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Re M0331
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15 Sep 2014, 23:21
Official Solution: 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 if \(b\) (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 \(2*5^3\). 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 a terminating decimal. (We need reducing in case we have a 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.) BACK TO THE ORIGINAL QUESTION: (1) \(s = 225\). Hence \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\). We don't know whether \(9^2\) can be reduced, so we cannot say whether this fraction will be a terminating decimal. (2) \(r = 81\). Not sufficient on its own. (1)+(2) \(\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}\), since denominator has only 5 as prime, then this fraction is a terminating decimal. Answer: C
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Re M0331
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08 Jan 2015, 09:49
I think this question is good and helpful.



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Bunuel wrote: Official Solution:
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 if \(b\) (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 \(2*5^3\). 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 a terminating decimal. (We need reducing in case we have a 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.) BACK TO THE ORIGINAL QUESTION: (1) \(s = 225\). Hence \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\). We don't know whether \(9^2\) can be reduced, so we cannot say whether this fraction will be a terminating decimal. (2) \(r = 81\). Not sufficient on its own. (1)+(2) \(\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}\), since denominator has only 5 as prime, then this fraction is a terminating decimal.
Answer: C Bunuel can you explain in more detail please ? Does this mean as long as reduced form of the fraction has positive power of either 2 or 5 it will be a terminating fraction ?
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Re: M0331
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17 Jun 2016, 05:17
rishi02 wrote: Bunuel wrote: Official Solution:
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 if \(b\) (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 \(2*5^3\). 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 a terminating decimal. (We need reducing in case we have a 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.) BACK TO THE ORIGINAL QUESTION: (1) \(s = 225\). Hence \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\). We don't know whether \(9^2\) can be reduced, so we cannot say whether this fraction will be a terminating decimal. (2) \(r = 81\). Not sufficient on its own. (1)+(2) \(\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}\), since denominator has only 5 as prime, then this fraction is a terminating decimal.
Answer: C Bunuel can you explain in more detail please ? Does this mean as long as reduced form of the fraction has positive power of either 2 or 5 it will be a terminating fraction ? If a fraction no matter whether it's reduced or not has only positive integer powers of 2's or 5's in the denominator the fraction will be terminating. To practice more check Terminating and Recurring Decimals Problems in our Special Questions Directory. Hope it helps.
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I got this right in below method. Is \(\frac{r}{s^{2}}\) a terminating decimal? ( Ending ) 1, s = 225 r = ? Insufficient 2, r = 81 s=? Insufficient 3, Combine \(\frac{81}{225 x 225}\) On simplifying 1/625 0.016 C Bunuel is this a right strategy. Pls reply. Thanks
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Re: M0331
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09 Feb 2017, 00:36



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Re: M0331
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05 Apr 2017, 03:30
I think this is a very simple question. In order to know the answer we simply need to know the values of r and s. 1) gives us only s, not sufficient. 2) only r, not sufficient. Both 1) and 2) give the required answer. It's basically a nobrainer. Since this is a data sufficiency, there is no need to actually calculate and give the answer whether it is or isn't a terminating decimal, but we know that by having two values we will be able to find that out.



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Re: M0331
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09 May 2017, 09:59
In terminating decimal 2 and 5 can only be there at the end .No other number should be there?Is it true?



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Re: M0331
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10 Oct 2017, 12:07
Bunuel,
In the denominator , when we have \(2^x 5^y\) only then it is considered as terminating decimal. From stat 1. 225 = 15*15 and this doesn't contain 2 power..so eliminated it.
But in the solution we have only 1/\(5^4\)  though it is terminating decimal ( this is not forever kind). This solution is contradicting the theory since it doesn't contain 2 as one of the factors. Please let me know what I am missing here.



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Re: M0331
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10 Oct 2017, 20:00



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Re M0331
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01 Sep 2018, 22:24
I think this is a highquality question and I agree with explanation.










