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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
23 May 2012, 23:11
1
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Expert's post
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sdpkind wrote:
1.S>T
S=1,T=-5;S+T=1-5=-4 and since R>S+T,R should be greater than -4, which can be positive or negative. Not sufficient.
2. R/(S+T) > 1
For R/(S+T) to be greater than 1, R should be greater than S+ T. Two cases: 1. R and S+T should be positive.this case is valid as R > S + T(from stem) 2. R and S+T should be negative. This case is not valid as the fraction would be less than 1 because of R > S + T and it would violate R/(S+T) >1
So,B
The red part is not correct: \(\frac{r}{s+t}>1\) can be true for \(r>s+t\) (consider \(r=2\) and \(s+t=1\)) as well for \(r<s+t\) (consider \(r=-2\) and \(s+t=-1\)).
If r > s + t , is r positive?
(1) s > t. This does not tell us much, consider \(r=2\), \(s=1\), \(t=0\) and \(r=-2\), \(s=-1\), \(t=-2\). Not sufficient.
(2) r/(s+t) > 1 --> first of all notice that this means that \(r\) and \(s+t\) must be either both positive or both negative. If they are both negative, then we can multiply the given inequality by negative \(s+t\), flip the sign because multiplication by negative value and get \(r<s+t\), which contradicts given info that \(r>s+t\). So, the assumption that both \(r\) and \(s+t\) are negative is wrong, which leaves us only one case: both \(r\) and \(s+t\) are positive --> \(r>0\). Sufficient.
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
23 May 2012, 21:48
1.S>T
S=1,T=-5;S+T=1-5=-4 and since R>S+T,R should be greater than -4, which can be positive or negative. Not sufficient.
2. R/(S+T) > 1
For R/(S+T) to be greater than 1, R should be greater than S+ T. Two cases: 1. R and S+T should be positive.this case is valid as R > S + T(from stem) 2. R and S+T should be negative. This case is not valid as the fraction would be less than 1 because of R > S + T and it would violate R/(S+T) >1
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
24 May 2012, 01:55
It's not quite clear. I don't understand your explanation about the second statement. How can I be sure that r is positive? Since I don't know the signs of the variables I cannot perform any action in the equation (so, I cannot tell with certainty whether r is positive or negative). Please, try to explain again. Thank you! _________________
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
24 May 2012, 02:03
Expert's post
Stiv wrote:
It's not quite clear. I don't understand your explanation about the second statement. How can I be sure that r is positive? Since I don't know the signs of the variables I cannot perform any action in the equation (so, I cannot tell with certainty whether r is positive or negative). Please, try to explain again. Thank you!
\(\frac{r}{s+t}>1\) means that either both \(r\) and \(s+t\) are positive or both \(r\) and \(s+t\) are negative.
Suppose they are both negative. In this case if we multiply both parts by negative \(s+t\) we'll get \(r<s+t\) (flip the sign when multiplying by a negative value), which contradicts given info that \(r>s+t\).
So, the assumption that both \(r\) and \(s+t\) are negative is wrong, which leaves us only one case: both \(r\) and \(s+t\) are positive --> \(r>0\).
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
09 Oct 2012, 11:29
kuttingchai wrote:
If r > s + t , is r positive?
(1) s > t (2) r/(s+t) > 1
not sure why the answer in book is "B"
I think most of us will face problem with statement 2. So i am explaining statement 2 only
From stem r>s+t ---> r-(s+t) >0 2) r/(s+t) >1 ----> r/(s+t) -1>0--->[r- (s+t)]/(s+t) >0 ----equation (A) From stem , its given that r-(s+t) >0 Thus the Numerator of equation (A) is positive, which means Denominator has to be positive as well because the ratio of Numerator/denominator is positive i.e. (s+t)>0
Now see the stem which says r > s + t r>0 (see the red color highlighted portion) r is positive Sufficient Answer B
I hope this will help many. _________________
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(2) \(\frac{r}{s+t}>1\) is equivalent to \(\frac{r}{s+t}-1>0\) or \(\frac{r-(s+t)}{s+t}>0\). Since the numerator is positive (from the stem, \(r > s + t\)), the fraction is positive only if the denominator is also positive, which means \(s + t > 0.\) Since \(r>s+t>0,\) it follows that \(r>0.\) Sufficient.
Answer B. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
02 Sep 2014, 06:39
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