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Set A comprises all 3-digit numbers that are multiples of 6. Set B com 20 Apr 2021, 05:00
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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 (the union of sets)?

A. 224
B. 225
C. 263
D. 265
E. 300
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B
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Set A comprises all 3-digit numbers that are multiples of 6. Set B com 20 Apr 2021, 12:47
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Bunuel
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 (the union of sets)?

A. 224
B. 225
C. 263
D. 265
E. 300

Set B contains every other multiple of 4, 100, 108, 116, etc. Hence it contains all 3-digit numbers that are a multiple of 8 minus 4. The LCM of 8 and 6 is 24, so we can observe the following:

Set A: {108, 114, 120, 126, 132}
Set B: {108, 116, 124, 132}

So for every 4 numbers in A (or every 3 numbers in B), there will be one number that overlaps with that in the other set. In other words, for each range of 24 after 108, we will have only 6 unique numbers.

We may count the number of such ranges of 24, starting from 108 and add all the way to near 1000. Add 720 to reach 828 which is 30 ranges, add 120 (5 ranges) to reach 948, add two more ranges to reach 996.

There are 37 of these ranges in total, in each range we have 6 numbers so \(37*6 = 180 + 42 = 222\) numbers from these. Finally include 996, 100 from set B, 102 from set A and we would have 225 numbers.

Ans: B
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Set A comprises all 3-digit numbers that are multiples of 6. Set B com 16 May 2021, 07:46
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Bunuel
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 (the union of sets)?

A. 224
B. 225
C. 263
D. 265
E. 300
Solution:

The number of elements in set A is (996 - 102)/6 + 1 = 150. The number of elements in set B is (996 - 100)/8 + 1 = 113. The elements that are common in both sets (i.e., A ∩ B) are 108, 132, …, 996. The number of these elements is (996 - 108)/24 + 1 = 38. Therefore, N(A U B) = N(A) + N(B) - N(A ∩ B) = 150 + 113 - 38 = 225.

Answer: B
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Set A comprises all 3-digit numbers that are multiples of 6. Set B com 18 May 2021, 11:39
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Total elements in Set A = Number of Elements divisible by 4 - Number of Elements divisible by 8
ie ((996 - 100)/4 + 1) - ((992 - 104)/8 + 1) = 225 - 112 = 113

Total elements in Set B = Number of Elements divisible by 6
ie ((996 - 102)/6 + 1) = 150

Now
Number of Elements in (A ∪ B) = Number of Elements in A + Number of Elements in B - Number of Common Elements
Number of Elements in (A ∪ B) = 113 + 150 - Number of Common Elements

We now need to find Number of Common Elements :
It will be the count of Elements in Set B that are divisible 4, and because B is already full of even numbers half of them will be divisible by 4 ie 75 (out of 150, we can divide by two because 150 itself i even). But wait ! The 75 here also includes the numbers that are divisible by 8 and those are not in they SET A in the first place, so they cannot be common, hence they need to be removed from 75. The number of Elements divisible by 8 is half of those divisible by 4 is [75/2] = 37 (the reason it is 37 and not 38 is that neither the lowest(108) nor highest(996) common numbers is divisible by 8) - (This part is error prone and can lead to wrong calculation particularly because of the 224 option in the answer choices, GCD approach might be better here)
Hence Number of Common Elements is : 75-37 = 38

Number of Elements in (A ∪ B) = 113 + 150 -38 = 225

Answer B
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Set A comprises all 3-digit numbers that are multiples of 6. Set B com 26 Jun 2024, 00:21
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