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

Question

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

The OA is B

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.

yezz · Answer

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

Sunchaser20 · Answer

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.

skpMatcha · Answer

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.

Sunchaser20 · Answer

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): x1 , 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 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..

skpMatcha · Answer

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.

GMAT TIGER · Answer

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.

yezz · Answer

guys am i going right or wrong??

GMAT TIGER · Answer

Your first statement is correct but little messy. Your second statement is not detail enough. Using capital letter is not a good idea. :oops:

Your answer is correct. 8-)

yezz · Answer

Thanks GT, so the content is good but the presentation is not

GMAT TIGER · Answer

To a little extent. :oops:

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.

yezz · Answer

Thanks GT, i appreciate

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

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