1. pq < 0
with p<q, p is neg and q is positive, no info about r (r can be either positive or negative), therefore insuff.

2. pr<0
with p<r, p is neg and r is positive, no info about q (q can be either positive or negative), therefore insuff.

1+2. pq<0 and pr<0, making p negative and both r & q positive, however if p = -1 and q and r both = 1, then (p)(q)(r) = p, therefore insuff.

(E)

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

E. if q=r=1,any -ve value for p makes p = pqr.
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GT

getting E

Each statement individually doesn't give information about the third variable. So they are insufficient individually.

Combined... two scenarios to consider

if p < 0, then q,r > 0 => pqr < p (-ve on both sides)
if p > 0, then q,r < 0 => pqr > p (+ve on both sides)

So insufficient.

answer E for me as well.

I wrote an eq'n: p(qr-1) < 0 ... so either p<0 and qr>1 or p>0 and qr<1

from stat 1, i know that p<0 and q>0, since p<q, but i dont know anything about r to see if it satisfies whether qr>1. insuff

from stat 2, i know that p<0 and r>, but nothing about q to see if qr>1 ... insuff.

together, all i know is that q and r are positive, but they could both be fractions, i.e. qr <1, or they could be whole numbers so that qr>1

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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I think It is C.

I am agree with that Options A and B are not possible.

Let's combined both options

we have pq < 0 and pr < 0 .

this possible only when p is negative and q and r are positive.

Also, p, q and r are not zero either.

so, (p) (q) (r) < p.

It is sufficient to say that it is pqr are always less than p. So answer is C.

C is not correct, take p = -6, q = 1, r = 1, S1 and S2 are met, and stem is also met. -> -6x1x1 is not < -6

OA is E from MGMAT

Here's OE they provide.

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?

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-p-q-and-p-r-is-pqr-p-1-pq-0-2-pr-126285.html
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