RRB Group D Official Paper (Held On: 09 Feb, 2026 Shift 2) (Previous Year Paper)

The correct answer is Hydrogenation of ethene.

Key Points

• An addition reaction is a type of organic reaction where two or more molecules combine to form a single larger molecule, typically occurring in compounds with multiple bonds (double or triple bonds).
• Hydrogenation of ethene involves the addition of Hydrogen (H2) across the double bond of Ethene (C2H4) to form Ethane (C2H6) in the presence of a catalyst such as Nickel (Ni) or Palladium (Pd). Hence, the statement is correct.
• Chlorination of methane is a classic example of a substitution reaction, where a hydrogen atom in the Methane (CH4) molecule is replaced by a Chlorine (Cl) atom. Hence, the statement is incorrect.
• Dehydration of alcohol (e.g., ethanol to ethene) is an elimination reaction, where a molecule of water is removed from the substrate to form an alkene. Hence, the statement is incorrect.
• Hydrolysis of acetic acid (or reaction with water) generally refers to its dissociation into ions or its formation from an ester, but it does not represent a standard addition reaction across a carbon-carbon double bond. Hence, the statement is incorrect.

Additional Information

• Unsaturated Hydrocarbons: Compounds like alkenes and alkynes are classified as unsaturated because they contain double or triple bonds. These bonds serve as active sites for addition reactions.
• Saturated Hydrocarbons: Alkanes like methane and propane are saturated because they contain only single bonds. They typically undergo substitution reactions because there is no space to "add" more atoms without removing existing ones.
• Industrial Application: The hydrogenation process is widely used in the food industry to convert liquid vegetable oils (which are unsaturated) into solid fats (which are saturated), such as Vanaspati ghee.
• Role of Catalysts: Transition metals like Nickel, Platinum (Pt), and Palladium are essential in hydrogenation as they provide a surface for hydrogen gas to adsorb and react with the hydrocarbon, lowering the activation energy of the reaction.

Important Points

• Reaction Mechanism: The addition reaction specifically converts a pi bond into sigma bonds. In the case of ethene, the pi bond between the two carbon atoms breaks to accommodate two new Hydrogen atoms.
• Chlorination Conditions: For substitution to occur in methane, the presence of sunlight or ultraviolet (UV) light is usually required to break the Chlorine-Chlorine bond into free radicals.
• Health Context: While hydrogenation is useful for food stability, the process can sometimes lead to the formation of trans-fats, which are associated with health risks such as heart disease.
• Bromine Water Test: An addition reaction is often used to test for unsaturation. When bromine water is added to an alkene, the reddish-brown color disappears (decolorizes) as the bromine adds to the double bond.

The correct answer is Reddish colour of sun at sunrise / sunset.

Key Points

• Atmospheric Refraction is the phenomenon of bending of light as it passes through the layers of the earth's atmosphere which have different optical densities.
• Sun appearing higher in the sky: Due to atmospheric refraction, when the sun is near the horizon, the rays of light coming from it have to pass through layers of air with increasing refractive index. This causes the light to bend towards the normal, making the sun appear at a higher position than its actual geometric position. Hence, statement 1 is caused by atmospheric refraction.
• Reddish colour of the sun at sunrise/sunset: This phenomenon occurs due to the Scattering of Light (specifically Rayleigh Scattering). During sunrise and sunset, sunlight travels a longer distance through the atmosphere. The shorter wavelengths (blue and violet) are scattered away by the air molecules, while the longer wavelengths (red and orange) reach our eyes. Hence, statement 2 is NOT caused by atmospheric refraction.
• Twinkling of stars: Stars act as point sources of light located very far away. As the light enters the atmosphere, it undergoes continuous refraction due to changing temperature and density of air layers. This causes the apparent position and brightness of the star to fluctuate rapidly, creating the twinkling effect. Hence, statement 3 is caused by atmospheric refraction.
• Advanced sunrise and delayed sunset: The sun is visible to us about 2 minutes before the actual sunrise and 2 minutes after the actual sunset because of the bending of light. When the sun is slightly below the horizon, the atmosphere refracts the light downward toward the earth, allowing us to see the sun. Hence, statement 4 is caused by atmospheric refraction.

Additional Information

• Scattering of Light: This is the process where light rays are deviated from their straight path upon striking obstacles like dust particles, water droplets, or gas molecules in the atmosphere.
• Tyndall Effect: This is the scattering of light by colloidal particles. It explains why the path of a light beam becomes visible when passing through smoke or fog.
• Blue Color of Sky: During the day, the fine particles in the atmosphere scatter the blue light (shorter wavelength) more strongly than red light, which is why the sky appears blue to an observer on Earth.
• Refractive Index: It is a measure of how much the speed of light is reduced inside a medium. In the atmosphere, the refractive index increases as we move from the rarer upper layers to the denser lower layers near the Earth's surface.

Important Points

• Speed of Light: Light travels at approximately 3 × 10&sup8; m/s in a vacuum, but its speed decreases when it enters the Earth's atmosphere, causing it to refract.
• Mirage: Another famous example of atmospheric refraction (combined with Total Internal Reflection) is a mirage, commonly seen in deserts or on hot tar roads during summer.
• Visible Spectrum: The human eye can perceive wavelengths between approximately 400 nm (Violet) and 700 nm (Red). Red light has the longest wavelength and is scattered the least, which is why it is used for danger signals.

The correct answer is - 120 cm.

Key Points

• According to the New Cartesian Sign Convention, the focal length (f) of a concave mirror is always considered negative. In this problem, the distance between the pole and focus is given as 30 cm, so f = -30 cm.
• A real and inverted image is formed on a screen, which means the image is formed in front of the mirror. Therefore, the image distance (v) is also negative. Given the image is at 40 cm, we have v = -40 cm.
• The Mirror Formula relates the focal length, image distance, and object distance: 1/f = 1/v + 1/u. Hence, the formula application is essential for this calculation.
• Substituting the known values into the formula: 1/(-30) = 1/(-40) + 1/u. Rearranging for the object distance (u), we get 1/u = -1/30 - (-1/40), which simplifies to 1/u = -1/30 + 1/40.
• Finding the common denominator (120): 1/u = (-4 + 3) / 120 = -1 / 120. Therefore, u = -120 cm. The negative sign indicates that the object is placed 120 cm in front of the mirror. Hence, the statement regarding the object distance being -120 cm is correct.

Additional Information

• Concave Mirror: A concave mirror, or converging mirror, has a reflecting surface that is recessed inward. It reflects light inward to one focal point. These mirrors are used to focus light and are common in telescopes, shaving mirrors, and headlights.
• Nature of Images:
• If the object is placed beyond the Center of Curvature (C), the image is real, inverted, and diminished.
• If the object is between C and Focus (F), the image is real, inverted, and magnified.
• If the object is placed between the Pole (P) and F, the image is virtual, erect, and magnified.
• Cartesian Sign Convention:
• The object is always placed to the left of the mirror, making u always negative.
• Distances measured in the direction of incident light are positive, and those opposite are negative.
• The height measured upwards from the principal axis is positive, while downwards is negative.

Important Points

• Magnification (m): It is the ratio of the height of the image to the height of the object. It is also given by the formula m = -v/u. For real images, m is negative, and for virtual images, m is positive.
• Radius of Curvature (R): The distance from the pole to the center of curvature. There is a specific relationship where R = 2f. In this problem, the radius of curvature would be 60 cm.
• Real vs Virtual Image: A real image can be obtained on a screen because the light rays actually intersect. A virtual image cannot be obtained on a screen as the light rays only appear to diverge from a point behind the mirror.
• The unit of measurement must remain consistent throughout the calculation. Since the focal length and image distance were in cm, the result for the object distance is also in cm.

The correct answer is Gas.

Key Points

• In the gaseous state, the intermolecular space between the particles is maximum. This is because the intermolecular forces of attraction between the particles are very weak or almost negligible compared to solids and liquids.
• Particles in a gas possess very high kinetic energy, which causes them to move randomly at high speeds in all directions. This rapid motion constantly pushes the particles away from each other, resulting in large empty spaces.
• Due to these maximum spaces, gases do not have a definite shape or a definite volume; they expand to fill whatever container they are placed in.
• The large distance between particles also makes gases highly compressible. For example, LPG (Liquefied Petroleum Gas) and CNG (Compressed Natural Gas) are stored by compressing large volumes of gas into small cylinders. Hence, option 2 is correct.
• In contrast, the space between particles is minimum in solids because of very strong cohesive forces that hold the particles in fixed positions. Hence, option 4 is incorrect.
• In liquids, the space between particles is intermediate—more than solids but significantly less than gases. Hence, option 1 is incorrect.

Additional Information

• Solid State:
• Particles are arranged in a closely packed manner, often in a regular crystal lattice.
• They have a definite shape, distinct boundaries, and fixed volumes.
• The diffusion rate in solids is extremely low due to the lack of space for particle movement.
• Liquid State:
• Particles have more freedom of movement than in solids, allowing liquids to flow and change shape.
• They have a fixed volume but take the shape of the container they occupy.
• The intermolecular force is strong enough to keep the particles together but not strong enough to keep them in a fixed position.
• Kinetic Theory of Matter:
• This theory states that all matter is made up of tiny particles that are in constant motion.
• As the temperature increases, the kinetic energy of the particles increases, which can lead to a change in the state of matter (e.g., melting or evaporation) by increasing the space between particles.
• Other States of Matter:
• Plasma: Often considered the fourth state of matter, it consists of super energetic and super excited particles in the form of ionized gases. It is found in stars and fluorescent tubes.
• Bose-Einstein Condensate (BEC): The fifth state of matter, formed by cooling a gas of extremely low density to super-low temperatures (near Absolute Zero).

Important Points

• Density: Usually, the density of a substance is maximum in its solid state and minimum in its gaseous state because density is inversely proportional to the volume occupied by the particles.
• Diffusion: The rate of diffusion is highest in gases because the particles move rapidly and there is a lot of space between them to mix with other particles.
• Brownian Motion: This refers to the zig-zag motion of particles, which is most observable in gases and liquids due to the available space for movement.

The correct answer is Each fragment possessing the capability to perform basic life functions.

Key Points

• Spirogyra is a genus of filamentous green algae belonging to the order Zygnematales. It is widely known for its characteristic spiral or helical arrangement of chloroplasts.
• In Spirogyra, asexual reproduction occurs primarily through a process known as fragmentation. In this process, the vegetative body (filament) simply breaks into smaller pieces or fragments upon reaching maturity.
• The fundamental condition for successful fragmentation is that each fragment must contain the necessary cellular machinery to sustain itself. Because Spirogyra is a simple multicellular organism, each cell within the filament is relatively autonomous and capable of performing metabolism, growth, and cell division.
• Once a fragment is separated, it undergoes repeated mitotic cell divisions to develop into a full-sized, new filament. This allows the organism to multiply rapidly in favorable environmental conditions.
• Unlike higher plants, Spirogyra does not require specialized reproductive organs like flowers or seeds for this type of reproduction. As a member of the Thallophyta group, it also lacks a vascular system (xylem and phloem).

Additional Information

• Morphology and Structure: Spirogyra is often called "water silk" or "pond scum" due to its bright green, slimy appearance. The slipperiness is due to a thick mucilage coating on the outer cell wall, which protects the filament from epiphytes and pathogens.
• Habitat: It is typically found in freshwater environments such as ponds, lakes, and slow-moving streams. It thrives in clear, stagnant water where sunlight is abundant for photosynthesis.
• Sexual Reproduction: While fragmentation is the common asexual method, Spirogyra also reproduces sexually through a process called conjugation. This involves the fusion of two non-flagellated gametes (isogamy), resulting in the formation of a thick-walled zygospore that can survive harsh conditions.
• Chloroplasts and Pyrenoids: The chloroplasts are ribbon-shaped and contain small round bodies called pyrenoids. These pyrenoids are responsible for the storage of starch and the synthesis of proteins.
• Cell Wall Composition: The cell wall is composed of two distinct layers: an inner layer made of cellulose and an outer layer made of pectose. The pectose layer eventually turns into pectin and dissolves in water to form the slimy sheath.

Important Points

• Classification: It belongs to the Kingdom Plantae and the division Thallophyta. The name is derived from the Greek words "spira" (spiral) and "gyros" (circle).
• Regeneration Capability: The success of fragmentation is a testament to the high regenerative capacity of simple multicellular organisms. As long as a fragment contains at least one healthy cell with a nucleus and cytoplasm, it can regenerate an entire organism.
• Ecological Role: As a primary producer, Spirogyra plays a vital role in aquatic ecosystems by producing oxygen through photosynthesis and serving as a food source for various aquatic insects and small fish.

The correct answer is Tracheids and vessels.

Key Points

• Xylem is a complex permanent tissue in vascular plants primarily responsible for the conduction of water and dissolved minerals from the roots to the rest of the plant.
• Tracheids and vessels are the two main types of conducting elements in the xylem. They are both dead cells at maturity and lack protoplasm, creating a hollow path for efficient liquid movement. Hence, option 3 is correct.
• Tracheids are elongated, tube-like cells with thick, lignified walls and tapering ends. Water moves between tracheids through pits in their cell walls. They are found in all vascular plants, including Pteridophytes and Gymnosperms.
• Vessels are long, cylindrical structures that result from the fusion of several cells called vessel members. These members are interconnected through perforation plates in their common walls, allowing for a more streamlined and high-volume vertical transport compared to tracheids.
• While Tracheids provide a degree of mechanical support, their primary function, alongside Vessels, is the ascent of sap. Most Angiosperms (flowering plants) possess both, whereas vessels are typically absent in most gymnosperms.

Additional Information

• Xylem Fibres: These cells have highly thickened walls and obliterated central lumens. Their primary function is to provide mechanical strength and structural support to the plant body, rather than active transport. Hence, option 1 is incorrect.
• Xylem Parenchyma: These are the only living cells in the xylem tissue. They are thin-walled and primarily involved in the storage of food materials like starch and fats, and the radial conduction of water.
• Sieve Tubes: These are the main conducting elements of the Phloem. They are long, tube-like structures arranged longitudinally, responsible for the transport of organic nutrients (food) from leaves to other parts. Hence, option 4 is incorrect.
• Companion Cells: These are specialized parenchymatous cells associated with sieve tube elements in the phloem. They help maintain the pressure gradient in the sieve tubes for food transport. Hence, option 2 is incorrect.

Important Points

• Lignification: The cell walls of tracheids and vessels are heavily impregnated with lignin, a complex polymer that provides rigidity and prevents the cells from collapsing under the negative pressure (tension) created by transpiration.
• Unidirectional Flow: Unlike the phloem, where transport can be multidirectional, the transport of water and minerals in the xylem is strictly unidirectional, moving from roots to shoots.
• Ascent of Sap: The movement of water through tracheids and vessels is driven by physical forces such as transpiration pull, capillary action, and root pressure. Transpiration pull is the primary driver in tall trees, pulling water upward through the continuous xylem columns.

The correct answer is Copper.

Key Points

• The heating effect of electric current, also known as Joule heating, is governed by the formula H = I^2Rt, where H is the heat produced, I is the current, R is the resistance, and t is the time for which the current flows.
• According to this law, the amount of heat generated in a conductor is directly proportional to its electrical resistance. Therefore, a material with lower resistance will produce less heat when the same amount of current passes through it.
• Copper is known for having very low electrical resistivity (approximately 1.68 x 10^-8 Ohm-m at 20 degrees Celsius), which makes it an excellent conductor of electricity. Because its resistance is minimal, the heating effect is least in copper compared to the other options provided. Hence, statement 1 is correct.
• Iron has a significantly higher resistivity (about 1.0 x 10^-7 Ohm-m) than copper. This means an iron wire will offer more resistance to the flow of electrons and consequently generate more heat than a copper wire of the same dimensions. Hence, statement 2 is incorrect.
• Nichrome is an alloy of Nickel and Chromium designed specifically to have high resistance and a high melting point. It is used in heating elements like toasters and room heaters because it produces a large amount of heat. Hence, statement 3 is incorrect.
• Tungsten has a resistivity of approximately 5.6 x 10^-8 Ohm-m. While it is a good conductor, its resistance is still higher than that of copper, and it is famously used in incandescent bulb filaments because it can withstand high temperatures while producing light and heat. Hence, statement 4 is incorrect.

Additional Information

• Electrical Resistivity: This is a fundamental property of a material that quantifies how strongly it resists or conducts electric current. It is denoted by the Greek letter rho (ρ). Metals like Silver and Copper have the lowest resistivity, making them the best conductors.
• Applications of Copper: Due to its high conductivity and low heat dissipation, copper is the primary material used in domestic electrical wiring, power cables, and printed circuit boards (PCBs). Using copper ensures that energy loss in the form of heat is minimized during transmission.
• Heating Elements (Nichrome): In appliances where heat is the desired output, materials with high resistivity are preferred. Nichrome is ideal for this because it does not oxidize easily at high temperatures, allowing it to remain functional even when glowing red hot in a heater.
• Tungsten in Bulbs: Tungsten is chosen for light bulb filaments not because it has the least heating effect, but because of its extremely high melting point (about 3422 degrees Celsius). This allows the filament to reach temperatures high enough to emit visible light without melting.
• Factors Influencing Resistance: The resistance of a conductor depends on its length, area of cross-section, the nature of the material, and temperature. For most metals, resistance increases as the temperature rises.

Important Points

• Silver is actually the best conductor of electricity, with even lower resistivity than copper. however, Copper is more widely used in electrical applications because it is much more cost-effective and abundant.
• The Joule effect is utilized in safety devices like Electric Fuses. A fuse wire is made of an alloy with a low melting point; when the current exceeds a safe limit, the heat produced melts the wire and breaks the circuit to prevent fire or damage.
• Power Loss: In large-scale power distribution, the heat generated in wires (I^2R loss) is a major concern. To reduce this, electricity is transmitted at high voltages and low currents over long distances using transformers.

The correct answer is 250 m.

Key Points

• To find the total distance, we first need to determine the acceleration of the race car using the formula for distance covered in the nth second of motion.
• The formula for the distance covered in the nth second is given by: Sn = u + a/2 (2n - 1), where u is the initial velocity, a is the uniform acceleration, and n is the time in seconds.
• Given that the car starts from rest, the initial velocity (u) is 0 m/s. The car covers 22.5 m in the 5th second (n = 5). Substituting these values: 22.5 = 0 + a/2 (2 * 5 - 1).
• Solving for acceleration: 22.5 = a/2 (9), which simplifies to 22.5 = 4.5a. Therefore, the acceleration (a) is 5 m/s².
• Now, to find the total distance covered at the end of 10 seconds, we use the second equation of motion: S = ut + 1/2 at².
• Substituting the known values (u = 0, a = 5, t = 10): S = (0 * 10) + 1/2 * 5 * (10)². This results in S = 1/2 * 5 * 100, which equals 250 m. Hence, the correct answer is 250 m.

Additional Information

• Equations of Motion:
• These equations describe the behavior of a physical system in terms of its motion as a function of time. They are valid only when the acceleration is constant (uniform).
• The three primary equations are: v = u + at, S = ut + 1/2 at², and v² = u² + 2aS.
• Uniform Acceleration:
• This occurs when the velocity of an object changes by an equal amount in every equal time interval. In this problem, the race car's acceleration remains 5 m/s² throughout the 10 seconds.
• Initial Velocity (u):
• In physics problems, the phrase "starting from rest" is a key indicator that the initial velocity is 0. If an object comes to a stop, the final velocity (v) becomes 0.
• Displacement vs. Distance:
• While often used interchangeably in straight-line motion without a change in direction, displacement is a vector quantity (shortest path), whereas distance is a scalar quantity (total path covered).

Important Points

• Nth second calculation: It is important to distinguish between the distance covered "in 10 seconds" and the distance covered "in the 10th second". The former is the cumulative sum from 0 to 10, while the latter is only the movement between second 9 and second 10.
• Units: Always ensure that the units are consistent. In this problem, distance is in meters (m) and time is in seconds (s), so the resulting acceleration is in meters per second squared (m/s²).
• Linear Relationship: For an object starting from rest with uniform acceleration, the total distance covered is directly proportional to the square of the time (S ∝ t²).

The correct answer is Iron (III) oxide.

Key Points

• A thermite reaction is a highly exothermic chemical reaction between a metal powder and a metal oxide. The most common and typical version involves Aluminium powder and Iron (III) oxide (Fe2O3).
• In this process, Aluminium acts as a strong reducing agent because it has a higher affinity for oxygen than iron. It reduces the Iron (III) oxide to produce molten iron and Aluminium oxide (Al2O3).
• The reaction is represented by the chemical equation: Fe2O3 + 2Al → Al2O3 + 2Fe + Heat. The heat generated is so intense that the iron produced is in a liquid (molten) state.
• Because of the production of molten iron, this reaction is extensively used for welding broken railway tracks and heavy machinery parts on-site where traditional welding is difficult.
• While other metal oxides like Copper oxide or Chromium oxide can technically undergo similar reactions with aluminium, Iron (III) oxide is the standard "thermite" component used in industrial and civil engineering applications. Hence, the correct answer is Iron (III) oxide.

Additional Information

• History and Discovery: The thermite process was discovered and patented by the German chemist Hans Goldschmidt in the year 1893. It is often referred to as the Goldschmidt reaction or the aluminothermic process.
• Temperature: The reaction can reach temperatures of approximately 2500°C (4500°F), which is sufficient to melt almost any metal.
• Ignition: Thermite is stable at room temperature and requires a very high activation energy to start. Typically, a Magnesium ribbon or a mixture of Potassium Permanganate and Glycerin is used as a fuse to provide the initial heat required for ignition.
• Applications:
• Railway Industry: Used for Thermite Welding (or Exothermic Welding) to join rails together, ensuring a smooth surface for trains.
• Military: Used in incendiary weapons and grenades to destroy enemy equipment, as the molten iron can melt through thick armor or engine blocks.
• Metallurgy: Employed in the reduction of ores for metals that are difficult to extract using carbon, such as Chromium and Manganese.

Important Points

• Redox Nature: The thermite reaction is a classic example of a Redox (Reduction-Oxidation) reaction and a displacement reaction, where a more reactive metal (Aluminium) displaces a less reactive metal (Iron) from its oxide.
• Reactivity Series: The success of the reaction depends on Aluminium being positioned higher than Iron in the reactivity series of metals.
• Safety: Due to the extreme temperatures and the fact that it is nearly impossible to extinguish once started (as it carries its own supply of oxygen within the metal oxide), it poses significant fire hazards and requires specialized safety protocols.
• States of Matter: The reactants are usually in powdered form to increase the surface area, ensuring a faster and more consistent reaction rate upon ignition.

The correct answer is Sulphide.

Key Points

• Pyrite, chemically known as iron disulfide with the formula FeS2, is the most common mineral in the sulfide group. Hence, Statement 2 is correct.
• It is widely distributed across various geological environments, including sedimentary, metamorphic, and igneous rocks.
• The mineral is characterized by its metallic luster and a pale brass-yellow color, which often leads to it being mistaken for gold.
• Due to this resemblance and its deceptive appearance, it is famously nicknamed Fool's Gold.
• The name is derived from the Greek word "pyrites lithos", which means "stone which strikes fire", referring to the sparks it produces when hit against steel or iron.
• Pyrite typically crystallizes in the isometric (cubic) system, frequently forming distinct cubic crystals with striated faces.

Additional Information

• Nitrate Ores:
• These are minerals where nitrogen and oxygen form the NO3 ion. Because they are highly water-soluble, they are rarely found in large deposits except in arid regions.
• An example is Chile Saltpeter (Sodium Nitrate, NaNO3), found in the Atacama Desert. Pyrite does not contain nitrate groups. Hence, Statement 1 is incorrect.
• Halide Ores:
• These minerals are defined by the presence of halogen elements such as fluorine, chlorine, bromine, or iodine as the primary anions.
• Examples include Halite (Rock Salt, NaCl), Fluorite (CaF2), and Cryolite (Na3AlF6). Pyrite is not a halide mineral. Hence, Statement 3 is incorrect.
• Oxide Ores:
• In these ores, oxygen is the main anion combined with one or more metals.
• Common examples include Hematite (Fe2O3), Magnetite (Fe3O4), and Bauxite (Al2O3.nH2O). Although Pyrite contains iron, its primary anion is sulfur, not oxygen. Hence, Statement 4 is incorrect.

Important Points

• Industrial Uses: Despite being an iron mineral, pyrite is not used as a primary source of iron for steel production because the sulfur content is difficult and expensive to remove. It is primarily used for the commercial production of sulfuric acid (H2SO4).
• Physical Properties: Pyrite has a Mohs hardness of 6 to 6.5, whereas real gold is much softer with a hardness of 2.5 to 3. Its streak is greenish-black to brownish-black, unlike gold's golden-yellow streak.
• Environmental Impact: When pyrite is exposed to air and water during mining, it reacts to form sulfuric acid and ferric hydroxide, leading to a phenomenon known as Acid Mine Drainage (AMD), which can severely pollute local waterways.
• Detection: In mineralogy, pyrite is often used as a pathfinder mineral in the exploration of gold, copper, and zinc deposits.

p data-end="57" data-start="0">The correct answer is 5 m/s².

 

Key Points

• Acceleration is defined as the rate of change of velocity of an object with respect to time.
• To find the acceleration in this problem, we use the standard formula: a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken.
• Based on the question, the Initial Velocity (u) is 20 m/s and the Final Velocity (v) is 40 m/s.
• The Time interval (t) given for this change is 4 seconds.
• Substituting the values into the formula: a = (40 - 20) / 4. This results in 20 / 4, which equals 5.
• The unit of velocity is m/s and time is s, therefore the SI unit for acceleration is m/s² (meters per second squared). Hence, the correct answer is 5 m/s².

Additional Information

• Concept of Acceleration:
• Acceleration is a vector quantity, meaning it has both magnitude and direction.
• If the velocity of an object increases, the acceleration is positive and acts in the direction of motion.
• If the velocity of an object decreases, it is known as deceleration or retardation, and the value is negative.
• In uniform circular motion, even if the speed is constant, the object is accelerating because its direction is constantly changing.
• Velocity and Speed:
• Velocity is the displacement of an object per unit time. It is different from speed because speed is a scalar quantity and does not account for direction.
• The average velocity can be calculated by dividing the total displacement by the total time taken.
• Newton's Second Law of Motion:
• The force (F) acting on an object is equal to the mass (m) of that object times its acceleration (a). The formula is written as F = ma.
• This law highlights that acceleration is directly proportional to the net force and inversely proportional to the mass.

Important Points

• Units of Measurement:
• It is crucial to distinguish between m/s (unit of velocity/speed) and m/s² (unit of acceleration).
• Option 3 mentions 5 m/s, which is a velocity value, not an acceleration value. Always check the numerical value along with its unit.
• Equations of Motion:
• The formula used (v = u + at) is the first equation of motion for objects moving with uniform acceleration.
• Other equations include s = ut + 1/2 at² (for distance) and v² = u² + 2as (relating velocity, acceleration, and distance).
• These equations were formulated to describe the kinematics of particles moving in a straight line.

The correct answer is Two or more reactants combine to form a single product.

Key Points

• A combination reaction, which is also frequently referred to as a synthesis reaction, is a fundamental chemical process where two or more simple substances (either elements or compounds) react together to generate a single, more complex product. Hence, statement 1 is correct.
• The general symbolic representation for a combination reaction is A + B → AB, where the reactants A and B merge their atomic or molecular structures to form the new substance AB.
• A well-known example of this reaction is the burning of coal in the presence of oxygen. When carbon (C) reacts with oxygen (O2), it produces carbon dioxide (CO2) as the sole product.
• Another classic demonstration involves magnesium ribbon. When magnesium (Mg) is heated in air, it combines with oxygen to form a white powder known as magnesium oxide (MgO).
• The reaction between quicklime (calcium oxide, CaO) and water (H2O) to form slaked lime (calcium hydroxide, Ca(OH)2) is a vigorous combination reaction that releases a significant amount of thermal energy.

Additional Information

• Exothermic Nature: A significant number of combination reactions are exothermic in nature, meaning they involve the evolution of heat. This happens because the energy released during the formation of new bonds in the product is greater than the energy required to break the bonds in the reactants.
• Displacement Reaction: This occurs when a more reactive element removes or displaces a less reactive element from its compound. For example, when iron (Fe) is placed in a copper sulfate (CuSO4) solution, it displaces copper. Hence, statement 2 describes a displacement reaction and is incorrect for this context.
• Double Displacement Reaction: This type of reaction involves an exchange of ions between two different compounds to form two new compounds. These often result in the formation of an insoluble solid called a precipitate. Hence, statement 3 describes a double displacement reaction and is incorrect for this context.
• Decomposition Reaction: This is essentially the reverse of a combination reaction. In a decomposition reaction, a single reactant breaks down into two or more simpler substances, usually requiring energy in the form of heat, light, or electricity. Hence, statement 4 relates to a decomposition process and is incorrect for this context.

Important Points

• Classification of Combination Reactions:
• Between two elements: For instance, hydrogen (H2) gas combining with chlorine (Cl2) gas to form hydrogen chloride (HCl).
• Between two compounds: For example, the reaction between ammonia (NH3) and hydrogen chloride (HCl) to form ammonium chloride (NH4Cl).
• Between an element and a compound: Such as carbon monoxide (CO) reacting with oxygen (O2) to produce carbon dioxide (CO2).
• Industrial Significance: Combination reactions are the backbone of many industrial processes, such as the Haber Process for the synthesis of ammonia (NH3) from nitrogen and hydrogen, which is vital for fertilizer production.
• Bond Formation: In these reactions, the chemical potential energy of the system typically decreases as the atoms achieve a more stable configuration in the single product formed.

The correct answer is Environmental harm and global criticism.

Key Points

• The continued use of Chlorofluorocarbons (CFCs) leads to the severe depletion of the Ozone Layer located in the stratosphere. This layer acts as a shield, absorbing the majority of the Sun's harmful ultraviolet (UV) radiation. Hence, the statement regarding environmental harm is correct.
• The Montreal Protocol, an international treaty signed in September 1987, mandates the phase-out of Ozone-Depleting Substances (ODS). Any country that continues to use CFCs violates this global consensus, leading to diplomatic isolation and global criticism. Hence, the statement regarding global criticism is correct.
• Exposure to increased UV-B radiation due to ozone thinning is linked to higher incidences of skin cancer, cataracts, and weakened immune systems in humans.
• CFCs are also highly potent greenhouse gases. Their continued emission would accelerate global warming, as they have a Global Warming Potential (GWP) thousands of times higher than carbon dioxide (CO2).
• Non-compliance with the Montreal Protocol can result in trade sanctions and the loss of financial assistance from the Multilateral Fund, which helps developing nations transition to ozone-friendly technologies.

Additional Information

• Chemical Composition of CFCs:
• CFCs are organic compounds that contain carbon, chlorine, and fluorine. They were widely used as refrigerants, propellants in aerosol cans, and solvents.
• Because they are non-flammable and non-toxic at the ground level, they were considered "miracle chemicals" until their impact on the stratosphere was discovered.
• The Ozone Hole:
• The most significant depletion of ozone occurs over Antarctica during the spring, leading to what is commonly known as the Ozone Hole.
• In 1985, scientists from the British Antarctic Survey first reported the dramatic thinning of the ozone layer, which spurred the creation of the Vienna Convention.
• Mechanism of Destruction:
• When CFCs reach the stratosphere, UV radiation breaks them down, releasing chlorine atoms.
• A single chlorine atom can destroy over 100,000 ozone molecules (O3) before it is removed from the atmosphere.
• Alternatives to CFCs:
• Initial replacements were Hydrochlorofluorocarbons (HCFCs), which also deplete ozone but to a lesser extent.
• The current industry standard involves Hydrofluorocarbons (HFCs); however, while they do not deplete ozone, they are strong greenhouse gases. This led to the Kigali Amendment in 2016 to phase down HFCs.

Important Points

• Atmospheric Lifetime: CFCs are extremely stable and can remain in the atmosphere for 50 to 100 years, meaning the damage caused today will persist for several generations.
• Ecological Impact: Increased UV radiation harms phytoplankton, the foundation of the marine food chain, and reduces the yield of sensitive crops like rice, wheat, and soybeans.
• Vienna Convention (1985): This was the first international framework for the protection of the ozone layer, which later led to the legally binding Montreal Protocol.
• World Ozone Day: Celebrated annually on September 16 to commemorate the date of the signing of the Montreal Protocol and to spread awareness about ozone preservation.

The correct answer is A force must act on an object, and the object must be displaced.

Key Points

• In the field of Physics, the term Work describes a specific process where energy is transferred. For work to be scientifically recognized as "done," two essential conditions must be simultaneously satisfied: an external Force must be applied, and the object must undergo Displacement in the direction of the force.
• A force must act on an object to initiate or change its state of motion. If there is no force, no scientific work can be attributed to an agent. Hence, statement 1 is incorrect.
• The object must move through a certain distance, known as displacement. If an individual exerts a massive amount of force against a stationary wall and the wall does not move, the displacement is zero, and thus no work is done. Hence, statement 3 and statement 4 are incorrect.
• The correct scientific condition requires both the presence of an acting force and a resulting displacement. Hence, statement 2 is correct.
• Furthermore, for work to be non-zero, the force must have a component that is not perpendicular to the direction of motion. If the force is applied at 90 degrees to the displacement, the work done is zero.

Additional Information

• Mathematical Formula: The amount of work done is calculated using the formula W = Fs cos θ, where W is work, F is the magnitude of the force, s is the displacement, and θ is the angle between the force and the displacement vectors.
• SI Unit: The standard unit of work is the Joule (J). One Joule is defined as the work done when a force of one Newton (N) displaces an object by one meter (m).
• Nature of Quantity: Work is a scalar quantity. While both force and displacement are vector quantities, their dot product (work) only has magnitude and no direction.
• Types of Work:
• Positive Work: When the force and displacement are in the same direction (angle 0°), such as an object falling under gravity.
• Negative Work: When the force acts opposite to the direction of motion (angle 180°), such as the force of friction acting on a moving vehicle.
• Zero Work: When there is no displacement, or when the force is perpendicular to the displacement (angle 90°), such as a satellite orbiting Earth in a circular path.

Important Points

• Work-Energy Theorem: This principle states that the net work done on an object is equal to the change in its Kinetic Energy. This establishes a direct link between motion and the work performed by forces.
• CGS Unit: In the Centimetre-Gram-Second system, the unit of work is the Erg. One Joule is equivalent to 10^7 Ergs.
• Historical Context: The concept of work was developed by mathematicians and physicists like Gaspard-Gustave de Coriolis in the 1820s to describe the "weight lifted through a height."
• Examples in Daily Life: Pulling a trolley, lifting a book from the floor, or stretching a spring are all examples where work is done because a force results in displacement. However, standing still while holding a heavy suitcase is not considered work in physics because the displacement is zero.

The correct answer is It facilitates the exchange of nutrients and wastes between mother and embryo.

Key Points

• The placenta is a specialized, temporary organ that develops in the uterus of a human female during pregnancy. It acts as a structural and functional unit between the developing embryo (fetus) and the maternal body.
• Its primary function is to facilitate the exchange of essential substances; it provides oxygen and nutrients such as glucose and amino acids from the mother to the embryo. Hence, statement 1 is correct.
• The placenta is also responsible for the removal of carbon dioxide and excretory waste materials produced by the embryo, ensuring a healthy internal environment for growth.
• Structurally, it contains villi on the embryonic side, which are surrounded by blood spaces on the maternal side. This configuration provides a massive surface area for the diffusion of nutrients and waste.
• The placenta is connected to the embryo through the umbilical cord, which helps in the transport of substances to and from the embryo.

Additional Information

• Fertilisation Site:
• Fertilisation, the fusion of egg and sperm, occurs in the Fallopian tube (also known as the oviduct), specifically in the ampullary region. The placenta does not play a role in the fertilisation process itself. Hence, statement 2 is incorrect.
• Uterus Structure:
• The uterus is the muscular organ where the embryo is implanted and developed for 9 months. While the uterus receives the two oviducts, the placenta is an organ attached to the uterine wall and not the structure that unites the oviducts. Hence, statement 3 is incorrect.
• Production of Gametes:
• The ovaries are the primary reproductive organs responsible for producing eggs (ova) and hormones like estrogen and progesterone. The placenta does not produce eggs. Hence, statement 4 is incorrect.
• Endocrine Function:
• The placenta also functions as an endocrine tissue and produces several hormones like Human Chorionic Gonadotropin (hCG), Human Placental Lactogen (hPL), estrogens, and progestogens, which are essential for maintaining the pregnancy.

Important Points

• The gestation period in humans is approximately 280 days or 40 weeks.
• During the first trimester, the placenta becomes the main source of progesterone, taking over the role from the corpus luteum in the ovary.
• The umbilical cord typically contains three blood vessels: two umbilical arteries and one umbilical vein.
• Antibodies (specifically IgG) can cross the placenta from the mother to the fetus, providing passive immunity to the newborn.

The correct answer is 20 m/s.

Key Points

• To find the initial velocity of the object, we use the equations of motion under uniform acceleration. In this case, the acceleration is the acceleration due to gravity (g).
• Given data from the problem:
• The maximum height (s) reached by the object is 20 m.
• The final velocity (v) at the maximum height is 0 m/s (since the object momentarily stops before falling back).
• The acceleration (a) is equal to -g because the object is moving upwards against gravity. Given g = 10 m/s², so a = -10 m/s².
• We use the Third Equation of Motion: v² = u² + 2as.
• Substituting the values: 0² = u² + 2(-10)(20).
• 0 = u² - 400.
• u² = 400.
• u = √400 = 20 m/s.
• Therefore, the initial velocity (u) with which the object was thrown is 20 m/s. Hence, the calculation confirms option 4 is correct.

Additional Information

• Equations of Motion:
• These equations describe the relationship between velocity, displacement, acceleration, and time for an object moving with constant acceleration.
• First Equation: v = u + at (Relates velocity and time).
• Second Equation: s = ut + ½at² (Relates displacement and time).
• Third Equation: v² = u² + 2as (Relates velocity and displacement).
• Acceleration due to Gravity:
• The standard value of gravity on Earth is approximately 9.8 m/s², but for competitive exams and ease of calculation, it is often taken as 10 m/s².
• It is a vector quantity directed towards the center of the Earth.
• Sign Convention in Kinematics:
• When an object is thrown vertically upwards, the direction of motion is positive, but gravity acts downwards, making acceleration negative.
• When an object falls downwards, both the displacement and acceleration can be considered in the same direction (usually taken as positive or negative depending on the chosen frame of reference).

Important Points

• Time of Ascent: The time taken to reach the maximum height can be calculated using the formula t = u / g. For this problem, t = 20 / 10 = 2 seconds.
• Maximum Height Formula: A shortcut formula derived from the third equation is H = u² / 2g.
• By plugging in the values: 20 = u² / (2 × 10) leads to u² = 400, which gives u = 20 m/s.
• Symmetry of Motion: In the absence of air resistance, the time of ascent is equal to the time of descent, and the object will hit the ground with the same speed it was thrown with (20 m/s).

The correct answer is CO2.

Key Points

• Baking soda is chemically known as Sodium Bicarbonate or Sodium Hydrogen Carbonate, represented by the chemical formula NaHCO3.
• When Sodium Bicarbonate is subjected to heating, it undergoes a process called thermal decomposition. This reaction typically becomes significant at temperatures above 50 degrees Celsius (122 degrees Fahrenheit).
• The thermal decomposition reaction of baking soda produces three distinct products: Sodium Carbonate (Na2CO3), Water vapor (H2O), and Carbon Dioxide (CO2).
• The chemical equation for this process is: 2 NaHCO3 + heat → Na2CO3 + H2O + CO2.
• The release of Carbon Dioxide (CO2) gas is the fundamental reason baking soda is used as a leavening agent in the food industry. As the gas is released during baking, it gets trapped in the dough or batter, causing it to expand and rise, resulting in a fluffy and porous texture.

Additional Information

• Physical Properties: Sodium Bicarbonate is a white crystalline solid, though it usually appears as a fine powder. It has a slightly alkaline, bitter taste and is monoclinic in its crystalline form.
• Industrial Production: The Solvay process is the primary industrial method for producing Sodium Bicarbonate. This process, established in the 1860s, involves the reaction of sodium chloride, ammonia, and carbon dioxide in water to precipitate NaHCO3.
• Medical Applications: It is widely used as an antacid to treat acidosis and heartburn. When ingested, it reacts with Hydrochloric acid (HCl) in the stomach to neutralize pH levels, producing NaCl, H2O, and CO2.
• Fire Extinguishers: Baking soda is a common component in dry chemical fire extinguishers (BC rated). When discharged onto a fire, the heat of the flames triggers the release of CO2, which acts to displace oxygen and smother the fire.
• Cleaning and Deodorizing: Due to its mild abrasive nature and ability to neutralize odors, it is used extensively in household cleaning, laundry, and even in personal care products like toothpaste and deodorants.

Important Points

• pH and Nature: Sodium Bicarbonate is amphoteric, meaning it can react with both acids and bases. However, in an aqueous solution, it is mildly basic with a pH of approximately 8.3.
• Reaction with Acids: In addition to heat, baking soda releases CO2 instantly when it comes into contact with acids like acetic acid (vinegar) or citric acid (lemon juice). This is a popular experiment used to demonstrate chemical reactions in science education.
• Baking Soda vs. Baking Powder: Baking soda requires an acidic ingredient (like yogurt, buttermilk, or molasses) to activate its leavening power at room temperature. In contrast, baking powder contains both sodium bicarbonate and a dry acidic salt (like cream of tartar), requiring only moisture and heat to react.
• Environmental and Safety: It is considered environmentally friendly and non-toxic. For storage, it should be kept in a cool, dry place, as humidity can cause it to lose its potency over time by initiating slow decomposition.

p data-end="57" data-start="0">The correct answer is Collenchyma.

 

Key Points

• Collenchyma is a specialized simple permanent tissue that provides mechanical support and flexibility (elasticity) to various developing plant parts such as the young stem and petiole of a leaf.
• The cells of collenchyma are living, elongated, and characteristically thickened at the corners due to the deposition of cellulose, hemicellulose, and pectin.
• Its primary function is to allow easy bending in various parts of a plant (like tendrils and stems of climbers) without breaking. Hence, the correct answer is Collenchyma.
• This tissue is generally absent in monocotyledonous plants and is typically found in the hypodermis of dicotyledonous stems.
• The intercellular spaces in collenchyma tissue are either very small or totally absent because of the heavy pectic deposits at the cell corners.

Additional Information

• Parenchyma and Chlorenchyma:
• Parenchyma is the most common simple permanent tissue consisting of relatively unspecialized cells with thin cell walls.
• When parenchyma contains chlorophyll and performs photosynthesis, it is called Chlorenchyma. In aquatic plants, large air cavities are present in parenchyma to help them float, known as aerenchyma.
• Sclerenchyma:
• Unlike collenchyma, sclerenchyma is a dead tissue at maturity. It makes the plant hard and stiff.
• The walls of sclerenchyma cells are thickened due to lignin, a chemical substance which acts as cement and hardens them. It is found in husks of coconuts and vascular bundles.
• Complex Permanent Tissues (Xylem and Phloem):
• Xylem and Phloem are examples of complex tissues because they consist of more than one type of cell working together as a unit.
• Xylem is responsible for the conduction of water and minerals from roots to leaves. It also provides mechanical strength to the plant body.
• Phloem transports food materials (sucrose) prepared in the leaves to other parts of the plant. Unlike xylem, materials can move in both directions in phloem.

Important Points

• Types of Collenchyma:
• Based on the pattern of pectin deposition, it can be classified into Angular collenchyma (thickening at corners), Lacunar collenchyma (thickening around intercellular spaces), and Lamellar collenchyma (thickening in tangential walls).
• Difference between Support Tissues:
• Collenchyma provides flexible support to living and growing regions.
• Sclerenchyma provides rigid support to non-growing or mature plant parts.

The correct answer is Heterogeneous mixture with fine particles.

Key Points

• A colloid is a heterogeneous mixture in which one substance consisting of microscopically dispersed insoluble particles (the dispersed phase) is suspended throughout another substance (the dispersion medium).
• The particle size in a colloidal solution is intermediate between a true solution and a suspension, generally ranging from 1 nanometer to 1000 nanometers. These fine particles are small enough that they do not settle out upon standing but large enough to scatter light. Hence, option 4 is correct.
• In a suspension, the particles are large (typically greater than 1000 nm) and will eventually settle due to gravity. Colloids do not settle, which distinguishes them from suspensions. Hence, option 1 is incorrect.
• A colloidal solution is usually translucent or opaque because the particles are large enough to scatter light (the Tyndall effect). It is not a transparent mixture limited only to gases; it can involve solids and liquids as well. Hence, option 2 is incorrect.
• In a true solution, the solute particles fully dissolve and are molecular or ionic in size (less than 1 nm). Because colloidal particles are insoluble and stay suspended without settling, they do not follow the behavior of a true solution or a settling suspension. Hence, option 3 is incorrect.

Additional Information

• The Tyndall Effect: This is a defining feature of colloidal solutions where a beam of light becomes visible as it passes through the mixture. The fine particles scatter the light rays, a phenomenon not seen in true solutions because their particles are too small.
• Brownian Motion: Colloidal particles exhibit a continuous, random, zig-zag motion known as Brownian motion. This motion is caused by the constant bombardment of the dispersed particles by the molecules of the dispersion medium, which helps keep the particles in suspension and prevents them from settling.
• Classification of Colloids: Colloids are categorized based on the state of the dispersed phase and dispersion medium:
• Aerosol: Liquid or solid in gas (e.g., fog, mist, smoke).
• Emulsion: Liquid in liquid (e.g., milk, face cream, butter).
• Sol: Solid in liquid (e.g., mud, paints, milk of magnesia).
• Gel: Liquid in solid (e.g., jelly, cheese, curd).
• Foam: Gas in liquid (e.g., shaving cream, whipped cream).
• Historical Context: The term "colloid" was first coined by the Scottish scientist Thomas Graham in 1861. He derived the name from the Greek word "kolla," meaning glue, as many early colloids studied had glue-like properties.
• Stability and Charge: Colloidal particles often carry an electric charge (either positive or negative), which is uniform across all particles in a specific colloid. This like-charge creates repulsion between particles, preventing them from aggregating and contributing to the stability of the solution.

Important Points

• Filtration: Unlike suspensions, colloidal particles cannot be separated from the mixture by ordinary filtration because they are small enough to pass through the pores of filter paper. However, they can be separated using ultra-filtration or centrifugation.
• Applications: Colloids are vital in various industries and biological processes. For example, the blood in our bodies is a colloidal system. They are also used in the purification of water (using alum to coagulate impurities), the tanning of leather, and the manufacture of medicines and cosmetics.
• Blue Color of the Sky: The blue appearance of the sky is actually a result of the scattering of light by colloidal dust particles and water molecules present in the atmosphere, illustrating the Tyndall effect on a massive scale.

The correct answer is Planaria.

Key Points

• Regeneration is a type of asexual reproduction in which an organism has the ability to give rise to a new individual from its body parts. If the individual is somehow cut or broken into many pieces, many of these pieces grow into separate individuals.
• Planaria is a genus of flatworms belonging to the phylum Platyhelminthes. They are widely cited in biological studies for their remarkable ability to regenerate lost body parts. Hence, the option Planaria is correct.
• This phenomenon is carried out by specialized cells called neoblasts. These are pluripotent stem cells that have the capacity to proliferate and differentiate into any of the 30 to 40 different cell types found in a planarian's body.
• In Planaria, even a tiny fragment (as small as 1/279th of the original worm) can regenerate into a complete, well-proportioned organism within a span of roughly one to two weeks.
• Unlike many other animals, Planaria maintain a large population of these stem cells throughout their adult life, which allows them to constantly renew their tissues and organs.

Additional Information

• Insects (House fly, Honey bee, Mosquito):
• These organisms belong to the phylum Arthropoda and class Insecta. While they undergo metamorphosis (either complete or incomplete), their regenerative capacity is extremely limited compared to Planaria.
• In insects like the House fly or Mosquito, regeneration is usually restricted to the repair of specific tissues or, in some species, the regrowth of limbs during larval stages or nymphal stages. However, an adult insect cannot regenerate its entire body from a fragment. Hence, options 1, 2, and 3 are incorrect.
• Hydra:
• Similar to Planaria, Hydra (phylum Cnidaria) also exhibits a high degree of regeneration. It uses regenerative cells for reproduction through budding and can regrow from small pieces of its body.
• Types of Regeneration:
• Morphallaxis: This involves the reorganization of existing tissues into new structures. It is primarily seen in Hydra.
• Epimorphosis: This involves the proliferation of new cells (blastema formation) to replace lost parts, commonly seen in the regeneration of a Salamander's limb or a Lizard's tail.
• Evolutionary Significance:
• Regeneration is an important survival mechanism. While simple organisms like Planaria and sponges show extensive regeneration, higher vertebrates have largely lost this ability, focusing instead on wound healing and tissue repair.

Important Points

• Platyhelminthes: Also known as flatworms, these are acoelomate, triploblastic, and bilaterally symmetrical organisms. Planaria is often used as a model organism in genetics and developmental biology.
• Stem Cell Research: The study of Planaria regeneration provides vital insights into human regenerative medicine, as scientists attempt to understand how stem cells can be triggered to repair human organs or spinal cord injuries.
• Asexual Reproduction: In addition to regeneration following injury, some Planaria species use binary fission (splitting themselves in half) as a natural method of population growth.

The correct answer is Kidney → Ureter → Urinary bladder → Urethra.

Key Points

• The human excretory system consists of a pair of kidneys, a pair of ureters, a urinary bladder, and a urethra. Its primary role is to filter out nitrogenous wastes and maintain osmoregulation.
• Kidney: Urine formation starts here through the filtration of blood. The kidneys extract waste products such as urea and excess salts from the bloodstream to form urine.
• Ureter: These are long, muscular tubes that carry urine from the renal pelvis of each kidney to the urinary bladder. Urine moves through these tubes via peristalsis.
• Urinary bladder: This is a hollow, muscular sac-like organ that stores urine temporarily. It expands as it fills and sends signals to the brain when it needs to be emptied.
• Urethra: This is the final canal through which urine is expelled from the body. The process of releasing urine is scientifically known as micturition.
• The correct physiological sequence for the passage of urine is Kidney (formation) → Ureter (conduction) → Urinary bladder (storage) → Urethra (excretion). Hence, the sequence Kidney → Ureter → Urinary bladder → Urethra is correct.

Additional Information

• The Kidneys:
• Each kidney contains approximately 1 million functional units called nephrons.
• The nephron is responsible for three main processes: glomerular filtration, tubular reabsorption, and tubular secretion.
• The kidneys also produce hormones like erythropoietin, which stimulates the production of red blood cells.
• The Ureters:
• These tubes are approximately 25 to 30 cm in length.
• They enter the bladder at an oblique angle to prevent the backflow (reflux) of urine toward the kidneys.
• The Urinary Bladder:
• An average adult bladder can hold between 400 ml to 600 ml of urine before the urge to urinate becomes strong.
• The bladder wall contains a specialized muscle called the detrusor muscle, which contracts during urination.
• The Urethra:
• In females, the urethra is much shorter (about 4 cm), while in males, it is longer (about 20 cm) and also serves as a passage for semen.
• The flow of urine is controlled by internal and external urethral sphincters.

Important Points

• Urine Composition: Healthy human urine typically consists of about 95% water and 5% organic/inorganic wastes like urea, creatinine, ammonia, and uric acid.
• Daily Output: An average adult human excretes about 1 to 1.5 liters of urine per 24 hours, though this varies based on fluid intake and environmental conditions.
• pH Level: Urine is generally slightly acidic, with an average pH of 6.0, but it can range from 4.5 to 8.0 depending on diet and health status.
• Diseases: Malfunctions in this sequence can lead to conditions like Kidney Stones (calculi), Urethritis (inflammation of the urethra), or Cystitis (inflammation of the bladder).

The correct answer is Killing weeds.

Key Points

• Herbicides, also commonly known as weedkillers, are a specific type of pesticide used to kill or inhibit the growth of undesired plants, known as weeds.
• The primary function of herbicides in agriculture is to eliminate competition; weeds compete with crops for vital resources such as water, soil nutrients, sunlight, and spatial area.
• By removing these competitors, herbicides ensure that the primary crop receives the maximum available resources, which directly leads to an increase in crop yield and overall quality of the harvest. Hence, the statement "Killing weeds" is correct.
• Herbicides are classified into two main types: Selective herbicides, which control specific weed species while leaving the desired crop relatively unharmed, and Non-selective herbicides (often called total weedkillers), which kill all plant material they contact.
• The application of these chemicals is a critical component of modern industrial agriculture, allowing for large-scale monoculture and reducing the need for manual weeding or intensive tillage.

Additional Information

• Fungicides: These are chemical compounds or biological organisms used to kill or inhibit fungi or fungal spores. Fungal diseases can cause devastating losses in crops like wheat, rice, and potatoes (e.g., Late Blight). Common examples include Bordeaux mixture and Sulfur-based sprays.
• Insecticides: These are substances used to kill insects. They include ovicides and larvicides used against insect eggs and larvae, respectively. These are used to protect crops from pests like aphids, locusts, and bollworms. A well-known historical example is DDT, though many modern versions like Neonicotinoids are now more common.
• Bactericides: These are chemicals used specifically to control bacterial plant diseases. While less common than fungicides, they are used to treat conditions like citrus canker or fire blight in orchards. Many bactericides are copper-based or involve antibiotics like streptomycin.
• Rodenticides: These are a category of pest control chemicals intended to kill rodents (rats and mice) that might consume crops in the field or damage stored grains in warehouses.

Important Points

• Modes of Action: Herbicides work through various biological mechanisms, such as inhibiting photosynthesis, disrupting cell division, or interfering with the synthesis of essential amino acids.
• Timing of Application: They can be applied as Pre-planting (incorporated into the soil before sowing), Pre-emergence (applied before the weed or crop sprouts), or Post-emergence (applied after the plants are visible).
• Popular Examples: 2,4-D is one of the oldest and most widely used selective herbicides in the world, primarily used to control broadleaf weeds. Glyphosate is a famous non-selective systemic herbicide used extensively in no-till farming.
• Environmental Impact: Excessive use of herbicides can lead to soil degradation, water contamination, and the development of herbicide-resistant weeds (often called superweeds), which poses a challenge for sustainable agriculture.

The correct answer is parallel to the wave direction.

Key Points

• In a longitudinal wave, the particles of the medium vibrate or oscillate in a direction that is parallel to the direction of energy transport. This means if the wave moves from left to right, the particles of the medium oscillate back and forth along the same left-to-right axis.
• These waves are characterized by the formation of compressions and rarefactions. Compressions are regions where the particles are closest together (high pressure), while rarefactions are regions where the particles are farthest apart (low pressure).
• A classic example of a longitudinal wave is a sound wave traveling through a medium like air, water, or steel. When a sound wave travels, the air molecules move back and forth, creating pressure variations that our ears detect.
• Longitudinal waves can propagate through solids, liquids, and gases. This is because these waves depend on the elasticity of volume of the medium, and all three states of matter possess this property.
• In seismology, the P-waves (Primary waves) generated during an earthquake are longitudinal waves. They are the fastest seismic waves and can travel through both the liquid outer core and the solid mantle of the Earth.

Additional Information

• Transverse Waves: Unlike longitudinal waves, in a transverse wave, the particles of the medium move perpendicular to the direction of the wave motion. Examples include light waves, radio waves, and waves produced in a plucked guitar string.
• Mechanical Waves: Both longitudinal and transverse waves (when traveling through a medium) are types of mechanical waves. These waves require a material medium (solid, liquid, or gas) for their propagation and cannot travel through a vacuum.
• Wave Parameters:
• Wavelength: In a longitudinal wave, it is the distance between two consecutive compressions or two consecutive rarefactions.
• Frequency: This refers to the number of oscillations per unit time. It is measured in Hertz (Hz).
• Amplitude: In these waves, amplitude refers to the maximum displacement of the particles from their mean equilibrium position.
• Speed of Sound: The speed of a longitudinal sound wave depends on the density and elasticity of the medium. Sound travels fastest in solids, followed by liquids, and is slowest in gases. For instance, the speed of sound in steel is approximately 5960 m/s, while in air at 20 degrees Celsius, it is about 343 m/s.

Important Points

• Wave Equation: The relationship between the velocity (v), frequency (f), and wavelength (λ) is expressed by the fundamental formula v = f × λ. This applies to all wave types, including longitudinal ones.
• Slinky Example: A common classroom demonstration of a longitudinal wave involves a Slinky spring. By pushing and pulling the end of the spring along its length, a pulse of compression moves down the spring, demonstrating parallel particle motion.
• Ultrasound: These are longitudinal waves with frequencies higher than the upper limit of human hearing (greater than 20,000 Hz). They are widely used in medical imaging and industrial non-destructive testing.
• Infrasound: These are longitudinal waves with frequencies below 20 Hz. They are often produced by large-scale natural phenomena like volcanic eruptions or avalanches.

The correct answer is The live wire is red, the neutral wire is black, and the earth wire is green.

Key Points

• In a typical domestic electrical circuit, three separate wires are utilized to ensure the safe and functional delivery of electricity: the live wire, the neutral wire, and the earth wire.
• The live wire (also known as the phase wire) is traditionally color-coded red in India. This wire is at a high potential, usually 220V relative to the ground. Hence, the statement describing the live wire as red is correct.
• The neutral wire is traditionally color-coded black. This wire completes the circuit by providing a return path for the current and is generally at zero potential. Hence, the statement describing the neutral wire as black is correct.
• The earth wire is color-coded green (or sometimes a combination of green and yellow in modern contexts). It is a safety wire used to connect the metallic casing of appliances to the ground. Hence, the statement describing the earth wire as green is correct.
• The standard potential difference (voltage) between the live and neutral wires in Indian households is 220 Volts with a frequency of 50 Hz.

Additional Information

• Functions of the Wires:
• Live Wire: It is the source of current coming into the house from the power station or transformer. It is the wire that carries the voltage and is connected to fuses or Circuit Breakers (MCB).
• Neutral Wire: It serves as the exit point for the current, maintaining the flow of electricity. While it is ideally at 0V, it can carry current if the circuit is active.
• Earth Wire: It provides a low-resistance path for leakage current. If the insulation of the live wire fails and touches the metal body of an appliance, the current flows into the ground, preventing a fatal electric shock to the user.
• Modern International Color Coding:
• In recent years, many countries, including India, have started adopting the international standard (IEC 60446).
• In this new system, the Live wire is Brown, the Neutral wire is Blue, and the Earth wire remains Green-Yellow.
• However, for most traditional installations and general examinations, the Red-Black-Green convention remains the standard reference.
• Safety Components:
• The Electric Fuse is a critical safety device connected in series with the live wire to protect against short-circuits or overloading.
• A Miniature Circuit Breaker (MCB) is a modern alternative to fuses that automatically switches off the circuit during abnormal current conditions.

Important Points

• Short Circuiting: This occurs when the live wire and the neutral wire come into direct contact due to damaged insulation, leading to a sudden, massive surge in current and potential fires.
• Overloading: This happens when too many high-power appliances (like air conditioners, heaters, and geysers) are connected to a single circuit, exceeding the wire's current-carrying capacity.
• Safety Tip: Switches should always be connected to the live wire so that when the switch is off, the appliance is completely disconnected from the high-voltage supply.

The correct answer is Continuous movement of particles.

Key Points

• The observation of ink spreading in water is a classic demonstration of the continuous movement of particles of matter. According to the Kinetic Molecular Theory, particles of matter are never at rest; they are in constant, random motion.
• This spontaneous intermixing of particles of two different types of matter on their own is called diffusion. In this case, the ink particles move and distribute themselves among the water particles.
• The movement happens because particles possess Kinetic Energy. As the ink is added, its particles collide with the water molecules, leading to a uniform distribution over time, even without stirring. Hence, statement 1 is correct.
• While there are spaces between particles, the act of "spreading" or "traveling" throughout the liquid specifically highlights the kinetic nature (movement) rather than just the existence of gaps.
• Particles in liquids have enough kinetic energy to slide past one another, which allows for the fluidity and the diffusion process observed in the experiment.

Additional Information

• Characteristics of Particles of Matter:
• Particles of matter are very small, often beyond our imagination.
• They have spaces between them (intermolecular spaces), which is why we can dissolve salt or sugar in water without a significant increase in volume.
• They attract each other due to intermolecular forces. The strength of this attraction varies from one substance to another (strongest in solids, weakest in gases).
• They are continuously moving, and this movement increases with an increase in temperature.
• Diffusion in Different States:
• Gases: Diffusion is fastest in gases because particles move at high speeds and have large spaces between them. An example is the smell of perfume or cooking food spreading in a room.
• Liquids: Diffusion is slower than in gases but faster than in solids. The spreading of ink or potassium permanganate in water are primary examples.
• Solids: Diffusion in solids is extremely slow. An example is chalk marks left on a blackboard for a long time becoming difficult to erase because particles diffuse into the surface.

Important Points

• Effect of Temperature: The rate of diffusion is directly proportional to temperature. On heating, particles gain more Kinetic Energy and move faster, causing the ink or salt to spread more quickly in hot water compared to cold water.
• Brownian Motion: This phenomenon is closely related to the continuous movement of particles. It was first observed by Robert Brown in 1827 while looking at pollen grains in water. It describes the zig-zag, random motion of microscopic particles.
• Concentration Gradient: Diffusion always occurs from an area of higher concentration to an area of lower concentration until a uniform distribution (equilibrium) is reached.
• Fixed Positions: Statement 2 is incorrect because if particles had fixed positions, the matter would be a rigid solid at absolute zero temperature, and the ink would stay exactly where it was dropped.
• Attraction Between Particles: While attraction (Statement 3) exists, it actually acts against diffusion by trying to keep the particles of the same substance together. Diffusion occurs because the kinetic energy of the particles is sufficient to overcome some of these attractive forces.

Shortcut Trick

Use complementary angle relations: cot 72° = tan 18° and tan 72° = cot 18°.

Group terms: (sec² 18° − tan² 18°) + (cot² 18° − cosec² 18°) + sin² 30° + cosec² 60°.

Apply identities: (1) + (−1) + sin² 30° + cosec² 60°.

Value = 0 + (1/2)² + (2/√3)² = 1/4 + 4/3.

Total = (3 + 16) / 12 = 19/12.

∴ The correct answer is 19/12.

Alternate Method

Given: sec² 18° − cot² 72° + sin² 30° + cosec² 60° − cosec² 18° + tan² 72°

Formula Used: cot(90° − θ) = tan θ | tan(90° − θ) = cot θ | sec² θ − tan² θ = 1 | cosec² θ − cot² θ = 1

⇒ sec² 18° − tan²(90° − 72°) + sin² 30° + cosec² 60° − cosec² 18° + cot²(90° − 72°)

⇒ sec² 18° − tan² 18° + sin² 30° + cosec² 60° − cosec² 18° + cot² 18°

⇒ (sec² 18° − tan² 18°) + sin² 30° + cosec² 60° − (cosec² 18° − cot² 18°)

⇒ 1 + (1/2)² + (2/√3)² − 1

⇒ 1/4 + 4/3

⇒ (3 + 16) / 12

⇒ 19/12

∴ The correct answer is 19/12.

Additional Information

Complementary Angle Identities

For angles adding to 90°, ratios convert: sin ↔ cos, tan ↔ cot, sec ↔ cosec.

Pythagorean Identities

The three basic identities are sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Reciprocal Trigonometric Values

Standard values for common angles: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3, and their reciprocals.

Shortcut Trick

Mode is the observation that occurs most frequently in the data set.

Count occurrences: 68 appears 4 times.

Count occurrences: 85 appears 3 times.

Count occurrences: 64 appears 3 times.

Other values (72, 76, etc.) occur only 1 or 2 times.

∴ The correct answer is 68.

Alternate Method

Given data: 85, 73, 89, 82, 67, 61, 78, 69, 86, 68, 64, 76, 64, 85, 68, 85, 68, 81, 68, 72, 64, 72, 76, 84, 88

Formula Used: Mode = Value with the maximum frequency.

⇒ Frequency of 61 = 1

⇒ Frequency of 64 = 3

⇒ Frequency of 68 = 4

⇒ Frequency of 72 = 2

⇒ Frequency of 76 = 2

⇒ Frequency of 85 = 3

⇒ All other values (67, 69, 73, 78, 81, 82, 84, 86, 88, 89) occur 1 time each.

⇒ The value 68 occurs the most number of times (4).

∴ The correct answer is 68.

Additional Information

Empirical Formula

For a moderately skewed distribution, the relationship between Mean, Median, and Mode is: Mode = (3 × Median) − (2 × Mean).

Bimodal and Multimodal Data

If two or more values have the same maximum frequency, the data set is said to be bimodal (two modes) or multimodal (multiple modes).

Mean and Median

Mean is the average calculated by Σx ÷ n. Median is the middle value of the data set when arranged in ascending or descending order.

Shortcut Trick

The average of the first 'n' odd natural numbers is always equal to 'n'.

Here, the total number of terms (n) = 16.

Average = 16.

∴ The correct answer is 16.

Alternate Method

Given: n = 16 odd natural numbers.

Formula: Average of an Arithmetic Progression (A.P.) = (First term + Last term) ÷ 2

⇒ First odd natural number (a) = 1

⇒ 16^th odd natural number (l) = a + (n − 1) × d = 1 + (16 − 1) × 2 = 31

⇒ Total Sum = (n ÷ 2) × (a + l) = (16 ÷ 2) × (1 + 31) = 8 × 32 = 256

⇒ Average = Total Sum ÷ n = 256 ÷ 16 = 16

∴ The correct answer is 16.

Additional Information

Average of Even Numbers

The average of the first n even natural numbers is always (n + 1).

Average of Consecutive Natural Numbers

The average of the first n natural numbers is given by (n + 1) ÷ 2.

Sum of First n Odd Numbers

The sum of the first n odd natural numbers is n².

Average of Squares

The average of squares of first n natural numbers is [(n + 1)(2n + 1)] ÷ 6.

Shortcut Trick

Let the total number of registered voters be X.

Apply the successive percentage factors: X × (Cast%) × (Valid%) × (Candidate Share%) = 9261

⇒ X × 75/100 × 98/100 × 75/100 = 9261

⇒ X × 3/4 × 49/50 × 3/4 = 9261

⇒ X × 441/800 = 9261 ⇒ X = 21 × 800 = 16800

∴ The correct answer is 16,800.

Alternate Method

Given:

Votes cast = 75% of Registered Voters

Invalid votes = 2% of cast votes ⇒ Valid votes = 98% of cast votes

Candidate share = 75% of valid votes = 9261

Formula: Registered Voters = (Winner Votes × 100 × 100 × 100) ÷ (Cast% × Valid% × Share%)

⇒ Total Voters = (9261 × 100 × 100 × 100) ÷ (75 × 98 × 75)

⇒ Total Voters = (9261 × 4 × 50 × 4) ÷ (3 × 49 × 3)

⇒ Total Voters = (9261 × 800) ÷ (9 × 49)

⇒ Total Voters = (9261 × 800) ÷ 441

⇒ Total Voters = 21 × 800

⇒ Total Voters = 16800

∴ The correct answer is 16,800.

Additional Information

Election Vote Distribution

Total Registered = Cast + Not Cast; Cast = Valid + Invalid. Calculations are always performed on Valid votes unless specified otherwise.

Successive Percentage Method

When multiple changes occur on a base value, the final value = Initial × (a/100) × (b/100) × (c/100). This is the fastest way to solve election and population problems.

Valid vs Invalid Ratio

If p% are invalid, then valid votes = Total Cast × (100 − p)/100.

Shortcut Trick

Effective successive discount of 30% and 20% = 30 + 20 − (30 × 20)/100 = 44%

Difference in discount percentages = 46% − 44% = 2%

Given that 2% of the Marked Price = ₹256

Marked Price (100%) = (256 ÷ 2) × 100 = ₹12,800

∴ The correct answer is ₹12,800.

Alternate Method

Given:

Single discount = 46%

Successive discounts = 30% and 20%

Difference in discount amount = ₹256

Formula Used:

Equivalent discount for two successive discounts a% and b% = (a + b − ab/100)%

Selling Price (SP) = Marked Price (MP) × (100 − Discount%)/100

⇒ Effective discount of 30% and 20% = (30 + 20 − (30 × 20)/100)%

⇒ Effective discount = 50 − 6 = 44%

⇒ Difference between first and second discount scenarios = 46% − 44% = 2%

⇒ According to the problem: 2% of Marked Price = 256

⇒ Marked Price = (256 × 100) ÷ 2

⇒ Marked Price = 128 × 100 = 12,800

∴ The correct answer is ₹12,800.

Additional Information

Single Equivalent Discount Formula

For two successive discounts of x% and y%, the single equivalent discount is (x + y − xy/100)%.

Relationship between MP and SP

The Selling Price after a discount d% on Marked Price is calculated as: SP = MP × (100 − d)/100.

Successive Discounts (Three or more)

For three successive discounts d₁, d₂, and d₃, the final SP is MP × (1 − d₁/100) × (1 − d₂/100) × (1 − d₃/100).

Shortcut Trick

When the distance is the same for both journeys, use the Harmonic Mean formula:

Average Speed = (2 × s₁ × s₂) / (s₁ + s₂)

⇒ Average Speed = (2 × 10 × 40) / (10 + 40)

⇒ Average Speed = 800 / 50 = 16 km/hr.

∴ The correct answer is 16 km/hr.

Alternate Method

Given:

Speed while going (s₁) = 10 km/hr

Speed while returning (s₂) = 40 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Let the distance between home and school be the LCM of 10 and 40, which is 40 km.

⇒ Distance (D) = 40 km

⇒ Time taken while going (t₁) = Distance / s₁ = 40 / 10 = 4 hours

⇒ Time taken while returning (t₂) = Distance / s₂ = 40 / 40 = 1 hour

⇒ Total Distance = 40 + 40 = 80 km

⇒ Total Time = 4 + 1 = 5 hours

⇒ Average Speed = 80 / 5 = 16 km/hr

∴ The correct answer is 16 km/hr.

Additional Information

Average Speed for different distances

When distances are not equal, use the general formula: Average Speed = (d₁ + d₂ + ... + d_n) / (t₁ + t₂ + ... + t_n).

Three equal distances

If a journey is divided into three equal parts covered at speeds x, y, and z, the average speed is (3xyz) / (xy + yz + zx).

Speed Conversion

To convert speed from km/hr to m/s, multiply the value by 5/18. To convert from m/s to km/hr, multiply by 18/5.

Shortcut Trick

Total deduction = 29 + 70 + 28 = ₹127

Remaining amount = 13,525 − 127 = ₹13,398

Sum of ratio parts = 19 + 6 + 8 = 33 units

1 unit = 13398 ÷ 33 = 406

Difference (B & C) = |(8 × 406 + 28) − (6 × 406 + 70)| = 812 − 42 = 770

∴ The correct answer is ₹770.

Alternate Method

Given:

Total amount = ₹13,525

Deductions from A, B, C = ₹29, ₹70, ₹28

Ratio after deduction = 19 : 6 : 8

Sum of remaining amount = 13,525 − (29 + 70 + 28)

⇒ 13,525 − 127 = 13,398

⇒ Let the shares after deduction be 19x, 6x, and 8x.

⇒ 19x + 6x + 8x = 13,398

⇒ 33x = 13,398

⇒ x = 13,398 ÷ 33 = 406

⇒ B's original share = 6x + 70 = 6 × 406 + 70 = 2,436 + 70 = 2,506

⇒ C's original share = 8x + 28 = 8 × 406 + 28 = 3,248 + 28 = 3,276

⇒ Difference = 3,276 − 2,506 = 770

∴ The correct answer is ₹770.

Additional Information

Ratio and Proportion Basics

A ratio a : b indicates that for every 'a' units of the first quantity, there are 'b' units of the second. To find individual values, divide the total sum by the sum of ratio parts (a + b) to find the value of one unit.

Handling Deductions in Ratios

When fixed amounts are deducted from shares to reach a specific ratio, subtract the total deduction from the grand total before applying the ratio distribution formula.

Unitary Method in Ratios

Value of 'n' parts = (Total Value ÷ Total Ratio Parts) × n. This is the fastest way to calculate individual components in banking exams.

Shortcut Trick

Ratio of quantities = 10 : 15 = 2 : 3

Weighted Average CP = [(2 × 20) + (3 × 30)] ÷ (2 + 3)

⇒ 130 ÷ 5 = ₹ 26 per kg

Selling price for 10% profit = 26 × 110% = ₹ 28.60

∴ The correct answer is ₹ 28.60.

Alternate Method

Given:

Quantity₁ = 10 kg, Price₁ = ₹ 20/kg

Quantity₂ = 15 kg, Price₂ = ₹ 30/kg

Desired Profit = 10%

Formula Used:

Total CP = (Q₁ × P₁) + (Q₂ × P₂)

Selling Price (SP) = CP_avg × (100 + Profit%)/100

⇒ Total cost of 10 kg wheat = 10 × 20 = ₹ 200

⇒ Total cost of 15 kg wheat = 15 × 30 = ₹ 450

⇒ Total CP of 25 kg mixture = 200 + 450 = ₹ 650

⇒ Cost price per kg of mixture = 650 ÷ 25 = ₹ 26

⇒ To earn 10% profit, SP = 26 + (10% of 26)

⇒ SP = 26 + 2.60 = ₹ 28.60

∴ The correct answer is ₹ 28.60.

Additional Information

Weighted Average

The average price of a mixture is calculated as (n₁c₁ + n₂c₂) ÷ (n₁ + n₂), where n is quantity and c is cost.

Profit Percentage

Profit is always calculated on the Cost Price (CP). Selling Price = CP × (100 + Profit%) ÷ 100.

Alligation Rule

It is a shorter way to find the ratio in which two ingredients are mixed: (Higher Price − Mean Price) : (Mean Price − Lower Price).

Shortcut Trick

Total capacity = LCM of (15, 35, 60) = 420 units.

Efficiencies: A = 420/15 = 28 | B = 420/35 = 12 | C = −420/60 = −7 (Outlet).

Combined Efficiency (A + B + C) = 28 + 12 − 7 = 33 units/min.

Required time = 420 ÷ 33 = 140/11 minutes.

Time = 12⁸/₁₁ minutes.

∴ The correct answer is 12⁸/₁₁ minutes.

Alternate Method

Given:

Time for Pipe A = 15 mins

Time for Pipe B = 35 mins

Time for Pipe C (Emptying) = 1 hour = 60 mins

Formula used: Net Efficiency = Efficiency of A + Efficiency of B − Efficiency of C

Combined efficiency of A + B + C = 28 + 12 − 7 = 33 units/min

⇒ Total capacity = LCM(15, 35, 60) = 420 units

⇒ Time to fill cistern = Capacity ÷ Combined Efficiency

⇒ Time = 420 ÷ 33

⇒ Time = 140 ÷ 11

⇒ Time = 12⁸/₁₁ minutes

∴ The correct answer is 12⁸/₁₁ minutes.

Additional Information

Inlet and Outlet Pipes

Work done by an inlet pipe is always taken as positive, whereas work done by an outlet pipe is taken as negative.

Net Work Formula

If pipe A fills in x hours and pipe B empties in y hours (y > x), then the part filled in 1 hour is (1/x − 1/y).

Efficiency Relation

Efficiency is inversely proportional to the time taken to complete the work (Efficiency ∝ 1/Time).

Shortcut Trick

Convert both values to a common form (percentage).

66% of 1^st Number = (3/5 × 100)% of 2^nd Number

66% of 1^st Number = 60% of 2^nd Number

Ratio (1^st : 2^nd) = Inverse of their coefficients = 60 : 66

Simplifying by dividing by 6 ⇒ 10 : 11

∴ The correct answer is 10 : 11.

Alternate Method

Given:

Percentage of first number = 66%

Fraction of second number = 3/5

Formula:

Value = (Percentage/100) × Number | Ratio A : B = A/B

Let the first number be x and the second number be y.

⇒ 66% of x = (3/5) of y

⇒ (66/100) × x = (3/5) × y

⇒ (33/50) × x = (3/5) × y

⇒ x/y = (3/5) ÷ (33/50)

⇒ x/y = (3/5) × (50/33)

⇒ x/y = (1 × 10) / (1 × 11)

⇒ x/y = 10/11

∴ The correct answer is 10 : 11.

Additional Information

Percentage to Fraction Conversion

To convert x% to a fraction, divide by 100: x/100. Conversely, multiply a fraction by 100 to get the percentage.

Ratio Balancing Rule

If m × A = n × B, then the ratio A : B is equal to n : m. This is known as the inverse coefficient rule.

Simplifying Ratios

A ratio a : b can be simplified by dividing both terms by their Highest Common Factor (HCF) to reach the simplest form.

Shortcut Trick

Income ratio = 8 : 5 and Expenditure ratio = 2 : 1

Savings are same (₹160), use cross-product method:

| (8 × 1) − (5 × 2) | units = | 160 × (1 − 2) |

⇒ | 8 − 10 | units = | −160 | ⇒ 2 units = 160

⇒ 1 unit = 80

Difference in income = (8 − 5) = 3 units = 3 × 80 = ₹240

∴ The correct answer is ₹240.

Alternate Method

Given: Ratio of Incomes = 8 : 5, Ratio of Expenditures = 2 : 1, Individual Savings = ₹160.

Formula: Income − Expenditure = Savings

Let the incomes be 8x and 5x.

Let the expenditures be 2y and y.

⇒ 8x − 2y = 160 ... (i)

⇒ 5x − y = 160 ⇒ y = 5x − 160 ... (ii)

Substituting the value of y from equation (ii) into equation (i):

⇒ 8x − 2(5x − 160) = 160

⇒ 8x − 10x + 320 = 160

⇒ −2x = 160 − 320

⇒ −2x = −160 ⇒ x = 80

⇒ Monthly income of first person = 8 × 80 = ₹640

⇒ Monthly income of second person = 5 × 80 = ₹400

⇒ Difference = 640 − 400 = ₹240

∴ The correct answer is ₹240.

Additional Information

Income, Expenditure and Savings

The basic relationship is Income = Expenditure + Savings. If savings are equal for both parties, the difference in income units is proportional to the difference in expenditure units after scaling.

Ratio Balancing

To solve without algebra, multiply the expenditure ratio by a factor so that the difference (Income Unit − Expenditure Unit) is equal for both persons.

Proportion Rule

If A:B is the ratio of two quantities, the difference is expressed as |A − B| units. To find the real value, multiply this unit difference by the value of 1 unit.

Shortcut Trick

Total Surface Area (TSA) ratio units = 2 × (4×3 + 3×2 + 2×4) = 52 units²

Scaling factor squared (x²) = Total Area ÷ Ratio units = 832 ÷ 52 = 16

Linear scaling factor (x) = √16 = 4 cm

Volume ratio units = 4 × 3 × 2 = 24 units³

Actual Volume = 24 × (x)³ = 24 × 4³ = 24 × 64 = 1536 cm³

∴ The correct answer is 1536 cm³.

Alternate Method

Given:

Ratio of length (L) : breadth (B) : height (H) = 4 : 3 : 2

Total Surface Area (TSA) = 832 cm²

Formula Used:

Total Surface Area of cuboid = 2(LB + BH + HL)

Volume of cuboid = L × B × H

Calculations:

⇒ Let Length (L) = 4x, Breadth (B) = 3x, and Height (H) = 2x

⇒ 2[(4x × 3x) + (3x × 2x) + (2x × 4x)] = 832

⇒ 2[12x² + 6x² + 8x²] = 832

⇒ 2[26x²] = 832

⇒ 52x² = 832

⇒ x² = 832 ÷ 52 = 16

⇒ x = √16 = 4

⇒ L = 4 × 4 = 16 cm

⇒ B = 3 × 4 = 12 cm

⇒ H = 2 × 4 = 8 cm

⇒ Volume = 16 × 12 × 8

⇒ Volume = 1536 cm³

∴ The correct answer is 1536 cm³.

Additional Information

Lateral Surface Area (LSA)

The surface area of the four vertical walls of the cuboid is 2h(l + b).

Diagonal of a Cuboid

The length of the longest rod that can fit in a cuboid is √(l² + b² + h²).

Properties of a Cube

A cube is a special cuboid where l = b = h = a. TSA = 6a² and Volume = a³.

Shortcut Trick

Ratio (A : B) 9 yrs ago = 3 : 1 (Difference = 2)

Ratio (A : B) 7 yrs hence = 2 : 1 × 2 = 4 : 2 (Difference = 2)

Total time gap = 9 + 7 = 16 years

Unit difference in B’s ratio = 2 − 1 = 1 unit

⇒ 1 unit = 16 years

Bhawna’s age 9 yrs ago = 1 unit = 16 years

Bhawna’s present age = 16 + 9 = 25 years

∴ The correct answer is 25 years.

Alternate Method

Given: 9 years ago Asha = 3 × Bhawna. 7 years hence Asha = 2 × Bhawna.

Let the present age of Bhawna be B and Asha be A.

⇒ (A − 9) = 3(B − 9)

⇒ A − 9 = 3B − 27

⇒ A = 3B − 18 — (i)

⇒ (A + 7) = 2(B + 7)

⇒ A + 7 = 2B + 14

⇒ A = 2B + 7 — (ii)

Equating (i) and (ii):

⇒ 3B − 18 = 2B + 7

⇒ 3B − 2B = 7 + 18

⇒ B = 25 years

∴ The correct answer is 25 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals always remains the same, regardless of how many years pass.

Ratio Balancing

In age problems involving ratios, multiply the ratios by the difference of the other ratio to equalize the unit gaps between years.

Average Age Property

If the average age of a group of n people is X, then after k years, the new average age will be X + k.

Shortcut Trick

To minimize sections, maximize students per section using HCF.

Students per section = HCF(156, 143)

156 = 13 × 12 and 143 = 13 × 11 ⇒ HCF = 13

Number of sections = (156 ÷ 13) + (143 ÷ 13)

Number of sections = 12 + 11 = 23

∴ The correct answer is 23.

Alternate Method

Given:

Number of boy students = 156

Number of girl students = 143

Condition: Each section must have same number of students and contain only boys or girls.

Formula Used:

To have the fewest sections, the size of each section must be the Highest Common Factor (HCF) of the number of boys and girls.

Total Sections = (Number of Boys ÷ HCF) + (Number of Girls ÷ HCF)

Calculations:

⇒ Prime factors of 156 = 2 × 2 × 3 × 13

⇒ Prime factors of 143 = 11 × 13

⇒ Common factor (HCF) = 13

⇒ Number of boy sections = 156 ÷ 13 = 12

⇒ Number of girl sections = 143 ÷ 13 = 11

⇒ Total number of sections = 12 + 11

⇒ Total sections = 23

∴ The correct answer is 23.

Additional Information

Highest Common Factor (HCF)

It is the largest positive integer that divides each of the integers without a remainder. It is used to find the maximum size for equal grouping.

Least Common Multiple (LCM)

LCM is the smallest positive integer that is divisible by both a and b. It is used for problems involving cycles or meeting times.

HCF × LCM Property

For two numbers a and b, Product of numbers = HCF(a, b) × LCM(a, b).

Shortcut Trick

Unit digit calculation: Unit digit of 13³ is 7 (3³ = 27).

Unit digit of 8³ is 2 (8³ = 512).

Difference in unit digits = 7 − 2 = 5.

Comparing options, only 1685 and 1725 end in 5.

Since 13³ − 8³ = 2197 − 512 = 1685.

∴ The correct answer is 1685.

Alternate Method

Given: 13³ − 8³

Formula used: a³ − b³ = (a − b)(a² + ab + b²)

Here, a = 13 and b = 8

⇒ (13 − 8) × (13² + 13 × 8 + 8²)

⇒ 5 × (169 + 104 + 64)

⇒ 5 × 337

⇒ 1685

∴ The correct answer is 1685.

Additional Information

Sum of Cubes Identity

The sum of two cubes is given by a³ + b³ = (a + b)(a² − ab + b²).

Difference of Squares

The difference of two squares is given by a² − b² = (a − b)(a + b).

Common Cubes

11³ = 1331, 12³ = 1728, 13³ = 2197, 14³ = 2744, 15³ = 3375.

Shortcut Trick

Surface area of a sphere = 4 × Area of its great circle

Area of great circle = π × (radius)² = π × 3² = 9π

Surface Area = 4 × 9π = 36π cm²

∴ The correct answer is 36 π cm².

Alternate Method

Given: Radius of the sphere (r) = 3 cm

Formula Used: Total Surface Area of a Sphere = 4πr²

⇒ Surface Area = 4 × π × (3)²

⇒ Surface Area = 4 × π × 9

⇒ Surface Area = 36π cm²

∴ The correct answer is 36 π cm².

Additional Information

Volume of a Sphere

The amount of space occupied by a sphere is given by V = (4/3)πr³.

Surface Area of a Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Ratio of Surface Area to Volume

For a sphere, the ratio of its surface area to its volume is 3/r.

Shortcut Trick

Single equivalent discount formula for two successive discounts a% and b%:

Equivalent Discount = [a + b − (a × b / 100)]%

⇒ Equivalent Discount = [28 + 16 − (28 × 16 / 100)]%

⇒ Equivalent Discount = [44 − 4.48]% = 39.52%

∴ The correct answer is 39.52%.

Alternate Method

Given: First discount (d₁) = 28%, Second discount (d₂) = 16%

Formula: Final Price = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Let the Marked Price (MP) be 100 units.

⇒ Price after 1^st discount = 100 − (28% of 100) = 100 − 28 = 72 units.

⇒ Price after 2^nd discount = 72 − (16% of 72)

⇒ Second discount amount = (16 × 72) / 100 = 1152 / 100 = 11.52 units.

⇒ Final Selling Price (SP) = 72 − 11.52 = 60.48 units.

⇒ Total Discount = MP − SP = 100 − 60.48 = 39.52 units.

⇒ Single Equivalent Discount Percentage = (39.52 / 100) × 100 = 39.52%

∴ The correct answer is 39.52%.

Additional Information

Three Successive Discounts

If three discounts a%, b%, and c% are given, the equivalent discount is 1 − [(1 − a/100)(1 − b/100)(1 − c/100)] × 100.

Successive Percentage Increase

For two successive increases of x% and y%, the effective increase is [x + y + (xy/100)]%.

Relation between SP, MP and Discount

Selling Price (SP) = Marked Price (MP) × (100 − Discount%)/100.

Shortcut Trick

Effective Rate of interest for 2 years = R + R + (R × R)/100

Effective Rate = 15 + 15 + (15 × 15)/100 = 32.25%

Total Interest = 15000 × 32.25%

Total Interest = 150 × 32.25 = 4837.50

∴ The correct answer is ₹4,837.50.

Alternate Method

Given:

Principal (P) = ₹15,000

Rate (R) = 15% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Compound Interest (CI) = A − P

⇒ Amount (A) = 15000 × (1 + 15/100)²

⇒ A = 15000 × (1.15)²

⇒ A = 15000 × 1.3225

⇒ A = 19837.50

⇒ Interest = 19837.50 − 15000

⇒ Interest = 4837.50

∴ The correct answer is ₹4,837.50.

Additional Information

Compound Interest Formula

CI = P[(1 + R/100)ⁿ − 1], where P is principal, R is rate, and n is time in years.

Difference between CI and SI

The difference for 2 years is given by CI − SI = P(R/100)².

Half-Yearly Compounding

If interest is compounded half-yearly, the rate becomes R/2 and the time becomes 2n.

Shortcut Trick

Solve following BODMAS order from inner to outer brackets:

Parentheses: (21 − 11 + 9) = 19

Curly Braces: {31 − 19} = 12

Square Brackets: [25 − 12] = 13

Final calculation: 35 − 13 = 22

∴ The correct answer is 22.

Alternate Method

Given expression: 35 − [25 − {31 − (21 − 11 + 9)}]

Formula used: BODMAS Rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

⇒ 35 − [25 − {31 − (10 + 9)}]

⇒ 35 − [25 − {31 − 19}]

⇒ 35 − [25 − 12]

⇒ 35 − 13

⇒ 22

∴ The correct answer is 22.

Additional Information

BODMAS / VBODMAS Rule

It defines the order of operations: V (Vinculum/Bar), B (Brackets), O (Of), D (Division), M (Multiplication), A (Addition), S (Subtraction).

Types of Brackets

Brackets must be opened in this specific order: Vinculum (line), then Circular brackets (), then Curly braces {}, and finally Square brackets [].

Addition and Subtraction Priority

When only addition and subtraction remain, operations are performed strictly from left to right to ensure accuracy.

Shortcut Trick

Net Discount Formula for two successive discounts a% and b%:

Net Discount = [a + b − (a × b / 100)]%

Here, a = 40 and b = 15

Net Discount = [40 + 15 − (40 × 15 / 100)]% = [55 − 6]% = 49%

∴ The correct answer is 49%.

Alternate Method

Given:

First discount (d₁) = 40%

Second discount (d₂) = 15%

Formula: Final Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100)

Let the Marked Price (MP) be 100 units.

⇒ Price after 1st discount = 100 − (40% of 100)

⇒ Price after 1st discount = 100 − 40 = 60 units

⇒ Price after 2nd discount = 60 − (15% of 60)

⇒ Price after 2nd discount = 60 − (15/100 × 60)

⇒ Price after 2nd discount = 60 − 9 = 51 units

⇒ Single Equivalent Discount = (MP − Final Price) / MP × 100

⇒ Single Equivalent Discount = (100 − 51) / 100 × 100

⇒ Single Equivalent Discount = 49%

∴ The correct answer is 49%.

Additional Information

Three Successive Discounts

For three successive discounts a%, b%, and c%, calculate the net of a and b first, then use that result with c using the formula x + y − xy/100.

Marked Price and Selling Price Relation

Selling Price = MP × (100 − d₁)/100 × (100 − d₂)/100. This is useful for large number calculations.

Successive Percentage Increase

If values increase successively, use the formula: a + b + (ab / 100)% for the overall percentage change.

Shortcut Trick

Angle between bisectors of ∠C and ∠D in a quadrilateral = (∠A + ∠B) ÷ 2

Given: ∠A = 36° and ∠B = 93°

∠DOC = (36° + 93°) ÷ 2

∠DOC = 129° ÷ 2 = 64.5°

∴ The correct answer is 64.5°.

Alternate Method

Given: ∠A = 36°, ∠B = 93°, CO and DO are bisectors of ∠C and ∠D.

Formula: Sum of angles in a quadrilateral = 360° | Sum of angles in a triangle = 180°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 36° + 93° + ∠C + ∠D = 360°

⇒ 129° + ∠C + ∠D = 360°

⇒ ∠C + ∠D = 360° − 129° = 231°

⇒ Divide by 2: (∠C + ∠D) / 2 = 231° / 2 = 115.5°

⇒ In ΔDOC, ∠DOC + ∠OCD + ∠ODC = 180°

⇒ ∠DOC + (∠C / 2) + (∠D / 2) = 180°

⇒ ∠DOC + 115.5° = 180°

⇒ ∠DOC = 180° − 115.5° = 64.5°

∴ The correct answer is 64.5°.

Additional Information

Sum of Interior Angles

For any n-sided polygon, the sum of interior angles is (n − 2) × 180°.

Angle Bisector Relation

In a quadrilateral, the angle between the internal bisectors of two opposite angles is half the difference of the other two angles.

Cyclic Quadrilateral

In a cyclic quadrilateral, opposite angles are supplementary, meaning their sum is always 180°.

Shortcut Trick

Smallest two-digit multiple of 6 = 12

Largest two-digit multiple of 6 = 96

Total numbers (n) = [(Last Term − First Term) ÷ Common Difference] + 1

Total numbers (n) = [(96 − 12) ÷ 6] + 1

n = [84 ÷ 6] + 1 = 14 + 1 = 15

∴ The correct answer is 15.

Alternate Method

Given: Two-digit numbers range from 10 to 99. We need to find multiples of 6 in this range.

Formula used: n^th term of an Arithmetic Progression (A.P.), a_n = a + (n − 1)d

⇒ First term (a) = 12

⇒ Last term (a_n) = 96

⇒ Common difference (d) = 6

⇒ 96 = 12 + (n − 1) × 6

⇒ 96 − 12 = (n − 1) × 6

⇒ 84 = (n − 1) × 6

⇒ n − 1 = 84 ÷ 6

⇒ n − 1 = 14

⇒ n = 14 + 1 = 15

∴ The correct answer is 15.

Additional Information

Arithmetic Progression (AP)

A sequence of numbers where the difference between consecutive terms is constant, denoted as 'd'.

Sum of n terms of AP

The sum S_n can be calculated using S_n = (n ÷ 2)[2a + (n − 1)d] or S_n = (n ÷ 2)[a + l].

Divisibility Rule of 6

A number is divisible by 6 if it is divisible by both 2 (ends in even digit) and 3 (sum of digits is divisible by 3).

Shortcut Trick

Efficiency ratio of Rishi : Shan = 100 : 110 = 10 : 11.

Since Efficiency is inversely proportional to Time, Time ratio = 11 : 10.

Rishi takes 11 units of time, which is given as 11 days.

∴ 1 unit = 1 day.

Shan takes 10 units of time = 10 × 1 = 10 days.

∴ The correct answer is 10.

Alternate Method

Given: Rishi completes work in 11 days. Shan is 10% more efficient than Rishi.

Let efficiency of Rishi = 10 units/day.

⇒ Efficiency of Shan = 10 + (10% of 10) = 11 units/day.

⇒ Total Work = Efficiency of Rishi × Time taken by Rishi

⇒ Total Work = 10 × 11 = 110 units.

⇒ Time taken by Shan = Total Work ÷ Efficiency of Shan

⇒ Time = 110 ÷ 11 = 10 days.

∴ The correct answer is 10.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiency is x : y, the ratio of time taken is y : x.

Work Formula

Total Work = Efficiency × Time. This remains constant for a specific piece of work regardless of who performs it.

Percentage Efficiency Increase

If A is k% more efficient than B, then Efficiency_A = Efficiency_B × (1 + k/100).

Shortcut Trick

Total remaining fraction = (1 − 1/5) × (1 − 1/4) = 4/5 × 3/4 = 3/5

Here, 3 units represent the remaining amount ₹525.

1 unit = 525 ÷ 3 = 175

Original amount (5 units) = 175 × 5 = ₹875

∴ The correct answer is ₹875.

Alternate Method

Given:

Loss percentage = 20%

Expenditure from remainder = 25%

Final amount left = ₹525

Formula Used:

Final Value = Initial Value × (100 − R₁)/100 × (100 − R₂)/100

Calculations:

Let the original amount be ₹ x.

⇒ Amount left after 20% loss = x − (20/100) × x = 0.8x

⇒ Amount spent from remainder = 25% of 0.8x = (25/100) × 0.8x = 0.2x

⇒ Final amount left = 0.8x − 0.2x = 0.6x

⇒ 0.6x = 525

⇒ x = 525 ÷ 0.6

⇒ x = 5250 ÷ 6

⇒ x = 875

∴ The correct answer is ₹875.

Additional Information

Successive Percentage Change

If a value changes by a% and then b%, the net effective change is given by (a + b + ab/100)%.

Successive Discounts

Two successive discounts of x% and y% are equivalent to a single discount of (x + y − xy/100)%.

Percentage to Fraction Conversion

To simplify calculations, use 20% = 1/5 and 25% = 1/4 to find the multipliers directly.

Shortcut Trick

When distance is constant, the ratio of time is the inverse of the ratio of speed.

Ratio of speeds = 2 : 3 : 5

Ratio of times = (3 × 5) : (2 × 5) : (2 × 3)

Ratio of times = 15 : 10 : 6

∴ The correct answer is 15 : 10 : 6.

Alternate Method

Given:

Ratio of speeds of three cars (S₁ : S₂ : S₃) = 2 : 3 : 5

Distance (D) = Constant for all three cars

Formula Used:

Time (T) = Distance (D) ÷ Speed (S)

Calculations:

⇒ T₁ : T₂ : T₃ = (D/S₁) : (D/S₂) : (D/S₃)

⇒ T₁ : T₂ : T₃ = (1/2) : (1/3) : (1/5)

⇒ To simplify, multiply each term by the LCM of 2, 3, and 5.

⇒ LCM(2, 3, 5) = 30

⇒ T₁ : T₂ : T₃ = (1/2 × 30) : (1/3 × 30) : (1/5 × 30)

⇒ T₁ : T₂ : T₃ = 15 : 10 : 6

∴ The correct answer is 15 : 10 : 6.

Additional Information

Inverse Proportionality

When the product of two variables is constant (Distance = Speed × Time), the variables are inversely proportional; as speed increases, time decreases.

Average Speed

For two equal distances covered at speeds x and y, Average Speed = 2xy ÷ (x + y).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

The logic followed here is based on the alphabetical position pattern:

Option 1: OY - TD

Option 2: FS - KY

Option 3: BK - GP

Option 4: MP - RU

Thus, FS - KY does not follow this uniform shift.

Hence, the odd one out is Option 2.

Given: (Left) 5 4 3 6 7 8 2 1 9 3 3 2 4 5 (Right)

Now,

Numbers which are immediately preceded by an even number and also immediately followed by an even number.

So, Needed pattern is: Even - Odd - Even.

(Left) 5 4 3 6 7 8 2 1 9 3 3 2 4 5 (Right)

Only two such patterns exist.

Hence, the correct answer is 'Option 1'.

BODMAS Table: 

Decoding the given information:

<br>

Now, replacing the sign from left to right using BODMAS Rule, we get:

⇒ 66 ÷ 6 × 3 + 128 × 2 − 15 ÷ 3 + 16 × 2 − 17 = ?

⇒ 33 + 256 − 5 + 32 − 17 = ?

⇒ 289 − 5 + 32 − 17 = ?

⇒ 284 + 32 − 17 = ?

⇒ 316 − 17 = ?

⇒ 299

Hence, the correct option is "Option 3".

Given:

Varun’s rank from the top = 24th, rank from the bottom = 47th.

Total number of students = 24 + 47 – 1 = 70.

So, there are 70 students in the class.

Hence, the correct answer is "Option 1".

Given: A, B, C, D, E, F and G are sitting around a circular table facing the centre.

F sits third to the right of G.

A sits third to the right of F.

D sits third to the right of A.

C is the immediate neighbour of both B and A.

Total people between E and D = 5

Hence, the correct answer is "Option 4".

The logic followed here is:

Given:

Mohak drives 12 km South

He turns left → 11 km East

He turns left → 15 km North

He turns left → 14 km West

He turns left → 3 km South

So, To reach Point A, he must go 3 km East.

Hence, the correct answer is Option 4.

Given: A, B, C, D, E, F and G are sitting in a straight line facing North.

Only five people sit between D and C.

E sits to the immediate left of C.

Only one person sits between E and A.

G sits at some place to the left of F but at some place to the right of B.

Now, people to the right of A: F, E, C → 3 people

Hence, the correct answer is "Option 3".

Given Statements:

All shells are mats.

All mats are lotions.

The least possible Venn diagram for the given statements is:

 

Conclusions:

I. Some lotions are shells → Follows (Since all shells are lotions, there must be some overlap between lotions and shells.)

II. No lotion is a shell → Does not follow  (This contradicts the fact that all shells are included within lotions.)

So, only conclusion I follows.

Hence, the correct answer is "Option 2".

The logic followed here is:

For, JLIU → EKGV

For, MWQE → HVOF

Now, applying the same pattern to CBTJ:

So, CBTJ → XARK

Hence, the correct answer is Option 4.

The logic followed here is:

For, YOBD → IYLN

For, MCPR → WMZB

Now, applying the same pattern to QGTV:

So, QGTV → AQDF

Hence, the correct answer is Option 4.

The logic followed here is based on the alphabetical position pattern:

Option 1: DGK

Option 2: FIN

Option 3: PSW

Option 4: JMQ

Thus, FIN does not follow the uniform pattern of +3, +4.

Hence, the odd one out is Option 2.

Given:

Alok is 12th from the right end, Pyush is 18th from the right end,

and Pyush is exactly midway between Alok and Priyanka.

Priyanka is 5th from the left end.

Distance between Alok and Pyush = 18 − 12 = 6

So, Priyanka’s position from the right end = 18 + 6 = 24

Total number of people = 5 + 24 − 1 = 28

So, there are 28 people in the row.

Hence, the correct answer is "Option 2".

The logic followed here is:

thus, "88" will replace the question mark in the given series.

Hence, the correct option is "Option 4".

The logic followed here is:

Given:

Rajesh drives 13 km North

He turns right → 6 km East

He turns right → 5 km South

He turns left → 3 km East

He turns right → 10 km South

He turns right → 9 km West

So, To reach Point M, he must go 2 km North.

Hence, the correct answer is Option 4.

The logic followed here is:

Given:

STAR → S(19) + T(20) + A(1) + R(18) = 58

HERO → H(8) + E(5) + R(18) + O(15) = 46

Now,

ACTOR → A(1) + C(3) + T(20) + O(15) + R(18) = 57

Hence, the correct answer is "Option 4".

Given number: 8239847

Logic:

Add 1 to each even digit

Subtract 2 from each odd digit

Original number: 8 2 3 9 8 4 7

New number: 9 3 1 7 9 5 5 → 9317955

Digits appearing more than once: 9 and 5

Hence, the correct answer is Option 2.

The logic followed here is:

For, BONZ → SFEQ

For, REDP → IVUG

Now, applying the same pattern to QDCO:

So, QDCO → HUTF

Hence, the correct answer is Option 4.

The logic followed here is:

So, 'NKN 91' will replace the question mark in the given series.

Hence, the correct answer is "Option 3".

Given: # 5 1 & & 5 + 8 3 & * < \ 9 $ > > * 5 / 7

Now,

We need to find symbols that are immediately preceded by another symbol and immediately followed by a number.

So, Needed pattern is: Symbol - Symbol - Number.

(Left) # 5 1 & & 5 + 8 3 & * < \ 9 $ > > * 5 / 7 (Right)

Total valid occurrences = 3.

Hence, the correct answer is 'Option 1.

Given Statements:

All notepads are diaries.

All diaries are organisers.

Some organisers are planners.

The least possible Venn diagram for the given statements is:

 

Conclusions:

I. All notepads are organisers → Follows (Since Notepads ⊆ Diaries and Diaries ⊆ Organisers, therefore Notepads ⊆ Organisers)

II. Some planners are notepads → Does not follow  (There is no definite overlap between planners and notepads; planners only partially overlap with organisers)

So, only conclusion I follows.

Hence, the correct answer is "Option 1".

Given: (Left) 8 7 4 1 5 7 4 8 6 6 6 2 5 5 8 4 3 9 4 3 1 7 7 5 1 2 5 (Right)

Now,

An odd number must be immediately preceded by an even number and immediately followed by an even number.

So, Needed pattern is: Even - Odd - Even.

(Left) 8 7 4 1 5 7 4 8 6 6 6 2 5 5 8 4 3 9 4 3 1 7 7 5 1 2 5 (Right)

Only one such pattern exists in the entire series.

Hence, the correct answer is 'Option 3'.

The logic followed here is:

Logic: Third Number = First Number × Second Number.

For, 15 - 3 - 45

⇒ 15 × 3 = 45

For, 12 - 4 - 48

⇒ 12 × 4 = 48

Now, checking the options using the same pattern:

Option 1) 5 - 6 - 36

⇒ 5 × 6 = 30 ≠ 36

Option 2) 41 - 3 - 113

⇒ 41 × 3 = 123 ≠ 113

Option 3) 10 - 2 - 30

⇒ 10 × 2 = 20 ≠ 30

Option 4) 31 - 2 - 62

⇒ 31 × 2 = 62

Hence, the correct answer is Option 4.

Given: A, B, C, D, E, F and G are sitting around a circular table facing the centre.

D sits third to the right of C.

E sits third to the left of C.

B sits second to the left of D.

G sits to the immediate left of A.

Total people between F and G = 3

Hence, the correct answer is "Option 2".

The logic followed here is:

So, 'XLP' will replace the question mark in the given series.

Hence, the correct answer is "Option 1".

Given:

 “touch snow gently” → “lu ek ag”

“snow falls light” → “ph lu yn”

So, the code for snow = lu

Hence, the correct answer is Option 4.

The logic followed here is:

thus, "225" will replace the question mark in the given series.

Hence, the correct option is "Option 4".

BODMAS Table: 

Decoding the given information:

<br>

Now, replacing the sign from left to right using BODMAS Rule, we get:

⇒ 37 ÷ 5 − 91 + 16 × 4 = ?

⇒ 7.4 − 91 + 64 = ?

⇒ 71.4 − 91 = ?

⇒ 272

Hence, the correct option is "Option 1".

Given: (Left) 3 1 2 9 8 4 1 5 4 2 1 5 8 1 3 4 7 5 3 6 5 7 6 6 1 1 2 (Right)

Now,

An odd number must be immediately preceded by an odd number and immediately followed by an odd number.

So, Needed pattern is: Odd - Odd - Odd.

(Left) 3 1 2 9 8 4 1 5 4 2 1 5 8 1 3 4 7 5 3 6 5 7 6 6 1 1 2 (Right)

Only one valid triplet satisfies the condition.

Hence, the correct answer is 'Option 2'.

Family Chart:

Decoding the Symbols:

Given: L @ O + V # E × R

It means: L is the father of O, O is the sister of V, V is the brother of E, and E is the wife of R.

 

So, L is the wife’s father of R.

Hence, the correct answer is "Option 3".

The logic followed here is:

thus, "231" will replace the question mark in the given series.

Hence, the correct option is "Option 4".

The correct answer is Agni-Prime.

Key Points

• The Defence Research and Development Organisation (DRDO) successfully flight-tested the Agni-Prime (also known as Agni-P) missile on 24 September 2025.
• This specific test was significant as it was conducted from a rail-based mobile launcher, enhancing the strategic flexibility and survivability of the missile system by allowing it to be moved across the country’s vast railway network.
• The launch was carried out under full operational conditions, validating the reliability of the entire weapon system, including the propulsion, control systems, and navigation technologies.
• The mission met all objectives, confirming the high degree of accuracy and precision of the missile. This test represents a major milestone in India’s transition toward modern canisterized missile systems.
• The flight test took place at the Dr. APJ Abdul Kalam Island located off the coast of Odisha, which is the primary site for India's strategic missile testing.

Additional Information

• Agni-Prime Missile Features:
• Type: It is a two-stage solid-propellant ballistic missile that belongs to the new generation of the Agni series.
• Range: It has a strike range of 1,000 to 2,000 kilometers, making it a medium-range ballistic missile (MRBM).
• Canisterization: The missile is canisterized, meaning it is stored in a sealed tube. This allows for easier transport, longer shelf life, and significantly reduces the time required for launch.
• Navigation: It incorporates advanced Ring Laser Gyro-based Inertial Navigation Systems (RINS) and Micro Navigation Systems (MINS) for superior guidance.
• Strategic Significance:
• Agni-Prime is designed to replace the aging Agni-I and Agni-II missiles in the Indian inventory.
• The rail-based launcher capability ensures that the missile can be hidden and moved quickly, making it harder for enemy reconnaissance to track and target India's nuclear deterrent.
• It contributes to India’s policy of Credible Minimum Deterrence and its "No First Use" nuclear doctrine.
• Evolution of Agni Series:
• The Agni missiles were developed under the Integrated Guided Missile Development Programme (IGMDP), spearheaded by Dr. APJ Abdul Kalam.
• The family includes variants ranging from the short-range Agni-I (700 km) to the Intercontinental Ballistic Missile (ICBM) Agni-V (over 5,000 km).
• Unlike the older liquid-fuel missiles, modern Agni-Prime uses composite motor casings and advanced materials to reduce weight and increase efficiency.

Important Points

• DRDO: Headquartered in New Delhi, it is the R&D wing of the Ministry of Defence. It was formed in 1958 by merging the Technical Development Establishment and the Directorate of Technical Development and Production.
• Operational Readiness: Full operational condition tests are usually conducted by the Strategic Forces Command (SFC) in coordination with DRDO to ensure that the user (the military) is capable of handling the equipment in real-world scenarios.
• Mobility: While road-mobile launchers are common, rail-mobile versions provide an extra layer of strategic depth because of India's extensive railway infrastructure, allowing for rapid redeployment across thousands of kilometers.

The correct answer is Studying biomarkers of healthy aging, resilience, adversity, and transitions.

Key Points

• The BHARAT study (Biomarkers of Healthy Aging, Resilience, Adversity, and Transitions) was launched by the Centre for Brain Research (CBR), an autonomous center of the Indian Institute of Science (IISc), Bengaluru, in 2023.
• The primary focus of this study is to identify biological markers (biomarkers) that correlate with healthy brain aging and to understand how certain individuals remain resilient against cognitive decline despite various adversities.
• It is a longitudinal study, meaning it tracks the same group of individuals over an extended period to observe changes in cognitive health, brain structure, and lifestyle factors.
• The initiative aims to address the growing challenge of neurodegenerative disorders such as Alzheimer’s disease and dementia within the specific context of the Indian population. Hence, the correct answer is option 1.
• By studying the resilience and transitions of the aging brain, the researchers hope to develop better diagnostic tools and preventative strategies tailored to Indian genetics and lifestyle.

Additional Information

• Centre for Brain Research (CBR):
• The CBR was established at IISc Bengaluru through a generous philanthropic contribution from Mr. Kris Gopalakrishnan, the co-founder of Infosys.
• The center focuses on finding solutions for brain-related disorders that affect the elderly, particularly those leading to cognitive impairment.
• Target Demographic:
• The BHARAT study typically recruits participants aged 45 years and above to capture the early transitions into the aging phase.
• This demographic is critical because pathological changes in the brain often begin decades before clinical symptoms of dementia appear.
• Methodology and Technology:
• The study employs state-of-the-art technologies, including 3-Tesla MRI scanners for neuroimaging, genomic sequencing, and proteomic analysis from blood samples.
• Participants undergo rigorous neuropsychological assessments to evaluate memory, attention, and executive functions.
• Significance for India:
• With India’s aging population expected to reach nearly 20% of the total population by 2050, research into healthy aging is a national priority.
• The study accounts for unique Indian socio-economic factors, dietary habits, and environmental exposures that differ significantly from the Western cohorts usually studied in global health research.
• Longitudinal Cohort Studies:
• Similar to the Srinivaspura Aging, Neuro Epidemiology and Genetics (SAGE) study, BHARAT creates a massive database of clinical and biological data to serve as a resource for future drug development and public health policy.

Important Points

• The IISc was founded in 1909 and is consistently ranked as one of India's premier institutes for scientific research.
• According to the Dementia in India 2020 report, an estimated 5.3 million people above age 60 had dementia in India, a number expected to double by 2050.
• The BHARAT study contributes to the Global Brain Health initiative by providing data from the Global South, which is currently underrepresented in genomic and aging research.
• The term "Biomarkers" in this context refers to measurable indicators such as blood proteins or brain volume changes that can predict disease risk.