Register    Login    Search    Rss Feeds

Forums Home » Ask An Instructor » GMAT Math » GMAT Prep Math


 Page 1 of 1 [ 7 posts ] 



  Print view Print view Previous topic Previous topic | Next topic Next topic
 

Advanced Math

Author Message
morrowgsm
 Post subject: Advanced Math
 Post Posted: Fri Aug 21, 2009 2:02 am 
Offline
Course Students


Posts: 2
I read the explanation - however, I don't believe that 'A' could be the answer because what if the consecutive numbers were:
-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.

06/22/09
Question
If p, q, r, and s are consecutive integers, with p < q < r < s, is pr < qs?

(1) pq < rs
(2) ps < qr
Top 
lalitkc
 Post subject: Re: Advanced Math
 Post Posted: Sun Aug 30, 2009 12:34 pm 
Offline
Students


Posts: 11
Take -4,-3,-2-,-1 as the 4 consecutive integers. The answer is A
Top 
morrowgsm
 Post subject: Re: Advanced Math
 Post Posted: Sun Aug 30, 2009 4:22 pm 
Offline
Course Students


Posts: 2
the numbers can be -4,-3,-2,-1 as well but do we know that those are the numbers?

-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.
Top 
shaji
 Post subject: Re: Advanced Math
 Post Posted: Mon Aug 31, 2009 3:45 am 
Offline
Students


Posts: 26
If p, q, r, and s are consecutive integers, with p < q < r < s and pq < rs;then r>0.5 and cannot be 0. In fact pr<qs making statement 1 sufficient.
Statement 2 give no information other than the universal truth 2>0.
The correct answer is indeed A.
Top 
Kweku.Amoako
 Post subject: Re: Advanced Math
 Post Posted: Mon Aug 31, 2009 9:24 am 
Offline


Posts: 47
P<q<r<s
If they are consecutive
Then q = p+1 , r = p+2 and s = p+3
The question is pr <qs let plug in
Is p(p+2) < (p+1)(p+3) lets simplify
Is p^2 + 2p < P^2 +4p +3 --> -2p < 3 --> p>-1.5

So we can rephrase the question to --> is p >-1.5

(1) pq <rs
p(p+1) < (p+2)(p+3)
p^2 +p < p^2 +5p+6
-4p <6
p>-1.5 SUFFICIENT

(2) ps <qr
P(P+3) < (P+1)(P+2)
P^2+3P < P^2+3P+2
0 <2
This tells us nothing. Insufficient

Answer = A
Top 
RonPurewal
 Post subject: Re: Advanced Math
 Post Posted: Fri Sep 25, 2009 11:19 pm 
Offline
ManhattanGMAT Staff


Posts: 7146
Kweku.Amoako wrote:
P<q<r<s
If they are consecutive
Then q = p+1 , r = p+2 and s = p+3
The question is pr <qs let plug in
Is p(p+2) < (p+1)(p+3) lets simplify
Is p^2 + 2p < P^2 +4p +3 --> -2p < 3 --> p>-1.5

So we can rephrase the question to --> is p >-1.5

(1) pq <rs
p(p+1) < (p+2)(p+3)
p^2 +p < p^2 +5p+6
-4p <6
p>-1.5 SUFFICIENT

(2) ps <qr
P(P+3) < (P+1)(P+2)
P^2+3P < P^2+3P+2
0 <2
This tells us nothing. Insufficient

Answer = A


well done.
Top 
RonPurewal
 Post subject: Re: Advanced Math
 Post Posted: Fri Sep 25, 2009 11:20 pm 
Offline
ManhattanGMAT Staff


Posts: 7146
morrowgsm wrote:
I read the explanation - however, I don't believe that 'A' could be the answer because what if the consecutive numbers were:
-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.


your first example is out-of-bounds for statement (1), because it does not satisfy the condition pq < rs. (using those numbers, that would be 2 < 0.)
Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 7 posts ] 





Who is online

Users browsing this forum: No registered users and 1 guest

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: