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:

Re: If x is an integer, is y an integer? [#permalink]

Show Tags

28 Apr 2012, 22:50

1)the average arithmatic mean is (x+y+y-2)/3 = x which simplies to x= y-1 => y is an integer. 1) if x=2 , y=1 mean = 1.5 not an integer, x=2 , y=2.5 mean = 2.25 not an integer => not sufficient

(1) The average (arithmetic mean) of x, y and y – 2 is x --> \(\frac{x+y+(y-2)}{3}=x\) --> \(y=x+1=integer+integer=integer\). Sufficient.

(2) The average (arithmetic mean) of x and y is not an integer --> if \(x=1\) and \(y=2\) the answer is YES but \(x=1\) and \(y=\frac{1}{2}\) the answer is NO. not sufficient.

Guys As i try addressing statement #1, my approach was: x+y+y-2 /3 = x thus, x+2y-2 = 3x i.e. ( integer + integer - integer = integer) Now we know that 2y is an integer however, y may or may not be an integer eg: y=2.5, thus 2y = 5 (integer)

I therefore concluded that option E ( neither statements make sense) Obviously i am wrong. Can some one tell me where my thought process is flawed?

from stmt 1: x+y+y-2 =3x or 2y-2=2x or y=x+1. since X is an integer Y = X+1 will be an integer. Sufficient.

from stmt 2:

let x=2 and y=1 then mean of x+y si 1.5 which is not an integer, while y is an integer. let x=2 and y=1.4 then mean of x and y is 1.7 which is again not an integer, and y is not an integer.

Guys As i try addressing statement #1, my approach was: x+y+y-2 /3 = x thus, x+2y-2 = 3x i.e. ( integer + integer - integer = integer) Now we know that 2y is an integer however, y may or may not be an integer eg: y=2.5, thus 2y = 5 (integer)

I therefore concluded that option E ( neither statements make sense) Obviously i am wrong. Can some one tell me where my thought process is flawed?

What is wrong in the logic above is that for 2y=5, we get x = 1.5,a non-integral value for x, which is not valid for the given problem.

From F.S 1, we know that (x+y+y-2)/3 = x

or 2x = 2y-2 or y = x+1. Thus, as x is an integer, y IS an integer.Sufficient.

From F.S 2, for x=1,y=2, we get the mean as a non-integral value. Thus, we get a YES for the question stem. However for x=1, y=2.5 also, we get a non-integral value of mean and a NO for the question stem. Insufficient.

1. The average (mean) of x, y & y-2 is equal to x. 2. The average (mean) of x and y is not an integer.

For me, Just follow the basics.....

Is Y an integer ?? when x is an Integer ...

As Given in statement 1 , The average (mean) of x, y & y-2 is equal to x .

Therefore, \(\frac{x+y+y-2}{3}=x\) \(\Rightarrow\) y-1 = x .

\(\Rightarrow\) y = x+1 & as we know x is an integer & Integer + Integer = Integer.

Hence, Y is an Integer.

Statement 2 :: The average (mean) of x and y is not an integer. This clearly means that the value of Both x & Y can be integer or non-integer to yield a Non- integer as their average.

Hence, A alone is sufficient.
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

mdbharadwaj, that's exactly the solution given in the QR. However, Whats wrong in my line of analysis of stmt 1?

Vinay has succinctly put it in his post. your assumption of values for Y is going against the information given in the question stem. X has to be an integer.

So i guess it will be easier and correct if you assume values of X first instead of Y,

Guys As i try addressing statement #1, my approach was: x+y+y-2 /3 = x thus, x+2y-2 = 3x i.e. ( integer + integer - integer = integer) Now we know that 2y is an integer however, y may or may not be an integer eg: y=2.5, thus 2y = 5 (integer)

I therefore concluded that option E ( neither statements make sense) Obviously i am wrong. Can some one tell me where my thought process is flawed?

From Stmnt 1 we can conclude easily that y-2, x and y are consecutive integers since mean =x --> x =y-1 Therefore sufficient.

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...