Updated On : September 15, 2023

Class 11 marks the start of a new cornerstone in a student’s life. It is the result of your class 11 that plays a vital role in motivating you for the next grade – class 12. If you have a good result, then you feel more confident about facing the 12th Boards. The leap between classes 10 and 11 is massive, and as such, adequate preparation must be done to ensure that you score well.

One of the main subjects that deserve special attention is mathematics. If you are eager to establish a career in any of the science fields, it is without a doubt that you have to at least have a basic understanding of the subject. The **11****th**** Maths Syllabus** is designed in a way to aid the students in understanding the topics better in class 12.

It is unwise to ignore Class 11 by thinking that questions wouldn't come from here on the boards. However, it is only by making your foundations stronger, can you hope to excel in the 12th Boards.

You can adopt certain preparation tips to ensure that your integration into class 11 is smooth. Also, this way you can start the preparation for the boards as well. Check out the preparation tips, which will make class 12 easier for you.

- Focus and determination: The feeling of achievement when you pass the first ever board exam in your life is irreplaceable. However, it is important to remember that the ICSE and ISC boards differ a lot. It is vital that you remind yourself that the next board is much tougher than the previous one, and hence, extreme determination, focus, and perseverance are essential. If you can keep yourself motivated, then you will find that you have passed the 12th boards with flying colors.
- Solve the sample papers: The best way to prepare for the 12th boards is to practice the ISC Class 11 Maths Sample Papers repeatedly. When you start solving them regularly, you will find that it acts as a revision. You will also be able to manage your time and increase your speed. Also, the more you practice, the more you are going to score in the upcoming examinations.
- Self-study: Apart from solving the ISC Class 11 Maths Sample Papers, you also need to allocate sufficient time for self-studying. It is vital that you analyze the knowledge that you have gained from school and private coaching. Only by processing them on your own, can you assess if you have any doubts or not. If you do have, you can clarify them in the next class to ensure that you have a strong foundation.

**Check Out - ISC Class 11 Maths Syllabus**

**ISC Class 11 Maths Sample Papers** are provided to offer aid to the students in preparing for the upcoming boards. You can check out the sample papers of different years given below.

Year | Question Papers |

2019 | Download PDF |

2020 | Download PDF |

2021 | Download PDF |

**Question 1 [10x2] **

(i) Let π: π → π be a function defined by (π₯) = π₯−π π₯−π , where π ≠ π. Then show that π is one-one but not onto.

(ii) Find the domain and range of the function (π₯) = [ππππ₯].

(iii) Find the square root of complex number 11 − 60π.

(iv) For what value of π will the equations π₯ 2 − ππ₯ − 21 = 0 πππ π₯ 2 − 3ππ₯ + 35 = 0 have one common root.

(v) In a βπ΄π΅πΆ, show that ∑(π + π) cos π΄ = 2π where π = π+π+π 2

(vi) Find the number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together.

(vii) Prove that sin 20° sin 40° sin 80° = √3 8

(viii) If two dice are thrown simultaneously, find the probability of getting a sum of 7 or 11.

(ix) Show that limπ₯→2 |π₯−2| π₯−2 does not exist.

(x) Find the point on the curve π¦ 2 = 4π₯, the tangent at which is parallel to the straight line π¦ = 2π₯ +4.

Question 2 [4]

Draw the graph of the function π¦ = |π₯ − 2| + |π₯ − 3|.

Question 3 [4]

Prove that cot π΄ + cot(60 + π΄) + cot(120 + π΄) = 3 cot 3π΄.

OR

In a βπ΄π΅πΆ prove that π cos πΆ + π cosπ΅ = π

Question 4 [4] Find the locus of a complex number, Z= x+iy, satisfying the relation | π§−3π π§+3π | ≤ √2. Illustrate the locus of Z in the organd plane.

Question 5 [4]

Find the number of words which can be formed by taking four letters at a time from the word “ COMBINATION”.

OR

A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls

(ii) at least 3 girls and

(iii) atmost three girls.

Question 6 [4]

Prove by the method of induction. 1 1.2 + 1 2.3 + 1 3.4 + … … … … … … . π’ππ‘π π π‘ππππ = π π + 1 π€βπππ π ∈ π.

Question 7 [4]

Find the term independent of π₯ and its value in the expansion of (√ π₯ 3 − √3 2 ) 12 .

OR

Find the sum of the terms of the binomial expansion to infinity: 1 + 2 4 + 2.5 4.8 + 2.5.8 4.8.12 + β― … … .π‘π ∞.

Question 8 [4]

Differentiate from first principle: (π₯) = √3π₯ + 4.

Question 9 [4]

Reduce the equation π₯ + π¦ + √2 = 0 to the normal form ( +π¦ ππππΌ = π) and find the values of π πππ πΌ.

Question 10 [4]

Write the equation of the circle having radius 5 and tangent as the line 3π₯ − 4π¦ + 5 = 0 at (1, 2).

Question 11 [6]

In a βπ΄π΅πΆ prove that cot π΄ + cot π΅ + cot πΆ = π 2+π 2+π 2 4β

Question 12 [6]

Find the nth term and deduce the sum to n terms of the series: 4 + 11 + 22 + 37 + 56 + β― … …

OR

If (p+q)th term and (p-q)th terms of G.P are a and b respectively, prove that pth term is √ππ.

Question 13 [6]

If π₯ is real, prove that the value of the expression (π₯−1)(π₯+3) (π₯−2)(π₯+4) cannot be between 4/9 and 1.

OR

If π₯ π occurs in the expansion of (π₯ 2 + 1 ) 2π , prove that its coefficient is (2π)! [ 1 3 (4π−π)!] [ 1 3 (2π+π)!] .

Question 14 [6]

Calculate the standard deviation of the following distribution:

Age | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 |

No. of persons | 170 | 110 | 80 | 45 | 40 | 35 |

Question 15

(a) Find the focus and directrix of the conic represented by the equation 5π₯ 2 = −12π¦. [2]

(b) Construct the truth table (~ π β ∼ π)β(π β ∼ π). [2]

(c) Write the converse, contradiction and contrapositive of statement “ If π₯ + 3 = 9,π‘βππ π₯ = 6.” [2]

Question 16 [4]

Show that the point (1, 2, 3) is common to the lines which join A(4, 8, 12) to B(2, 4, 6) and C(3, 5, 4) to D(5, 8, 5).

OR

Calculate the Cosine of the angle A of the triangle with vertices A(1, -1, 2) B ( 6, 11, 2) and C( 1, 2, 6).

Question 17 [4]

Find the equation of the hyperbola whose focus is (1, 1), the corresponding directrix 2π₯ + π¦ − 1 = 0 πππ π = √3.

OR

Find the equation of tangents to the ellipse 4π₯ 2 + 5π¦ 2 = 20 which are perpendicular to the line 3π₯ + 2π¦ − 5 = 0.

Question 18 [6]

Show that the equation 16π₯ 2 − 3π¦ 2 − 32π₯ − 12π¦ − 44 = 0 represents a hyperbola. Find the lengths of axes and eccentricity.

SECTION C (20 Marks)

Question 19

(i) Two sample sizes of 50 and 100 are given .The mean of these samples respectively are 56 and 50 Find the mean of size 150 by combining the two samples [2]

(ii) Calculate π95, for the following data : [4]

Marks | 0-10 | 10- 20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 3 | 7 | 11 | 12 | 23 | 4 |

OR

Calculate Mode for the following data: [4]

C.I. | 17-19 | 14-16 | 11-13 | 8-10 | 5-7 | 2-4 |

Frequency | 12 | 4 | 8 | 16 | 11 | 4 |

Question 20

(i) Find the covariance between X and Y when ∑ π = 50, ∑ π = −30, and ∑ ππ = 115. [2]

(ii) Calculate Spearman’s Rank Correlation for the following data and interpret the result: [4]

Marks in Mathematics | 36 | 48 | 27 | 36 | 29 | 30 | 36 | 39 | 42 | 48 |

Marks in Statistics | 27 | 45 | 24 | 27 | 31 | 33 | 35 | 45 | 41 | 45 |

OR

Find Karl Pearson’s Correlation Coefficient from the given data: [4]

x | 21 | 24 | 26 | 29 | 32 | 43 | 25 | 30 | 35 | 37 |

y | 120 | 123 | 125 | 128 | 131 | 142 | 124 | 129 | 134 | 136 |

Question 21

Find the consumer price index for 2007 on the basis of 2005 from the following data using weighted average of price relative method: [4]

Items | Food | Rent | Cloth | Fuel |

Price in 2005(Rs) | 200 | 100 | 150 | 50 |

Price in 2007(Rs) | 280 | 200 | 120 | 100 |

Weighted | 30 | 20 | 20 | 10 |

Question 22

Using the following data. Find out the trend using Quaterly moving average and plot them on graph: [4]

Year/ Quarter | Q1 | Q2 | Q3 | Q4 |

1994 | 29 | 37 | 43 | 34 |

1995 | 90 | 42 | 55 | 43 |

1996 | 47 | 51 | 63 | 53 |

You can check out the **ISC Class 11 Maths Sample Papers** PDF given here. Take a look at the table below which provides the PDF file of sample papers of different years.

SL No. | Sample Papers | |

1 | ISC Class 11 Maths Sample Paper 2020-21 | |

2 | ISC Class 11 Maths Sample Paper 2019 | |

3 | ISC Class 11 Maths Sample Paper 2018 |

The structure of ISC class 11 is done in a way that the "Algebra" section holds the maximum marks. It comprises of 34 marks. Under this section, there are six chapters. Important questions along with answers to these six chapters are provided here. Apart from the **ISC Class 11 Maths Sample Papers**, you can also check it out to enhance yourself with more knowledge before appearing for the examinations.

Chapter: Principle of Mathematical Induction

Question 1

Using the principle of mathematical induction, prove that

1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1} for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}.

Putting n =1 in the given statement, we get

LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}.

Now, 1² + 2² + 3² + ......... + k² + (k + 1)²

= (1/6) {k(k + 1)(2k + 1) + (k + 1)²

= (1/6){(k + 1).(k(2k + 1)+6(k + 1))}

= (1/6){(k + 1)(2k² + 7k + 6})

= (1/6){(k + 1)(k + 2)(2k + 3)}

= 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)²

= (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 2:

Prove that 1 + 3 + 5 + … + (2n – 1) = n2 using the principle of Mathematical induction.

Solution:

Given Statement: 1 + 3 + 5 + … + (2n – 1) = n2

Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n2 , for n ∈ N

Note that P(1) is true, since

P(1) : 1 = 12

Let P(k) is true for some k ∈ N,

It means that,

P(k) : 1 + 3 + 5 + … + (2k – 1) = k2

To prove that P(k + 1) is true, we have

1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k2 + (2k + 1)

= k2 + 2k + 1

By using the formula, the above form can be written as:

= (k + 1)2

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

Question 3

Prove that 2n > n for all positive integers n by the Principle of Mathematical Induction

Solution:

Assume that P(n): 2n > n

If n =1, 21>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

2k > k …(1)

We shall now prove that P(k +1) is true whenever P(k) is true.

Now, multiplying both sides of the equation (1) by 2, we get

- 2k > 2k

Now by using the property,

i.e., 2k+1> 2k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Chapter: Complex Numbers

Question 1:

Express the given complex number (-3) in the polar form.

Solution:

Given complex number is -3

Let r cos θ = -3 …(1)

and rsinθ = 0 …(2)

Add both the terms:

rcosθ +rsinθ = -3+0

rcosθ +rsinθ = -3

Take square on both the sides, we get

r2cos2θ +r2sin2θ = (-3)2

Take r2 outside from L.H.S, we get

r2(cos2θ+sin2θ) = 9

We know that, cos2θ+sin2θ = 1, then the above equation becomes

r2 = 9

r = 3 (Conventionally, r > 0)

Now, subsbtitute the value of r in (1) and (2)

3 cos θ = -3 and 3 sinθ = 0

Cos θ = -1 and sinθ = 0

Therefore, θ = π

Hence, the polar representation is,

-3 = rcos θ + i rsin θ

3cosπ+ 3sin π = 3(cosπ+isin π)

Thus the required polar form is 3cosπ+ i3sin π = 3(cosπ+isin π)

Question 2:

Find the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]

Solution:

Given: [(1+i)/(1-i)] – [(1-i)/(1+i)]

Simplify the given expression, we get:

[(1+i)/(1-i)] – [(1-i)/(1+i)] = [(1+i)2– (1-i)2]/ [(1+i)(1-i)]

= (1+i2+2i-1-i2+2i)) / (12+12)

Now, cancel out the terms,

= 4i/2

= 2i

Now, take the modulus,

| [(1+i)/(1-i)] – [(1-i)/(1+i)]| =|2i| = √22 = 2

Therefore, the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)] is 2.

Question 3:

For any two complex numbers z1 and z2, show that Re(z1z2) = Rez1 Rez2– Imz1Imz2

Solution:

Given: z1 and z2 are the two complex numbers

To prove: Re(z1z2) = Rez1 Rez2– Imz1Imz2

Let z1 = x1+iy1 and z2 = x2+iy2

Now, z1 z2 =(x1+iy1)(x2+iy2)

Now, split the real part and the imaginary part from the above equation:

⇒ x1(x2+iy2) +iy1(x2+iy2)

Now, multiply the terms:

= x1x2+ix1y2+ix2y1+i2y1y2

We know that, i2 = -1, then we get

= x1x2+ix1y2+ix2y1-y1y2

Now, again seperate the real and the imaginary part:

= (x1x2 -y1y2) +i (x1y2+x2y1)

From the above equation, take only real part:

⇒ Re (z1z2) =(x1x2 -y1y2)

It means that,

⇒ Re(z1z2) = Rez1 Rez2– Imz1Imz2

Hence, the given statement is proved.

Chapter: Quadratic Equations

Question 1

Solve the quadratic equation 2x2+ x – 300 = 0 using factorisation.

Solution: 2x2+ x – 300 = 0

2x2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

So,

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of given equation.

Question 2

Solve the quadratic equation 2x2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax2+bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

x=−b±b2−4ac√2a

Now putting the values of a,b and c.

x=−1±1+4(2)(528)√4=−1±4225√4=−1±654

x=64/4 or x=-66/4

x=16 or x=-33/2

Question 3

Find the discriminant of the equation: 3x2-2x+β = 0.

Solution: Here, a = 3, b=-2 and c=β

Hence, discriminant, D = b2 – 4ac

D = (-2)2-4.3.(β )

D = 4-4

D=0

Chapter: Permutations and Combinations

Question 1

Evaluate the following

(i) 6 ! (ii) 5 ! – 2 !

Solution:

(i) 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720

(ii) 5! = 1 × 2 × 3 × 4 x 5 = 120

As 2! = 1 × 2 = 2

Question 2

How many words can be formed using all the letters of the word Dinomite, using each letter exactly one time?

Solution:

As there are 7 different letters in the word – Dinomite. So the number of different words formed using these 7 letters will be 8p7 = 8! / (8 -7)! = 8

Question 3

It is needed to seat 5 boys and 4 girls in a row so that the girl gets the even places. How many such arrangements are possible?

Solution:

5 boys and 4 girls are to be seated in a row so that the girl gets the even places.

The 5 boys can be seated in 5! Ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! Ways.

Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Chapter: Binomial Theorem

Question 1

Evaluate (101)4 using the binomial theorem

Solution:

Given: (101)4.

Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.

Therefore, 101 = 100+1

Hence, (101)4 = (100+1)4

Now, by applying the binomial theorem, we get:

(101)4 = (100+1)4 = 4C0(100)4 +4C1 (100)3(1) + 4C2(100)2(1)2 +4C3(100)(1)3 +4C4(1)4

(101)4 = (100)4+4(100)3+6(100)2+4(100) + (1)4

(101)4 = 100000000+ 4000000+ 60000+ 400+1

(101)4 = 104060401

Hence, the value of (101)4 is 104060401.

Question 2

Find the value of r, If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal.

Solution:

For the given condition, the coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)34 are 34Cr-6 and 34C2r-2 respectively.

Since the given terms in the expansion are equal,

34Cr-6 = 34C2r-2

From this, we can write it as either

r-6=2r-2

(or)

r-6=34 -(2r-2) [We know that, if nCr = nCp , then either r = p or r = n – p]

So, we get either r = – 4 or r = 14.

We know that r being a natural number, the value of r = – 4 is not possible.

Hence, the value of r is14.

Question 3

Expand the expression (2x-3)6 using the binomial theorem.

Solution:

Given Expression: (2x-3)6

By using the binomial theorem, the expression (2x-3)6 can be expanded as follows:

(2x-3)6 = 6C0(2x)6 –6C1(2x)5(3) + 6C2(2x)4(3)2 – 6C3(2x)3(3)3 + 6C4(2x)2(3)4 – 6C5(2x)(3)5 + 6C6(3)6

(2x-3)6 = 64x6 – 6(32x5 )(3) +15(16x4 )(9) – 20(8x3 )(27) +15(4x2 )(81) – 6(2x)(243) + 729

(2x-3)6 = 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

Thus, the binomial expansion for the given expression (2x-3)6 is 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.

Chapter: Sequence and Series

Question 1

Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Solution:

Assume that A1, A2, A3, A4, and A5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A1, A2, A3, A4, A5, 26

Here, a= 8, b =26,n= 7

Therefore, 26= 8+(7-1)d

Hence it becomes,

26 = 8+6d

6d = 26-8 = 18

6d= 18

d = 3

A1= a+d = 8+ 3 =11

A2= a+2d = 8+ 2(3) =8+6 = 14

A3= a+3d = 8+ 3(3) =8+9 = 17

A4= a+4d = 8+ 4(3) =8+12 = 20

A5= a+5d = 8+ 5(3) =8+15 = 23

Hence, the required five numbers between the number 8 and 26 are11, 14, 17, 20, 23

Question 2

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Solution:

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the kth term of an A.P. is given by

ak = a +(k -1)d

Therefore, am+n = a +(m+n -1)d

am-n = a +(m-n -1)d

am = a +(m-1)d

Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:

am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

= 2 [a + (m-1)d]

= 2 am {since am = a +(m-1)d}

Therefore, the sum of (m + n)thand (m – n)th terms of an A.P. is equal to twice the mth term.

Hence, proved.

Question 3

The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms

Solution:

Let a1, a2 and d1, d2 be the first terms and the common difference of the first and second

arithmetic progression respectively.

Then,

(Sum of n terms of the first A.P)/(Sum of n terms of the second A.P) = (5n+4)/(9n+6)

⇒ [ (n/2)[2a1+ (n-1)d1]]/ [(n/2)[2a2+ (n-1)d2]]= (5n+4)/(9n+6)

Cancel out (n/2) both numerator and denominator on L.H.S

⇒[2a1+ (n-1)d1]/[2a2+ (n-1)d2]= (5n+4)/(9n+6) …(1)

Now susbtitute n= 35 in equation (1), {Since (n-1)/2 = 17}

Then equation (1) becomes

⇒[2a1+ 34d1]/[2a2+ 34d2]= (5(35)+4)/(9(35+6)

⇒[a1+ 17d1]/[a2+ 17d2]= 179/321 …(2)

Now, we can say that.

18th term of first AP/ 18th term of second AP = [a1+ 17d1]/[a2+ 17d2]….(3)

Now, from (1) and (2), we can say that,

18th term of first AP/ 18th term of second AP = 179/321

Hence, the ratio of 18th term of both the AP are 179:321

ISC Class 11 Maths Sample Papers are essential for the examinations that lay ahead. However, it is normal for students to have doubts regarding the ISC Class 11 Maths Sample Papers. The most common one is – are they really helpful? Well, sample papers are vital and can aid in preparing the students for the boards as well as any other examinations.

Take a look at the reasons as to why practising ISC Class 11 Maths Sample Papers are so important.

- Acts as revision: Once you have completed the entire syllabus, you can solve the ISC Class 11 Maths Sample Papers as revision. In this way, you get to enhance your knowledge while practicing more problems. The more problems you solve, the more will it become easier for you. Also, by solving them, you can assess the areas that are your weak points. You can then work on improving them.
- Eliminates exam fear: Keep in mind that the ISC Class 11 Maths Sample Papers are a reflection of the actual examination papers. By solving then, you gain an insight as to how the paper will be. As such, you can lay your worries and fears to rest. After solving these papers repeatedly, you will find that you have nothing to fear anymore. Moreover, you are prepared as much as it is possible for the examination.
- Enhances speed: While solving numerous ISC Class 11 Maths Sample Papers, you will realize that your speed has increased. It is because you are now adept at finding the solutions than before. Also, solving them is the perfect way to ensure that you complete them within a particular timeframe. It is vital that you learn time management simultaneously. With time, you will also find that the number of mistakes you make has dwindled.
- Aids in analyzing question pattern: Practicing the ISC Class 11 Maths Sample Papers on a daily basis will aid you in understanding the question format and pattern. After some days, you will be able to identify which topics are given more preference than the others. You will also understand which ones need more work on your part than others.
- Helps to assess the marks allocation: One of the most beneficial aspects of solving the ISC Class 11 Maths Sample Papers is the assessment of the marks allocation. After solving an adequate number of papers, you will realize which chapters hold more weightage than the others. It also means that these are the chapters that you must focus on the most.
- Increases self-confidence: The more you keep on solving the ISC Class 11 Maths Sample Papers, the more will your self-confidence grow. You will not only identify the areas that need work, but you will also find the sections that you are good at. You can focus on them to ensure that all the questions from these sections are attempted and solved correctly.

Frequently Asked Questions

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September 15, 2023

Class 11 marks the start of a new cornerstone in a student’s life. It is the result of your class 11 that plays a vital role in motivating you for the next grade – class 12. If you have a good result, then you feel more confident about facing the 12th Boards. The leap between classes 10 and 11 is massive, and as such, adequate preparation must be done to ensure that you score well.

One of the main subjects that deserve special attention is mathematics. If you are eager to establish a career in any of the science fields, it is without a doubt that you have to at least have a basic understanding of the subject. The **11****th**** Maths Syllabus** is designed in a way to aid the students in understanding the topics better in class 12.

It is unwise to ignore Class 11 by thinking that questions wouldn't come from here on the boards. However, it is only by making your foundations stronger, can you hope to excel in the 12th Boards.

You can adopt certain preparation tips to ensure that your integration into class 11 is smooth. Also, this way you can start the preparation for the boards as well. Check out the preparation tips, which will make class 12 easier for you.

- Focus and determination: The feeling of achievement when you pass the first ever board exam in your life is irreplaceable. However, it is important to remember that the ICSE and ISC boards differ a lot. It is vital that you remind yourself that the next board is much tougher than the previous one, and hence, extreme determination, focus, and perseverance are essential. If you can keep yourself motivated, then you will find that you have passed the 12th boards with flying colors.
- Solve the sample papers: The best way to prepare for the 12th boards is to practice the ISC Class 11 Maths Sample Papers repeatedly. When you start solving them regularly, you will find that it acts as a revision. You will also be able to manage your time and increase your speed. Also, the more you practice, the more you are going to score in the upcoming examinations.
- Self-study: Apart from solving the ISC Class 11 Maths Sample Papers, you also need to allocate sufficient time for self-studying. It is vital that you analyze the knowledge that you have gained from school and private coaching. Only by processing them on your own, can you assess if you have any doubts or not. If you do have, you can clarify them in the next class to ensure that you have a strong foundation.

**Check Out - ISC Class 11 Maths Syllabus**

**ISC Class 11 Maths Sample Papers** are provided to offer aid to the students in preparing for the upcoming boards. You can check out the sample papers of different years given below.

Year | Question Papers |

2019 | Download PDF |

2020 | Download PDF |

2021 | Download PDF |

**Question 1 [10x2] **

(i) Let π: π → π be a function defined by (π₯) = π₯−π π₯−π , where π ≠ π. Then show that π is one-one but not onto.

(ii) Find the domain and range of the function (π₯) = [ππππ₯].

(iii) Find the square root of complex number 11 − 60π.

(iv) For what value of π will the equations π₯ 2 − ππ₯ − 21 = 0 πππ π₯ 2 − 3ππ₯ + 35 = 0 have one common root.

(v) In a βπ΄π΅πΆ, show that ∑(π + π) cos π΄ = 2π where π = π+π+π 2

(vi) Find the number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together.

(vii) Prove that sin 20° sin 40° sin 80° = √3 8

(viii) If two dice are thrown simultaneously, find the probability of getting a sum of 7 or 11.

(ix) Show that limπ₯→2 |π₯−2| π₯−2 does not exist.

(x) Find the point on the curve π¦ 2 = 4π₯, the tangent at which is parallel to the straight line π¦ = 2π₯ +4.

Question 2 [4]

Draw the graph of the function π¦ = |π₯ − 2| + |π₯ − 3|.

Question 3 [4]

Prove that cot π΄ + cot(60 + π΄) + cot(120 + π΄) = 3 cot 3π΄.

OR

In a βπ΄π΅πΆ prove that π cos πΆ + π cosπ΅ = π

Question 4 [4] Find the locus of a complex number, Z= x+iy, satisfying the relation | π§−3π π§+3π | ≤ √2. Illustrate the locus of Z in the organd plane.

Question 5 [4]

Find the number of words which can be formed by taking four letters at a time from the word “ COMBINATION”.

OR

A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls

(ii) at least 3 girls and

(iii) atmost three girls.

Question 6 [4]

Prove by the method of induction. 1 1.2 + 1 2.3 + 1 3.4 + … … … … … … . π’ππ‘π π π‘ππππ = π π + 1 π€βπππ π ∈ π.

Question 7 [4]

Find the term independent of π₯ and its value in the expansion of (√ π₯ 3 − √3 2 ) 12 .

OR

Find the sum of the terms of the binomial expansion to infinity: 1 + 2 4 + 2.5 4.8 + 2.5.8 4.8.12 + β― … … .π‘π ∞.

Question 8 [4]

Differentiate from first principle: (π₯) = √3π₯ + 4.

Question 9 [4]

Reduce the equation π₯ + π¦ + √2 = 0 to the normal form ( +π¦ ππππΌ = π) and find the values of π πππ πΌ.

Question 10 [4]

Write the equation of the circle having radius 5 and tangent as the line 3π₯ − 4π¦ + 5 = 0 at (1, 2).

Question 11 [6]

In a βπ΄π΅πΆ prove that cot π΄ + cot π΅ + cot πΆ = π 2+π 2+π 2 4β

Question 12 [6]

Find the nth term and deduce the sum to n terms of the series: 4 + 11 + 22 + 37 + 56 + β― … …

OR

If (p+q)th term and (p-q)th terms of G.P are a and b respectively, prove that pth term is √ππ.

Question 13 [6]

If π₯ is real, prove that the value of the expression (π₯−1)(π₯+3) (π₯−2)(π₯+4) cannot be between 4/9 and 1.

OR

If π₯ π occurs in the expansion of (π₯ 2 + 1 ) 2π , prove that its coefficient is (2π)! [ 1 3 (4π−π)!] [ 1 3 (2π+π)!] .

Question 14 [6]

Calculate the standard deviation of the following distribution:

Age | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 |

No. of persons | 170 | 110 | 80 | 45 | 40 | 35 |

Question 15

(a) Find the focus and directrix of the conic represented by the equation 5π₯ 2 = −12π¦. [2]

(b) Construct the truth table (~ π β ∼ π)β(π β ∼ π). [2]

(c) Write the converse, contradiction and contrapositive of statement “ If π₯ + 3 = 9,π‘βππ π₯ = 6.” [2]

Question 16 [4]

Show that the point (1, 2, 3) is common to the lines which join A(4, 8, 12) to B(2, 4, 6) and C(3, 5, 4) to D(5, 8, 5).

OR

Calculate the Cosine of the angle A of the triangle with vertices A(1, -1, 2) B ( 6, 11, 2) and C( 1, 2, 6).

Question 17 [4]

Find the equation of the hyperbola whose focus is (1, 1), the corresponding directrix 2π₯ + π¦ − 1 = 0 πππ π = √3.

OR

Find the equation of tangents to the ellipse 4π₯ 2 + 5π¦ 2 = 20 which are perpendicular to the line 3π₯ + 2π¦ − 5 = 0.

Question 18 [6]

Show that the equation 16π₯ 2 − 3π¦ 2 − 32π₯ − 12π¦ − 44 = 0 represents a hyperbola. Find the lengths of axes and eccentricity.

SECTION C (20 Marks)

Question 19

(i) Two sample sizes of 50 and 100 are given .The mean of these samples respectively are 56 and 50 Find the mean of size 150 by combining the two samples [2]

(ii) Calculate π95, for the following data : [4]

Marks | 0-10 | 10- 20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 3 | 7 | 11 | 12 | 23 | 4 |

OR

Calculate Mode for the following data: [4]

C.I. | 17-19 | 14-16 | 11-13 | 8-10 | 5-7 | 2-4 |

Frequency | 12 | 4 | 8 | 16 | 11 | 4 |

Question 20

(i) Find the covariance between X and Y when ∑ π = 50, ∑ π = −30, and ∑ ππ = 115. [2]

(ii) Calculate Spearman’s Rank Correlation for the following data and interpret the result: [4]

Marks in Mathematics | 36 | 48 | 27 | 36 | 29 | 30 | 36 | 39 | 42 | 48 |

Marks in Statistics | 27 | 45 | 24 | 27 | 31 | 33 | 35 | 45 | 41 | 45 |

OR

Find Karl Pearson’s Correlation Coefficient from the given data: [4]

x | 21 | 24 | 26 | 29 | 32 | 43 | 25 | 30 | 35 | 37 |

y | 120 | 123 | 125 | 128 | 131 | 142 | 124 | 129 | 134 | 136 |

Question 21

Find the consumer price index for 2007 on the basis of 2005 from the following data using weighted average of price relative method: [4]

Items | Food | Rent | Cloth | Fuel |

Price in 2005(Rs) | 200 | 100 | 150 | 50 |

Price in 2007(Rs) | 280 | 200 | 120 | 100 |

Weighted | 30 | 20 | 20 | 10 |

Question 22

Using the following data. Find out the trend using Quaterly moving average and plot them on graph: [4]

Year/ Quarter | Q1 | Q2 | Q3 | Q4 |

1994 | 29 | 37 | 43 | 34 |

1995 | 90 | 42 | 55 | 43 |

1996 | 47 | 51 | 63 | 53 |

You can check out the **ISC Class 11 Maths Sample Papers** PDF given here. Take a look at the table below which provides the PDF file of sample papers of different years.

SL No. | Sample Papers | |

1 | ISC Class 11 Maths Sample Paper 2020-21 | |

2 | ISC Class 11 Maths Sample Paper 2019 | |

3 | ISC Class 11 Maths Sample Paper 2018 |

The structure of ISC class 11 is done in a way that the "Algebra" section holds the maximum marks. It comprises of 34 marks. Under this section, there are six chapters. Important questions along with answers to these six chapters are provided here. Apart from the **ISC Class 11 Maths Sample Papers**, you can also check it out to enhance yourself with more knowledge before appearing for the examinations.

Chapter: Principle of Mathematical Induction

Question 1

Using the principle of mathematical induction, prove that

1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1} for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}.

Putting n =1 in the given statement, we get

LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}.

Now, 1² + 2² + 3² + ......... + k² + (k + 1)²

= (1/6) {k(k + 1)(2k + 1) + (k + 1)²

= (1/6){(k + 1).(k(2k + 1)+6(k + 1))}

= (1/6){(k + 1)(2k² + 7k + 6})

= (1/6){(k + 1)(k + 2)(2k + 3)}

= 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)²

= (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 2:

Prove that 1 + 3 + 5 + … + (2n – 1) = n2 using the principle of Mathematical induction.

Solution:

Given Statement: 1 + 3 + 5 + … + (2n – 1) = n2

Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n2 , for n ∈ N

Note that P(1) is true, since

P(1) : 1 = 12

Let P(k) is true for some k ∈ N,

It means that,

P(k) : 1 + 3 + 5 + … + (2k – 1) = k2

To prove that P(k + 1) is true, we have

1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k2 + (2k + 1)

= k2 + 2k + 1

By using the formula, the above form can be written as:

= (k + 1)2

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

Question 3

Prove that 2n > n for all positive integers n by the Principle of Mathematical Induction

Solution:

Assume that P(n): 2n > n

If n =1, 21>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

2k > k …(1)

We shall now prove that P(k +1) is true whenever P(k) is true.

Now, multiplying both sides of the equation (1) by 2, we get

- 2k > 2k

Now by using the property,

i.e., 2k+1> 2k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Chapter: Complex Numbers

Question 1:

Express the given complex number (-3) in the polar form.

Solution:

Given complex number is -3

Let r cos θ = -3 …(1)

and rsinθ = 0 …(2)

Add both the terms:

rcosθ +rsinθ = -3+0

rcosθ +rsinθ = -3

Take square on both the sides, we get

r2cos2θ +r2sin2θ = (-3)2

Take r2 outside from L.H.S, we get

r2(cos2θ+sin2θ) = 9

We know that, cos2θ+sin2θ = 1, then the above equation becomes

r2 = 9

r = 3 (Conventionally, r > 0)

Now, subsbtitute the value of r in (1) and (2)

3 cos θ = -3 and 3 sinθ = 0

Cos θ = -1 and sinθ = 0

Therefore, θ = π

Hence, the polar representation is,

-3 = rcos θ + i rsin θ

3cosπ+ 3sin π = 3(cosπ+isin π)

Thus the required polar form is 3cosπ+ i3sin π = 3(cosπ+isin π)

Question 2:

Find the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]

Solution:

Given: [(1+i)/(1-i)] – [(1-i)/(1+i)]

Simplify the given expression, we get:

[(1+i)/(1-i)] – [(1-i)/(1+i)] = [(1+i)2– (1-i)2]/ [(1+i)(1-i)]

= (1+i2+2i-1-i2+2i)) / (12+12)

Now, cancel out the terms,

= 4i/2

= 2i

Now, take the modulus,

| [(1+i)/(1-i)] – [(1-i)/(1+i)]| =|2i| = √22 = 2

Therefore, the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)] is 2.

Question 3:

For any two complex numbers z1 and z2, show that Re(z1z2) = Rez1 Rez2– Imz1Imz2

Solution:

Given: z1 and z2 are the two complex numbers

To prove: Re(z1z2) = Rez1 Rez2– Imz1Imz2

Let z1 = x1+iy1 and z2 = x2+iy2

Now, z1 z2 =(x1+iy1)(x2+iy2)

Now, split the real part and the imaginary part from the above equation:

⇒ x1(x2+iy2) +iy1(x2+iy2)

Now, multiply the terms:

= x1x2+ix1y2+ix2y1+i2y1y2

We know that, i2 = -1, then we get

= x1x2+ix1y2+ix2y1-y1y2

Now, again seperate the real and the imaginary part:

= (x1x2 -y1y2) +i (x1y2+x2y1)

From the above equation, take only real part:

⇒ Re (z1z2) =(x1x2 -y1y2)

It means that,

⇒ Re(z1z2) = Rez1 Rez2– Imz1Imz2

Hence, the given statement is proved.

Chapter: Quadratic Equations

Question 1

Solve the quadratic equation 2x2+ x – 300 = 0 using factorisation.

Solution: 2x2+ x – 300 = 0

2x2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

So,

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of given equation.

Question 2

Solve the quadratic equation 2x2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax2+bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

x=−b±b2−4ac√2a

Now putting the values of a,b and c.

x=−1±1+4(2)(528)√4=−1±4225√4=−1±654

x=64/4 or x=-66/4

x=16 or x=-33/2

Question 3

Find the discriminant of the equation: 3x2-2x+β = 0.

Solution: Here, a = 3, b=-2 and c=β

Hence, discriminant, D = b2 – 4ac

D = (-2)2-4.3.(β )

D = 4-4

D=0

Chapter: Permutations and Combinations

Question 1

Evaluate the following

(i) 6 ! (ii) 5 ! – 2 !

Solution:

(i) 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720

(ii) 5! = 1 × 2 × 3 × 4 x 5 = 120

As 2! = 1 × 2 = 2

Question 2

How many words can be formed using all the letters of the word Dinomite, using each letter exactly one time?

Solution:

As there are 7 different letters in the word – Dinomite. So the number of different words formed using these 7 letters will be 8p7 = 8! / (8 -7)! = 8

Question 3

It is needed to seat 5 boys and 4 girls in a row so that the girl gets the even places. How many such arrangements are possible?

Solution:

5 boys and 4 girls are to be seated in a row so that the girl gets the even places.

The 5 boys can be seated in 5! Ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! Ways.

Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Chapter: Binomial Theorem

Question 1

Evaluate (101)4 using the binomial theorem

Solution:

Given: (101)4.

Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.

Therefore, 101 = 100+1

Hence, (101)4 = (100+1)4

Now, by applying the binomial theorem, we get:

(101)4 = (100+1)4 = 4C0(100)4 +4C1 (100)3(1) + 4C2(100)2(1)2 +4C3(100)(1)3 +4C4(1)4

(101)4 = (100)4+4(100)3+6(100)2+4(100) + (1)4

(101)4 = 100000000+ 4000000+ 60000+ 400+1

(101)4 = 104060401

Hence, the value of (101)4 is 104060401.

Question 2

Find the value of r, If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal.

Solution:

For the given condition, the coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)34 are 34Cr-6 and 34C2r-2 respectively.

Since the given terms in the expansion are equal,

34Cr-6 = 34C2r-2

From this, we can write it as either

r-6=2r-2

(or)

r-6=34 -(2r-2) [We know that, if nCr = nCp , then either r = p or r = n – p]

So, we get either r = – 4 or r = 14.

We know that r being a natural number, the value of r = – 4 is not possible.

Hence, the value of r is14.

Question 3

Expand the expression (2x-3)6 using the binomial theorem.

Solution:

Given Expression: (2x-3)6

By using the binomial theorem, the expression (2x-3)6 can be expanded as follows:

(2x-3)6 = 6C0(2x)6 –6C1(2x)5(3) + 6C2(2x)4(3)2 – 6C3(2x)3(3)3 + 6C4(2x)2(3)4 – 6C5(2x)(3)5 + 6C6(3)6

(2x-3)6 = 64x6 – 6(32x5 )(3) +15(16x4 )(9) – 20(8x3 )(27) +15(4x2 )(81) – 6(2x)(243) + 729

(2x-3)6 = 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

Thus, the binomial expansion for the given expression (2x-3)6 is 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.

Chapter: Sequence and Series

Question 1

Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Solution:

Assume that A1, A2, A3, A4, and A5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A1, A2, A3, A4, A5, 26

Here, a= 8, b =26,n= 7

Therefore, 26= 8+(7-1)d

Hence it becomes,

26 = 8+6d

6d = 26-8 = 18

6d= 18

d = 3

A1= a+d = 8+ 3 =11

A2= a+2d = 8+ 2(3) =8+6 = 14

A3= a+3d = 8+ 3(3) =8+9 = 17

A4= a+4d = 8+ 4(3) =8+12 = 20

A5= a+5d = 8+ 5(3) =8+15 = 23

Hence, the required five numbers between the number 8 and 26 are11, 14, 17, 20, 23

Question 2

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Solution:

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the kth term of an A.P. is given by

ak = a +(k -1)d

Therefore, am+n = a +(m+n -1)d

am-n = a +(m-n -1)d

am = a +(m-1)d

Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:

am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

= 2 [a + (m-1)d]

= 2 am {since am = a +(m-1)d}

Therefore, the sum of (m + n)thand (m – n)th terms of an A.P. is equal to twice the mth term.

Hence, proved.

Question 3

The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms

Solution:

Let a1, a2 and d1, d2 be the first terms and the common difference of the first and second

arithmetic progression respectively.

Then,

(Sum of n terms of the first A.P)/(Sum of n terms of the second A.P) = (5n+4)/(9n+6)

⇒ [ (n/2)[2a1+ (n-1)d1]]/ [(n/2)[2a2+ (n-1)d2]]= (5n+4)/(9n+6)

Cancel out (n/2) both numerator and denominator on L.H.S

⇒[2a1+ (n-1)d1]/[2a2+ (n-1)d2]= (5n+4)/(9n+6) …(1)

Now susbtitute n= 35 in equation (1), {Since (n-1)/2 = 17}

Then equation (1) becomes

⇒[2a1+ 34d1]/[2a2+ 34d2]= (5(35)+4)/(9(35+6)

⇒[a1+ 17d1]/[a2+ 17d2]= 179/321 …(2)

Now, we can say that.

18th term of first AP/ 18th term of second AP = [a1+ 17d1]/[a2+ 17d2]….(3)

Now, from (1) and (2), we can say that,

18th term of first AP/ 18th term of second AP = 179/321

Hence, the ratio of 18th term of both the AP are 179:321

ISC Class 11 Maths Sample Papers are essential for the examinations that lay ahead. However, it is normal for students to have doubts regarding the ISC Class 11 Maths Sample Papers. The most common one is – are they really helpful? Well, sample papers are vital and can aid in preparing the students for the boards as well as any other examinations.

Take a look at the reasons as to why practising ISC Class 11 Maths Sample Papers are so important.

- Acts as revision: Once you have completed the entire syllabus, you can solve the ISC Class 11 Maths Sample Papers as revision. In this way, you get to enhance your knowledge while practicing more problems. The more problems you solve, the more will it become easier for you. Also, by solving them, you can assess the areas that are your weak points. You can then work on improving them.
- Eliminates exam fear: Keep in mind that the ISC Class 11 Maths Sample Papers are a reflection of the actual examination papers. By solving then, you gain an insight as to how the paper will be. As such, you can lay your worries and fears to rest. After solving these papers repeatedly, you will find that you have nothing to fear anymore. Moreover, you are prepared as much as it is possible for the examination.
- Enhances speed: While solving numerous ISC Class 11 Maths Sample Papers, you will realize that your speed has increased. It is because you are now adept at finding the solutions than before. Also, solving them is the perfect way to ensure that you complete them within a particular timeframe. It is vital that you learn time management simultaneously. With time, you will also find that the number of mistakes you make has dwindled.
- Aids in analyzing question pattern: Practicing the ISC Class 11 Maths Sample Papers on a daily basis will aid you in understanding the question format and pattern. After some days, you will be able to identify which topics are given more preference than the others. You will also understand which ones need more work on your part than others.
- Helps to assess the marks allocation: One of the most beneficial aspects of solving the ISC Class 11 Maths Sample Papers is the assessment of the marks allocation. After solving an adequate number of papers, you will realize which chapters hold more weightage than the others. It also means that these are the chapters that you must focus on the most.
- Increases self-confidence: The more you keep on solving the ISC Class 11 Maths Sample Papers, the more will your self-confidence grow. You will not only identify the areas that need work, but you will also find the sections that you are good at. You can focus on them to ensure that all the questions from these sections are attempted and solved correctly.

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