Category: Class 9

Students can download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Which of the following expressions are polynomials. If not give reason:
(i) \(\frac{1}{x^2}\) + 3x – 4
Solution:
(i) \(\frac{1}{x^2}\) + 3x – 4 is not a polynomial. Since the exponent of x² is not a whole number, but it is (\(\frac{1}{x^2}\) = x^-2) negative number.

(ii) x² (x – 1)
Solution:
x² (x – 1) is a polynomial.

(iii) \(\frac{1}{x}\) (x + 5)
Solution:
\(\frac{1}{x}\) (x + 5) is not a polynomial. Since the exponent of x is not a whole number, but it is (\(\frac{1}{x}\) = x^-1) negative number.

(iv) \(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7
Solution:
\(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7 is a polynomial. (\(\frac{1}{x^{-2}}\) = x² and \(\frac{1}{x^{-1}}\) = x)

(v) √5x² + √3x + √2
Solution:
√5x² + √3x + √2 is a polynomial.

(vi) m² – \(\sqrt[3]{m}\) + 7m – 10
m² –\(\sqrt[3]{m}\) + 7m – 10 is not a polynomial. Since the exponent of m is not a whole number.
(\(\sqrt[3]{m}\) = m^1/3)

Question 2.
Write the coefficient of x² and x in each of the following polynomials.
(i) 4 + \(\frac{2}{5}\) x² – 3x
Solution:
Coefficient of x² is \(\frac{2}{5}\) and coefficient of x is -3.

(ii) 6 – 2x² + 3x³ – √7x
Solution:
Coefficient of x² is -2 and coefficient of x is -√7

(iii) π x² – x + 2
Solution:
Coefficient of x² is π and coefficient of x is -1.

(iv) √3x² + √2x + 0.5
Solution:
Coefficient of x² is √3 and coefficient of x is √2

(v) x² – \(\frac{7}{2}\) x + 8
Solution:
Coefficient of x² is 1 and coefficient of x is –\(\frac{7}{2}\)

Question 3.
Find the degree of the following polynomials.
(i) 1 – √2 y² + y⁷
(ii) \(\frac{x^{3}-x^{4}+6 x^{6}}{x^{2}}\)
(iii) x³ (x² + x)
(iv) 3x⁴ + 9x² + 27x⁶
(v) 2√5p⁴ \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
Solution:
(i) 1 – √2 y² + y⁷
The degree of the polynomial is 7.

(ii)
= x – x² + 6x⁴
The degree of the polynomial is 4.

(iii) x³ (x² + x) = x⁵ + x⁴
The degree of the polynomial is 5.

(iv) 3x⁴ + 9x² + 27x⁶
The degree of the polynomial is 6.

(v) 2√5p⁴ \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
The degree of the polynomial is 4.

Question 4.
Rewrite the following polynomial in standard form.
(i) x – 9 + √7x³ + 6x²
Solution:
The standard form is √7x³ + 6x² – x – 9
(or) – 9 + x + 6x² + √7x³

(ii) √2x² – \(\frac{7}{2}\) x⁴ + x – 5x³
Solution:
The standard form is – \(\frac{7}{2}\) x⁴ – 5x³ + √2x² + x
(or) x + √2x² – 5x³ – \(\frac{7}{2}\) x⁴

(iii) 7x³ – \(\frac{6}{5}\) x² + 4x – 1
Solution:
The given polynomial is in standard form (or) – 1 + 4x – \(\frac{6}{5}\) x² + 7x³

(iv) y² + √5y³ – 11 – \(\frac{7}{3}\) y + 9y⁴
Solution:
The standard form is 9y⁴ + √5y³ + y² – \(\frac{7}{3}\) y – 11
(or) – 11 – \(\frac{7}{3}\) y + y² + √5y³ + 9y⁴

Question 5.
Add the following polynomials and find the degree of the resultant polynomial
(i) p(x) = 6x² – 7x + 2; q(x) = 6x³ – 7x + 15
Solution:
p(x) + q(x) = 6x² – 7x + 2 + 6x³ – 7x + 15
= 6x³ + 6x² – 7x – 7x + 2 + 15
= 6x³ + 6x² – 14x + 17
The degree of the polynomial is 3.

(ii) h(x) = 7x³ – 6x + 1; f(x) = 7x² + 17x – 9
Solution:
h(x) + f(x) = 7x³ – 6x + 1 + 7x² + 17x – 9
= 7x³ + 7x² + 11x – 8
The degree of the polynomial is 3.

(iii) f(x) = 16x⁴ – 5x² + 9; g(x) = -6x³ + 7x – 15
Solution:
f(x) + g(x) = 16x⁴ – 5x² + 9 – 6x³ + 7x – 15
= 16x⁴ – 6x³ – 5x² + 7x + 9 – 15
= 16x⁴ – 6x³ – 5x² + 7x – 6
The degree of the polynomial is 4.

Question 6.
Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x² + 6x – 1; q(x) = 6x – 9
Solution:
p(x) – q(x) = 7x² + 6x – 1 – (6x – 9)
= 7x² + 6x – 1 – 6x + 9
= 7x² + 6x – 6x – 1 + 9
= 7x² + 8
The degree of the polynomial is 2.

(ii) f(y) = 6y² – 7y + 2; g(y) = 7y + y³
Solution:
f(y) – g(y) = 6y² – 7y + 2 – (7y + y³)
= 6y² – 7y + 2 – 7y – y³
= -y³ + 6y² – 7y – 7y + 2
= -y³ + 6y² – 14y + 2
The degree of the polynomial is 3.

(iii) h(z) = z⁵ – 6z⁴ + z; f(z) = 6z² + 10z – 7
Solution:
h(z) – f(z) = z⁵ – 6z⁴ + z – (6z² + 10z – 7)
= z⁵ – 6z⁴ + z – 6z² – 10z + 7
= z⁵ – 6z⁴ – 6z² + z – 10z + 7
= z⁵ – 6z⁴ – 6z² – 9z + 7
The degree of the polynomial is 5.

Question 7.
What should be added to 2x³ + 6x² – 5x + 8 to get 3x³ – 2x² + 6x + 15?
Solution:
3x³ – 2x² + 6x + 15 – (2x³ + 6x² – 5x + 8)
= 3x³ – 2x² + 6x + 15 – 2x³ – 6x² + 5x – 8
= 3x³ – 2x³- 2x² – 6x² + 6x + 5x + 15 – 8
= x³ – 8x² + 11x + 7
x³ – 8x² + 11x + 7 must be added to get 3x³ – 2x² + 6x + 15.

Question 8.
What must be subtracted from 2x⁴ + 4x² – 3x + 7 to get 3x³ – x² + 2x + 1?
Solution:
2x⁴ + 4x² – 3x + 7 – (3x³ – x² + 2x + 1)
= 2x⁴ + 4x² – 3x + 7 – 3x³ + x² – 2x – 1
= 2x⁴ – 3x³ + 4x² + x² – 3x – 2x + 7 – 1
= 2x⁴ – 3x³ + 5x² – 5x + 6
2x⁴ – 3x³ + 5x² – 5x + 6 must be subtracted to get 3x³ – x² + 2x + 1.

Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x² – 9, q(x) = 6x² + 7x – 2
Solution:
p(x) × q(x) = (x² – 9) (6x² + 7x – 2)
= 6x⁴ + 7x³ – 2x² – 54x² – 63x + 18
= 6x⁴ + 7x³ – 56x² – 63x + 18
The degree of the polynomial is 4.

(ii) f(x) = 7x + 2, g(x) = 15x – 9
Solution:
f(x) × g(x) = (7x + 2) (15x – 9)
= 105x² – 63x + 30x – 18
= 105x² – 33x – 18
The degree of the polynomial is 2.

(iii) h(x) = 6x² – 7x + 1, f(x) = 5x – 7
Solution:
h(x) × f(x) = (6x² – 7x + 1) (5x – 7)
= 30x³ – 42x² – 35x² + 49x + 5x – 7
= 30x³ – 77x² + 54x – 7
The degree of the polynomial is 3.

Question 10.
The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x and y. If x = 10, y = 5 find the amount paid by him.
Solution:
The cost of a chocolate = (x + y)
Number of chocolates bought by Amir = x + y
Total amount paid by him = (x + y) (x + y)
= x² + xy + xy + y²
= x² + 2xy + y²
When x = 10 and y = 5
The total amount paid by him = (10)² + 2(10)(5) + (5)²
= 100 + 100 + 25 = 225

Question 11.
The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution:
Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x – 2 units
Area of the rectangle = (3x + 2) (3x – 2)
= 9x² – 6x + 6x – 4
= 9x² – 4
When x = 20
Area of the rectangle = 9(20)² – 4
= 9(400) – 4
= 3600 – 4
= 3596 sq.units.

Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial is p(x) × q(x)?
Solution:
Degree of the polynomial p(x) = 1
Degree of the polynomial q(x) = 2
Degree of p(x) × q(x) = 3
The polynomial is a cubic polynomial (or) Polynomial of degree 3.

Students can download Maths Chapter 2 Real Numbers Ex 2.9 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 1.
If n is a natural number then √n is……….
(a) always a natural number
(b) always an irrational number
(c) always a rational number
(d) may be rational or irrational
Solution:
(d) may be rational or irrational

Question 2.
Which of the following is not true?
(a) Every rational number is a real number
(b) Every integer is a rational number
(c) Every real number is an irrational number
(d) Every natural number is a whole number
Solution:
(c) Every real number is an irrational number

Question 3.
Which one of the following, regarding sum of two irrational numbers, is true?
(a) always an irrational number
(b) may be a rational or irrational number
(c) always a rational number
(d) always an integer
Solution:
(b) may be a rational or irrational number

Question 4.
Which one of the following has a terminating decimal expansion?
(a) \(\frac{5}{64}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{14}{15}\)
(d) \(\frac{1}{12}\)
Solution:
(a) \(\frac{5}{64}\)
Hint:
\(\frac{5}{64}\) = \(\frac{5}{2^{6}}\)

Question 5.
Which one of the following is an irrational number?
(a) \(\sqrt{25}\)
(b) \(\sqrt{\frac{9}{4}}\)
(c) \(\frac{7}{11}\)
(d) π
Solution:
(d) π
Hint:
We take frequently π as \(\frac{22}{7}\) (which gives the value of 3.1428571428571…….) to be its correct value, but in reality these are only approximations

Question 6.
An irrational number between 2 and 2.5 is………
(a) \(\sqrt{11}\)
(b) √5
(c) \(\sqrt{2.5}\)
(d) √8
Solution:
(b) √5
Hint:
√5 = 2.236, it lies between 2 and 2.5

Question 7.
The smallest rational number by which \(\frac{1}{3}\) should be multiplied so that its decimal expansion terminates with one place of decimal is ………
(a) \(\frac{1}{10}\)
(b) \(\frac{3}{10}\)
(c) 3
(d) 30
Solution:
(b) \(\frac{3}{10}\)
Hint:
\(\frac{1}{3}\) × \(\frac{3}{10}\) = \(\frac{1}{10}\) = \(\frac{1}{2×5}\) it has terminating decimal expansion.

Question 8.
If \(\frac{1}{7}\) = 0.\(\overline { 142857 }\) then the value of \(\frac{5}{7}\) is ………..
(a) 0.\(\overline { 142857 }\)
(b) 0.\(\overline { 714285 }\)
(c) 0.\(\overline { 571428 }\)
(d) 0.714285
Solution:
(b) 0.\(\overline { 714285 }\)

Question 9.
Find the odd one out of the following.
(a) \(\sqrt{32}×\sqrt{2}\)
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
(c) \(\sqrt{72}×\sqrt{8}\)
(d) \(\frac{\sqrt{54}}{\sqrt{18}}\)
Solution:
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
Hint:
\(\frac{\sqrt{27}}{\sqrt{3}}\) = \(\frac{\sqrt{27}}{\sqrt{3}}\) = √9 = 3. It is an odd number

Question 10.
0.\(\overline { 34 }\) + 0.3\(\overline { 4 }\) = ……….
(a) 0.6\(\overline { 87 }\)
(b) 0.\(\overline { 68 }\)
(c) 0.6\(\overline { 8 }\)
(d) 0.68\(\overline { 7 }\)
Solution:
(a) 0.6\(\overline { 87 }\)
Hint:
0.34343434
0.34444444
0.68787878

Question 11.
Which of the following statement is false?
(a) The square root of 25 is 5 or -5
(b) \(-\sqrt{25}\) = -5
(c) \(\sqrt{25}\) = 5
(d) \(\sqrt{25}\) = ±5
Solution:
(d) \(\sqrt{25}\) = ±5

Question 12.
Which one of the following is not a rational number?
(a) \(\sqrt{\frac{8}{18}}\)
(b) \(\frac{7}{3}\)
(c) \(\sqrt{0.01}\)
(d) \(\sqrt{13}\)
Solution:
(d) \(\sqrt{13}\)

Question 13.
\(\sqrt{27}\) + \(\sqrt{12}\) = ……….
(a) \(\sqrt{39}\)
(b) 5√6
(c) 5√3
(d) 3√5
Solution:
(d) 3√5
Hint:
\(\sqrt{27}\) + \(\sqrt{12}\) = \(\sqrt{9×3}\) + \(\sqrt{3×4}\) = 3√3 + 2√3 = 5√3

Question 14.
If \(\sqrt{80}\) = k√5, then k = ………
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
Hint:
\(\sqrt{80}\) = k√5
\(\sqrt{16×5}\) = k√5
4√5 = k√5
∴ k = 4

Question 15.
4√7 × 2√3 = ……….
(a) 6\(\sqrt{10}\)
(b) 8\(\sqrt{21}\)
(c) 8\(\sqrt{10}\)
(d) 6\(\sqrt{21}\)
Solution:
(b) 8\(\sqrt{21}\)
Hint:
4√7 × 2√3 = 4 × 2\(\sqrt{7×3}\) = 8\(\sqrt{21}\)

Question 16.
When written with a rational denominator, the expression \(\frac {2\sqrt{3}}{3\sqrt{2}}\) can be simplified as……..
(a) \(\frac {\sqrt{2}}{3}\)
(b) \(\frac {\sqrt{3}}{2}\)
(c) \(\frac {\sqrt{6}}{3}\)
(d) \(\frac {2}{3}\)
Solution:
(c) \(\frac {\sqrt{6}}{3}\)

Question 17.
When (2√5 – √2)² is simplified, we get………
(a) 4√5 + 2√2
(b) 22 – 4\(\sqrt{10}\)
(c) 8 – 4\(\sqrt{10}\)
(d) 2\(\sqrt{10}\) – 2
Solution:
(b) 22 – 4\(\sqrt{10}\)
Hint:
(2√5 – √2)² = (2√5)² + (√2)² – 2 × 2√5 × √2
= 20 – 4\(\sqrt{10}\) + 2
= 22 – 4\(\sqrt{10}\)

Question 18.
\((0.000729)^\frac{-3}{4}\) × \((0.09)^\frac{-3}{4}\) = ……..
(a) \(\frac {10^3}{3^3}\)
(b) \(\frac {10^5}{3^5}\)
(c) \(\frac {10^2}{3^2}\)
(d) \(\frac {10^6}{3^6}\)
Solution:
(d) \(\frac {10^6}{3^6}\)
Hint:

Question 19.
If √9^x = \(\sqrt[3]{9^2}\), then x = …….
(a) \(\frac {2}{3}\)
(b) \(\frac {4}{3}\)
(c) \(\frac {1}{3}\)
(d) \(\frac {5}{3}\)
Solution:
(b) \(\frac {4}{3}\)
Hint:

Question 20.
The length and breadth of a rectangular plot are 5 × 10⁵ and 4 × 10⁴ metres respectively. Its area is ……….
(a) 9 × 10¹ m²
(b) 9 × 10⁹ m²
(c) 2 × 10¹⁰ m²
(d) 20 × 10²⁰ m²
Solution:
(c) 2 × 10¹⁰ m²
Hint:
Area of a rectangle = l × b = 5 × 10⁵ × 4 × 10⁴
= 5 × 4 × 10⁵⁺⁴
= 20 × 10⁹
= 2.0 × 10 × 10⁹
= 2 × 10¹⁰ m²

Students can download Maths Chapter 1 Set Language Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Additional Questions

Choose the correct answer

Question 1.
A = {set of odd natural numbers}, B = {set of even natural numbers}, then A and B are……….
(a) equal set
(b) equivalent sets
(c) overlapping sets
(d) disjoint sets
Solution:
(d) disjoint sets

Question 2.
Number of subsets in set A = {1, 2, 3} is
(a) 3
(b) 6
(c) 8
(d) 9
Solution:
(c) 8

Question 3.
The set does not have a proper subset is
(a) Finite set
(b) Infinite set
(c) Null set
(d) Singleton set
Solution:
(c) Null set

Question 4.
Sets having the same number of elements are called
(a) overlapping sets
(b) disjoints sets
(c) equivalent sets
(d) equal sets
Solution:
(c) equivalent sets

Question 5.
The set (A – B) ∪ (B – A) is
(a) AΔB
(b) A∪B
(c) A∩B
(d) A’∪B’
Solution:
(a) AΔB

Question 6.
The set of (A∪B) – (A∩B) is
(a) (A∪B)’
(b) AΔB
(c) (A∩B)’
(d) A’∪B’
Solution:
(b) AΔB

Question 7.
The set {x : x ∈ A, x ∈ B, x ∉ A∩B} is
(a) A∩B
(b) A∪B
(c) A – B
(d) AΔB
Solution:
(d) AΔB

Question 8.
The number of elements of the set {x : x ∈ Z , x² = 1} is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 9.
If A is a proper subset of B, then A∩B =…………..
(a) A
(b) B
(c) 0
(d)A∪B
Solution:
(a) A

Question 10.
The shade region with adjoint diagram represents ……….

(a) A – B
(b) B – A
(c) A’
(d) B’
Solution:
(c) A’

Question 11.
From the given venn diagram (A∪B)’ is ………..

(a) {5, 6}
(b) {1, 2, 3, 4, 7}
(c) {1, 2, 3, 4, 5, 6, 7}
(d) {8, 9}
Solution:
(d) {8, 9}

Question 12.
If n(A∪B∪C) = 73, n(A) = 2x, n(B) = 3x, n(C) = 5x, n(A∩B) = 10, n(B∩C) = 15, n(A∩C) = 5 and n(A∩B∩C) = 3, then the value of x is ………
(a) 9
(b) 10
(c) 5
(d) 18
Solution:
(b) 10

Question 13.
For any three sets, n(A∪B∪C) = 60, n(A) = 25, n(B) = 20, n(C) = 15, n(A∩B) = 10, n(B∩C) = 7, n(A∩C) = 3, then n(A∩B∩C) is……….
(a) 10
(b) 15
(c) 20
(d) 25
Solution:
(c) 20

Question 14.
If n(U) = 70, n(A) = 25, n(B) = 30, n(A∩B) = 5, then n(A∪B)’ is……….
(a) 5
(b) 10
(c) 15
(d) 20
Solution:
(d) 20

Question 15.
Which of the following is not correct?
(a) A – (B∪C) = (A – B) ∩ (A – C)
(b) A – (B∩C) = (A – B) ∪ (A – C)
(c) (A∪B)’ = A’∩B’
(d) A’∪B’ = (A – B)’
Solution:
(d) A’∪B’ = (A – B)’

Answer the following questions.

Question 1.
Write the following in “Roster” form?
(a) A = set of the months having 31 days.
(b) B = {x : x is a natural number of 2 digits divisible by 13}
(c) C = {set of vowels in the word “father”}
(d) D = {x : 5 < x ≤ 10; x ∈ N}
(e) E = {x : x is a square natural number less than 16}
Solution:
(a) A = {Jan, March, May, July, Aug, Oct, Dec}
(b) B = {13, 26, 39, 52, 65, 78, 91}
(c) C = {a, e}
(d) D = {6, 7, 8, 9, 10}
(e) E = {1, 4, 9}

Question 2.
Given that A = {1, 3, 5, 7} B = {1, 2, 4, 6, 8}. Find
(i) AΔB and
(ii) BΔA
Solution:
(i) A = {1, 3, 5, 7}; B = {1, 2, 4, 6, 8}
A – B = {1, 3, 5, 7} – {1, 2, 4, 6, 8}
= {3, 5, 7}
B – A = {1, 2, 4, 6, 8} – {1, 3, 5, 7}
= {2, 4, 6, 8}
AΔB = (A – B) ∪ (B – A)
= {3, 5, 7} ∪ {2, 4, 6, 8}
= {2, 3, 4, 5, 6, 7, 8}
(ii) BΔA = (B – A) ∪ (A – B)
= {2, 4, 6, 8} ∪ {3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}

Question 3.
From the venn-diagram, list the following:

(i) A
(ii) B
(iii) A∩B
(iv)A∪B
(v) A – B
(vi) B – A
(vii) (A – B) ∩ (B – A)
Solution:
(i) A = {1, 2, 5, 6, 7}
(ii) B = {3, 4, 5, 6}
(iii) A∩B = {5, 6}
(iv) A∪B = {1, 2, 3, 4, 5, 6, 7}
(v) A – B = {1, 2, 7}
(vi) B – A = {3, 4}
(vii) (A – B) ∩ (B – A) = { }

Question 4.
In a class there are 40 students. 26 have opted for Mathematics and 24 have opted for Science. How many student have opted for Mathematics and Science.
Solution:
Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M∪S) = 40, n (M) = 26, n(S) = 24
n(M∪S) = n (M) + n (S)- n(M∩S)
40 = 26 + 24 – n(M∩S)
n (M∩S) = 26 + 24 – 40 = 50 – 40 = 10
∴ Number of students opted for Mathematics and Science = 10.
Another Method:
Let “x” be the number of students opted for Mathematics and Science.
Let M and S represent students opting Mathematics and Science.
n(M∪S) = 40, n(M) = 26, n(S) = 24

By venn-diagram, number of students in a class = 26 – x + x + 24 – x
40 = 50 – x
x = 50 – 40 = 10
x = 10
∴ Number of students opted for Mathematics and Science = 10.

Question 5.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {4, 5, 7, 9}, B = {1, 3, 5, 7, 8}, Verify De Morgan’s Laws for complementation.
De Morgan’s Laws (i) (A∪B)’ = A’∩B’ (ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {4, 5, 7, 9} ∪ {1, 3, 5, 7, 8}
= {1, 3, 4, 5, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 4, 5, 7, 8, 9}
= {2, 6}……….(1)
A’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 5, 7, 9}
= {1, 2, 3, 6, 8}
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 8}
= {2, 4, 6, 9}
A’∩B’ = {1, 2, 3, 6, 8} ∩ {2, 4, 6, 9}
= {2, 6}………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’.

(ii) A∩B = {4, 5, 7, 9} ∩ {1, 3, 5, 7, 8}
= {5, 7}
(A∩B)’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 7}
= {1, 2, 3, 4, 6, 8, 9}………(1)
A’ = {1, 2, 3, 6, 8}
B’ = {2, 4, 6, 9}
A’∪B’ = {1, 2, 3, 6, 8} ∪ {2, 4, 6, 9}
= {1, 2, 3, 4, 6, 8, 9}………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’.

Students can download Maths Chapter 1 Set Language Ex 1.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.7

I. Multiple Choice Questions.

Question 1.
Which of the following is correct?
(a) {7} ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(c) 7 ∉ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(d) {7} ⊄ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Question 2.
The set P = {x | x ∈ Z, -1 < x < 1} is a ……….
(a) Singleton set
(b) Power set
(c) Null set
(d) Subset
Solution:
(a) Singleton set
Hint: P = {0}

Question 3.
If U = {x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A’)’ is………..
{a) {1, 6, 7, 8, 9}
(b) {1, 2, 3, 4}
(c) {2, 3, 4, 5}
(d) { }
Solution:
(c) {2, 3, 4, 5}
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A= {2, 3, 4, 5}; A’ = {1, 6, 7, 8, 9}
(A’)’ = U – A
(A’)’ = {2, 3, 4, 5}

Question 4.
If B⊆A then n(A∩B) is………..
(a) n(A – B)
(b) n(B)
(c) n(B – A)
(d) n(A)
Solution:
(b) n(B)

Question 5.
If A = {x, y, z} then the number of non- empty subsets of A is…………..
(a) 8
(b) 5
(c) 6
(d) 7
Solution:
(d) 7
Hint:
n(A) = 3; P(A) = 2^m = 2³ = 8
Non-empty subsets of A = 8 – 1 = 7

Question 6.
Which of the following is correct?
(a) Ø ⊆ {a, b}
(b) Ø ∈ {a, b}
(c) {a} ∈ {a, b}
(d) a ⊆ {a, b}
Solution:
(a) Ø ⊆ {a, b}
Hint:
‘Q’ is a subset of every set.

Question 7.
If A∪B = A∩B then ……………
(a) A ≠ B
(b) A = B
(c) A ⊂ B
(d) B ⊂ A
Solution:
(b) A = B

Question 8.
If B – A is B, then A∩B is………….
(a) A
(b) B
(c) U
(d) Ø
Solution:
(d) Ø

Question 9.
From the adjacent diagram n[P(AΔB)] is………….

(a) 8
(b) 16
(c) 32
(d) 64
Solution:
(c) 32
Hint:
A – B = {60, 85, 75}
B – A = {90, 70}
AΔB = (A – B) ∪ (B – A)
= {60, 70, 75, 85, 90}
n[P(AΔB)] = 2⁵ = 32

Question 10.
If n(A) = 10 and n(B) = 15 then the minimum and maximum number of elements in A∩B is …………
(a) (10, 15)
(b) (15, 10)
(c) (10, 0)
(d) (0, 10)
Solution:
(d) (0, 10)

Question 11.
Let A = {Ø} and B = P(A) then A∩B is………..
(a) {Ø, {Ø}}
(b) {Ø}
(c) Ø
{d) {0}
Solution:
(b) {Ø}
Hint:
A = {Ø}, B = {{ }, {Ø}}
A∩B = {Ø}

Question 12.
In a class of 50 boys, 35 boys play carrom and 20 boys play chess then the number of boys play both games is……..
(a) 5
(b) 30
(c) 15
(d) 10
Solution:
(a) 5
Hint:
n(A∪B) = 50, n(A) = 35, n(B) = 20
n(A∩B) = n(A) + n(B) – n(A∪B)
= 35 + 20 – 50 = 5

Question 13.
If U = {x : x ∈ N and x < 10}, A = {1, 2, 3, 5, 8} and B = (2, 5, 6, 7, 9}, then n [(A∪B)’] is
(a) 1
(b) 2
(c) 4
(d) 8
Solution:
(a) 1
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A∪B ={1, 2, 3, 5, 8} ∪ {2, 5, 6, 7, 9}
= {1, 2, 3, 5, 6, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 5, 6, 7, 8, 9}
= {4}
n(A∪B)’ = 1

Question 14.
For any three sets P, Q and R, P – (Q∩R) is ………
(a) P – (Q∪R)
(i)(P∩Q) – R
(c) (P – Q) ∪ (P – R)
(d) (P – Q) ∩ (P – R)
Solution:
(c)(P – Q) ∪ (P – R)

Question 15.
Which of the following is true?
(a) A – B = A∩B
(b) A – B = B – A
(c) (A∪B)’ = A’∪B’
(d) (A∩B)’ = A’∪B’
Solution:
(d) (A∩B)’ = A’∪B’

Question 16.
If n(A∪B∪C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A∩B) = 20, n(B∩C) = 15 and n(A∩C) = 25, then n(A∩B∩C) = 10, then the value of x is……….
(a) 10
(b) 15
(c) 25
(d) 30
Solution:
(b) 15
Hint:
n(A∪B∪C) = n( A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
100 = 4x + 6x + 5x – 20 – 15 – 25 + 10
100 = 15x – 50 ⇒ 150 = 15x ⇒ x = \(\frac{150}{15}\) = 10

Question 17.
For any three sets A, B and C, (A – B) ∩ (B – C) is equal to
(a) A only
(b) B only
(c) C only
(d) \(\phi \)
Solution:
(d) \(\phi \)

Question 18.
If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of shapes with right angle, then J∩K∩L is………
(a) Set of isosceles triangles
(b) Set of equilateral triangles
(c) Set of isosceles right triangles
(d) Set of right angled triangles
Solution:
(c) Set of isosceles right triangles
J = Set of three sided shapes; It is a triangle
K = Set of shapes with two equal sides (Isosceles triangle)
L = Set of shapes with right angle (One angle is right angle)
∴ J∩K∩L = Set of isosceles right triangles

Question 19.
The shaded region in the Venn diagram is

(a) Z – (X∪Y)
(b) (X∪Y) ∩ Z
(c) Z – (X∩Y)
(d) Z ∪ (X∩Y)
Solution:
(c) Z – (X∩Y)

Question 20.
In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?
(a) 5
(b) 8
(c) 10
(d) 15
Solution:
(a) 5

Students can download Maths Chapter 1 Set Language Ex 1.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5

Question 1.
Using the adjacent Venn diagram, find the following sets:

(i) A – B
(ii) B – C
(iii) A’∪B’
(iv) A’∩B’
(v) (B∪C)’
(vi) A – (B∪C)
(vii) A – (B∩C)
Solution:
From the diagram we get
U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8},
A= {-2,-1, 3, 4, 6}, B = {-2,-1, 5, 7, 8}
C = {-3, -2, 0, 3, 8}
A’ = U – A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 3, 4, 6}
= {-3, 0, 1, 2, 5, 7, 8}
B’ = U – B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 5, 7, 8}
= {-3, 0, 1, 2, 3, 4, 6}
B∪C = {-2, -1, 5, 7, 8} ∪ {-3, -2, 0, 3, 8} = {-3, -2, -1, 0, 3, 5, 7, 8}
B∩C = {-2, -1, 5, 7, 8} ∩ {-3, -2, 0, 3, 8} = {-2, 8}

(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’∪B’= {-3, 0, 1, 2, 5, 7, 8} ∪ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A’∩B’ = {-3, 0, 1, 2, 5, 7, 8} ∩ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2}
(v) (B∪C)’ = U – (B∪C)= {-3,-2,-1,0, 1,2, 3,4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
= {1, 2, 4, 6}
(vi) A – (B∪C) = {-2, -1, 3, 4, 6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
(vii) A – (B∩C) = {-2,-1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

Question 2.
If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L) ∩ (K∪M)
(iv) (K∩L) ∪ (K∩M)
and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(i) K∪(L∩M)
(L∩M) = {b, c, d, g} ∩ [a, b, c, d, h}
= {b, c, d}
K∪(L∩M) = {a, b, d, e, f} ∪ {b, c, d}
= {a, b, c, d, e, f}

(ii) K∩(L∪M)
(L∪M) = {b, c, d, g} ∪ {a, b, c, d, h}
= {a, b, c, d, g, h}
K∩(L∪M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h}
= {a, b, d }

(iii) (K∪L) ∩ (K∪M)
(K∪L) = {a, b, d, e, f} ∪ {b, c, d, g}
= {a, b, c, d, e, f, g}
(K∪M) = {a, b, d, e, f} ∪ {a, b, c, d, h}
= {a, b, c, d, e, f, h}
(K∪L) ∩ (K∪M) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}

(iv) (K∩L) ∪ (K∩M)
(K∩L) = {a, b, d, e, f) ∩ {b, c, d, g}
= {b, d}
(K∩M) = {a, b, d, e, f} ∩ {a, b, c, d, h}
= {a, b, d}
(K∩L) ∪ (K∩M) = {b, d} ∪ [a, b, d}
= {a, b, d}
From (ii) & (iv) we get, K∩(L∪M) = (K∩L) ∪ (K∩M)
From (i) & (iii) we get, K∪(L∩M) = (K∪L) ∩ (K∪M)

Question 3.
For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}
verify A∪(B∩C) = (A∪B) ∩ (A∪C).
Solution:
A = {-1, 0, 1, 2, 3, 4}, B = {0, 1, 2, 3, 4, 5} and C = {-4, -1, 0, 2, 3, 4}
B∩C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4}
= {0, 2, 3, 4}
A∪(B∩C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(1)
A∪B = {-1, 0, 1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
= {-1, 0, 1, 2, 3, 4, 5}
A∪C = {-1, 0, 1, 2, 3, 4} ∪ {-4, -1, 0, 2, 3, 4}
= {-4, -1, 0, 1, 2, 3, 4}
(A∪B) ∩ (A∪C) = {-1, 0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 1, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(2)
From (1) and (2) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 4.
Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.
Solution:

From (ii) and (v) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h} ; B = {a, c, d, g, f}; C = {a, d, e, g, h}
B∩C = {a, c, d, g, i} ∩ {a, d, e, g, h}
= {a, d, g}
A – (B∩C) = {b, c, e, g, h} – {a, d, g}
= {b, c, e, h}…….(1)
A – B = {b, c, e, g, h} – {a, c, d, g, i}
= {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h}
= {b, c}
(A – B) ∪ (A – C) = {b, e, h} ∪ {b, c}
= {b, c, e, h)……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C)

Question 6.
If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and
C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)
Solution:
A = {0, 6, 12, 18, 24, 30}; B = {6, 8, 10, 12, 14, 16, 18}; C = {12, 15, 18, 21, 24, 27}
B∩C = {6, 8, 10, 12, 14, 16, 18} ∩ {12, 15, 18, 21, 24, 27}
= {12, 18}
A – (B∩C) = {0, 6, 12, 18, 24, 30} – {12, 18}
= {0, 6, 24, 30}………(1)
A – B = {0, 6, 12, 18, 24, 30} – {6, 8, 10, 12, 14, 16, 18}
= {0, 24, 30}
A – C = {0, 6, 12, 18, 24, 30} – {12, 15, 18, 21, 24, 27}
= {0, 6, 30}
(A – B) ∪ (A – C) = {0, 24, 30} ∪ {0, 6, 30}
= {0, 6, 24, 30}……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C).

Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that
A – (B∪C) = (A – B) ∩ (A – C).
Solution:
A= {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6}, C = {-1, 2, 5, 6, 7}
B∪C = {-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}
= {-1, 0, 2, 5, 6, 7}
A – (B∪C) = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6, 7}
= {-2, 1, 3} ………(1)
A – B = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6}
= {-2, 1, 3}
A – C = {-2, 0, 1, 3, 5}- {-1, 2, 5, 6, 7}
= {-2, 0, 1, 3}
(A- B) ∩ (A- C) = {-2, 1, 3} ∩ {-2, 0, 1, 3}
= {-2, 1, 3} ….(2)
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 8.
IF A = {y : y = \(\frac{a + 1}{2}\), a ∈ W and a ≤ 5}, B = {y : y = \(\frac{2n – 1}{2}\), n ∈ W and n < 5} and C = {-1, \(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).
Solution:

From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 9.
Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:


From (ii) and (v) we get A- (B∩C) = (A – B) ∪ (A – C).

Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U= {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {7, 8, 11, 12} ∪ {4, 8, 12, 15}
= {4, 7, 8, 11, 12, 15}
(A∪B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10,16} ………(1)
A’ = {4, 7, 8, 10, 11, 12, 15, 16} – {7, 8, 11, 12}
= {4, 10, 15, 16}
B’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 8, 12, 15}
= {7, 10, 11, 16}
A’∩B’ = {4, 10, 15, 16} ∩ {7, 10, 11, 16}
= {10,16} ………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’

(ii) A∩B = {7, 8, 11, 12} ∩ {4, 8, 12, 15}
= {8, 12}
(A∩B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {8, 12}
= {4, 7, 10, 11, 15, 16} ………(1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’∪B’ = {4, 10, 15, 16} ∪ {7, 10, 11, 16}
= {4, 7, 10, 11, 15, 16} ………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’

Question 11.
Verify (A∩B)’ = A∪B’ using Venn diagrams.
Solution:

From (ii) and (i) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 1 Set Language Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6

Question 1.
(i) If n(A) = 25, n(B) – 40, n(A∪B) = 50 and n(B’) = 25, find n(A∩B) and n(U).
Solution:
Given, n(A) = 25, n(B) = 40, n(A∪B) = 50 and n(B’) = 25 n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 25 + 40 – 50
= 65 – 50
= 15
n(U) = n(B) + n(B)’
= 40 + 25
= 65
∴ n(A∩B) = 15 and n(U) = 65

(ii) If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
Given, n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350
n(A∪B) = n(A) + n(B) – n(A∩B)
500 = 300 + n(B) – 50
500 = 250 + n(B)
500 – 250 = n(B)
250 = n(B)
∴ n(B) = 250
n(U) = n(B) + n(B)’
250 + 350 = 600
∴ n(B) = 250 and n(U) = 600

Question 2.
If U = {x : x ∈ N, x ≤ 10}, A = { 2, 3, 4, 8, 10} and B = {1, 2, 5, 8, 10}, then verify that n(A∪B) = n(A) + n(B) – n(A∩B)
Solution:
U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 3, 4, 8, 10}; B = {1, 2, 5, 8, 10}
n(U) = 10, n(A) = 5, n(B) = 5
(A∪B) = {2, 3, 4, 8, 10} ∪ {1, 2, 5, 8, 10}
= {1, 2, 3, 4, 5, 8, 10}
∴ n(A∪B) = 7 ……..(1)
(A∩B) = {2, 3, 4, 8, 10} ∩ {1, 2, 5, 8, 10}
= {2, 8, 10}
n(A∩B) = 3
n(A) + n(B) – n(A∩B) = 5 + 5 – 3
= 10 – 3
= 7 ……(2)
From (1) and (2) we get,
n(A∪B) = n(A) + n(B) – n(A∩B)

Question 3.
Verify n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
Solution:
A∩B = {a, c, e, f, h} ∩ {c, d, e, f}
= {c, e, f}
B∩C = {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
A∩C = {a, c, e, f, h} ∩ {a, b, c, f}
= {c, f}
(A∩B∩C) = {a, c, e, f, h} ∩ {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
(A∪B∪C) = {a, c, e, f, h} ∪ {c, d, e, f} ∪ {a, b, c, f}
= {a, b, c, d, e, f, h}
n(A∩B) = 3, n(B∩C) = 2, n(A∩C) = 3, n(A∩B∩C) = 2
n(A∪B∪C) = 7……….(1)
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n( A∩C) + n(A∩B∩C)
= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8
= 7 ……….(2)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

(ii) A= {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}
A∩B = {1, 3, 5} ∩ {2, 3, 5, 6}
= {3, 5}
B∩C = {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5, 6}
A∩C = {1, 3, 5} ∩ {1, 5, 6, 7}
= {1, 5}
A∩B∩C = {1, 3, 5} ∩ {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5}
A∪B∪C = {1, 3, 5} ∪ {2, 3, 5, 6} ∪ {1, 5, 6, 7}
= {1, 2, 3, 5, 6, 7}
n(A) = 3, n(B) = 4, n(C) = 4
n(A∩B) = 2, n(B∩C) = 2, n(A∩C) = 2
n(A∩B∩C) = 1
n(A∪B∪C) = 6……….(1)
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 – 2 – 2 – 2 + 1
= 12 – 6
= 6………(2)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

Question 4.
In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Solution:

Let M be the set of all students take part in music.
Let D be the set of all students take part in drama.
n( M) = 25, n(D) = 30 and n(M∩D) = 8
By using venn-diagram
From the venn – diagram we get.
(i) Number of students take part in only music = 17
(ii) Number of students take part in only drama = 22
(iii) Total number of students in the class = 17 + 8 + 22 = 47

Question 5.
In a party of 45 people, each one likes Tea or Coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both Tea and Coffee.
(ii) do not like Tea.
(iii) do not like Coffee.
Solution:

Let’T’ be the set of people likes Tea
Let ‘C’ be the set of people likes Coffee
n(T∩C) = 45, n(T) = 35 and n(C) = 20
Let X be the number of people likes both Tea and Coffee.
By using venn diagram
From the venn – diagram we get.
35 – x + x + 20 – x = 45
55 – x = 45
55 – 45 = x
10 = x
(i) People like both tea and coffee = 10
(ii) People do not like tea = 20 – x
= 20 – 10 = 10
(iii) People do not like coffee = 35 – x
= 35 – 10 = 25

Question 6.
In an examination 50% of the students passed in mathematics and 70% of students passed in science while 10% students failed in both subjects. 300 students passed in both the subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let M and S represent the student failed in Mathematics and Science.
Given: Number of students passed in Mathematics is 50%

∴ Number of students failed in Mathematics = 100 – 50% = 50%
n(M) = 50%
Number of students passed in Science is 70%
∴ Number of students failed in Science = 100 – 70% = 30%
n(S) = 30%
Number of students failed in both the subjects is 10%
n(M∩S) = 10%
n(M∪S)= n(M) + n(S) – n(M∩S)
= 50 + 30 – 10 = 80 – 10 = 70
Given: 70% of the students failed in atleast any one of the subject
∴ 30% of the students passed in atleast any one of the subjects.
30 students passed mean, the total number of students is 100.
∴ 300 students passed means, the total number of students = \(\frac{100 × 300}{30}\)
Total number of students appeared in the examination = 1000

Question 7.
A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A∩B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B). Calculate the value of x.
Solution:
n(A – B) = 32 + x, n(B – A) = 5x
n(A∩B) = x
From the Venn diagram:

Given n( A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x – 2x
32 = 4x
x = \(\frac{32}{4}\) = 8
The value of x = 8

Question 8.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Solution:
Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A∩B) = 50
n(A∪B) = 500………..(1)
n(A) + n(B) – n(A∩B) = 400 + 200 – 50
= 600 – 50
= 550………(2)
From (1) and (2) we get
n(A∪B) ≠ n(A) + n(B) – n(A∩B)
∴ The given data is not correct.

Question 9.
In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Solution:
Let T, E and H represent families buying Tamil newspaper, English newspaper and Hindi newspaper respectively.
n(T) = 275, n(E) = 150, n(H) = 45
n(T∩E) = 125, n(E∩H) = 17, n(T∩H) = 5
n(T∩E∩H) = 3
Let us represent the given data in Venn diagrams.

(i) Number of families buy only one news paper = 148 + 11 + 26
= 185
(ii) Number of families buy atleast two news paper = 122 + 2 + 14 + 3
= 141
(iii) Total number of families in the colony = 148 + 122 + 11 + 14 + 3 + 2 + 26
= 326

Question 10.
A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Solution:
Let P, R and C represent sets of farmers grew paddy, ragi and com respectively.
n(P∪R∪C) = 1000, n(P) = 600, n(R) = 350, n(C) = 280
n(P∩R) = 120, n(R∩C) = 100, w(P∩C) = 80 Let the number of farmers who grew all the three be “x”
n(P∪R∪C ) = n(P) + n( R) + n( C) – n(P∩R) – n(R∩C) – n(P∩C) + n(P∩R∩C )
1000 = 600 + 350 + 280 – 120 – 100 – 80 + x = 1230 – 300 + x.
1000 = 930 + x
1000 – 930 = x
70 = x
Number of farmers who grew all the three = 70.

Question 11.
In the adjacent diagram, if n(U) = 125, y is two times of x and z is 10 more than x, then find the value of x, y and z.

Solution:
n(U) = 125
y = 2x and z = x + 10
n(U) = x + 4 + y + 17 + 3 + 6 + z + 5
125 = x + 4 + 2x + 17 + 3 + 6 + x + 10 + 5
125 = 4x + 45
125 – 45 = 4x
80 = 4x
x = 80/4 = 20
y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
∴ The value of x = 20, y = 40 and z = 30.

Question 12.
Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Solution:
Let A, B and C represent students play chess, carrom and table tennis.
n(A) = 22, n(B) = 21 , n(C) = 15
n(A∩C) = 10 , n(B∩C) = 8 , n(A∩B∩C) = 6
Let “x” represent student play chess and carrom but not table tennis.
Let us represent the data in Venn diagram.

From the Venn diagram we get,
Number of students play atleast one game = 35
12 – x + x + 13 – x + 2 + 6 + 4 + 3 = 35
40 – 35 = x
5 = x
(i) Number of students who play chess and carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 – x
= 12 – 5 = 7
(iii) Number of students who play only carrom = 13 – x
= 13 – 5 = 8

Question 13.
In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Solution:
Let B, C and D represent students come to school by bus, bicycle and foot respectively.
n(B∪C∪D) = 50 , n(B) = 25 , n(C) = 20 , n(D) = 30, n(B∩C∩D) = 10
Let x, y and z represent the students come to school exactly by two modes of transport.
Let us represent the given data in Venn diagrams.
<
Total number of students in the class = 50
15 – x – z + x + 10 – x – y + y + 10 + z + 20 – z – y = 50
55 – x – y – z = 50
55 – 50 = x + y + z
5 = x + y + z
Number of students come to school exactly by two modes of transport = 5

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Samacheer Kalvi 9th English Book Solutions Prose

• Chapter 1 Learning the Game
• Chapter 2 I Can’t Climb Trees Anymore
• Chapter 3 Old Man River
• Chapter 4 Seventeen Oranges
• Chapter 5 Water – The Elixir of Life
• Chapter 6 From Zero to Infinity
• Chapter 7 A Birthday Letter

Samacheer Kalvi 9th English Book Solutions Poem

• Chapter 1 Stopping by Woods on a Snowy Evening
• Chapter 2 A Poison Tree
• Chapter 3 On Killing a Tree
• Chapter 4 The Spider and the Fly
• Chapter 5 The River
• Chapter 6 The Comet
• Chapter 7 The Stick-together families

Samacheer Kalvi 9th English Book Solutions Supplementary

• Chapter 1 The Envious Neighbour
• Chapter 2 The Fun They Had
• Chapter 3 Earthquake
• Chapter 4 The Cat and the Pain-killer
• Chapter 5 Little Cyclone: The Story of a Grizzly Cub
• Chapter 6 Mother’s Voice
• Chapter 7 The Christmas Truce

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• Prefix / Suffix
• Anagrams
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• Prepositional Verbs
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• General Comprehension
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• Chapter 1 Evolution of Humans and Society – Prehistoric Period
• Chapter 2 Ancient Civilisations
• Chapter 3 Early Tamil Society and Culture
• Chapter 4 Intellectual Awakening and Socio-Political Changes
• Chapter 5 The Classical World
• Chapter 6 The Middle Ages
• Chapter 7 State and Society in Medieval India
• Chapter 8 The Beginning of the Modern Age
• Chapter 9 The Age of Revolutions
• Chapter 10 Industrial Revolution
• Chapter 11 Colonialism in Asia and Africa

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• Chapter 1 Lithosphere – I Endogenetic Processes
• Chapter 2 Lithosphere – II Exogenetic Processes
• Chapter 3 Atmosphere
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• Chapter 8 Disaster Management: Responding to Disasters

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• Chapter 2 Election, Political Parties and Pressure Groups
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• Chapter 1 Understanding Development: Perspectives, Measurement and Sustainability
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• Chapter 1 மனிதப் பரிணாம வளர்ச்சியும் சமூகமும்: வரலாற்றுக்கு முந்தைய காலம்
• Chapter 2 பண்டைய நாகரிகங்கள்
• Chapter 3 தொடக்ககாலத் தமிழ் சமூகமும் பண்பாடும்
• Chapter 4 அறிவு மலர்ச்சியும், சமூக – அரசியல் மாற்றங்களும்
• Chapter 5 செவ்வியல் உலகம்
• Chapter 6 இடைக்காலம்
• Chapter 7 இடைக்கால இந்தியாவில் அரசும் சமூகமும்
• Chapter 8 நவீன யுகத்தின் தொடக்கம்
• Chapter 9 புரட்சிகளின் காலம்
• Chapter 10 தொழிற்புரட்சி
• Chapter 11 ஆசிய ஆப்பிரிக்க நாடுகளில் காலனியாதிக்கம்

9th Std Social Science Book Back Answers Geography

• Chapter 1 நிலக்கோளம் – I புவி அகச்செயல்பாடுகள்
• Chapter 2 நிலக்கோளம் – II புவி புறச்செயல்பாடுகள்
• Chapter 3 வளிமண்டலம்
• Chapter 4 நீர்க்கோளம்
• Chapter 5 உயிர்க்கோளம்
• Chapter 6 மனிதனும் சுற்றுச் சூழலும்
• Chapter 7 நிலவரைபடத் திறன்கள்
• Chapter 8 பேரிடர் மேலாண்மை – பேரிடரை எதிர்கொள்ளுதல்

9th Std Social Science Guide Civics

• Chapter 1 அரசாங்க அமைப்புகள் மற்றும் மக்களாட்சி
• Chapter 2 தேர்தல், அரசியல் கட்சிகள் மற்றும் அழுத்தக் குழுக்கள்
• Chapter 3 மனித உரிமைகள்
• Chapter 4 அரசாங்கங்களின் வகைகள்
• Chapter 5 உள்ளாட்சி அமைப்புகள்
• Chapter 6 சாலை பாதுகாப்பு

9th Standard Social Science Guide Economics

• Chapter 1 மேம்பாட்டை அறிவோம்: தொலைநோக்கு, அளவீடு மற்றும் நிலைத் தன்மை
• Chapter 2 இந்தியா மற்றும் தமிழ்நாட்டில் வேலைவாய்ப்பு
• Chapter 3 பணம் மற்றும் கடன்
• Chapter 4 தமிழகத்தில் வேளாண்மை
• Chapter 5 இடம்பெயர்தல்

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• Chapter 3 இயற்கணிதம் Ex 3.2
• Chapter 3 இயற்கணிதம் Ex 3.3
• Chapter 3 இயற்கணிதம் Ex 3.4
• Chapter 3 இயற்கணிதம் Ex 3.5
• Chapter 3 இயற்கணிதம் Ex 3.6
• Chapter 3 இயற்கணிதம் Ex 3.7
• Chapter 3 இயற்கணிதம் Ex 3.8
• Chapter 3 இயற்கணிதம் Ex 3.9
• Chapter 3 இயற்கணிதம் Ex 3.10
• Chapter 3 இயற்கணிதம் Ex 3.11
• Chapter 3 இயற்கணிதம் Ex 3.12
• Chapter 3 இயற்கணிதம் Ex 3.13
• Chapter 3 இயற்கணிதம் Ex 3.14
• Chapter 3 இயற்கணிதம் Ex 3.15

9th Standard Maths Answer Key Chapter 4 வடிவியல்

• Chapter 4 வடிவியல் Ex 4.1
• Chapter 4 வடிவியல் Ex 4.2
• Chapter 4 வடிவியல் Ex 4.3
• Chapter 4 வடிவியல் Ex 4.4
• Chapter 4 வடிவியல் Ex 4.5
• Chapter 4 வடிவியல் Ex 4.6
• Chapter 4 வடிவியல் Ex 4.7

9th Class Maths Guide State Syllabus Pdf ஆயத்தொலை வடிவியல்

• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.1
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.2
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.3
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.4
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.5
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.6

9th Samacheer Maths Solution Book Chapter 6 முக்கோணவியல்

• Chapter 6 முக்கோணவியல் Ex 6.1
• Chapter 6 முக்கோணவியல் Ex 6.2
• Chapter 6 முக்கோணவியல் Ex 6.3
• Chapter 6 முக்கோணவியல் Ex 6.4
• Chapter 6 முக்கோணவியல் Ex 6.5

Samacheer 9th Maths Guide Tamil Medium Pdf Free Download Chapter 7 அளவியல்

• Chapter 7 அளவியல் Ex 7.1
• Chapter 7 அளவியல் Ex 7.2
• Chapter 7 அளவியல் Ex 7.3
• Chapter 7 அளவியல் Ex 7.4

Samacheer Kalvi Class 9 Maths Solutions Chapter 8 புள்ளியியல்

• Chapter 8 புள்ளியியல் Ex 8.1
• Chapter 8 புள்ளியியல் Ex 8.2
• Chapter 8 புள்ளியியல் Ex 8.3
• Chapter 8 புள்ளியியல் Ex 8.4

9th Standard Samacheer Kalvi Maths Guide Chapter 9 நிகழ்தகவு

• Chapter 9 நிகழ்தகவு Ex 9.1
• Chapter 9 நிகழ்தகவு Ex 9.2
• Chapter 9 நிகழ்தகவு Ex 9.3

We hope these Tamilnadu State Board Samacheer Kalvi Class 9th Maths Book Solutions and Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

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Samacheer Kalvi 9th Science Book Back Answers

Tamilnadu State Board Samacheer Kalvi 9th Science Book Back Answers Solutions Guide Term 1, 2, 3.

Samacheer Kalvi 9th Science Physics Book Solutions

• Chapter 1 Measurement
• Chapter 2 Motion
• Chapter 3 Fluids
• Chapter 4 Electric Charge and Electric Current
• Chapter 5 Magnetism and Electromagnetism
• Chapter 6 Light
• Chapter 7 Heat
• Chapter 8 Sound
• Chapter 9 Universe

Samacheer Kalvi 9th Science Chemistry Book Solutions

• Chapter 10 Matter Around Us
• Chapter 11 Atomic Structure
• Chapter 12 Periodic Classification of Elements
• Chapter 13 Chemical Bonding
• Chapter 14 Acids, Bases and Salts
• Chapter 15 Carbon and its Compounds
• Chapter 16 Applied Chemistry

Samacheer Kalvi 9th Science Biology Book Solutions

• Chapter 17 Animal Kingdom
• Chapter 18 Organization of Tissues
• Chapter 19 Plant Physiology
• Chapter 20 Organ Systems in Animals
• Chapter 21 Nutrition and Health
• Chapter 22 World of Microbes
• Chapter 23 Economic Biology
• Chapter 24 Environmental Science

Samacheer Kalvi 9th Science Computer Science Book Solutions

• Chapter 25 Computer – An Introduction
• Chapter 26 Parts of Computer
• Chapter 27 Hardware and Software

9th New Science Book Back Questions Tamil Medium with Answers

9th Science Book Back Questions Tamil Medium with Answers, 9th Std Science Guide Pdf Download 2021 Tamil Medium, 9th New Science Book Back Answers in Tamil 2021-2022.

Samacheer Kalvi 9th Science Physics Book Solutions

• Chapter 1 அளவீடு
• Chapter 2 இயக்கம்
• Chapter 3 பாய்மங்கள்
• Chapter 4 மின்னூட்டமும் மின்னோட்டமும்
• Chapter 5 காந்தவியல் மற்றும் மின்காந்தவியல்
• Chapter 6 ஒளி
• Chapter 7 வெப்பம்
• Chapter 8 ஒலி
• Chapter 9 அண்டம்

Samacheer Kalvi 9th Science Chemistry Book Solutions

• Chapter 10 நம்மைச் சுற்றியுள்ள பொருட்கள்
• Chapter 11 அணு அமைப்பு
• Chapter 12 தனிமங்களின் வகைப்பாட்டு அட்டவணை
• Chapter 13 வேதிப்பிணைப்பு
• Chapter 14 அமிலங்கள், காரங்கள் மற்றும் உப்புகள்
• Chapter 15 கார்பனும் அவற்றின் சேர்மங்களும்
• Chapter 16 பயன்பாட்டு வேதியியல்

Samacheer Kalvi 9th Science Biology Book Solutions

• Chapter 17 விலங்குலகம்
• Chapter 18 திசுக்களின் அமைப்பு
• Chapter 19 தாவர உலகம் – தாவர செயலியல்
• Chapter 20 விலங்குகளின் உறுப்பு மண்டலங்கள்
• Chapter 21 ஊட்டச்சத்து மற்றும் ஆரோக்கியம்
• Chapter 22 நுண்ணுயிரிகளின் உலகம்
• Chapter 23 பொருளாதார உயிரியல்
• Chapter 24 சூழ்நிலை அறிவியல்

Samacheer Kalvi 9th Science Computer Science

• Chapter 25 கணினி – ஓர் அறிமுகம்
• Chapter 26 கணினியின் பாகங்கள்
• Chapter 27 வன்பொருளும் மென்பொருளும்

We hope these Tamilnadu State Board Samacheer Kalvi Class 9th Science Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 9th Standard Science Guide Pdf of Text Book Back Questions and Answers Term 1, 2, 3, Chapter Wise Important Questions, Study Material, Question Bank, Notes, drop a comment below and we will get back to you as soon as possible.