Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

As I understand, in order to be a non-terminating decimal we should be able to convert a number into X/99 format. If b>d then there is no way we can get 99 in the denominator and hence it will always be a terminating decimal. Thus, B is an answer.

If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

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.

BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d?

You mean for (2)? In this case the denominator will have only 2's in it, and if the denominator has only 2's or only 5's in it, it still will be a terminating decimal.
_________________

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

28 Jul 2012, 08:10

1

This post received KUDOS

enigma123 wrote:

If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

Of all the theories.

Among 1/2, 1/3 and 1/5, only 1/3 is non terminating. So if we don't have 3 in the denominator then only p/q will be terminating. b>d, ensures we have no "3" left in the denominator, hence the decimal is terminating. (it holds true for 7,11,13....)
_________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks!

The question here is, whether b >= d. Why is that? p and q are given in their prime factorization. If q has more twos and/or fives in its prime factorisation than p, it won't result in a non-terminating decimal, Remainder of 2 can only be 1: 1/2=0.5 and remainders of 5 result in: 1/5=0.2, 2/5=0.4 3/5=0.6 and 4/5=0.8.

However, this is not the case with the 3. If q has more threes than p, you can cancel all of the threes in the numerator, but there will remain some threes in the denominator, resulting in a non-terminating decimal, because 1/3=0.33333 and 2/3=0.666666

Statement (1) gives us no information about b and d. Statement (2) does. There are fewer threes in the denominator. They will cancel with some of the threes in the numerator. Therefore, this statement is sufficient. We know that p/q will be a terminating decimal.

I hope my explanation is good enough.

Last edited by Zinsch123 on 02 Sep 2012, 05:29, edited 2 times in total.

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

05 Nov 2012, 10:44

2

This post received KUDOS

himanshuhpr wrote:

what if b = -2 & d = -3 , then we have a case for terminating decimal ?? because the denominator now would be in 2^m * 5^n form.

Yes, \(p/q\) will be a terminating decimal. For \(b = -2\) and \(d = -3, b > d.\)

Since \(p/q = 2^{a-c}3^{b-d}5^{-e}\), the given ratio is a terminating decimal if and only if \(b-d\geq{0}\) or \(b\geq{d}.\) Which means there is no factor of 3 in the denominator, only factors of 2 and/or 5, if at all. If in addition \(a\geq{c}\) and \(e\leq{0}\), the given ratio is in fact an integer, which is a terminating decimal.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

24 Oct 2014, 07:20

Hello from the GMAT Club BumpBot!

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.
_________________

If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

25 Oct 2014, 22:58

Hi Bunuel,

Quick question on this rule. How about 1/15? it can be written as 1/2^0 * 3 * 5. The denominator has 5, but the fraction is not a terminating decimal. Can you please explain why?

Bunuel wrote:

enigma123 wrote:

If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

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.

BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d?

Quick question on this rule. How about 1/15? it can be written as 1/2^0 * 3 * 5. The denominator has 5, but the fraction is not a terminating decimal. Can you please explain why?

Bunuel wrote:

enigma123 wrote:

If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

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.

BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d?

(1) a > c. Not sufficient. (2) b > d. Sufficient.

Answer: B.

Hope it helps.

1/15 = 1/(3*5). For a reduced fraction to be terminating, the denominator of the fraction should NOT have any prime but 2 or/and 5.

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

05 Dec 2015, 07:15

Hello from the GMAT Club BumpBot!

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.
_________________

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

(1) a > c (2) b > d

We can derive from p/q=2^a3^b/2^c3^d5^e, that b>=d as the denominator has to be of only 2 or 5 out of the prime factors, so 3 is eliminated and (B) hence becomes the answer.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...