AimMent

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

Jump straight to your paper

Class 11 Physics · +1Class 11 Chemistry · +1Class 11 Maths · +1Class 11 Biology · +1Class 12 Physics · +2Class 12 Chemistry · +2Class 12 Maths · +2Class 12 Biology · +2
ClassBoth11 · +1 (400)12 · +2 (400)
SubjectAll (800)Physics (200)Chemistry (200)Maths (200)Biology (200)
MarksAny1-mark2-mark3-mark5-mark

400 questions · clear filters

MathsClass 113 marksmedium

Permutations and Combinations

How many 4-digit numbers can be formed using the digits 0 to 9 if no digit is repeated?

Reveal model answer + marking points▾

The thousands place cannot be 0, so it can be filled in 9 ways (digits 1 to 9). After using one digit, the hundreds place can be filled in 9 ways (including 0 now), the tens place in 8 ways and the units place in 7 ways. Total = 9 x 9 x 8 x 7 = 4536.

Marking-scheme points

  • ✓Thousands place: 9 ways (cannot be 0)
  • ✓Then 9, 8, 7 ways for remaining places
  • ✓9 x 9 x 8 x 7 = 4536
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
MathsClass 112 marksmedium

Permutations and Combinations

Find the value of r if 10Cr = 10C(r + 2).

Reveal model answer + marking points▾

We use the property that nCa = nCb implies either a = b or a + b = n. Since r is not equal to r + 2, we must have r + (r + 2) = 10, so 2r + 2 = 10, giving 2r = 8 and r = 4.

nCa = nCb => a = b or a + b = n

Marking-scheme points

  • ✓nCa = nCb => a = b or a + b = n
  • ✓Here r + (r + 2) = 10
  • ✓r = 4
Still unsure? Ask the AI tutor →
MathsClass 112 markseasy

Binomial Theorem

State the binomial theorem for (a + b)^n where n is a positive integer, and write the general term.

Reveal model answer + marking points▾

The binomial theorem states that (a + b)^n = nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + ... + nCn b^n = sum over r from 0 to n of nCr a^(n-r) b^r. The general term (the (r+1)th term) is T(r+1) = nCr a^(n-r) b^r.

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

Marking-scheme points

  • ✓(a + b)^n = sum of nCr a^(n-r) b^r, r = 0 to n
  • ✓There are (n + 1) terms
  • ✓General term T(r+1) = nCr a^(n-r) b^r
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
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 →
MathsClass 113 markshard

Binomial Theorem

Find the term independent of x in the expansion of (x^2 - 1/x)^6.

Reveal model answer + marking points▾

General term T(r+1) = 6Cr (x^2)^(6-r) (-1/x)^r = 6Cr (-1)^r x^(12 - 2r) x^(-r) = 6Cr (-1)^r x^(12 - 3r). For the term independent of x, the power of x must be zero: 12 - 3r = 0, so r = 4. The term = 6C4 (-1)^4 = 15 x 1 = 15.

set power of x to 0

Marking-scheme points

  • ✓General term = 6Cr (-1)^r x^(12 - 3r)
  • ✓Independent of x: 12 - 3r = 0 -> r = 4
  • ✓Term = 6C4 = 15
Still unsure? Ask the AI tutor →
MathsClass 112 marksmedium

Binomial Theorem

Find the middle term in the expansion of (x + 2y)^6.

Reveal model answer + marking points▾

Here n = 6 is even, so there is one middle term, the (n/2 + 1) = 4th term. T4 = 6C3 x^3 (2y)^3 = 20 x^3 x 8 y^3 = 160 x^3 y^3.

middle term = (n/2 + 1)th term when n is even

Marking-scheme points

  • ✓n = 6 even -> single middle term = 4th term
  • ✓T4 = 6C3 x^3 (2y)^3
  • ✓= 20 x 8 x^3 y^3 = 160 x^3 y^3
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Binomial Theorem

Using the binomial theorem, evaluate (sqrt2 + 1)^4 + (sqrt2 - 1)^4.

Reveal model answer + marking points▾

For (a + b)^4 + (a - b)^4, the odd-power terms cancel and we get 2[a^4 + 6 a^2 b^2 + b^4]. With a = sqrt2 and b = 1: a^4 = (sqrt2)^4 = 4, a^2 b^2 = 2 x 1 = 2 so 6 a^2 b^2 = 12, and b^4 = 1. Sum = 2[4 + 12 + 1] = 2 x 17 = 34.

(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2 b^2 + b^4]

Marking-scheme points

  • ✓(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2 b^2 + b^4]
  • ✓a = sqrt2, b = 1: 4 + 12 + 1 = 17
  • ✓Answer = 2 x 17 = 34
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Binomial Theorem

Find the coefficient of x^6 y^3 in the expansion of (x + 2y)^9.

Reveal model answer + marking points▾

General term T(r+1) = 9Cr x^(9-r) (2y)^r = 9Cr 2^r x^(9-r) y^r. For x^6 y^3 we need 9 - r = 6 and r = 3, which is consistent (r = 3). Coefficient = 9C3 x 2^3 = 84 x 8 = 672.

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

Marking-scheme points

  • ✓General term = 9Cr 2^r x^(9-r) y^r
  • ✓For x^6 y^3: r = 3
  • ✓Coefficient = 9C3 x 2^3 = 84 x 8 = 672
Still unsure? Ask the AI tutor →
MathsClass 112 markseasy

Sequences and Series

Find the 20th term of the arithmetic progression 3, 7, 11, 15, ...

Reveal model answer + marking points▾

Here the first term a = 3 and common difference d = 7 - 3 = 4. The nth term is a_n = a + (n - 1)d. So a_20 = 3 + (20 - 1) x 4 = 3 + 76 = 79.

a_n = a + (n - 1)d

Marking-scheme points

  • ✓a = 3, d = 4
  • ✓a_n = a + (n - 1)d
  • ✓a_20 = 3 + 19 x 4 = 79
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Sequences and Series

Find the sum of the first 20 terms of the arithmetic progression 3, 7, 11, 15, ...

Reveal model answer + marking points▾

Here a = 3, d = 4, n = 20. The sum of n terms is S_n = (n/2)[2a + (n - 1)d]. So S_20 = (20/2)[2 x 3 + 19 x 4] = 10[6 + 76] = 10 x 82 = 820.

S_n = (n/2)[2a + (n - 1)d]

Marking-scheme points

  • ✓a = 3, d = 4, n = 20
  • ✓S_n = (n/2)[2a + (n - 1)d]
  • ✓S_20 = 10 x 82 = 820
Still unsure? Ask the AI tutor →
MathsClass 113 markshard

Sequences and Series

The sum of three numbers in AP is 24 and their product is 440. Find the numbers.

Reveal model answer + marking points▾

Let the three numbers be a - d, a, a + d. Their sum = 3a = 24, so a = 8. Their product = (a - d)(a)(a + d) = a(a^2 - d^2) = 440. So 8(64 - d^2) = 440, giving 64 - d^2 = 55, so d^2 = 9 and d = 3 (or -3). The numbers are 5, 8, 11.

AP terms: a - d, a, a + d

Marking-scheme points

  • ✓Take terms as a - d, a, a + d; sum 3a = 24 -> a = 8
  • ✓Product 8(64 - d^2) = 440 -> d^2 = 9
  • ✓Numbers are 5, 8, 11
Still unsure? Ask the AI tutor →
MathsClass 112 markseasy

Sequences and Series

Find the 8th term of the geometric progression 2, 6, 18, 54, ...

Reveal model answer + marking points▾

Here the first term a = 2 and common ratio r = 6/2 = 3. The nth term of a GP is a_n = a r^(n-1). So a_8 = 2 x 3^7 = 2 x 2187 = 4374.

a_n = a r^(n-1)

Marking-scheme points

  • ✓a = 2, r = 3
  • ✓a_n = a r^(n-1)
  • ✓a_8 = 2 x 3^7 = 4374
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Sequences and Series

Find the sum of the first 6 terms of the geometric progression 5, 15, 45, ...

Reveal model answer + marking points▾

Here a = 5, r = 3, n = 6. Since r > 1, S_n = a(r^n - 1)/(r - 1). So S_6 = 5(3^6 - 1)/(3 - 1) = 5(729 - 1)/2 = 5 x 728/2 = 5 x 364 = 1820.

S_n = a(r^n - 1)/(r - 1), r > 1

Marking-scheme points

  • ✓a = 5, r = 3, n = 6
  • ✓S_n = a(r^n - 1)/(r - 1)
  • ✓S_6 = 5 x 728/2 = 1820
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Sequences and Series

Find the sum to infinity of the geometric progression 1 + 1/3 + 1/9 + 1/27 + ...

Reveal model answer + marking points▾

Here a = 1 and r = 1/3. Since |r| < 1, the sum to infinity exists and is S_infinity = a/(1 - r) = 1/(1 - 1/3) = 1/(2/3) = 3/2.

S_infinity = a/(1 - r), |r| < 1

Marking-scheme points

  • ✓a = 1, r = 1/3, |r| < 1 so sum exists
  • ✓S_infinity = a/(1 - r)
  • ✓= 1/(2/3) = 3/2
Still unsure? Ask the AI tutor →
MathsClass 112 markseasy

Sequences and Series

Find the arithmetic mean (AM) and geometric mean (GM) of 4 and 16, and verify AM >= GM.

Reveal model answer + marking points▾

AM = (4 + 16)/2 = 20/2 = 10. GM = sqrt(4 x 16) = sqrt64 = 8. Since 10 > 8, we have AM > GM, which agrees with the general result AM >= GM for positive numbers.

AM = (a + b)/2; GM = sqrt(ab)

Marking-scheme points

  • ✓AM = (a + b)/2 = 10
  • ✓GM = sqrt(ab) = 8
  • ✓AM (10) >= GM (8) verified
Still unsure? Ask the AI tutor →
MathsClass 113 marksmedium

Sequences and Series

Find the sum of the squares of the first 10 natural numbers, that is 1^2 + 2^2 + ... + 10^2.

Reveal model answer + marking points▾

The sum of squares of the first n natural numbers is given by n(n + 1)(2n + 1)/6. For n = 10: sum = 10 x 11 x 21 / 6 = 2310/6 = 385.

sum of squares = n(n + 1)(2n + 1)/6

Marking-scheme points

  • ✓Formula: sum = n(n + 1)(2n + 1)/6
  • ✓n = 10: 10 x 11 x 21
  • ✓= 2310/6 = 385
Still unsure? Ask the AI tutor →
MathsClass 115 markshard

Sequences and Series

Find the sum of the series 1 x 2 + 2 x 3 + 3 x 4 + ... up to n terms.

Reveal model answer + marking points▾

The rth term is a_r = r(r + 1) = r^2 + r. So the sum S = sum over r of (r^2 + r) = sum of r^2 + sum of r. Using sum of r^2 = n(n + 1)(2n + 1)/6 and sum of r = n(n + 1)/2: S = n(n + 1)(2n + 1)/6 + n(n + 1)/2. Take n(n + 1)/6 common: S = [n(n + 1)/6][(2n + 1) + 3] = [n(n + 1)/6](2n + 4) = [n(n + 1) x 2(n + 2)]/6 = n(n + 1)(n + 2)/3.

S = n(n + 1)(n + 2)/3

Marking-scheme points

  • ✓rth term = r(r + 1) = r^2 + r
  • ✓Sum = sum of r^2 + sum of r
  • ✓= n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3
Still unsure? Ask the AI tutor →
MathsClass 112 markseasy

Straight Lines

Find the slope of the line passing through the points (2, 3) and (5, 9).

Reveal model answer + marking points▾

The slope of a line through (x1, y1) and (x2, y2) is m = (y2 - y1)/(x2 - x1). Here m = (9 - 3)/(5 - 2) = 6/3 = 2.

m = (y2 - y1)/(x2 - x1)

Marking-scheme points

  • ✓m = (y2 - y1)/(x2 - x1)
  • ✓= (9 - 3)/(5 - 2)
  • ✓m = 2
Still unsure? Ask the AI tutor →
← PrevPage 13 of 20Next →

You are more ready than you feel.

One question at a time is how every topper started. Bookmark this, revise a few each day, and watch the fear shrink. And if a friend is stressing about boards — send this their way. You both win.

Share with a friendAsk the AI tutor