niteshwaghray wrote:
If \(ap > aq\), is \(\frac{a}{pq}>0\)?
(1) \(p < q\)
(2) \(\frac{1}{p}<\frac{1}{q}\)
\(ap > aq = a(p-q)>0\):
[case 1] \(a>0…p-q>0…p>q\)
[case 2] \(a<0…p-q<0…p<q\)
(1) \(p < q\): [case 1] \(a<0\)
if \({p,q}>0\) then \(\frac{a=positive}{pq=positive}>0\)
if \(p>0…q<0\) then \(\frac{a=positive}{pq=negative}<0\)
different answers, insuf.
(2) \(\frac{1}{p}<\frac{1}{q}:…\frac{q-p}{pq}<0\) means that \(q-p\) and \(pq\) are different signs;
if \(pq<0\), then \(q-p>0…q>p\); [case 2], so \(a<0\): \(\frac{a=negative}{pq=negative}>0\)
if \(pq>0\), then \(q-p<0…q<p\); [case 1], so \(a>0\): \(\frac{a=positive}{pq=positive}>0\)
same answers, sufic.
Answer (B).