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

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If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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23 May 2012, 20:59
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If r > s + t , is r positive?

(1) s > t
(2) r/(s+t) > 1
[Reveal] Spoiler: OA
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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24 May 2012, 00:11
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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.

Hope it's clear.
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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18 Mar 2016, 05:22
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Good one. Basics of positive and negative numbers and sign change.
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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23 May 2012, 22:48
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

So,B
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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24 May 2012, 02:55
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!
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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24 May 2012, 03:03
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 --> $$r>0$$.

Hope it's clear.
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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09 Oct 2012, 12:29
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

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]

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09 Oct 2012, 16:26
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.

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

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

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18 Mar 2016, 04:43
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Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1 [#permalink]

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02 Apr 2016, 03:28
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.
Re: If r > s + t , is r positive? (1) s > t (2) r/(s+t) > 1   [#permalink] 02 Apr 2016, 03:28
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