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?

- 224
- 225
- 263
- 265
- 300

From INR 2999

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}

Elements in set A are in AP,

First term, a_{1} = 102; Last term, a_{n} = 996; Common difference, d = 6

nth term of an AP, a_{n} = a_{1} + (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**

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}

Elements in set B are in AP with,

First term, a_{1} = 100; Last term, a_{n} = 996; Common difference, d = 8

nth term of an AP : a_{n} = a_{1} + (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**

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 }

nth term of an AP : a_{n} = a_{1} + (n -1)d

996 = 108 + ( (n - 1) × 24 )

n - 1 = \\frac{996}{108}) = 37 → n = 38

Therefore, **n(A ∩ B) = 38**

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**

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