DS: If y is an integer

[cesar.rodriguez.blanco]

If y is an integer and y=|x|+x, is y=0?
1) x<0
2) y<1

I do not understand why the solution is D.
It is obvious that 1) is SUF, but what about 2)?

[gulatin2]

As you wrote , Statement 1 is sufficient.

Stmt 2 is y < 1 , since y is an integer , it can be 0 only if it is less than 1.

So Stmt 2 is sufficient too.

Both statements are sufficient enough to deduce if y = 0 .

Answer is D

[vrajesh.dave]

the explaination does not make sense. Y could be a negative interger, such as -1, -2, etc...

Please explain..

[kevinmarmstrong]

Remember that y= x +|x|, i.e. y= 2x if x>0 and y = 0 otherwise

[vrajesh.dave]

I do agree with y = x + |x|
then y = 2x if x > 0
and y = 0 if x < 0

However, I do not understand why
2) y < 1 is sufficient

since y is an integer and integer can be negative or positive

i.e. -10, ..., -2, -1, 0

So how can 2) be Sufficient.

Please clarify why 'y' can only be 0

[himadribora]

For Y to remain an integer, X must be an integer or have .5 to it i.e. 0.5, 1.5, 2.5, whether positive or negative.

Stmt 2 says that Y<1 i.e. it can be 0, -1, -2, -3. But for Y to be less than 1, x has to be negative. Because positive x will invariably give a result of 1 or higher. If x is negative, then it will never let Y to be anything else but 0. So X < 0.

We can rephrase the question stem to Is x < 0, which statement 2 answers. Hence, Stmt 2 is SUFFICIENT

[vrajesh.dave]

Thanks!
it makes sense now

since Y = |x| + x
Stmt 2 == Y < 1

Hence, X has to be -0.5 < X < 0.5

Therefore stmt 2 is sufficient.

[RonPurewal]

If y is an integer and y=|x|+x, is y=0?
1) x<0
2) y<1

I do not understand why the solution is D.
It is obvious that 1) is SUF, but what about 2)?

the posters above me have submitted algebraic (or at least pseudo algebraic) solutions - which are workable - so i'll just weigh in with a non-algebraic, NUMBER PROPERTIES-based solution.

the expression |x| is:
* EQUAL to x, if x is positive (or 0)
* the OPPOSITE of x, if x is negative (or 0)
(note that 0 fits into both these categories, because its equal and opposite are the same.)

therefore, x + |x| must be positive (if x is positive) or zero (if x is zero or negative).

therefore, if this quantity is an integer less than 1, it MUST be 0 (since negative possibilities are out of the question).
so, "yes" to the question.
sufficient.