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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0

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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 [#permalink] New post 16 Jul 2007, 18:22
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A
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E

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Question Stats:

44% (01:48) correct 56% (00:32) wrong based on 133 sessions
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-p-q-and-p-r-is-pqr-p-1-pq-0-2-pr-126285.html

iamba 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jan 2012, 02:56, edited 4 times in total.
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 [#permalink] New post 16 Jul 2007, 21:06
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] New post 17 Jul 2007, 02:14
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] New post 03 May 2008, 15:07
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] New post 03 May 2008, 21:34
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] New post 03 May 2008, 22:49
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] New post 04 May 2008, 06:37
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 DS-type 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] New post 04 May 2008, 08:42
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] New post 15 Jul 2008, 10:01
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] New post 15 Jul 2008, 10:18
answer E for me as well.

I wrote an eq'n: p(qr-1) < 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] New post 15 Jul 2008, 14:52
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] New post 16 Sep 2008, 05:54
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] New post 16 Sep 2008, 05:55
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] New post 16 Sep 2008, 11:54
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] New post 18 Sep 2008, 01:16
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] New post 18 Sep 2008, 09:09
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] New post 27 Nov 2008, 02:03
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, 02:06, edited 1 time in total.
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Re: No. prop Q [#permalink] New post 27 Nov 2008, 04:15
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] New post 27 Jan 2012, 19:07
+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] New post 28 Jan 2012, 02:41
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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(qr-1)<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 qr-1>0 (in p(qr-1) first multiple is negative - p<0, so in order the product to be negative second multiple must be positive - qr-1>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 qr-1>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: if-p-q-and-p-r-is-pqr-p-1-pq-0-2-pr-126285.html
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