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I selected A as the answer (saying statement 1 is sufficient). The actual answer is C (saying both statments are required). Now my thought process was, any number divided by a square of 225 will always be terminating at some point. Its not like dividing 10 by 3 where you can go on indefientely with 3.333333..... My guess is any number when divided by 225 always gets resolved (i think the simplicity with 5 makes it happen). I tried in excel and couldn't think of any number which is not terminating on dividing with 225 square. Am i missing something? Any help will be much appreciated.

i would have said A as well. I thought that if the prime factorization of the denominator included only 2s or only 5s, then the decimal would terminate ...

The question is: Is r/s(square) a terminating decimal?

1. s=225 2. r=81

I selected A as the answer (saying statement 1 is sufficient). The actual answer is C (saying both statments are required). Now my thought process was, any number divided by a square of 225 will always be terminating at some point. Its not like dividing 10 by 3 where you can go on indefientely with 3.333333..... My guess is any number when divided by 225 always gets resolved (i think the simplicity with 5 makes it happen). I tried in excel and couldn't think of any number which is not terminating on dividing with 225 square. Am i missing something? Any help will be much appreciated.

How could it be?

225 is square of 15, which has factor of 3.

Keep aside 25, you still have 9...1/9=0.111111111111 indefinite.mutliply it by .04.

Square it , it wont change the fact that it is indefinite...

Hey guys i had thot that i had finally gotten the terminating decimal thing down.Apparently not.

I distinctly remember that to find out whether a fraction is terminating or nor all we need is the denominator.The numerator has no role in it. So from the first stmt we know that since 225 does not contain ONLY 2's and 5's it will not result in a terminationg decimal...Suff.

Hey guys i had thot that i had finally gotten the terminating decimal thing down.Apparently not.

I distinctly remember that to find out whether a fraction is terminating or nor all we need is the denominator.The numerator has no role in it. So from the first stmt we know that since 225 does not contain ONLY 2's and 5's it will not result in a terminationg decimal...Suff.

Stmt 2 is insuff. Please help

To know whether r/s^2 is a terminating decimal, we must know the value of r and s. For sure if we know that s is either 2 or 5 or their combination and or multiples, then only we do not need r. Otherwise, we also need to know r too.

From 1; only the value of s (225) is given. S has 3 as one of its factor. When ever the denominator has 3, then the fraction is not terminating one unless the numerator has equal number of 3's as factor/s. NSF...

From 1 and 2: r/s^2 = 81/(225^2) = 1/25^2. Now only it is clear that r/s^2 is a terminating decimal as the 9 in denominator is cancelled out by r in numerator. Sufff...

Basically, is x/15 a terminating decimal? Put in 5 for x, and all of a sudden your fraction = 1/3, which doesn't terminate. The numerator can factor out the 5's or 2's in your denominator. Watch out.

I selected A as the answer (saying statement 1 is sufficient). The actual answer is C (saying both statments are required). Now my thought process was, any number divided by a square of 225 will always be terminating at some point. Its not like dividing 10 by 3 where you can go on indefientely with 3.333333..... My guess is any number when divided by 225 always gets resolved (i think the simplicity with 5 makes it happen). I tried in excel and couldn't think of any number which is not terminating on dividing with 225 square. Am i missing something? Any help will be much appreciated.

This question was posted in PS forum too. Below is my post from there:

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 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.)

BACK TO THE ORIGINAL QUESTION: Is \(\frac{r}{s^2}\) a terminating decimal?

(1) \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\) --> we don't know whether 9^2 can be reduced, so we can not say whether this fraction will be terminating, .

(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 selected A as the answer (saying statement 1 is sufficient). The actual answer is C (saying both statments are required). Now my thought process was, any number divided by a square of 225 will always be terminating at some point. Its not like dividing 10 by 3 where you can go on indefientely with 3.333333..... My guess is any number when divided by 225 always gets resolved (i think the simplicity with 5 makes it happen). I tried in excel and couldn't think of any number which is not terminating on dividing with 225 square. Am i missing something? Any help will be much appreciated.

This question was posted in PS forum too. Below is my post from there:

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 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.)

BACK TO THE ORIGINAL QUESTION: Is \(\frac{r}{s^2}\) a terminating decimal?

(1) \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\) --> we don't know whether 9^2 can be reduced, so we can not say whether this fraction will be terminating, .

(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.

Respect ! Bunuel you are a legend - not for explaining/answering the question but for providing a background/knoweldge that can be used to similar and/or other questions

Nothing has been said for r and s in the main line.So it can be possible that s itself is a nonterminating decimal.

Eventhough we know that (1/225)^2 is 16*10^-6 and any integer multiplied with it can be a terminating integer.But the whole computation dependes on r too and r can be nonterminating.

So i am going for C where both 1 and 2 are required.

Hey even if you consider that just to know that the denominator only has 2's and 5's as primes suffices to say that the fraction terminates, our denominator here, 225, doesn't satisfy this misconception. In addition, even when the denominator has other factors like 3, as in this case, the fraction still can terminate if the nominator cancel out those irrelevant factors ! That is the case with 3/12. 12 has a prime 3 but since the the nominator cancels out with that 3, leaving the denominator with only two 2's, the fraction terminates. HIH
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I selected A as the answer (saying statement 1 is sufficient). The actual answer is C (saying both statments are required). Now my thought process was, any number divided by a square of 225 will always be terminating at some point. Its not like dividing 10 by 3 where you can go on indefientely with 3.333333..... My guess is any number when divided by 225 always gets resolved (i think the simplicity with 5 makes it happen). I tried in excel and couldn't think of any number which is not terminating on dividing with 225 square. Am i missing something? Any help will be much appreciated.

This question was posted in PS forum too. Below is my post from there:

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 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.)

BACK TO THE ORIGINAL QUESTION: Is \(\frac{r}{s^2}\) a terminating decimal?

(1) \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\) --> we don't know whether 9^2 can be reduced, so we can not say whether this fraction will be terminating, .

(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.

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