GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
pqr < 0 < pqs, andpr > ps
[#permalink]
06 Aug 2019, 07:04
06 Aug 2019, 07:04
11
Bookmarks
00:00
Question Stats:
70% (01:40) correct
29% (01:17) wrong based on 130 sessions
Hide Show timer Statistics
\(pqr < 0 < pqs\), and \(pr > ps\)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
q |
0 |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
pqr < 0 < pqs, andpr > ps
[#permalink]
06 Aug 2019, 07:24
06 Aug 2019, 07:24
3
1
Bookmarks
GreenlightTestPrep wrote:
\(pqr < 0 < pqs\), and \(pr > ps\)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
q |
0 |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Take: \(pr > ps\)
Subtract ps from both sides to get: \(pr - ps > 0\)
In other words, pr - ps is POSITIVE
Now take: \(pqr < 0 < pqs\)
Rewrite as: \(pqr < pqs\)
Subtract pqs from both sides to get: \(pqr - pqs < 0\)
Factor out the q to get: \(q(pr - ps) < 0\)
Since we already know that pr - ps is POSITIVE, we can safely divide both sides of the inequality by \((pr - ps)\)
When we do so, we get: \(q < 0\)
Answer: B
RELATED VIDEO
Re: pqr < 0 < pqs, andpr > ps
[#permalink]
05 Aug 2020, 10:01
05 Aug 2020, 10:01
So we get two things by solving this --> r > s but since pqr < 0 < pqs it is more likely |r| > |s|
now solving pqr < 0 < pqs
0< (+)pqs+ pqr [by bringing from -ve to positive side the sign changes thus pqs is + now]
thus taking pq common --> pq(r +s)> 0 hence p or q is negative
thus B. and q > 0
now solving pqr < 0 < pqs
0< (+)pqs+ pqr [by bringing from -ve to positive side the sign changes thus pqs is + now]
thus taking pq common --> pq(r +s)> 0 hence p or q is negative
thus B. and q > 0
