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# Is pqr > 0 ?

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Intern
Joined: 26 Apr 2012
Posts: 5
Is pqr > 0 ?  [#permalink]

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Updated on: 25 Oct 2013, 07:25
1
00:00

Difficulty:

45% (medium)

Question Stats:

69% (01:27) correct 31% (01:03) wrong based on 240 sessions

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Is pqr > 0 ?

(1) pq > 0
(2) q/r < 0

1 alone insufficient
2 alone insufficient
combined sufficient to state that pqr < 0

where do i make a mistake?

Originally posted by kobi90 on 05 May 2012, 12:35.
Last edited by Bunuel on 25 Oct 2013, 07:25, edited 1 time in total.
Edited the question.
Intern
Joined: 12 Mar 2011
Posts: 3
Re: Is pgr > 0 ?  [#permalink]

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05 May 2012, 12:59
1). pq > 0 --- It doesn't say anything about r,which can be 0, -ve , +ve . So insuff to figure whether pqr > 0 or not.

2) q/r < 0
Multiply both side by r
(q/r)*r < 0*r
q<0
So 2nd statement tell that q<0 but still we can not predict anything about r and hence insuff to figure whether pqr > 0 or not.

Combining both also doesn't tell anything about r.
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Re: Is pgr > 0 ?  [#permalink]

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05 May 2012, 13:45
yeah got it now!
however you've made a mistake here
Quote:
Multiply both side by r
(q/r)*r < 0*r
q<0
as we don't know if r < 0 or > 0 and therefore may need to change the sign while multiplying.
hevertheless, thanks!
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Re: Is pgr > 0 ?  [#permalink]

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05 May 2012, 13:55
one more thought!

1). if pq>0 then p & q are both negative or both positive (still insufficient)
2). q/r<0. Lest multiply both sides by r^2 (no sign change) then we have qr<0 (still insufficient)

but combined we have 2 sets of variables and their signs:
1. p+ q+ r- then pqr<0
2. p- q- r+ then pgr>0

only now having 2 possibilities can state that E is correct!
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Posts: 54493
Re: Is pqr > 0 ? (1) pq > 0 (2) q/r < 0  [#permalink]

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06 May 2012, 01:35
Is pqr > 0 ?

(1) pq > 0 --> p and q have the same sign, but we now nothing about r, so this statement is insufficient.

(2) q/r < 0 --> q and r have the opposite signs, but we now nothing about p, so this statement is insufficient.

(1)+(2)If p and q are both positive and r is negative then pqr=positive*positive*negative=negative BUT if p and q are both negative and r is positive then pqr=negative*negative*positive=positive. Not sufficient.

Hope it's clear.
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Re: Is pqr > 0 ? (1) pq > 0 (2) q/r < 0  [#permalink]

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06 May 2012, 18:55
1+2,

p q r pqr
- - + +
+ + - -

hence E, pqr can be + or - that is pqr can be >0 or <0.
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Is pqr > 0 ?  [#permalink]

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Updated on: 15 Mar 2016, 22:31
Bunuel wrote:
Is pqr > 0 ?

(1) pq > 0 --> p and q have the same sign, but we now nothing about r, so this statement is insufficient.
(2) q/r < 0 --> q and r have the opposite signs, but we now nothing about p, so this statement is insufficient.

(1)+(2)If p and q are both positive and r is negative then pqr=positive*positive*negative=negative BUT if p and q are both negative and r is positive then pqr=negative*negative*positive=positive. Not sufficient.

Hope it's clear.

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Originally posted by honchos on 18 Oct 2013, 00:01.
Last edited by honchos on 15 Mar 2016, 22:31, edited 1 time in total.
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Re: Is pqr > 0 ? (1) pq > 0 (2) q/r < 0  [#permalink]

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25 Oct 2013, 05:20
Pls put ans in the spoiler option dont show up.....
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Re: Is pqr > 0 ? (1) pq > 0 (2) q/r < 0  [#permalink]

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25 Oct 2013, 07:25
1
monirjewel wrote:
Pls put ans in the spoiler option dont show up.....

Done. Thank you for the suggestion.
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Re: Is pqr > 0 ?  [#permalink]

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15 Mar 2016, 21:02
1 alone = p+ and q+ or p- and q-
2 alone = p+ and r- or p- and r+
alone insufficient either of them
1+2:
P+ q+ r- => negative or
p- q- r+ -> positive
so insufficient.

E.
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Re: Is pqr > 0 ?  [#permalink]

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30 Apr 2016, 13:51
pqr > 0?

stmt1. pq>0 this means same sign => (+) (+) or (-) (-) for p and q but nothing about r. NS.

stmt2. q/r<0 this means differen sign => (+) (-) or (+) (-) for p and r but nothing about p. NS.

combined. you can have (+) (+) (-) for p, q, r and (-) (-) (+) for p, q, r => this gives you negative answer or positive answer. So this is not sufficient.

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Re: Is pqr > 0 ?  [#permalink]

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06 Jul 2017, 20:23
kobi90 wrote:
Is pqr > 0 ?

(1) pq > 0
(2) q/r < 0

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

This question is about an inequality. We count 1 inequality in conditions as 1 equation for inequality questions.

There are 3 variables and 0 equation. Thus the answer E is most likely.

When we consider both conditions 1) and 2) together, we have inequalities $$pq > 0$$ and $$qr < 0$$ which is equivalent to $$q/r < 0$$.
We can't figure our $$pqr > 0$$ from those inequalities.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
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Re: Is pqr > 0 ?   [#permalink] 06 Jul 2017, 20:23
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