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who said PR quant is easy...man I got blown away after the

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who said PR quant is easy...man I got blown away after the [#permalink] New post 24 Jan 2005, 17:30
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A
B
C
D
E

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who said PR quant is easy...man I got blown away after the 15 th quesiton had 5 Probability and 3 permutation question...not easy..will post em soon!

If 0 < x < 100, what is the value of x?



(1) sqrt(x) is a multiple of 8.



(2) sqrt(x) and cubert(x) are even integers.

Last edited by FN on 24 Jan 2005, 18:43, edited 2 times in total.
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 [#permalink] New post 24 Jan 2005, 18:15
fresinha12 can u restate Statement II
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 [#permalink] New post 24 Jan 2005, 18:58
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D
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 [#permalink] New post 24 Jan 2005, 22:12
I don t think it is D.
I would vote for A.

II alone does not give me a clue :

if n=16
then n^-2= 4, so even
then n^-3=2, so even

so n can be 64 & 16

what do you think?
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 [#permalink] New post 24 Jan 2005, 22:15
oops sorry guys , i forgot to note that D means I need both to find out ? is nit .. anyway I think we need both assumptions to answer
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 [#permalink] New post 24 Jan 2005, 22:24
in the meantime,
n^-2 belongs to [8,16, 24....96]

so
n belongs to [8^2,16^2....96^2]

n belongs also to [0,100}

there is only 8^2 which belongs to both conditions.
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 [#permalink] New post 25 Jan 2005, 01:03
it has to be D. 64 is the only sqrt of x that will be a multiple of X and its cube root and sqaure root will be even integers.

D.
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 [#permalink] New post 25 Jan 2005, 07:25
The OA is A...go figure!
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 [#permalink] New post 25 Jan 2005, 16:49
please see what I wrote above for noting that there is only A as possible answer:
with 1.

if n^2 is multiple of 8 then
n^-2 belongs to [8,16, 24....96]

so
n belongs to [8^2,16^2....96^2]

n belongs also to [0,100}

there is only 8^2 which belongs to both conditions.

with 2 , as explained there are two solutions : 16 or 64. so not enough
hope I explained properly though my choice of A, B,C,D,E was not very clear at first
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 [#permalink] New post 25 Jan 2005, 17:01
mdf2 wrote:
please see what I wrote above for noting that there is only A as possible answer:
with 1.

if n^2 is multiple of 8 then
n^-2 belongs to [8,16, 24....96]

so
n belongs to [8^2,16^2....96^2]

n belongs also to [0,100}

there is only 8^2 which belongs to both conditions.

with 2 , as explained there are two solutions : 16 or 64. so not enough
hope I explained properly though my choice of A, B,C,D,E was not very clear at first


x = 16 can't be a solution for statement 2 as it doesn't satisfy statement 2.
sqrt (16) = 4....and (16)^1/3 = not even an integer
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 [#permalink] New post 25 Jan 2005, 17:11
Yep I know, I screwed up.. so I am lost too... II can be a solution.. so it should be D.
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 [#permalink] New post 26 Jan 2005, 11:00
why cant it be 24 for statement 1, the question stem never stated that the X is an integer?

nocilis wrote:
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D
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 [#permalink] New post 26 Jan 2005, 11:03
fresinha12 wrote:
why cant it be 24 for statement 1, the question stem never stated that the X is an integer?

nocilis wrote:
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D


if x = 24, sqrt of 24 is not an integer, hence can't be a mutiple of 8.
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 [#permalink] New post 26 Jan 2005, 12:56
banarjee what I meant was what if the sqrt(x) is 24 which is a multiple of 8? in that case X is not an integer ???

banerjeea_98 wrote:
fresinha12 wrote:
why cant it be 24 for statement 1, the question stem never stated that the X is an integer?

nocilis wrote:
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D


if x = 24, sqrt of 24 is not an integer, hence can't be a mutiple of 8.
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 [#permalink] New post 26 Jan 2005, 13:08
I say B is sufficient. The only number where the square root and the cube root are even integers between 0 and 100 is 64. The reason A is not sufficient is because any number can be a multiple of 8 unless it is specifically states integer multiple...IMHO.
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 [#permalink] New post 26 Jan 2005, 13:51
fresinha12 wrote:
banarjee what I meant was what if the sqrt(x) is 24 which is a multiple of 8? in that case X is not an integer ???

banerjeea_98 wrote:
fresinha12 wrote:
why cant it be 24 for statement 1, the question stem never stated that the X is an integer?

nocilis wrote:
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D


if x = 24, sqrt of 24 is not an integer, hence can't be a mutiple of 8.


In that case X = 24^2 which is > 100....stem already says that 0 < x < 100....so sqrt(x) can't be 24
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 [#permalink] New post 26 Jan 2005, 13:59
Multiple implies integer multiple by definition, IIRC.
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 [#permalink] New post 27 Jan 2005, 04:04
If 0 < x < 100, what is the value of x?



(1) sqrt(x) is a multiple of 8.



(2) sqrt(x) and cubert(x) are even integers.

Statement 1: x can be 64 only.
Statement 2: x can be 64 as root 64 = 8 and cuberoot 64 = 4

isnt that it? both are sufficient.. whats the official answer?
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 [#permalink] New post 28 Jan 2005, 06:45
I would go with D . whats the OA by the way...
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 [#permalink] New post 29 Jan 2005, 12:12
I thought I already posted the OA...its A!
  [#permalink] 29 Jan 2005, 12:12
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