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Is X even?

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by anuptvm » Wed Dec 29, 2010 6:12 am
IF X and Y are positive integers, is X an even integer?

1. X(Y+5) is an even integer

2. 6y^2 + 41y +25 is an even integer


Stmt 1 is insufficient as X can be ODD or EVEN depending on Y+5 being ODD or EVEN

Stmt 2 6y^2 + 41y +25 is EVEN => 6y^2 + 41y is ODD (EVEN -ODD = ODD)

IF y is EVEN then 6y^2 = ODD => y^2 is not an integer

IF y is ODD then 6y^2 = ODD => y^2 is not an integer

Is my reasoning correct?
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Source: — Data Sufficiency |

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by Anurag@Gurome » Wed Dec 29, 2010 6:27 am
anuptvm wrote:IF X and Y are positive integers, is X an even integer?

1. X(Y+5) is an even integer
2. 6y^2 + 41y +25 is an even integer
Statement 1: x(y +5) is even integer.
If y is even, x must be even.
If y is odd, x may or may not be even.

Not sufficient.

Statement 2: 6y² + 41y +25 is an even integer.
=> (6y² + 41y) is odd
=> y(6y + 41) is odd
=> y and (6y + 41) both are odd

No information about x.

Not sufficient.

1 & 2 Together: y is odd from statement 2. From statement 1, x may or may not be even.

Not sufficient.

The correct answer is E.

Stmt 2 6y^2 + 41y +25 is EVEN => 6y^2 + 41y is ODD (EVEN -ODD = ODD)

IF y is EVEN then 6y^2 = ODD => y^2 is not an integer
IF y is ODD then 6y^2 = ODD => y^2 is not an integer

Is my reasoning correct?
Whether y is even or odd, 6y² is always even.
And what do you mean/conclude by saying "y² is not an integer"?
Last edited by Anurag@Gurome on Wed Dec 29, 2010 6:32 am, edited 1 time in total.
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by anshumishra » Wed Dec 29, 2010 6:29 am
anuptvm wrote:IF X and Y are positive integers, is X an even integer?

1. X(Y+5) is an even integer

2. 6y^2 + 41y +25 is an even integer


Stmt 1 is insufficient as X can be ODD or EVEN depending on Y+5 being ODD or EVEN

Stmt 2 6y^2 + 41y +25 is EVEN => 6y^2 + 41y is ODD (EVEN -ODD = ODD)

IF y is EVEN then 6y^2 = ODD => y^2 is not an integer

IF y is ODD then 6y^2 = ODD => y^2 is not an integer

Is my reasoning correct?
X, Y > 0 and Integer
Is X Even ?

Statement 1:
X(Y+5) = Even
Y+5 can be Even and then doesn't matter whether X is odd or Even, the multiplied value would be even -- Not Sufficient

Statement 2:
6y^2 + 41y + 25 = even
Even + 41y + Odd = even
=> 41y + Odd = even
=> 41y = odd
=> y = odd ---> Not sufficient as no info about x.

Combining Statement 1 and 2 :
y = odd
x(y+5) = even
=> x*even = even

So x can be either even or odd, doesn't matter.
Answer is E.
Thanks
Anshu

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by anuptvm » Wed Dec 29, 2010 6:47 am
Anurag@Gurome wrote:
Stmt 2 6y^2 + 41y +25 is EVEN => 6y^2 + 41y is ODD (EVEN -ODD = ODD)

IF y is EVEN then 6y^2 = ODD => y^2 is not an integer
IF y is ODD then 6y^2 = ODD => y^2 is not an integer

Is my reasoning correct?
Whether y is even or odd, 6y² is always even.
And what do you mean/conclude by saying "y² is not an integer"?
Thanks, my assumption was basically wrong.

IF y is even or odd then 6y^2 is always even.

I said y^2 is not an integer because I assumed that 6y^2 is an odd number and there is no such number which can be divided by 6 to give an integer
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