Let A and B be two disjoint sets. If the cardinality of A is equal to square of 3 and cardinality of B is equal to square root of 16, then what is the cardinality of A $\cup$ B?

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Solution

The correct option is A 13
A and B are disjoint sets. Therefore, n(A $\cap$ B) = 0

n(A) = square of 3 = 9

n(B) = square root of 16 = 4

Therefore, n(A $\cup$ B) = n(A) + n(B) - n(A $\cap$ B)

= 9 + 4

= 13

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Let A and B be two disjoint sets. If the cardinality of A is equal to square of 3 and cardinality of B is equal to square root of 16, then what is the cardinality of A ∪ B?

The correct option is A 13A and B are disjoint sets. Therefore, n(A ∩ B) = 0 n(A) = square of 3 = 9 n(B) = square root of 16 = 4 Therefore, n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 9 + 4 = 13

Let A and B be two disjoint sets. If the cardinality of A is equal to square of 3 and cardinality of B is equal to square root of 16, then what is the cardinality of A $\cup$ B?

The correct option is A 13
A and B are disjoint sets. Therefore, n(A $\cap$ B) = 0

n(A) = square of 3 = 9

n(B) = square root of 16 = 4

Therefore, n(A $\cup$ B) = n(A) + n(B) - n(A $\cap$ B)

= 9 + 4

= 13