# Samacheer Kalvi Class 9 Maths Chapter 1 Set Language Solutions

## Samacheer Kalvi Class 9 Maths Chapter 1 Set Language Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 1 Set Language. Here students can easily find all the solutions for Set Language Exercise 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7 and 1.8. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus. #### Set Language Exercise 1.1 Solutions :

(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) The collection of prime numbers up to 100 is set.

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97}

(ii) There are lots of rich people in India. So, it’s not a set.

(iii) There are lots of rivers in India. It’s not set.

(iv) The collection of good Hockey players. It’s not set.

(4) Represent the following sets in Roster form.

(i) A = The set of all even natural numbers less than 20.

(ii) B = {y : y = 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) Roster form : A = {2, 4, 6, 8, 10, 12, 14, 16, 18}

(ii) Roster form : B = {1/2, 1/4, 1/6, 1/8, 1/10}

Y = 1/2n n = N, n ≤ 5

If n = 1, y = 1/2

n = 2, y = 1/4

n = 3, y = 1/6

n = 4, y = 1/8

n = 5, y = 1/10

(iii) Roster form : C = {64, 125}

4 × 4 × 4 = 60,

5 × 5 × 5 = 125

(iv) Roster form : D = {-4, -3, -2, -1, 0, 1, z)

(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 = {1/2, 2/3, 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 :

Set builder form :

(i) B = {A : A is an Indian player who scored double centuries in one Day internationals}

(ii) C = {B : B = n/n+1 n ∈ N}

(iv) D = {C : C is a Tamil month in a year}

(v) E = {D : D is an odd number and D < 9}

(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 :

Descriptive form :

(i) P is the set of months storing starting with j

(ii) Q is the set of all prime number where 5 < Q < 31.

(iii) R is the set of all natural numbers less than 5

(iv) S is a set of consonants in English alphabets

#### Set Language Exercise 1.2 Solutions :

(1) Find the cardinal number of the following sets.

(i) M = {p, q, r, s, t, u}

(ii) P = {x : x = 3n+2, n W and x < 15}

(iii) Q = {y : y = 4/3n, n N and 2 < n ≤ 5}

(iv) R = {x : x is an integers, x Z and –5 ≤ x <5}

(v) S = The set of all leap years between 1882 and 1906.

Solution :

(i) The cardinal number of M = {p, q, r, s, t, u} = n (M) = P.

(ii) The cardinal number of P = {x : x = 3n+2, n ∈ W and x < 15}

= n (P) = 5. n = 0,

x = z …… n = 4

x = 14

(iii) The cardinal number of Q = {y : y = 4/3n, n ∈ N and 2 < n ≤ 5}

= 3, y = 4/3×3 = 4/9

n = 4, y = 4/3×3 = 1/3

n = 5, y = 4/3×5 = 4/15

n (Q) = 3

(iv) The cardinal number of R = {x : x is an integers, x ∈ Z and –5 ≤ x <5}

= R = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4}

= n (R) = 10

(v) The cardinal number of S = the set of all leap years between 1882

and 1906
S = {1884, 1888, 1892, 1896, 1904}

n (s) = 5

(2) Identify the following sets as finite or infinite.

(i) X = The set of all districts in Tamilnadu.

(ii) Y = The set of all straight lines passing through a point.

(iii) A = { x : x Z and x <5}

(iv) B = { x : x2–5x + 6 = 0, x N }

Solution :

(i) Finite set.

(ii) Infinite set.

(iii) Infinite set.

(iv) Finite set.

x² – 5x + 6 = 0

= (x – 3) (x – 2) = 0

∴ B = 2, 3

(3) Which of the following sets are equivalent or unequal or equal sets?

(i) A = The set of vowels in the English alphabets.

B = The set of all letters in the word “VOWEL”

(ii) C = {2, 3, 4, 5}

D = { x : x W, 1< x < 5}

(iii) X = { x : x is a letter in the word “LIFE”}

Y = { F, I, L, E}

(iv) G = { x : x is a prime number and 3 < x < 23}

H = { x : x is a divisor of 18}

Solution :

(i) A = {a, e, I, o, u} B = {v, o, w, E, L}

∴ They are equivalent set

(ii) It is a Unequal set.

(iii) It is a Equal set.

(4) Identify the following sets as null set or singleton set.

(i) A = {x : x N , 1 < x < 2}

(ii) B = The set of all even natural numbers which are not divisible by 2

(iii) C = {0}.

(iv) D = The set of all triangles having four sides.

Solution :

(i) It is a null set.

(ii) It is a null set.

(iii) It is a singleton set.

(iv) It is a null set.

(5) State which pairs of sets are disjoint or overlapping?

(i) A = {f, i, a, s} and B = {a, n, f, h, s}

(ii) C = {x : x is a prime number, x >2} and D = {x : x is an even prime number}

(iii) E = {x : x is a factor of 24} and  F = {x : x is a multiple of 3, x < 30}

Solution :

(i) A∩B = {f, i, a, s} ∩ {a, n, f, n, s} = {f, a, s}

∴ They are overlapping.

C∩D = {3, 5, 7, 11 ….} ∩ {z} = {0}

∴ They are non-overlapping on disjoint set.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24

F = {3, 6, 9, 12, 15, 18, 21, 24, 27}

E∩F = {3, 6, 12, 24}

∴ They are overlapping.

(7) If A = {a, {a, b}}, write all the subsets of A.

Solution :

All the subsets of A are, {a}, {a, b}, {a, {a, b}}

(8) Write down the power set of the following sets:

(i) A = {a, b}

(ii) B = {1, 2, 3}

(iii) D = {p, q, r, s}

(iv) E =

Solution :

(i) All the subsets of A are, {a}, {a, b}, {a, {a, b}}

(9) Find the number of subsets and the number of proper subsets of the following sets.

(i) W = {red, blue, yellow}

(ii) X = { x2 : x N , x2 ≤ 100}.

Solution :

(i) W = {red , blue, yellow}

N (w) = 3

∴ The number of set,

= 2n = 2³

= 8

∴ Then proper subsets,

= 8 -1

= 7

(ii) x = {1, 2, 3}

x² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Number of subsets

= 2^10

= 1024

∴ The number of proper subsets

= 1024 – 1

= 1023

(10) – (i) If n(A) = 4, find n[P(A)].

Solution :

n (A) = 4

n [P(A)] = 2n

= 2⁴

= 16

(ii) If n(A)=0, find n[P(A)].

Solution :

n (A) = 0

∴ n [P(A)] = 2° = 1

(iii) If n[P(A)] = 256, find n(A).

Solution :

n [P(A)] = 256

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

n [P(A)] = 28

Therefore, n (A) = 8

#### Set Language Exercise 1.3 Solutions :

(1) Using the given Venn diagram, write the elements of

(i) A

(ii) B

(iii) AB

(iv) A∩B

(v) A–B

(vi) B–A

(vii) A′

(viii) B′

(ix) U

Solution :

(i) The elements of A = {2, 4, 7, 8, 10}

(ii)The elements of B = {3, 4, 7, 6, 9, 11}

(iii) The elements of A∪B = {2, 3, 4, 6, 7, 8, 9, 10, 11}

(iv) The elements of A∩B = {4, 7}

(v) The elements of A – B = {8, 2, 10}

(vi)The elements of B – A = {3, 9, 6, 11}

(vii) The elements of A′ = {1, 6, 3, 9, 11, 12}

(viii) The elements of B′ = {1, 2, 8, 10, 12}

(ix) The elements of U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

(2) Find AB, 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, 6, 5, 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}

(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}

= {m, t, e}

A∪B

= {m, a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y}

= {m, a, t, h, e, i, c, s, g, o, r, y}

A – B

= {m, a, t, h, e, i, c, s} – {g, e, o, m, t, r, y}

= {a, h, i, c, s}

B – A

= {g, e, o, m, t, r, y} – {m, a, t, h, e, i, c, s}

= {g, o, r, y}

(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′

(ii) B′

(iii) A′B′

(iv) A′∩B′

(v) (AB)′

(vi) (A∩B)′

(vii) (A′)′

(viii) (B′)′

Solution :

(i) U = {a, b, c, d, e, f, g, h}

A = {b, d, f, h}

B = {a, d, e, h}

Therefore,

= A′ = U – A

= {a, c, e, g}

(ii) Given, U = {a, b, c, d, e, f, g, h}

A = {b, d, f, h}

B = {a, d, e, h}

∴ (B′)′ = U – B′

= {a, b, c, d, e, f, g, h} – {b, c, f, g}

= {a, d, e, h}

(iii) A′ ∩ B′ = {a, c, e, g} ∩ {b, c, f, g}

= {c, g}

(iv) (A∩B)′ = ∪ – (A∩B)

= {a, b, c, d, e, f, g, h} – {d, h}

= {a, b, c, e, f, g}

(v) A′∪B′ = {a, c, e, g} ∪ {b, c, f, g}

= {a, b, c, e, f, g}

(vi) (A∪B)′ = ∪ (A∪B)

= {a, b, c, d, e, f, g, h} – {a, b, d, e, f, h}

= {a, b, c, e, f, g}

(vii) (A′)′ = U – A′

= {a, b, c, d, e, f, g, h} – {a, c, e, g}

= {b, d, f, h}

(viii) (B′)′ = U – B′

= {a, b, c, d, e, f, g, h} – {b, c, f, g}

= {a, d, e, h}

(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′

(ii) B′

(iii) A′B′

(iv) A′∩B′

(v) (AB)′

(vi) (A∩B)′

(vii) (A′)′

(viii) (B′)′

Solution :

(i) Given, ∪ = {0, 1, 2, 3, 4, 5, 6, 7}

A = {1, 3, 5, 7}

B = {0, 2, 3, 5, 7}

∴ A′ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 3, 5, 7}

= {0, 2, 4, 6}

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

= {1, 4, 6}

(iii) A′ ∪ B′ = {0, 2, 4, 6} ∪ {1, 4, 6}

= {0, 1, 4, 6}

(iv) A′ ∩ B′

= {0, 2, 4, 6} ∩ {1, 4, 6}

= {4, 6}

(v) (A∪B)′ = ∪ – (A∪B) = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7}

= {4, 6}

(vi) (A∩B)′ = ∪ – (A∩B)

= {0, 1, 2, 3, 4, 5, 6, 7} – {3, 5, 7}

= {0, 1, 2, 4, 6}

(5) Find the symmetric difference between the following sets.

(i) P = {2, 3, 5, 7, 11} and Q={1, 3, 5, 11}

(ii) R = {l, m, n, o, p} and S = {j, l, n, q}

(iii) X = {5, 6, 7} and Y={5, 7, 9, 10}

Solution :

(i) P = {2, 3, 5, 7, 11} Q = {1, 3, 5, 11}

P – Q = {2, 7} Q – P = {1}

P△Q = (P – Q) ∪ (Q – P)

= {2, 7} ∪ {1} = {1, 2, 7}

(ii) R = {l, m, n, o, p} S = {j, i, n, q}

R – S = {m, o, p} S – R = {j, q}

R△S = (R – S) ∪ (S – R) = {m, o, p} ∪ {j, q} = {j, m, o, p, q}

(iii) x = {5, 6, 7} y= {5, 7, 9, 10}

x – y = {6}, y – x = {9, 10}

x△y = (x-y) ∪ (y – x) = {6} ∪ {9, 10} = {6, 9, 10}

#### Set Language Exercise 1.4 Solutions :

(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 (PQ)R

(i) (PQ)R

Solution :

[{1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}] ∪ {3, 4, 5, 7, 9}

= {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}

= {1, 2, 3, 4, 5, 7, 9, 11}

(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 a rational number between 2 and 7}.

Solution :

P = {3, 4, 5, 6} Q = {√3, √5, √6}

P∪Q = {4, 3, 4, 5, 6} ∪ {√3, √5, √6} = {3, 4, 5, 6, √3, √5, √6} …… (i)

Q∪P = {√3, √5, √6} ∪ {3, 4, 5, 6} = {√3, √5, √6, 3, 4, 5, 6} …… (ii)

Comparing (i) and (ii)

We get P∪Q = Q∪P (Proved)

(P∩Q) = {3, 4, 5, 6} ∩ {√3, √5, √6} = {} …… (iii)

(Q∩P), {√3, √5, √6} ∩ {3, 4, 5, 6} = {} ……. (iv)

From (iii) and (iv) we get

P∩Q = Q∩P ….(Proved)

(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 :

(A∪B) ∪ C

= [{P, q, r, s} ∪ {m, n, q, s, t}] ∪ {m, n, p, q, s}

= {p, q, r, s, m, n, t} ∪ {m, n, p, q, s}

= {p, q, r, s, m, n, t} ………..(i)

A ∪ (B∪C)

= {P, q, r, s} ∪ [{m, n, q, s, t} ∪ {m, n, p, q, s}]

= {p, q, r, s} ∪ {m, n, p, q, s, t}

= {m, n, p, q, r, s, t} ……..(2)

Comparing (i) and (2) we get

(A∪B) ∪ c = A∪ (B∪C) (Proved)

(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 :

A∩ (B∩C) = {-11, √2, √5, 7} ∩ [{√3, √5, 6, 13} ∩ {√2, √3, √5, 9}

= {-11, √2, √5, 7} ∩ {√3, √5}

= {√5} ………(1)

(A∩B) ∩c = [{-11, √2, √5, 7} ∩ {√3, √5, 6, 13}] ∩ {√2, √3, √5, 9}

= {√5} ∩ {√2, √3, √5, 9}

= {√5} ………..(2)

Comparing (1) and (2) we get

A∩ (B∩C) = (A∩B) ∩C ….(Proved)

#### Set Language Exercise 1.5 Solutions :

(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 :

(i) A – B = {3, 4, 6}

(ii) B – C = {-1, 5, 7}

(iii) A′∪B′

= {1, 2, 0, -3, 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}

(B∪C)′ = ∪ – (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}

(2) If If K = {a, b, d, e, f}, L = {b, c, d, g} and M = 2a, b, c d, h} then find the following

(i)  K (L ∩ M)

(ii) K ∩ (L M)

(iii) (K L) ∩ (K M)

and verify distributive laws

Solution :

Given, K = {a, b, d, e, f},

L = {b, c, d, g}

M = 2a, b, c d, h}

(i) K ∪ (L ∩ M)

= K∪ [{b, c, d, g} ∩ 2a, b, c, d, h}]

= 2 a, b, d, e, h} ∪ 2b, c, d}

= 2a, b, c, d, e, f}

(ii) K∩ (L ∪ M)

= K∩ [2b, c, d, g} ∪ {a, b, c, d, h}

= {a, b, d, e, f} ∪ {a, b, c, d, g, h]

= {a, b, d}

(iii) (K ∪ L) ∩ (K ∪ M)

= [{a b, c, d, e, f, g} ∩ {a b, c, d, e, f, h}

= {a, b, c, e, f}

(5) If A = {b, c, e, g, h}, B = {a, c, d, g, i} 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, I,}

C = {a, d, e, g, h}

A – (B ∩ C) = A- [{c, d, g, I, a} ∩ {a, d, e, g, h}]

= {b, c, e, g, h} – {a, d, g}

= {b, c, e, h} …..(i) no

A-B = {b, c, e, g, h} – 2a, 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) no

From (1) and (2) we get

A (B ∩ c) = (A-B) ∪ (A-c)

(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 – (BC) = (A – B)∩(A – C)

Solution :

A = {-2, 0, 1, 3, 5}

B = {-1, 0, 2, 5, 6}

C = {-1, 2, 5, 6, 7}

A – (B ∪ C) = A – [{-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}]

= {-2, 0, 1, 3, 3, 5} – {-1, 0, 2, 5, 6, 7}

= {-2, 1, 3} …. (i)

(A-B) ∩ (A-C)

= [{-2, 0, 13, 5} – {-1, 0, 2, 5, 6}] – [{-2, 0, 1, 3, 5} – {-1, 2, 5, 6, 7}]

= {-2, 1, 3} ∩ {-2, 0, 1, 3}

= {-2, 1, 3} …. (2) no

Complaining (ii) and (ii) we get

A – (B ∪ C) = (A-B) ∩ (A-C) ….Proved

(8) If A = {y:y = a+1/2, aW and a ≤ 5}, B = {y:y = 2n-1/2, nW an n < 5} and C = {-1,-1/2,1,3/2, 2}, then show that A – (BC) = (A-B)∩(A-C)

Solution :

A = {y – y = a+1/2 a Î W and a ≤ 5}

So, a = {0, 1, 2, 3, 4, 5}

Y = 0+1/2 = 1/2 when a = 0

Y = 1+1/2 = 1, when a = 1

Y = 3/2, when a = 2

Y = 4/2, when a = 3

Y = 5/2, when a = 4

Y = 6/2, when a = 5

Because, A = { 1/2, 1, 3/2, 2, 5/2, 3}

B = {y:y = 2n-1/2 n Î W and n < 5}

So, y = 2 × 0 – 1/2 = -1/2 when n = 0

Y = 2 × 1 – 1/2 = 1/2 when n = 1

Y = 3/2, when n = 2

Y = 5/2, when n = 3

Y = 7/2, when n = 4

Therefore,

B = { -1/2, 1/2, 3/2, 5/2, 7/2}

C = { -1, -1/2, 1, 3/2, 2}

A = {(B∪ C} = { 1/2, 1, 3/2, 2, 5/2, 3} – {-1, -1/2, 1, 1/2, 2, 3/2, 2 5/2, 7/2}

= {3} …..(i)

A – ( B∪ C) = (A-B) ∩ (A – C)

= {1,2,3} ∩ { 1/2,5/2,3}

= {3} …. (i)

Complaining (i) and (2) we get A-( B∪ C) = (A-B) ∩ (A – C)

(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

Solution :

U = {4, 7, 8, 10, 11, 12, 15, 16}

A = {7, 8, 11, 12}

B = {4, 8, 12,15}

Be De Morgan’s law

(A U B) = {4, 7, 8, 11, 12, 15}

(A U B)′ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}

= {10, 16} – (i) no

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}

= {16, 16} – (2) no

Complaining (i) and (ii) we get

(A ∪ B)′ = A′ ∩ B′ proved

#### Set Language Exercise 1.6 Solutions :

(1) (i) If n(A) = 25, n(B) = 40, n(AB) = 50 and n(B′) = 25 , find n(A∩B) and n(U)

Solution :

n(A) = 25, n(B) = 40 n(A ∪ B) = 50 n(B′) = 25

n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 25 + 40 – 50 = 15

n(∪) = n (B) + n(B)

= 40 + 25 = 65

(ii) If n(A) = 300, n(AB) = 500, n(A∩B) = 50 and n(B′) = 350, find n(B) and n(U)

Solution :

n(A) = 300 n (A ∩ B) = 500 n(A ∩B) = 50 n(B′) = 350

= n (A ∩ B) = n(A) + n (B) – n (A∩ B)

= n(B) = x (A∩ B) – n(A) + n(A∪ B)

= 50- 300 + 500

= 250

n (∪) = n (B) + n(B)

= 250 + 350 = 600

(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(AB) = n(A) + n(B) – n(A∩B)

Solution :

So, n(A) = 5 and x(B) = 5

A ∪ B = {1, 2,3, 4, 5, 8, 10}

A ∩ B = {2, 8, 10}

n(A ∪ B) = 7 …..(i)

Again n(A) + n(b) – n(A∩ B)

= 5+5-3

= 7 ……(i)

Complaining (i) and (ii) we get

n(A ∪ B) = n(A) + n(B) –n(A ∩ B)

(3) Verify n(ABC) = 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}

(ii) A = {1,3,5}, B = {2,3,5,6} and C = {1,5,6,7}

Solution :

(i) n(A) = 5, n(B) = 4, n(C) = 4

n(A∩B)= 3, n(b ∩ C) = 2, n((A ∩ C) = 3 and n(A ∩ B ∩ C) = 2

A ∩ B = {c, e, f} B ∩ C = {c, f} A ∩ C = {a, c, f}

A ∩ B ∩ C = {c, f}

A ∪ B ∪ C = {a, c, d, e, f, b, h}

Therefore, n(A ∪ B ∪ C) = 7 —-(i)

So, n (A) + n (B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)

= 5+4+4-3-2-3+2

= 7 …..(ii)

From (i) and (ii) we get the prove

(ii) 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

A B C = {1, 2, 3, 5, 6, 7}

n( A ∪ B ∪ C) = 6 …..(i)

From (i) and (ii) we get the prove

Again, n(A) + n(B) + n(c) – n(A ∩ B) – x (A ∩ c) – n(B ∩ c) + n(A ∩ B ∩ c)

= 3 + 4 + 4 – 2 – 2 – 2 +1

= 6 ….(i)

From (i) and (ii) we get the prove.

(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, we like tea A

We like coffee B

N(A) = 35 n(B) = 20 n(A ∪ B) = 45

(i) Number of people who like both tea and coffee

= n(A) + n(B) – x (A ∪ B)

= 35 + 20 – 45 = 10

(ii) Number of people who don’t like coffee

= N(∪) – n (A)

= 45 – 35 = 10

(iii) Number of people who don’t like coffee

= n (∪) – n(B) = 45 – 20 = 25

(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

Given, n(A) = n(B)

Since, n(A) and n(B)

So, 32 + x + x = 5x + x

= 4x = 32

= x = 8

So, the value of x is 8

(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 :

Given n(A B) = 500

n(A) = 400 and n(B) = 200

n(A ∩ B) = 50

Therefore,

= n(A ∪ B) = n(A) + n(B) – x( A ∩ B)

= 400+200-50

= 550

No, this data is not correct.

(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 be, Tamil newspaper readers n(A) = 275

English newspaper readers n(B) = 150

Hindi newspaper readers n(C) = 45

Tamil and English newspaper readers n(A∩ B) = 125

English and Hindi newspaper readers n(B ∩ C) = 17

Tamil and Hindi news paper readers n(A∩ C) = 5

Tamil, English and Hindi newspaper readers n(A ∩ B ∩ C) = 3

(i) Number of families who buy only are newspaper

= 148 + 11 + 26

= 185

(ii) Number of families who buy one two newspaper

= 122+14+2+3

= 141

(iii) Total number of families in the colony

= 148 + 11+ 26 + 122 + 14 + 2 +3

= 326

(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(∪) = 125

Y = 2x, z = x+10

So, x + y + z + 4 + 17 + 6 + 3 + 5 = 125

= x + 2x + x +1 0 + 4 + 17 + 6 + 3 + 5 = 125

= 4x = 125-45

= 4x = 80

= X = 20

Y = 2 × 20 = 40

Z = 20 + 10 = 30

(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 be, The students who come by bus = A

The students who come by bicycle = B

The students who come by foot= C

n(A) = 25 n(B) = 20 n(c) = 30

n(A ∩ B ∩ C) = 10

We know, n(A B C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)

= 50 = 25 + 20 + 30 – (10 + x) – (10 + y) – (10 + z) + 10

= 50 = 25 + 20 + 30 – 10 – x – 10 – y – 10 – z + 10

= x + y + z = 55 – 50 = 5

= x + y + z = 5

Therefore, the number of student who come to school by two modes of transport = 5

#### Set Language Exercise 1.7 Solutions :

(1) Which of the following is correct?

(1) {7} {1,2,3,4,5,6,7,8,9,10}

(2) 7 {1,2,3,4,5,6,7,8,9,10}

(3) 7 {1,2,3,4,5,6,7,8,9,10}

(4) {7} M {1,2,3,4,5,6,7,8,9,10}

Solution :

Correct option → (2)

7 ∈ {1,2,3,4,5,6,7,8,9,10} is the correct one.

(2) The set P = {x | x Z , –1< x < 1} is a

(1) Singleton set

(2) Power set

(3) Null set

(4) Subset

Solution :

Correct option → (1)

The set P = {x | x ∈ Z , –1< x < 1} is a Singleton set.

(3) If U = {x | x N , x < 10} and A = {x | x N , 2 ≤ x < 6} then (A′)′ is

(1) {1, 6, 7, 8, 9}

(2) {1, 2, 3, 4}

(3) {2, 3, 4, 5}

(4) { }

Solution :

Correct option → (3)

If U ={x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A′)′ is {2, 3, 4, 5}

(4) If B A then n(A∩B) is

(1) n(A–B)

(2) n(B)

(3) n(B – A)

(4) n(A)

Solution :

Correct option → (2)

If B ⊆ A then n(A∩B) is n(B)

(5) If A = {x, y, z} then the number of non- empty subsets of A is

(1) 8

(2) 5

(3) 6

(4) 7

Solution :

Correct option → (4)

If A = {x, y, z} then the number of non-empty subsets of A is 7.

(6) Which of the following is correct?

(1) {a, b}

(2) {a, b}

(3) {a} {a, b}

(4) a {a, b}

Solution :

Correct option → (1)

∅ ⊆ {a, b} is the correct one.

(7) If AB = A∩B, then

(1) A ≠ B

(2) A = B

(3) A B

(4) B A

Solution :

Correct option → (2)

If A∪B = A∩B, then A = B

(8) If B – A is B, then A∩B is

(1) A

(2) B

(3) U

(4)

Solution :

Correct option → (4)

If B – A is B, then A∩B is ∅

(9) From the adjacent diagram n[P(A∆B)] is

(1) 8

(2) 16

(3) 32

(4) 64

Solution :

Correct option → (3)

The adjacent diagram n[P(A∆B)] is 32

(10) If n(A) = 10 and n(B) = 15, then the minimum and maximum number of elements in A ∩ B is

(1) 10,15

(2) 15,10

(3) 10,0

(4) 0,10

Solution :

Correct option → (4)

If n(A) = 10 and n(B) = 15, then the minimum and maximum number of elements in A ∩ B is 0,10

(11) Let A = {} and B = P(A), then A∩B is

(1) { , {} }

(2) {}

(3)

(4) {0}

Solution :

Correct option → (2)

Then A∩B is {∅}

(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

(1) 5

(2) 30

(3) 15

(4) 10.

Solution :

Correct option → (1)

The number of boys play both games will be 5.

(13) If U = {x:x N and x < 10}, A = {1,2,3,5,8} and B = {2,5,6,7,9}, then n[(AB)′] is

(1) 1

(2) 2

(3) 4

(4) 8

Solution :

Correct option → (1)

Then n[(A∪B)′] is 1

(14) For any three sets P, Q and R, P− (Q ∩ R) is

(1) P – (Q R)

(2) (P∩Q) – R

(3) (P – Q) (P – R)

(4) (P – Q) ∩ (P – R)

Solution :

Correct option → (3)

For any three sets P, Q and R, P− (Q ∩ R) is (P – Q) ∪ (P – R)

(16) If n(A B C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A ∩ B) = 20, n(B ∩ C) = 15, n(A ∩ C) = 25 and n (A ∩ B ∩ C) = 10, then the value of x is

(1) 10

(2) 15

(3) 25

(4) 30

Solution :

Correct option → (1)

Hence, the value of x is 10

(17) For any three sets A, B and C,(A – B) ∩ (B – C) is equal to

(1) A only

(2) B only

(3) C only

(4) Ø

Solution :

Correct option → (4)

For any three sets A, B and C,(A – B) ∩ (B – C) is equal to Ø

(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

(1) Set of isoceles triangles

(2) Set of equilateral triangles

(3) Set of isoceles right triangles

(4) Set of right angled triangles

Solution :

Correct option → (3)

Then J ∩ K ∩ L is Set of isoceles right triangles.

(19) The shaded region in the Venn diagram is

(1) Z – (X Y)

(2) (X Y) ∩ Z

(3) Z – (X ∩ Y)

(4) Z (X ∩ Y)

Solution :

Correct option → (3)

The shaded region in the Venn diagram is Z → (X ∩ Y).

(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?

(1) 5

(2) 8

(3) 10

(4) 15

Solution :

Correct option → (1)

5% people do not like any one of the above three fruits.

Next Chapter Solution :

Updated: August 1, 2023 — 4:05 pm