UGC NET CS 2014 Dec-Paper-2

Method-1:
Given data,
— Set A={1, 2, 3, …….., 1000}
— Set A shall be divisible by 3=?
— Set A shall be divisible by 5=?
— Set A shall be divisible by 3 and 5=?
Step-1: To find divisible by 3 numbers are
=⌊1000/3⌋
= 333
Step-2: To find divisible by 5 numbers are
=⌊1000/5⌋
= 200
Step-3: To find divisible by 3 and 5 numbers are
=⌊1000/(3*5)⌋
= 66
These 66 is already part of 333 and 200. So, we have to exclude it from the list.
Total= 333+200-66
= 467
Note: We are using floor because excluding fraction value.
Method-2:
The above problem is in the form of (AUB) = (A)+(B)-(A∩B)
A=1000/3
B=1000/5
(A∩B)=66
(AUB)=467
Note: Getting this idea in exam hall is very difficult. So, better follow method-1 ratherhan method-2