It is currently 21 Oct 2017, 11:05

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 23 Aug 2008
Posts: 63

Kudos [?]: 54 [0], given: 0

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

### Show Tags

03 Nov 2008, 20:33
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 54 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 708 [0], given: 1

### Show Tags

03 Nov 2008, 20:51
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 ....
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 708 [0], given: 1

Manager
Joined: 23 Aug 2008
Posts: 63

Kudos [?]: 54 [0], given: 0

### Show Tags

03 Nov 2008, 21:25
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...

Kudos [?]: 54 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3350

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

03 Nov 2008, 22:58
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?

Kudos [?]: 319 [0], given: 2

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

04 Nov 2008, 00:29
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.

Kudos [?]: 279 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 708 [0], given: 1

### Show Tags

04 Nov 2008, 01:01
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 ....
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 708 [0], given: 1

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

04 Nov 2008, 01:39
amitdgr wrote:
what if p is a positive fraction < 1 ....

I read "is p>1" as "can p be greater than 1".

Kudos [?]: 279 [0], given: 0

SVP
Joined: 05 Jul 2006
Posts: 1750

Kudos [?]: 431 [0], given: 49

### Show Tags

04 Nov 2008, 03:36
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

2p+r+s>r+s

2p>0......insuff

from 2

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, 08:14, edited 1 time in total.

Kudos [?]: 431 [0], given: 49

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 708 [0], given: 1

### Show Tags

04 Nov 2008, 04:09
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

2p+r+s>r+s

2p>0......insuff

from 2

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?
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 708 [0], given: 1

Manager
Joined: 23 Aug 2008
Posts: 63

Kudos [?]: 54 [0], given: 0

### Show Tags

04 Nov 2008, 09:25
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.

Kudos [?]: 54 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3350

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

04 Nov 2008, 09:32
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.

Kudos [?]: 319 [0], given: 2

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

04 Nov 2008, 09:33
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.

Kudos [?]: 279 [0], given: 0

Manager
Joined: 23 Aug 2008
Posts: 63

Kudos [?]: 54 [0], given: 0

### Show Tags

04 Nov 2008, 09:59
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!
Attachments

mcgraw.jpg [ 86.11 KiB | Viewed 1167 times ]

Kudos [?]: 54 [0], given: 0

Re: DS: Inequality   [#permalink] 04 Nov 2008, 09:59
Display posts from previous: Sort by