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}
Let the number of students be 100. Number of students who failed in Hindi is 30%.
n(H) = 30
Number of students who failed in English is 45%.
? n(E) = 45
Number of students who failed in both the subjects is 20%. n(H ? E) = 20
Applying the rule,
n(H ? E) = n(H) + n(E) - n(H ? E)
= 30 + 45 - 20 = 55
Percentage of students who failed in Hindi or English or both the subjects = 55%
Number of students who passed in both the subjects = 100 - 55 = 45%
Given in the question,
20% of 80 = 20 * 80 / 100 = 16
Remaining 50%
= (80 ? 16) × 50 / 100 = 32
No. of families not owning any vehicle
= 80 ? (32 + 16) = 80 ? 48 = 32
NA
NA
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.
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