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Source: McGraw Hill's GMAT Prep 2008 If r + s + p > 1, is

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Source: McGraw Hill's GMAT Prep 2008 If r + s + p > 1, is [#permalink] New post 03 Nov 2008, 19:33
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Source: McGraw Hill's GMAT Prep 2008

If r + s + p > 1, is p > 1?

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

Could you please explain your logic? I'm having trouble thinking this one though. Thanks
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Re: DS: Inequality [#permalink] New post 03 Nov 2008, 19:51
prince13 wrote:
Source: McGraw Hill's GMAT Prep 2008

If r + s + p > 1, is p > 1?

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

Could you please explain your logic? I'm having trouble thinking this one though. Thanks


E ?

I am not very sure though ..... I got confused half way through

stem --> If r + s + p > 1 --> p > [1 - (r+s)] --- (X)
stmt1) p > r + s - 1 -----> p > - [1- (r+s)] ---(Y)

add (X) + (Y) ===> 2p > 0 or p > 0 ....... insufficient to prove if p>1

stmt2) 1 - (r + s) > 0
p > [1 - (r+s)] > 0 ==========> p>0 ... insufficient to prove if p>1

combined also no definite result ....

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Re: DS: Inequality [#permalink] New post 03 Nov 2008, 20:25
OA is not E though ... I'll give a bit more time for others to try this challenge before posting the explanation and OA - although to be honest I don't understand the explanation... :oops:
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Re: DS: Inequality [#permalink] New post 03 Nov 2008, 21:58
prince13 wrote:
Source: McGraw Hill's GMAT Prep 2008
If r + s + p > 1, is p > 1?
(1) p > r + s - 1
(2) 1 - (r + s) > 0
Could you please explain your logic? I'm having trouble thinking this one though. Thanks



r+s-1>-p

1) p>r+s-1

r+s-1>-r-s+1
r+s>-r-s+2
0>-2r-2s+2
2r+2s-2<0
r+s<1 which implies p>1 since p>r+s-1 we dont know if P>1 or not..

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


these 2 statements contradict each other..so this leads me to believe the r+s=0 there P>1 so is C the ans if E isnt?
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Re: DS: Inequality [#permalink] New post 03 Nov 2008, 23:29
D.

From stmt1: r+s < 1+p
or, r+s+p < 1+2p
or, 1 < 1+2p
or, p > 0.
And, since p >0 hence, p>1, sufficient.

From stmt2:
r+s < 1
or, r+s+p < 1+p
or, 1 < 1+p
or, p > 0 and hence p>1, sufficient.
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 00:01
scthakur wrote:
D.

From stmt1: r+s < 1+p
or, r+s+p < 1+2p
or, 1 < 1+2p
or, p > 0.
And, since p >0 hence, p>1, sufficient.

From stmt2:
r+s < 1
or, r+s+p < 1+p
or, 1 < 1+p
or, p > 0 and hence p>1, sufficient.


what if p is a positive fraction < 1 ....

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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 00:39
amitdgr wrote:
what if p is a positive fraction < 1 ....


I read "is p>1" as "can p be greater than 1".
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 02:36
Source: McGraw Hill's GMAT Prep 2008

If r + s + p > 1, is p > 1?

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

from 1

add

2p+r+s>r+s

2p>0......insuff

from 2

add

1-r-s+r+s+p>1

1+p>1
, p>0..........insuff

i see E is the right answer

Last edited by yezz on 04 Nov 2008, 07:14, edited 1 time in total.
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 03:09
yezz wrote:
Source: McGraw Hill's GMAT Prep 2008

If r + s + p > 1, is p > 1?

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

from 1

add

2p+r+s>r+s

2p>0......insuff

from 2

add

1-r-s+r+s+p>1

1-p>1

-p>0 , p<0..........insuff

i see E is the right answer


Yezz....... see the portion in red
1-r-s+r+s+p>1 ---- this will simplify as 1+p>1


prince13, what is the OA ? B?

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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 08:25
OA is B.
I don't understand it. I don't see how they arrive at the part in red; and I strongly do not agree with the final part in blue (what if r+s=0.5; then p=0.6 is a solution, so p<1 and answer would then be NO).

Anyway, the official explanation:

Statement (1) alone is insufficient. The fastest approach to this problem is probably to treat (r+s) as a single variable and plug in values for p.
Note that if (r+s)<1, then it has to be true that p>1; p cannot be 0 or a negative number, because then both r+s+p>1 and p>r+s-1 cannot be true.
If p=1 and (r+s)=1, then both conditions can be true, and then the answer to the question is NO;
If p=2 and (r+s)=1, then both conditions can be true, and then the answer to the question is YES

Statement (2) alone is Sufficient. You can restate the inequality as -(r+s)>-1, then multiply -1 times both sides (which reverses the direction of the inequality sign) and you get r+s<1, and if both r+s<1 and r+s+p>1 are true, then p>1 and the answer is YES.
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 08:32
THIS OA is wrong and OE is wronger!!! suppose if r+s=0.9 and p=0.2 ?? r+s+p>1 but p<1

we are not told if P, r and s are integers..

i say throw this Mccgraw hill book away..

prince13 wrote:
OA is B.
I don't understand it. I don't see how they arrive at the part in red; and I strongly do not agree with the final part in blue (what if r+s=0.5; then p=0.6 is a solution, so p<1 and answer would then be NO).

Anyway, the official explanation:

Statement (1) alone is insufficient. The fastest approach to this problem is probably to treat (r+s) as a single variable and plug in values for p.
Note that if (r+s)<1, then it has to be true that p>1; p cannot be 0 or a negative number, because then both r+s+p>1 and p>r+s-1 cannot be true.
If p=1 and (r+s)=1, then both conditions can be true, and then the answer to the question is NO;
If p=2 and (r+s)=1, then both conditions can be true, and then the answer to the question is YES

Statement (2) alone is Sufficient. You can restate the inequality as -(r+s)>-1, then multiply -1 times both sides (which reverses the direction of the inequality sign) and you get r+s<1, and if both r+s<1 and r+s+p>1 are true, then p>1 and the answer is YES.
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 08:33
prince13 wrote:
OA is B.
I don't understand it. I don't see how they arrive at the part in red; and I strongly do not agree with the final part in blue (what if r+s=0.5; then p=0.6 is a solution, so p<1 and answer would then be NO).

Anyway, the official explanation:

Statement (1) alone is insufficient. The fastest approach to this problem is probably to treat (r+s) as a single variable and plug in values for p.
Note that if (r+s)<1, then it has to be true that p>1; p cannot be 0 or a negative number, because then both r+s+p>1 and p>r+s-1 cannot be true.
If p=1 and (r+s)=1, then both conditions can be true, and then the answer to the question is NO;
If p=2 and (r+s)=1, then both conditions can be true, and then the answer to the question is YES

Statement (2) alone is Sufficient. You can restate the inequality as -(r+s)>-1, then multiply -1 times both sides (which reverses the direction of the inequality sign) and you get r+s<1, and if both r+s<1 and r+s+p>1 are true, then p>1 and the answer is YES.



Prince, looking at the OE, I feel some portion in your question is incomplete. I do agree with OE provided r,s and p are integers. Does the question explicitly say that r,s and p are integers? If not, OE does not make any sense.

For example, for stmt2: if r+s = 0.9 and p = 0.2, r+s+p > 1, but p < 1.
However, if r,s and p are integers, then if r+s < 1 the only next value could be 0 and in such a case, p > 1.
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Re: DS: Inequality [#permalink] New post 04 Nov 2008, 08:59
Good point... I agree that it works if they are integers, but the question doesn't say that. Screenshot shown below to prove I'm not going crazy. This OE is definitely wrong.

Actually I've already found a number of typos in the text part - but first time I've found problems on the CDROM too.
Just now though, I googled the book, and found out the Amazon reviews of it explicitly mention the number of typos in the text! So warning to future bookbuyers: Forget McGraw Hill! :evil: :evil:

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Re: DS: Inequality   [#permalink] 04 Nov 2008, 08:59
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