Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Apr 2015, 00:56

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

who said PR quant is easy...man I got blown away after the [#permalink]  24 Jan 2005, 17:30
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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.
Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
Followers: 3

Kudos [?]: 270 [0], given: 22

fresinha12 can u restate Statement II
Director
Joined: 19 Nov 2004
Posts: 563
Location: SF Bay Area, USA
Followers: 3

Kudos [?]: 57 [0], given: 0

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

II) x =64
Ok
D
Manager
Joined: 11 Jan 2005
Posts: 101
Followers: 1

Kudos [?]: 0 [0], given: 0

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
Followers: 1

Kudos [?]: 0 [0], given: 0

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
Followers: 1

Kudos [?]: 0 [0], given: 0

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
Followers: 0

Kudos [?]: 10 [0], given: 0

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: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

The OA is A...go figure!
Manager
Joined: 11 Jan 2005
Posts: 101
Followers: 1

Kudos [?]: 0 [0], given: 0

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: 1442
Followers: 2

Kudos [?]: 20 [0], given: 0

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
Followers: 1

Kudos [?]: 0 [0], given: 0

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: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

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: 1442
Followers: 2

Kudos [?]: 20 [0], given: 0

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: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

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: 296
Location: California
Followers: 1

Kudos [?]: 2 [0], given: 0

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: 1442
Followers: 2

Kudos [?]: 20 [0], given: 0

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: 2250
Followers: 13

Kudos [?]: 216 [0], given: 0

Multiple implies integer multiple by definition, IIRC.
Intern
Joined: 26 Jan 2005
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

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?
Senior Manager
Joined: 25 Oct 2004
Posts: 250
Followers: 1

Kudos [?]: 6 [0], given: 0

I would go with D . whats the OA by the way...
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

I thought I already posted the OA...its A!

Go to page    1   2    Next  [ 21 posts ]

Similar topics Replies Last post
Similar
Topics:
Careers after Said 1 07 Feb 2014, 07:14
After 3 months of Studying, I got a 610 4 19 Oct 2010, 11:34
4 got 640 even after getting 47 in quant 4 25 May 2008, 00:32
2 The substance in the pile is gradually blown away by wind. 2 25 Jan 2008, 15:06
Is 750 98 or 99%? I got the score and it said 98% 2 04 Oct 2006, 18:27
Display posts from previous: Sort by