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Re: if r > s + t , is r positive? [#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? [#permalink]
23 May 2012, 23:11

1

This post received KUDOS

Expert's post

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]
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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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]
09 Oct 2012, 15:26

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"

(1) Take t = -2, \,\,s = -1,and r = -1. \,\,-1 > -1 + (-2) = -3. Or, t = 0, \,\,s = 1, \,\,r = 2. \,\,2 > 1 + 0. Not sufficient.

(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.
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1
[#permalink]
09 Oct 2012, 15:26