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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
_________________

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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.

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.