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Find P. (1) |P|P||=3 (2) 2|P|=3

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SVP
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Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]

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New post 09 Jun 2003, 23:49
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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Find P.

(1) |P–|P||=3
(2) 2|P|=3

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Manager
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DS: find P [#permalink]

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New post 26 Jun 2003, 07:59
Statement 1 gives the value of P as -3/2

Statement 2 gives values of P as {-3/2, 3/2}

Therefore, statement 1 is sufficient alone to answer the question but statement 2 alone is not sufficient

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 [#permalink]

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New post 04 May 2006, 06:13
I am getting E

1) 4P^2 = 9
P = +-3/2

2) P = +-3/2

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Director
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 [#permalink]

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New post 04 May 2006, 06:24
I am also getting E

First eq also gives values of P as +3/2, -3/2.

Edit ........

Ok I got it. Silly me .. :wall
It is A.

Last edited by remgeo on 04 May 2006, 07:48, edited 1 time in total.

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 [#permalink]

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New post 04 May 2006, 07:25
I am getting A.

(1) |P–|P||=3
if P is +ve then |p-p| = 0 = 3 which makes no sense as 0 != 3
so P must be -ve |-p-p|= 3
-2p = 3 -> p = -2/3 sufficient

(2) 2|P|=3

this means p = +3/2 or -3/2

hence A is suff

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 [#permalink]

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New post 04 May 2006, 07:50
minu wrote:
I am getting A.

(1) |P–|P||=3
if P is +ve then |p-p| = 0 = 3 which makes no sense as 0 != 3
so P must be -ve |-p-p|= 3
-2p = 3 -> p = -2/3 sufficient


Not that it matters in this case, but still..

| -2p | = 3
ie. 2p = 3
ie. p = 3/2.

Such mistakes can kill you :)

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 [#permalink]

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New post 04 May 2006, 08:25
Don't understand why from 1) p can't be +-3/2.

Can somebody explain.

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 [#permalink]

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New post 04 May 2006, 13:52
Natalya Khimich wrote:
Don't understand why from 1) p can't be +-3/2.
Can somebody explain.


Well just plug in numbers. If p>0, then the expression will always be 0.
The graph of |P–|P|| is
-2x, for x<0, and
0, for x>= 0

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 [#permalink]

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New post 05 May 2006, 04:06
It is A indeed:
The first equation has only one solution –P-P=3P=-3/2

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  [#permalink] 05 May 2006, 04:06
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Find P. (1) |P|P||=3 (2) 2|P|=3

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