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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 [#permalink]
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16 Jul 2007, 19:22
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OPEN DISCUSSION OF THIS QUESTION IS HERE: ifpqandprispqrp1pq02pr126285.htmliamba wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0
a) (p)(q)(r) < p b) (p)(q)(r)/p < p/p c) (q)(r)<1 rephrased: if p<q and p<r is (q)(r)<1 1) we know that p or q is negative. this tells us nothing about whether q*r < 1 2) same thing here. 1 and 2 together still tells us nothing. 1 and 2 tells us that p is ve but doesn't tell us whether q or r is 0. I get E
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Last edited by Bunuel on 28 Jan 2012, 03:56, edited 4 times in total.
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1. pq < 0
with p<q, p is neg and q is positive, no info about r (r can be either positive or negative), therefore insuff.
2. pr<0
with p<r, p is neg and r is positive, no info about q (q can be either positive or negative), therefore insuff.
1+2. pq<0 and pr<0, making p negative and both r & q positive, however if p = 1 and q and r both = 1, then (p)(q)(r) = p, therefore insuff.
(E)



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Re: DS  P, Q, R [#permalink]
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17 Jul 2007, 03:14
iamba wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0
Stmt1: pq < 0
p < 0 and q > 0
as q < r, so r > 0
But we can't say pqr < p , as p, q and r can be the fractions.
So insuff
Stmt2:
pr < 0
p < 0 and r > 0
we don't know whether q < 0 or > 0
But we can't say pqr < p ,
INSUFF
Taking them together:
Still INSUFF
Hence 'E'



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Re: MGMAT DS Problem [#permalink]
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03 May 2008, 16:07
Answer is C
Explanation : p<q and p<r and we need to find if pqr<p
Statement 1 : pq<0 means both have opposite signs and therefore p must be ve and q must be +ve as p<q as given pqr depends on sign of r. If r is +ve pqr<p If r is ve pqr>p NOT SUFFICIENT
Statement 2 : pr<0 . As above both must have opposite signs and p must be negative and r must be +ve as p<r given again we cannot determine pqr<p as if r is +ve its true if r is ve its not. NOT SUFFICIENT
Both together
pq<0 and pr<0 , therefore p is _ve q and r +ve which means pqr is negative but product will be less than p



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Re: MGMAT DS Problem [#permalink]
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03 May 2008, 22:34
seongbae wrote: if p is 2 and q and r are both 1, then pqr=2 which is equal to p. so shouldn't the answer be e? The question asks if pqr < p or not, so all you need to know is yes/no. If pqr = 2 and p = 2, the answer will just be no, and solves the question.



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Re: MGMAT DS Problem [#permalink]
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03 May 2008, 23:49
jbpayne wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0
Does anyone have a quicker method to determine positives and negatives? I can get this one right, but it just takes me too long. E. if q=r=1,any ve value for p makes p = pqr.
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Re: MGMAT DS Problem [#permalink]
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04 May 2008, 07:37
GMAT TIGER wrote: jbpayne wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0
Does anyone have a quicker method to determine positives and negatives? I can get this one right, but it just takes me too long. E. if q=r=1,any ve value for p makes p = pqr. The poster a few posts above you proved that C is sufficient to prove that p is NOT greater than pqr. This always confuses me too, but a DStype question doesn't ask if the statement p > pqr is true; it asks if we can figure out, in all cases, whether it is true or it is false.



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Re: MGMAT DS Problem [#permalink]
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04 May 2008, 09:42
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getting E
Each statement individually doesn't give information about the third variable. So they are insufficient individually.
Combined... two scenarios to consider
if p < 0, then q,r > 0 => pqr < p (ve on both sides) if p > 0, then q,r < 0 => pqr > p (+ve on both sides)
So insufficient.



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Re: p,q and r [#permalink]
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15 Jul 2008, 11:01
nirimblf wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0 1 & 2) tells me that q > 0 and r > 0 but we don't know whether both q and r greater than 1 or whether both are fractions. So, E.



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Re: p,q and r [#permalink]
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15 Jul 2008, 11:18
answer E for me as well.
I wrote an eq'n: p(qr1) < 0 ... so either p<0 and qr>1 or p>0 and qr<1
from stat 1, i know that p<0 and q>0, since p<q, but i dont know anything about r to see if it satisfies whether qr>1. insuff
from stat 2, i know that p<0 and r>, but nothing about q to see if qr>1 ... insuff.
together, all i know is that q and r are positive, but they could both be fractions, i.e. qr <1, or they could be whole numbers so that qr>1



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Re: p,q and r [#permalink]
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15 Jul 2008, 15:52
yeah I get E..too
p<q, q<r given..pqr<p?
1) pq<0
clearly p<0 q>0..could be 1/2 or 2 etc; however dont know about r..insuff
2) pr<0 clearly p<0 r>0 could 1/2 or 2..dont know anything about q..insuff
together we know that q*r is POSITIVE.. p(q*r)<p?
well if q, r=2 then and say p=2 the 2*2<2 YES but if q, r=1/2 and p=2 the 2(1/4)>2 NO..
INSUFF E it is



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Re: Zumit DS 026 [#permalink]
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16 Sep 2008, 06:54
dancinggeometry wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0 1) p is ve and q is +ve r can be ve or +ve insuffcient  multiple solutions possible 2)p is ve and r is +ve q can be ve or +ve insuffcient  multiple solutions possible combined p ve and q and r are postive. insuffcient. qr <1 or >1 two solutions possible. pqr<p or pqr>p E
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Re: Zumit DS 026 [#permalink]
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16 Sep 2008, 06:55
dancinggeometry wrote: If p < q and p < r, is (p)(q)(r) < p?
(1) pq < 0 (2) pr < 0 from 1 ) there are 2 cases p and q have opposite signs. p has to be ve in order to satisfy inequality p< q ve * +ve * r < ve if r is positive this will be true but we dont know anything abt r. so insuff. from 2) p and r have opposite signs. p has to be ve in order to satisfy inequality p< r ve * q * +ve < ve if q is positive this will be true but we dont know anything abt q. so insuff. together we know that q and r are positive. ve * +ve * +ve < ve sound suff, but what if p and q are fractions  0.25 * 0.1 * 0.2 is not less than 0.25 so insuff. E



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Re: Zumit DS 026 [#permalink]
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16 Sep 2008, 12:54
I think It is C.
I am agree with that Options A and B are not possible.
Let's combined both options
we have pq < 0 and pr < 0 .
this possible only when p is negative and q and r are positive.
Also, p, q and r are not zero either.
so, (p) (q) (r) < p. It is sufficient to say that it is pqr are always less than p. So answer is C.



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Re: Zumit DS 026 [#permalink]
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18 Sep 2008, 02:16
Twoone wrote: I think It is C.
I am agree with that Options A and B are not possible.
Let's combined both options
we have pq < 0 and pr < 0 .
this possible only when p is negative and q and r are positive.
Also, p, q and r are not zero either.
so, (p) (q) (r) < p. It is sufficient to say that it is pqr are always less than p. So answer is C. As x2suresh has explained, since p < 0, if qr >1 then pqr < p. However, if 0 < qr < 1 then pqr > p....hence, C cannot be the answer.



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Re: Zumit DS 026 [#permalink]
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18 Sep 2008, 10:09
scthakur wrote: Twoone wrote: I think It is C.
I am agree with that Options A and B are not possible.
Let's combined both options
we have pq < 0 and pr < 0 .
this possible only when p is negative and q and r are positive.
Also, p, q and r are not zero either.
so, (p) (q) (r) < p. It is sufficient to say that it is pqr are always less than p. So answer is C. As x2suresh has explained, since p < 0, if qr >1 then pqr < p. However, if 0 < qr < 1 then pqr > p....hence, C cannot be the answer. C is not correct, take p = 6, q = 1, r = 1, S1 and S2 are met, and stem is also met. > 6x1x1 is not < 6



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Re: No. prop Q [#permalink]
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27 Nov 2008, 03:03
pqr can be less than p only if p is negative and pq positive, so, p and q must be both either positive or negative
1) pq < 0 => p < 0, q > 0, but we don't know whether r is also positive. Ex: p=2, q=2, r=1, or p=2, q=2, r=1. Insuff. 2) The same, Insuff
Together, qr > 0, Suff
C
Last edited by atletikos on 27 Nov 2008, 03:06, edited 1 time in total.



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Re: No. prop Q [#permalink]
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27 Nov 2008, 05:15
OA is E from MGMAT
Here's OE they provide.
The question tells us that p < q and p < r and then asks whether the product pqr is less than p. Statement (1) INSUFFICIENT: We learn from this statement that either p or q is negative, but since we know from the question that p < q, p must be negative. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet only 2 conditions: p must be negative and q must be positive. p q r pqr Is pqr < p? 2 5 10 100 YES 2 5 10 100 NO
Statement (2) INSUFFICIENT: We learn from this statement that either p or r is negative, but since we know from the question that p < r, p must be negative. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet only 2 conditions: p must be negative and r must be positive. p q r pqr Is pqr < p? 2 10 5 100 NO 2 10 5 100 YES If we look at both statements together, we know that p is negative and that both q and r are positive. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet 3 conditions: p must be negative, q must be positive, and r must be positive. p q r pqr Is pqr < p? 2 10 5 100 YES 2 7 4 56 YES At first glance, it may appear that we will always get a "YES" answer. But don't forget to test out fractional (decimal) values as well. The problem never specifies that p, q, and r must be integers. p q r pqr Is pqr < p? 2 .3 .4 .24 NO Even with both statements, we cannot answer the question definitively. The correct answer is E.



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Re: If p < q and p < r, is (p)(q)(r) < p? (1) pq < 0 [#permalink]
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27 Jan 2012, 20:07
+1 E Don't forget that q and r can be fractions (0 < q,r < 1) That possibility lower the value of p.
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 [#permalink]
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28 Jan 2012, 03:41
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If p < q and p < r, is pqr < p?Given: \(p<q\) and \(p<r\). Question: is \(pqr<p\)? > is \(p(qr1)<0\)? (1) pq < 0 > \(p\) and \(q\) have opposite signs, as given that \(p<q\) then \(p<0\) and \(q>0\) > as \(p<0\) then the question becomes whether \(qr1>0\) (in \(p(qr1)\) first multiple is negative  \(p<0\), so in order the product to be negative second multiple must be positive  \(qr1>0\)) > is \(qr>1\)? We know that \(q>0\) but all we know about \(r\) is that \(p<r\). Not sufficient. (2) pr < 0 > the same here: \(p\) and \(r\) have opposite signs, as given that \(p<r\) then \(p<0\) and \(r>0\) > as \(p<0\) then the question becomes whether \(qr1>0\) > is \(qr>1\)? We know that \(r>0\) but all we know about \(q\) is that \(p<q\). Not sufficient. (1)+(2) we have that: \(p<0\), \(q>0\) and \(r>0\). Again question becomes: is \(qr>1\)? Though both \(q\) and \(r\) are positive their product still can be more than 1 (for example \(q=1\) and \(r=2\)) as well as less then 1 (for example \(q=1\) and \(r=\frac{1}{3}\)) or even equal to 1. Not sufficient. Answer: E. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifpqandprispqrp1pq02pr126285.html
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0
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