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Is r/s^2 a terminating decimal?

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Math Expert
Joined: 02 Sep 2009
Posts: 43896
Re: Is r/s^2 a terminating decimal? [#permalink]

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16 Nov 2014, 11:25
davidfrank wrote:
Bunuel wrote:
SudiptoGmat wrote:
Is r/s^2 a terminating decimal?

1. s=225
2. r=81f t 5r y t5

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient. Knowing only without is not enough information to answer the question.

Statement (2) by itself is insufficient. Knowing only without is not enough information to answer the question.

Statements (1) and (2) combined are sufficient. We know both and , so we can calculate the given expression.

But I think answer is A. ST 1 is sufficient. Any comment ??

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 non-negative 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 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 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.

Now:

For (1) $$\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}$$, we can not say whether this fraction will be terminating, as 9^2 can be reduced or not.

(2) is clearly insufficient.

(1)+(2) $$\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}$$, as denominator has only 5 as prime, hence this fraction is terminating decimal.

Hey Bunuel,

As per the theory if the denominator can be expressed in the form of 2^m*5^n, then the numerator is a terminating decimal. However, in the question above after reducing the fraction we are left only with 5^n and hence not sure why in theory we are mentioning 2^n when any integer when divided by 5 will always be terminating. I might be asking a very stupid question but really now want to understand what is the missing link here.

The denominator should be in the form of 2^m*5^n, where $$m$$ and $$n$$ are non-negative integers. Thus n and m could be 0, meaning that the denominator could have only 2's, only 5's or both.

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

Hope it helps.
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Re: Is r/s^2 a terminating decimal? [#permalink]

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27 Aug 2016, 08:49
Hi, I had a doubt in this question. Using the same logic of denominator needs to be in a ratio of 2^n * 5^m, with option 1 we clearly know its not the case regardless of the numerator.

s^2 = 225 ^ 2 = 3^ 4 * 5^4

So how does it matter what the numerator is? It will never be a terminating decimal and the answer should be A. Brunel can you please explain this
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Posts: 43896
Re: Is r/s^2 a terminating decimal? [#permalink]

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27 Aug 2016, 08:54
varunjoshi31 wrote:
Hi, I had a doubt in this question. Using the same logic of denominator needs to be in a ratio of 2^n * 5^m, with option 1 we clearly know its not the case regardless of the numerator.

s^2 = 225 ^ 2 = 3^ 4 * 5^4

So how does it matter what the numerator is? It will never be a terminating decimal and the answer should be A. Brunel can you please explain this

Let me ask you: what id the numerator is 3^4?
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Re: Is r/s^2 a terminating decimal? [#permalink]

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27 Aug 2016, 18:19
Thanks Brunel for the clarification, missed out that 2^m could be 1 where m is 0 i.e. non negative. Always thought that the decimal to be terminating should be a multiple of 10, but now realize it could be a multiple of 5 or even 2 (2^m * 5^n) where m and n are non negative.
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Re: Is r/s^2 a terminating decimal? [#permalink]

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26 Sep 2016, 06:29
C

(1) s=225 --> NOT SUFFICIENT - we don't know the value of r to make that determination

(2) r=81 --> NOT SUFFICIENT - we don't know value of s

A,B,D eliminated

(1) + (2) - SUFFICIENT - we can determine if 81/(225^2) is a terminating decimal or not
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Re: Is r/s^2 a terminating decimal? [#permalink]

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27 Apr 2017, 21:32
davidfrank wrote:
Bunuel wrote:
SudiptoGmat wrote:
Is r/s^2 a terminating decimal?

1. s=225
2. r=81f t 5r y t5

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient. Knowing only without is not enough information to answer the question.

Statement (2) by itself is insufficient. Knowing only without is not enough information to answer the question.

Statements (1) and (2) combined are sufficient. We know both and , so we can calculate the given expression.

But I think answer is A. ST 1 is sufficient. Any comment ??

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 non-negative 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 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 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.

Now:

For (1) $$\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}$$, we can not say whether this fraction will be terminating, as 9^2 can be reduced or not.

(2) is clearly insufficient.

(1)+(2) $$\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}$$, as denominator has only 5 as prime, hence this fraction is terminating decimal.

Hey Bunuel,

As per the theory if the denominator can be expressed in the form of 2^m*5^n, then the numerator is a terminating decimal. However, in the question above after reducing the fraction we are left only with 5^n and hence not sure why in theory we are mentioning 2^n when any integer when divided by 5 will always be terminating. I might be asking a very stupid question but really now want to understand what is the missing link here.

You can always do like in this case: 5^4*2^0
Re: Is r/s^2 a terminating decimal?   [#permalink] 27 Apr 2017, 21:32

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