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If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r +

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maulikmajithia
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If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 11:30
If r+s+p > 1 , Is P > 1 ?

1) P > r + s -1
2) 1-(r + s) > 0

According to me the answer is E - Neither is suffcient , but Mc grawhill solves it for B

It says in option 2 , if ( r + s ) 1

take this scenario ... If r + s = .3 and p = .8 then r + s + p > 1 yet p is not > 1

Am I overlooking something here ?

opinions pls ?



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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 11:52
@ above

but the 2nd option does satisfy the values i am assuming for

r + s = .3
and p = .8

in that case it is insufficient right ?

u think i am taking wrong values for r + s and p ?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:08
Original Question wrote:
If r+s+p > 1 , Is P > 1 ?

1) P > r + s -1
2) 1-(r + s) > 0


The numbers you chose, r+s=.3 and p = .8 do work for the question stem. .3 + .8 = 1.1 which is > 1. Don't forget that when you move to the statements, you have to find values that work for both the inequality in the statement as well as the inequality in the question stem.

Here, is r+s = .3 and p = .8, it doesn't work for Statement 1). Substituting the numbers you chose in for the inequality provided in Statement 1).
P > (r+s)-1
.8 > .3 - 1
.8 > -.7 - this is true, but it is not enough to answer the question. You might think "Ok, P can be .8, which is less than 1 so I can answer the question, but that's not how we answer DS questions.

We must continue and find a number where P < 1 and then a number for r+s, use it in this inequality, make sure this inequaltiy is true, then use those same values for the inequality in the stem. If that is true also, and P < 1, then there are too many options for P so we cannot say for certain that P MUST BE greater than 1.

In the question stem, with addition of 3 variables, there is an infinite number of possibilties of values that will make r+s+p > 1. The key is with the restrictions in the statements. If you restrict the value of P so that the value of P > r+s-1, AND r+s+p>1 still, can you find values of P that are both greater than 1 and less than 1 that satisfy both inequalities. The reason #1 statement is insufficient is because you CAN find values for r,s and p that satisfy both statements, and in the values that solve both, P can be either greater than 1 or less than 1.

When you approach some question like this ask yourself "In the sets of numbers that satisfy both statement #1 and the stem (r+s+p>1), is P required to be greater than 1?" If you find any set of numbers for r, s, and p that include p being less than 1, that statement is insufficient.

maulikmajithia wrote:
@ above
but the 2nd option does satisfy the values i am assuming for
r + s = .3
and p = .8
in that case it is insufficient right ?
u think i am taking wrong values for r + s and p ?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:16
jallenmorris wrote:
maulikmajithia wrote:
If r+s+p > 1 , Is P > 1 ?

1) P > r + s -1
1 is insufficient because P could be -2 and r = -3, s= -4, so you'd have -2 > -8, which is true. or you could have P = 2, r = -2, s = -3, so 2 > -6. We don't know for sure if P > 1. Insufficient.

2) 1-(r + s) > 0

1-(r+2) > 0 is the same as 0 > (r + s) - 1. If this is true, AND r+s+P is greater than 1, then to get a sum greater than 1 when adding to any negative number, P must be greater than 1, therefore #2 is sufficient.

Let me clear this up. R could = .49 and S could = .5, then add them together to get 0.99, and substract 1. you get -0.01. Because (r+s)-1 is less than 0 (i.e., negative), in order to add P to it and get something greater than 1, P must be greater than 1.



According to me the answer is E - Neither is suffcient , but Mc grawhill solves it for B

It says in option 2 , if ( r + s ) < 1 then P has to be > 1

take this scenario ... If r + s = .3 and p = .8 then r + s + p > 1 yet p is not > 1

Am I overlooking something here ?

opinions pls ?



In 2), didn't you you solve the original statement for (r + s) -1 + P > 1 ? I mean, you said (for R = .49 and S = .5) that r + s - 1 = -.01. And if you add P to this in an effort to get a value larger than 1, P must be larger than 1. But you're considering r + s -1 + P in the original statement when the original statement only includes r + s + P.

Does that make sense?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:29
@jallen morris

again ,
I take your point of view completely , but i will try and answer to your last couple of lines for option # 2

"When you approach some question like this ask yourself "In the sets of numbers that satisfy both statement #1 and the stem (r+s+p>1), is P required to be greater than 1?" If you find any set of numbers for r, s, and p that include p being less than 1, that statement is insufficient." - this is what u had to say

If we take statement #2 and the stem r+s+p>1 , P is not required to be greater than 1 , however it can also be greater than 1 . I can have numerous sets for r , s and p which includes p being less than 1 and also p being greater than 1.

set 1 - r = .15 s = .15 p = .8
set 2 - r = .2 s= .3 p = .6 etc

Set 1 and set 2 satisfy both the stem and statement # 2 yet p < 1

now set 3 = r = -2 , s = -1 , p = 4
here P needs to be greater than 1 but then as we see P can be either less than 1 or more than 1 so statement #2 is insufficient too ....

Now my problem is , whereas i see where ur explanation comes from , i cant see a problem with the values i am assuming either cos they satisfy both the stem and statement #2 ..... statement #1 we both agree is insufficient
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:32
@smkrn

For statement # 2 - u r asking me to solve for this equation "(r + s) -1 + P > 1"? how did u get it , can u pl explain

if p + r + s > 1 then P + r + s -1 will be > 0 and not > 1
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:37
Original Post wrote:
In 2), didn't you you solve the original statement for (r + s) -1 + P > 1 ? I mean, you said (for R = .49 and S = .5) that r + s - 1 = -.01. And if you add P to this in an effort to get a value larger than 1, P must be larger than 1. But you're considering r + s -1 + P in the original statement when the original statement only includes r + s + P.

Does that make sense?


I think what you're asking makes sense. I did use -0.01 which includes subtracting 1. If we start out with r+s+p>1

this can be changed up to find P in terms of r and s.

r+s+p>1
p> 1 - r -s (factor out a negative 1 from r -s)
p > 1 - (r+s)
See similarities between
0 > 1 - (r+s) ?

I think this means that 2 is insufficient also and the time I spent writing my prior post is moot!
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:38
maulikmajithia wrote:
If r+s+p > 1 , Is P > 1 ?

1) P > r + s -1
2) 1-(r + s) > 0



Ok consider these cases, where we consider ONLY statement 2):

Case A:
r+s= .3
p= .8
From original statement, r+s+p = .3 + .8 = 1.1 > 1 so the original statement is met.
From statement 2), 1-(r+s) = 1- .3 = .7 > 0 so statement 2) is met.
Now is p>1? NO, p= .8

Case B:
r+s= .3
p= 1.1
From original statement, r+s+p = .3 + 1.1 = 1.4 > 1 so the original statement is met.
From statement 2), 1-(r+s) = 1 - (.3) = .8 > 0 so statement 2) is met.
Now is p>1? YES, p=1.1

So unless I'm missing something obvious, I say that statement 2) is INSUFFICIENT. In which case I think that the answer should be E. Am I missing something simple?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:46
yeah and so we prove the mcgrawhill solved answer as wrong ... ha ha .. this one better be a test question in the actual GMAT :)
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:51
I used mcgraw hill and I don't recommend it, for this exact reason. There were too many typos. Not a ton but enough to make studying enfuriating.

Their CATs are okay just to get practice on format and timing, don't overanalyze the answers though. I noticed some questions didn't even have enough answers???? wtf
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:54
Ya, no kidding!

By the way, combining the statements, I can't get anything better than p>0. Anyone else get this?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:54
Dude , how do u rate the difficulty of the power prep tests ?

how close is it to actual gmat ?

I gave 2 CATs from power prep .. scored 710 and 700 ... does this mean i am doing decent ?
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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink] New post   09 Jun 2008, 12:59
Not sure if that was directed at me, but I'll answer anyway. I haven't taken any of the PowerPrep tests, so I don't know. I've only taken MGMAT practice tests so far.



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Re: If r+s+p > 1 , Is P > 1 ? 1) P > r + s -1 2) 1-(r + [#permalink]
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