GRE® Quant Practice | Set Theory Q1

The GRE maths sample question given below is a problem solving question in Set Theory. This GRE sample question is an easy question which tests the concept of union of two sets and elementary concepts in arithmetic progression and divisibility of numbers.

Question 1: Set A comprises all 3-digit numbers that are multiples of 6. Set B comprises all 3-digit numbers that are multiples of 4 but are not multiples of 8. How many elements does (A ∪ B) comprise?

1. 224
2. 225
3. 263
4. 265
5. 300

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Video Explanation

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Explanatory Answer | GRE Set Theory Questionbank Q1

Step 1: Identify key elements of Set A

Set A comprises all 3-digit multiples of 6.
First term of set A = 6 × 17 = 102
Second term of set A = 6 × 18 = 108
Last term of set A = 996 (6 × 166)
Therefore, set A = {102, 108, 114, ..., 990, 996}

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

Elements in set A are in AP,
First term, a₁ = 102; Last term, a_n = 996; Common difference, d = 6
nth term of an AP, a_n = a₁ + (n - 1) d
996 = 102 + ((n - 1) × 6)
n - 1 = \\frac{894}{6}) = 149 → n = 150
Therefore, number of elements in set A = n(A) = 150

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Step 2: Identify key elements of Set B

Set B comprises all 3-digit numbers that are multiples of 4 but are not multiples of 8.
First term of set B = 100 (4 × 25)
Second term of set B = 108 (4 × 27)
Last term of set B = 996 (4 × 249)
Therefore, set B = {100, 108, 116, ..., 988, 996}

Step 2A: Compute the number of terms in set B

Elements in set B are in AP with,
First term, a₁ = 100; Last term, a_n = 996; Common difference, d = 8
nth term of an AP : a_n = a₁ + (n - 1) d
996 = 100 + ((n - 1) × 8)
n - 1 = \\frac{896}{8}) = 112 → n = 113
Therefore, number of terms in set B = n(B) = 113

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Step 3: List elements common to both the sets

Set A is an AP with common difference 6.
Set B is an AP with common difference 8.
So, common difference of elements common to A and B is LCM (6, 8) = 24

Compare set A = { 102, 108, 114, ..., 990, 996 } and set B = { 100, 108, 116, ..., 988, 996 }
First term common to both sets = 108
Last term common to both sets = 996
Let n(A ∩ B) = Number of terms common to both sets
Elements common to both sets = { 108, 132, 156, ..., 996 }

Step 3A: Compute number of elements common to both the sets

nth term of an AP : a_n = a₁ + (n -1)d
996 = 108 + ( (n - 1) × 24 )
n - 1 = \\frac{996}{108}) = 37 → n = 38
Therefore, n(A ∩ B) = 38

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Step 4: Compute UNION OF TWO SETS

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) = 150 + 113 - 38 = 225

Choice B is the correct answer