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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.

The correct answer is C.

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

Hi dude... long time.....

Yes correct is C.....

Any fraction can only be a terminating decimal if it has the denominator in the form of \(2^m*5^n\) where \(m\geq 0\) & \(n\geq 0\)

Ques: Is \(\frac{r}{s^2}\) a terminating decimal?

S1: s = 225 = \(3^2 * 5^2\)... Therefore deno = \(3^4 * 5^4\).. Not SUff.... as If the numerator is divisible by \(3^4\).. then it would be terminating... else it would not be!

S2: r = 81... gives no idea about denominator... so... Not SUFF...

Together....

SUFF.... as now the denominator \(3^4\) cancels out the numerator 81... and hence the fraction can be terminating.

Hence C... _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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.

The correct answer is C.

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

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

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.

I chose A but the OA is C...This question is from gmatclub test # 02 in which I got 35/37 correct. I think any fraction with denominator ending with 2, 4, 5, 10 always is a terminating decimal, so I chose A. Pls help me to understand.

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

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 \(2*5^2\). 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.

Answer: C.

Hi Bunuel, From what I can understand from the above, The denominator should have only 2 and or 5 for the fraction to be terminating. Please correct me if I am wrong.

If the fraction is already reduced to its lowest term then yes.

For example, \(\frac{6}{15}\) has extra 3 in the denominator but this fraction will still be terminating decimal, since that 3 can be reduced: \(\frac{6}{15}=\frac{2}{5}\).

to have a terminating decimal, the denominator should be of hte form 2^n * 5^m after the given expression is broken down.

st 1) s = 225 = 3^2 * 5^2 we dont know about r and where it has 3^4 as a factor ... only in this case, it would be terminating or else it wont not sufficient st 2) r=81=3^4 dont know about denominator .. so cannot say not sufficient

combinng the expression will become 1/5^4 and this is terminating

the denominator is factored to ( 3^2 * 5^2 ) ^2 which is 3^4 *5^4. numerator is 81 , 81 is 3^4 , so the fraction reduced to 1 / 5^4 , powers of 5 is a terminating decimel.

Isn't any number ending in 5 a terminating decimal? I can't think of number that is divided by 5 that doesn't terminate. Can someone explain why the numerator is necessary?

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.

The correct answer is C.

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

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

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 \(2*5^2\). 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.

Answer: C.

Hi Bunuel, From what I can understand from the above, The denominator should have only 2 and or 5 for the fraction to be terminating. Please correct me if I am wrong. _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

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.

The correct answer is C.

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

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

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.

Answer: C.

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.

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.

The correct answer is C.

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

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

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.

Answer: C.

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 are not reading carefully...

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

I spent a good chunk of yesterday binge-watching installments of “ Handicapping Your MBA Odds ” that John Byrne from Poets and Quants and Sandy Kreisberg with HBS Guru...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...