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

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

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19 Jul 2009, 18:53
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If r > (s+t),is r positive?
1. s > t
2. r / (s+t) >1

[Reveal] Spoiler:
The OA is B

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.

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19 Jul 2009, 22:48
[quote="tejal777"]If r>s+t,is r positive?
1. s>t
2. r/s+t >1
FOR R TO BE +VE

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
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19 Jul 2009, 23:13
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.
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20 Jul 2009, 13:20
Quote:
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.

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20 Jul 2009, 14:28
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skpMatcha wrote:
Quote:
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.

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.
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21 Jul 2009, 04:02
hey thanks a bunch ! i got it clearly.

Initially I missed the point in the stem r > s+t and was wondering , how could you conclude .

but now its clear.
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21 Jul 2009, 06:15
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

[Reveal] Spoiler:
The OA is B

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.

_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

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21 Jul 2009, 06:43
yezz wrote:
tejal777 wrote:
If r>s+t,is r positive?
1. s>t
2. r/s+t >1
FOR R TO BE +VE

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??
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21 Jul 2009, 08:48
yezz wrote:
yezz wrote:
tejal777 wrote:
If r>s+t, is r positive?
1. s>t
2. r/s+t >1

FOR R TO BE +VE

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.
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21 Jul 2009, 09:32
Thanks GT, so the content is good but the presentation is not
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21 Jul 2009, 20:10
yezz wrote:
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.
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22 Jul 2009, 00:45
GMAT TIGER wrote:
yezz wrote:
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
Re: is r positive?   [#permalink] 22 Jul 2009, 00:45
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