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# D01-23

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Math Expert
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16 Sep 2014, 00:12
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49% (01:09) correct 51% (01:41) wrong based on 199 sessions

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List A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the list less than 1/5?

(1) Reciprocal of the median is a prime number.

(2) The product of any two terms of the list is a terminating decimal.
[Reveal] Spoiler: OA

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16 Sep 2014, 00:12
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Official Solution:

List A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the list less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the list is a terminating decimal. This statement implies that the list must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

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24 Nov 2014, 09:19
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Bunuel wrote:
Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number.

(2) The product of any two terms of the set is a terminating decimal.

Hi again Bunuel!

Prompt: A set consists of reciprocals of 10 different prime numbers (1/2, 1/3 ..... 1/101....). Is the sum of the 5 and 6th term less than 1.5?

Statement 1: Reciprocal of the average of the 5th and 6th term is also a prime number. I understand your explanation and this is clearly not sufficient.

Statement 2: 1/2 and 1/5 are the only numbers which are a part of this set (squares or cubes of these numbers also can't be a part of this set as the question states that the numbers are reciprocals of primes only). The 10 numbers could be 9 1/2's and 1 1/5 as well. So we don't know have sufficiency?

Is this understanding correct? IMHO answer is E
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JA
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24 Nov 2014, 09:37
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joseph0alexander wrote:
Bunuel wrote:
Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number.

(2) The product of any two terms of the set is a terminating decimal.

Hi again Bunuel!

Prompt: A set consists of reciprocals of 10 different prime numbers (1/2, 1/3 ..... 1/101....). Is the sum of the 5 and 6th term less than 1.5?

Statement 1: Reciprocal of the average of the 5th and 6th term is also a prime number. I understand your explanation and this is clearly not sufficient.

Statement 2: 1/2 and 1/5 are the only numbers which are a part of this set (squares or cubes of these numbers also can't be a part of this set as the question states that the numbers are reciprocals of primes only). The 10 numbers could be 9 1/2's and 1 1/5 as well. So we don't know have sufficiency?

Is this understanding correct? IMHO answer is E

For (2) 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}
....

Let me ask you: could the median of any of the sets above be less than 1/5?
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24 Nov 2014, 09:42
Bunuel wrote:
Let me ask you: could the median of any of the sets above be less than 1/5?

Thanks Bunuel! I got it!
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JA
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22 Dec 2014, 15:10
Hello.. i did not understand the solution . Does not a Set contains all unique elements ?

Based on that, All these are invalid combinations :

{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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23 Dec 2014, 19:50
Read the wording of the question stem very carefully. It never says DISTINCT. Therefore, the elements of the set could be the same.

1) So, the reciprocal of the median is prime. All the elements of the set are prime, so the median is could be an element of the set, right. If the set is all 1/2, the answer is "No." If the set is all 1/7, the answer is "Yes." INSUFFICIENT
2) This limits the set elements to some combination of the numbers 1/2 and 1/5. Any set with these elements will have a median higher than 1/5. So the answer is "Always No." SUFFICIENT
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23 Dec 2014, 23:30
Thanks TobyTutor for yout input.

By Defination , a "Set" means distinct element.

So, based on the standard definition of set, above combination does not seem valid.
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23 Dec 2014, 23:53
True for "official" set theory, not true for the GMAT -- if it were, the concept of "MODE" would never come up. The GMAT will refer to groups of numbers, repeated or distinct, as a "set."
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05 Jan 2015, 04:31
Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=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 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Hi Bunuel,
How can we get to be sure while solving the problem that ONLY 1/2 or 1/5 or combination of both results in terminating decimal.
Because there can be infinite # of cases of reciprocal of prime numbers in a set.
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05 Jan 2015, 05:55
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Ankur9 wrote:
Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=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 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Hi Bunuel,
How can we get to be sure while solving the problem that ONLY 1/2 or 1/5 or combination of both results in terminating decimal.
Because there can be infinite # of cases of reciprocal of prime numbers in a set.

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 non-negative 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 2-s and/or 5-s 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.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
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18 Dec 2015, 06:39
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I was just about right, but made the usual mistake. The question asks, if mean is less than 1/5. And I deduced that it could be 1/5 too and marked E :'(
Great question!
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14 Jul 2016, 12:59
I think this is a high-quality question and I agree with explanation. Great question!!
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15 Jul 2016, 23:26
Hi Expert, I can't understand why in condition 2 we came to conclusion that set can have 1/2 or 1/5 or both 1/2 & 1/5. Kindly explain in detail logic behind it. Thanks.

Regards,
Vivek Tripathi
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15 Jul 2016, 23:31
vivek16 wrote:
Hi Expert, I can't understand why in condition 2 we came to conclusion that set can have 1/2 or 1/5 or both 1/2 & 1/5. Kindly explain in detail logic behind it. Thanks.

Regards,
Vivek Tripathi

This is explained in the discussion above.

Hope it helps.
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24 Sep 2016, 05:53
I think this is a high-quality question and I agree with explanation.
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17 Dec 2016, 03:11
Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=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 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Hi Bunuel !

I have a doubt in the question stem of this problem.

"Set A consist of 10 terms, each of which is a reciprocal of a prime number"

Question stem says that each term of set A is a reciprocal of a prime number. Doesn't it imply that each term is reciprocal a unique prime number ?
such as : (1/p1, 1/p1, 1/p1,1/p1,.....) --------------(1)

(1/p1, 1/p2,1/p3,1/p3,1/p4.....)----------(2)

where p1,p2,p3,p4 are all prime numbers.

for scenario (2) , I believe a more apt verbose would be:
"Set A consist of 10 terms, each of which is a reciprocal of prime number"

Regards
SR
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20 Feb 2017, 10:58
Hi,

I didnt understand why we eliminate A. When we choose 1/2 for all the median is 1/2 and reciprocal of median is 2. 2 is also a prime number. So why did we say No for this?
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20 Feb 2017, 11:30
dyg wrote:
Hi,

I didnt understand why we eliminate A. When we choose 1/2 for all the median is 1/2 and reciprocal of median is 2. 2 is also a prime number. So why did we say No for this?

(1) says "Reciprocal of the median is a prime number".

If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.
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12 Jun 2017, 06:45
I think this is a poor-quality question and I don't agree with the explanation. In the second statement, the median can be 1/2 so we can say it is less than1/5 and the median can also be 1/5 then it is not less than 1/5. so we can't be sure.
Re: D01-23   [#permalink] 12 Jun 2017, 06:45

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