Answer: (A) Greenwashing

Greenwashing means creating a misleading impression that an organisation, product or activity is environmentally responsible. The key clue is the gap between the public claim and the underlying environmental performance. A claim may use vague language, selective disclosure or unsupported labels. Ecocriticism is an approach to studying literature and the environment; it is not the marketing practice described here. Remember: a green claim needs evidence, not merely green branding.

Source: United Nations: greenwashing

Answer: (C) Photons of solar radiation

Chlorophyll and associated pigments absorb photons. The absorbed light energy drives the light-dependent reactions, producing ATP and reducing power that support carbon fixation. Temperature affects the rate of enzyme-controlled reactions, but it is not the original energy input. The speed of light is relevant to the photon-energy relationship E = hc/λ; the actual energy arriving at the photosynthetic system is carried by photons.

Answer: (C) K = ½Iω²

For a small mass element, kinetic energy is ½mv². In rotation, v = rω, so summing over all mass elements gives K = ½(Σmr²)ω² = ½Iω². Moment of inertia plays the role that mass plays in translational kinetic energy. A common error is to forget the square on angular velocity. Doubling ω, while I remains fixed, makes the stored energy four times as large.

Answer: (B) The description of Inceptisols is incorrect

Inceptisols show relatively early or weak soil-profile development. Calling them old soils typical of dry conditions reverses their defining developmental character. Entisols have still less horizon development; Histosols contain substantial organic material; Vertisols have pronounced shrink-swell behaviour associated with clay. In soil-taxonomy questions, identify the diagnostic property of the order rather than assigning every order to a single climate.

Source: USDA NRCS: Inceptisols

Answer: (B) Kalpavriksh — Belgaum

Kalpavriksh’s published contact address is in Pune, Maharashtra, making the Belgaum pairing incorrect. The other listed locations correspond to the institutions named in the question. Institution-location MCQs are best revised as matched pairs: organisation, main location and field of work. Do not confuse the location of an institution’s project, field camp or occasional event with its principal contact location.

Source: Kalpavriksh: official contact address

Answer: (C) Living organisms and their environment

Ecology examines interactions among organisms and between organisms and their physical surroundings. It therefore includes biotic relationships, such as competition and predation, and abiotic influences, such as temperature, water and soil. Each of the other options selects only a narrow part of this wider field. The phrase “organisms and environment” is the most complete description.

Answer: (A) Lethality

The familiar four characteristics are ignitability, corrosivity, reactivity and toxicity. Flammability corresponds to ignitability in the choices. Lethality can describe a toxic effect or enter toxicity assessment, but it is not a separate heading in this four-characteristic framework. This is a terminology question: the answer does not mean that a lethal waste is harmless or cannot be hazardous.

Source: US EPA: hazardous-waste characteristics

Answer: (B) Sulphuric acid, H₂SO₄

Sulphur dioxide can be oxidised in the atmosphere and contribute to sulphuric acid formation. Nitrogen oxides contribute nitric acid, the other major acid-rain component. Ammonia is basic and can partly neutralise acids. The percentage in the question is a conventional textbook description, not a universal proportion: the sulphuric-to-nitric-acid contribution varies with emissions, location and atmospheric chemistry. The intended choice is H₂SO₄.

Answer: (C) Eliminate non-value-adding activities

Sustainable cost leadership removes avoidable waste while protecting activities that deliver customer value. Cutting training, delaying necessary replacement or weakening suppliers may reduce immediate spending but increase defects, downtime or lifecycle cost. Value analysis asks whether an activity contributes to the required function and customer benefit. The best option therefore targets non-value-adding work rather than applying indiscriminate spending cuts.

Answer: (B) Quadratic functions

The classical HMMS aggregate-planning model approximates relevant costs with quadratic functions. These costs can include inventory-related and workforce-adjustment costs. Optimising a quadratic objective can produce linear decision rules, which explains a common trap: the decision rule may be linear even though the cost functions are quadratic. The question asks specifically about the representation of costs.

Source: INFORMS: quadratic production-inventory planning

Answer: (A) Sales revenue analysis

A make-or-buy choice compares the relevant cost and implications of internal production against external purchase. Simple cost comparison, broader economic analysis and break-even analysis can all help. Sales revenue alone usually does not distinguish the alternatives when the same final product is sold in either case. Capacity, quality, supplier reliability and opportunity cost still matter; the answer concerns the named analytical approaches, not a claim that revenue is never relevant.

Answer: (A) Line of Balance

Line of Balance links planned deliveries of repetitive units with progress at key production or completion stages. A cumulative delivery schedule shows how many completed units are required by successive dates. Assembly-line balancing allocates tasks among workstations; PERT studies activity networks and project timing; Six Sigma focuses on variation and defects. The word “cumulative delivery” points to Line of Balance.

Answer: (A) 60 units

Use EOQ = √(2DS/H), where D is annual demand, S is ordering cost per order and H is annual holding cost per unit. Substitution gives √[(2 × 600 × 30)/10] = √3600 = 60 units. At this quantity the basic model balances annual ordering and holding costs. Check the units: an annual demand requires a holding cost expressed for the same annual period.

Answer: (B) Aggregate planning precedes MPS

An aggregate plan sets a broad output, workforce and inventory framework, often for product families. The master production schedule disaggregates that framework into specific end items, quantities and dates. Thus the logical planning sequence is aggregate plan followed by MPS, which then supports more detailed material and capacity planning. Actual organisations use feedback between levels, but this does not reverse the basic hierarchy.

Answer: (B) Minimise material-handling costs

CRAFT improves a facility layout by evaluating exchanges of departments or activity areas. Its objective commonly reflects material flow multiplied by travel distance and handling cost. Placing strongly connected departments closer together can reduce total handling cost. Labour cost and utilisation may be important to management, but they are not the specific objective identified by CRAFT in this question.

Answer: (B) Tactical decision

In the standard production-management classification, improving an existing product through value analysis is a tactical decision. It translates broad competitive objectives into a practical improvement programme. Strategic decisions define major long-term directions, such as a new plant or product domain; operational decisions concern routine execution and daily scheduling. The focus here is an improvement initiative within an existing operating framework.

Answer: (A) Approximately 0.68

A BCC unit cell contains two atoms effectively. The body diagonal gives √3a = 4r. Therefore APF = [2 × (4πr³/3)]/a³ = π√3/8 ≈ 0.680. FCC and ideal HCP structures have an APF of about 0.74. An APF of 1 would mean no unoccupied space, which is not achieved by these arrangements of equal spheres.

Answer: (D) Slip casting

The clay suspension is called a slip. A porous mould absorbs liquid, leaving a consolidated layer of clay particles against its wall. Excess slip may be drained to produce a hollow component. Die casting involves forcing material into a die, centrifugal casting uses rotational effects, and investment casting uses an expendable pattern. The slurry-plus-porous-mould description is characteristic of slip casting.

Answer: (C) Alumina

Specific modulus is elastic modulus divided by density, E/ρ. It measures stiffness per unit mass, which matters in lightweight structures. Typical dense alumina combines a high elastic modulus with substantially lower density than steel, giving a higher specific modulus than the listed alternatives. Do not select steel solely because it is strong: strength, modulus and modulus-to-density ratio are different properties. Exact values depend on grade and porosity.

Answer: (B) Strain hardening

Cold working plastically deforms metal below its recrystallisation temperature. Dislocation density increases, and dislocations obstruct one another, increasing strength and hardness while generally reducing ductility. This is strain hardening, also called work hardening. Recrystallisation can later restore ductility during appropriate heating, but it is not the strengthening mechanism that characterises cold deformation.

Answer: (A) Alumina

Modulus of rupture is a flexural-strength measure: the maximum bending stress calculated at fracture in a specified test. In standard comparisons of these engineering ceramics and glass, dense alumina has the highest typical flexural strength among the choices. It should not be confused with Young’s modulus, which measures stiffness. Ceramic strength depends strongly on flaws, porosity, specimen preparation and test conditions, so this is a typical-material comparison rather than an absolute ranking of every commercial grade.

Answer: (C) Martensitic transformation

Rapid cooling can suppress the diffusion-controlled formation of pearlite and allow austenite to transform into martensite. This is a diffusionless transformation involving a coordinated change in crystal structure. Carbon trapped in the transformed structure contributes to high hardness and internal strain. The usual exam association is rapid quenching → martensite; the actual structure also depends on composition, cooling rate and section size.

Answer: (B) Covalent bonding

When electronegativities are similar, neither atom overwhelmingly attracts the bonding electrons, so electron sharing is favoured. This is the basic description of covalent bonding. A large electronegativity difference favours ionic character. Metallic bonding is described by delocalised electrons in a metal, while secondary bonding includes weaker interactions between atoms or molecules. The clue is the small difference and resulting sharing tendency.

Answer: (D) 91.88%

One formula unit contains five copper atoms and one silicon atom. Copper contributes 5 × 63.54 = 317.70 mass units; the total is 317.70 + 28.09 = 345.79. Hence copper percentage = (317.70/345.79) × 100 ≈ 91.88%. The subscript 5 is essential. Using a 1:1 Cu:Si ratio would solve a different chemical formula.

Answer: (C) ROM

The BIOS must retain its firmware when power is switched off, so the textbook answer is non-volatile ROM. Modern motherboards generally use rewritable flash memory for firmware updates, but flash is still non-volatile firmware storage. Ordinary RAM and cache lose their contents without power, and the operating-system disk is not the conventional location of motherboard BIOS firmware.

Answer: (B) IEEE 802.11

IEEE 802.11 specifies wireless local-area networking, the standards family underlying Wi-Fi. IEEE 802.3 is associated with Ethernet, 802.15 with wireless personal-area networking, and 802.16 with broadband wireless metropolitan-area networking. New Wi-Fi generations add or update capabilities within the 802.11 family; they do not change the basic family number asked here.

Answer: (D) Wardriving

Wardriving involves moving through an area, traditionally in a vehicle, while detecting wireless networks. In this question the search is specifically for unsecured Wi-Fi. Phishing deceives users into revealing information; spoofing impersonates an identity or address; packet switching is a method of transmitting data. Recognise the terminology without confusing network discovery with those separate activities.

Answer: (B) Resolution

A larger pixel grid can represent finer spatial detail, which is the idea of resolution in this question. Brightness concerns intensity, contrast concerns differences between tones, and refresh rate concerns display updates over time. More pixels do not automatically make a poor source image sharper: optics, focus, sensor quality and processing also matter. Nevertheless, resolution is the property directly associated with pixel count.

Answer: (C) Demand basis

In packet-switched networks, traffic is divided into packets that use shared links as needed. Resources are statistically multiplexed rather than permanently reserved for one communication. This makes packet switching suitable for bursty data traffic. A dedicated end-to-end resource reservation is more characteristic of the basic circuit-switching model. Demand-based sharing can also create queues when packets compete for a busy link.

Answer: (B) Conjugate symmetry

For a real-valued time signal x(t), its Fourier transform satisfies X(−f) = X*(f), where the star denotes complex conjugation. Consequently, the magnitude spectrum is even and the phase spectrum is odd where phase is well defined. A real time-domain signal does not have to be even in time. Conjugate symmetry is therefore the general frequency-domain property.

Answer: (C) 22,300 miles

A geostationary satellite orbits at an altitude of roughly 35,786 km, or about 22,236 miles, conventionally rounded to 22,300 miles in this option. Its orbital period matches Earth’s sidereal rotation, and the orbit is circular and equatorial. Distinguish altitude above Earth’s surface from distance measured from Earth’s centre, which is about 42,164 km.

Answer: (C) Uplink about 14 GHz; downlink about 12 GHz

The conventional Ku-band pairing is approximately 14 GHz for uplink and 12 GHz for downlink. The familiar C-band pairing is 6/4 GHz, and the common Ka-band pairing is 30/20 GHz. Actual assigned frequency ranges depend on region and service. The question seeks the standard approximate band pairing, not one universal operating frequency for every satellite.

Answer: (D) Simplicity

The commonly taught five values in this framework are truth, right conduct, peace, love and non-violence. Simplicity may be a valuable personal quality, but it is not one of these five named headings. Such questions test a specified educational classification rather than judging whether an excluded quality is ethically undesirable. Truth, right conduct and non-violence are all present in the conventional list.

Answer: (D) Industrial standardisation

Standards provide shared specifications, terminology, test methods and quality expectations. They can improve compatibility, reduce unnecessary variety and help buyers compare products. Manufacturers benefit from clearer production requirements, while users gain greater predictability and safety. Competition alone does not establish common technical requirements. Industrial standardisation is therefore the best match for the combined benefits described.

Answer: (C) Balanced outlook on law

A balanced outlook recognises that law is necessary for safety, accountability and orderly professional practice, while also recognising the engineer’s responsibility to exercise ethical judgment. Compliance is a foundation, not a substitute for all professional reasoning. Standardisation concerns common technical requirements, whereas the question directly asks about the role of rules and regulations.

Answer: (B) Destructive testing

Destructive testing deliberately loads, stresses or otherwise tests a specimen until damage or failure occurs. It can reveal actual failure loads, failure mechanisms and margins, but the tested specimen may be unusable afterwards. Prototype testing is a wider category and need not destroy the prototype. Simulation predicts behaviour through a model rather than necessarily producing a physical failure.

Answer: (A) Failure Mode and Effects Analysis

FMEA proceeds systematically through components or functions, identifies how each may fail and examines the effects of those failures. It is often described as a bottom-up approach. A fault tree starts from an undesirable top event and works backwards through combinations of causes; an event tree follows outcomes after an initiating event. The component-by-component emphasis identifies FMEA.

Answer: (C) Fault-tree Analysis

Fault-tree analysis represents an undesirable top event using logical combinations of contributing events, commonly with AND and OR gates. Basic events can include equipment malfunctions and human errors. This makes it useful for examining how different categories of failure can jointly produce an accident. FMEA is primarily organised around individual failure modes; the question’s focus is a combined causal logic.

Answer: (D) Gaylord Nelson

US Senator Gaylord Nelson is associated with initiating the April 22 Earth Day environmental teach-in movement, first held in 1970. The other options are important environmental writers or conservation figures but are not the intended answer. Historically, an equinox Earth Day proposal is also associated with John McConnell; he is not among the choices. The standard April 22 context makes Gaylord Nelson the correct selection.

Answer: (C) Unqualified increased resource use

The intended distinction is between meeting essential needs sustainably and treating increasing resource consumption itself as the goal. The Brundtland framework stresses equity, participation, a sustainable population and conservation of the resource base. It does not forbid development or all increases in resource use in poorer countries. Read option C as an unqualified resource-consumption objective, which omits the ecological limits central to sustainable development.

Review note: Interpretation: option C is the best available answer; meeting essential needs in developing countries remains a central Brundtland objective.

Source: Brundtland Report, Chapter 2: needs, equity and ecological limits

Answer: (D) −cos t + e^t + te^t

Resolve the rational function into partial fractions: 2s²/[(s² + 1)(s − 1)²] = −s/(s² + 1) + 1/(s − 1) + 1/(s − 1)². The standard inverse transforms are cos t, e^t and te^t respectively, with the first term carrying a minus sign. Adding them gives −cos t + e^t + te^t. The repeated factor (s − 1)² is what creates the te^t term.

Answer: (B) y = t³/6 − t + (e^t − e^(−t))/2

Set z = y″. Then z′ + z = e^t + t + 1. Multiplication by the integrating factor e^t and integration give z = e^t/2 + t + Ce^(−t). The condition z(0) = 0 gives C = −1/2. Integrating twice and applying the remaining zero initial conditions yields y = t³/6 − t + e^t/2 − e^(−t)/2. As a quick check, y″ = t + sinh t and y‴ = 1 + cosh t, whose sum is t + 1 + e^t.

Answer: (A) ½ + Σ from n = 1 to ∞ of 2((−1)^n − 1)cos(nπx)/(n²π²)

Here f(x) = |x| is even, so all sine coefficients vanish. On −1 ≤ x ≤ 1, a₀ = ∫|x| dx = 1, giving constant term ½. For n ≥ 1, aₙ = 2∫₀¹ x cos(nπx) dx = 2[((−1)^n − 1)/(n²π²)]. Thus only odd harmonics remain, and the equivalent series is |x| = ½ − (4/π²)Σ cos((2k + 1)πx)/(2k + 1)², with k starting at zero. This matches option A.

Answer: (B) A row replacement Rᵢ → Rᵢ + kRⱼ leaves the determinant unchanged

Three rules should be kept distinct. Swapping two rows reverses the determinant’s sign. Multiplying one row by a scalar multiplies the determinant by that scalar. Adding a multiple of one row to a different row leaves the determinant unchanged. Also det(I) = 1. Only option B correctly states a standard determinant property; subtraction as a row replacement is not automatically a sign-changing operation.

Answer: (D) 0.006, or 0.6%

Logarithmic differentiation gives du/u = dx/x + 2dy/y − 3dz/z. For a maximum-error estimate, add the absolute contributions: |Δu/u| ≈ |Δx/x| + 2|Δy/y| + 3|Δz/z|. At x = y = z = 1, this is (1 + 2 + 3) × 0.001 = 0.006. If all errors had specified positive signs, cancellation could occur; the printed options indicate the usual maximum-error interpretation.

Review note: Convention: maximum first-order relative error, not the signed change when all three errors are equal and positive.

Answer: (D) 43 m

Integrate acceleration: v(t) = C − 4t³. Since v(2) = 76, C = 108, so v(t) = 108 − 4t³. The car stops when 108 − 4t³ = 0, giving t = 3 s. Distance after braking is ∫₂³(108 − 4t³)dt = [108t − t⁴]₂³ = (324 − 81) − (216 − 16) = 43 m. Do not reset the given time variable to zero without also changing the acceleration expression.

Answer: (B) Statement 2 only

The characteristic equation is ar² + br + c = 0. Its discriminant Δ = b² − 4ac determines the roots: Δ < 0 gives a complex-conjugate pair; Δ = 0 gives a repeated real root; Δ > 0 gives two distinct real roots. Statement 1 is correct, but statement 2 uses the condition for distinct roots. The question asks for the incorrect statement, so select 2 only.

Calculated answer: 455/969 ≈ 0.46956; no printed option matches

This is a hypergeometric selection. Choose one damaged component in C(5,1) ways and three undamaged components in C(15,3) ways; choose the full sample in C(20,4) ways. Therefore P = C(5,1)C(15,3)/C(20,4) = (5 × 455)/4845 = 455/969. None of A–D equals this value. Even a with-replacement binomial approximation gives 4(¼)(¾)³ = 27/64, but that changes the sampling model. We do not replace the exact answer by that approximation merely to fit an option. Any official key or deletion decision should be checked separately.

Review note: Printed-option mismatch: exact answer is 455/969. A = 27/64 is the with-replacement/binomial result, not the exact answer for the stated finite sample.

Answer: (A) √68 = 2√17

The displacement vector is B − A = (−3 − 5, 2 − 4) = (−8,−2). Its magnitude is √[(−8)² + (−2)²] = √(64 + 4) = √68. Squaring removes the signs of the coordinate differences. Do not add the absolute horizontal and vertical changes; that would measure a different, grid-based distance rather than straight-line Euclidean distance.

Answer: (D) (1 + se^(−πs/4))/(s² + 1)

Write f(t) = sin t + H(t − a)cos(t − a), where a = π/4 and H is the unit-step function. The second-shifting theorem gives L{H(t − a)g(t − a)} = e^(−as)G(s). Since L{sin t} = 1/(s² + 1) and L{cos t} = s/(s² + 1), the required transform is [1 + se^(−πs/4)]/(s² + 1). The added delayed term has a plus sign.

Answer: (A) ∛2

The density increases throughout its support, so its mode is at x = θ. Integrating gives F(x) = x³/θ³. The median m satisfies F(m) = ½, so m = θ/∛2. Dividing mode by median gives θ/(θ/∛2) = ∛2. The parameter θ cancels, which is a useful check: changing the scale should not change this dimensionless ratio.

Answer: (A) 2.9(0.9)^19

Model X as Binomial(n = 20, p = 0.1). “At most one” includes zero and one: P(X ≤ 1) = (0.9)²⁰ + C(20,1)(0.1)(0.9)¹⁹. Factoring out (0.9)¹⁹ gives (0.9 + 2)(0.9)¹⁹ = 2.9(0.9)¹⁹. The expression is not P(X = 1) alone, so remember to include the zero-success term.

Answer: (D) 17/65

Conditional probability restricts the sample space to X = 0,1,2,3,4. Thus the numerator is P(2) + P(3) + P(4) and the denominator is P(0) + P(1) + P(2) + P(3) + P(4). Every Poisson term contains e^(−1), which cancels. The remaining numerator is 1/2! + 1/3! + 1/4! = 17/24; the denominator is 1 + 1 + 17/24 = 65/24. Their ratio is 17/65.

Answer: (A) 1.5435e^(−2.1)

For independent Poisson breakdown processes, the total weekly count has mean λ = 35 × 0.06 = 2.1. Hence P(N = 3) = e^(−2.1)(2.1)³/3! = 1.5435e^(−2.1), approximately 0.189. The printed wording says “three machines,” while the offered expressions model three breakdown events. If distinct machines with at least one breakdown were counted exactly, a different model would be required. Option A follows the clearly intended aggregate Poisson calculation.

Review note: Modelling note: the options count breakdown events, not exactly three distinct machines experiencing at least one breakdown.

Answer: (A) (220/27)(2/3)^9

For each ball, the probability of entering the first box is 1/3; the probability of entering either other box is 2/3. Therefore X is Binomial(12,1/3). P(X = 3) = C(12,3)(1/3)³(2/3)⁹ = (220/27)(2/3)⁹. This uses the standard interpretation that balls are assigned independently and boxes are equally likely; it does not assign equal probabilities to all possible occupancy triples.

Answer: (C) 30/61

A obtains sum 6 with probability p = 5/36; B obtains sum 7 with probability q = 6/36. If both fail, the game returns to the same situation with A to play. Let W be A’s eventual winning probability. Then W = p + (1 − p)(1 − q)W. Thus W = (5/36)/[1 − (31/36)(30/36)] = 30/61. A’s first-roll success probability alone is not the probability of eventually winning.

Answer: (C) 13/41

An initial plus ends as plus after an even number of changes. That probability is (1/3)³ + C(3,2)(2/3)²(1/3) = 13/27. An initial minus becomes plus after an odd number of changes, with probability 14/27. Bayes’ theorem gives P(initial + | final +) = [(1/3)(13/27)]/[(1/3)(13/27) + (2/3)(14/27)] = 13/41. The prior probabilities 1/3 and 2/3 must be included.

Answer: (A) 1/3

The first-white probabilities for B1, B2 and B3 are 1/3, 2/3 and 1/2. After observing white, the posterior box probabilities are therefore 2/9, 4/9 and 1/3. The corresponding next-white probabilities are 0, 1/2 and 1/3. Multiply and add: (2/9)0 + (4/9)(1/2) + (1/3)(1/3) = 2/9 + 1/9 = 1/3. The observed first colour changes how likely each box is.

Answer: (C) 3/4

The density is a triangle with base 10 and height 5A. Its area must equal one: ½ × 10 × 5A = 1, so A = 1/25. Convert the sales limits: 250 and 750 pounds correspond to X = 2.5 and 7.5. Each excluded tail is a small triangle with base 2.5 and height 0.1, hence area 0.125. The required central probability is 1 − 0.125 − 0.125 = 0.75 = 3/4.

Answer: (D) Approximately 711.5 litres

Let c be tank capacity measured in thousands of litres. Exhaustion occurs when X > c. The tail probability is ∫c¹ 5(1 − x)⁴ dx = (1 − c)⁵. Set (1 − c)⁵ = 0.002, giving c = 1 − (0.002)^(1/5) ≈ 0.711460. Convert back to litres by multiplying by 1000: approximately 711.46 litres, rounded to 711.5. The complement and the unit conversion are both essential.

Answer: (A) FTPBJ

Compare successive letter positions: K→P is +5, A→G is +6, V→C is +7 with alphabet wraparound, Y→G is +8 and A→J is +9. Apply these position-specific shifts to ANITA: A+5 = F, N+6 = T, I+7 = P, T+8 = B and A+9 = J. This gives FTPBJ. The same original letter may receive different codes at different positions.

Answer: (B) 10 km

Track coordinates from (0,0). After 4 km north he is at (0,4). Turning left means west, so he reaches (−3,4). From west, a right turn points north, taking him to (−3,8). The final right turn points east; after 9 km he is at (6,8). Straight-line distance from home is √(6² + 8²) = 10 km. The 20 km total walked is not the displacement.

Answer: (C) 6 days

Their daily work rates are 1/10 and 1/15 of the job. Together the rate is 1/10 + 1/15 = 3/30 + 2/30 = 1/6. Taking the reciprocal gives six days for the whole job. This assumes both work at their stated constant rates on a divisible task. Adding completion times would be incorrect because it is work rates, not durations, that combine.

Answer: (D) (2,9,16)

Read the given set as (n, (n + 1)², (n + 2)²). For n = 3 this produces (3,4²,5²) = (3,16,25). For n = 2 the corresponding set is (2,3²,4²) = (2,9,16). The first term is not itself squared. A useful analogy technique is to explain every term in the given set with one consistent rule before checking the alternatives.

Answer: (C) D

D is below A, and A is below B and therefore below E’s shared flat. C is explicitly above D, while F is on top. Thus every named person is either directly or indirectly placed above D. The precise position of C among the other middle floors is unnecessary: the available relations already identify D as the bottom occupant.

Answer: (A) Mother, under the standard family-tree assumption

In the conventional exam interpretation, “my brother’s nephew” is taken to be Avinash’s son, Mohan. The woman is Mohan’s grandmother, so she is Avinash’s mother. The wording alone does not uniquely establish this: the grandmother could be on the child’s other parental side, or the nephew could be another sibling’s child. Mother is the intended listed choice, but it relies on the usual simplified family-tree assumption.

Review note: Family-tree ambiguity: Mother is the intended listed answer, not a uniquely forced relationship from the sentence alone.

Answer: (B) 9 times

The common days are Monday and Friday. With October 1 on Thursday, Fridays fall on 2,9,16,23 and 30: five Fridays. Mondays fall on 5,12,19 and 26: four Mondays. Therefore they meet on 5 + 4 = 9 scheduled walking days. First find the intersection of the weekly schedules, then count those weekdays in the month.

Answer: (A) soc

The first and second phrases share “and,” and their shared code is bac, so bac = and. The second and third phrases share “orange,” and their shared code is en, so en = orange. In the second phrase, bac and en are now identified, leaving soc for tomato. The order of words need not match the order of codes; shared occurrences provide the reliable deductions.

Answer: (C) 171

The alphabet positions are D = 4, H = 8, L = 12, N = 14 and S = 19. Their sum is 4 + 8 + 12 + 14 + 19 = 57. Since every position is multiplied by three, the required total is 3 × 57 = 171. Add the positions first and multiply once to reduce arithmetic and avoid accidentally skipping a letter.

Answer: (D) 11 May 2026

From January 22 to February 22 is 31 days; then to March 22 is 28 days, to April 22 is 31 days, and to May 11 is 19 days. The total is 109 days, with remainder 4 on division by 7. Four weekdays after Thursday is Monday. Therefore May 11 is the required date. February has 28 days because 2026 is not a leap year.

Answer: (C) 26th and 20th

For a class with N students and no ties, rank from bottom = N + 1 − rank from top. Ravi’s rank is 34 − 8 = 26; Vishal’s is 34 − 14 = 20. The extra one appears because the ranked student is included when counting from either end. A quick check is that top rank plus bottom rank must equal N + 1 = 34.

Answer: (D) 45 matches

Each match corresponds to an unordered pair of teams. The count is C(10,2) = 10 × 9/2 = 45. Multiplying 10 by 9 alone counts every match twice, once from each team’s perspective. Dividing by two removes that duplication. For n teams in a single round-robin competition, the general formula is n(n − 1)/2.

Answer: (D) A

One valid clockwise order before the swaps is C, B, E, F, A, D. For people facing the centre, a person’s immediate left is clockwise in this representation. Swapping D with E gives C, B, D, F, A, E. Swapping A with B gives C, A, D, F, B, E. A is now immediately to C’s left. Alternatively, C’s original left-hand neighbour B is replaced by A in the second swap.

Answer: (B) 125:27

Sphere volume is V = (4/3)πr³. When taking the ratio, the common factor (4/3)π cancels, leaving the cube of the radius ratio: 5³:3³ = 125:27. Squaring would give the surface-area ratio, not volume. Remember the dimensional relationship: lengths scale as k, areas as k² and volumes as k³.

Answer: (C) 46%

Contradiction occurs when A tells the truth and B lies, or A lies and B tells the truth. Under independence these mutually exclusive cases have probabilities (0.4)(0.7) and (0.6)(0.3). Their sum is 0.28 + 0.18 = 0.46, or 46%. The calculation assumes the usual true/false reporting framework; unrestricted different false accounts of a complex incident would require more information.

Answer: (D) 2/7

Total committees = C(8,3) = 56. The favourable cases are exactly two women and one man, or all three women. Their counts are C(3,2)C(5,1) = 15 and C(3,3) = 1. Therefore P = (15 + 1)/56 = 16/56 = 2/7. “At least two” includes three; overlooking that final committee changes the answer.

Answer: (A) 2/7

A leap year has 366 days = 52 complete weeks + 2 extra days. Those extra days are consecutive weekdays. There will be 53 Sundays if the extra pair is Saturday–Sunday or Sunday–Monday. Two of the seven possible starting weekdays satisfy this condition. Assuming the seven starting weekdays are equally likely in the question’s standard model, the probability is 2/7.

Answer: (B) 19

A consistent rule is “square the previous term and add 3.” Starting with 1 gives 1² + 3 = 4. Next, 4² + 3 = 19. Finally, 19² + 3 = 361 + 3 = 364. The last step confirms the pattern and selects option B. Checking the rule against every transition is more reliable than choosing a number from only the first two terms.

Answer: (C) 98

Subtract seven from each candidate. The implied n values are 32,85,91 and 81 respectively. Only 91 lies outside the stated interval, so 98 cannot be n + 7. This remains true whether “between” is interpreted strictly or with endpoints included. Directly recovering n from each option avoids unnecessary argument over endpoint conventions.

Answer: (A) 11

To maximise X, minimise the sum Y + Z. With natural numbers interpreted as 1,2,3,…, the smallest choices are Y = Z = 1. Then X = 13 − 1 − 1 = 11. Some mathematical conventions include zero among natural numbers, which would allow X = 13. The intended school-exam convention here is positive natural numbers; this assumption is stated so the reasoning remains explicit.

Review note: Convention: natural numbers are taken as positive integers. If zero were allowed, option C (13) would be possible.

Answer: (C) Bhopal

The Department of Atomic Energy’s parliamentary reply, published on 29 January 2026, identifies Bhopal as the location of the Rare Earth and Titanium Theme Park recycling plant. The plant recovers magnetic rare earths from end-of-life magnets. The same release distinguishes this facility from the samarium-cobalt permanent-magnet plant at Vizag. This distinction is the useful revision point: do not interchange the recycling-plant location and the magnet-manufacturing location.

Source: DAE/PIB, 29 January 2026: rare-earth minerals

Answer: (A) New Delhi

The Ministry of Agriculture and Farmers’ Welfare held the workshop at Sushma Swaraj Bhawan, New Delhi. AgriStack was discussed as digital public infrastructure for agriculture, connecting farmer identification, land information and service delivery. To retain this current-affairs fact, link the event to its organising ministry, date and exact venue. The question asks for the city, so New Delhi is the required level of detail.

Source: Agriculture Ministry/PIB, 29 January 2026: AgriStack workshop

Answer: (D) Yotta Network Services Private Limited

The 29 January 2026 parliamentary reply lists OneWeb, Jio Satellite Communications and Starlink among the companies issued the relevant authorisation. It lists Yotta and Nelco among rejected applications. Of the four choices, Yotta is therefore correct. GMPCS means Global Mobile Personal Communication by Satellite. This answer is tied to the dated government update; licensing positions should not be treated as permanently fixed.

Source: Communications Ministry/PIB, 29 January 2026: satellite services

Answer: (A) Phuket Deep Sea Port, Thailand

The Defence Ministry’s release states that the squadron’s Phuket visit concluded with a passage exercise involving HTMS Huahin on 28 January 2026. The release itself was published the following day. This is a useful date-reading lesson: an event date and its publication date need not be identical. Associate Phuket with Thailand, which is the country specified in the correct option.

Source: Defence Ministry/PIB, 29 January 2026: Phuket visit

Answer: (A) 28 January 2026

The government release dated 29 January 2026 records the inauguration as taking place on 28 January. The question asks when the project was inaugurated, so the event date, not the press-release date, must be selected. The project concerned civil aviation infrastructure at Air Force Station Leh. Keeping a separate “event date” column in a current-affairs notebook prevents this common one-day error.

Source: Defence Ministry/PIB, 29 January 2026: Leh infrastructure

Answer: (C) December 2028

The Cabinet approved the restructured JJM framework on 10 March 2026, extending the mission to December 2028. The updated approach emphasises reliable water-service delivery and sustainability alongside infrastructure creation. The month matters because all four choices are expressed as December deadlines. The answer should be read in the context of this 2026 approval, rather than an earlier mission target.

Source: Cabinet/PIB, 10 March 2026: JJM extension

Answer: (C) 2047

The Budget announcement identifies 2047 as the horizon for the specified foreign cloud-service providers using data-centre services from India. This is a targeted policy provision, not a blanket tax exemption for every technology company or every data-centre activity. For this MCQ, connect three elements: foreign provider, global cloud services and India-based data-centre services. The year completing that statement is 2047.

Source: Finance Ministry/PIB: Budget 2026–27 cloud-services provision

Answer: (B) Skilling for AI Readiness

The President of India’s official release expands SOAR as Skilling for AI Readiness. It records the launch of the #SkilltheNation challenge under this initiative on 1 January 2026. The initiative is associated with the Ministry of Skill Development and Entrepreneurship.

Source: President of India, 1 January 2026: SOAR and #SkilltheNation

Answer: (C) Viksit Bharat@2047

The Ministry of Information and Broadcasting describes the campaign as aligned with Viksit Bharat@2047. It invited citizens to express experiences of India’s transformation through creative formats, in collaboration with MyGov. The question tests the overarching vision rather than a specific technology or manufacturing scheme. Remember the campaign name together with its public-participation format and the 2047 development horizon.

Source: Information & Broadcasting Ministry/PIB, 3 January 2026: campaign winners

Answer: (B) Current AI and Kalpa Impact

The launch release of 2 June 2026 names Current AI and Kalpa Impact as BHASHINI’s partners. The challenge promotes multilingual, voice-first AI that can work in offline or low-connectivity environments. The context is edge AI and linguistic access, which helps distinguish it from unrelated space, telecom or general digital-governance partnerships appearing in the distractors.

Source: MeitY/PIB, 2 June 2026: VYOMA launch

Answer: (B) Agharkar Research Institute (ARI), Pune

The Department of Science and Technology’s release attributes the reported work to the Nanobioscience Group at Agharkar Research Institute, Pune. The platform is designed to deliver a targeted gene-silencing approach using biodegradable nanocarriers. This is a research-development report; it should not be read as an announcement of a routinely available or clinically proven treatment. The exam’s factual target is the institute and city.

Source: DST/PIB, 3 June 2026: nanocarrier research

Answer: (B) AI, Digital Infrastructure and Advanced Manufacturing

The 3 June 2026 government announcement explicitly foregrounds these three fields as key priorities. It also describes a wider science-and-technology relationship. The question’s wording asks for the highlighted trio, not every field in which the countries cooperate. Revise the announcement as a linked set: partner country South Africa; future-technology focus; AI, digital infrastructure and advanced manufacturing.

Source: DST/PIB, 3 June 2026: bilateral technology priorities

Answer: (D) Quantum Computing

AI, digital infrastructure and advanced manufacturing are the three headline priorities. Quantum computing is therefore the intended exclusion from that particular list. However, the same release also discusses quantum technologies in the wider partnership context. The correct inference is “not one of the named headline trio,” not “the countries never discussed quantum cooperation.” This wording distinction prevents an overbroad conclusion.

Review note: Wording note: quantum technologies are discussed in the wider release, although not in its three-item headline priority list.

Source: DST/PIB, 3 June 2026: full announcement and wider context

Answer: (C) Kathmandu University’s DPI-AI Centre

The 6 June 2026 release identifies Kathmandu University’s Centre for Digital Public Infrastructure and Artificial Intelligence as the partner. The collaboration concerns language AI, multilingual public infrastructure and digital inclusion across India and Nepal. The institutional sub-unit matters: the option names the centre within Kathmandu University, rather than merely a broad bilateral government arrangement.

Source: MeitY/PIB, 6 June 2026: BHASHINI–Kathmandu University MoU

Answer: (C) Indian National Science Academy (INSA)

The official 9 June 2026 release states that INSA convened the first meeting as part of India’s BRICS Presidency. The meeting addressed responsible AI, sustainable development and Global South cooperation. Distinguish the convening academy from the government department that published or supported the announcement. The presence of the Science and Technology Ministry’s name on a release does not make DST the convenor.

Source: DST/PIB, 9 June 2026: INSA convenes BRICS academies

Answer: (C) 90.7 million light-years

The DST report on supernova SN 2023zcu gives the host galaxy’s approximate distance as 90.7 million light-years. A light-year is a distance: the distance light travels in one year. Astronomical distance estimates can vary with measurement methods; the question refers to the rounded figure in this report. Associate NGC 2139 with SN 2023zcu and the reported 90.7-million-light-year scale.

Source: DST/PIB, 12 June 2026: NGC 2139 and SN 2023zcu

Answer: (D) Red supergiant

Type IIP supernovae are typically core-collapse explosions of massive red supergiants that retain a substantial hydrogen envelope. The P refers to a plateau in the light curve, linked to the evolving hydrogen-rich envelope after the explosion. A white dwarf is the standard association for Type Ia supernovae, not Type IIP. The source report explains this red-supergiant origin in connection with the study of SN 2023zcu.

Source: DST/PIB, 12 June 2026: Type IIP supernova explanation

Answer: (C) Avataar AI

The 12 June 2026 MeitY release identifies Avataar as Varya’s developer. It describes support through the IndiaAI Mission and a distillation approach intended to make video generation more efficient and relevant to Indian contexts. In machine learning, distillation transfers capabilities from a larger teacher model to a more compact student model. For this MCQ, retain the direct association Varya → Avataar → IndiaAI Mission support.

Source: MeitY/PIB, 12 June 2026: Varya launch

Answer: (C) Rajasthan

The 15 May 2026 MeitY release identifies Sahasra Semiconductors’ facility at Bhiwadi, Rajasthan. The question’s distinguishing phrase is “SME-led,” so it should not be confused with separate large semiconductor investments in other states. In revision, connect the state Rajasthan, the location Bhiwadi and the company Sahasra. Avoid enlarging the claim into a statement that this was India’s first semiconductor activity of any kind.

Source: MeitY/PIB, 15 May 2026: Rajasthan semiconductor plant

Answer: (B) Conservation Breeding Centre, Jaisalmer

The 13 March 2026 environment-ministry update reports two new chicks at Rajasthan’s Conservation Breeding Centre, corresponding to option B. It distinguishes one chick from natural mating and one from artificial insemination. The key distinction is a managed conservation-breeding facility rather than simply a national park or a zoo. Read the question as referring to that dated two-chick update; later programme totals do not change the event being asked about.

Source: Environment Ministry/PIB, 13 March 2026: two GIB chicks