranes of the cells lining the alimentary canal to be absorbed into the bloodstream.

(b) The villi in the small intestine are highly adapted for absorption:
1. Microvilli: The epithelial cells on the surface of the villi have microscopic folds called microvilli, which greatly increase the surface area for diffusion and active transport.
2. One-cell thick epithelium: The barrier between the lumen of the gut and the capillaries is extremely thin, minimizing the distance over which nutrients must diffuse.
3. Capillary network: A rich blood supply inside each villus continuously carries away water-soluble nutrients (glucose and amino acids), maintaining a steep concentration gradient between the lumen and the blood.
4. Lacteal: A central lymphatic vessel (lacteal) in each villus absorbs and transports lipids (fatty acids and glycerol) away from the intestine.

(c)(i) Substrate: starch; Product: maltose.
(ii) The stomach secretes hydrochloric acid, which creates a highly acidic environment (around pH 1.5 to 2.0). Amylase is adapted to work in the neutral to slightly alkaline conditions of the mouth (around pH 7). The extreme acidity in the stomach denatures the amylase enzyme, altering the specific three-dimensional shape of its active site so that the starch substrate can no longer bind to it.

(a)
- breakdown of large, insoluble molecules; [1]
- into small, soluble molecules; [1]
- so they can be absorbed / pass through the wall of the gut into the blood; [1]

(b) Max 4 marks from:
- microvilli; [1]
- increases surface area (for diffusion/absorption); [1]
- thin wall / epithelium is only one cell thick; [1]
- provides a short diffusion distance; [1]
- blood capillaries / rich blood supply; [1]
- maintains concentration gradient (by carrying away glucose/amino acids); [1]
- lacteal; [1]
- absorbs / transport of fats / fatty acids and glycerol; [1]

(c)(i)
- substrate: starch AND product: maltose; [1] (both required for the mark)

(c)(ii)
- stomach is acidic / has low pH; [1]
- (amylase / enzyme) denatures / active site changes shape (so substrate no longer fits); [1]

(a)(i) Ethyl propanoate contains an 'ethyl' group from ethanol and a 'propanoate' group from propanoic acid. (a)(ii) The reaction requires a concentrated sulfuric acid catalyst and heating. (b)(i) Addition polymerisation involves unsaturated monomers (with C=C double bonds) reacting to form a single product (the polymer). Condensation polymerisation involves monomers with two functional groups reacting to form a polymer and a small molecule (such as water). (b)(ii) The reaction between a dicarboxylic acid and a diol forms an ester linkage, eliminating water. (b)(iii) Nylon is another common synthetic condensation polymer (a polyamide).

(a)(i) ethanol [1], propanoic acid [1]. (a)(ii) concentrated sulfuric acid [1], heating / reflux [1] (accept: warm). (b)(i) addition polymerisation uses unsaturated monomers / alkenes containing C=C double bonds whereas condensation polymerisation uses monomers with two functional groups [1]; addition polymerisation forms a single product / polymer only [1]; condensation polymerisation forms the polymer and a small molecule (e.g. water or HCl) [1]. (b)(ii) ester linkage [1], water [1] (accept: \(\text{H}_2\text{O}\)). (b)(iii) nylon / polyamide [1].

(a) Heating solid copper(II) carbonate produces solid copper(II) oxide and carbon dioxide gas: \(\text{CuCO}_3\text{(s)} \rightarrow \text{CuO(s)} + \text{CO}_2\text{(g)}\). (b)(i) \(M_r = 64 + 12 + (3 \times 16) = 124\). (b)(ii) \(\text{Moles} = 6.20 \div 124 = 0.05\text{ mol}\). (b)(iii) \(\text{Volume} = 0.05 \times 24 = 1.2\text{ dm}^3\). (c)(i) \(\text{Percentage yield} = (0.96 \div 1.2) \times 100\% = 80\%\). (c)(ii) Gas could have escaped before the bung was inserted, or the reaction did not go to completion, or some gas dissolved in the water used to collect it.

(a) Correct formulas for reactants and products: \(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\) [1]; Correct state symbols: \(\text{(s)}\), \(\text{(s)}\), \(\text{(g)}\) [1]; Reactants and products balanced [1]. (b)(i) \(64 + 12 + 48 = 124\) [1]. (b)(ii) 0.05 mol [1]. (b)(iii) Use of 24: \(0.05 \times 24\) [1]; Correct answer: 1.2 \(\text{dm}^3\) [1]. (c)(i) Calculation: \((0.96 \div 1.2) \times 100\) [1]; Correct percentage: 80% [1]. (c)(ii) Any one valid reason: carbon dioxide gas escaped before the bung was secured / reaction was incomplete / copper(II) carbonate was impure / some gas dissolved in water [1].

(a)(i) At the anode, hydroxide ions from water are discharged to form oxygen gas, which is observed as bubbles. (a)(ii) At the cathode, copper(II) ions are discharged to form copper metal, which is observed as a pink/brown coating. (a)(iii) The blue colour of the solution fades as copper(II) ions are removed. (b)(i) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). (b)(ii) \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\) (or equivalent water oxidation equation). (c) Hydrogen ions (\(\text{H}^+\)) and sulfate ions (\(\text{SO}_4^{2-}\)) remain in solution, forming sulfuric acid, which is acidic. (d) With copper electrodes, the anode dissolves (decreases in size) instead of bubbles forming.

(a)(i) bubbles / effervescence of a colourless gas [1]. (a)(ii) pink / brown / red-brown solid deposited [1]. (a)(iii) blue colour of solution fades / becomes paler [1]. (b)(i) \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\): correct species [1], correct state symbols [1]. (b)(ii) \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\) (or equivalent): correct species [1], correct state symbols [1]. (c) copper(II) ions and hydroxide ions are discharged / removed [1], leaving hydrogen ions / \(\text{H}^+\) and sulfate ions / \(\text{SO}_4^{2-}\) in solution (forming sulfuric acid / increasing \(\text{H}^+\) concentration) [1]. (d) copper anode dissolves / decreases in mass / shrinks [1] (Do not accept: bubbles form).

(a) Decane is cracked to yield propene, a second alkene, and an alkane. Balancing the remaining atoms (7 carbons and 16 hydrogens) gives one alkene and one alkane, such as pentene and ethane: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_5\text{H}_{10} + \text{C}_2\text{H}_6\). Other combinations such as butene and propane (\(\text{C}_4\text{H}_8 + \text{C}_3\text{H}_8\)) or hexene and methane (\(\text{C}_6\text{H}_{12} + \text{CH}_4\)) are also acceptable. (b)(i) Alkenes react together via addition polymerization because they contain a double bond. (b)(ii) The repeat unit is drawn by changing the carbon-carbon double bond of propene to a single bond with extending bonds at each end: \(-(\text{CH}_2-\text{CH}(\text{CH}_3))-\). (c)(i) The catalytic hydration of ethene requires a phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst and a temperature of \(300\text{ }^\circ\text{C}\) (accept \(250\text{ }^\circ\text{C}\) to \(350\text{ }^\circ\text{C}\)). (c)(ii) Fermentation of glucose uses renewable resources (crops/plants) and runs at low temperatures (approx \(30-40\text{ }^\circ\text{C}\)), which saves energy compared to the fossil-fuel derived ethene and high temperatures/pressures of hydration. (d) Add bromine water (aqueous bromine). Propane (an alkane) does not react and the mixture remains orange/brown. Propene (an alkene) reacts quickly and decolourises the bromine water (turns from orange to colourless).

(a) [2 marks] - 1 mark for correct molecular formulas of an alkene and an alkane that together sum to \(\text{C}_7\text{H}_{16}\). - 1 mark for the fully balanced equation. (b)(i) [1 mark] - Addition (polymerization). (b)(ii) [2 marks] - 1 mark for two carbon atoms linked by a single covalent bond, with continuation bonds shown at both ends. - 1 mark for correct substituents showing all bonds: three hydrogen atoms and one methyl group (with the carbon of the methyl group bonded to three hydrogens). (c)(i) [2 marks] - 1 mark for phosphoric acid (or \(\text{H}_3\text{PO}_4\)). - 1 mark for temperature in the range \(250\text{ }^\circ\text{C}\) to \(350\text{ }^\circ\text{C}\). (c)(ii) [1 mark] - Uses renewable resources OR requires less energy / lower temperature. (d) [2 marks] - 1 mark for using aqueous bromine / bromine water. - 1 mark for stating that propane remains orange/yellow/brown AND propene decolourises / turns colourless.

**(a) (i)**
Using the acceleration formula:
\(a = \frac{v - u}{t}\)
\(a = \frac{8.0 \text{ m/s} - 0 \text{ m/s}}{5.0 \text{ s}} = 1.6 \text{ m/s}^2\)

**(a) (ii)**
Using Newton's second law:
\(F = m \times a\)
\(F = 80 \text{ kg} \times 1.6 \text{ m/s}^2 = 128 \text{ N}\)

**(b) (i)**
Using the formula for distance at constant speed:
\(d = v \times t\)
\(d = 8.0 \text{ m/s} \times 15 \text{ s} = 120 \text{ m}\)

**(b) (ii)**
Using the kinetic energy formula:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = 0.5 \times 80 \text{ kg} \times (8.0 \text{ m/s})^2 = 40 \times 64 = 2560 \text{ J}\) (or \(2.56 \text{ kJ}\))

**(a) (i)**
- Formula: \(a = \frac{v-u}{t}\) or substitution \(\frac{8.0}{5.0}\) [1]
- Correct numerical value: \(1.6\) [1]
- Correct unit: \(\text{m/s}^2\) (or \(\text{m s}^{-2}\)) [1]

**(a) (ii)**
- Formula: \(F = ma\) or substitution \(80 \times 1.6\) [1]
- Correct answer: \(128 \text{ N}\) (accept error carried forward from (a)(i)) [1]

**(b) (i)**
- Formula/Substitution: \(8.0 \times 15\) [1]
- Correct answer: \(120 \text{ m}\) [1]

**(b) (ii)**
- Formula: \(E_k = \frac{1}{2}mv^2\) [1]
- Correct substitution: \(0.5 \times 80 \times 8.0^2\) [1]
- Correct answer with unit: \(2560 \text{ J}\) or \(2.56 \text{ kJ}\) [1]

**(a) (i)**
Using the parallel resistance formula:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{12.0} + \frac{1}{6.0} = \frac{1}{12.0} + \frac{2}{12.0} = \frac{3}{12.0}\)
Therefore, \(R_p = \frac{12.0}{3} = 4.0 \ \Omega\).
Alternatively, using the product-over-sum formula:
\(R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{12.0 \times 6.0}{12.0 + 6.0} = \frac{72.0}{18.0} = 4.0 \ \Omega\).

**(a) (ii)**
Total resistance in series:
\(R_{\text{total}} = R_{\text{series}} + R_p = 4.0 \ \Omega + 4.0 \ \Omega = 8.0 \ \Omega\).

**(b) (i)**
Using Ohm's law for the total circuit:
\(I = \frac{V}{R_{\text{total}}} = \frac{12.0 \text{ V}}{8.0 \ \Omega} = 1.5 \text{ A}\).

**(b) (ii)**
Potential difference across the series 4.0 \(\Omega\) resistor:
\(V = I \times R = 1.5 \text{ A} \times 4.0 \ \Omega = 6.0 \text{ V}\).

**(c)**
The potential difference across the parallel branch is: \(V_p = 12.0 \text{ V} - 6.0 \text{ V} = 6.0 \text{ V}\).
Therefore, the potential difference across the 12.0 \(\Omega\) resistor is 6.0 V.
Now calculate the power dissipated:
\(P = \frac{V^2}{R} = \frac{6.0^2}{12.0} = \frac{36.0}{12.0} = 3.0 \text{ W}\).
Alternatively, the current through the 12.0 \(\Omega\) resistor is \(I = \frac{6.0 \text{ V}}{12.0 \ \Omega} = 0.5 \text{ A}\).
Then, \(P = I^2 R = 0.5^2 \times 12.0 = 0.25 \times 12.0 = 3.0 \text{ W}\).

**(a) (i)**
- Correct formula for parallel resistors, e.g., \(\frac{1}{R_p} = \frac{1}{12} + \frac{1}{6}\) or \(\frac{12 \times 6}{12 + 6}\) [1]
- Correct calculation leading to 4.0 \(\Omega\) [1]

**(a) (ii)**
- Correct total resistance of 8.0 \(\Omega\) (addition of 4.0 + parallel value) [1]

**(b) (i)**
- Formula: \(I = \frac{V}{R}\) or substitution \(\frac{12}{8.0}\) [1]
- Correct value of 1.5 A [1]

**(b) (ii)**
- Formula: \(V = IR\) or substitution \(1.5 \times 4.0\) [1]
- Correct value of 6.0 V (accept error carried forward from (b)(i)) [1]

**(c)**
- Identification of the voltage across the 12.0 \(\Omega\) resistor as 6.0 V (or calculation of current through it as 0.5 A) [1]
- Substitution into a power formula, e.g., \(P = \frac{V^2}{R}\) or \(P = I^2 R\) [1]
- Correct answer of 3.0 W [1]

**(a) (i)**
Using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
Rearranging to solve for \(\sin r\):
\(\sin r = \frac{\sin i}{n} = \frac{\sin 40^\circ}{1.5}\)
\(\sin r = \frac{0.6428}{1.5} \approx 0.4285\)
\(r = \arcsin(0.4285) \approx 25.4^\circ\) (or \(25^\circ\))

**(a) (ii)**
- **Speed:** decreases (light slows down as it enters a denser medium).
- **Wavelength:** decreases.
- **Frequency:** remains unchanged (frequency of wave source is constant).

**(b) (i)**
Using the refractive index and speed relation:
\(n = \frac{v_{\text{air}}}{v_{\text{glass}}}\)
\(v_{\text{glass}} = \frac{3.0 \times 10^8 \text{ m/s}}{1.5} = 2.0 \times 10^8 \text{ m/s}\)

**(b) (ii)**
The light ray is in glass (the denser medium) attempting to cross into air (the rarer medium).
The angle of incidence (\(45^\circ\)) is greater than the critical angle (\(42^\circ\)).
Therefore, total internal reflection occurs, and all the light is reflected back inside the glass.

**(a) (i)**
- Correct formula: \(n = \frac{\sin i}{\sin r}\) or rearranged form [1]
- Substitution: \(\sin r = \frac{\sin 40^\circ}{1.5}\) [1]
- Correct calculation: \(25^\circ\) or \(25.4^\circ\) (accept range \(25^\circ - 25.4^\circ\)) [1]

**(a) (ii)**
- Speed decreases [1]
- Frequency remains unchanged / constant [1]
- Wavelength decreases [1]

**(b) (i)**
- Formula/substitution: \(v = \frac{c}{n}\) or \(\frac{3.0 \times 10^8}{1.5}\) [1]
- Correct value with unit: \(2.0 \times 10^8 \text{ m/s}\) [1]

**(b) (ii)**
- States that angle of incidence is greater than critical angle (or \(45^\circ > 42^\circ\)) [1]
- States that total internal reflection occurs [1]

(a) An alternating current in the primary coil creates a continuously changing magnetic field around it. This magnetic field is guided by the soft iron core to the secondary coil. As the changing magnetic field cuts through the turns of the secondary coil, a changing magnetic flux linkage is established, which induces an alternating electromotive force (e.m.f.) across the secondary coil.

(b) Using the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{240}{12} = \frac{800}{N_s}\)
\(N_s = \frac{12 \times 800}{240} = 40\) turns

(c) (i) Using the power equation for the secondary circuit:
\(P = I_s \times V_s\)
\(36 = I_s \times 12\)
\(I_s = \frac{36}{12} = 3.0\text{ A}\)

(ii) Since the transformer is \(100\%\) efficient, the input power equals the output power:
\(I_p \times V_p = I_s \times V_s = 36\text{ W}\)
\(I_p \times 240 = 36\)
\(I_p = \frac{36}{240} = 0.15\text{ A}\)

(d) High voltage means a lower current is needed to transmit the same electrical power. Since power loss in transmission cables is given by \(P = I^2 R\), reducing the current significantly reduces the energy wasted as thermal energy in the cables.

(a) Max 3 marks:
- Alternating current produces a changing / alternating magnetic field [1]
- Iron core links / guides the magnetic field to the secondary coil [1]
- Changing magnetic field cuts secondary coil / change in magnetic flux linkage induces an alternating e.m.f. [1]

(b) Max 2 marks:
- Correct formula or substitution: \(\frac{240}{12} = \frac{800}{N_s}\) [1]
- Correct calculation of \(N_s = 40\) [1]

(c)(i) Max 2 marks:
- Correct formula: \(P = I \times V\) or substitution: \(36 = I_s \times 12\) [1]
- Correct current: \(3.0\text{ A}\) (unit required) [1]

(c)(ii) Max 2 marks:
- Correct formula / relationship: \(I_p V_p = I_s V_s\) or \(I_p V_p = 36\text{ W}\) [1]
- Correct primary current: \(0.15\text{ A}\) (unit required) [1]

(d) Max 1 mark:
- To reduce current so that energy / power / heat loss in the transmission cables is reduced [1]

(a) An alternating current in the primary coil creates a continuously changing magnetic field around it. This magnetic field is guided by the soft iron core to the secondary coil. As the changing magnetic field cuts through the turns of the secondary coil, a changing magnetic flux linkage is established, which induces an alternating electromotive force (e.m.f.) across the secondary coil.

(b) Using the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{240}{12} = \frac{800}{N_s}\)
\(N_s = \frac{12 \times 800}{240} = 40\) turns

(c) (i) Using the power equation for the secondary circuit:
\(P = I_s \times V_s\)
\(36 = I_s \times 12\)
\(I_s = \frac{36}{12} = 3.0\text{ A}\)

(ii) Since the transformer is \(100\%\) efficient, the input power equals the output power:
\(I_p \times V_p = I_s \times V_s = 36\text{ W}\)
\(I_p \times 240 = 36\)
\(I_p = \frac{36}{240} = 0.15\text{ A}\)

(d) High voltage means a lower current is needed to transmit the same electrical power. Since power loss in transmission cables is given by \(P = I^2 R\), reducing the current significantly reduces the energy wasted as thermal energy in the cables.

(a) Max 3 marks:
- Alternating current produces a changing / alternating magnetic field [1]
- Iron core links / guides the magnetic field to the secondary coil [1]
- Changing magnetic field cuts secondary coil / change in magnetic flux linkage induces an alternating e.m.f. [1]

(b) Max 2 marks:
- Correct formula or substitution: \(\frac{240}{12} = \frac{800}{N_s}\) [1]
- Correct calculation of \(N_s = 40\) [1]

(c)(i) Max 2 marks:
- Correct formula: \(P = I \times V\) or substitution: \(36 = I_s \times 12\) [1]
- Correct current: \(3.0\text{ A}\) (unit required) [1]

(c)(ii) Max 2 marks:
- Correct formula / relationship: \(I_p V_p = I_s V_s\) or \(I_p V_p = 36\text{ W}\) [1]
- Correct primary current: \(0.15\text{ A}\) (unit required) [1]

(d) Max 1 mark:
- To reduce current so that energy / power / heat loss in the transmission cables is reduced [1]