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Re: MGMAT DS Problem [#permalink]
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
Re: MGMAT DS Problem [#permalink]
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.
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 _________________
Your attitude determines your altitude Smiling wins more friends than frowning
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.
If p<q and p<r, is pqr<p? (1) pq<0 (2) pr<0 [#permalink]
28 Jan 2012, 02:41
4
This post received KUDOS
Expert's post
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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