Maths NCERT solutions Class 10

Maths NCERT solutions Class 10

Maths NCERT solutions class 10 for CBSE and other board examination is an important milestone in your life as you take some serious decisions about your future based on your performance. Indeed, Mathematics forms a vital component of Class 6th to 12th as well as every competitive exams’ syllabus.

Chapter 1 Real Number

• 1.1 Introduction
• 1.2 Euclid’s Division Lemma
• 1.3 The Fundamental Theorem of Arithmetic
• 1.4 Revisiting Irrational Numbers
• 1.5 Revisiting Rational Numbers and Their Decimal Expansions
• 1.6 Summary

Chapter 1 real numbers

Important Formula

(i) 135 and 225

Solution:1 (i) 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 × 1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 × 1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 × 45+ 0
As there are no remainder so divisor 45 is our HCF.

(ii) 196 and 38220

(ii) 38220 > 196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 × 195 + 0
As there is no remainder so divisor 196 is our HCF.

(iii) 867 and 255

(iii) 867 > 255 we always divide greater number with smaller one.
Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 × 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 × 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0
As there is no remainder so divisor 51 is our HCF.

Q.2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:2
Let take a as any positive integer and b = 6.
Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6
So total possible forms will 6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5
6q + 0
6 is divisible by 2 so it is a even number
6q + 1
6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number
6q + 2
6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6q +3
6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6q + 4
6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6q + 5
6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

Q.3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Q.4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Solution:4
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

Q.5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.

Chapter 2 Polynomials

• 2.1 Introduction
• 2.2 Geometrical Meaning of the Zeroes of a Polynomial
• 2.3 Relationship between Zeroes and Coefficients of a Polynomial
• 2.4 Division Algorithm for Polynomials
• 2.5 Summary

Chapter 2 Polynomials

Important formula

Chapter 3 Pairs of linear equations in two variables

• 3.1 Introduction
• 3.2 Pair of Linear Equations in Two Variables
• 3.3 Graphical Method of Solution of a Pair of Linear Equations
• 3.4 Algebraic Methods of Solving a Pair of Linear Equations
•• 3.4.1 Substitution Method
•• 3.4.2 Elimination Method
•• 3.4.3 Cross-Multiplication Method
• 3.5 Equations Reducible to a Pair of Linear Equations in Two Variables
• 3.6 Summary

Chapter 3 Pairs of linear equations in two variables

Important formula

Exercise 3.4

• 4.1 Introduction
• 4.3 Solution of a Quadratic Equation by Factorisation
• 4.4 Solution of a Quadratic Equation by Completing the Square
• 4.5 Nature of Roots
• 4.6 Summary
Important formula

Chapter 5 Arithmetic Progressions

•5.1 Introduction
• 5.2 Arithmetic Progression
• 5.3 nth Term of an AP
• 5.4 Sum of First n Terms of an AP
• 5.5 Summary
Important formula

Chapter 5 Arthimatic progression

Chapter 6 Triangles

• 6.1 Introduction
• 6.2 Similar Figures
• 6.3 Similarity of Triangles
• 6.4 Criteria for Similarity of Triangles
• 6.5 Areas of Similar Triangles
• 6.6 Pythagoras Theorem
• 6.7 Summary

Important formula

Chapter 6 Triangles

Chapter 7 Coordinate Geometry

• 7.1 Introduction
• 7.2 Distance Formula
• 7.3 Section Formula
• 7.4 Area of a Triangle
• 7.5 Summary

Chapter 7 Coordinate Geometry

Important formula

Chapter 8 Introduction to Trigonometry

• 8.1 Introduction
• 8.2 Trigonometric Ratios
• 8.3 Trigonometric Ratios of Some Specific Angles
• 8.4 Trigonometric Ratios of Complementary Angles
• 8.5 Trigonometric Identities
• 8.6 Summary

Chapter 8 Introduction to Trigonometry

Important formula

Chapter 9 Some Applications of Trigonometry

• 9.1 Introduction
• 9.2 Heights and Distances
• 9.3 Summary

Chapter 9 Some Applications of Trigonometry

Important formula

Chapter 10 Circles

• 10.1 Introduction
• 10.2 Tangent to a Circles
• 10.3 Number of Tangents from a Point on a Circle
• 10.4 Summary

Chapter 10 Circles

Important formula

Chapter 11 Constructions

• 11.1 Introduction
• 11.2 Division of a Line Segment
• 11.3 Construction of Tangents to a Circle
• 11.4 Summary

Chapter 11 Constructions

Important formula

Chapter 12 Areas Related to Circles

• 12.1 Introduction
• 12.2 Perimeter and Area of a Circle — A Review
• 12.3 Areas of Sector and Segment of a Circle
• 12.4 Areas of Combinations of Plane Figures
• 12.5 Summary

Chapter 12 Areas Related to Circles

Important formula

Chapter 13 Surface Areas and Volumes

• 13.1 Introduction
• 13.2 Surface Area of a Combination of Solids
•13.3 Volume of a Combination of Solids
• 13.4 Conversion of Solid from One Shape to Another
• 13.5 Frustum of a Cone
• 13.6 Summary

Chapter 13 Surface Areas and Volumes

Important formula

Chapter 14 Statistics

• 14.1 Introduction
• 14.2 Mean of Grouped Data
• 14.3 Mode of Grouped Data
• 14.4 Median of Grouped Data
• 14.5 Graphical Representation of Cumulative Frequency Distribution
• 14.6 Summary

Chapter 14 Statistics

Important formula

Chapter 15 Probability

• 15.1 Introduction
• 15.2 Probability — A Theoretical Approach
• 15.3 Summary

Chapter 15 Probability

Important formula

Maths NCERT solutions Class 10

Maths NCERT solutions list

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