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

 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 = {i,g,p,e,n}, B = {o,m,q} and C = {i,n,o}

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

 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 = {o,z,l,y,p}, B = {g,z,f} and C = {o,p,g}

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

 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 = {e,c,w,q,n}, B = {x,d,l} and C = {e,n,x}

 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 = {3,4,5}, B = {4,5,6,7} and C = {5,6,7,8}

 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 = {k,e,y,c,r}, B = {k,p,f} and C = {k,r,k}

 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 = {r,b,q,c,r}, B = {k,s,d} and C = {r,r,k}

 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 = {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. 

 



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 = {l,f,t,d,a}, B = {b,h,z} and C = {l,a,b}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {l,f,t,d,a}, B = {b,h,z} and C = {l,a,b}

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

(A∩B) =  {l,f,t,d,a} ∩ {b,h,z}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {b,h,z} ∩ {l,a,b}

           =  {b}

⇒     n(B∩C)  = 1

 (AC)  = {l,f,t,d,a} ∩ {l,a,b}

            = {l,a}

⇒       n(A∩C) = 2

(A∪B∪C) = [{l,f,t,d,a} ∪ {b,h,z}∪{l,a,b}

               = {l,f,t,d,a,b,h,z}∪{l,a,b}

        = {l,f,t,d,a,b,h,z}

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

(A∩B∩C) = [{l,f,t,d,a} ∩ {b,h,z} ∩ {l,a,b}]

                  = { } ∩ {t,d,a,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. 

 



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 = {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. 

 



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 = {l,m,f,a,s}, B = {z,p,r} and C = {l,s,z}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {l,m,f,a,s}, B = {z,p,r} and C = {l,s,z}

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

(A∩B) =  {l,m,f,a,s} ∩ {z,p,r}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {z,p,r} ∩ {l,s,z}

           =  {z}

⇒     n(B∩C)  = 1

 (AC)  = {l,m,f,a,s} ∩ {l,s,z}

            = {l,s}

⇒       n(A∩C) = 2

(A∪B∪C) = [{l,m,f,a,s} ∪ {z,p,r}∪{l,s,z}

               = {l,m,f,a,s,z,p,r}∪{l,s,z}

        = {l,m,f,a,s,z,p,r}

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

(A∩B∩C) = [{l,m,f,a,s} ∩ {z,p,r} ∩ {l,s,z}]

                  = { } ∩ {f,a,s,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. 

 



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 = {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. 

 



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 = {d,n,c,k,p}, B = {o,n,y} and C = {d,p,o}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {d,n,c,k,p}, B = {o,n,y} and C = {d,p,o}

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

(A∩B) =  {d,n,c,k,p} ∩ {o,n,y}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {o,n,y} ∩ {d,p,o}

           =  {o}

⇒     n(B∩C)  = 1

 (AC)  = {d,n,c,k,p} ∩ {d,p,o}

            = {d,p}

⇒       n(A∩C) = 2

(A∪B∪C) = [{d,n,c,k,p} ∪ {o,n,y}∪{d,p,o}

               = {d,n,c,k,p,o,n,y}∪{d,p,o}

        = {d,n,c,k,p,o,n,y}

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

(A∩B∩C) = [{d,n,c,k,p} ∩ {o,n,y} ∩ {d,p,o}]

                  = { } ∩ {c,k,p,o}

                  = { }

⇒ 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 = {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

 (AC)  = {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. 

 



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 = {x,i,r,p,i}, B = {b,u,w} and C = {x,i,b}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {x,i,r,p,i}, B = {b,u,w} and C = {x,i,b}

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

(A∩B) =  {x,i,r,p,i} ∩ {b,u,w}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {b,u,w} ∩ {x,i,b}

           =  {b}

⇒     n(B∩C)  = 1

 (AC)  = {x,i,r,p,i} ∩ {x,i,b}

            = {x,i}

⇒       n(A∩C) = 2

(A∪B∪C) = [{x,i,r,p,i} ∪ {b,u,w}∪{x,i,b}

               = {x,i,r,p,i,b,u,w}∪{x,i,b}

        = {x,i,r,p,i,b,u,w}

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

(A∩B∩C) = [{x,i,r,p,i} ∩ {b,u,w} ∩ {x,i,b}]

                  = { } ∩ {r,p,i,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. 

 



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

 (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. 

 



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 = {l,i,a,q,b}, B = {s,c,j} and C = {l,b,s}

 n(A∪B∪C) = Answer: 8


SOLUTION 1 :

  Given : 

A = {l,i,a,q,b}, B = {s,c,j} and C = {l,b,s}

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

(A∩B) =  {l,i,a,q,b} ∩ {s,c,j}

           = { }

⇒       n(A∩B) = 0

 (B∩C) = {s,c,j} ∩ {l,b,s}

           =  {s}

⇒     n(B∩C)  = 1

 (AC)  = {l,i,a,q,b} ∩ {l,b,s}

            = {l,b}

⇒       n(A∩C) = 2

(A∪B∪C) = [{l,i,a,q,b} ∪ {s,c,j}∪{l,b,s}

               = {l,i,a,q,b,s,c,j}∪{l,b,s}

        = {l,i,a,q,b,s,c,j}

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

(A∩B∩C) = [{l,i,a,q,b} ∩ {s,c,j} ∩ {l,b,s}]

                  = { } ∩ {a,q,b,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.