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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

483 questions · clear filters

MathsClass 112 markseasy

Introduction to Three Dimensional Geometry

Find the distance between the points (2, 3, 5) and (4, 3, 1).

Reveal model answer + marking points

The distance between (x1, y1, z1) and (x2, y2, z2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Here = sqrt[(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2] = sqrt[4 + 0 + 16] = sqrt20 = 2 sqrt5.

d = sqrt[(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2]

Marking-scheme points

  • Distance = sqrt[(dx)^2 + (dy)^2 + (dz)^2]
  • = sqrt[4 + 0 + 16] = sqrt20
  • = 2 sqrt5 units
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MathsClass 112 markseasy

Introduction to Three Dimensional Geometry

Find the distance of the point (1, -2, 3) from the origin.

Reveal model answer + marking points

The distance of a point (x, y, z) from the origin is sqrt(x^2 + y^2 + z^2). Here = sqrt(1^2 + (-2)^2 + 3^2) = sqrt(1 + 4 + 9) = sqrt14 units.

d = sqrt(x^2 + y^2 + z^2)

Marking-scheme points

  • Distance from origin = sqrt(x^2 + y^2 + z^2)
  • = sqrt(1 + 4 + 9)
  • = sqrt14 units
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MathsClass 112 markseasy

Introduction to Three Dimensional Geometry

Find the midpoint of the line segment joining the points (2, 3, 4) and (6, 7, 8).

Reveal model answer + marking points

The midpoint of the segment joining (x1,y1,z1) and (x2,y2,z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2). = ((2+6)/2, (3+7)/2, (4+8)/2) = (4, 5, 6).

M = ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)

Marking-scheme points

  • Midpoint = average of endpoints
  • = ((2+6)/2, (3+7)/2, (4+8)/2)
  • = (4, 5, 6)
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MathsClass 112 markseasy

Limits and Derivatives

Evaluate the limit: lim as x approaches 2 of (x^2 - 4)/(x - 2).

Reveal model answer + marking points

Direct substitution gives 0/0, so factorise the numerator: x^2 - 4 = (x - 2)(x + 2). Then (x^2 - 4)/(x - 2) = x + 2 for x not equal to 2. Taking the limit as x approaches 2 gives 2 + 2 = 4.

a^2 - b^2 = (a - b)(a + b)

Marking-scheme points

  • 0/0 form -> factorise
  • (x^2 - 4)/(x - 2) = x + 2
  • Limit = 4
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MathsClass 112 markseasy

Limits and Derivatives

State and use the standard limit to evaluate lim as x approaches 0 of (sin x)/x.

Reveal model answer + marking points

The standard limit is lim as x approaches 0 of (sin x)/x = 1 (with x in radians). This is a fundamental result used throughout limits and derivatives of trigonometric functions.

lim x->0 (sin x)/x = 1

Marking-scheme points

  • Standard result (x in radians)
  • lim x->0 (sin x)/x = 1
  • Used for trigonometric derivatives
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MathsClass 112 marksmedium

Limits and Derivatives

Find the derivative of f(x) = x^2 from first principles.

Reveal model answer + marking points

By definition f'(x) = lim as h approaches 0 of [f(x + h) - f(x)]/h = lim [ (x + h)^2 - x^2 ]/h = lim [x^2 + 2xh + h^2 - x^2]/h = lim [2xh + h^2]/h = lim (2x + h) = 2x. So f'(x) = 2x.

f'(x) = lim h->0 [f(x+h) - f(x)]/h

Marking-scheme points

  • f'(x) = lim h->0 [f(x+h) - f(x)]/h
  • (x+h)^2 - x^2 = 2xh + h^2
  • Limit = 2x
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MathsClass 112 markseasy

Limits and Derivatives

Differentiate y = x^3 + 3x^2 - 5x + 2 with respect to x.

Reveal model answer + marking points

Using the power rule d/dx(x^n) = n x^(n-1) term by term: dy/dx = 3x^2 + 3(2x) - 5 + 0 = 3x^2 + 6x - 5.

d/dx(x^n) = n x^(n-1)

Marking-scheme points

  • Apply power rule to each term
  • Derivative of constant = 0
  • dy/dx = 3x^2 + 6x - 5
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MathsClass 112 marksmedium

Limits and Derivatives

Find the derivative of y = sin x cos x.

Reveal model answer + marking points

Using the product rule with u = sin x (u' = cos x) and v = cos x (v' = -sin x): dy/dx = cos x cos x + sin x (-sin x) = cos^2 x - sin^2 x = cos 2x. (Alternatively, y = (1/2) sin 2x, whose derivative is cos 2x.)

cos^2 x - sin^2 x = cos 2x

Marking-scheme points

  • Product rule: cos x cos x + sin x(-sin x)
  • = cos^2 x - sin^2 x
  • = cos 2x
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MathsClass 112 markseasy

Statistics

Find the mean of the first 10 natural numbers.

Reveal model answer + marking points

The first 10 natural numbers are 1, 2, ..., 10. Their sum = n(n + 1)/2 = 10 x 11/2 = 55. Mean = sum/number of observations = 55/10 = 5.5.

mean = (sum of observations)/n

Marking-scheme points

  • Sum of first 10 natural numbers = 55
  • Mean = sum/n
  • = 55/10 = 5.5
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MathsClass 112 markseasy

Statistics

Define variance and standard deviation of a set of observations.

Reveal model answer + marking points

Variance is the arithmetic mean of the squares of the deviations of the observations from their mean; for n observations, variance = (1/n) sum of (xi - mean)^2. Standard deviation is the positive square root of the variance; it has the same units as the observations and measures the spread or dispersion of the data about the mean.

SD = sqrt(variance)

Marking-scheme points

  • Variance = mean of squared deviations from the mean
  • SD = positive square root of variance
  • Both measure dispersion; SD has same units as data
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MathsClass 112 markseasy

Probability

A fair die is thrown once. Find the probability of getting a prime number.

Reveal model answer + marking points

The sample space is {1, 2, 3, 4, 5, 6}, so total outcomes = 6. The prime numbers on a die are 2, 3 and 5, which is 3 favourable outcomes. Probability = favourable/total = 3/6 = 1/2.

P(E) = favourable outcomes / total outcomes

Marking-scheme points

  • Total outcomes = 6
  • Primes on die: 2, 3, 5 (3 outcomes)
  • P = 3/6 = 1/2
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BiologyClass 112 markseasy

The Living World

What is taxonomy? Define the terms taxon and taxonomic hierarchy.

Reveal model answer + marking points

Taxonomy is the branch of biology dealing with the identification, nomenclature and classification of organisms based on their characteristics. A taxon is a unit or group of organisms of any rank in classification (for example, mammals, insects, dogs). Taxonomic hierarchy is the arrangement of taxonomic categories in a definite descending order, from kingdom down to species.

Marking-scheme points

  • Taxonomy = identification, nomenclature and classification
  • Taxon = a group/rank in classification (e.g. mammals)
  • Taxonomic hierarchy = ordered ranks from kingdom to species
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BiologyClass 112 markseasy

The Living World

What is binomial nomenclature? State two rules of it.

Reveal model answer + marking points

Binomial nomenclature is the system of naming organisms with two words, given by Carolus Linnaeus, where the first word is the genus and the second is the species (for example, Mangifera indica for mango). Rules: (1) the generic name begins with a capital letter and the specific name with a small letter; (2) the biological name is printed in italics (or underlined separately when handwritten) and is usually in Latin.

Marking-scheme points

  • Two-word Latin name: genus + species (Linnaeus)
  • Genus starts capital, species small letter
  • Name written in italics or underlined; e.g. Mangifera indica
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BiologyClass 112 markseasy

Biological Classification

State the main characteristics of Kingdom Monera.

Reveal model answer + marking points

Kingdom Monera includes bacteria, which are prokaryotic (they lack a true membrane-bound nucleus and membrane-bound organelles). They are microscopic, unicellular organisms with a cell wall (except mycoplasma). Their nutrition may be autotrophic (photosynthetic or chemosynthetic) or heterotrophic (saprophytic or parasitic). They are grouped as Archaebacteria (live in extreme habitats) and Eubacteria (true bacteria).

Marking-scheme points

  • Prokaryotic, unicellular, microscopic (bacteria)
  • Cell wall present (except mycoplasma)
  • Nutrition autotrophic or heterotrophic; Archaebacteria and Eubacteria
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BiologyClass 112 marksmedium

Biological Classification

State the characteristics of Kingdom Protista with examples.

Reveal model answer + marking points

Kingdom Protista includes single-celled eukaryotes. They have a well-defined nucleus and membrane-bound organelles. Some have flagella or cilia for locomotion, and they may be autotrophic or heterotrophic. This kingdom forms a link between plants, animals and fungi. Examples: Chrysophytes (diatoms), dinoflagellates, Euglenoids (Euglena), slime moulds and protozoans (Amoeba, Paramecium).

Marking-scheme points

  • Unicellular eukaryotes with true nucleus and organelles
  • Locomotion by cilia/flagella; autotrophic or heterotrophic
  • Examples: diatoms, dinoflagellates, Euglena, Amoeba
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BiologyClass 112 marksmedium

Biological Classification

What are viruses? Briefly define viroids and lichens.

Reveal model answer + marking points

Viruses are non-cellular infectious agents that are inert outside the host but reproduce inside a living host cell. A virus has a protein coat called a capsid enclosing genetic material of either DNA or RNA (never both). Viroids are the smallest infectious agents, made of free naked RNA without a protein coat (e.g. potato spindle tuber disease). Lichens are symbiotic associations between an alga (phycobiont) and a fungus (mycobiont) and are good indicators of air pollution.

Marking-scheme points

  • Virus: non-cellular, protein capsid + DNA or RNA (not both)
  • Viroid: smallest agent, free naked RNA, no protein coat
  • Lichen: symbiosis of alga + fungus; pollution indicator
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BiologyClass 112 marksmedium

Plant Kingdom

Why are bryophytes called the amphibians of the plant kingdom? Give two examples.

Reveal model answer + marking points

Bryophytes are called the amphibians of the plant kingdom because they can live in soil but need water for sexual reproduction, as the male gametes (antherozoids) require a film of water to swim to the female sex organ. They lack true vascular tissue and true roots (they have rhizoids). Examples: Funaria (moss), Marchantia and Riccia (liverworts).

Marking-scheme points

  • Live on land but need water for sexual reproduction
  • Male gametes swim in water to reach the egg
  • Examples: Funaria (moss), Marchantia (liverwort)
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BiologyClass 112 marksmedium

Plant Kingdom

State the main characteristics of pteridophytes with examples.

Reveal model answer + marking points

Pteridophytes are the first terrestrial plants to have true vascular tissues (xylem and phloem). The plant body is a sporophyte differentiated into true root, stem and leaves. They reproduce by spores produced in sporangia, and water is required for fertilisation. Examples: Selaginella, Equisetum and ferns such as Pteris and Dryopteris.

Marking-scheme points

  • First plants with true vascular tissue (xylem, phloem)
  • Body (sporophyte) has true root, stem and leaves; reproduce by spores
  • Examples: Selaginella, Equisetum, ferns
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BiologyClass 112 marksmedium

Animal Kingdom

Distinguish between diploblastic and triploblastic animals, and define a coelomate animal.

Reveal model answer + marking points

Diploblastic animals have cells arranged in two embryonic germ layers, an outer ectoderm and an inner endoderm (e.g. coelenterates). Triploblastic animals have three germ layers: ectoderm, mesoderm and endoderm (e.g. from flatworms upward). A coelomate animal is one that has a true body cavity (coelom) lined by mesoderm, for example annelids, arthropods and chordates.

Marking-scheme points

  • Diploblastic: two layers (ectoderm, endoderm)
  • Triploblastic: three layers (ectoderm, mesoderm, endoderm)
  • Coelomate: true body cavity lined by mesoderm
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BiologyClass 112 marksmedium

Animal Kingdom

State two characteristics each of Phylum Porifera and Phylum Coelenterata (Cnidaria).

Reveal model answer + marking points

Porifera (sponges): (1) they are primitive multicellular animals with a cellular level of organisation and body with pores (ostia) and a canal system; (2) the body is supported by spicules and they show intracellular digestion (e.g. Sycon, Spongilla). Coelenterata (Cnidaria): (1) they are diploblastic with tissue-level organisation and radial symmetry; (2) they possess stinging cells called cnidoblasts and show two body forms, polyp and medusa (e.g. Hydra, jellyfish).

Marking-scheme points

  • Porifera: pores/canal system, spicules, intracellular digestion (Sycon)
  • Coelenterata: diploblastic, radial symmetry, cnidoblasts
  • Coelenterata show polyp and medusa forms (Hydra)
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