Educational

If n(A∩B) = 10, n(B∩C) = 20 and n(A∩C) = 30, then the greater possible value of n(A ∩ B ∩ C) is….. Solution \[ n(A \cap B)=10 \] \[ n(B \cap C)=20 \] \[ n(A \cap C)=30 \] Since \[ A \cap B \cap C \subseteq A \cap B \] therefore, \[ n(A \cap […]

Let S = {x : x is a positive multiple of 3 less than 100}, P = {x : x is a prime number less than 20}, Then, n(S) + n(P) = _____ Solution \[ S = \{3,6,9,\ldots,99\} \] Number of multiples of 3 less than 100: \[ \frac{99}{3} = 33 \] Therefore, \[ n(S)=33

For any three sets A, B and C, (A∪B∪C) ∩ (A∩B’∩C’) ∩ C’ is equal to __________ Solution \[ (A \cup B \cup C) \cap (A \cap B’ \cap C’) \cap C’ \] \[ = (A \cap B’ \cap C’) \cap (A \cup B \cup C) \] Using distributive law, \[ = (A \cap B’

If A and B are two sets, then (A ∩ B’)’ ∪ (B ∩ C) is equal to _________ Solution \[ (A \cap B’)’ \cup (B \cap C) \] \[ = (A’ \cup B) \cup (B \cap C) \] \[ = A’ \cup [B \cup (B \cap C)] \] \[ = A’ \cup B \]

For any three sets A, B and C, (A – B) ∩ (C – B) is equal to __________ Solution \[ (A – B) \cap (C – B) \] \[ = (A \cap B’) \cap (C \cap B’) \] \[ = A \cap C \cap B’ \] \[ = (A \cap C) – B \]

For any three sets A, B and C, (A – B) – (B – C) is equal to __________ Solution \[ (A – B) – (B – C) \] \[ = (A \cap B’) – (B \cap C’) \] \[ = (A \cap B’) \cap (B \cap C’)’ \] \[ = (A \cap B’) \cap

For any two sets A and B, [B’ ∪ (B’ – A)]’ is equal to……. Solution \[ [B’ \cup (B’ – A)]’ \] Using \[ X – Y = X \cap Y’ \] \[ = [B’ \cup (B’ \cap A’)]’ \] Using absorption law, \[ B’ \cup (B’ \cap A’) = B’ \] Therefore, \[

If A and B are two sets, then ((A’ ∪ B’) – A)’ is equal to………. Solution \[ ((A’ \cup B’) – A)’ \] Using \[ X – Y = X \cap Y’ \] we get \[ = ((A’ \cup B’) \cap A’)’ \] Since \[ A’ \cap (A’ \cup B’) = A’ \] therefore,

If A and B are two sets, then A ∩ (A ∪ B)’ is equal to ___________ Solution \[ A \cap (A \cup B)’ \] By De Morgan’s law, \[ = A \cap (A’ \cap B’) \] Using associative law, \[ = (A \cap A’) \cap B’ \] \[ = \Phi \cap B’ \] \[

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}, Then (i) \((B \cup C)’\) = …. (ii) \((C – A)’\) = …. Solution \[ U = \{1,2,3,4,5,6,7,8,9,10\} \] \[ B = \{2,4,6,7\}