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 500,000 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: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr [#permalink]

Show Tags

19 Nov 2014, 12:08

2

This post received KUDOS

Bunuel wrote:

Tough and Tricky questions: Inequalities.

If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)?

(1) \(pq \lt rs\)

(2) \(ps \lt qr\)

Kudos for a correct solution.

let p=k-1, q=k, r=k+1, s=k+2

now question states is pr<qs

(k-1)(k+1)<k(k+2) k^2 - 1 < k^2 + 2k or k>-1/2 or the question becomes is k>-1/2

st.1 pq<rs (k-1)k<(k+1)((k+2) k^2 -k < k^2 + 3k + 2 or 4k>-2 or k>-1/2

hence statement 1 alone is sufficient

st.2

ps<qr

actually this statement doesn't provide us any new information. because for any 4 set of consecutive integers the above inequality will always hold true. for e.g. let 4 consecutive integers be -5,-4,-3,-2. as can be seen (-5)*(-2)<(-4)*(-3).

Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr [#permalink]

Show Tags

19 Nov 2014, 21:05

1

This post received KUDOS

For the set of consecutive integers, p,q,r,s, we have find out whether pr < qs is true Statement 1: pq < rs This statement is true if a) All the integers are positive. In that case, pr < qs is true b) When the numbers are p= -1, q= 0, r= 1, s= 2. In this case also, pr < qs is true

There are not other sets of consecutive integers that fulfills the condition of pq<rs. Since, from this condition we always get pr < qs as true, this statement is sufficient.

Statement 2 : ps < qr Here are the cases: 1. p= -1, q= 0, r= 1, s= 2 ( pr < qs is true ) 2. p= -2, q= -1, r= 0, s= 1 ( pr < qs is false) 3. p= -3, q= -2, r= -1, s= 0 ( pr < qs is false) ....... Since, we get both true and false from this condition. This condition is insufficient.

If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)?

(1) \(pq \lt rs\)

(2) \(ps \lt qr\)

Kudos for a correct solution.

Official Solution:

If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)?

We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep \(p\) as the basic variable, then \(q = p + 1\), \(r = p + 2\), and \(s = p + 3\).

Now we can rephrase the question:

Is \(pr \lt qs\)?

Is \(p(p + 2) \lt (p + 1)(p + 3)\)?

Is \(p^2 + 2p \lt p^2 + 4p + 3\)?

Is \(2p \lt 4p + 3\)?

Is \(0 \lt 2p + 3\)?

Is \(-3 \lt 2p\)?

Is \(-\frac{3}{2} \lt p\)?

Since \(p\) is an integer, the question is answered "yes" if \(p = -1\) or greater, and "no" if \(p = -2\) or less.

Since 0 is always less than 2, no matter the value of \(p\), the statement is always true. Thus, we do not gain any information that would help us answer the question.

Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr [#permalink]

Show Tags

03 Nov 2016, 06:51

Bunuel wrote:

Tough and Tricky questions: Inequalities.

If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)?

(1) \(pq \lt rs\)

(2) \(ps \lt qr\)

Kudos for a correct solution.

picked A...didn't think that much.. if the consecutive numbers start with negative values and reach positive values, then we can't know for sure anything... 1. clearly states that all the numbers are positive - sufficient. 2. not much offered.

Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr [#permalink]

Show Tags

05 Sep 2017, 10:17

I have a question about how to approach it at an actual exam.

The above explanations are very mathematical and accurate, but I don't think I will have time to really think it through like that.

My method of getting through this problem was 1. noticing that they could be negative or positive (all or in part) 2. plug in two sets of numbers to explore sufficiency of each option.

Am I putting myself at a risk of missing something crucial by plugging it in?

I have a question about how to approach it at an actual exam.

The above explanations are very mathematical and accurate, but I don't think I will have time to really think it through like that.

My method of getting through this problem was 1. noticing that they could be negative or positive (all or in part) 2. plug in two sets of numbers to explore sufficiency of each option.

Am I putting myself at a risk of missing something crucial by plugging it in?

Hi..

for saving time, one should exactly do the way you have mentioned...

the MAIN point of these Q where consecutive integers are involved is they could be positive or negative and plugging 2-3 sets of number should tell you.. the center TWo would always be greater than product of !st and 4th, so statement II would not tell us whether the numbers are positive or negative..

a set of number in each positive integers and negative integers would tell you that statement I will give you a clearcut answer.

Having said that, it is not always that such method will work. If these were not given as consecutive integers, there would have been many options possible. there may have been some other info then in statements to get to your answer _________________