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