Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 09:04

### 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
Your Progress

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Current Student
Joined: 28 Dec 2004
Posts: 3361
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 16

Kudos [?]: 297 [0], given: 2

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

### Show Tags

24 Jan 2005, 18:30
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Followers: 7

Kudos [?]: 794 [0], given: 22

### Show Tags

24 Jan 2005, 19:15
fresinha12 can u restate Statement II
Director
Joined: 19 Nov 2004
Posts: 557
Location: SF Bay Area, USA
Followers: 4

Kudos [?]: 210 [0], given: 0

### Show Tags

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
Followers: 1

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

### Show Tags

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
Followers: 1

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

### Show Tags

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
Followers: 1

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

### Show Tags

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
Followers: 0

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

### Show Tags

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: 3361
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 16

Kudos [?]: 297 [0], given: 2

### Show Tags

25 Jan 2005, 08:25
The OA is A...go figure!
Manager
Joined: 11 Jan 2005
Posts: 101
Followers: 1

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

### Show Tags

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: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

### Show Tags

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
Followers: 1

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

### Show Tags

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: 3361
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 16

Kudos [?]: 297 [0], given: 2

### Show Tags

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: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

### Show Tags

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: 3361
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 16

Kudos [?]: 297 [0], given: 2

### Show Tags

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: 295
Location: California
Followers: 1

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

### Show Tags

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: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

### Show Tags

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: 2236
Followers: 16

Kudos [?]: 342 [0], given: 0

### Show Tags

26 Jan 2005, 14:59
Multiple implies integer multiple by definition, IIRC.
Intern
Joined: 26 Jan 2005
Posts: 11
Followers: 0

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

### Show Tags

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
Followers: 1

Kudos [?]: 29 [0], given: 0

### Show Tags

28 Jan 2005, 07:45
I would go with D . whats the OA by the way...
Current Student
Joined: 28 Dec 2004
Posts: 3361
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 16

Kudos [?]: 297 [0], given: 2

### Show Tags

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

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.