Given, n(U)=1000, n(F)=700, n(C)=600, n(F∩C) =100
n (F∪C) = n (F) + n (C) ? n(F∩C)
n (F∪C) = 700+600-100=1200
We know, n (A∪O) ≤ n(U). Here this is not valid so, data is incorrect.

Intersection of set A and B is a set that contains elements which is common to both set A and set B.
A = {a, e, i, o, u}
B = {a, e, u}
A∩B = {a, e, u} = B.

Here, A={1,2,3}, B={2,3,4}, C={2,3,5}, D={1,3,5,7}
A∩B = {1,2,3}∩{2,3,4} = {2,3} and C∩D = {2,3,5}∩{1,3,5,7} = {3,5}.
(A∩B) ∪ (C∩D) = {2,3} ∪ {3,5} = {2,3,5} = C.

Complement of a set A is a set which contains elements of U which are not elements of B. So, B? = U – B = {1, 2, 3, 4, 5, 6} – {2,3,5} = {1,4,6}.

Element that lie inside boundary of B but outside boundary of A is 3. So, element that belongs to B but not A is 3. Set is {3}.

Since all people read newspaper so, n (T ∪ H) = 1000.
n(H) = 600 and n(T) = 700.
n (T ∪ H) = n (T) + n (H) ? n (T ∩ H)
1000 = 700 + 600 ? n (T ∩ H)
n (T ∩ H) = 1300-1000 = 300.
n (T? ∩ H) = n(H)- n (T ∩ H) = 600-300 = 300.
300 persons read only Hindustan Times.

Since denominator is one more than numerator and only 5 terms are present. So, above set can be written as {x : \(x=\frac{n}{n+1}\) where n is a natural number less than 6}.

If A={1,2,3} and B={3,4,5}, n A-B={1,2,3} – {3,4,5} = {1,2} and B-A={3,4,5} – {1,2,3} = {4,5}. They are not equal so,
A-B and B-A are not always equal.

The given sequence is an odd number sequence of format 2n+1 or 2n-1, but since n is given as a natural number so 2n+1 is not valid as 0 is not a natural number hence 2n-1 is correct answer.

X={A,L,O,Y} and Y={L,O,Y,A} .
Since both sets have same elements so y are equal.