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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 non-negative 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 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 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.
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 non-negative 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 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 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 ?
_________________
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 non-negative 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 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 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.
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 no-brainer. 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.
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.
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.
You should read carefully. Every word counts: "...is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers". What will happen if n or/and m are 0?
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57% (00:41) correct 
