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

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Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
who said PR quant is easy...man I got blown away after the [#permalink]

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24 Jan 2005, 18:30
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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, 19:43, edited 2 times in total.
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

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24 Jan 2005, 19:15
fresinha12 can u restate Statement II
Director
Joined: 19 Nov 2004
Posts: 556
Location: SF Bay Area, USA

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24 Jan 2005, 19:58
I) only x between 0 and 100 is x=64
OK

II) x =64
Ok
D
Manager
Joined: 11 Jan 2005
Posts: 101

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24 Jan 2005, 23:12
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?
Manager
Joined: 11 Jan 2005
Posts: 101

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24 Jan 2005, 23:15
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
Manager
Joined: 11 Jan 2005
Posts: 101

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24 Jan 2005, 23:24
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.
Manager
Joined: 08 Oct 2004
Posts: 224

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25 Jan 2005, 02:03
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.
_________________

Believe in yourself.

Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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25 Jan 2005, 08:25
The OA is A...go figure!
Manager
Joined: 11 Jan 2005
Posts: 101

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25 Jan 2005, 17:49
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
VP
Joined: 18 Nov 2004
Posts: 1433

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25 Jan 2005, 18:01
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
Manager
Joined: 11 Jan 2005
Posts: 101

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25 Jan 2005, 18:11
Yep I know, I screwed up.. so I am lost too... II can be a solution.. so it should be D.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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26 Jan 2005, 12:00
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
VP
Joined: 18 Nov 2004
Posts: 1433

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26 Jan 2005, 12:03
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.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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26 Jan 2005, 13:56
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.
Senior Manager
Joined: 30 Dec 2004
Posts: 294
Location: California

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26 Jan 2005, 14:08
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.
_________________

"No! Try not. Do. Or do not. There is no try.

VP
Joined: 18 Nov 2004
Posts: 1433

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26 Jan 2005, 14:51
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
SVP
Joined: 03 Jan 2005
Posts: 2233

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26 Jan 2005, 14:59
Multiple implies integer multiple by definition, IIRC.
Intern
Joined: 26 Jan 2005
Posts: 11

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27 Jan 2005, 05:04
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?
Manager
Joined: 25 Oct 2004
Posts: 247

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28 Jan 2005, 07:45
I would go with D . whats the OA by the way...
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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29 Jan 2005, 13:12
I thought I already posted the OA...its A!
29 Jan 2005, 13:12

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