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If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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Updated on: 02 Nov 2012, 23:42
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If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits? (1) x is greater than 100 (2) x is divisible by 21 Source: Jamboree
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Originally posted by Marcab on 02 Nov 2012, 23:40.
Last edited by Bunuel on 02 Nov 2012, 23:42, edited 1 time in total.
Renamed the topic and edited the question.






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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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02 Nov 2012, 23:56
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 \(2*5^3\). 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 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 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.htmlifrandsarepositiveintegers141000.htmlBACK TO THE ORIGINAL QUESTION:If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?First of all \(\frac{5}{28}*3.02*\frac{9}{10}*x=\frac{5*302*9*x}{28*100*10}=\frac{5*302*9*x}{7*(4*100*10)}=\frac{5*302*9*x}{7*(2^2*2^2*5^2*2*5)}\). Now, according to the theory above, in order this number to be termination decimal, 7 must be reduced by a factor of x (no other number in the numerator has 7 as a factor and all other numbers in the denominator have only 2's and 5's), so it'll be a terminating decimal if x is a multiple of 7. (1) x is greater than 100. Not sufficient. (2) x is divisible by 21. Sufficient. Answer: B. Hope it's clear.
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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02 Nov 2012, 23:54
If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits? (1) x is greater than 100 (2) x is divisible by 21 Ok. WHat do we have at the denominator in fraction. 100 and 28. To be non recurring or finite decimal, Denominator should be made of multiple 2 or 5 or both. 100 is not a problem. 28 has 7 which is an issue. B says X is divisible by 21. 7 goes in denominator. Hence Sufficient.
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finite number of nonzero digits
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10 Jan 2013, 11:10
If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?
(1) x is greater than 100 (2) x is divisible by 21



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Re: finite number of nonzero digits
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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21 Jun 2013, 02:21
Bunuel wrote: 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 \(2*5^3\). 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 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 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.htmlifrandsarepositiveintegers141000.htmlBACK TO THE ORIGINAL QUESTION:If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?First of all \(\frac{5}{28}*3.02*\frac{9}{10}*x=\frac{5*302*9*x}{28*100*10}=\frac{5*302*9*x}{7*(4*100*10)}=\frac{5*302*9*x}{7*(2^2*2^2*5^2*2*5)}\). Now, according to the theory above, in order this number to be termination decimal, 7 must be reduced by a factor of x (no other number in the numerator has 7 as a factor and all other numbers in the denominator have only 2's and 5's), so it'll be a terminating decimal if x is a multiple of 7. (1) x is greater than 100. Not sufficient. (2) x is divisible by 21. Sufficient. Answer: B. Hope it's clear. "Brilliant Bunuel"



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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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11 Dec 2014, 03:20
What happens if x has many 000s in the end so that there is no decimal left e.g. if x= 21000000  there will no finite number of nonzero decimals. I think (b) is also not sufficient. Please help!!



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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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21 Apr 2015, 14:38
Marcab wrote: If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?
(1) x is greater than 100 (2) x is divisible by 21
Source: Jamboree So when the question says CAN the number be a terminating decimal, it's actually asking is it ALWAYS a terminating decimal? Posted from GMAT ToolKit



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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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21 Apr 2015, 15:29
dqi2016 wrote: Marcab wrote: If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?
(1) x is greater than 100 (2) x is divisible by 21
Source: Jamboree So when the question says CAN the number be a terminating decimal, it's actually asking is it ALWAYS a terminating decimal? Posted from GMAT ToolKitHello dqi. It can't be always or not always it's or can be represent or cannot be represent. If X doesn't contain 7 than this number can't be represent as terminate decimal. (always) If X contain 7 than this number can be represent as terminate decimal. (always)
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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21 Apr 2015, 19:52
But according to statement 1, if x>100, then if x is 280, that number would be a terminating decimal. But if x=101, then it's non terminating. So technically, abusing by statement 1, x "can" be a number to make it terminating, but isn't necessarily?



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If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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22 Apr 2015, 00:37
dqi2016 wrote: But according to statement 1, if x>100, then if x is 280, that number would be a terminating decimal. But if x=101, then it's non terminating. So technically, abusing by statement 1, x "can" be a number to make it terminating, but isn't necessarily? I think now I correctly understand your first question about "always" Yes we should find statement (or both statements) that give us condition when X always make this number terminating. If statement give us variants  this is wrong statement. In this task in second statement X always make this number terminating. And in first statement sometimes yes (x =280)  this number will be terminating and sometimes no (x = 101). This is general rule for any DS task. We should find condition (or combination) which always will give us needed result without any execeptions.
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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02 Jun 2016, 10:15
Marcab wrote: If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?
(1) x is greater than 100 (2) x is divisible by 21
Source: Jamboree A number can be represented as infinite decimal if the denominator is in the form 2^n 5^m Looking at the expression, (5/28)(3.02)(90%)(x), we are fine with all numbers except 28 as it is a multiple of 7. If any statement can tell us that either the expression gets rid of this 7 or it doesn't, we can conclude if the expression is terminating or not. (Basically look for statement that tells us if x is or not a multiple of 7) (1) x is greater than 100. x may or may not be multiple of 7. Not sufficient (2) x is divisible by 21. That means x is a factor of 7; and hence it will leave 4 as the decimal. we will get a terminating decimal. Sufficient B is the answer
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be
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05 Oct 2018, 10:01
Marcab wrote: If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of nonzero decimal digits?
(1) x is greater than 100 (2) x is divisible by 21
Source: Jamboree \(\frac{5}{28} * \frac{302}{100} * \frac{90}{100} * x\) for this to be a terminating decimal the denominator must be in the form of \(\frac{x}{2^a * 5^b}\) Rewriting the question stem numbers to be \(\frac{5}{7*2^2} * \frac{302}{2^2 * 5^2} * \frac{90}{2^2 * 2^5} * x\) from statement 1) if we x > 100 it might include a multiple of 7 and it might not. so insufficient. 2) x is divisible by 21, meaning it could be 21,42,63 etc.. it is a multiple of 21 and 21 = 7 * 3 This will cancel the 7 in the denominator and will lead to the terminating decimal form, sufficient. Answer choice B




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