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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, 19:53
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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) 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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Re: is r positive? [#permalink]
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19 Jul 2009, 23:48
[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
ST>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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Re: is r positive? [#permalink]
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20 Jul 2009, 00:13
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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Re: is r positive? [#permalink]
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20 Jul 2009, 14:20
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. kindly please explain.



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Re: is r positive? [#permalink]
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20 Jul 2009, 15:28
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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. 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<yRE 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..



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Re: is r positive? [#permalink]
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21 Jul 2009, 05:02
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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Re: is r positive? [#permalink]
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21 Jul 2009, 07:15
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 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.
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Re: is r positive? [#permalink]
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21 Jul 2009, 07:43
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
ST>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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Re: is r positive? [#permalink]
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21 Jul 2009, 09:48
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: ST>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............suffB 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.
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Re: is r positive? [#permalink]
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21 Jul 2009, 10:32
Thanks GT, so the content is good but the presentation is not



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Re: is r positive? [#permalink]
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21 Jul 2009, 21:10
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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Re: is r positive? [#permalink]
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22 Jul 2009, 01:45
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




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