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

It is currently 29 Sep 2014, 22:03

Close

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

I am not able to find out whether y is perfect square. Is

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
User avatar
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 4 [0], given: 0

I am not able to find out whether y is perfect square. Is [#permalink] New post 22 Dec 2003, 08:12
I am not able to find out whether y is perfect square. Is the answer E ?
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 08:37
to be continuous perfect sq, ain't there sq roots need to be continuous intergers too?

D ??
Senior Manager
Senior Manager
User avatar
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 22 Dec 2003, 08:52
dj wrote:
to be continuous perfect sq, ain't there sq roots need to be continuous intergers too?

D ??


Can you please give example and elaborate ?
Manager
Manager
avatar
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

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

 [#permalink] New post 22 Dec 2003, 08:53
B

Actually I put answer to wrong question before. The answer shouldn't be C.

(1) More than one value of x and y that satisfy this condition.
(2) Way way many values of x and y that satisfy this condition.
(combined) x = 25, y = 16 (just another subset of what Stmnt2 gives us, so Stmnt2 is sufficient)

Last edited by wonder_gmat on 22 Dec 2003, 09:59, edited 1 time in total.
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 09:05
oops.. m sorry.
this should be C only.
Director
Director
User avatar
Joined: 14 Oct 2003
Posts: 592
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 22 Dec 2003, 09:47
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 10:37
i need to do this from the beginning...

A) in terms of Z:
Z^2 - 8Z - 1 = -Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = -5812. nope
for X=0, Z=0 --> Y = 1. Yes

A, not sufficient.

B) Z^2 -2Z +1 = Y
(Z-1)^2 = Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = 6400. Yes
for X=0, Z=0 --> Y = 1. Yes

B suffice.
furthermore, (sqrt(X)-1)^2 = Y says it all, ain't it?? X and Y will always be consecutive.

ans, B

....it's difficult to work on, this kinda stuff, from office :wink:

Last edited by dj on 22 Dec 2003, 11:03, edited 1 time in total.
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 801
Location: BULGARIA
Followers: 1

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

 [#permalink] New post 22 Dec 2003, 10:48
In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 10:58
BG wrote:
In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?


yes, (sqrt(X)-1)^2 or (Z-1)^2 = Y, will hold good, always.
CEO
CEO
avatar
Joined: 15 Aug 2003
Posts: 3470
Followers: 60

Kudos [?]: 670 [0], given: 781

 [#permalink] New post 22 Dec 2003, 11:27
dj wrote:
i need to do this from the beginning...

A) in terms of Z:
Z^2 - 8Z - 1 = -Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = -5812. nope
for X=0, Z=0 --> Y = 1. Yes

A, not sufficient.

B) Z^2 -2Z+1 = Y
(Z-1)^2 = Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = 6400. Yes
for X=0, Z=0 --> Y = 1. Yes

B suffice.
furthermore, (sqrt(X)-1)^2 = Y says it all, ain't it?? X and Y will always be consecutive.

ans, B

....it's difficult to work on, this kinda stuff, from office :wink:


x y z are positive integers.

and x>y

I think its D.

the only solution is x = 25 , y =16
CEO
CEO
avatar
Joined: 15 Aug 2003
Posts: 3470
Followers: 60

Kudos [?]: 670 [0], given: 781

 [#permalink] New post 22 Dec 2003, 11:31
Titleist wrote:
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B


titliest,

could you check the bold part. the solution does not equal 0.

x= 25 , y =16 is the only solution.
Director
Director
User avatar
Joined: 14 Oct 2003
Posts: 592
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 22 Dec 2003, 11:38
praetorian123 wrote:
Titleist wrote:
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B


titliest,

could you check the bold part. the solution does not equal 0.

x= 25 , y =16 is the only solution.


OOPS! Yes - you're right Praetorian = X=25, Y=16 is the only solution - I agree with DJ - it is tough doing this in the office. :shock:

I stand corrected again! The answer is still B - I was right by being wrong! As DJ pointed out, X=36, Z=6, Y=13 (This is a solution but X and Y are not consecutive squares).

Last edited by Titleist on 22 Dec 2003, 12:10, edited 2 times in total.
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 801
Location: BULGARIA
Followers: 1

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

 [#permalink] New post 22 Dec 2003, 11:42
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 11:43
what if X=36
Z=6

Z^2 - 8Z - 1 = -Y
36-48-1=-Y
Y=13
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 21 [0], given: 0

 [#permalink] New post 22 Dec 2003, 11:44
BG wrote:
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?


this, actually, is the first hit. but, without using two eqs also you can solve the problem.
Director
Director
User avatar
Joined: 14 Oct 2003
Posts: 592
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 22 Dec 2003, 12:24
BG wrote:
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?


No. This is a "Yes/No" problem. You have both yes and no for (1) and only yes for (2).
  [#permalink] 22 Dec 2003, 12:24
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic Confused whether I am ready or not msvel2304 3 20 Jun 2012, 05:36
Experts publish their posts in the topic If x and y are perfect squares, then which one of the shreya717 7 05 Jun 2012, 02:48
1 If x and y are perfect squares, which of the following is Celesta 2 03 Oct 2011, 09:49
Why am I not able to chat ? amitdgr 4 23 Oct 2008, 19:09
i am not been able to download the practise test aarti 1 07 Oct 2005, 23:28
Display posts from previous: Sort by

I am not able to find out whether y is perfect square. Is

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.