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