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is 1/p > r/(r^2 + 2) ? 1. r=p 2. r>0 why isn't first

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is 1/p > r/(r^2 + 2) ? 1. r=p 2. r>0 why isn't first [#permalink]

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New post 01 Nov 2007, 21:56
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is 1/p > r/(r^2 + 2) ?

1. r=p
2. r>0

why isn't first statement enough...please explain.
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New post 01 Nov 2007, 22:41
is 1/p > r/(r^2 + 2)

Statement 1

r=p

1/r > r/(r^2 + 2)

this is insufficient because you are assuming that r > 0 ---> then:

r^2+2 > r^2

if r < 0 then

r^2+2 < r^2

insufficient

Statement 2

state that r > 0

insufficient

both Statements

sufficient

so the answer is (C)

:)

Last edited by KillerSquirrel on 01 Nov 2007, 23:25, edited 1 time in total.
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New post 01 Nov 2007, 23:14
KillerSquirrel wrote:
is 1/p > r/(r^2 + 2)

Statement 1

r=p

1/r > r/(r^2 + 2)

this is insufficient because you are assuming that (r^2 + 2) > 0 ---> then:

r^2+2 > r^2

if (r^2 + 2) < 0 then

r^2+2 <r> 0

insufficient

so the answer is (C)

:)


Hi squirrel,

Can you give any example in which (r^2 + 2) < 0???

r^2 is always positive, whether r is positive, negative or fraction.

Thus statement (1) should be sufficient.

Answer should be (A)
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New post 01 Nov 2007, 23:24
LM wrote:
KillerSquirrel wrote:
is 1/p > r/(r^2 + 2)

Statement 1

r=p

1/r > r/(r^2 + 2)

this is insufficient because you are assuming that (r^2 + 2) > 0 ---> then:

r^2+2 > r^2

if (r^2 + 2) < 0 then

r^2+2 <r> 0

insufficient

so the answer is (C)

:)


Hi squirrel,

Can you give any example in which (r^2 + 2) < 0???

r^2 is always positive, whether r is positive, negative or fraction.

Thus statement (1) should be sufficient.

Answer should be (A)


(r^2 + 2) cannot be less then 0 but r can - you multiply r*(r^2+2) both sides of the equation !

thanks for the heads up ! I will correct my post, the answer is still (C)

:)
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New post 01 Nov 2007, 23:40
KillerSquirrel wrote:
LM wrote:
KillerSquirrel wrote:
is 1/p > r/(r^2 + 2)

Statement 1

r=p

1/r > r/(r^2 + 2)

this is insufficient because you are assuming that (r^2 + 2) > 0 ---> then:

r^2+2 > r^2

if (r^2 + 2) < 0 then

r^2+2 <r> 0

insufficient

so the answer is (C)

:)


Hi squirrel,

Can you give any example in which (r^2 + 2) < 0???

r^2 is always positive, whether r is positive, negative or fraction.

Thus statement (1) should be sufficient.

Answer should be (A)


(r^2 + 2) cannot be less then 0 but r can - you multiply r*(r^2+2) both sides of the equation !

thanks for the heads up ! I will correct my post, the answer is still (C)

:)


But you don't have to multiply both sides with "r".

The equation is r^2 + 2 > r^2
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New post 01 Nov 2007, 23:55
Hi squirrel, Thanks for your explanation.
I still second LM's point. why should we consider multiplying any positive expression with a negative variable ?
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New post 02 Nov 2007, 00:13
(C) as well :)

is 1/p > r/(r^2 + 2) ?
<=> (r^2+2)/p > r ? as r^2 + 2 > 0

Stat1
p=r

implies that (r^2+2)/p = (r^2+2)/r = r + 2/r

r + 2/r > r only if 2/r >0 and so r > 0.

INSUFF.

Stat2
r > 0 : p can be anything...

INSUFF.

Both 1 and 2
Bingo... We know have r > 0.... So r + 2/r > r <=> 1/p > r/(r^2 + 2)

SUFF.
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New post 02 Nov 2007, 05:22
LM wrote:
KillerSquirrel wrote:
LM wrote:
KillerSquirrel wrote:
is 1/p > r/(r^2 + 2)

Statement 1

r=p

1/r > r/(r^2 + 2)

this is insufficient because you are assuming that (r^2 + 2) > 0 ---> then:

r^2+2 > r^2

if (r^2 + 2) < 0 then

r^2+2 <r> 0

insufficient

so the answer is (C)

:)


Hi squirrel,

Can you give any example in which (r^2 + 2) < 0???

r^2 is always positive, whether r is positive, negative or fraction.

Thus statement (1) should be sufficient.

Answer should be (A)


(r^2 + 2) cannot be less then 0 but r can - you multiply r*(r^2+2) both sides of the equation !

thanks for the heads up ! I will correct my post, the answer is still (C)

:)


But you don't have to multiply both sides with "r".

The equation is r^2 + 2 > r^2


how do you know the sign of r??

if r>1/r it doesnt mean r^2>1

simply because the equation changes as per the sign of r.
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New post 02 Nov 2007, 05:56
Fig wrote:
(C) as well :)

is 1/p > r/(r^2 + 2) ?
<r> r ? as r^2 + 2 > 0

Stat1
p=r

implies that (r^2+2)/p = (r^2+2)/r = r + 2/r

r + 2/r > r only if 2/r >0 and so r > 0.

INSUFF.

Stat2
r > 0 : p can be anything...

INSUFF.

Both 1 and 2
Bingo... We know have r > 0.... So r + 2/r > r <1> r/(r^2 + 2)

SUFF.


Good explanation Fig.
Problem with equation like 1/r> r/r2+2 is that one can get tempt to go for full - r^2 + 2 = r^2- but the best way to solve this to stop at r^2 + 2 / r = r.With this answer become clear.
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New post 02 Nov 2007, 19:16
My Take is C. But I worked out in a different way.

1. Is not sufficient because, when r=p, it appears to be sufficient but what if r=p=0? Hence, in my view it is not sufficient

2. Not sufficient

Together makes sense.

Do you agree?
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New post 02 Nov 2007, 19:24
appuvar wrote:
My Take is C. But I worked out in a different way.

1. Is not sufficient because, when r=p, it appears to be sufficient but what if r=p=0? Hence, in my view it is not sufficient

2. Not sufficient

Together makes sense.

Do you agree?


Sounds reasonable, at least I can't deny it.
  [#permalink] 02 Nov 2007, 19:24
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