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:

92. If a, b, c, and d, are positive numbers, is a/b < c/d?

1) 0 < (c-a) / (d-b)

2) (ad/bc)^2 < (ad)/(bc)

Hi catty,

We're looking for whether a/b < c/d. Fortunately, we're told a useful bit of info in the question stem. All four terms are positive. That's very important with inequalities, because it means that we can multiply and divide without having to worry about the direction of the inequality signs. In this case, we could rephrase the question to whether ad < bc by cross-multiplying. This will be useful laters.

Statement 1) is not useful, however. (c-a) and (d-b) could both be positive or negative; that means that when me multiply to get rid of a term, we might or might not have to flip the terms. Since any of the variables could be greater or less than any of the other variables, this statement is insufficient.

Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get ad/bc < 1, and can cross-multiply to get ad < bc. That answers our question with a definite yes, so it's sufficient and the answer is (B) _________________

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

19 Sep 2012, 15:56

First of all, I have rephrased the statement. I have become "a/b < c/d" to "ad < cb".

Then, I have answered (B) due to the fact that I know that the result of a proper fraction to the power of 2 is always less than the result of the proper fraction. Thus, in this case (ad/bc)^2 < (ad)/(bc), the fraction ad/bc must be a proper fraction and therefore it must be true that ad<bc.

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

19 Sep 2012, 21:02

4

This post received KUDOS

3

This post was BOOKMARKED

catty2004 wrote:

If a, b, c, and d, are positive numbers, is a/b < c/d?

(1) 0 < (c-a) / (d-b)

(2) (ad/bc)^2 < (ad)/(bc)

We know that a,b,c and d are positive numbers. This is a Yes/No DS question type - is a/b < c/d. Since we are certain that we have no negative values, we can manipulate the inequality question to - is ad < bc? It's much easier to look at.

(1) (c-a)/(d-b) - a positive fraction or whole number

Say c=d=5 and a=2 and b=1 for 3/4, then ad < bc is false Say c=d=5 and a=1 and b=2 for 4/3, then ad < bc is true thus (1) is INSUFFICIENT

(2) Thus, YES! SUFFICIENT. See attachment.

Answer: B

Attachments

photo.JPG [ 259.84 KiB | Viewed 14230 times ]

_________________

Impossible is nothing to God.

Last edited by mbaiseasy on 20 Sep 2012, 07:21, edited 2 times in total.

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

20 Sep 2012, 01:51

I agree with your approach to the problem. But is not it easier to realize about the rule of proper fraction to the power of 2 instead of manipulate the ecuation in the stem 2?

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

20 Sep 2012, 07:13

racingip wrote:

I agree with your approach to the problem. But is not it easier to realize about the rule of proper fraction to the power of 2 instead of manipulate the ecuation in the stem 2?

You are right. That's what I did. I did cancelling of of the powers of . Sorry my explanation is not clear. haha! I just summarized that when you start cancelling out, it's like multiplying that fraction I put up. _________________

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

26 May 2014, 08:36

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. _________________

92. If a, b, c, and d, are positive numbers, is a/b < c/d?

Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get \(ad/bc\) < 1, and can cross-multiply to get \(ad < bc\). That answers our question with a definite yes, so it's sufficient and the answer is (B)

Hi could you please explain the part on cross multiplication? I am getting \(a/b\) > \(b/c\).

92. If a, b, c, and d, are positive numbers, is a/b < c/d?

Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get \(ad/bc\) < 1, and can cross-multiply to get \(ad < bc\). That answers our question with a definite yes, so it's sufficient and the answer is (B)

Hi could you please explain the part on cross multiplication? I am getting \(a/b\) > \(b/c\).

\((\frac{ad}{bc})^2 < \frac{ad}{bc}\) --> reduce by ad/bc: \(\frac{ad}{bc} <1\) --> multiply by bc: \(ad<bc\).

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

21 May 2015, 09:05

Is a/b<c/d or (because all the values are >0) ad<bc ? Prethinking: c/d>a/b if ca-same and d<b or db - same and c>a...

1) (c-a)/(d-b) --> c>a and d>b Not suff. yes and no.. see explanation above...

2) (ad/bc)^2 < ad/bc this means we have a proper fraction here (1/2^2 < 1/2) this also means that ad<bcSufficient --> see underlined part above _________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

09 Nov 2015, 08:46

Bunuel,

2) This quite clearly sufficient 1) However, it was not very clear to me how this one is insufficient. Is there any way we know this choice is not sufficient w/o resorting to number picking?

Thanks! _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

10 Nov 2015, 11:49

1

This post received KUDOS

Expert's post

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.

If a, b, c, and d, are positive numbers, is a/b < c/d?

(1) 0 < (c-a) / (d-b)

(2) (ad/bc)^2 < (ad)/(bc)

If we modify the question, the sign of the inequality does not change as a,b,c,d are positive integers. Hence, we want to know whether a/b < c/d?, or ad<bc. From condition 1, 0 < (c-a) / (d-b), and if we multiply (d-b)^2 on both sides, 0<(c-a)(d-b). We cannot know whether ad<bc, so this is insufficient. From condition 2, we can divide both sides by (ad)/(bc), which gives us (ad/bc)^2 < (ad)/(bc), or (ad/bc)<1, and when we multiply bc on both sides, (ad/bc)<1, or ad<bc. This answers the question 'yes' so this is sufficient, and the answer becomes (B).

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

If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

10 Dec 2015, 07:07

Please correct me if I'm wrong.

1) not sufficient

2) sufficient

Let ab=4 and bc=2 so (ab/bc)^2=4, which cannot be less than (ab/bc)=2. We can try any values such as ab=4 and bc=3.

Therefore we can conclude that ab has to be less than bc.

So 2) is sufficient.

My only problem is that how would I be able to think of such an approach in the main exam , which requires that every question be solved in less than 2 mins ?

Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]

Show Tags

26 Dec 2015, 16:59

1

This post received KUDOS

NoHalfMeasures wrote:

Bunuel,

2) This quite clearly sufficient 1) However, it was not very clear to me how this one is insufficient. Is there any way we know this choice is not sufficient w/o resorting to number picking?

Thanks!

I know your question is to Bunuel, but I am just sharing how I eliminated the first statement:

Statement 1 tells us that \(\frac{c-a}{d-b}>0\), i.e. \(c-a\) and \(d-b\) have same signs (either both are +ve or both are -ve). However that doesn't tells us anything about their individual values, which makes this statement insufficient.

Hope it helps.

gmatclubot

Re: If a, b, c, and d, are positive numbers, is a/b < c/d?
[#permalink]
26 Dec 2015, 16:59

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...