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

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<y
RE 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..

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.

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.

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.

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

To a little extent.



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