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:

rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator

then p/q < (p+a)/(q+a)

similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always

Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3

You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values. for x,y = 6,1;...... x/y = 6; for x,y = 10, 5 ......x/y = 2

stmt 2: sufficient. -5x < -14 y or x/y > 14/5

Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3

Last edited by rashminet84 on 12 Jul 2009, 20:18, edited 1 time in total.

rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator

then p/q < (p+a)/(q+a)

similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always

Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3

You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values. for x,y = 6,1;...... x/y = 6; for x,y = 5,10 ......x/y = 2

stmt 2: sufficient. -5x < -14 y or x/y > 14/5

Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3

No inconvenience, at all. We are all helping each other!

Oh yes, ofcourse

but sometimes people try a lot to make sense of a wrong post, thereby wasting time and effort. It might lead to frustration also when one comes to know that it was the mistake of the one who posted.

I think ans is D.What is the OA? Both stmnts are sufficient.

Ankur,for proving the sufficiency of stmnt1 you can plug in the below values y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true

You can also plug in fractions to double check the ans. Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6 This time also 3x>7y

I think ans is D.What is the OA? Both stmnts are sufficient.

Ankur,for proving the sufficiency of stmnt1 you can plug in the below values y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true

You can also plug in fractions to double check the ans. Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6 This time also 3x>7y

Yes Shruti, OA is D. Somehow plugging doesnt work for me sometimes in DS, hence I was trying to solve stmt 1 to check whether it is sufficient. Anyway thanks a lot for explanation. Plugging will work for this problem. Thanks again.

given x >y + 4 . Can any one verify if we can substitute x > y+4 in the to be proved inequation ??

can we say this ? 3(y+4) > 7(y) --> stmt 1

if yes then A makes sense..because stmt 1 will end up with y < 3 and for all values of 0 < y < 3 , 3x > 7y.

or is it wrong to approach this problem that way ? Please help !

The problem with your approach is that it is already assuming both conditions to be true ie, 3x>7y and that x > y+4

and after doing this, you get that these two conditions will be satisfied only when y<3, which is true. As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y. But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.

The problem with your approach is that it is already assuming both conditions to be true ie, 3x>7y and that x > y+4

and after doing this, you get that these two conditions will be satisfied only when y<3, which is true. As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y. But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.

OA is D but I'm getting B after solving both the statements Thats why I logged this question in the forum So far I havnt got the conclusive solution why statement 1 is sufficient. Or is it the OA is wrong?

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

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...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...