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

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28 Apr 2012, 23:50

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.

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