D01-23

Official Solution:


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

(1) The reciprocal of the median is a prime number.

If all the terms in the list are 1/2, then the median is 1/2, which is not less than 1/5. On the other hand, if all the terms are 1/7, then the median is 1/7, which is less than 1/5. Not sufficient.

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

Since the terms in the list are reciprocals of prime numbers, the product of any two terms will be a reduced fraction \(\frac{1}{p*q}\), where \(p\) and \(q\) are prime numbers. For a reduced fraction, in its decimal form, to have a finite number of digits after the decimal point, it must have only 2's and/or 5's in its denominator. Therefore, the list can only consist of 1/2's, 1/5's, or a combination of 1/2's and 1/5's. It cannot include the reciprocal of any other prime, such as 1/3, because no pair with 1/3 will produce a product that is a terminating decimal.

If the two middle terms are both 1/2, then the median is 1/2, which is NOT less than 1/5.

If the two middle terms are both 1/5, then the median is 1/5, which is NOT less than 1/5.

Finally, if the two middle terms are 1/5 and 1/2, then the median is (1/5 + 1/2)/2 = 7/20, which is also NOT less than 1/5.

Since none of the possible medians are less than 1/5, the answer to the question is no. Therefore, statement (2) alone is sufficient to answer the question.


Answer: B
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Hi again Bunuel!

This is how I am interpreting this question. Based on this interpretation when I read your explanation, I'm getting totally lost. Please help.

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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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:

A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) 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 the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of 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 a terminating decimal.

(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)

Questions testing this concept:
https://gmatclub.com/forum/does-the-dec ... 89566.html
https://gmatclub.com/forum/any-decimal- ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-a ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the ... 88937.html
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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!

I think this is a high-quality question and I agree with explanation. Great question!!

I think this is a high-quality question and I agree with explanation.

I can certainly see the logic in the explanation with this question, but I answered E. My basis for E is that all the terms could be 1/5 in which case the median is 1/5. My understanding of the wording of the question is the median has to be less than 1/5 to give a hard yes, but if you can have 1/5 as well as 1/2 and 7/20 as answers for #2, there is no clear yes or no. Am I overlooking something on this one?

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

The question asks: is the median of the list less than 1/5?

From (2) the median could be 1/2, 1/5 or 7/20. None of the possible values is less than 1/5. So, we have a definite NO answer to the question for all three possible values of the median. Sufficient.

Hope it's clear.
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I think this is a high-quality question and I agree with explanation.

Why only 1/2 and 1/5 for statement two? It can also be 1/3, 1/7, 1/11 etc., right?

No. If there are reciprocals of other primes than 2 and 5, then the product of ANY two terms of the list won't be a terminating decimal. For example, if the list is {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/5, 1/5, 1/5, 1/3}, then no pair with 1/3 will produce the product which is a terminating decimal: 1/2*1/3 = 1/6 = 0.1666666.... and 1/5*1/6 = 0.0333333............... (reduced fraction to be terminating its denominator must have only 2's and/or 5 in it).
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I think this is a high-quality question. Loved this questions!!

Amazing question!! One that just blows your mind off!! Hats off Bunuel

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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Poorly written question. The stem needs to make it clear that the terms can be repetitive.

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