If p,q,r are non-zero numbers, is pq/r0?

Question

If p,q,r are non-zero numbers, is pq/r>0?

1. p+q>0
2. rq>0

Dare Devil · Answer

If p,q,r are non-zero numbers, is pq/r>0?

For pq/r to be greater than zero any either all should be positive or two of then can be negative (given non-zeros)
We need to know sign of all p,q,r to answer the question

1. p+q>0
We can't predict the sign of p and q,
1) p,q both may be positive
2) p >q and p is positive and q is negative
3) p < q and p - Negative , q - positive
Insufficient

2. \(rq>0\)
q,r have the same sign either positive OR negative, but resulting in a positive values
Still if P is negative,\(pq/r < 0\)
if P is positive, \(pq/r > 0\)
Insufficient

(1) and (2) together,
Still insufficient, as we don't have enough information about the sign of all the variables

Option E is correct

akurathi12 · Answer

Thanks for the detailed explanation.

fskilnik · Answer

p,q,r\,\, 
e 0

\frac{{pq}}{r}\,\,\mathop > \limits^? \,\,0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\,\,?\,\,:\,\,pq\,\,{	ext{and}}\,\,r\,\,{	ext{have}}\,\,{	ext{same}}\,\,{	ext{signs}}\,\,}

\left( {1 + 2} ight)\,\,\left\{ \matrix{
 \,p + q > 0 \hfill \cr 
 \,r\,\,{m{and}}\,\,q\,\,{m{same}}\,\,{m{signs}} \hfill \cr} ight.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,\left( {p,q,r} ight) = \left( {1,1,1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{m{YES}}} ightangle \,\, \hfill \cr 
 \,{m{Take}}\,\,\left( {p,q,r} ight) = \left( { - 1,2,2} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{m{NO}}} ightangle \,\, \hfill \cr} ight.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {m{E}} ight)

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

bumpbot · Answer

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If p,q,r are non-zero numbers, is pq/r>0?

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