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 = {3,4,5}, B = {4,5,6,7} and C = {5,6,7,8} 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 = {u,c,z,r,h}, B = {j,i,u} and C = {u,h,j} 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 = {1,2,3}, B = {2,3,4,5} and C = {3,4,5,6} 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 = {f,z,t,c,o}, B = {v,n,b} and C = {f,o,v} 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 = {5,6,7}, B = {6,7,8,9} and C = {7,8,9,10} 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 = {g,j,q,r,g}, B = {i,d,l} and C = {g,g,i} 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 = {d,e,c,j,q}, B = {s,p,a} and C = {d,q,s} 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 = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9} 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 = {q,t,f,n,r}, B = {k,l,u} and C = {q,r,k} 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 = {3,4,5}, B = {4,5,6,7} and C = {5,6,7,8} n(A∪B∪C) = Answer: 6 SOLUTION 1 : Given : A = {3,4,5}, B = {4,5,6,7} and C = {5,6,7,8} ⇒ n(A) = 3, n(B) = 4, n(C) = 4 (A∩B) = {3,4,5} ∩ {4,5,6,7} = {4,5} ⇒ n(A∩B) = 2 (B∩C) = {4,5,6,7} ∩ {5,6,7,8} = {5,6,7} ⇒ n(B∩C) = 3 (A∩C) = {3,4,5} ∩ {5,6,7,8} = {5} ⇒ n(A∩C) = 1 (A∪B∪C) = [{3,4,5} ∪ {4,5,6,7}∪ {5,6,7,8}] = {3,4,5,6,7}∪{5,6,7,8} = {3,4,5,6,7,8} ⇒ n(A∪B∪C) = 6 ............... (1) (A∩B∩C) = [{3,4,5} ∩ {4,5,6,7} ∩ {5,6,7,8}] = {4,5} ∩ {5,6,7,8} = {5} ⇒ 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 = {e,g,i,f,s}, B = {y,l,x} and C = {e,s,y} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {e,g,i,f,s}, B = {y,l,x} and C = {e,s,y} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {e,g,i,f,s} ∩ {y,l,x} = { } ⇒ n(A∩B) = 0 (B∩C) = {y,l,x} ∩ {e,s,y} = {y} ⇒ n(B∩C) = 1 (A∩C) = {e,g,i,f,s} ∩ {e,s,y} = {e,s} ⇒ n(A∩C) = 2 (A∪B∪C) = [{e,g,i,f,s} ∪ {y,l,x}∪{e,s,y} = {e,g,i,f,s,y,l,x}∪{e,s,y} = {e,g,i,f,s,y,l,x} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{e,g,i,f,s} ∩ {y,l,x} ∩ {e,s,y}] = { } ∩ {i,f,s,y} = { } ⇒ 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 = {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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| 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 = {w,p,j,n,c}, B = {s,y,i} and C = {w,c,s} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {w,p,j,n,c}, B = {s,y,i} and C = {w,c,s} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {w,p,j,n,c} ∩ {s,y,i} = { } ⇒ n(A∩B) = 0 (B∩C) = {s,y,i} ∩ {w,c,s} = {s} ⇒ n(B∩C) = 1 (A∩C) = {w,p,j,n,c} ∩ {w,c,s} = {w,c} ⇒ n(A∩C) = 2 (A∪B∪C) = [{w,p,j,n,c} ∪ {s,y,i}∪{w,c,s} = {w,p,j,n,c,s,y,i}∪{w,c,s} = {w,p,j,n,c,s,y,i} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{w,p,j,n,c} ∩ {s,y,i} ∩ {w,c,s}] = { } ∩ {j,n,c,s} = { } ⇒ 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 = {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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| 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 = {w,y,f,z,u}, B = {s,r,o} and C = {w,u,s} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {w,y,f,z,u}, B = {s,r,o} and C = {w,u,s} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {w,y,f,z,u} ∩ {s,r,o} = { } ⇒ n(A∩B) = 0 (B∩C) = {s,r,o} ∩ {w,u,s} = {s} ⇒ n(B∩C) = 1 (A∩C) = {w,y,f,z,u} ∩ {w,u,s} = {w,u} ⇒ n(A∩C) = 2 (A∪B∪C) = [{w,y,f,z,u} ∪ {s,r,o}∪{w,u,s} = {w,y,f,z,u,s,r,o}∪{w,u,s} = {w,y,f,z,u,s,r,o} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{w,y,f,z,u} ∩ {s,r,o} ∩ {w,u,s}] = { } ∩ {f,z,u,s} = { } ⇒ 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 = {i,p,a,y,w}, B = {r,n,h} and C = {i,w,r} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {i,p,a,y,w}, B = {r,n,h} and C = {i,w,r} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {i,p,a,y,w} ∩ {r,n,h} = { } ⇒ n(A∩B) = 0 (B∩C) = {r,n,h} ∩ {i,w,r} = {r} ⇒ n(B∩C) = 1 (A∩C) = {i,p,a,y,w} ∩ {i,w,r} = {i,w} ⇒ n(A∩C) = 2 (A∪B∪C) = [{i,p,a,y,w} ∪ {r,n,h}∪{i,w,r} = {i,p,a,y,w,r,n,h}∪{i,w,r} = {i,p,a,y,w,r,n,h} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{i,p,a,y,w} ∩ {r,n,h} ∩ {i,w,r}] = { } ∩ {a,y,w,r} = { } ⇒ 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 = {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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| 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 = {p,z,k,y,h}, B = {x,y,a} and C = {p,h,x} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {p,z,k,y,h}, B = {x,y,a} and C = {p,h,x} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {p,z,k,y,h} ∩ {x,y,a} = { } ⇒ n(A∩B) = 0 (B∩C) = {x,y,a} ∩ {p,h,x} = {x} ⇒ n(B∩C) = 1 (A∩C) = {p,z,k,y,h} ∩ {p,h,x} = {p,h} ⇒ n(A∩C) = 2 (A∪B∪C) = [{p,z,k,y,h} ∪ {x,y,a}∪{p,h,x} = {p,z,k,y,h,x,y,a}∪{p,h,x} = {p,z,k,y,h,x,y,a} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{p,z,k,y,h} ∩ {x,y,a} ∩ {p,h,x}] = { } ∩ {k,y,h,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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