Question
Question: If \(n\left( A \right) = 1000,n\left( B \right) = 500\) and if \(n\left( {A \cap B} \right) \geqslan...
If n(A)=1000,n(B)=500 and if n(A∩B)⩾1 and n(A∪B)=p, then
A. 500⩽p⩽1000
B. 1001⩽p⩽1498
C. 1000⩽p⩽1498
D. 999⩽p⩽1499
E. 1000⩽p⩽1499
Solution
With the given details we can use them in the formula n(A∩B)=n(A)+n(B)−n(A∪B)and using the given condition we get that p needs to be less than or equal to 1499 and as the cardinality of union of the sets is greater than or equal to the cardinality of the sets we get that p must be greater than or equal to 1000. combing the conditions we get the required range.
Step by step solution :
We are given that n(A)=1000,n(B)=500 and n(A∩B)⩾1and n(A∪B)=p
We know that
⇒n(A∩B)=n(A)+n(B)−n(A∪B)
Now using the given values in this we get
⇒n(A∩B)=1000+500−p ⇒n(A∩B)=1500−p
Since n(A∩B)⩾1the value of p should be less than 1500 that is
⇒p⩽1499……….(1)
As we have n(A)=1000,n(B)=500
Since the union of A and B consists of all the elements of A and B
It is clear that n(A∪B)⩾n(A)
Hence we get
⇒p⩾1000………(2)
Combining (1) and (2) we get the range of p to be
⇒1000⩽p⩽1499
Therefore the correct answer is option E.
Note :
In English, the union of two sets A and B is the set containing elements that are either in A or in B.
An element belongs to the intersection of two sets if the element is in both set A and in set B.
In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.