Board Boosters

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 questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

98 questions · clear filters

MathsClass 113 marksmedium

Permutations and Combinations

From 7 men and 5 women, in how many ways can a committee of 5 members be formed consisting of exactly 3 men and 2 women?

Reveal model answer + marking points▾

Choose 3 men from 7 in 7C3 ways and 2 women from 5 in 5C2 ways. 7C3 = (7 x 6 x 5)/(3 x 2 x 1) = 35 and 5C2 = (5 x 4)/(2 x 1) = 10. By the multiplication principle, total ways = 35 x 10 = 350.

total = 7C3 x 5C2

Marking-scheme points

  • ✓Choose men: 7C3 = 35
  • ✓Choose women: 5C2 = 10
  • ✓Total = 35 x 10 = 350
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MathsClass 113 marksmedium

Binomial Theorem

Expand (2x - 3)^4 using the binomial theorem.

Reveal model answer + marking points▾

(2x - 3)^4 = 4C0 (2x)^4 + 4C1 (2x)^3(-3) + 4C2 (2x)^2(-3)^2 + 4C3 (2x)(-3)^3 + 4C4 (-3)^4 = 16x^4 + 4(8x^3)(-3) + 6(4x^2)(9) + 4(2x)(-27) + 81 = 16x^4 - 96x^3 + 216x^2 - 216x + 81.

(a + b)^4 expansion

Marking-scheme points

  • ✓Use T(r+1) = 4Cr (2x)^(4-r)(-3)^r
  • ✓Compute each of the 5 terms
  • ✓= 16x^4 - 96x^3 + 216x^2 - 216x + 81
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MathsClass 113 marksmedium

Binomial Theorem

Find the coefficient of x^5 in the expansion of (x + 3)^8.

Reveal model answer + marking points▾

The general term is T(r+1) = 8Cr x^(8-r) 3^r. For the term in x^5, we need 8 - r = 5, so r = 3. The coefficient = 8C3 x 3^3 = 56 x 27 = 1512.

T(r+1) = nCr a^(n-r) b^r

Marking-scheme points

  • ✓General term = 8Cr x^(8-r) 3^r
  • ✓For x^5: 8 - r = 5 -> r = 3
  • ✓Coefficient = 8C3 x 3^3 = 56 x 27 = 1512
Still unsure? Ask the AI tutor →
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