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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, 01:45