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tinki
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 Updated on: 28 Apr 2016, 14:30
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
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54% (02:17) correct 46% (02:02) wrong based on 819 sessions

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

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


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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
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E
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Bunuel
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 07 Feb 2011, 11:49
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tinki
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
Show more

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

The above means that \(p\) and \(q\) have opposite signs and since given that \(p<q\), then \(p<0\) and \(q>0\). Since \(p<0\) then the question becomes:

    Is \(qr-1>0\) (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 \(r>0\).

Since \(p<0\) then the question becomes:

    Is \(qr-1>0\)?

    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, the question becomes: is \(qr>1\)? Even 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.
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 07 Feb 2011, 12:03
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Bunuel
tinki
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.
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aha, Got it. i was making some stupid assumptions. (as always :-D )
+ Kudo for the best explanation and swift response
thanks again
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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 05 Apr 2011, 20:21
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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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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 05 Apr 2011, 22:07
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The question asks :

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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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 04 Sep 2011, 17:07
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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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If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 06 Sep 2011, 13:46
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If p < q and p < r, is pqr < p?

(1) pq < 0

(2) pr < 0
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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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