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UGC-NET | UGC NET CS 2014 Dec – II | Question 1

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Consider a set A = {1, 2, 3, …….., 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5?
(A) 533
(B) 599
(C) 467
(D) 66


Answer: (C)

Explanation: From set A numbers {3,6,9,……..999} which are divisible by 3 are 999 / 3 (A)= 333 .
From set A numbers {5,10,……995,1000} which are divisible by 5 are 1000 / 5 (B)= 200.
From set A numbers {15, 30, …990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66.
So, numbers divisible by 3 or by 5 or by both 3 and 5:
(A ∨ B) = A + B – (A ∧ B)
(A ∨ B) = 333 + 200 – 67.
(A ∨ B) = 467.
So, option (C) is correct.


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Last Updated : 27 Jun, 2018
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