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