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 500,000 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:
OA is B. But I got stuck after a while. Please see below my solution.
From question stem
z = \sqrt{x} i.e. x = \(z^2\) ---------------------------------------------(1)
Considering Statement 1
y = 8z-x+1 -----------------------------------------------------(2)
Putting some values of x and z
When x =2 , z =4
Substituting these values in Equation 2 we will get y = 13. But this cannot be applied as x >y. So I tried another values which are
x = 25 then z =5
Substituting these values in Equation 2 we will get y = 16. Therefore , YES x and y are perfect squares.
I tried one more value which was
when x = 36, z =6
Substituting these values in Equation 2 we will get y = 13. Therefore , NO x and y are NOT perfect squares. Two conclusions, therefore this statement alone is NOT sufficient.
Considering Statement 2
After simplifying we will get y = \((z-1)^2\) Therefore z-1 = \sqrt{y}
Re: Consecutive Perfect Sqaures [#permalink]
31 Jan 2012, 16:41
22
This post received KUDOS
Expert's post
8
This post was BOOKMARKED
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?
If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?
(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.
(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.
Re: If x, y, and z are positive integers, where x > y and z = √x [#permalink]
10 Mar 2012, 02:13
Isn't this question a good candidate for plugging in simple numbers? I plugged in x=4 and y=1 and got the answer pretty quickly. _________________
If you like it, Kudo it!
"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."
Re: If x, y, and z are positive integers, where x > y and z = √x [#permalink]
12 Jun 2013, 22:33
3
This post received KUDOS
enigma123 wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?
(1) x + y = 8z +1 (2) x – y = 2z – 1
Excellent explanation by Bunuel, but i will try to give some alternative solution. Since the question askes whether x and y are consecutive perfect squares, if yes they must be perfect numbers in all cases, so we can plug some numbers and check. While plugging in our task is to find numbers when x and y are consecutive and when they are not.
Lets take 4, 9, 16 and 25 - consecutive perfect squares.
(1) if x=9, y=4, z=3, then according to the statement we have 9+4=24+1 which is not true, in case of x=16, y=9 and z=4 we have 16-9=8-1 which is again true. Lets try x=25, y=16, z=5 so we have 25-16=10-1, true again. So in all cases have the same conclusions based on the statement. Sufficient - B.
Hope that helps.
(2) if x=9, y=4, z=3, then according to the statement we have 9-4=6-1 which is true, in case of x=16, y=9 and z=4 we have 16-9=+1 which is again not true. When plug the numbers it is advisable to check at least 3 numbers. So lets try x=25, y=16, z=5 so we have 25+16=40+1 - Bingo! In this case the statement supports that X and Y are consecutive. So we have two different conclusions based on the statement. Not sufficient. _________________
If you found my post useful and/or interesting - you are welcome to give kudos!
Re: If x, y, and z are positive integers, where x > y and z = √x [#permalink]
03 Sep 2013, 09:36
ziko wrote:
enigma123 wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?
(1) x + y = 8z +1 (2) x – y = 2z – 1
Excellent explanation by Bunuel, but i will try to give some alternative solution. Since the question askes whether x and y are consecutive perfect squares, if yes they must be perfect numbers in all cases, so we can plug some numbers and check. While plugging in our task is to find numbers when x and y are consecutive and when they are not.
Lets take 4, 9, 16 and 25 - consecutive perfect squares.
(1) if x=9, y=4, z=3, then according to the statement we have 9+4=24+1 which is not true, in case of x=16, y=9 and z=4 we have 16-9=8-1 which is again true. Lets try x=25, y=16, z=5 so we have 25-16=10-1, true again. So in all cases have the same conclusions based on the statement. Sufficient - B.
Hope that helps.
(2) if x=9, y=4, z=3, then according to the statement we have 9-4=6-1 which is true, in case of x=16, y=9 and z=4 we have 16-9=+1 which is again not true. When plug the numbers it is advisable to check at least 3 numbers. So lets try x=25, y=16, z=5 so we have 25+16=40+1 - Bingo! In this case the statement supports that X and Y are consecutive. So we have two different conclusions based on the statement. Not sufficient.
Solution is truly great but the way it has been explained is confusing !Also there are typos !
in the highlighted part above we are testing statement 2 so 16-9=8-1 = 7 which is true !
More simply :
1) x + y = 8z +1
Work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )
If z=4 ,x= 16 we get y = 17 cannot take this case as condition is that x>y ( for all positive z<5, y>x , hence we cannot take those cases ) If z=5, x=25,y=16, does satisfy, here x and y are consecutive perfect squares If z=6,x=36, y=13, also does satisfy, here x and y are not consecutive perfect squares.In fact y is not even a perfect square. If z=7, x=49,y=9 also does satisfy, here x and y are not consecutive perfect squares.
Two cases hence insufficient.
(2) x – y = 2z – 1
Again work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )
if Z=1 then x=1 on solving we get y =0 cannot take this case as y is supposed to be positive.
if Z= 2 then x=4 on solving and finding y we get y =1 , here 4 and 1 are consecutive perfect squares if Z=3 then x=9 then y = 4 , here also x and y are consecutive perfect squares if z=4 then x= 16 then y = 9 also satisfies the equation and x and y are perfect squares . ........ Take any positive integer for z and find x and y we will see both x and y are consecutive perfect squares Lets take z= 10 then x= 100 and we get y =81 here also x and y are consecutive perfect squares,
Re: If x, y, and z are positive integers, where x > y and z = √x [#permalink]
27 Oct 2014, 07:32
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
If x, y, and z are positive integers, where x > y and z = √x [#permalink]
27 Jun 2015, 07:40
1
This post received KUDOS
Can we just consider the consecutive perfect squares as \(n^2\) and \((n-1)^2\) and the difference could then be expressed as \((n^2)\)- \((n^2-2n+1)\)which is the same as 2n-1 which indicates that Statement 2 is sufficient?
OA is B. But I got stuck after a while. Please see below my solution.
From question stem
z = \sqrt{x} i.e. x = \(z^2\) ---------------------------------------------(1)
Considering Statement 1
y = 8z-x+1 -----------------------------------------------------(2)
Putting some values of x and z
When x =2 , z =4
Substituting these values in Equation 2 we will get y = 13. But this cannot be applied as x >y. So I tried another values which are
x = 25 then z =5
Substituting these values in Equation 2 we will get y = 16. Therefore , YES x and y are perfect squares.
I tried one more value which was
when x = 36, z =6
Substituting these values in Equation 2 we will get y = 13. Therefore , NO x and y are NOT perfect squares. Two conclusions, therefore this statement alone is NOT sufficient.
Considering Statement 2
After simplifying we will get y = \((z-1)^2\) Therefore z-1 = \sqrt{y}
But from statement 1 we can say z = \sqrt{x}
Therefore, \sqrt{x} - 1 = \sqrt{y}
\sqrt{x} = \sqrt{y} + 1
I am stuck after this. Can someone please help?
_________________
Kudos to you, for helping me with some KUDOS.
gmatclubot
Re: If x, y, and z are positive integers, where x > y and z = √x
[#permalink]
28 Jun 2015, 07:50
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...