The GRE maths sample question given below is a problem solving question in Set Theory. This question tests the concept of the union of 2 sets and basics of Number Systems - divisibility of numbers and number of multiples of a number - and Arithmetic Progression.

Question 2: Set A comprises all multiples of 3 less than 200. Set B comprises of all odd multiples of 5 less than 200. How many elements does A∪B have?

- 73
- 86
- 82
- 79
- 83

From INR 2999

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A) → Number of multiples of 3 less than 200 = {3, 6, 9,..., 195, 198}

n(B) → Number of odd multiples of 5 less than 200 = {5, 15, 25, ...185, 195}

n(A∩B) → Number of integers less than 200 that are divisible by 3 and are odd multiples of 5.

The easiest way to compute the number of terms in set A is to find the quotient when 200 is divided by 3.

The quotient when 200 is divided by 3 is 66.

The 66^{th} multiple of 3 is 198 which is less than 200.

n(A) = 66.

Elements in set B are in AP with,

First term, a_{1} = 5; Last term, a_{n} = 195; Common difference, d = 10

n^{th} term of an AP, a_{n} = a_{1} + (n - 1) d

195 = 5 + ((n - 1) × 10)

n - 1 = \\frac{\text{190}}{\text{10}}) = 19 → n = 20

Therefore, number of elements in set B = **n(B) = 20**

Positive integers lesser than 200 that are multiples of 3 and odd multiples of 5.

→15, 45, 75,...,195

The number of terms are not many. So, listing down the terms of the set is the easiest option.

A∩B = {15, 45, 75, 105, 135, 165, 195}

Therefore, number of terms in A∩B = **n(A∩B) = 7**

n(A∪B) = n(A) + n(B) - n(A ∩ B)

Substitute the value of n(A), n(B) and n(A ∩ B):

n (A ∪ B) = 66 + 20 - 7 = **79**

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