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If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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24 Jun 2013, 02:20
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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.
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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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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 nonnegative 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 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 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: doesthedecimalequivalentofpqwherepandqare89566.htmlanydecimalthathasonlyafinitenumberofnonzerodigits101964.htmlifabcdandeareintegersandp2a3bandq2c3d5eispqaterminatingdecimal125789.html700question94641.htmlisrs2isaterminatingdecimal91360.htmlplexplain89566.htmlwhichofthefollowingfractions88937.htmlHope it helps.
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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 nonnegative 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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28 Jul 2013, 11:18
Can someone explain me what is the meaning of terminating decimal.
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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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trafficspinners wrote: Can someone explain me what is the meaning of terminating decimal. A decimal number that has digits that do not go on forever. Examples: 0.25 (it has two decimal digits) 0.123456789 (it has nine decimal digits) In contrast a Recurring Decimal has digits that go on forever Example of a Recurring Decimal: 1/3 = 0.333... (the 3 repeats forever)
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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. Put up for guidance please.



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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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06 Sep 2013, 11:58
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?



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



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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 May 2016, 13:35
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. Given information : 0 < x < 1 (x is a fraction between 0 and 1)Question asked: x as a terminating decimal? x can only be written as terminating decimal when the denominator can be written in the form of 2^n or 5^n (1) 24x is an integer x can be any factor of 24 1, 2, 4, 3, 6, 8, 12, 24 So x may or may not be a terminating decimal. (2) 28x is an integer x can be any factor of 28 1, 2, 4, 7, 14, 28 So x may or may not be a terminating decimal Combining both statements, only 4 is common between these two. Hence, combining the statements we get one digit that could be the denominator 4. C is the answer
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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.



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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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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 nonnegative 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 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 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: http://gmatclub.com/forum/doesthedeci ... 89566.htmlhttp://gmatclub.com/forum/anydecimalt ... 01964.htmlhttp://gmatclub.com/forum/ifabcdan ... 25789.htmlhttp://gmatclub.com/forum/700question94641.htmlhttp://gmatclub.com/forum/isrs2isa ... 91360.htmlhttp://gmatclub.com/forum/plexplain89566.htmlhttp://gmatclub.com/forum/whichofthe ... 88937.htmlHope it helps. hi man since 0<x<1, say x is a proper fraction .... taking 2 statements together, 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 ...



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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, 23:49
gmatcracker2017 wrote: 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 nonnegative 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 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 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: http://gmatclub.com/forum/doesthedeci ... 89566.htmlhttp://gmatclub.com/forum/anydecimalt ... 01964.htmlhttp://gmatclub.com/forum/ifabcdan ... 25789.htmlhttp://gmatclub.com/forum/700question94641.htmlhttp://gmatclub.com/forum/isrs2isa ... 91360.htmlhttp://gmatclub.com/forum/plexplain89566.htmlhttp://gmatclub.com/forum/whichofthe ... 88937.htmlHope it helps. hi man since 0<x<1, say x is a proper fraction .... taking 2 statements together, 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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Re: If 0 < x < 1, is it possible to write x as a terminating dec [#permalink]
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01 Sep 2017, 02:42
Bunuel wrote: gmatcracker2017 wrote: 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 nonnegative 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 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 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: http://gmatclub.com/forum/doesthedeci ... 89566.htmlhttp://gmatclub.com/forum/anydecimalt ... 01964.htmlhttp://gmatclub.com/forum/ifabcdan ... 25789.htmlhttp://gmatclub.com/forum/700question94641.htmlhttp://gmatclub.com/forum/isrs2isa ... 91360.htmlhttp://gmatclub.com/forum/plexplain89566.htmlhttp://gmatclub.com/forum/whichofthe ... 88937.htmlHope it helps. hi man since 0<x<1, say x is a proper fraction .... taking 2 statements together, 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. thanks man .. you are great ..




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