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# Is (p/q) > ( /s) 1) (p/q) > (r/s) 2) r = 1

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Is (p/q) > ( /s) 1) (p/q) > (r/s) 2) r = 1 [#permalink]

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16 Jun 2008, 18:58
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Is (p/q) > ([r^2]/s)

1) (p/q) > (r/s)
2) r = 1
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Re: DS Inequalities [#permalink]

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16 Jun 2008, 19:15
1. It gives different (r^2)/s and r/s values for different numbers +ve, -ve and fractions. - Insufficient.
2. It simply says that p/q > 1/s - Insufficient.

Combine 1 & 2 it should be sufficient.

if r = 1 & p/q > r/s it means p/q > (r^2)/s

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Re: DS Inequalities [#permalink]

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16 Jun 2008, 19:16
is it C ?

reason :
if $$r = 1$$ then $$\frac{r}{s}= \frac{r^2}{s}$$

Hence when we combine A and B
$$\frac{p}{q} > \frac{r}{s}$$ and hence$$\frac{p}{q} > \frac{r^2}{s}$$
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Re: DS Inequalities [#permalink]

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16 Jun 2008, 19:26
It is C, but can you prove it by eliminating statement 1 and 2 first, before selecting C?
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Re: DS Inequalities [#permalink]

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16 Jun 2008, 19:41
B is easy to eliminate - it provides no info what so ever on p and q.

A states that
$$\frac{p}{q}> \frac{r}{s}$$

consider $$p = 1, q = 1, r = -1, and, s = 1$$
in this case $$\frac{p}{q} = 1 > \frac{r}{s}$$ (which is -1)
however, $$\frac{r^2}{s}$$= 1 which is EQUAL to $$\frac{p}{q}$$

hence A is insufficient.

jimmyjamesdonkey wrote:
It is C, but can you prove it by eliminating statement 1 and 2 first, before selecting C?
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Re: DS Inequalities [#permalink]

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17 Jun 2008, 04:12
C.

1. p/q > r/s. INSUF. (BSD has good explanation.)
2. r=1. INSUF

==> Both: SUF
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Re: DS Inequalities [#permalink]

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17 Jun 2008, 05:27
jimmyjamesdonkey wrote:
Is (p/q) > ([r^2]/s)

1) (p/q) > (r/s)
2) r = 1

1: insuf
For all p,q,s > 0 and r < 0 condition 1 is satisfied but not necessarily (p/q) > (r^2/s)
2: insuf
(p,q,s) = (1,1,2) or (1,2,1)

1&2 suf (p/q) > (r/s) = (1*r)/s = r^2/s
Re: DS Inequalities   [#permalink] 17 Jun 2008, 05:27
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# Is (p/q) > ( /s) 1) (p/q) > (r/s) 2) r = 1

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