CSS With A CSP

General Science and Ability · CSS 2025

CSS 2025 General Science and Ability Paper - solved

By CSP Qasim Farooq

This set carries the seven subjective questions of the CSS 2025 General Science and Ability paper, Q2 through Q8, each solved and written from scratch for this platform. Q2 to Q5 are the General Science questions of Section A; Q6 to Q8 are the General Ability questions of Section B. The objective part of the paper is not covered here.

The science answers are kept to the length a five-mark response earns, and each one carries a compact illustration of the process rather than a wall of prose. The ability answers show the method without a calculator: what is given, the formula used, the substitution, the working line by line, the units, and the final answer, with a check where one is available.

New to solved papers? How to actually use CSS past papers explains the difference between reading them and working them.

Section A: General Science

Q2 Earth's rotation and revolution, electromagnetic radiation, hurricanes, and nitrogen-containing fertilizers and insecticides

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Question 2 asks four separate short answers on the Earth's two motions, the electromagnetic spectrum, hurricane formation and damage, and nitrogen inputs. The examiner is marking accurate mechanism rather than length, so each part below states the process and then its consequences.

(a) What effects are produced due to Rotation & Revolution of Earth? Explain briefly.

Rotation is the Earth's spin from west to east around its tilted axis, completed in about 24 hours. Its most visible effect is the alternation of day and night as different parts of the Earth face the Sun. Rotation also creates the apparent daily movement of the Sun and stars from east to west and provides the basis for local time and time zones. The Coriolis effect produced by rotation deflects winds and ocean currents to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. Rotation also contributes to the Earth's equatorial bulge and slight polar flattening.

Rotation, about 24 hours
day and night
apparent movement of Sun and stars
time zones
Coriolis deflection

Revolution is the Earth's yearly movement around the Sun in an elliptical orbit, completed in about 365.25 days. Revolution combined with the constant 23.5° axial tilt produces seasons: a hemisphere tilted toward the Sun receives longer days and more direct rays, while the other experiences shorter days and weaker heating. It also changes the length of day and night through the year, producing solstices and equinoxes. The extra quarter-day explains the addition of a leap day approximately every four years. The Earth-Sun distance changes slightly, but axial tilt, not distance, is the principal cause of seasons.

Revolution, about 365.25 days
23.5° axial tilt
seasons
solstices and equinoxes
leap day
Earth's revolution, axial tilt and the seasonsThe Sun is at the centre of an oval orbit, which is a nearly circular orbit seen at an angle. Earth is drawn at four positions, each with its axis tilted 23.5 degrees and pointing the same way, with N marking the north end. On the left, the June solstice: the North Pole leans toward the Sun, northern summer. On the right, the December solstice: the North Pole leans away from the Sun, northern winter. At the far side is the March equinox and at the near side the September equinox, where neither pole leans toward the Sun. Arrows on the orbit show Earth moving from December to March to June to September. Not to scale.SunNNNNtilt 23.5°June solsticenorthern summerDecember solsticenorthern winterMarch equinoxSeptember equinox
Earth's revolution and the seasons, with Northern Hemisphere names. The axis keeps its 23.5° tilt and points the same way all year, so the North Pole leans toward the Sun in June and away from it in December. Not to scale: the orbit is seen at an angle and is nearly circular, and Earth is in fact slightly closer to the Sun in early January.

(b) Describe Electromagnetic Radiations. Give its types and explain their applications.

Electromagnetic radiation is energy transmitted through space as mutually oscillating electric and magnetic fields. It needs no material medium and travels in vacuum at approximately 3 × 10⁸ m/s. All electromagnetic waves have the same basic nature but differ in wavelength, frequency and photon energy; frequency increases as wavelength decreases.

The spectrum and major uses are:

  1. Radio waves: broadcasting, mobile communication, navigation and radio astronomy.
  2. Microwaves: radar, satellite links, Wi-Fi and heating food, because water-rich material absorbs part of their energy.
  3. Infrared: thermal imaging, remote controls, night vision and heat sensing.
  4. Visible light: human vision, photography, illumination and optical-fibre communication.
  5. Ultraviolet: sterilizing water and instruments, detecting fluorescent substances and helping the skin form vitamin D; excessive exposure damages skin and eyes.
  6. X-rays: medical and dental imaging, security scanning and examination of crystal structure; excessive ionizing exposure can damage tissue.
  7. Gamma rays: cancer radiotherapy, sterilization of medical equipment, industrial inspection and nuclear research.
Radio
Microwave
Infrared
Visible
Ultraviolet
X-ray
Gamma

Read that order from left to right: wavelength decreases while frequency, photon energy and biological hazard increase. Longer-wavelength radiation is therefore mainly useful for communication and heating, while shorter-wavelength, higher-energy radiation is valuable for imaging and sterilization but requires stricter shielding.

(c) What are Hurricanes? How they are formed and what level of damage they can produce?

A hurricane is a mature tropical cyclone in the Atlantic or north-eastern Pacific with sustained winds of at least 119 km/h. The same storm type is called a typhoon in the north-western Pacific and a cyclone in the Indian Ocean.

Formation requires ocean water generally warmer than about 26.5°C, moist unstable air, a pre-existing low-pressure disturbance, weak vertical wind shear and sufficient Coriolis force, so hurricanes rarely form very near the equator. Warm water evaporates; moist air rises and cools; condensation releases latent heat, which warms the storm core and lowers surface pressure. More air spirals inward, rises in thunderstorms and organizes into rainbands around a calm eye and violent eyewall. The system strengthens while it remains over warm water but weakens over land, cooler water or strong wind shear.

Warm ocean above 26.5°C
evaporation
rising moist air
condensation releases latent heat
surface pressure falls
rotating inflow
eye, eyewall and rainbands

Damage comes from several interacting hazards. Extreme winds destroy roofs, trees, power lines and weak buildings. Storm surge, the wind-driven rise of sea level, can inundate low-lying coasts and is often the deadliest coastal threat. Torrential rain causes flash floods and landslides far inland; waves erode shorelines; and embedded tornadoes may add local destruction. The Saffir-Simpson scale ranks hurricanes from Category 1 to 5 by sustained wind, but even a lower-category storm may cause catastrophic flooding and surge.

(d) Explain the advantages and disadvantages of nitrogen containing fertilizers and insecticides.

Nitrogen fertilizers such as urea, ammonium nitrate and ammonium sulphate supply nitrogen needed to form chlorophyll, amino acids, proteins and nucleic acids. They rapidly correct deficiency, increase leaf growth and raise crop yield, making intensive food production possible. However, excess application produces weak, overly lush plants, soil acidification and nitrate accumulation. Nitrate leaching can contaminate groundwater, while runoff causes eutrophication, algal blooms and oxygen depletion. Production and field use also release greenhouse gases, especially carbon dioxide and nitrous oxide.

The phrase "nitrogen-containing insecticides" is not a standard agricultural class, but many insecticides contain nitrogen in their molecules. Their advantage is effective control of insects that damage crops or transmit disease, reducing yield loss and improving food quality. Their disadvantages may include poisoning of farm workers, residues on food, killing of pollinators and natural predators, contamination of soil and water, pest resistance and secondary pest outbreaks. Risks depend on the particular chemical, dose and persistence, not simply on the presence of nitrogen.

Correct dose
better growth and pest control
higher usable yield
Excess or misuse
runoff and residues
eutrophication or poisoning
resistance and ecosystem damage

Both inputs should therefore be used through soil testing, precise timing and placement, protective equipment, observation of withholding periods and integrated pest management. Biological control, crop rotation and threshold-based spraying can reduce reliance on chemicals while retaining agricultural benefits.

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Q3 Temperature and light intensity in photosynthesis, the structure and functions of capillaries, hepatitis, and biomass compared with biogas

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Question 3 moves from plant physiology to human circulation, then to liver disease and renewable energy. Each part rewards a stated mechanism rather than a list, and the last one turns on a single distinction: biomass is the feedstock, biogas is one carrier obtained from it.

(a) Explain how temperature and light intensity affect the rate of photosynthesis in plants.

Photosynthesis converts carbon dioxide and water into carbohydrates using light energy. Its rate is controlled by limiting factors, especially light intensity, temperature and carbon-dioxide supply.

At low light intensity, light is the limiting factor. Increasing intensity supplies more photons to chlorophyll, so the light-dependent reactions generate ATP and NADPH faster and the photosynthetic rate rises almost proportionally. At the light-saturation point, the chloroplast machinery or another factor such as carbon dioxide becomes limiting, so further light gives little increase. Extremely intense light may damage photosystems and reduce the rate, a process called photoinhibition. Below the light-compensation point, respiration uses as much or more carbohydrate than photosynthesis produces.

Low light
rate rises with intensity
light saturation
plateau
photoinhibition

Temperature mainly affects enzyme-controlled reactions. At low temperature, molecules move slowly and enzymes involved in carbon fixation work less rapidly. As temperature rises, the rate increases to an optimum that varies with plant type and adaptation. Beyond the optimum, enzymes lose efficiency, stomata may close to conserve water, carbon-dioxide entry falls and photorespiration may increase. At very high temperature, enzymes and membranes can be damaged, causing a sharp decline.

Low temperature
rate rises
optimum
enzyme and stomatal stress
rate falls

The two factors interact: raising temperature will not increase photosynthesis when light is insufficient, and stronger light has limited effect when temperature or carbon dioxide is already limiting.

(b) Describe the structure and functions of capillaries in the human blood circulatory system.

Capillaries are the smallest blood vessels and form dense networks between arterioles and venules. Their diameter is roughly that of a red blood cell, so cells often pass through one at a time. A capillary wall consists mainly of a single layer of flattened endothelial cells resting on a thin basement membrane. It contains no thick muscular or elastic coat. Tiny intercellular gaps and, in some organs, pores make the wall selectively permeable. Precapillary sphincters regulate how much blood enters particular capillary beds.

Arteriole
capillary wall one cell thick
exchange with tissue fluid
venule
A capillary between an arteriole and a venule, and exchange across its wallA narrow capillary runs from an arteriole on the left to a venule on the right, surrounded by tissue fluid. The capillary wall is drawn as a single row of thin, flat cells, and red blood cells pass along it in single file. Above the capillary, near the arteriole end, arrows show water being pushed out; near the venule end, arrows show water being drawn back in. Below, near the arteriole end, arrows show oxygen and glucose moving out to the tissues; near the venule end, arrows show carbon dioxide and wastes moving in. Not to scale.tissue fluidarteriolevenulewater pushed outwater drawn back inO₂ and glucose outCO₂ and wastes incapillary wall: one layer of flat cells
A capillary between an arteriole and a venule. Its wall is a single layer of flat cells and red cells pass in single file, so substances cross a very short distance. Water is pushed out near the arteriole end and drawn back in near the venule end; the small excess drains into lymph vessels (not shown). Not to scale.

This structure is suited to exchange. The wall is only one cell thick, giving a short diffusion distance. The combined cross-sectional area of millions of capillaries slows blood flow, allowing time for transfer. Oxygen and nutrients diffuse from blood into tissue fluid and cells, while carbon dioxide and other wastes move toward the blood. Hydrostatic pressure near the arterial end filters water and small solutes outward; much fluid returns by osmotic forces near the venous end, while excess enters lymphatic vessels. Capillaries also exchange hormones, distribute heat and permit white blood cells to move into tissues during inflammation.

Their permeability varies by organ: brain capillaries form tight barriers, kidney capillaries are fenestrated for filtration, and liver sinusoids are wider and more permeable.

(c) What is hepatitis? Give its types and briefly explain its common symptoms along with preventions.

Hepatitis means inflammation of the liver. It may be caused by viruses, alcohol, medicines, toxins, metabolic disorders or autoimmune disease. The principal viral forms are A, B, C, D and E. Hepatitis A and E usually spread by contaminated food or water and commonly cause acute illness. Hepatitis B spreads through infected blood, sexual contact and mother-to-child transmission. Hepatitis C spreads mainly through infected blood, while hepatitis D occurs only in a person infected with hepatitis B. B, C and D may become chronic and lead to cirrhosis or liver cancer.

Some infected people have no early symptoms. Common features include fatigue, fever, poor appetite, nausea, vomiting, discomfort in the right upper abdomen, dark urine, pale stools, joint pain and jaundice, which makes the skin and eyes yellow. Severe cases may produce bleeding, confusion or liver failure.

Prevention depends on route. Safe water, sanitation, handwashing and properly cooked food reduce A and E. Effective vaccines are available for A and B; preventing B also prevents D. Screening blood, using sterile needles and medical instruments, avoiding shared razors, practising safer sex and testing pregnant women reduce B and C. There is no widely used vaccine for C, so early testing and antiviral treatment are important. Alcohol and unnecessary hepatotoxic medicines should be avoided during liver illness.

Hepatitis A and E
faecal-oral route
safe water, sanitation and cooked food
Hepatitis B, C and D
blood and body fluids
vaccination for B, sterile equipment, screening, safer sex

(d) Differentiate between biomass and biogas. Briefly explain various methods for obtaining energy from biomass.

Biomass is organic material of recent biological origin, including wood, crop residues, animal manure and biodegradable municipal waste. It is the solid or liquid feedstock itself. Biogas is a combustible gas produced when microorganisms break down wet biomass without oxygen. It usually contains methane and carbon dioxide, with small amounts of other gases. Biogas is therefore one energy carrier obtained from biomass, not an alternative to it.

Energy can be recovered from biomass by several methods:

  1. Direct combustion: dry wood, bagasse or crop residue is burned to produce heat or steam for electricity. It is simple but smoky and inefficient in poor stoves.
  2. Anaerobic digestion: manure, sewage or food waste decomposes in a sealed digester. Methane-rich biogas fuels cooking, heating, engines or generators; the digestate can be used as fertilizer.
  3. Gasification: limited oxygen converts solid biomass into synthesis gas containing mainly carbon monoxide and hydrogen, which can fuel boilers or engines.
  4. Pyrolysis: heating without oxygen produces bio-oil, combustible gas and biochar.
  5. Fermentation: sugars and starches are converted into ethanol; oils can be processed into biodiesel.

These five are parallel routes from one feedstock, not stages of one process. Each begins with biomass and ends in a different carrier:

  • Combustion gives heat and electricity.
  • Anaerobic digestion gives biogas and a fertilizer digestate.
  • Gasification gives synthesis gas.
  • Pyrolysis gives bio-oil, gas and biochar.
  • Fermentation gives ethanol or biodiesel.

Biomass can be renewable when regrowth and waste recovery match use. Unsustainable harvesting, however, causes deforestation, smoke and competition for land.

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Q4 Deforestation and its environmental damage, how water pollution arises and the part played by bacteria, examples of human pollution, and ten effects of air pollution

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Question 4 is a four-part environmental question whose highest-scoring part asks what bacteria actually do after sewage enters a river. That part rewards a mechanism, not a general account of pollution, so each answer below gives the process first and the control measures second.

(a) What is deforestation? State the disadvantages caused to the environment due to this.

Deforestation is the large-scale, long-term removal of natural forest cover and conversion of the land to agriculture, settlements, roads, mining or other non-forest uses. It differs from carefully managed harvesting followed by effective regeneration.

Its environmental disadvantages are extensive. Forests store carbon in wood and soil; cutting or burning them releases carbon dioxide and removes a major carbon sink, intensifying climate change. Habitat loss fragments ecosystems, reduces genetic diversity and may drive plants and animals toward extinction. Without roots and leaf litter, rainfall strikes bare soil directly, increasing erosion, landslides and loss of fertile topsoil. Sediment then enters rivers and reservoirs, damaging aquatic habitats and reducing storage capacity.

Forest removal
carbon released and sink lost
bare soil and lost habitat
erosion, floods and biodiversity decline
disturbed rainfall

Deforestation also disturbs the water cycle. Reduced transpiration can lower atmospheric moisture and regional rainfall, while rapid surface runoff increases flash flooding. In dry regions, continued loss of vegetation promotes desertification. Forest-dependent communities lose fuel, food, medicines and livelihoods, and closer contact among people, livestock and displaced wildlife may increase the risk of zoonotic disease.

Control requires protected forests, enforcement against illegal clearing, sustainable forestry, agroforestry, restoration with native species and higher productivity on existing farmland. Tree planting helps, but a plantation cannot immediately replace the biodiversity and ecological functions of an old natural forest.

(b) Explain how water pollution is produced? Describe the part played by bacteria after a river has been polluted by sewage.

Water pollution occurs when physical, chemical or biological substances enter water in quantities that damage organisms or make it unsafe for use. Major sources include untreated domestic sewage, industrial effluent, agricultural fertilizers and pesticides, oil, plastics, mining drainage, hot-water discharge and eroded sediment. Pollutants may enter through a visible pipe, called a point source, or through diffuse runoff from a broad area.

After sewage enters a river, decomposer bacteria feed on its organic matter. Their population increases rapidly and they consume dissolved oxygen during aerobic respiration. This raises the biochemical oxygen demand (BOD). If oxygen is used faster than it is replenished from the air and photosynthesis, dissolved-oxygen concentration falls. Fish and oxygen-dependent invertebrates become stressed or die. Farther downstream, once much organic material has been decomposed and water mixes with air, oxygen may gradually recover through natural self-purification.

Sewage adds organic matter
decomposer bacteria multiply
respiration raises BOD
dissolved oxygen falls
fish and invertebrates die
anaerobic decay and foul gases

If sewage loading is very high, water becomes anaerobic. Anaerobic bacteria then produce foul-smelling gases such as hydrogen sulphide and methane. Sewage may also carry pathogens that cause diseases and nutrients that stimulate algal blooms; decay of the algae consumes still more oxygen.

Treatment should remove solids and organic matter, disinfect effluent and, where necessary, remove nitrogen and phosphorus before discharge.

(c) State some examples of pollution caused by human activity and discuss the effects of each example.

Human activities produce pollution in several connected forms. These are parallel categories, not a sequence: one activity commonly produces several of them at once.

  1. Air pollution: vehicles, power stations and industry emit particulate matter, nitrogen oxides, sulphur dioxide and carbon monoxide. These cause asthma and cardiovascular disease, reduce visibility, form smog and acid rain, and damage crops and buildings. Carbon dioxide and methane intensify climate change.
  2. Water pollution: sewage, industrial chemicals, oil and farm runoff contaminate rivers and groundwater. Effects include water-borne disease, toxicity, eutrophication, oxygen depletion and death of aquatic organisms.
  3. Soil pollution: pesticides, heavy metals, poorly managed landfill and industrial waste reduce soil organisms and fertility. Toxins can enter crops and become concentrated along food chains.
  4. Plastic pollution: discarded packaging and fishing gear entangle or are swallowed by wildlife. Larger items fragment into microplastics that persist in water and soil.
  5. Noise pollution: traffic, aircraft, construction and machinery cause stress, sleep disturbance, hearing loss and reduced concentration; wildlife communication and breeding may also be disrupted.
  6. Thermal and light pollution: heated industrial water lowers dissolved oxygen and alters aquatic communities, while excessive night lighting disturbs insects, birds, turtles and human sleep cycles.

Because pollutants cross environmental boundaries, control works best as an ordered hierarchy rather than as cleanup alone:

Prevent at source
cleaner production
treat the waste
enforce emission standards
monitor and report

(d) What does air pollution mean? State ten serious effects of air pollution on the environment and briefly explain how they can be reduced.

Air pollution is the presence of gases, particles or biological material in the atmosphere at concentrations harmful to health, ecosystems, property or climate.

Ten serious effects are: (1) respiratory illness from fine particles and ozone; (2) heart disease and premature death; (3) photochemical smog and poor visibility; (4) acid rain from sulphur and nitrogen oxides; (5) crop damage and lower yield from ground-level ozone; (6) forest decline and soil acidification; (7) contamination of water and food chains by deposited mercury and other toxins; (8) corrosion of buildings and monuments; (9) climate warming from greenhouse gases and black carbon; and (10) stratospheric ozone depletion by ozone-destroying chemicals, increasing harmful ultraviolet radiation.

Reduction needs action at source. Cleaner fuels, renewable electricity, energy efficiency and reliable public transport reduce combustion. Vehicles and factories require emission standards, inspection, catalytic converters, filters and scrubbers. Open burning of waste and crop residue should be restricted while providing practical collection and residue-management alternatives. Low-sulphur fuel and control of industrial nitrogen oxides reduce acid-forming emissions. Methane leaks from oil, gas, landfills and agriculture should be captured. Dust at construction sites must be suppressed, and harmful refrigerants should be recovered and replaced under ozone-protection rules. Monitoring, public reporting and urban green space support these measures, although trees cannot substitute for emission control.

Cleaner fuels and efficiency
capture and control emissions
enforce standards and inspection
monitor and report air quality
protect health, crops and climate
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Q5 Food contaminants compared with adulterants, the major food groups, artificial intelligence and its branches, and the advantages and structure of optical fibres

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Question 5 pairs two food-science parts with two technology parts, and each turns on one clean distinction: intent separates a contaminant from an adulterant, and learned patterns separate artificial intelligence from ordinary programming.

(a) Differentiate between food contaminants and food adulterants.

A food contaminant is an unwanted biological, chemical or physical substance that enters food unintentionally during production, processing, transport or storage. Examples include Salmonella from poor hygiene, pesticide residues, heavy metals absorbed from soil, cleaning chemicals, glass fragments and toxins formed by mould. Contamination can also be natural, such as aflatoxin in badly stored grain. Its defining feature is accidental or incidental presence, although negligence may allow it.

A food adulterant is a substance deliberately added, removed or substituted to make food appear better, increase quantity, reduce cost or conceal poor quality. Examples are water added to milk, cheap oil mixed with expensive oil, artificial colour added to spices, starch mixed with dairy products, or removal of valuable fat followed by sale as full-quality food. Deliberate mislabelling and substitution are forms of economic fraud even where the added material is not immediately poisonous.

Unintended entry
contaminant
hygiene, process control and testing
Deliberate addition or substitution
adulterant
authenticity testing and enforcement

Both can make food unsafe or nutritionally inferior, but intent is the main distinction: contamination is ordinarily unintended; adulteration is intentional deception. Control also differs. Contamination is reduced through sanitation, hazard analysis, temperature control, residue limits, testing and traceability. Adulteration requires these measures plus supply-chain verification, authenticity testing, labelling enforcement, inspections and penalties. A substance may be both an adulterant and a health hazard if it is knowingly added despite toxicity.

(b) Enlist major food groups based on nutrients, their functions and sources.

Foods can be grouped by their dominant nutrients and functions:

  1. Carbohydrate-rich foods: cereals, rice, wheat, maize, potatoes and other starchy foods provide glucose, the body's principal immediate energy source. Whole grains also supply fibre and B vitamins.
  2. Protein-rich foods: meat, fish, eggs, milk, pulses, beans and nuts provide amino acids for growth, tissue repair, enzymes, hormones and antibodies.
  3. Fat-rich foods: vegetable oils, ghee, butter, nuts and seeds provide concentrated energy, essential fatty acids, insulation and absorption of vitamins A, D, E and K.
  4. Milk and dairy foods: milk, yoghurt and cheese supply protein, calcium, phosphorus and often vitamin D for bones, teeth, muscles and nerves.
  5. Fruits and vegetables: citrus, guava, carrots, leafy vegetables and many others supply vitamins, minerals, antioxidants, water and fibre. They support immunity, vision, blood formation, digestion and normal cell function.
  6. Water and fibre: water maintains blood volume, temperature and chemical reactions; fibre from whole grains, legumes, fruit and vegetables promotes bowel health and helps regulate blood glucose and cholesterol.

Grouped by the job each does in the body, these are parallel functions rather than steps:

  • Carbohydrates and fats supply energy.
  • Proteins supply growth and repair.
  • Fats also carry the fat-soluble vitamins.
  • Vitamins, minerals, fibre and water regulate and protect.

A balanced diet combines groups rather than relying on one. Required proportions vary with age, health and activity, while foods high in added sugar, salt and trans fat should remain limited.

(c) What is Artificial Intelligence and how does it differ from traditional programming? Describe the main branches of Artificial Intelligence.

Artificial Intelligence (AI) is the field of creating computer systems that perform tasks associated with human intelligence, such as learning, recognizing patterns, understanding language, reasoning and making decisions.

In traditional programming, a developer writes explicit rules and the computer applies them to input data to produce an output. In much of modern AI, especially machine learning, the system is trained on examples and discovers statistical patterns; the model can then predict outputs for new data.

Traditional programming: rules and data
program applies the rules
output
Machine learning: data and correct examples
training
learned model
prediction on new data

Traditional programs are usually deterministic and easier to trace; AI systems can adapt to complex patterns but may make probabilistic errors and require careful testing for bias, safety and explainability. Not every AI system learns: expert systems may use hand-written knowledge and inference rules.

Main branches include machine learning, with deep learning and neural networks; natural-language processing for text and speech; computer vision for images and video; knowledge representation and expert systems; reasoning, planning and search; robotics combining perception, control and action; and speech recognition and synthesis. Reinforcement learning trains an agent through rewards and penalties.

AI therefore extends conventional software where fixed rules are inadequate, but it does not remove the need for human-defined objectives, reliable data and oversight.

(d) What are the advantages of optical fibers in the communication system? Draw the basic structure of optical fiber and explain propagation of light through it.

An optical fibre is a thin transparent strand, usually of glass, that carries information as pulses of light. Its core has a slightly higher refractive index than the surrounding cladding; an outer coating, strength layer and jacket provide mechanical protection.

Optical fibres offer high bandwidth and can carry very large data volumes. Their low attenuation allows longer distances between repeaters. They are thin, light and immune to electromagnetic interference, so nearby motors, lightning and power cables do not corrupt the signal. Fibres do not radiate ordinary electromagnetic signals, improving security, and they provide electrical isolation, avoiding sparks and ground loops. They are also resistant to corrosion and permit many channels through wavelength-division multiplexing.

Read the structure as a cross-section, working from the outside of the cable inwards to the centre:

Outer jacket
Buffer coating
Cladding
Core
Optical fibre: construction and the path of lightTwo drawings. Top, a cross-section: four concentric circles, from the centre outwards the core (refractive index n1), the cladding (n2), the buffer coating and the outer jacket. Bottom, a side view along the fibre: the same layers as horizontal bands, with a ray of light travelling from left to right inside the core. Each time the ray meets the boundary between core and cladding it is reflected back into the core, so it zigzags along the fibre. At one reflection a dashed normal is drawn and the angle of incidence, theta, is marked; theta is greater than the critical angle and the core's index n1 is greater than the cladding's n2, which is why the reflection is total. Not to scale.Cross-sectionOuter jacketBuffer coatingCladding (n₂)Core (n₁)Side view: light guided along the coreθtotal internal reflectionn₁ > n₂, and θ is greater than the critical angle
Optical fibre construction, and how light travels in it. The core has a higher refractive index than the cladding (n₁ > n₂), so light meeting the core-cladding boundary at more than the critical angle is totally internally reflected and stays in the core. Not to scale: a real core is far thinner, and its light runs almost parallel to the axis.

Light launched into the core within the fibre's acceptance angle reaches the core-cladding boundary at an angle greater than the critical angle. Because the core has the higher refractive index, the light undergoes total internal reflection and remains guided along the core through repeated reflections.

Light enters within the acceptance angle
strikes the core-cladding boundary
angle exceeds the critical angle
total internal reflection
guided along the core to the receiver

In practical fibres, the guided energy is better described as modes rather than a sharp zigzag ray. Single-mode fibre minimizes modal dispersion over long distances; multimode fibre is easier to couple but is more suited to shorter links. Disadvantages include delicate joining, specialized equipment and signal loss at poor bends or connections.

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Section B: General Ability

Q6 Finding the pre-sale price of a fridge, a 36-instalment payment, dividing Rs. 600 among three people, and correcting five jumbled spellings

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Question 6 tests reverse-percentage work, instalment arithmetic, a three-way share written as equal expressions, and word building. In every money part the trap is the same: the discounted or taxed figure is a percentage of the original, so divide rather than add the percentage back.

(a) In an annual sale, Saima buys a Fridge and a Freezer. The sale offers 15% off everything and she pays a total of Rs. 57,120/-. If the price of a freezer, before the sale, is Rs. 40,000/-, what was the price of the Fridge before the sale?

Given: the fridge and freezer together sold for Rs. 57,120 after a 15% discount, and the freezer's price before the sale was Rs. 40,000.

Method: a 15% discount means the buyer pays 85% of the original combined price, so the original price is the sale price divided by 0.85.

Working:

57,120 ÷ 0.85 = 67,200

67,200 - 40,000 = 27,200

Answer: the price of the fridge before the sale was Rs. 27,200.

Check: 15% of 67,200 = 10,080, and 67,200 - 10,080 = 57,120.

(b) Mr. Rasheed pays a deposit of Rs. 60,000/- followed by 36 equal monthly payments. The total amount that he pays is 127% of the basic price of Rs. 3,360,000/-. Calculate Rasheed's monthly payment.

Given: a basic price of Rs. 3,360,000, a total payable of 127% of that price, a deposit of Rs. 60,000, and 36 monthly instalments.

Method: total payable = 1.27 × basic price; subtract the deposit; divide the balance by 36.

Working:

1.27 × 3,360,000 = 4,267,200

4,267,200 - 60,000 = 4,207,200

4,207,200 ÷ 36 = 116,866.666...

Answer: Rasheed's monthly payment is approximately Rs. 116,866.67, or Rs. 116,867 to the nearest rupee.

Check: 36 × 116,866.67 is approximately Rs. 4,207,200, and adding the Rs. 60,000 deposit gives Rs. 4,267,200.

(c) Divide Rs. 600/- among A, B, and C so that Rs. 40 more than 2/5 of A's share, Rs. 20 more than 2/7 of B's share, and Rs. 10 more than 9/17 of C's share may be equal.

Given: A + B + C = 600, and the three expressions (2/5)A + 40, (2/7)B + 20 and (9/17)C + 10 are all equal.

Method: set the three equal expressions to a single value k, make A, B and C the subject, substitute into the total and solve for k.

Working:

(2/5)A + 40 = (2/7)B + 20 = (9/17)C + 10 = k

A = (5/2)(k - 40), B = (7/2)(k - 20), C = (17/9)(k - 10)

(5/2)(k - 40) + (7/2)(k - 20) + (17/9)(k - 10) = 600

Multiplying every term by 18 clears the denominators:

45(k - 40) + 63(k - 20) + 34(k - 10) = 10,800

142k - 3,400 = 10,800

142k = 14,200, so k = 100

A = (5/2)(60) = 150, B = (7/2)(80) = 280, C = (17/9)(90) = 170

Answer: A receives Rs. 150, B receives Rs. 280, and C receives Rs. 170.

Check: 60 + 40 = 80 + 20 = 90 + 10 = 100, and 150 + 280 + 170 = 600.

(d) Find out the correct words from the jumbled spellings given below:

Method: rebuild each word using every supplied letter exactly once, checking the letter count before committing to an answer.

Answers:

  1. TEANAIRM is MARINATE
  2. VAGNACEXETRA is EXTRAVAGANCE
  3. VADIETLIONC is VALEDICTION
  4. PKUNYBARTC is BANKRUPTCY
  5. TAESRONSC is ANCESTORS

Each answer uses every supplied letter exactly once, which is the check to run before writing a word down.

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Q7 Marked price under successive discounts, sales tax inside a total charge, the area and perimeter of a rectangular campsite, and arranging fractions in descending order

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Question 7 turns on one idea that examiners repeatedly catch candidates on: successive discounts multiply, they do not add. The geometry part is a Pythagoras equation in disguise, and the fractions part is fastest by cross-multiplication.

(a) Calculate the marked price of the sculpture if it is sold for Rs. 79,950/- after a discount of 18%. On another day, if 10% discount is given on the marked price of the same sculpture before it is sold at a further discount of 8%, would the sale price still be Rs. 79,950/-?

Given: an 18% discount produced a selling price of Rs. 79,950; on another day the discounts offered were 10% followed by a further 8%.

Method: after an 18% discount the selling price is 82% of the marked price, so divide to recover the marked price; then apply the two later discounts one after the other as multipliers.

Working:

Marked price = 79,950 ÷ 0.82 = 97,500

After the 10% discount: 97,500 × 0.90 = 87,750

After the further 8% discount: 87,750 × 0.92 = 80,730

Answer: the marked price was Rs. 97,500, and the second sale price would be Rs. 80,730, not Rs. 79,950.

It would be Rs. 780 higher. The combined successive discount is 1 - (0.90 × 0.92) = 17.2%, not 18%.

(b) Majid pays a total of Rs. 51,520/- for his car insurance. The total is made up of a basic charge plus 15% sales tax. Calculate the amount of sales tax that Majid pays.

Given: Majid paid Rs. 51,520 in total, which is the basic charge plus 15% sales tax.

Method: the total is 115% of the basic charge, so divide by 1.15 to recover the basic charge, then subtract.

Working:

Basic charge = 51,520 ÷ 1.15 = 44,800

Sales tax = 51,520 - 44,800 = 6,720

Answer: Majid pays Rs. 6,720 as sales tax.

Check: 15% of 44,800 = 6,720.

(c) A camp site in the shape of a rectangle has sides (3x + 6) m and (x + 1) m and the length of the diagonal is (4x + 1) m. Find the area and the perimeter of the camp site.

The campsite rectangle and its diagonalA rectangle, wider than it is tall. Its long side is labelled (3x + 6) metres and its short side (x + 1) metres. A diagonal joins the bottom-left corner to the top-right corner and is labelled (4x + 1) metres. The triangle below the diagonal is shaded, with a right angle marked at the bottom-right corner: the diagonal is the hypotenuse of that right-angled triangle. Not to scale.(3x + 6) m(x + 1) m(4x + 1) m
The diagonal and two sides of the rectangle form a right-angled triangle, with the diagonal as its hypotenuse, which is why Pythagoras' theorem applies. Not to scale.

Given: the sides are (3x + 6) m and (x + 1) m, and the diagonal is (4x + 1) m.

Method: the diagonal of a rectangle is the hypotenuse of a right triangle formed by the two sides, so apply Pythagoras and solve the resulting quadratic in x. Note that x here is the unknown length variable, never a multiplication sign.

Working:

(3x + 6)² + (x + 1)² = (4x + 1)²

9x² + 36x + 36 + x² + 2x + 1 = 16x² + 8x + 1

x² - 5x - 6 = 0

(x - 6)(x + 1) = 0

x = 6; x = -1 is rejected because it gives an invalid side.

Dimensions: 3(6) + 6 = 24 m and 6 + 1 = 7 m

Area = 24 × 7 = 168 m²

Perimeter = 2(24 + 7) = 62 m

Answer: area = 168 m²; perimeter = 62 m.

Check: the diagonal is 4(6) + 1 = 25 m, and 7² + 24² = 49 + 576 = 625 = 25².

(d) Arrange the following fractions in descending order:

Given, as printed:

  1. Set (a): 7/12, 4/7, 9/14, 25/42
  2. Set (b): 5/8, 7/12, 5/9, 13/24
  3. Set (c): 6/35, 5/21, 2/7, 4/15
  4. Set (d): 2/5, 6/11, 7/15, 9/20, 13/25
  5. Set (e): 1/5, 2/7, 3/8, 4/13, 5/17

Method: compare two fractions by cross-multiplying across positive denominators; the larger cross-product marks the larger fraction. For example, 9/14 > 25/42 because 9 × 42 = 378 while 25 × 14 = 350.

Answers:

  1. Set (a): 9/14 > 25/42 > 7/12 > 4/7
  2. Set (b): 5/8 > 7/12 > 5/9 > 13/24
  3. Set (c): 2/7 > 4/15 > 5/21 > 6/35
  4. Set (d): 6/11 > 13/25 > 7/15 > 9/20 > 2/5
  5. Set (e): 3/8 > 4/13 > 5/17 > 2/7 > 1/5

A common denominator gives the same orders and is worth using as a check when two fractions are close.

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Q8 Total distance and average speed of a train, the height of a cuboid from its surface area, relationships in a six-member family, and counting primes between square roots

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Question 8 mixes a two-stage speed problem, a surface-area equation, a family-relationship puzzle and prime counting between square roots. The speed part is the one to slow down on, because the two halves of the journey are given in different units.

(a) A train in first five successive minutes from its start runs 68 m, 127 m, 208 m, 312 m and 535 m and for next 5 minutes maintains an average speed of 33 km/hr. Find the whole distance covered and the average speed in covering this total distance.

Given: distances of 68 m, 127 m, 208 m, 312 m and 535 m in the first five minutes, then an average speed of 33 km/h for the next five minutes. The fifth figure is the paper's printed value of 535 m.

Method: add the first five distances, convert the second stage to metres using distance = speed × time, then divide total distance by total time.

Working:

68 + 127 + 208 + 312 + 535 = 1,250 m

Time for the second stage = 5/60 hour = 1/12 hour

Distance = 33 × 1/12 = 2.75 km = 2,750 m

Total distance = 1,250 + 2,750 = 4,000 m = 4 km

Total time = 10 minutes = 1/6 hour

Average speed = 4 ÷ (1/6) = 4 × 6 = 24 km/h

Answer: the train covers 4 km in total, at an overall average speed of 24 km/h.

(b) A solid cuboid with base 10 cm by 6 cm is available. The height of the cuboid is 'y' centimetres and the total surface area of the cuboid is given to be 376 cm². Find the height of the cuboid.

Given: length 10 cm, breadth 6 cm, total surface area 376 cm², height y cm.

Method: apply the total surface area formula for a cuboid, 2(lb + lh + bh), substitute the known values and solve the linear equation for y.

Working:

2(lb + lh + bh) = 376

2[(10 × 6) + 10y + 6y] = 376

2(60 + 16y) = 376

120 + 32y = 376

32y = 256

y = 8

Answer: the height of the cuboid is 8 cm.

Check: 2[(10 × 6) + (10 × 8) + (6 × 8)] = 2(60 + 80 + 48) = 376 cm².

(c) In a six-member family (A, B, C, D, E and F), there are two fathers, three brothers and a mother. If C is the sister of F, B is the brother of E's husband, D is the father of A and grandfather of F. Who is E's husband?

Given, as printed:

  • D is the father of A.
  • D is the grandfather of F, so one of D's children is a parent of F.
  • C is the sister of F, so C and F are siblings.
  • B is the brother of E's husband.
  • The family contains two fathers, three males described as brothers, and one mother.

Method: fix the generations first from the grandfather statement, then place the married pair, then attach the siblings. The wording is compressed, so the assumptions are kept visible rather than folded into the answer.

Working: the arrangement that satisfies all statements is:

  • D is the father of A and B, and the grandfather of C and F.
  • A is married to E.
  • A and E are the parents of C and F.
  • B is A's brother.
  • F is C's brother.
Generation 1: D
Generation 2: A and B, with A married to E
Generation 3: C and F

This gives two fathers, D and A; one mother, E; and three males who are brothers within their own sibling groups, A, B and F.

Answer: E's husband is A.

(d) How many prime numbers are between each of the following pairs of numbers?

Method: approximate each square-root boundary to two decimal places, then list the primes strictly between the two values and count them.

Answers:

  1. Pair (a), √3 and √120. √3 is about 1.73 and √120 is about 10.95. Primes: 2, 3, 5, 7. Count = 4.
  2. Pair (b), √10 and √410. √10 is about 3.16 and √410 is about 20.25. Primes: 5, 7, 11, 13, 17, 19. Count = 6.
  3. Pair (c), √10 and √999. √10 is about 3.16 and √999 is about 31.61. Primes: 5, 7, 11, 13, 17, 19, 23, 29, 31. Count = 9.
  4. Pair (d), √28 and √120. √28 is about 5.29 and √120 is about 10.95. Prime: 7. Count = 1.
  5. Pair (e), √8 and √400. √8 is about 2.83 and √400 = 20. Primes: 3, 5, 7, 11, 13, 17, 19. Count = 7.

The boundaries are exclusive, so a prime equal to a boundary would not count; here only √400 is exact, and 20 is not prime in any case.

Working a solved paper is different from reading one. How to actually use CSS past papers sets out the difference.

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Each of the seven questions is solved here, including ones a candidate would not attempt. In the examination hall you follow the paper's own instructions on how many questions to attempt and from which section; this set is a study resource, so none of the seven is left unanswered.

The questions are reproduced from the Federal Public Service Commission's CSS 2025 General Science and Ability paper for study purposes. Every answer was written from scratch for this platform and is offered as one good attempt, not as an official key. Verify any figure against a primary source before reproducing it in an examination.

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