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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

fresinha12 can u restate Statement II



Director
Joined: 19 Nov 2004
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Location: SF Bay Area, USA

I) only x between 0 and 100 is x=64
OK
II) x =64
Ok
D



Manager
Joined: 11 Jan 2005
Posts: 101

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

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
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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

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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Current Student
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The OA is A...go figure!



Manager
Joined: 11 Jan 2005
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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
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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
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Yep I know, I screwed up.. so I am lost too... II can be a solution.. so it should be D.



Current Student
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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
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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
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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
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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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VP
Joined: 18 Nov 2004
Posts: 1433

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
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Multiple implies integer multiple by definition, IIRC.



Intern
Joined: 26 Jan 2005
Posts: 11

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

I would go with D . whats the OA by the way...



Current Student
Joined: 28 Dec 2004
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I thought I already posted the OA...its A!







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