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