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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. 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.

Answer: B.

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.

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Thanks!

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Thanks!

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Hope it' clear.

Just want to confirm the following:

1. (1/5+1/2)/2 = 7/10 and not 7/20 (as its mentioned twice)

2. If the theory part of 2^n*5^m was not known than it would not had been possible to answer the question, right? I mean is there any other way?
_________________

“Today I will do what others won't, so tomorrow I can accomplish what others can't”

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Thanks!

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Hope it' clear.

Just want to confirm the following:

1. (1/5+1/2)/2 = 7/10 and not 7/20 (as its mentioned twice)

2. If the theory part of 2^n*5^m was not known than it would not had been possible to answer the question, right? I mean is there any other way?

1. (1/5+1/2)/2 = 7/20, not 7/10. Check your math. 2. Yes, you need to know this property to solve the question.
_________________

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Thanks!

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Hope it' clear.

hi i did not understand why statement b is right??

Could you elaborate your work for statement 2? I'm confused how (1/5+1/2)/2=7/20 comes into play. The statement mentions product of two terms (which would be 1/10), how does that play a role?

Thanks!

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Hope it' clear.

hi i did not understand why statement b is right??

Can you please specify which part is unclear? Thank you.
_________________

From (2) it follows that the set must consists of 1/2 or/and 1/5. It cannot have any other reciprocal of a prime since in this case the product of ANY two terms in the set won't be a terminating decimal (refer to Theory part in the post). For example if there is 1/3 and say 1/2 in the set, then 1/3*1/2=1/6, which is not a terminating decimal.

Now, if the set has only 1/2 or/and 1/5 in it, then the median of the set can be 1/5 (if the two middle terms are 1/5), 1/2 (if the two middle terms are 1/2) or (1/5+1/2)/2=7/20 (if the two middle terms are 1/5 and 1/2).

Hope it' clear.[/quote]

hi i did not understand why statement b is right?? [/quote]

Can you please specify which part is unclear? Thank you.[/quote]

WHY THE ANSWER IS B AND NOT A..I WAS NOT ABLE TO UNDERSTAND THE REASONING BEHIND STATEMENT B