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16 Sep 2014, 00:17
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If \(qs \neq 0\), is \(\frac{p}{q} \gt \frac{r^2}{s}\)?
(1) \(\frac{p}{q} \gt \frac{r}{s}\)
(2) \(r = 1\)
(1) \(\frac{p}{q} \gt \frac{r}{s}\)
(2) \(r = 1\)
16 Sep 2014, 00:17
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Official Solution:
If \(qs \neq 0\), is \(\frac{p}{q} \gt \frac{r^2}{s}\)?
(1) \(\frac{p}{q} \gt \frac{r}{s}\).
If \(\frac{p}{q}=2\) and \(r=s=1\), then the answer is YES. However, if for example \(\frac{p}{q}=2\) and \(r=s=2\), then the answer is NO. This statement is not sufficient.
(2) \(r = 1\).
This statement alone is clearly insufficient.
(1)+(2) Given that \(r=1\) from statement (2), the question simplifies to whether \(\frac{p}{q} \gt \frac{1}{s}\). Statement (1) directly answers this question, so the combination of statements is sufficient.
Answer: C
If \(qs \neq 0\), is \(\frac{p}{q} \gt \frac{r^2}{s}\)?
(1) \(\frac{p}{q} \gt \frac{r}{s}\).
If \(\frac{p}{q}=2\) and \(r=s=1\), then the answer is YES. However, if for example \(\frac{p}{q}=2\) and \(r=s=2\), then the answer is NO. This statement is not sufficient.
(2) \(r = 1\).
This statement alone is clearly insufficient.
(1)+(2) Given that \(r=1\) from statement (2), the question simplifies to whether \(\frac{p}{q} \gt \frac{1}{s}\). Statement (1) directly answers this question, so the combination of statements is sufficient.
Answer: C
24 Apr 2018, 17:12
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Q and S cannot be zero IS P/Q > R^2/S ?? I simplified the stem to is PS > QR^2 ??
1) P/Q > R/S
Simplified the fraction by crossing multiply PS > QR
P=2 S=1 Q=1 R=1 then 2 > 1
Is 2 > 1 ? Yes
P= -2 S=1 Q=1 R=-3 then -2 > -3
Is -2 > 9 No
INSUFFICIENT
2) R=1
Then the question stem becomes....Is PS>Q ??
INSUFFICIENT
Taking 1 and 2 together
PS > QR and R=1
It becomes PS > Q ALWAYS YES SUFFICIENT
Thanks!
Alecita
1) P/Q > R/S
Simplified the fraction by crossing multiply PS > QR
P=2 S=1 Q=1 R=1 then 2 > 1
Is 2 > 1 ? Yes
P= -2 S=1 Q=1 R=-3 then -2 > -3
Is -2 > 9 No
INSUFFICIENT
2) R=1
Then the question stem becomes....Is PS>Q ??
INSUFFICIENT
Taking 1 and 2 together
PS > QR and R=1
It becomes PS > Q ALWAYS YES SUFFICIENT
Thanks!
Alecita
