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M03-31

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M03-31 [#permalink]

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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 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
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Re M03-31 [#permalink]

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New post 08 Jan 2015, 10:49
I think this question is good and helpful.

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M03-31 [#permalink]

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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 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 ?
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New post 17 Jun 2016, 06:17
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 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.

To practice more check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.
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M03-31 [#permalink]

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New post 08 Feb 2017, 20:37
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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New post 09 Feb 2017, 01:36
kanusha wrote:
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


If you understand why exactly the first statement is not sufficient, then yes, it's a right approach.
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Re: M03-31 [#permalink]

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New post 05 Apr 2017, 04:30
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 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.

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New post 09 May 2017, 10:59
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: M03-31 [#permalink]

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New post 10 Oct 2017, 13:07
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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New post 10 Oct 2017, 21:00
msk0657 wrote:
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.


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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Re: M03-31   [#permalink] 10 Oct 2017, 21:00
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