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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.

Hope the explanation will help many.
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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.

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?

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.
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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.

Hope it helps.
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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.

Hope the explanation will help many.

indeed i am benefited by this solution . Kudos to you !!
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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?".

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20 May 2016, 11:45

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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.

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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18 Jun 2017, 10:35

Hello from the GMAT Club BumpBot!

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

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31 Aug 2017, 23:37

Bunuel wrote:

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.

x cannot be 1/3, as 3 can divide 24, but cannot divide 28... in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24... So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....

please say to me whether the reasoning is okay....

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.

x cannot be 1/3, as 3 can divide 24, but cannot divide 28... in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24... So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....

please say to me whether the reasoning is okay....

thanks in advance, man ...

Yes, x, when reduced to its simplest form must have 2 or 2^2 in the denominator.
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