Question 1.
\nIf A = {0, 1, 2, 3, …….., 8}, B = {3, 5, 7, 9, 11} and C = {0, 5, 10, 20}, find
\n(i) A \u222a B
\n(ii) A \u222a C
\n(iii) B \u222a C
\n(iv) A \u2229 B
\n(v) A \u2229 C
\n(vi) B \u2229 C
\nAlso find the cardinal number of the sets B \u222a C, A \u222a B, A \u2229 C and B \u2229 C.
\nSolution:
\nA = {0, 1, 2, 3, …, 8}
\nB = {3, 5, 7, 9, 11}
\nC = {0, 5, 10, 20}
\n(i) A \u222a B = {0,1, 2, 3, 4, 5, 6, 7, 8, 9,11},
\nn(A \u222a B) = 11
\n(ii) A \u222a C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20},
\nn(A \u222a C) = 11
\n(iii) B \u222a C = {0, 3, 5, 7, 9, 10, 11, 20},
\nn(B \u222a C) = 8
\n(iv) A \u2229 B = {3, 5, 7}, n(A \u2229 B) = 3
\n(v) A \u2229 C = (0, 5), n(A \u2229 C) = 2
\n(vi) B \u2229 C = {5}, n(B \u2229 C) = 1<\/p>\n
Question 2.
\nFind A’ when
\n(i) A= {0, 1, 4, 7} and E, = {x | x \u03f5 W, x \u2264 10}
\n(ii) A = {consonants} and \u03be = {alphabets of English}
\n(iii) A = boys in class VIII of all schools in Bengaluru} and \u03be = {students in class VIII of all schools in Bengaluru}
\n(iv) A = {letters of KALKA} and \u03be = {letters of KOLKATA}
\n(v) A = {odd natural numbers} and \u03be = {whole numbers}.
\nSolution:
\nFind A’
\n(i) A = {0, 1,4,7} and \u03be= {x|x \u03f5 W, x \u226410}
\n\u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\n\u2234 A’ = {2, 3, 5, 6, 8, 9, 10}<\/p>\n
(ii) A = {consonants} and \u03be = {alphabets of English}
\n\u2234 A’ = {Vowels}<\/p>\n
(iii) A = boys in class VIII of all schools in Bengaluru} and
\n\u03be = {students in class VIII of all schools in Bengaluru}
\n\u2234 A’ = {Girls in class VIII of all schools in Bengaluru}<\/p>\n
(iv) A = {letters of KALKA}
\n\u03be = {letters of KOLKATA}
\n= {KAL} and \u03be = {KOLAT}
\n\u2234 A’ = {O, T}<\/p>\n
(v) A = {odd natural numbers}
\n\u03be = {whole numbers}
\n\u2234 A’ = {0, even whole numbers}<\/p>\n
Question 3.
\nIf A {x : x \u03f5 N and 3 < x < 1} and B = {x : x \u03f5 Wand x \u2264 4}, find
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A
\nSolution:
\nA = {x : x \u03f5 N and 3 < x < 7}
\n= {4, 5, 6}
\nB = {x: x \u03f5 W and x \u2264 4}
\n= {0, 1, 2, 3, 4}
\n(i) A \u222a B = {0, 1, 2, 3, 4, 5, 6}
\nor {x : x \u03f5 W and x \u2264 4}
\n(ii) A \u2229 B = {4}
\n(iii) A – B = {5, 6}
\n(iv) B – A = {0, 1, 2, 3}<\/p>\n
Question 4.
\nIf P = {x : x \u03f5 W and x < 6} and Q = {x : x \u03f5 N and 4 \u2264 x \u2264 9}, find
\n(i) P \u222a Q
\n(ii) P \u2229 Q
\n(iii) P – Q
\n(iv) Q – P
\nIs P \u222a Q a proper superset of P \u2229 Q ?
\nSolution:
\nP = {x: x \u03f5 W and x < 6}
\n= {0, 1, 2, 3, 4, 5}
\nQ = {x : x \u03f5 N and 4 \u2264 x \u2264 9}
\n= (4, 5, 6, 7, 8}
\n(i) P \u222a Q = {0, 1, 2, 3, 4, 5, 6, 7, 8}
\n{x : x \u03f5 W and x \u2264 8}
\n(ii) P \u2229 Q = {4, 5}
\n(iii) P – Q = {0, 1, 2, 3}
\n(iv) Q – P = {6, 7, 8}
\nYes, P \u222a Q is the super set of P \u2229 Q
\nbecause P \u2229 Q is subset of P \u222a Q.<\/p>\n
Question 5.
\nIf A = (letters of word INTEGRITY) and B = (letters of word RECKONING), find
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A
\nAlso verify that:
\n(a) n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n(b) n(A – B) = n(A \u222a B) – n(B)
\n= n(A) – n(A \u222a B)
\n(c) n(B – A) = n(A \u222a B) – n(A)
\n= n(B) – n(A \u2229 B)
\n(d) n(A \u222a B) = n(A – B) + n(B – A) + n(A \u2229 B).
\nSolution:
\nA = (I, N, T, E, G, R, Y}
\nB = {R, E, C, K, O, N, I, G}
\n(i) A \u222a B = {I, N, T, E, G, R, Y} \u222a {R, E, C, K, O, N, I, G}
\n= {I, N, T, E, G, R, Y, C, K, 0}<\/p>\n
(ii) A \u2229 B = {I, N, T, E, G, R, Y} \u2229 {R, E, C, K, O, N, I, G}
\n= {I, N, E, R, G}<\/p>\n
(iii) A – B = {R, E, C, K, O, N, I, G} – {1, N, T, E, G, R, Y}
\n= {T, Y}<\/p>\n
(iv) B – A = – {R, E, C, K, O, N, I, G} {I, N, T, E, G, R, Y}
\n= {C, K, O}
\nVerification:
\nHere n(A \u222a B) = 10
\nn(A) = 7
\nn(B) = 8
\nn(A \u2229 B) = 5
\nn(A – B) = 2
\nn(B – A) = 3
\n(a) n(A \u222a B) = 10
\nand n(A) + n(B) – n(A \u2229 B)
\n= 7 + 8 – 5 = 15 – 5 = 10
\n\u2234 n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)<\/p>\n
(b) n(A – B) = 2
\nn(A \u222a B) – n(B) = 10 – 8 = 2
\nand n(A) – n(A \u2229 B) = 7 – 5 = 2
\n\u2234 n(A – B) = n(A \u222a B) – n(B)
\n= n(A) – n(A \u2229 B)<\/p>\n
(c) n(B – A) = 3
\nn(A \u222a B) – n(A) = 10 – 7 = 3
\nand n(B) – n(A \u2229 B) = 8 – 5 = 3
\n\u2234 n(B – A) = n(A \u222a B) – n(A)
\n= n(B) – n(A \u2229 B)<\/p>\n
(d) n(A \u222a B) = 10
\nand n(A – B) + n(B – A) + n(A \u2229 B)
\n= 2 + 3 + 5 = 10
\n\u2234 n(A \u222a B) = n(A – B) + n(B – A) + n(A \u2229 B).<\/p>\n
Question 6.
\nIf \u03be = {natural numbers between 10 and 40}
\nA = {multiples of 5} and
\nB = {multiples of 6}, then
\n(i) find A \u222a B and A \u2229 B
\n(ii) verify that
\nn(A \u222a B) = B (A) + n(B) – n(A \u2229 B).
\nSolution:
\nHere, \u03be = {11, 12, 13, ……., 39}
\nIt is understood that A and B are subsets of \u03be {Universal set)
\nSo, the elements of A and B are to be taken only from \u03be.
\nA= {multiples of 5}
\n= multiples of 5 which belong to \u03be
\n= {15, 20, 25, 30, 35}
\nB = {multiples of 6}
\n= multiples of 6 which belong to \u03be
\n= {12, 18, 24, 30, 36}
\n(i) A \u222a B = {15, 20, 25, 30, 35, 12, 18, 24, 36}
\nA \u2229 B = {30}
\n(ii) n(A) = 5, n(B) = 5
\nn(A \u222a B) = 9 and n(A \u2229 B) = 1
\n\u2234 n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n9 = 5 + 5 – 1 = 10 – 1 = 9
\nHence, verified.<\/p>\n
Question 7.
\nIf \u03be ={1,2, 3, …. 9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8}, then find.
\n(i) A’
\n(ii) B’
\n(iii) A \u222a B
\n(iv) A \u2229 B
\n(v) A – B
\n(vi) B – A
\n(vii) (A \u2229 B)’
\n(viii) A’ \u222a B\u2019
\nAlso verify that:
\n(a) (A \u2229 B)’ = A’ \u222a B’
\n(b) n(A) + n(A’) = n(\u03be)
\n(c) n(A \u2229 B) + n((A \u2229 B)’) = n(\u03be)
\n(d) n(A – B) + n(B – A) + n(A \u2229 B)
\n= n(A \u222a B).
\nSolution:
\nThe given sets in the roster form are:
\n\u03be = {1,2, 3, …, 9}
\nA = {1,2, 3, 4, 6, 7,8} and B = {4, 6, 8}
\n(i) A’ = \u03be – A
\n= {1,2, 3, 4, 5, 6, 7, 8, 9} – {1,2, 3, 4, 6, 7, 8}
\n= {5, 8, 9}<\/p>\n
(ii) B’ = \u03be – B
\n= {1,2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
\n= {1,2,3, 5, 7, 9}<\/p>\n
(iii) A \u222a B = {1, 2, 3, 4, 6, 7, 8} \u222a {4, 6, 8}
\n= {1,2, 3, 4, 6, 7, 8}<\/p>\n
(iv) A \u2229 B = {1, 2, 3, 4, 6, 7, 8} \u2229 {4, 6, 8}
\n= {4, 6, 8}<\/p>\n
(v) A – B = {1, 2, 3, 4, 6, 7, 8} – {4, 6, 8}
\n= {1, 2, 3, 7}<\/p>\n
(vi) B – A = {4, 6, 8} – {1, 2, 3, 4, 6, 7, 8}
\n= {}<\/p>\n
(vii) (A \u2229 B)’ = \u03be – (A \u2229 B)
\n= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
\n= {1,2, 3, 5, 7, 9}<\/p>\n
(viii) A’ \u222a B’ = {5, 8, 9} \u222a {1, 2, 3, 5, 7, 9}
\n= {1,2, 3, 5, 7, 8, 9}
\nVerification :
\nHere n(A) = 6
\nn(A’) = 3
\nn(\u03be) = 9
\nn(A \u2229 B) = 3
\nn((A \u2229 B)’) = 6
\nn(A – B) = 4
\nn(B – A) = 0
\nn(A \u2229 B)’ = 6
\nn( A’ \u2229 B’) = 7
\nn(A \u222a B) = 8
\n(a) (A \u2229 B)’ = \u03be – (A \u2229 B)
\n= {1,2, 3, 4, 5, 6, 7, 8, 9} – (4, 6, 8}
\n= {1,2, 3, 5, 7, 9}
\nA’ \u222a B’ = {5, 7, 9} \u222a {1, 2, 3, 5, 7, 9}
\n= {1, 2, 3, 5, 7, 9}
\n(A \u2229 B)’ = A’ \u222a B’ (Verified)<\/p>\n
(b) n(A) + n(A’) = 6 + 3 = 9
\nn(\u03be) = 9
\n\u2234 n(A) + n(A’) = n(\u03be) (Verified)<\/p>\n
(c) n(A \u2229 B) + n((A \u2229 B)’) = 3 + 6 = 9
\nn(\u03be) = 9
\n\u2234 n(A \u2229 B) + n((A \u2229 B)’) = n(\u03be) (Verified)<\/p>\n
(d) n(A – B) + n(B – A) + n(A \u2229 B)
\n= 4 + 0 + 3 = 7
\nn(A \u222a B) = 7
\n\u2234 n(A – B) + n(B – A) + n(A \u2229 B)
\n= n(A \u222a B) (Verified)<\/p>\n
Question 8.
\nIf 4 = {x : x \u03f5 W, x \u2264 10}, A. = {x : x \u2265 5} and B = {x : 3 \u2264 x < 8}, then verify that:
\n(i) (A \u222a B)’ = A’ \u2229 B’
\n(ii) (A \u2229 B)’= A’ \u222a B’
\n(iii) A – B = A \u2229 B’
\n(iv) B – A = B \u2229 A’
\nSolution:
\nThe given sets in the roster form are :
\n\u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\nA = {5, 6, 7, 8, 9, 10}
\nB = {3, 4, 5, 6, 7}
\n(i) L.H.S. = (A \u222a B)’
\nA \u222a B = {5, 6, 7, 8, 9, 10} \u222a {3, 4, 5, 6, 7}
\n= {3, 4, 5, 6, 7, 8, 9, 10}
\n\u2234 (A \u222a B)’ = \u03be – (A \u222a B)
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7, 8, 9, 10}
\n= {0, 1, 2}
\nR.H.S. = A’ \u2229 B’
\nA’ = \u03be – A
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
\n= {0, 1, 2, 3, 4}
\nB’ = \u03be – B
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
\n= {0, 1, 2, 8, 9, 10}
\nA’ \u2229 B’ = {0, 1, 2, 3, 4} \u2229 {0, 1, 2, 8, 9, 10}
\n= {0, 1, 2}
\nHence (A \u222a B)’ = A’ \u2229 B’ (Verified)
\n(ii) L.H.S. = (A \u2229 B)’
\nA n B = {5, 6,7, 8,9, 10} n {3,4, 5,6, 7} = {5, 6, 7}
\n(A \u2229 B)’ = \u03be – (A \u2229 B)
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\n– {5, 6, 7}
\n= {0, 1, 2, 3, 4, 8, 9, 10}
\nR.H.S. = A’ u B’
\nA’ = \u03be – A
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
\n= {0, 1, 2, 3, 4}
\nB’ = \u03be – B
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
\n= {0, 1, 2, 8, 9, 10}
\n\u2234 A’ \u222a B’ = {0, 1, 2, 3, 4} \u222a {0, 1, 2, 8, 9, 10}
\n= {0, 1, 2, 3, 4, 8, 9, 10}
\nHence (A \u2229 B)’ =A’ \u222a B’ (Verified)<\/p>\n
(iii) L.H.S. = A – B
\n\u2234 A – B = {5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
\n= {8, 9, 10}
\nR.H.S. = A \u2229 B’
\nB’ = \u03be – B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
\n= {0, 1, 2, 8, 9, 10}
\n\u2234 A \u2229 B’ = {5, 6, 7, 8, 9, 10} \u2229 {0, 1, 2, 8, 9, 10}
\n= {8, 9, 10}
\nHence A – B = A \u2229 B’<\/p>\n
(iv) L.H.S. = B – A
\n\u2234 B – A = {3, 4, 5, 6, 7} – {5, 6, 7, 8, 9, 10}
\n= {3, 4}
\nR.H.S. = B \u2229 A’
\nA’ = \u03be – A
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
\n= (0, 1,2, 3, 4}
\nB \u2229 A’ = {3, 4, 5, 6, 7} \u2229 {0, 1, 2, 3, 4} = {3, 4}
\nHence B – A = B \u2229 A’.<\/p>\n
Question 9.
\nIf n(A) = 20, n(B) = 16 and n(A \u222a B) = 30, find n(A \u2229 B).
\nSolution:
\nGiven that,
\nn(A) = 20, n(B) = 16 and
\nn(A \u222a B) = 30 Then n(A \u2229 B) = ?
\nWe know that,
\nn(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n\u21d2 30 = 20 + 16 – n(A \u2229 B)
\n\u21d2 30 = 36 – n(A \u2229 B)
\n\u21d2 n(A \u2229 B) = 36 – 30
\n\u21d2 n(A \u2229 B) = 6
\nHence n(A \u2229 B) = 6.<\/p>\n
Question 10.
\nIf n \u03be = 20 and n(A’) = 7, then find n(A).
\nSolution:
\nWe know that,
\nn(A’) = n \u03be – n(A)
\n7 = 20 – n(A)
\n7 – 20 = – n(A)
\n-13 = -n(A) \u21d2 n(A) = 13.<\/p>\n
Question 11.
\nIf n(\u03be) = 40, n(A) = 20, n(B’) = 16 and n(A \u222a B) = 32, then find n(B) and n(A \u2229 B).
\nSolution:
\nWe know that,
\nn(B’) = n(\u03be) – n(B)
\n16 = 40 – n(B)
\n16 – 40 = – n(B)
\n-24 = -n(B)
\nn(B) = 24
\nand also, we know that,
\nn(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n32 = 20 + 24 – n(A \u2229 B)
\n32 = 44 – n(A \u2229 B)
\n32 – 44 = – n(A \u2229 B)
\n\u21d2 -12 = -n(A \u2229 B)
\n\u21d2 n(A \u2229 B) = 12<\/p>\n
Question 12.
\nIf n(\u03be) = 32, n(A) = 20, n(B) = 16 and n((A \u222a B)’) = 4, find :
\n(i) n(A \u222a B)
\n(ii) n(A \u2229 B)
\n(iii) n(A – B)
\nSolution:
\nGiven that,
\nn(\u03be) = 32, n(A) = 20, n(B) = 16,
\nn((A \u222a B)’) = 4
\n(i) n(A \u222a B) = n(\u03be) – n((A \u222a B)’)
\n= 32 – 4 = 28<\/p>\n
(ii) We know that,
\nn(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n\u21d2 28 = 20 + 16 – n(A \u2229 B)
\n\u21d2 28 = 36 – n(A \u2229 B)
\n\u21d2 n(A \u2229 B) = 36 – 28 \u21d2 n(A \u2229 B) = 8
\nHence n(A \u2229 B) = 8.<\/p>\n
(iii) n(A – B) = n(A) – n(A \u2229 B)
\n= 20 – 8 = 12.<\/p>\n
Question 13.
\nIf n(\u03be) = 40, n(A’) = 15, n(B) = 12 and n((A \u2229 B)’) = 32, find :
\n(i) n(A)
\n(ii) n(B’)
\n(iii) n(A \u2229 B)
\n(iv) n(A \u222a B)
\n(v) n(A – B)
\n(vi) n(B – A)
\nSolution:
\nGiven that,
\nn(\u03be) = 40
\nn(A’)= 15
\nn(B) = 12
\nn((A \u2229 B)’) = 32
\n(i) n(A) = n(\u03be) – n(A’)
\n= 40 – 15 = 25<\/p>\n
(ii) n(B’) = n(\u03be) – n(B) = 40 – 12 = 28
\n(iii) n(A \u2229 B) = n(\u03be) – n((A \u2229 B)’)
\n= 40 – 32 = 8<\/p>\n
(iv) n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n= 25 + 12 – 8 = 25 + 4 = 29<\/p>\n
(v) n(A – B) = n(A) – n(A\u2229B)
\n= 25 – 8 = 17<\/p>\n
(vi) n(B – A) = n(B) – n(A \u2229 B)
\n= 12 – 8 = 4<\/p>\n
Question 14.
\nIf n(A – B) = 12, n(B – A) = 16 and n(A \u2229 B) = 5, find:
\n(i) n(A)
\n(ii) n(B)
\n(iii) n(A \u222a B)
\nSolution:
\nGiven that n(A – B) = 12,
\nn(B – A)= 16 and n(A \u2229 B) = 5
\n(i) n(A) = n(A – B) + n(A \u2229 B)
\n{\u2235 n(A – B) = n(A) – n(A \u2229 B)}
\n= 12 + 5 = 17<\/p>\n
(ii) n(B) = n(B – A) + n(A \u2229 B)
\n{\u2235 n(B – A) = n(B) – n(A \u2229 B)}
\n= 16 + 5 = 21<\/p>\n
(iii) n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n= 17 + 21 – 5 = 38 – 5 =33<\/p>\n
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Ex 6.1 Question 1. If A = {0, 1, 2, 3, …….., 8}, B = {3, 5, 7, 9, 11} and C = {0, 5, 10, 20}, find (i) A \u222a B (ii) A \u222a C (iii) B \u222a C …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/44777"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=44777"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/44777\/revisions"}],"predecessor-version":[{"id":159184,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/44777\/revisions\/159184"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=44777"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=44777"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=44777"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}