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Re M2855 [#permalink]
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16 Sep 2014, 01:44



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Re: M2855 [#permalink]
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01 Oct 2014, 16:14
Hello Bunuel. Can you please explain why do you have 7/20 and not 1/10...1/2*1/5 = 1/10..why do you have to add 1/5 and 1/2 and then divide it by 2? I did not get the rational behind it....



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02 Oct 2014, 02:58



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Hi Bunuel, i could not understand the 2nd statement (2) The product of any two terms of the set is a terminating decimal
so what is the set looks like? Is it five 1/2 and five 1/5? Thanks for your help. Bunuel wrote: Official Solution:
(1) Reciprocal of the median is a prime number. If all the terms equal \(\frac{1}{2}\), then the \(median=\frac{1}{2}\) and the answer is NO but if all the terms equal \(\frac{1}{7}\), then the \(median=\frac{1}{7}\) and the answer is YES. Not sufficient. (2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of \(\frac{1}{2}\) or/and \(\frac{1}{5}\). Thus the median could be \(\frac{1}{2}\), \(\frac{1}{5}\) or \(\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}\). None of the possible values is less than \(\frac{1}{5}\). Sufficient.
Answer: B
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Re: M2855 [#permalink]
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02 Oct 2014, 15:26
sunita123 wrote: Hi Bunuel, i could not understand the 2nd statement (2) The product of any two terms of the set is a terminating decimal
so what is the set looks like? Is it five 1/2 and five 1/5? Thanks for your help. Bunuel wrote: Official Solution:
(1) Reciprocal of the median is a prime number. If all the terms equal \(\frac{1}{2}\), then the \(median=\frac{1}{2}\) and the answer is NO but if all the terms equal \(\frac{1}{7}\), then the \(median=\frac{1}{7}\) and the answer is YES. Not sufficient. (2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of \(\frac{1}{2}\) or/and \(\frac{1}{5}\). Thus the median could be \(\frac{1}{2}\), \(\frac{1}{5}\) or \(\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}\). None of the possible values is less than \(\frac{1}{5}\). Sufficient.
Answer: B The set could be any combination of 1/2's and 1//5: {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2} {1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5} {1/5, 1/5, 1/5, 1/5, 1/5, 1/2, 1/2, 1/2, 1/2, 1/2} ....
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so for example if set is {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2} then it does not satisfy this condition (2) The product of any two terms of the set is a terminating decimal sorry it does. Thanks Bunuel Bunuel wrote: sunita123 wrote: Hi Bunuel, i could not understand the 2nd statement (2) The product of any two terms of the set is a terminating decimal
so what is the set looks like? Is it five 1/2 and five 1/5? Thanks for your help. Bunuel wrote: Official Solution:
(1) Reciprocal of the median is a prime number. If all the terms equal \(\frac{1}{2}\), then the \(median=\frac{1}{2}\) and the answer is NO but if all the terms equal \(\frac{1}{7}\), then the \(median=\frac{1}{7}\) and the answer is YES. Not sufficient. (2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of \(\frac{1}{2}\) or/and \(\frac{1}{5}\). Thus the median could be \(\frac{1}{2}\), \(\frac{1}{5}\) or \(\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}\). None of the possible values is less than \(\frac{1}{5}\). Sufficient.
Answer: B The set could be any combination of 1/2's and 1//5: {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2} {1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5} {1/5, 1/5, 1/5, 1/5, 1/5, 1/2, 1/2, 1/2, 1/2, 1/2} ....
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Re: M2855 [#permalink]
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04 Nov 2014, 19:47
Bunuel wrote: Official Solution:
(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of \(\frac{1}{2}\) or/and \(\frac{1}{5}\). Thus the median could be \(\frac{1}{2}\), \(\frac{1}{5}\) or \(\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}\). None of the possible values is less than \(\frac{1}{5}\). Sufficient.
Hi Bunuel, When you say that the second statement implies that the set must consist of 1/2, 1/5, or a mix of both, what exactly implies this? Are 1/2 and 1/5 the only terminating reciprocals of prime numbers and that this is just a rule we should memorize? Or is there something else that gives this away?



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Re: M2855 [#permalink]
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05 Nov 2014, 05:41
OutOfTheHills wrote: Bunuel wrote: Official Solution:
(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of \(\frac{1}{2}\) or/and \(\frac{1}{5}\). Thus the median could be \(\frac{1}{2}\), \(\frac{1}{5}\) or \(\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}\). None of the possible values is less than \(\frac{1}{5}\). Sufficient.
Hi Bunuel, When you say that the second statement implies that the set must consist of 1/2, 1/5, or a mix of both, what exactly implies this? Are 1/2 and 1/5 the only terminating reciprocals of prime numbers and that this is just a rule we should memorize? Or is there something else that gives this away? The stem says that set A consist of 10 terms, each of which is a reciprocal of a prime number. Therefore the elements of A could be 1/2, 1/3, 1/5, 1/7, 1/11, 1/13, ... Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Note that if denominator already has only 2s and/or 5s then it doesn't matter whether the fraction is reduced or not. For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal. We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced. Therefore, for the product of any two terms of the set to be terminating decimal, it must consists of only of 1/2's, 1/5's, or both. Check Terminating and Recurring Decimals Problems in our Special Questions Directory. Hope it helps.
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Re: M2855 [#permalink]
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06 Nov 2014, 14:53
Bunuel wrote: I was still a bit confused at first, but the directory of problems is a huge, huge help. For anyone with similar problem is me, this is just a rule you should remember.



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Re M2855 [#permalink]
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24 Aug 2016, 02:54
I think this is a highquality question and I agree with explanation.



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Re: M2855 [#permalink]
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17 Nov 2016, 21:16
@Bunnel
For statement 2, the median is either 1/2 or 1/5, there are two different answers. either less or equal to 1/5. How can we take B to be sufficient?
The question asks if median is less than 1/5. It's either yes with median=1/2 or no with median=1/5.
This makes the answer E, as even after combining the two options, we wouldn't get a definite answer.
Please advice.



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17 Nov 2016, 21:41



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Re: M2855 [#permalink]
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22 Nov 2017, 06:56
what if the set consists of 1/2^6s and 1/5^6s? In that case, does statement 2 suffice?



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Re: M2855 [#permalink]
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22 Nov 2017, 07:09










