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Sets A and B each consist of three terms selected from the 5
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25 Dec 2012, 23:33
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Sets A and B each consist of three terms selected from the first five prime integers. No term appears more than once within a set, but any integer may be a term in both sets. If the average of the terms in Set A is 4 and the product of the terms in Set B is divisible by 22, how many terms are shared by both sets? (1) The product of the terms in Set B is not divisible by 5. (2) The product of the terms in Set B is divisible by 14.
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Re: Sets A and B each consist of three terms selected from the 5
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26 Dec 2012, 02:23
Sets A and B each consist of three terms selected from the first five prime integers. No term appears more than once within a set, but any integer may be a term in both sets. If the average of the terms in Set A is 4 and the product of the terms in Set B is divisible by 22, how many terms are shared by both sets?The first five primes are: 2, 3, 5, 7, and 11. The average of the terms in Set A is 4 > the sum of the terms is Set A is 4*3=12 > A={2, 3, 7}; The product of the terms in Set B is divisible by 22 > B={2, 11, x}. (1) The product of the terms in Set B is not divisible by 5. This implies that 5 is not in Set B, thus B={2, 11, 3} or B={2, 11, 7}. In both cases, A and B share two terms. Sufficient. (2) The product of the terms in Set B is divisible by 14. This implies that 7 must be in Set B, thus B={2, 11, 7}. Therefore, A and B share two terms. Sufficient. Answer: D.
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Re: Sets A and B each consist of three terms selected from the 5
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19 Dec 2014, 17:25
Nice Question.. The first five prime integers are 2, 3, 5, 7, and 11. Set A: The average of the terms in Set A is 4, so the sum of the terms is (4)(3) = 12. There is only one way for three of the first five primes to sum to 12: 2 + 3 + 7. Set A is {2, 3, 7}. Set B: The product of the terms in Set B is divisible by 22, so 2 and 11 are terms in Set B. Set B is {2, 11, x}, where x can be 3, 5, or 7. We know that Sets A and B share one term: the 2. If x is either 3 or 7, the sets will share two terms. If x is 5, the sets will only share one term. Thus, we can rephrase the question as “Is x = 5?” A definite Yes or No answer leads to a definite value answer for the number of shared terms (that is, Yes = 1, No = 2). (1) SUFFICIENT: If the product of the terms in Set B is not divisible by 5, x ≠ 5 and the answer to our rephrased question is a definite No. (2) SUFFICIENT: If the product of the terms in Set B is divisible by 14, then 2 and 7 are terms in B. Therefore, x = 7 and the answer to our rephrased question is a definite No.
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Re: Sets A and B each consist of three terms selected from the 5
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20 Dec 2014, 22:12
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
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08 Mar 2017, 20:34
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
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22 May 2017, 08:50
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
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22 May 2017, 09:48
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
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23 May 2018, 03:13
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Re: Sets A and B each consist of three terms selected from the 5 &nbs
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