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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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28 Jul 2013, 11:25
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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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20 Aug 2013, 05:57
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? Hope it's clear.
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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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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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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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21 May 2016, 03:59
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. Check Terminating and Recurring Decimals Problems in our Special Questions Directory. Hope it helps.
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