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20 Apr 2021, 05:00
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B
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Difficulty:
95%
(hard)
Question Stats:
40% (02:59) correct
60%
(03:38)
wrong
based on 100
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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
A. 224
B. 225
C. 263
D. 265
E. 300
20 Apr 2021, 12:47
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Bunuel
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
16 May 2021, 07:46
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Bunuel
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
