Written Instructions:
For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..
For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.
Leave your answers in the simplest form or correct to two decimal places.
| 1) 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 sets given below : (i) A = {1,2,3}, B = {2,3,4,5} and C = {3,4,5,6} n(A∪B∪C) = Answer:_______________ |
| 2) 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 sets given below : (i) A = {g,d,f,z,s}, B = {e,b,d} and C = {g,s,e} n(A∪B∪C) = Answer:_______________ |
| 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 sets given below : (i) A = {6,7,8}, B = {7,8,9,10} and C = {8,9,10,11} n(A∪B∪C) = Answer:_______________ |
| 4) 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 sets given below : (i) A = {c,p,a,z,v}, B = {o,d,e} and C = {c,v,o} n(A∪B∪C) = Answer:_______________ |
| 5) 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 sets given below : (i) A = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9} n(A∪B∪C) = Answer:_______________ |
| 6) 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 sets given below : (i) A = {a,n,d,z,c}, B = {o,t,f} and C = {a,c,o} n(A∪B∪C) = Answer:_______________ |
| 7) 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 sets given below : (i) A = {2,3,4}, B = {3,4,5,6} and C = {4,5,6,7} n(A∪B∪C) = Answer:_______________ |
| 8) 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 sets given below : (i) A = {s,n,e,q,m}, B = {f,b,a} and C = {s,m,f} n(A∪B∪C) = Answer:_______________ |
| 9) 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 sets given below : (i) A = {5,6,7}, B = {6,7,8,9} and C = {7,8,9,10} n(A∪B∪C) = Answer:_______________ |
| 10) 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 sets given below : (i) A = {u,c,u,k,b}, B = {m,d,v} and C = {u,b,m} n(A∪B∪C) = Answer:_______________ |
| 1) 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 sets given below : (i) A = {1,2,3}, B = {2,3,4,5} and C = {3,4,5,6} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {1,2,3}, B = {2,3,4,5} and C = {3,4,5,6} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {1,2,3} ∩ {2,3,4,5} = {2,3} ⇒ n(A∩B) = 2 (B∩C) = {2,3,4,5} ∩ {3,4,5,6} = {3,4,5} ⇒ n(B∩C) = 3 (A∩C) = {1,2,3} ∩ {3,4,5,6} = {3} ⇒ n(A∩C) = 1 (A∪B∪C) = [{1,2,3} ∪ {2,3,4,5}∪ {3,4,5,6}] = {1,2,3,4,5}∪{3,4,5,6} = {1,2,3,4,5,6} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{1,2,3} ∩ {2,3,4,5} ∩ {3,4,5,6}] = {2,3} ∩ {3,4,5,6} = {3} ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 - 2 - 3 - 1 + 1 = 12 - 6 = 6 ............... (2) From (1) and (2), It is true for given sets.
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| 2) 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 sets given below : (i) A = {h,m,u,v,n}, B = {u,z,v} and C = {h,n,u} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {h,m,u,v,n}, B = {u,z,v} and C = {h,n,u} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {h,m,u,v,n} ∩ {u,z,v} = { } ⇒ n(A∩B) = 0 (B∩C) = {u,z,v} ∩ {h,n,u} = {u} ⇒ n(B∩C) = 1 (A∩C) = {h,m,u,v,n} ∩ {h,n,u} = {h,n} ⇒ n(A∩C) = 2 (A∪B∪C) = [{h,m,u,v,n} ∪ {u,z,v}∪{h,n,u} = {h,m,u,v,n,u,z,v}∪{h,n,u} = {h,m,u,v,n,u,z,v} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{h,m,u,v,n} ∩ {u,z,v} ∩ {h,n,u}] = { } ∩ {u,v,n,u} = { } ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 5 + 3 + 3 - 0 - 1 - 2 + 0 = 11 - 3 = 8 ............... (2) From (1) and (2), It is true for given sets.
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| 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 sets given below : (i) A = {6,7,8}, B = {7,8,9,10} and C = {8,9,10,11} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {6,7,8}, B = {7,8,9,10} and C = {8,9,10,11} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {6,7,8} ∩ {7,8,9,10} = {7,8} ⇒ n(A∩B) = 2 (B∩C) = {7,8,9,10} ∩ {8,9,10,11} = {8,9,10} ⇒ n(B∩C) = 3 (A∩C) = {6,7,8} ∩ {8,9,10,11} = {8} ⇒ n(A∩C) = 1 (A∪B∪C) = [{6,7,8} ∪ {7,8,9,10}∪ {8,9,10,11}] = {6,7,8,9,10}∪{8,9,10,11} = {6,7,8,9,10,11} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{6,7,8} ∩ {7,8,9,10} ∩ {8,9,10,11}] = {7,8} ∩ {8,9,10,11} = {8} ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 - 2 - 3 - 1 + 1 = 12 - 6 = 6 ............... (2) From (1) and (2), It is true for given sets.
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| 4) 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 sets given below : (i) A = {h,x,q,z,c}, B = {l,f,m} and C = {h,c,l} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {h,x,q,z,c}, B = {l,f,m} and C = {h,c,l} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {h,x,q,z,c} ∩ {l,f,m} = { } ⇒ n(A∩B) = 0 (B∩C) = {l,f,m} ∩ {h,c,l} = {l} ⇒ n(B∩C) = 1 (A∩C) = {h,x,q,z,c} ∩ {h,c,l} = {h,c} ⇒ n(A∩C) = 2 (A∪B∪C) = [{h,x,q,z,c} ∪ {l,f,m}∪{h,c,l} = {h,x,q,z,c,l,f,m}∪{h,c,l} = {h,x,q,z,c,l,f,m} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{h,x,q,z,c} ∩ {l,f,m} ∩ {h,c,l}] = { } ∩ {q,z,c,l} = { } ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 5 + 3 + 3 - 0 - 1 - 2 + 0 = 11 - 3 = 8 ............... (2) From (1) and (2), It is true for given sets.
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| 5) 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 sets given below : (i) A = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {4,5,6} ∩ {5,6,7,8} = {5,6} ⇒ n(A∩B) = 2 (B∩C) = {5,6,7,8} ∩ {6,7,8,9} = {6,7,8} ⇒ n(B∩C) = 3 (A∩C) = {4,5,6} ∩ {6,7,8,9} = {6} ⇒ n(A∩C) = 1 (A∪B∪C) = [{4,5,6} ∪ {5,6,7,8}∪ {6,7,8,9}] = {4,5,6,7,8}∪{6,7,8,9} = {4,5,6,7,8,9} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{4,5,6} ∩ {5,6,7,8} ∩ {6,7,8,9}] = {5,6} ∩ {6,7,8,9} = {6} ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 - 2 - 3 - 1 + 1 = 12 - 6 = 6 ............... (2) From (1) and (2), It is true for given sets.
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| 6) 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 sets given below : (i) A = {m,y,k,i,c}, B = {t,c,h} and C = {m,c,t} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {m,y,k,i,c}, B = {t,c,h} and C = {m,c,t} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {m,y,k,i,c} ∩ {t,c,h} = { } ⇒ n(A∩B) = 0 (B∩C) = {t,c,h} ∩ {m,c,t} = {t} ⇒ n(B∩C) = 1 (A∩C) = {m,y,k,i,c} ∩ {m,c,t} = {m,c} ⇒ n(A∩C) = 2 (A∪B∪C) = [{m,y,k,i,c} ∪ {t,c,h}∪{m,c,t} = {m,y,k,i,c,t,c,h}∪{m,c,t} = {m,y,k,i,c,t,c,h} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{m,y,k,i,c} ∩ {t,c,h} ∩ {m,c,t}] = { } ∩ {k,i,c,t} = { } ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 5 + 3 + 3 - 0 - 1 - 2 + 0 = 11 - 3 = 8 ............... (2) From (1) and (2), It is true for given sets.
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| 7) 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 sets given below : (i) A = {2,3,4}, B = {3,4,5,6} and C = {4,5,6,7} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {2,3,4}, B = {3,4,5,6} and C = {4,5,6,7} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {2,3,4} ∩ {3,4,5,6} = {3,4} ⇒ n(A∩B) = 2 (B∩C) = {3,4,5,6} ∩ {4,5,6,7} = {4,5,6} ⇒ n(B∩C) = 3 (A∩C) = {2,3,4} ∩ {4,5,6,7} = {4} ⇒ n(A∩C) = 1 (A∪B∪C) = [{2,3,4} ∪ {3,4,5,6}∪ {4,5,6,7}] = {2,3,4,5,6}∪{4,5,6,7} = {2,3,4,5,6,7} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{2,3,4} ∩ {3,4,5,6} ∩ {4,5,6,7}] = {3,4} ∩ {4,5,6,7} = {4} ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 - 2 - 3 - 1 + 1 = 12 - 6 = 6 ............... (2) From (1) and (2), It is true for given sets.
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| 8) 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 sets given below : (i) A = {b,g,p,a,m}, B = {f,t,l} and C = {b,m,f} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {b,g,p,a,m}, B = {f,t,l} and C = {b,m,f} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {b,g,p,a,m} ∩ {f,t,l} = { } ⇒ n(A∩B) = 0 (B∩C) = {f,t,l} ∩ {b,m,f} = {f} ⇒ n(B∩C) = 1 (A∩C) = {b,g,p,a,m} ∩ {b,m,f} = {b,m} ⇒ n(A∩C) = 2 (A∪B∪C) = [{b,g,p,a,m} ∪ {f,t,l}∪{b,m,f} = {b,g,p,a,m,f,t,l}∪{b,m,f} = {b,g,p,a,m,f,t,l} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{b,g,p,a,m} ∩ {f,t,l} ∩ {b,m,f}] = { } ∩ {p,a,m,f} = { } ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 5 + 3 + 3 - 0 - 1 - 2 + 0 = 11 - 3 = 8 ............... (2) From (1) and (2), It is true for given sets.
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| 9) 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 sets given below : (i) A = {5,6,7}, B = {6,7,8,9} and C = {7,8,9,10} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {5,6,7}, B = {6,7,8,9} and C = {7,8,9,10} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {5,6,7} ∩ {6,7,8,9} = {6,7} ⇒ n(A∩B) = 2 (B∩C) = {6,7,8,9} ∩ {7,8,9,10} = {7,8,9} ⇒ n(B∩C) = 3 (A∩C) = {5,6,7} ∩ {7,8,9,10} = {7} ⇒ n(A∩C) = 1 (A∪B∪C) = [{5,6,7} ∪ {6,7,8,9}∪ {7,8,9,10}] = {5,6,7,8,9}∪{7,8,9,10} = {5,6,7,8,9,10} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{5,6,7} ∩ {6,7,8,9} ∩ {7,8,9,10}] = {6,7} ∩ {7,8,9,10} = {7} ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 - 2 - 3 - 1 + 1 = 12 - 6 = 6 ............... (2) From (1) and (2), It is true for given sets.
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| 10) 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 sets given below : (i) A = {h,d,z,g,q}, B = {x,t,u} and C = {h,q,x} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {h,d,z,g,q}, B = {x,t,u} and C = {h,q,x} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {h,d,z,g,q} ∩ {x,t,u} = { } ⇒ n(A∩B) = 0 (B∩C) = {x,t,u} ∩ {h,q,x} = {x} ⇒ n(B∩C) = 1 (A∩C) = {h,d,z,g,q} ∩ {h,q,x} = {h,q} ⇒ n(A∩C) = 2 (A∪B∪C) = [{h,d,z,g,q} ∪ {x,t,u}∪{h,q,x} = {h,d,z,g,q,x,t,u}∪{h,q,x} = {h,d,z,g,q,x,t,u} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{h,d,z,g,q} ∩ {x,t,u} ∩ {h,q,x}] = { } ∩ {z,g,q,x} = { } ⇒ n(A∩B∩C) = 1 ∴ n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C) = 5 + 3 + 3 - 0 - 1 - 2 + 0 = 11 - 3 = 8 ............... (2) From (1) and (2), It is true for given sets.
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