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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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24 Jun 2013, 02:32

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If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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.

Re: If 0 < x < 1, is it possible to write x as a terminating [#permalink]

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27 Jun 2013, 22:37

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If 0 < x < 1, is it possible to write x as a terminating decimal? (1) 24x is an integer. (2) 28x is an integer.

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

Statement 1- If 24x is an integer than x can take the following values 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/24 Some values of x can be reduced to a terminating decimal (1/2, 1/4, 1/8), while few can not be (1/3,1/6,1/12, 1/24) Insufficient

Statement 2- If 28x is an integer than x can take the following values 1/2, 1/4, 1/7, 1/14, 1/28 Some values of x can be reduced to a terminating decimal (1/2, 1/4), while few can not be (1/7, 1/14, 1/28) Insufficient

Statement 1& 2- If both 24x & 28x are integers than x can take the following values 1/2, 1/4 Both of these values of x can be reduced to a terminating decimal Sufficient

Ans C.

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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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20 Aug 2013, 04:16

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Rock750 wrote:

If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer.

(2) 28x is an integer.

I have a bit of difficulty in understanding the intended meaning of "is it possible" part of the question.

The answer can be yes and ofcourse no, but just that there is a possibility that the answer could be yes confuses me a bit. Had the question been framed like this " is x a terminating decimal?", then it would have been clearer. The use of the term "possible" makes it just a bit ambiguous.

Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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20 Aug 2013, 05:57

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agourav wrote:

Rock750 wrote:

If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer.

(2) 28x is an integer.

I have a bit of difficulty in understanding the intended meaning of "is it possible" part of the question.

The answer can be yes and ofcourse no, but just that there is a possibility that the answer could be yes confuses me a bit. Had the question been framed like this " is x a terminating decimal?", then it would have been clearer. The use of the term "possible" makes it just a bit ambiguous.

Put up for guidance please.

The question basically asks: if x is written as a decimal will it be a terminating decimal?

Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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07 Sep 2013, 05:30

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thebloke wrote:

since the question asks "is it possible", wouldn't the answer be D since .5 is a terminating decimal and 24*.5=12, and 28*.5=24?

You misinterpret the question. The question asks: if x is written as a decimal will it be a terminating decimal? Thus the correct answer is C, not D. _________________

Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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05 Jan 2014, 09:48

we know that x is a proper positive fraction. we need to check whether x has powers of 5 or 2 in the denominator or not.

1. 24(x)=INT ---> \(x=Int/24\) if our integer is 3 then x is can be written as a terminating decimal otherwise x will be a non-terminating decimal

2. 28(x)=INT ----> same story here if our int is 7 then x can be written as a terminating decimal, otherwise x will be a non-terminating decimal

1+2 \(Int/3(2^3)=Int/7(2^2)\) -----> 7(4)Int=8(3)Int the expression has to be equal on both sides thus on the right hand side we need a 7 and on the right hand side we need a 3 and a two. We now know that our integer a terminating decimal because we can get rid of both 7 and 3 in the denominator.

C.

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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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27 Feb 2015, 12:16

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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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03 Mar 2015, 03:34

fameatop wrote:

If 0 < x < 1, is it possible to write x as a terminating decimal? (1) 24x is an integer. (2) 28x is an integer.

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

Statement 1- If 24x is an integer than x can take the following values 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/24 Some values of x can be reduced to a terminating decimal (1/2, 1/4, 1/8), while few can not be (1/3,1/6,1/12, 1/24) Insufficient

Statement 2- If 28x is an integer than x can take the following values 1/2, 1/4, 1/7, 1/14, 1/28 Some values of x can be reduced to a terminating decimal (1/2, 1/4), while few can not be (1/7, 1/14, 1/28) Insufficient

Statement 1& 2- If both 24x & 28x are integers than x can take the following values 1/2, 1/4 Both of these values of x can be reduced to a terminating decimal Sufficient

Ans C.

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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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13 May 2015, 15:46

thebloke wrote:

since the question asks "is it possible", wouldn't the answer be D since .5 is a terminating decimal and 24*.5=12, and 28*.5=24?

i agree with this. The way it is phrased, it should be D. I understand how the answer C is achieved, but I don't think this question is worded well....

The fact that a terminating decimal is possible should be enough. The only way, in my mind that C is correct is if the question asks "is X a terminating decimal?".

Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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20 May 2016, 11:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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21 May 2016, 03:48

Bunuel I am sorry I can not get the combining statements. We do not need to prove that n is a multiple of 7 ? another question regarding the denominator, is it enough to be 2s or 5s to terminate X as decimal or 2s * 5s is a must.

Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]

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21 May 2016, 03:59

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hatemnag wrote:

Bunuel I am sorry I can not get the combining statements. We do not need to prove that n is a multiple of 7 ? another question regarding the denominator, is it enough to be 2s or 5s to terminate X as decimal or 2s * 5s is a must.

From (1) we have that \(x=\frac{m}{24}=\frac{m}{2^3*3}\). If m is a multiple of 3, then 3 in the denominator will be reduced and x will be a terminating decimal.

Similarly, from (2) we have that \(x=\frac{n}{28}=\frac{n}{2^2*7}\). If n is a multiple of 7, then 7 in the denominator will be reduced and x will be a terminating decimal.

The answer to your other question is yes, if a fraction has only 2's or 5's in the denominator it'll terminate.

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