# CLASS-8PROBLEM & SOLUTION BY APPLYING VENN DIAGRAM FORMULA

PROBLEM & SOLUTION BY APPLYING VENN DIAGRAM FORMULA

Formula of Venn-Diagram -

1)  n(A U B) + n(A B) = n(A) + n(B)

or,   n(A U B)  = n(A) + n(B) - n(A B)

2)   n( A U B) = n (A) + n (B)

3)  n(A) + n(A)’ = n(U)

4)  n(A U B) =  n (B) + n ( A – B )

5)   n(A) = n(A B) + n( A – B )

6)  n (A – B) + n ( B – A ) = n(A U B) - n(A ∩ B)

Example.-A) Let U be the universal set and P, Q be any two set , If n(U) = 100 , n(P) = 50 , n(Q) = 25 and  n(P Q)’ = 40, find the following

1)  n(P – Q) ,  2)  n(P ∩ Q) ,  3)  n ( P U Q ) , 4)  n ( Q – P )

1)    n(P – Q)

Ans.)    n( P – Q ) + n(Q) = n ( P U Q )

=>  n( P – Q ) = n ( P U Q ) - n(Q)

Now we have to find out the value of   n ( P U Q )

So, n(P U Q) = n(P) + n(Q) – n (P ∩ Q) =  15

Now as per the given condition and requirement –

n( P – Q ) = n ( P U Q ) - n(Q)

so,    n( P – Q ) = 15 – 25  =  - 10             (Ans.)

2)  n(P ∩ Q)

Answer)  we have to find the value of  n ( P ∩ Q )

As per the formula we know that n(A) + n(A’) = n(U)

Replacing the A by P ∩ Q, then we get –

n(P ∩ Q) + n(P ∩ Q)’ = n(U)

=>     n(P ∩ Q) + 40 = 100

n(P ∩ Q) = 100 – 40 = 60              (Ans.)

3)   n ( P U Q )

Answer)   we have to find the value of  n ( P U Q )

As per the formula we know that n(A U B) = n(A) + n(B) – n (A  ∩ B)

Replacing A & B by P & Q, we get n(P U Q) = n(P) + n(Q) – n (P ∩ Q

When n(P ∩ Q) + n(P ∩ Q)’ = n(U), then  n(P ∩ Q) = 60

So,  n(P U Q) = n(P) + n(Q) – n (P ∩ Q

=  50 + 25 – 60  =  15              (Ans.)

4)  n ( Q – P )

As per the formula we know that,

n(A – B ) + n(B – A) = n(A U B) – n(A ∩ B)

Now we will replace A & B with P & Q we get –

n(P – Q ) + n(Q – P) = n(P U Q) – n(P ∩ Q)

now,   n(Q – P)  =  n(P U Q) – n(P ∩ Q) - n(P – Q )

=  15 – 60 – (-10)

= 15 – 60 + 10

=  - 35             (Ans.)

Exercise-B)  Let U be the universal set and A & B any two sets. If n(U) = 20, n(A) = 6, n(B) = 4 and n(A U B )’ = 8 , then find the following

1)  n(A)’ ,  2) n(B)’ ,  3) n (AUB),   4) n (A ∩ B),   5)  n(A – B ),  6) n (B – A )

1)   n(A)’

Ans.)  according to the formula, we get  n(A) + n(A)’ =  n(U)

Now, replacing the value of n(A) & n(U) we can get,

6 + n(A)’ = 20

Or,  n(A)’ = 14           (Ans.)

2)  n(B)’

Ans.)  according to the formula, we get  n(B) + n(B)’ =  n(U)

Now,  replacing the value of n(B) & n(U) we can get,

So,  4  +  n(B)’ =  20

n(B)’ = 20 – 4 = 16                (Ans.)

3)    n (AUB)

Ans.)  according to the formula, we get  n(A) + n(A)’ =  n(U)

Now,  replacing A by n(AUB)

n(AUB) + n(AUB)’ = n(U)

replacing the value of n(U) = 20 &  n(A U B )’ = 8 we get,

n(AUB) + 8 = 20

now,  n(AUB) = 20 – 8 = 12             (Ans.)

4)    n (A ∩ B)

Ans.) according to the formula, we get

n(A U B) = n(A) + n(B) – n(A ∩ B)

First, we have to find the value of n(A U B)

Now,  replacing A by n(AUB)

n(AUB) + n(AUB)’ = n(U)

replacing the value of n(U) = 20 &  n(A U B )’ = 8 we get,

n(AUB) + 8 = 20

now,  n(AUB) = 20 – 8 = 12

Now, n(A U B) = n(A) + n(B) – n(A ∩ B)

Replacing the value n(A U B) = 12 , n(A) = 6, n(B) = 4

12  =  6 + 4 - n(A ∩ B)

n(A ∩ B) = 12 – 10 = 2               (Ans.)

5)    n ( A – B )

Ans.) According to the formula, we get  n(A) – n(A – B) =  n(A ∩ B)

we have to replace the value of n(A ∩ B) and  n(A)

n(A) – n(A – B) =  n(A ∩ B)

n(A) - n(A ∩ B)  =  n(A – B)

n(A – B) = n(A) - n(A ∩ B)  =  6 – 2  =  4         (Ans.)

6)  n (B – A )

Ans.)  according to the formula, we get

n(A – B) + n(B – A ) =  n( A U B ) - n(A ∩ B)

Now,  n(B – A )  =  n( A U B ) - n(A B) - n(A – B)

Replace the value of  n( A U B ),  n(A B)  &  n(A – B)

n(B – A )  =  12 – 2 – 4

=  6           (Ans.)

Example.-C)  Let, A & B be any two sets. If n(A – B) = 30, n(A U B) = 70,  n(A ∩ B) = 20 then find

1) n(B – A ),  2)   n(A),   3)  n(B)

1)  n (B – A)

n(A – B) + n(B – A) =  n(A U B) – n(A B)

Now, replace the value of  n(A – B) = 30, n( A U B ) = 50,  n(A ∩ B) = 20

We get,    n(A – B)  +  n(B – A)  =  n(A U B) –  n(A B)

=>     30 +  n(B – A)  =  70 – 20

=>     n(B – A)  =  50 – 30  =   20     (Ans.)

2)   n(A)

n(A) - n(A – B)  =  n(A B)

Now, replace the value of n(A – B) = 30, n( A U B ) = 50,  n(A ∩ B) = 20

n(A) - n(A – B)  =   n(A B)

=>         n(A)  =  n(A – B) + n(A B)

=   30 + 20  =  50                 (Ans.)

3)   n(B)