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Re: Sets A and B each consist of three terms selected from the 5 [#permalink]
Set B product divisible by 22 i.e should have 2 and 11.
Set A average =4 i.e sum =12 i.e should have 2,3 ,7
a/c to statement 1 set B product not divisible by 5 so set be does not contain 5 so what are the options either 3 or 7. So in Set be could be either of the following 2,3,11 or 2,7, 11. In both the cases it shares two integers with set b i.e 2,3 or 2,7. Hence sufficient.

A/c to statement 2 product of the terms in Set b divisible by 14 so must have 2 and 7 set B has 2,7 and 11 as the integers.Now Set be has two integers common with set A i.e 2 and 7. hence sufficient. Hope this explanation helps.
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Sets A and B each consist of three terms selected from the 5 [#permalink]
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1st 5 prime integers are 2, 3 , 5, 7, 11

Given: Average of terms in Set A is 4
=> Sum of terms in Set A is (4*3) = 12

2+3+7 = 12
Hence, terms in Set A are 2, 3, 7

Given: Product of terms in Set B is divisible by 22 which implies that 2 and 11 are definitely in Set B

Statement (1) Product of the terms in Set B are not divisible by 5
Means that 5 is definitely not a part of Set B

So, Set B could be either (2,3,11) or (2,7,11), so the two sets share 2 terms. Hence, statement is sufficient.

Statement (2) Product of the terms in Set B are divisible by 14
Means that 2 and 7 are definitely a part of Set B and since the question stem has already established that 2 and 11 are a part of Set B, the remaining term is 7. It completely discards 3 and 5.
Hence, Set B contains 2, 7, 11. Which means that 2 terms are shared by both sets, i.e. 2 and 7. Hence, statement is sufficient. (D)
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Sets A and B each consist of three terms selected from the 5 [#permalink]
I am not comprehending how each statement alone is sufficient . Without looking at the statements I can arrive at (7,3,2) & (2,3,11) . If I use statement one alone I can arrive at (2,3,11) to form 66 which is divisible by 22. I need statement two to arrive at the LCM of 22 and 14 ( 154) . If one was to use 66 or 154, set A and B will share two terms . Is the answer D because there are no other primes that can form a multiple of 22 that perhaps share one or 3 terms? I am really confused about this data sufficiency thing . Your help will be appreciated.
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Re: Sets A and B each consist of three terms selected from the 5 [#permalink]
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NanaA wrote:
Without looking at the statements I can arrive at (7,3,2) & (2,3,11)


Without looking at the statements, Set A has to be (2, 3, 7). But, there are actually some other possibilities for Set B. B could be (2,3,11), (2,7,11), OR (2,5,11).

Data Sufficiency questions can never be answered without any info from the statements. So, if you look at a question and you feel like you can narrow it down to a single scenario without using either of the statements, it means you've left something out.

Statement 1 narrows B down to either (2, 3, 11) or (2, 7, 11). (You've missed the second one there.) Make sure you go all the way back to the question and answer it! If B is (2, 3, 11), then the answer ("how many terms are shared by both sets?") is two. If B is (2, 7, 11), the answer is still two.

That means statement 1 lets you answer the question definitively. So, it's sufficient.

Statement 2 narrows it down to just (2, 7, 11). Go back to the question. How many terms are shared? The answer is again, two. That's a single answer, so it's sufficient.
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Re: Sets A and B each consist of three terms selected from the 5 [#permalink]
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Re: Sets A and B each consist of three terms selected from the 5 [#permalink]
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