eeakkan wrote:
If 2p not equal to -q, is (2p-q)/(2p+q)>1?
(1) p<0
(2) q>0
Please help me with this. According to me:
ıf we arrange question: 2p-q>2p+q
then -q>q and
so (B) should be ok. Because if q>0, -q will be always <q.
If 2p not equal to -q, is (2p-q)/(2p+q)>1?[/m]?Is \(\frac{2p-q}{2p+q}>1\)? --> is \(0>1-\frac{2p-q}{2p+q}\)? --> is \(0>\frac{2p+q-2p+q}{2p+q}\)? --> is \(0>\frac{2q}{2p+q}\)?
(1) \(p<0\). Not sufficient.
(2) \(q>0\). Not sufficient.
(1)+(2) \(p<0\) and \(q>0\) --> the numerator (2q) is positive, but we cannot say whether the denominator {negative (2p)+positive (q)} is positive or negative. Not sufficient.
Answer: E.
The problem with your solution is that when you are writing \(2p-q>2p+q\), you are actually multiplying both sides of inequality by \(2p+q\):
never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(2p+q>0\) you should write \(2p-q>2p+q\) BUT if \(2p+q<0\), you should write \(2p-q<2p+q\), (flip the sign when multiplying by negative expression).
Hope it helps.
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