# GRE® Math Practice | Set Theory Q2

#### GRE Sample Questions | Set Theory Question bank

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?

1. 73
2. 86
3. 82
4. 79
5. 83

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### Explanatory Answer | GRE Set Theory Questionbank Q2

#### Step 1: Identify key elements required to find out A∪B

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.

#### Step 1A: Compute the number of terms in set A

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 66th multiple of 3 is 198 which is less than 200.
n(A) = 66.

#### Step 1B: Compute the number of terms in Set B

Elements in set B are in AP with,
First term, a1 = 5; Last term, an = 195; Common difference, d = 10
nth term of an AP, an = a1 + (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

#### Step 1C: Compute the number of terms in n(A∩B)

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

#### Step 2: Number of Elements in A∪B

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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