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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.

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

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07 Feb 2011, 11:03

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.

Answer: E.

aha, Got it. i was making some stupid assumptions. (as always ) + Kudo for the best explanation and swift response thanks again

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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15 Apr 2012, 15:01

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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20 May 2013, 23:14

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?

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.

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

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29 Dec 2013, 02:16

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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21 Oct 2014, 12:02

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

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

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28 Apr 2016, 12:13

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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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15 Aug 2017, 07:34

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