NA
NA
Given: n(C) = 50, n(F) = 20, n(C ? F) = 10.
Number of students playing at least one of these two games = n (C ? F) = n (C) + n (F) ? n (C ? F)
= 50 + 20 ? 10 = 60.
(a) 1? A but 1? B and 6 ? B but 6 ? A.
? A and B are not comparable.
(b) A = {x: x ? N and x ? 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Clearly, A ? B ? A and B are comparable.
(c) {4, 5} ? A but {4, 5} ? B and 4 ? B but 4 ? A.
? A and B are not comparable.
NA
A new set having those elements which are in A but not in B is said to be the difference of sets A and B and it is denoted by A ? B.
Given that in question, A = {1, 2, 3, 4, 5} and B = {2, 4, 6}
So the numbers which are in A and but not in B = {1, 3, 5}
? A ? B = {1, 3, 5}
NA
NA
(a) {x: x is a natural number and 4 ? x ? 6} = {4, 5, 6}. Now, {1, 2, 3, 4} and {4, 5, 6} have one element 4 common. Therefore, the given two sets are not disjoint.
(b) The sets {a, e, i, o, u} and {c, d, e, f} have one element e as common. Then, the given two sets are not disjoint.
(c) The sets {x: x is an even integer} and {x: x is an odd integer} have no element as common and therefore they are disjoint sets.
Given in the question ,
n(X ?Y) = 18, n(X) = 8, n(Y) = 15.
Using the formula
n(X ? Y) = n(X) + n(Y) ? n(X ? Y), we get n(X ? Y) = 8 + 15 ? 18 = 5.
Given in the question ,
n(A) = 40, n(A ? B) = 60 and n(A ? B) = 10.
As we know the formula,
n(A ? B) = n(A) + n(B) ? n(A ? B)
Putting these values in the formula, we get
60 = 40 + n(B) ? 10
? n(B) = 30.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.