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 = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9} 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 = {l,i,h,m,a}, B = {n,r,w} and C = {l,a,n} 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 = {h,d,a,p,n}, B = {e,n,g} and C = {h,n,e} 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 = {2,3,4}, B = {3,4,5,6} and C = {4,5,6,7} 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 = {t,f,r,x,u}, B = {o,b,n} and C = {t,u,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 = {6,7,8}, B = {7,8,9,10} and C = {8,9,10,11} 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 = {y,e,t,k,h}, B = {q,b,z} and C = {y,h,q} 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 = {3,4,5}, B = {4,5,6,7} and C = {5,6,7,8} 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 = {g,y,h,x,z}, B = {y,d,u} and C = {g,z,y} 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 = {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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| 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 = {q,n,g,s,x}, B = {l,x,m} and C = {q,x,l} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {q,n,g,s,x}, B = {l,x,m} and C = {q,x,l} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {q,n,g,s,x} ∩ {l,x,m} = { } ⇒ n(A∩B) = 0 (B∩C) = {l,x,m} ∩ {q,x,l} = {l} ⇒ n(B∩C) = 1 (A∩C) = {q,n,g,s,x} ∩ {q,x,l} = {q,x} ⇒ n(A∩C) = 2 (A∪B∪C) = [{q,n,g,s,x} ∪ {l,x,m}∪{q,x,l} = {q,n,g,s,x,l,x,m}∪{q,x,l} = {q,n,g,s,x,l,x,m} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{q,n,g,s,x} ∩ {l,x,m} ∩ {q,x,l}] = { } ∩ {g,s,x,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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| 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 = {v,a,y,p,r}, B = {u,t,i} and C = {v,r,u} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {v,a,y,p,r}, B = {u,t,i} and C = {v,r,u} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {v,a,y,p,r} ∩ {u,t,i} = { } ⇒ n(A∩B) = 0 (B∩C) = {u,t,i} ∩ {v,r,u} = {u} ⇒ n(B∩C) = 1 (A∩C) = {v,a,y,p,r} ∩ {v,r,u} = {v,r} ⇒ n(A∩C) = 2 (A∪B∪C) = [{v,a,y,p,r} ∪ {u,t,i}∪{v,r,u} = {v,a,y,p,r,u,t,i}∪{v,r,u} = {v,a,y,p,r,u,t,i} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{v,a,y,p,r} ∩ {u,t,i} ∩ {v,r,u}] = { } ∩ {y,p,r,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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| 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 = {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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| 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,e,r,n}, B = {l,o,d} and C = {m,n,l} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {m,y,e,r,n}, B = {l,o,d} and C = {m,n,l} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {m,y,e,r,n} ∩ {l,o,d} = { } ⇒ n(A∩B) = 0 (B∩C) = {l,o,d} ∩ {m,n,l} = {l} ⇒ n(B∩C) = 1 (A∩C) = {m,y,e,r,n} ∩ {m,n,l} = {m,n} ⇒ n(A∩C) = 2 (A∪B∪C) = [{m,y,e,r,n} ∪ {l,o,d}∪{m,n,l} = {m,y,e,r,n,l,o,d}∪{m,n,l} = {m,y,e,r,n,l,o,d} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{m,y,e,r,n} ∩ {l,o,d} ∩ {m,n,l}] = { } ∩ {e,r,n,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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| 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 = {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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| 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 = {m,y,t,y,w}, B = {c,u,s} and C = {m,w,c} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {m,y,t,y,w}, B = {c,u,s} and C = {m,w,c} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {m,y,t,y,w} ∩ {c,u,s} = { } ⇒ n(A∩B) = 0 (B∩C) = {c,u,s} ∩ {m,w,c} = {c} ⇒ n(B∩C) = 1 (A∩C) = {m,y,t,y,w} ∩ {m,w,c} = {m,w} ⇒ n(A∩C) = 2 (A∪B∪C) = [{m,y,t,y,w} ∪ {c,u,s}∪{m,w,c} = {m,y,t,y,w,c,u,s}∪{m,w,c} = {m,y,t,y,w,c,u,s} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{m,y,t,y,w} ∩ {c,u,s} ∩ {m,w,c}] = { } ∩ {t,y,w,c} = { } ⇒ 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 = {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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| 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 = {z,y,o,e,h}, B = {a,s,q} and C = {z,h,a} n(A∪B∪C) = Answer: 8 SOLUTION 1 : Given : A = {z,y,o,e,h}, B = {a,s,q} and C = {z,h,a} ⇒ n(A) = 5, n(B) = 3, n(C) = 3 (A∩B) = {z,y,o,e,h} ∩ {a,s,q} = { } ⇒ n(A∩B) = 0 (B∩C) = {a,s,q} ∩ {z,h,a} = {a} ⇒ n(B∩C) = 1 (A∩C) = {z,y,o,e,h} ∩ {z,h,a} = {z,h} ⇒ n(A∩C) = 2 (A∪B∪C) = [{z,y,o,e,h} ∪ {a,s,q}∪{z,h,a} = {z,y,o,e,h,a,s,q}∪{z,h,a} = {z,y,o,e,h,a,s,q} ⇒ n(A∪B∪C) = 8 ............... (1) (A∩B∩C) = [{z,y,o,e,h} ∩ {a,s,q} ∩ {z,h,a}] = { } ∩ {o,e,h,a} = { } ⇒ 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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