Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Students can download Maths Chapter 1 Set Language Ex 1.1 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 1 Set Language Ex 1.1

Question 1.
Which of the following are sets?
(i) The collection of prime numbers upto 100
(ii) The collection of rich people in India
(iii) The collection of all rivers in India
(iv) The collection of good hockey players
Solution:
(i) It is a set
(ii) It is not a set (The word “rich” is not well defined)
(iii) It is a set
(iv) It is not a set (The word “good” is not well defined)

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Question 2.
List the set of letters of the following words in Roster form.
(i) INDIA
(ii) PARALLELOGRAM
(iii) MISSISSIPPI
(iv) CZECHOSLOVAKIA
Solution:
(i) A = {I, N, D, A}
(ii) B = {P, A, R, L , E, O, G, M}
(iii) C = {M, I, S, P}
(iv) D = {C, Z, E, H, O, S, L, V, A, K, I}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Question 3.
Consider the following sets A = {0, 3, 5, 8}, B = {2, 4, 6, 10} and C = {12, 14, 18, 20}.
(a) State whether True or False:
(i) 18 ∈ C
(if) 6 ∉ A
(iii) 14 ∉ C
(iv) 10 ∈ B
(v) 5 ∈ B
(vi) 0 ∈ B
Solution:
(i) True
(ii) True
(iii) False
(iv) True
(v) False
(vi) False

(b) Fill in the blanks:
(i) 3 ∈ …………
(ii) 14 ∈…………
(iii) 18 ……….. B
(iv) 4 ………. B
Solution:
(i) A
(ii) C
(iii) ∉
(iv) ∈

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Question 4.
Represent the following sets in Roster form.
(i) A = The set of all even natural numbers less than 20.
(ii) B = {y : y = \(\frac{1}{2n}\), n∈N, n ≤ 5}
(iii) C = {x : x is perfect cube, 27 < x < 216}
(iv) D = {x : x ∈Z, – 5 < x ≤ 2}
Solution:
(i) A= {2, 4, 6, 8, 10, 12, 14, 16, 18}
(ii) B = {\(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), \(\frac{1}{8}\), \(\frac{1}{10}\)}
(iii) C = {64, 125}
(iv) D = {-4, -3, -2, -1, 0, 1, 2}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Question 5.
Represent the following sets in set builder form.
(i) B = The set of all cricket players in India who scored double centuries in one day internationals.
(ii) C = {\(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), …….}
(iii) D = The set of all Tamil months in a year.
(iv) E = The set of odd Whole numbers less than 9.
Solution:
(i) B = {x : x is a set of all cricket players in India who scored double centuries in one day internationals}
(ii) C = {x : n ∈ N, x = \(\frac{n}{n + 1}\) }
(iii) D = {x : x ∈ set of all Tamil months in a year}
(iv) E = {x : x is an odd whole number and x < 9}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.1

Question 6.
Represent the following sets in descriptive form.
(i) P = { January, June, July}
(ii) Q = {7, 11, 13, 17, 19, 23, 29}
(iii) R = {x : x∈N, x < 5}
(iv) S = {x : x is a consonant in English alphabets}
Solution:
(i) P = The set of all months beginning with the letter “J”
(ii) Q = The set of all prime numbers between 5 and 31
(iii) R = The set of natural numbers less than 5
(iv) S = The set of consonants in English alphabets

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Students can download Maths Chapter 2 Real Numbers Ex 2.7 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 2 Real Numbers Ex 2.7

Question 1.
Rationalise the denominator:
(i) \( \frac{1}{\sqrt{50}}\)
(ii) \( \frac{5}{3\sqrt{5}}\)
(iii) \( \frac{\sqrt{75}}{\sqrt{18}}\)
(iv) \( \frac{3\sqrt{5}}{\sqrt{6}}\)
Solution:
(i) \( \frac{1}{\sqrt{50}}\) = \(\frac{1}{\sqrt{25 \times 2}}=\frac{1}{5 \sqrt{2}}=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{5 \times 2}=\frac{\sqrt{2}}{10}\)

(ii) \( \frac{5}{3\sqrt{5}}\) = \(\frac{5}{3 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{5 \sqrt{5}}{3 \times 5}=\frac{\sqrt{5}}{3}\)

(iii) \( \frac{\sqrt{75}}{\sqrt{18}}\) = \(\frac{\sqrt{3 \times 25}}{\sqrt{2 \times 9}}=\frac{5 \sqrt{3}}{3 \sqrt{2}}=\frac{5 \sqrt{3}}{3 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{5 \sqrt{6}}{3 \times 2}=\frac{5 \sqrt{6}}{6}\)

(iv) \( \frac{3\sqrt{5}}{\sqrt{6}}\) = \( \frac{3 \sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=\frac{3 \sqrt{30}}{6}=\frac{\sqrt{30}}{2} \)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Question 2.
Rationalise the denominator and simplify:
(i) \(\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 1

(ii) \(\frac{5\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 2

(iii) \(\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 3

(iv) \(\frac{\sqrt{5}}{\sqrt{6}+2} – \frac{\sqrt{5}}{\sqrt{6}-2}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 4

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Question 3.
Find the value of a and b if \(\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b\).
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 5

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Question 4.
If x = \(\sqrt{7}\) + 2, then find the value of x² + \(\frac{1}{x^2}\)
Solution:
\(\sqrt{7}\) + 2 ⇒ x² = \((\sqrt{5}+2)^{2}\)
= \((\sqrt{5})^{2}\) + 2 × 2 × \(\sqrt{5}\) + 2² = 5 + 4 \(\sqrt{5}\) + 4 = 9 + 4\(\sqrt{5}\)
\(\frac{1}{x}=\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}=\frac{\sqrt{5}-2}{(\sqrt{5})^{2}-2^{2}}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\)
\(\frac{1}{x^{2}}\) = (\(\sqrt{5} – 2)^{2}\)
= \((\sqrt{5})^{2}\) – 2 × \(\sqrt{5}\) × 2 + 2² = 5 – 4 \(\sqrt{5}\) + 4 = 9 – 4 \(\sqrt{5}\)
∴ x² + \(\frac{1}{x^{2}}\) = 9 + \(4\sqrt{5}\) + 9 – \(4\sqrt{5}\) = 18
The value of x² + \(\frac{1}{x^{2}}\) = 18

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Question 5.
Given \(\sqrt{2}\) = 1.414, find the value of \(\frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}}\) (to 3 places of decimals).
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7 6
= 4 + \(\sqrt{2}\) = 4 + 1.414 = 5.414

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.7

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

Students can download Maths Chapter 2 Real Numbers Ex 2.6 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 2 Real Numbers Ex 2.6

Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\)
(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\)
(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\)
(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)
Solution:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\) = (5 + 18 – 2)\(\sqrt{3}\)
= (23 – 2) \(\sqrt{3}\) = 21\(\sqrt{3}\)

(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\) = (4 + 2 – 3) \(\sqrt[3]{5}\)
= (6 – 3) \(\sqrt[3]{5}\) = 3\(\sqrt[3]{5}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\) = \(3\sqrt{5^{2}×3} + 5\sqrt{2^{4}×3} – \sqrt{3^{5}}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 1
= 3 × 5\(\sqrt{3}\) + 5 × 2²\(\sqrt{3}\) – 3²\(\sqrt{3}\) = 15\(\sqrt{3}\) + 20\(\sqrt{3}\) – 9\(\sqrt{3}\)
= (15 + 20 – 9)\(\sqrt{3}\)
= (35 – 9)\(\sqrt{3}\)
= 26 \(\sqrt{3}\)

(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 2
= 5\(\sqrt[3]{2^{3}×5} + 2\sqrt[3]{5^{3}×5} – 3\sqrt[3]{2^{3}×2^{3}×5}\)
5 × 2\(\sqrt[3]{5} + 2 × 5\sqrt[3]{5} – 3 × 2 × 2\sqrt[3]{5} \)
= 10\(\sqrt[3]{5} + 10\sqrt[3]{5} – 12\sqrt[3]{5} \)
= 20\(\sqrt[3]{5} – 12\sqrt[3]{5}\)
= 8\(\sqrt[3]{5}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

Question 2.
Simplify the following using multiplication and division properties of surds:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\)
(ii) \(\sqrt{35}\) ÷ \(\sqrt{7}\)
(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\)
(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)
Solution:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\) = \(\sqrt{3×5×2} = \sqrt{30}\)

(ii) \(\sqrt{37} ÷ \sqrt{7} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{\frac{35}{7}} = \sqrt{5}\)

(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\) = \(\sqrt[3]{27×8×125}\)
= \(\sqrt[3]{3^{3}×2^{3}×5^{3}}\) = 3 × 2 × 5 = 30

(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
[using a2 – b2 = (a + b) (a – b)]
(7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\)) = \((7\sqrt{a})^{2} – (5\sqrt{b})^{2}\) = 49a – 25b

(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 3
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 4
= \(\frac{5}{36}\) × \(\frac{9}{4}\)
= \(\frac{5×1}{4×4}\)
= \(\frac{5}{16}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

Question 3.
If \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732, \(\sqrt{5}\) = 2.236, \(\sqrt{10}\) = 3.162, then find the values of the following correct to 3 places of decimals.
(i) \(\sqrt{40}\) – \(\sqrt{20}\)
(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\)
Solution:
(i) \(\sqrt{40}\) – \(\sqrt{20}\) = \(\sqrt{4×10} – \sqrt{4×5} = 2\sqrt{10} – 2\sqrt{5}\)
= 2 × 3.162 – 2 × 2.236 = 6.324 – 4.472 = 1.852

(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\) = \(\sqrt{3×100} + \sqrt{9×10} – \sqrt{4×2}\)
= 10\(\sqrt{3}\) + 3\(\sqrt{10}\) – 2\(\sqrt{2}\)
= 10 × 1.732 + 3 × 3.162 – 2 × 1.414
= 17.32 + 9.486 – 2.828
= 26.806 – 2.828
= 23.978

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

Question 4.
Arrange surds in descending order
(i) \(\sqrt[3]{5}\), \(\sqrt[9]{4}\), \(\sqrt[6]{3}\)
Solution:
LCM of 3, 9 and 6 is 18
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 5
\(\sqrt[3]{5}\) = \(\sqrt[3×6]{5^{6}}\) = \(\sqrt[18]{15625}\)
\(\sqrt[9]{4}\) = \(\sqrt[2×9]{4^{2}}\) = \(\sqrt[18]{16}\)
\(\sqrt[6]{3}\) = \(\sqrt[3×6]{3^{3}}\) = \(\sqrt[18]{27}\)
\(\sqrt[18]{15625}\) > \(\sqrt[18]{27}\) > \(\sqrt[18]{16}\)
\(\sqrt[3]{5}\) > \(\sqrt[6]{3}\) > \(\sqrt[9]{4}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

(ii) \(\sqrt[2]{\sqrt[3]{5}}\), \(\sqrt[3]{\sqrt[4]{7}}\), \(\sqrt{\sqrt{3}}\)
Solution:
\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\); \(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\); \(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\)
LCM of 6, 12 and 4 is 12
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6 6
\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\) = \(\sqrt[12]{5^{2}}\) = \(\sqrt[12]{25}\)
\(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\) = \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\) = \(\sqrt[12]{3^{3}}\) = \(\sqrt[12]{27}\)
\(\sqrt[12]{27}\) > \(\sqrt[12]{25}\) > \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) > \(\sqrt[2]{\sqrt[3]{5}}\) > \(\sqrt[3]{\sqrt[4]{7}}\)

Question 5.
Can you get a pure surd when you find:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes we can get a surd.
Example:
(a) 3\(\sqrt{2}\) + 5\(\sqrt{2}\) = (3 + 5)\(\sqrt{2}\) = 8\(\sqrt{2}\)
(b) 3\(\sqrt{6}\) + 2\(\sqrt{6}\) = (3 + 2)\(\sqrt{6}\) = 5\(\sqrt{6}\)

(ii) Yes we can get a surd.
Example:
(a) \(\sqrt{75}\) – \(\sqrt{48}\) = \(\sqrt{25×3}\) – \(\sqrt{16×3}\) = (5 – 4) \(\sqrt{3}\) = \(\sqrt{3}\)
(b) \(\sqrt{98}\) – \(\sqrt{72}\) = \(\sqrt{49×2}\) – \(\sqrt{36×2}\) = (7 – 6) \(\sqrt{2}\) = \(\sqrt{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

(iii) Yes we can get a surd.
Example:
(a) \(\sqrt{8}\) × \(\sqrt{6}\) = \(\sqrt{8×6}\) = \(\sqrt{48}\)
(b) \(\sqrt{11}\) × \(\sqrt{3}\) = \(\sqrt{11×3}\) = \(\sqrt{33}\)

(iv) Yes we can get a surd.
Example:
(a) \(\sqrt{55}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{11×5}}{\sqrt{5}} = \sqrt{11}\)
(b) \(\sqrt{65}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{13×5}}{\sqrt{13}} = \sqrt{5}\)

Question 6.
Can you get a rational number when you compute:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes, the sum of two surds will give a rational number.
Example:
(a) (2 + \(\sqrt{3}\)) + (2 – \(\sqrt{3}\)) = 4
(b) (\(\sqrt{5}\) + 4) + (7 – \(\sqrt{5}\)) = 11

(ii) Yes, the difference of two surds will give a rational number.
Example:
(a) (5 + \(\sqrt{7}\)) – (- 5 + \(\sqrt{7}\)) = 10
(b) (\(\sqrt{11}\) + 5) – (-3 + \(\sqrt{11}\)) = 8

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.6

(iii) Yes, the product of two surds will give a rational number.
Example:
(a) \(\sqrt{125}\) × \(\sqrt{45}\) = \(\sqrt{25×5}\) × \(\sqrt{9×5}\) = 5\(\sqrt{5}\) × 3\(\sqrt{5}\) = 15 × 5 = 75
(b) \(\sqrt{150}\) × \(\sqrt{6}\) = \(\sqrt{25×6}\) × \(\sqrt{6}\) = 5\(\sqrt{6}\) × \(\sqrt{6}\) = 5 × 6 = 30

(iv) Yes. The quotient of two surds will give a rational number.
Example:
(a) \(\sqrt{32}\) ÷ \(\sqrt{8}\) = \(\frac{\sqrt{8×4}}{\sqrt{8}} = \sqrt{4}\) = 2
(b) \(\sqrt{50}\) ÷ \(\sqrt{2}\) = \(\frac{\sqrt{25×2}}{\sqrt{2}} = \sqrt{25}\) = 5

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Students can download Maths Chapter 1 Set Language Ex 1.4 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 1.4

Question 1.
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find
(i) (P∪Q)∪R
(ii) (P∩Q)∩S
(m) (Q∩S)∩R
Solution:
P = {1, 2, 5, 7, 9}; Q = {2, 3, 5, 9, 11}; R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
(i) P∪Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}
= {1, 2, 3, 5, 7, 9, 11}
(P∪Q)∪R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}
= {1, 2, 3, 4, 5, 7, 9, 11}

(ii) P∩Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}
= {2, 5, 9}
(P∩Q)∩S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}
= (2, 5}

(iii) Q∩S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}
= {2, 3, 5}
(Q∩S)∩R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}
= {3, 5}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
P is a real number set
Q is a set of irrational number
∴ Q⊂P
P∪Q= Q∪P = P
∴ Union of sets is commutative.
P∩Q = Q∩P = Q
∴ Intersection of sets is commutative.

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Solution:
When union of sets is associative
A∪(B∪C) = (A∪B)∪C
(B∪C) = {m, n, q, s, t) ∪ {m, n, p, q, s}
= {m, n, p, q, s, t}
A∪(B∪C) = {p, q, r, s} ∪ {m, n, p, q, s, t}
= {m, n, p, q, r, s, t} ……..(1)
(A∪B) = {p, q, r, s} ∪ {m, n, q, s, t}
= {m, n, p, q, r, s, t}
(A∪B)∪C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, r, s, t} ……….(2)
From (1) and (2) it is verified that A∪(B∪C) = (A∪B)∪C

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Question 4.
Verify the associative property of intersection of sets for A = {-11, √2, √5, 7},
B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}.
Solution:
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9}
= {√3, √5}
A∩(B∩C) = {-11, √2, √5, 7} ∩ {√3, √5}
{√5} ………(i)
(A∩B) = {-11, √2, √5, 7} ∩ {√3, √5 ,6, 13}
= {√5}
(A∩B)∩C = {√5} n {√2, √3, √5, 9}
= {√5}……..(2)
From (1) and (2) it is verified that A∩(B∩C) = (A∩B)∩C

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Question 5.
If A={ x : x = 2n, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2n, n ∈ W and n < 4}
A = {1, 2, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
B = {2, 4, 6, 8}
C ={0, 1, 2, 5, 6}
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {2, 4, 6, 8} ∩ {0, 1, 2, 5, 6}
= {2, 6}
A∩(B∩C)= {1 ,2, 4, 8} ∩ {2, 6}
= {2}……….(1)
(A∩B) = {1, 2, 4, 8} ∩ {2, 4, 6, 8}
= {2, 4, 8}
(A∩B)∩c= {2, 4, 8} n {0, 1, 2, 5, 6}
= {2}………..(2)
From (1) and (2) we get A∩(B∩C) = (A∩B)∩C

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.4

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Students can download Maths Chapter 1 Set Language Ex 1.3 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 1.3

Question 1.
Using the given venn diagram, write the elements of
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 1
(i) A
(ii) B
(iii) A∪B
(iv) A∩B
(v) A – B
(vi) B – A
(vii) A’
(viii) B’
(ix) U
Solution:
(i) A = {2, 4, 7, 8, 10}
(ii) B = {3, 4, 6, 7, 9, 11}
(iii) A∪B = {2, 3, 4, 6, 7, 8, 9, 10, 11}
(iv) A∩B = {4, 7}
(v) A – B = {2, 8, 10}
(vi) B – A = {3, 6, 9, 11}
(vii) A’ = {1, 3, 6, 9, 11, 12}
(viii) B’ = {1,2, 8, 10, 12}
(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Question 2.
Find A∪B, A∩B, A – B and B – A for the following sets
(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}
Solution:
A∪B = {2, 6, 10, 14} ∪ {2, 5, 14, 16}
= {2, 5, 6, 10, 14, 16}
A∩B = {2, 6, 10, 14} ∩ {2, 5, 14, 16}
= {2, 14}
A – B = {2, 6, 10, 14} – {2, 5, 14, 16}
= {6, 10}
B – A = {2, 5, 14, 16} – {2, 6, 10, 14}
= {5, 16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}
Solution:
A∪B = {a, b, c, e, u} ∪ {a, e, i, o, u}
= {a, b, c, e, i, o, u}
A∩B = {a, b, c, e, u} ∩ {a, e, i, o, u}
= {a, e, u}
A – B = {a, b, c, e, u} – {a, e, i, o, u}
= {b, c}
B – A = {a, e, i, o, u} – {a, b, c, e, u}
= {i, o}

(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6}
Solution:
A = {1, 2, 3, 4, 5, 6,7, 8, 9, 10} and B = {0, 1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5}
A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5}
= {6, 7, 8, 9, 10}
B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {0}

(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”
Solution:
A = {m, a, t, h, e, i, c, s}
B = {g, e, o, m, t, r, y}
A∪B = {m, a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y}
= {a, c, e, g, h, i, m, o, r, s, t, y}
A∩B= {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t, r, y}
= {e, m, t}
A – B = {m, a, t, h, e, i, c, 5} – {g, e, o, m, t, r, y}
= {a, c, h, i, s}
B -A = {g, e, o, m, t, r, y} – {m, a, t, h, e, i, c, s}
= {g, o, r, y}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Question 3.
If U = {a, b, c, d, e, f, g, h), A = {b, d, f, h} and B = {a, d, e, h}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {a, b, c, d, e, f, g, h} – {b, d, f, h)
= {a, c, e, g}

(ii) B’
Solution:
B’ = U – B
= {a, b, c, d, e, f, g, h} – {a, d, e, h}
= {b, c, f, g}

(iii) A’∪B’
Solution:
A’∪B’ = {a, c, e, g} ∪ {b, c,f, g}
= {a, b, c, e, f, g}

(iv) A’∩B’
Solution:
A’∩B’ = {a, c, e, g} ∩ {b, c, f, g}
= {c, g}

(v) (A∪B)’
Solution:
A∪B = {b, d, f, h} ∪ {a, d, e, h}
= {a, b, d, e, f, h}
(A∪B)’ = U – (A∪B)
= {a, b, c, d, e, f, g, h} – {a, b, d, e, f, h}
= {c, g}

(vi) (A∩B)’
Solution:
(A∩B) = {b, d, f, h) ∩ {a, d, e, h)
= {d, h}
(A∩B)’ = U – (A∩B)
= {a, b, c, d, e, f, g, h} – {d, h}
= {a, b, c, e, f, g}

(vii) (A’)’
Solution:
A’ = {a, c, e, g}
(A’)’ = U – A’
= {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(viii) (B’)’
Solution:
B’ = {b, c, f, g}
(B’)’ = U – B’
= {a, b, c, d, e, f, g, h) – {b, c, f, g)
= {a, d, e, h}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Question 4.
Let U = {0, 1, 2, 3, 4, 5, 6, 7} A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {0, 1, 2, 3, 4, 5, 6, 7} – {1, 3, 5, 7}
= {0, 2, 4, 6}

(ii) B’ = U – B
Solution:
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 3, 5, 7}
= {1, 4, 6}

(iii) A’∪B’
Solution:
A’∪B’ = {0, 2, 4, 6,}∪{1, 4, 6}
{0, 1, 2, 4, 6}

(iv) A’∩B’
Solution:
A’∩B’ = {0, 2, 4, 6,}∩{1, 4, 6}
{4, 6}

(v) (A∪B)’
Solution:
A∪B = {1, 3, 5, 7}∪{0, 2, 3, 5, 7}
= {0, 1, 2, 3, 5, 7}
(A∪B)’ = U – (A∪B)
{0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7}
{4, 6}

(vi) (A∩B)’
Solution:
(A∩B)= {1, 3, 5, 7}∩{0, 2, 3, 5, 7}
= {3, 5, 7}
(A∩B)’ = U – (A∩B)
= {0, 1, 2, 3, 4, 5, 6, 7} – {3, 5, 7}
= {0, 1, 2, 4, 6}

(vii) (A’)’
A’ = {0, 2, 4, 6}
(A’)’ = U – A’
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6}
= {1, 3, 5, 7}

(viii) (B’)’
B’ = {1, 4, 6}
(B’)’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6}
= {0, 2, 3, 5, 7}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Question 5.
Find the symmetric difference between the following sets.
(i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}
Solution:
Use anyone of the formula to find A & B
AΔB = (A – B)∪(B – A) or AΔB = (A∪B) – (A∩B)
P∪Q = {2, 3, 5, 7, 11} ∪ {1,3,5, 11}
= {1, 2, 3, 5, 7, 11}
P∩Q = {2, 3, 5, 7, 11}∩{1, 3, 5,11}
= {3, 5, 11}
PΔQ = (P∪Q) – (P∩Q)
= {1, 2, 3, 5, 7, 11} – {3, 5, 11}
= {1, 2, 7}
(OR)
P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11}
= {2,7}
Q – P = {1, 3, 5, 11} – {2, 3, 5, 7, 11}
= {1}
PΔQ = (P – Q)∪(Q – P)
= {2, 7} ∪ {1}
= {1, 2, 7}

(ii) R = {l, m, n, o, p} and S = {j, l, n, q}
Solution:
R- S = {l, m, n, o, p} – {j, l, n, q}
= {m, o, p}
S – R = {j, l, n, q} – {l, m, n, o, p}
= {j, q}
RΔS = (R – S)∪(S – R)
= {m, o, p} – {j, q} = {j, m, o, p, q)
(OR)
R∪S = {l, m, n, o, p} ∪ {j,l,n,q}
= {l, m, n, o, p, j, q}
R∩S = {l, m, n, o,p} ∩ {j, l, n, q}
= {l, n}
RΔS = (R∪S) – (R∩S)
= {l, m, n, o, p, j, q} – { l, n}
= {m, o, p, j, q}

(iii) X = {5, 6, 7} and Y = {5, 7, 9, 10}
Solution:
X∪Y = {5, 6, 7} ∪ {5, 7, 9, 10}
= {5 ,6, 7, 9, 10}
X∩Y = {5, 6, 7} ∩ {5, 7, 9, 10}
= {5, 7}
XΔY = (X∪Y) – (X∩Y)
= {5, 6, 7, 9, 10} – {5, 7}
= {6, 9, 10}
OR
X – Y = {5, 6, 7} – {5, 7, 9, 10} = {6}
Y – X = {5, 7, 9, 10} – {5, 6, 7} = {9, 10}
XΔY = (X – Y) ∪ (Y – X)
= {6}∪{9, 10}
= {6, 9, 10}

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

Question 6.
Using the set symbols, write down the expressions for the shaded region in the following
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 2
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 3

Question 7.
Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following,
(i) A∪B
(ii) A∩B
(iii) (A∩B)’
(iv) (B – A)’
(v) A’∪B’
(Vi) A’∩B’
(vii) What do you observe from the Venn diagram (iii) and (v)?
Solution:
(i) A∪B
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 4

(ii) A∩B
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 5

(iii) (A∩B)’
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 6

(iv) (B – A)’
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 7

(v) A’∪B’
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 8

(Vi) A’∩B’
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 9

Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3

(vii) What do you observe from the diagram (iii) and (v)?
From the diagram (iii) and (v) we get (A∩B)’ = A’∪B’
Samacheer Kalvi 9th Maths Guide Chapter 1 Set Language Ex 1.3 10

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2

Therefore, one Decimal is equal to four hundred and thirty five decimal point six Square Feet (sq ft) in Online.

Question 1.
Express the following rational numbers into decimal and state the kind of decimal expression.
(i) \(\frac{2}{7}\)
(ii) -5\(\frac{3}{11}\)
(iii) \(\frac{22}{3}\)
(iv) \(\frac{327}{200}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 1
(i) \(\frac{2}{7}\) = 0.2857142….
= 0.\(\overline {285714}\)
Non-terminating and recurring decimal expansion.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(ii) -5\(\frac{3}{11}\) = -5 + 0.272 = -5.272……..
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 2
= -5.\(\overline {27}\)
Non-terminating and recurring decimal expansion.

(iii) \(\frac{22}{3}\) = 7.333……..
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 3
= 7.\(\overline {3}\)
Non-terminating and recurring decimal expansion.

(iv) \(\frac{327}{200}\) = \(\frac{327}{2×100}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 4
= \(\frac{3.27}{2}\)
= 1.635
Terminating decimal expansion.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

How to convert 1/32 to decimal form? … In the fraction 1/32, 1 is the numerator and 32 is the denominator, the fraction bar means “divided by”.

Question 2.
Express \(\frac{1}{13}\) in decimal form. Find the length of the period of decimals.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 5
\(\frac{1}{13}\) = 0.07692307
= 0.\(\overline {076923}\)
Length of the period of decimal is 6.

9/20 as a decimal is 0.45

Question 3.
Express the rational number \(\frac{1}{33}\) in recurring decimal form by using the recurring decimal expansion of \(\frac{1}{11}\). Hence write \(\frac{71}{33}\) in recurring decimal form.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 6
\(\frac{1}{11}\) = 0.0909……… = 0.\(\overline {09}\)
∴ \(\frac{1}{33}\) = \(\frac{1}{3}\) × \(\frac{1}{11}\)
= \(\frac{1}{3}\) × 0.0909 ……..
= 0.0303 …… = 0.\(\overline {03}\)
\(\frac{71}{33}\) = 2\(\frac{5}{33}\) = 2 + \(\frac{5}{33}\) = 2 + 5 × \(\frac{1}{33}\)
= 2 + 5 × 0.\(\overline {03}\)
2 + (5 × 0.030303 ……..)
2 + 0.151515 ………
2+ 0.\(\overline {15}\)
2.\(\overline {15}\)

That’s literally all there is to it! 1/2 as a decimal is 0.5.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

Question 4.
Express the following decimal expression into rational numbers.
(i) 0.24
Solution:
Let x = 0.242424 ………. →(1)
100 x = 24.2424 ……… →(2)
(2) – (1) ⇒ 100 x – x = 24.2424 ……….. (-)
 0.2424 ……..
99 x = 24.0000
x = \(\frac{24}{99}\)
(or)
\(\frac{8}{33}\)

(ii) 2.327
Solution:
Let x = 2.327327327 ………. →(1)
1000 x = 2327.327327 ……… →(2)
(2) – (1) ⇒ 1000 x – x = 2327.327327 ……….. (-)
  2.327327 ……..
999 x = 2325.000
x = \(\frac{2325}{999}\)
(or)
\(\frac{775}{333}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

3/5 as a decimal is 0.6

(iii) – 5.132
Solution:
– 5.132 = -5 + \(\frac{1}{10}\) + \(\frac{3}{100}\) + \(\frac{2}{1000}\)
= \(\frac{-5000 + 100 +30 + 2}{1000}\) = \(\frac{-4868}{1000}\)
(or)
\(\frac{-1217}{250}\)

(iv) 3.17
Solution:
Let x = 3.1777 ………. →(1)
10 x = 31.777 ……… →(2)
100 x = 317.77 …….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 317.77 ……….. (-)
 31.777 ……..
90 x = 286.000
x = \(\frac{286}{90}\)
(or)
\(\frac{143}{45}\)

(v) 17.215
Solution:
Let x = 17.2151515 ………. →(1)
10 x = 172.151515 ……… →(2)
100 x = 17215.1515 …….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 17215.1515 ……….. (-)
 17215.1515 ……..
990 x = 17043
x = \(\frac{17043}{990}\)
(or)
\(\frac{5681}{330}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(vi) -21.2137
Solution:
Let x = -21.213777 ………. →(1)
1000 x = -21213.777 ……… →(2)
100 x = -212137.77 …….. →(3)
(3) – (2) ⇒ 10000 x – 1000 x = -21213.777 ……….. (-)
-21213.777 ……..
9000 x = -190924
x = \(\frac{-190924}{9000}\)
(or)
\(\frac{-47731}{2250}\)

That’s literally all there is to it! 11/16 as a decimal is 0.6875.

Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expression.
(i) \(\frac{7}{128}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 7
\(\frac{7}{128}\) = \(\frac{7}{2^{7}}\)
∴ \(\frac{7}{128}\) has terminating decimal expression.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(ii) \(\frac{21}{15}\)
Solution:
\(\frac{21}{15}\) = \(\frac{7}{5}\) = \(\frac{7}{5^1}\)
\(\frac{21}{15}\) has terminating decimal expression.

(iii) 4\(\frac{9}{35}\)
Solution:
4\(\frac{9}{35}\) = \(\frac{149}{35}\)
4\(\frac{149}{5×7}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ 4\(\frac{9}{35}\) has non-terminating recurring decimal expression.

(iv) \(\frac{219}{2200}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 8
\(\frac{219}{2200}\) = \(\frac{219}{2^{3} × 5^{2} × 11}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ \(\frac{219}{2200}\) has non-terminating recurring decimal expression.

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