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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: DS: Inequality [#permalink]
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...

Re: DS: Inequality [#permalink]
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?

Re: DS: Inequality [#permalink]
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.

Re: DS: Inequality [#permalink]
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.

Re: DS: Inequality [#permalink]
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.

Re: DS: Inequality [#permalink]
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!

Attachments

mcgraw.jpg [ 86.11 KiB | Viewed 808 times ]

gmatclubot

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