Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions - Future Study Point

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Basic maths preboard 2021 term 1

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

 

10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions is created for the class 10 Maths students who has chosen Maths -Basic, question paper of maths basic consists of 50 multiple questions. The students have to solve 40 questions. Each question carries 1 mark. The basic maths question paper with solutions is helpful for the students in their Term -1 exam CBSE Board 2021.The importance of studying these solutions of basic maths question paper preboard exam is to get the idea about the type of the questions in Term 1 CBSE Board exam 2021.

Basic maths preboard 2021 term 1

 

Class 10 Maths Sample Paper (Basic) with Solutions for Term 1 CBSE Board Exam 2021-22

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MCQ’s on Class 10 Maths Co-ordinate Geometry for Term 1 CBSE

Class 10 Maths MCQ’s on Trigonometry for Term 1 CBSE with Solutions

Class 10 MCQ’s questions with solutions-Polynomial

 

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

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Q1. After how many places of decimal will the decimal expansion of the number 27/(2²×5³) terminate ?

(a) 3           (b) 4            (c) 4           (d) 1

Ans. (a) 3

(27)/(2²×5³)

Multiplying the denominator and numerator by 2

54/(2×5)³= 0.054

Q2. A die is thrown once. The probability of getting an even number is :

(a) 1/3           (b) 1/6      (c) 1/4           (d) 1/2

Ans.(d) 1/2

When a die is thrown once,the total possible outcomes are = 6(1,2,3,4,5,6)

Favourable outcomes = 2,4 and 6

P(E) = favourable outcomes/total possible outcomes

P(even number) = 3/6 = 1/2

Q3. Sum of two numbers is 35 and their difference is 13,then the numbers are 

(a) 24,13        (b) 24,11          (c) 12,11           (d) 12,23

Ans  (b) 24,11

(Let the numbers are x and y

x + y = 35…….(i)  and x – y = 13……(ii)

Adding both equation

2x = 48 ⇒x = 24

Putting the value in equation (i)

24 + y = 35

y = 11

Q4. The value of sin 60°/cos 30° is

(a) √3/2           (b) 1/2           (c) 1           (d) 2

Ans. (c) 1

Q5. If a bag contains 3 red and 7 black balls, then what will be the probability of getting a black ball ?

(a) 3/10          (b) 4/10          (c) 7/10         (d) 5/10

Ans. (c) 7/10

Total number of balls are = 3 red balls + 7 black ball = 10 balls

The number of black balls = 7

P(black ball) = 7/10

Q6.The HCF of 513 and 26   will be

(a) 0        (b) 1         (c) 13        (d) 26

Ans.(b) 1

The numbers given 513 and 26are co-prime numbers(1 is a common factor between them)

  1. Q7. Equations a1x + b1y + c1= 0 and a2x + b2y + c2= 0 has infinite many solutions if :

(a) a1/ a2≠b1/ b2     (b) a1/ a2=b1/ b2     (c) a1/ a2= b1/ b2=c1/ c2     (d) a1/ a2= b1/ b2≠ c1/ c2

Ans.   (c) a1/ a2= b1/ b2=c1/ c2   

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q8. sin 2A = 2 sin A is true when A is equal to

(a) 0°          (b) 30°         (c) 45°          (d) 60°

Ans. (a) 0°

The value of LHS sin (2×0°) =sin 0° = 0

The value of RHS 2 sin A = 2× sin 0° = 2×0 = 0

Hence,at A = 0,sin 2A = 2sin A

Q9. If the ratio of areas of two circle are 4 : 9, then the ratio of their radii will be:

(a) 4 : 9         (b) 2 : 3         (c) 8 : 27         (d) 3 : 2

Ans. (b) 2 : 3

Area of the circle is = πr²

Let the radii of two circles are r1 and r2

πr1²/πr2² = 4/9

r1²/r2² = 4/9

r1/r2 = 2/3

Q10. Distance between the points (-1,3) and (2,-1) is

(a) 1 unit        (b) 6 units       (c) 5 units       (d) 7 units

Ans. (c) 5 units

The distance(d) between two points (x1,y1 ) and (x2,y2) is

d= √[(x2-x1)²+ (y2– y1)²]

The distance between (-1,3) and (2,-1) is

d = √[(2+1)²+ (-1- 3)²] =√[3² +(-4)²] =√(9 +16) =√25 = 5 units

Q11. How many rational numbers are there between any two rational numbers ?

(a) 1          (b) 2             (c) 3            (d) infiitely many

Ans. (d) infinitely many

There are numberless numbers and any sets of numbers are infinite, hence between two rational numbers, there are infinite numbers.

Q12. If coordinates of the diameter of a circle are (-6,3) and (6,9), then the coordinates of the centre is:

(a) (8,-8)       (b) (12,6)        (c) (0,6)        (d) (0,3)

Ans. (c) (0,6)

The mid point between  two points (x1,y1 ) and (x2,y2) is = (x1+x2 )/2,  (y1+y2)/2

The midpoints of the diameter is the centre of the circle

The endpoints of the diameter are (-6,3) and (6,9)

The coordinates of the centre of the circle are = (-6+6)/2,( 3+9)/2 = (0,6)

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q13.If quadratic polynomial p(x) = x² – x + 4 has roots α and β, then the value of α + β is :

(a) -1          (b) 1            (c) 4            (d) 0

Ans. (b) 1

In the standard quadratic equation ax² + bx + c, if the roots are   α and β

α + β = -b/a

The given quadratic polynomial is

p(x) = x² – x + 4 ,where a = 1 and b = -1

α  + β = -(-1)/1 = 1

Q14.   is a /an

(a) Integer         (b) Rational Number      (c) Irrational Number    (d) Natural Number

Ans.(b) Rational Number

= 2.353535……, which is a recurrent (repeating) infinite decimal numbers,it can be written in the form of P/Q,where P and Q are co-prime numbers

x = 2.353535…..(i)

Multiplying it by 100,we get equation (ii)

100 x = 235.3535….(ii)

Subtracting equation (i) from (ii)

99x = 233

x = 233/99

Q15. If tan (3x -15°) = 1,then the value of x is:

Ans.

tan (3x -15°) = 1

tan (3x -15°) = tan 45°

3x – 15° = 45°

3x = 60°

x = 20°

Q16.From a well shuffled cards,the probability of getting a red face card is:

(a) 3/26      (b) 6/13       (c) 3/52       (d) 1/13

Ans. Total number of cards are = 52

The number of red face cards are = 26

P( getting a red face card) = (number of red face cards)/(total number of cards) = 26/52 =1/2

Q17. In what ratio the point P(0,0) divides the line segment joining the points A(3,3) and B(-3,-3)?

(a) 1 : 2       (b) 1 : 3        (c) 1 : 4         (d) 1 : 1

Ans.  (d) 1 : 1

Applying the section formula

x = (mx2+nx1)(m+n), y  = (my2+ny1)(m+n)

We are given that the point P(0,0)  divides the line segment A(3,3)  and B(-3,-3)

0 = (m×-3 +n×3)/(m+n)

-3m + 3n = 0

-3m = -3n

m : n = 1 : 1

Q18. The pair of linear equation 6x – 7y = 1 and 3x – 4y = 5 has solution:

(a) Two solutions        (b) Infinitely many solutions   (c) Unique solution  (d) No solution

Ans. (c) Uniques solution

From the given equation 6x -7y =1 and 3x -4y =5,we have the coefficients

are a1=6,b1=-7, c1=-1,a2=3,b2=-4,c2= -5

a1/ a2= 6/3 = 2, b1/ b2=(-7)/(-4) = 7/4, c1/c2= (-1)/(-5) = 1/5

a1/ a2≠ b1/ b2

Therefore both equations has a uniques solution

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q19.If the distance between the points A (4,P) and B(1,0) is 5 units, then the value of P are :

(a) 4,0         (b) -4,0        (c) 4, -4         (d) 0, 1

Ans. (c) 4, -4

The distance(d) between two points (x1,y1 ) and (x2,y2) is

d= [(x2-x1)²+ √(y2– y1])²

The distance between the points A (4,P) and B(1,0) is 5 units

5 =  [(1-4)²+ √(0- P])²

25 = (-3)² + (-P)²

P² = 25 -9 = 16

P = ± 4

Q20. 225 can be expressed as :

(a) 5² × 3²         (b) 5²× 3       (c) 5²× 3²      (d) 5³× 3

Ans. 5²×3²

225 = 3×3×5×5 = 5²× 3²

Q21. In an isosceles triangle ABC, if AC = BC and AB² = 2AC², then value of C is:

(a) 30°                    (b) 45°                (c) 60°           (d) 90°

Ans.

The given triangle is isosceles triangle in which AC = BC and we are given AB² = 2AC²

AB² = 2AC² = AC² + AC²

Putting AC = BC

AB² = BC² + AC²

AB =Hyptenuse ,BC and AC are perpendicular to each other

∴ ∠C = 90°

Q22. In a triangle ABC ,DE parallel to BC, then the value of EC is

(a) 5 cm     (b) 10 cm       (c) 8 cm        (d) 10 cm

Ans.(c) 8 cm

In ΔABC and ΔADE

DE parallel to BC and AB is the transversal

∠ADE = ∠ABC (corresponding angle)

∠AED = ∠ACB (corresponding angle)

∴ΔABC ∼ΔADE (AA rule criteria)

Applying the rule of similar triangles

AD/AB = AE/AC = DE/BC

3/(3 +4) = 6/AC

3/7 = 6/AC

3AC = 42

AC = 14

EC = AC – AE = 14 – 6 = 8 cm

Q23. The parimeter of semicircular protector whose radius ‘r ’ is

(a) π + r         (b) πr         (c) π + 2r       (d) πr + 2r

Ans. The perimeter of semicircular protector = Length of the semicircular arc + Diameter

Length of semicircular arc = (2πr)/2 =πr

Diameter = 2r

Hence perimeter of semicircular protector = πr + 2r

Q24. What is the area of a sector of a circle with radius 14 cm and central angle 45°?

(a)76 cm²  (b) 77 cm²     (c)66 cm²      (d) 55 cm²

Ans. (b) 77 cm²

Area of sector of a circle is given as

Where θ = 45°, r = 14 cm

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q25. What will be radius of the circle whose circumference is 44 m ?

(a) 14 m       (b) 7 m        (c) 5 m       (d) 44 m

Ans.(b) 7 m

The circumerence of the circle =2πr

The circumference of the given circle is = 44 m

2πr = 44

2 (22/7) r = 44

r = (44×7)/(44) = 7 m

Q26. (1 +tan²A)/cosec²A will be equal to:

(a) sec²A               (b) cosec²A         (c) cot²A          (d) tan²A

Ans. (d) tan²A

The given expression is

(1 +tan²A)/cosec²A

Converting tan A and cosec A into sin A and cos A

(1 + sin²A/cos²A)

Q27. If p and q are co-prime numbers, then the HCF of p³q² and p²q will be:

(a) p³q²       (b) p²q         (c) p²q²      (d) pq

Ans. (b) p²q

Since p and q are co-prime numbers,therefore in p and q there is no common factor except to 1.

The HCF of  p³q²  and p²q is = the lower power of p in both terms × lower power of q in both terms =p²q

Q28. In equation x + 2y =9 ,if x = 5,then the value of y will be:

(a) 1         (b) -2        (c) 2           (d) 4

Ans. (c) 2

Putting x = 5 in given equation x + 2y =9

5 + 2y =9

2y = 9 -5 = 4

y = 4/2 = 2

Q29. If (2K -1,K) is the solution of 10 x -9y =12, then the value of K will be:

(a) 1            (b) 2           (c) 3         (d) 4

Ans. (b) 2

The solution of the given equation is 10x – 9y =12 is =(2K-1,K)

Therefore putting x = 2K-1 and y = k in the given equation

10(2K-1) – 9K =12

20 K -10 -9K = 12

11K = 10 + 12 = 22

K = 2

Q30. If x = a cosθ and y = b sinθ, then the value of b²x² +a²y² is :

(a) a² +b²        (b) a²/b²         (c) a²b²         (d) 1/a² + 1/b²

Ans. (c) a²b²

Putting the value of x = a cos θ and y = b sin θ in the expression b²x² + a²y²

b² a² cos²θ + a²b² sin²θ

a²b²( cos²θ + sin²θ) = a²b²

Q31. Distance of the point Q(3,-4) from the origin is :

(a)  3 units             (b) 4 units       (c) 5 units         (d) 7 units

Ans. (c) 5 units

The coordinates of the origins are (0,0) and the point given to us is Q(3, -4),distance(d) between two two points is given as

d=√ [(x2-x1)²+ √(y2– y1])²

= √[(3 -0)² + (-4 -0)²] = √(3² + 4²) = √(9 + 16) = √25 = 5 units

 

Q32. If the point C(K,4) divides the line segement joining the two points A(2,6) and B(5,1) in the ratio 2 : 3, then the value of K is:

(a) 4/5       (b) 16/5       (c) 11/5          (d) 19/5

Ans.(b) 16/5

Applying the section formula

x = (mx2+nx1)(m+n), y  = (my2+ny1)(m+n)

Putting the value , x = K , y = 4, x1

=2,x2 = 5, y1= 6,y2=1, m : n = 2: 3

K=(2×5 + 3×2)/(2+3) =(10 +6)/5 = 16 /5

Q33. An event A is very unlikely to happen ,then its probabilty will be closest to :

(a) 0.0001          (b) 0.001             (c) 0.01        (d) 0.1

Ans.

The probabilty of an event is very likely to happen =   P(A)  = 1

The sum of probabilty of an event and  not happening of the probabilty  is equal to 1

Q34.In a linear equation ax +by +c = 0,a,b and c are:

(a) Natural number             (b) Whole number       (c) Integers       (d) Real Number

Ans. (d) Real Number

As an example √2 x + 3y -4 = 0 ,3x + 3y – 5 = 0 ,both are linear equations,no matters if a,b and c are natural numbers,whole numbers,integers ,rational or irrational numbers ,so a,b and c should be real numbers.

Q35. If a pair of dice is thrown once,then probability of getting a sum of 8 will be :

(a) 1/6         (b) 5/36        (c) 1/9          (d) 1/12

Ans. (b) 5/36

If a pair of dice is thrown once then the total outcomes are = 6² = 36

The  outcomes in which sum of the digits in both dice is 8 are = (2,6),(3,5),(4,4),(5,3),(6,2) [i.e 5)

P(getting the sum of 8) = 5/36

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q36.The perimeter of rectangle is 44 cm.Its length exceeds twice its breadth by 4 cm. Find the area of rectangle:

(a) 46 cm²      (b) 49 cm²       (c) 96 cm²       (d) 69 cm²

Ans. (c) 96 cm²

Let the breadth of rectangle = x cm  then length = (2x + 4 ) cm

Perimeter of the rectangle  given to us = 44 cm

Perimeter of the rectangle is = 2 ( L+ B) = 2( 2x + 4+x) = 2(3x + 4)

According to question

2(3x + 4) = 44

3x + 4 = 22

3x = 22 – 4 = 18

x = 6

Therefore breadth =6 cm and the length of rectangle  = 2x + 4 = 2×6 + 4 =16 cm

Area of the rectangle = L × B = 16 ×6 = 96 cm²

Q37. If sin θ = cos θ, then the value of tan ²θ + cot²θ is:

(a) 2             (b) 4            (c) 1              (d) 10/3

Ans. We are given

sin θ = cos θ

sinθ/cos θ = 1

tan θ = 1 ⇒ cot θ = 1/tan θ = 1

The value of

tan²θ + cot²θ = 1² + 1² = 1 + 1 = 2

Te questions 38 -40 consists of Assertion (A) and Reason (R) . Answer these questions selecting the appropriate option given below:

(a) Both A and R are true and R is the correct exolanation of A.

(b) Both A and R are true but R is not the correct exolanation of A.

(c) A is true,but R is false

(d) A is false, but R is true

Q38.Assertion(A): The point (0,4) lies on the y-axis.

Reason(R): The x coordinate of the point on the y-axis is zero.

Ans.(a) Both A and R are true and R is the correct explanation of A.

The point (0,4) lies on the y -axis because x = 0

Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions

Q39. Assertion(A): If the circumference of two circles are in the ratio 2 : 3, then the ratio of their areas is 4 : 9.

Reason(R): The circumference of a circle of radius is 2πr and its area is πr²

Ans.(a) Both A and R are true and R is the correct explanation of A.

Let the radii of two circles are r1 and r2

The ratio of their circumference is given as 2 : 3

2πr1/2πr2= 2/3

r1/r2= 2/3

Ratio between their areas

= πr1²/πr2² = (r1/r2)² =(2/3)² = 4/9

Q40. Assertion(A):  For any two positive integers a and b, LCM (a,b) ×HCF(a,b) = a× b.

Reason (R) : Product of two numbers is 150 and HCF is 5,then their LCM is 40.

Ans.(a) A is true but R is false

The relationship between HCF(a,b) and LCM(a,b) and the product of numbers a and b  is given as

LCM (a,b) ×HCF(a,b) = ab

LCM×5 = 150

LCM = 150/5 =30

SECTION – C

(Case Study Based  Questions)

Section ‘C’ consists of 10 questions. Do any 8 questions.

Points A and B are on the opposite edges of a pond a shown in below figure. To find distance between the two points, Ramesh makes a right angled triangle using the rope. Distance between point B and C is 12 m, distance between  C and D is 40 m and  A and D is 30 m such that ∠ADC = 90°.

Q41 explan basic maths term 1

Q41. Which property of geometry is used to find the distance in the figure ?

(a) Similarity of triangles           (b) Thales theorem

(c) Pythogorus theorem                       (d) Converse of thales

Ans.Pythogorus theorem

Q42. What is total length of pond AB and rope BC in the figure ?

(a) 50 m        (b) 70 m       (c) 100 m       (d) 38 m

Ans. (a) 50 m

∠ADC = 90°

AD  and DC are perpendicular to each other

AC² = AD² + DC²

AC² = 30² + 40² = 900 + 1600 = 2500

AC = √(2500) = 50

AB = AC – BC = 50 – 12 = 38

Q43. Length of the total rope used is:

(a) 82 m       (b) 70 m       (c) 120 m        (d) 100 m

Ans.(a) 82 m

The lenghth of total rope = BC  +DC + AD =12 + 40 + 30 = 82 m

Q44. Which of the following doesn’t form Pythagoras triplet ?

(a) (20,21,28)      (b) (8,15,17)       (c) (5,12,13)      (d) (7,24,25)

Ans.(a) (20,21,28)

(a) 28² = 21² +20²

784 ≠ 441 + 400=841

(b) 17² = 15² + 8²

289 = 225 + 64 =289

(c) 13² = 12² + 5²

169 = 144 + 25 =169

(d)  25² = 24² + 7²

625 = 576 +49 = 625

(a) doesn’t show Pythogoras triplet because 28² ≠ 21²+20²

Q45. Length of the Pond AB is:

(a) 50 m         (b) 62 m        (c) 70 m        (d) 38 m

(d) 38 m

Case Study -II

The path moved by a group of ants has been traced on graph. Aditi is a student of class 10. She compared the ant’s path with the diagram she learned in maths class then she drew the sketch on the graph.

Case study ll class 10 maths preboard 2021

Q46. The shape of the path formed by ants is:

(a) Spiral       (b) Oval       (c) Parabola      (d) Linear

Ans. (c) Parabola

Q47. How many zeroes are possible for this shape ?

(a) 2         (b) 3        (c) 4       (d) 0

Ans. (a) 2

A parabola can intersects any of the axises in two points,so maximum possible zeroes of the parabolic polynomial is 2.

Q48. According to graph ,the zeroes of the polynomial are:

(a) -3,-1       (b) 3,-1        (c) -3, 1         (d) 0,0

Ans.(c) -3,1

The parabola is intersecting the x-axis at x = -3 and at x = 1,therefore its zeroes are -3,1

Q49. What will be the expression of the polynomial shown in figure?

(a) x² -2x +3    (b) x² -3x +2    (c) x² +3x +2     (d) x² +2x – 3

Ans. (d) x² +2x -3

The algebraic expression of a quadratic polynomial with zeroes α and β is given as

x² – (α +β) x + αβ

Here α = -3 and β = 1

x² -(-3 +1)x +(-3) 1

x² +2x -3

Q50. What will be the value of the polynomial if x = -1 ?

(a) -5         (b) 6         (c) -4        (d) 0

Ans.(c) -4

The polynomial,P(x) = x² +2x -3

P(-1)  = (-1)² + 2(-1) -3 = 1 -2 -3 = 1-5 = -4

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NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

NCERT Solutions of class 9 maths

Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3- Coordinate Geometry Chapter 11-Construction
Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles Chapter 14-Statistics
Chapter 7-Triangles Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure? Chapter 10- Gravitation
Chapter3- Atoms and Molecules Chapter 11- Work and Energy
Chapter 4-Structure of the Atom Chapter 12- Sound
Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ?
Chapter 6- Tissues Chapter 14- Natural Resources
Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources
Chapter 8- Motion Last years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions of class 10 maths

Chapter 1-Real number Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3-Linear equations Chapter 11- Construction
Chapter 4- Quadratic equations Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume
Chapter 6-Triangle Chapter 14-Statistics
Chapter 7- Co-ordinate geometry Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT solutions of class 10 science

Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds Chapter 12- Electricity
Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process Chapter 14-Sources of Energy
Chapter 7-Control and Coordination Chapter 15-Environment
Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

NCERT solutions of class 11 maths

Chapter 1-Sets Chapter 9-Sequences and Series
Chapter 2- Relations and functions Chapter 10- Straight Lines
Chapter 3- Trigonometry Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations Chapter 15- Statistics
Chapter 8- Binomial Theorem  Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

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NCERT solutions of class 12 maths

Chapter 1-Relations and Functions Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra
Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability Chapter 13-Probability
Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

 

 

 

 

 

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