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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
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Question Stats:
70% (01:37) correct
29% (01:16) wrong based on 123 sessions
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\(pqr < 0 < pqs\), 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
