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:

Guys,Please help me find out where I went wrong.My solution: Stmt. 1, s>t.Taking values, s=4,t=3 implies r is greater than 7.r is +ive. s=4,t=-3 implies r is greater than 1.r is +ive s=-3,t=-4 implies r is greater than -7.Can't say about r. Hence,Stmt 1 is insufficient.

Stmt 2. r/(s+t)>1 r>(s+t)....No new info.This is already mentioned in the ques.Hence insuff.so my answer came out (e)

GT: Please post the question properly.

"r/s+t>1" could be "(r/s)+t>1" or "r/(s+t)>1. _________________

Statement 2: \(\frac{r}{s+t}>1\) gives us 2 facts: 1) r and (s+t) are of the same sign. 2) |r|>|s+t| There are 2 different options: r and s+t are both positive or both negative. Combine it with given r>s+t, and we will have only option when both are positive.

Statement 2: gives us 2 facts: 1) r and (s+t) are of the same sign. 2) |r|>|s+t| There are 2 different options: r and s+t are both positive or both negative. Combine it with given r>s+t, and we will have only option when both are positive.

thats a bundle of info right there:) I am lost ..

I agree with your statement 1, clear. I got lost how you concluded statement 2 and how you narrowed down to positive case.

Statement 2: gives us 2 facts: 1) r and (s+t) are of the same sign. 2) |r|>|s+t| There are 2 different options: r and s+t are both positive or both negative. Combine it with given r>s+t, and we will have only option when both are positive.

thats a bundle of info right there:) I am lost ..

I agree with your statement 1, clear. I got lost how you concluded statement 2 and how you narrowed down to positive case.

kindly please explain.

Yes, sure. Just a little bit of theory: when dealing with inequalities such as \(\frac{x}{y}>1\) we can't just multiply both sides with y, as y can be either positive or negative. When y is positive, we will have x>y, but if y is negative, we will have to flip the sign of inequality (just a general rule when multiplying both sides of inequality with a negative number): x<y RE my 2nd statement: if \(\frac{x}{y}>1\), then \(\frac{|x|}{|y|}>1\), as this is the necessary condition to have 1 in the right part. |y|>=0, so we can safely multiply both sides of the equation by |y|, knowing that we don't have to flip the inequality sign: |x|>|y|. Or you can just consider 2 different possibilities for x and y: a) x and y are positive: \(\frac{x}{y}>1\) -> \(x>y\) b) x and y are negative: \(\frac{x}{y}>1\) -> \(x<y\) Just plug in some numbers, as I always do to understand some concept: x could be -3 and y=-2, not vice versa (to satisfy \(\frac{x}{y}>1\)) or x could be 3 and y=2, not vice versa. Narrowing down to positive case now should be clear: there are only 2 options: a) and b). Option a) satisfies the given \(x>y\), while option b) doesn't. Please feel free to ask any questions..

1. s > t tells us nothing about r. Insuff. 2. r / (s+t) > 1 tells two things: r and (s+t) are both either (i) +ve or (ii) -ve. However in each case, lrl > ls+tl.

(i) If r and (s+t) are both +ve, r is already +ve. (ii) If r and (s+t) are both -ve, r has to be smaller than (s+t) and this invalidates the statement that r > (s+t) given in the question.

Therefore, only r is +ve in (i) is correct. So that makes B as OA.

tejal777 wrote:

If r > (s+t),is r positive? 1. s > t 2. r / (s+t) >1

Guys,Please help me find out where I went wrong.My solution: Stmt. 1, s>t.Taking values, s=4,t=3 implies r is greater than 7.r is +ive. s=4,t=-3 implies r is greater than 1.r is +ive s=-3,t=-4 implies r is greater than -7.Can't say about r. Hence,Stmt 1 is insufficient.

Stmt 2. r/(s+t)>1 r>(s+t)....No new info.This is already mentioned in the ques.Hence insuff.so my answer came out (e)

S+T HAS TO BE >0, BOTH S,T ARE +VE OR OF DIFFERENT SIGNS AND +VE ONE HAS A GREATER // VALUE THAN THE -VE.

FROM 1: S-T>0, VALID WHEN , S,T -VE AND /T/>/S/ OR S+VE AND T -VE AND /S/>/T/ , BOTH ARE +VE AND S>T...........INSUFF

FROM2: R/S+T>1, ie: s+t as a doniminator thus >0............suff B

guys am i going right or wrong??

Your first statement is correct but little messy. Your second statement is not detail enough. Using capital letter is not a good idea. Your answer is correct.
_________________

Thanks GT, so the content is good but the presentation is not

To a little extent.

yezz wrote:

tejal777 wrote:

If r > (s+t), is r positive? 1. s > t 2. r / (s+t) > 1

Given that r>(s+t),

For r to be +ve, (s+t) has to be +ve or 0. This is possible if one of s and t is +ve and the other is -ve but the absolute value of -ve is equal or smaller than the +ve value. Then only, it is proved that r is +ve.

From 1: If s>t, (s - t) > 0. The +ve value of r is possible only when s and t are both +ve or lsl > ltl but thats not confirmed from the given info. However s>t is also valid when s and t, both, are -ve. If both are -ve, r could be -ve or +ve. Therefore, statement 1 is not suff.

From 2: Given that r/(s+t) > 1, r can only be +ve since r>(s+t). r/(s+t) > 1 possible only when lrl > ls+tl. If so, r cannot be -ve (or 0) and > (s+t). (I am little short cut here) Therefore, if r cannot be -ve (or 0), it must be +ve. Suff.

Therefore it is B. Hope that helps.
_________________

Thanks GT, so the content is good but the presentation is not

To a little extent.

yezz wrote:

tejal777 wrote:

If r > (s+t), is r positive? 1. s > t 2. r / (s+t) > 1

Given that r>(s+t),

For r to be +ve, (s+t) has to be +ve or 0. This is possible if one of s and t is +ve and the other is -ve but the absolute value of -ve is equal or smaller than the +ve value. Then only, it is proved that r is +ve.

From 1: If s>t, (s - t) > 0. The +ve value of r is possible only when s and t are both +ve or lsl > ltl but thats not confirmed from the given info. However s>t is also valid when s and t, both, are -ve. If both are -ve, r could be -ve or +ve. Therefore, statement 1 is not suff.

From 2: Given that r/(s+t) > 1, r can only be +ve since r>(s+t). r/(s+t) > 1 possible only when lrl > ls+tl. If so, r cannot be -ve (or 0) and > (s+t). (I am little short cut here) Therefore, if r cannot be -ve (or 0), it must be +ve. Suff.

Therefore it is B. Hope that helps.

Thanks GT, i appreciate

gmatclubot

Re: is r positive?
[#permalink]
22 Jul 2009, 00:45