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

Question

If r > s + t , is r positive?

(1) s > t
(2) r/(s+t) > 1

Bunuel · Accepted Answer

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 , 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 rs+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. Answer: B. Hope it's clear.

sdpkind · Answer

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

Stiv · Answer

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!

Bunuel · Answer

\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 .

Hope it's clear.

fameatop · Answer

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---> /(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.

EvaJager · Answer

(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.

MKS7 · Answer

Good one. Basics of positive and negative numbers and sign change.

Vinayprajapati · Answer

If r > s + t , is r positive?

(1) s > t
s+t>2t
r>s+t => r>2t but no inf about sign of t so we cnt predict sign of r. Insufficient.
(2) r/(s+t) > 1
(r-(s+t))/(s+t) >0 => +ve/(s+t) >0 => (s+t) is positive. r is >(s+t) => r is positive. Sufficient.

bumpbot · Answer

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

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