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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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?

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.

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.

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.