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Manager  Joined: 18 Aug 2010
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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Question Stats: 50% (02:13) correct 50% (02:06) wrong based on 327 sessions

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If p < q and p < r, is pqr < p?

(1) pq < 0
(2) pr < 0

i did it in this way:
Question : p(qr-1)<o
P<0 or qr< 1 ?

combined both statements say that p<0
So according to the logic above, why answer C (both ) is not sufficient ?

Anybody ?
Thanks

Originally posted by tinki on 07 Feb 2011, 11:18.
Last edited by Bunuel on 28 Apr 2016, 14:30, edited 2 times in total.
Edited the question and added the OA
Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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10
3
tinki wrote:
If p < q and p < r, is pqr < p?
(1) pq < 0

(2) pr < 0

i did it in this way:
Question : p(qr-1)<o
P<0 or qr< 1 ?

combined both statements say that p<0
So according to the logic above, why answer C (both ) is not sufficient ?

Anybody ?
Thanks

What does "C (both ) is not sufficient" mean? Anyway:

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.

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Manager  Joined: 18 Aug 2010
Posts: 75
Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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Bunuel wrote:
tinki wrote:
If p < q and p < r, is pqr < p?
(1) pq < 0

(2) pr < 0

i did it in this way:
Question : p(qr-1)<o
P<0 or qr< 1 ?

combined both statements say that p<0
So according to the logic above, why answer C (both ) is not sufficient ?

Anybody ?
Thanks

What does "C (both ) is not sufficient" mean? Anyway:

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.

aha, Got it. i was making some stupid assumptions. (as always )
+ Kudo for the best explanation and swift response
thanks again
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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I couldn't find it when I searched for it. Thank you, Bunuel!
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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1
1) Insufficient. Does not preclude r from being zero

2) Insufficient. Does not preclude q from being zero

1) + 2)

Consider p -ve and q and r +ve.

let p = -1. q = 2 r = 3. The answer is YES
let p = -1. q = 0.5 r = 0.6. The answer is NO

Insufficient. Hence E
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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2

pqr - p < 0

=> p(qr - 1) < 0

=> either p < 0 and qr > 1 or p > 0 and qr < 1

From (1) it can be seen that q > 0 and p < 0

But no information about r, hence insufficient

e.g. if 0 < r < 1 or if r < 0 and 0 < q < 1, then qr < 1

From (2), pr < 0 => r > 0 and p < 0

But no information about q, hence insufficient

e.g. if r > 1 and then qr > 1 and the expression is true, but may not be so otherwise

Combining (1) and (2) also, nothing is conclusive about qr, so answer is E.
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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2
Restate the question stem: "is p(qr - 1) < 0?"

Statement 1) pq < 0 implies that pq != 0 and that p and q have different signs. Since p < q, p < 0 < q.

Since p < 0, the question stem becomes "qr - 1 > 0?", or "qr > 1?"

We have deduced that q > 0, but nothing about r so the statement is insufficient.

Yes example: {p,q,r} = {-5, 2, 5} and pqr < p since (-5)(2)(5) < (-5)
No example: {p,q,r} = {-5, 1/2, 1} and pqr > p since (-5)(1/2)(1) > (-5)

Statement 2) Similar to statement 1, pr < 0 implies that pr != 0 and that p and r have different signs. Since p < r, p < 0 < r.

Since p < 0, the question stem becomes "qr - 1 > 0?", or "qr > 1?", which cannot be answered without more information about q.

The same yes/no examples from above can be used to illustrate this statement's insufficiency.

Combined) The combined statements tell us that p < 0 < q and that p < 0 < r. Even with this information, the inequality qr > 1 can be true or false.

Yes example: {p,q,r} = {-5, 2, 5} and pqr < p since (-5)(2)(5) < (-5)
No example: {p,q,r} = {-5, 1/2, 1} and pqr > p since (-5)(1/2)(1) > (-5)

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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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bschool83 wrote:
If p < q and p < r, is pqr < p?

(1) pq < 0

(2) pr < 0

If, from Statement 1, pq < 0, then one of p or q is negative, the other positive. Negative numbers are always smaller than positive numbers, so if p < q, then clearly p must be the negative number and q the positive number. We learn the same thing from Statement 2: p is negative, and r is positive.

Now that we know that p is negative, we can rephrase the question by dividing by p on both sides, reversing the inequality when we do (because we're dividing by a negative) :

Is pqr < p ?

Is qr > 1 ?

While we know q and r are positive, so qr > 0, we have no way to tell whether qr > 1. It might be that q = r = 2, and the answer is 'yes', or it might be that q = r = 1/2 and the answer is 'no'. So the answer is E.
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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I got C. Did not consider fractions...assumed them as integers...
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GMAT 1: 680 Q50 V32 Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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Straight E because 1) and 2) individually provide no information about one of the variables.
Combined they dont provide information on magnitude.
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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Great explanation provided..
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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we need to see if the statement pqr is less than p, that is only the case if qr is a fraction or if pqr is a larger negative number. it seems that the question is more asking about if the term is more negative.
stmt 1 . pq < 0 so p and q are opposite terms. nothing about r
stmt 2. pr < 0 so q and r are opposite terms. nothing about q

together we know that if p is positive q and r is negative which makes the term pqr a larger positive number or maybe even q and r are a fraction that makes it smaller we are unsure.
if p is negative then q and r are positive but still we dont know if the the term qr is a fraction.
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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tinki wrote:
If p < q and p < r, is pqr < p?

(1) pq < 0
(2) pr < 0

I did it quickly in following way... clearly A & B are not answers because we need info for all p,q,r

1) says p is -ve and q is +ve
2) says p is -ve and r is +ve

Now q and r can be >1 or <1 ( no restriction on them) so pqr can be < or > p, Easy E
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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qIf p < q and p < r, is (p)(q)(r) < p?

(1) pq < 0

(2) pr < 0

while solving this problem I simplified the quesiton to
pqr-p<0
taking p as a common factor
p(qr-1)<0.
so either p<0 or qr<1.

Now the question I have is - if I am able to prove one of the two i.e
p<0 or qr<1 , then would that suffice to say that either of the options work.
On these lines, I went and chose D as the answer. Mathematically is it correct to say p must be less than zero or qr must be less than one for pqr<p.

Logically when I try other values I do realize it does not fit in with the above line of thought and the answer is E. But can you help me understand what does it mean resolving from an algebra point of view?
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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gmatrant wrote:
qIf p < q and p < r, is (p)(q)(r) < p?

(1) pq < 0

(2) pr < 0

while solving this problem I simplified the quesiton to
pqr-p<0
taking p as a common factor
p(qr-1)<0.
so either p<0 or qr<1.

Now the question I have is - if I am able to prove one of the two i.e
p<0 or qr<1 , then would that suffice to say that either of the options work.
On these lines, I went and chose D as the answer. Mathematically is it correct to say p must be less than zero or qr must be less than one for pqr<p.

Logically when I try other values I do realize it does not fit in with the above line of thought and the answer is E. But can you help me understand what does it mean resolving from an algebra point of view?

Merging similar topics. Please refer to the solutions above.

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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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Hi,
First thing i would like to tell you that you can't subtract the inequalities. Now lets come to the combination of the fact statements.

In all these kind of Yes/ No D.S. questions always use numbers. lets plugin the numbers.

p q r pqr ( is pqr<p)

-1 1 1 -1 No

-2 1 2 -4 Yes

So even after combining both the fact statements we are not getting a definite answer, that's why the answers must be E.

Tarun Kaushik
Sr. Quant Faculty
Princeton Review

tinki wrote:
If p < q and p < r, is pqr < p?

(1) pq < 0
(2) pr < 0

i did it in this way:
Question : p(qr-1)<o
P<0 or qr< 1 ?

combined both statements say that p<0
So according to the logic above, why answer C (both ) is not sufficient ?

Anybody ?
Thanks
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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p<q and p<r is pqr<p?

1. pq<0 since from the stem p has to be less than q, q is therefore positive and p is negative. This statement does not provide any information about r.
Non sufficient

2. pr<0 same story here, the statement is conveying us once again that p is less than zero.
Non sufficient

1+2) I find myself pretty comfortable plugging some numbers in and figuring out the outcome if p=-2, q=2, r=1 ----> pqr<p. If p=-0,5, q=0,5, r=0,5 ----->pqr>p
Non sufficient. E.
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Re: If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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tinki wrote:
If p < q and p < r, is pqr < p?

(1) pq < 0
(2) pr < 0

i did it in this way:
Question : p(qr-1)<o
P<0 or qr< 1 ?

combined both statements say that p<0
So according to the logic above, why answer C (both ) is not sufficient ?

Anybody ?
Thanks

1. pq < 0 - Either of the two is negative and the other one positive. Two possible scenarios for p and q : (- +) and (+ -)
2. pr < 0 - Either of the two is negative and the other one positive. Two possible scenarios for p and r : (- +) and (+ -)

Combining the two: The value will be negative if p is negative and value will be positive if p is positive. But we don't know whether the product will be less than p, because we don't know the value of p, q, r

So, insufficient. E
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0  [#permalink]

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tinki wrote:
If p < q and p < r, is pqr < p?

(1) pq < 0
(2) pr < 0

Bunuel pls check whether my approach is right?

pqr<p. Dividing both the sides by p. In short, we need to find is qr<1?

In such questions, we need to prove that the statement is not sufficient especially by plugging in numbers.

Stat(1) pq<0
We know p<q

YES= -2*2=-4<0 NO= 2*4=8 and not less than 0. NOT SUFFICIENT

Stat (2) qr<1
We know p<r

Plugged in similar numbers as in stat (1). NOT SUFFICIENT

(1)+(2) Using the values that we used in both statements doesn't satisfy the question stem, qr<1 (that is pqr<p)

_________________ If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0   [#permalink] 28 Nov 2018, 22:24
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