Scroll:set and function >> Exercice 1.3 >> saq (4261)


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 = {1,2,3}, B = {2,3,4,5} and C = {3,4,5,6}

 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 = {g,d,f,z,s}, B = {e,b,d} and C = {g,s,e}

 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 = {6,7,8}, B = {7,8,9,10} and C = {8,9,10,11}

 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 = {c,p,a,z,v}, B = {o,d,e} and C = {c,v,o}

 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 = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9}

 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 = {a,n,d,z,c}, B = {o,t,f} and C = {a,c,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 = {2,3,4}, B = {3,4,5,6} and C = {4,5,6,7}

 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 = {s,n,e,q,m}, B = {f,b,a} and C = {s,m,f}

 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 = {5,6,7}, B = {6,7,8,9} and C = {7,8,9,10}

 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 = {u,c,u,k,b}, B = {m,d,v} and C = {u,b,m}

 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 = {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

 (AC)  = {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. 

 



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 = {h,m,u,v,n}, B = {u,z,v} and C = {h,n,u}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {h,m,u,v,n}, B = {u,z,v} and C = {h,n,u}

⇒          n(A) = 5, n(B) = 3, n(C) = 3

(A∩B) =  {h,m,u,v,n} ∩ {u,z,v}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {u,z,v} ∩ {h,n,u}

           =  {u}

⇒     n(B∩C)  = 1

 (AC)  = {h,m,u,v,n} ∩ {h,n,u}

            = {h,n}

⇒       n(A∩C) = 2

(A∪B∪C) = [{h,m,u,v,n} ∪ {u,z,v}∪{h,n,u}

               = {h,m,u,v,n,u,z,v}∪{h,n,u}

        = {h,m,u,v,n,u,z,v}

⇒     n(A∪B∪C) = 8     ...............  (1)

(A∩B∩C) = [{h,m,u,v,n} ∩ {u,z,v} ∩ {h,n,u}]

                  = { } ∩ {u,v,n,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. 

 



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 = {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

 (AC)  = {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. 

 



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,x,q,z,c}, B = {l,f,m} and C = {h,c,l}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {h,x,q,z,c}, B = {l,f,m} and C = {h,c,l}

⇒          n(A) = 5, n(B) = 3, n(C) = 3

(A∩B) =  {h,x,q,z,c} ∩ {l,f,m}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {l,f,m} ∩ {h,c,l}

           =  {l}

⇒     n(B∩C)  = 1

 (AC)  = {h,x,q,z,c} ∩ {h,c,l}

            = {h,c}

⇒       n(A∩C) = 2

(A∪B∪C) = [{h,x,q,z,c} ∪ {l,f,m}∪{h,c,l}

               = {h,x,q,z,c,l,f,m}∪{h,c,l}

        = {h,x,q,z,c,l,f,m}

⇒     n(A∪B∪C) = 8     ...............  (1)

(A∩B∩C) = [{h,x,q,z,c} ∩ {l,f,m} ∩ {h,c,l}]

                  = { } ∩ {q,z,c,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. 

 



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 = {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

 (AC)  = {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. 

 



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,k,i,c}, B = {t,c,h} and C = {m,c,t}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {m,y,k,i,c}, B = {t,c,h} and C = {m,c,t}

⇒          n(A) = 5, n(B) = 3, n(C) = 3

(A∩B) =  {m,y,k,i,c} ∩ {t,c,h}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {t,c,h} ∩ {m,c,t}

           =  {t}

⇒     n(B∩C)  = 1

 (AC)  = {m,y,k,i,c} ∩ {m,c,t}

            = {m,c}

⇒       n(A∩C) = 2

(A∪B∪C) = [{m,y,k,i,c} ∪ {t,c,h}∪{m,c,t}

               = {m,y,k,i,c,t,c,h}∪{m,c,t}

        = {m,y,k,i,c,t,c,h}

⇒     n(A∪B∪C) = 8     ...............  (1)

(A∩B∩C) = [{m,y,k,i,c} ∩ {t,c,h} ∩ {m,c,t}]

                  = { } ∩ {k,i,c,t}

                  = { }

⇒ 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 = {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

 (AC)  = {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. 

 



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,g,p,a,m}, B = {f,t,l} and C = {b,m,f}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {b,g,p,a,m}, B = {f,t,l} and C = {b,m,f}

⇒          n(A) = 5, n(B) = 3, n(C) = 3

(A∩B) =  {b,g,p,a,m} ∩ {f,t,l}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {f,t,l} ∩ {b,m,f}

           =  {f}

⇒     n(B∩C)  = 1

 (AC)  = {b,g,p,a,m} ∩ {b,m,f}

            = {b,m}

⇒       n(A∩C) = 2

(A∪B∪C) = [{b,g,p,a,m} ∪ {f,t,l}∪{b,m,f}

               = {b,g,p,a,m,f,t,l}∪{b,m,f}

        = {b,g,p,a,m,f,t,l}

⇒     n(A∪B∪C) = 8     ...............  (1)

(A∩B∩C) = [{b,g,p,a,m} ∩ {f,t,l} ∩ {b,m,f}]

                  = { } ∩ {p,a,m,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. 

 



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 = {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

 (AC)  = {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. 

 



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 = {h,d,z,g,q}, B = {x,t,u} and C = {h,q,x}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {h,d,z,g,q}, B = {x,t,u} and C = {h,q,x}

⇒          n(A) = 5, n(B) = 3, n(C) = 3

(A∩B) =  {h,d,z,g,q} ∩ {x,t,u}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {x,t,u} ∩ {h,q,x}

           =  {x}

⇒     n(B∩C)  = 1

 (AC)  = {h,d,z,g,q} ∩ {h,q,x}

            = {h,q}

⇒       n(A∩C) = 2

(A∪B∪C) = [{h,d,z,g,q} ∪ {x,t,u}∪{h,q,x}

               = {h,d,z,g,q,x,t,u}∪{h,q,x}

        = {h,d,z,g,q,x,t,u}

⇒     n(A∪B∪C) = 8     ...............  (1)

(A∩B∩C) = [{h,d,z,g,q} ∩ {x,t,u} ∩ {h,q,x}]

                  = { } ∩ {z,g,q,x}

                  = { }

⇒ 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.