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.

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+1/y = 3, is xy an integer? [#permalink]
06 May 2013, 23:55

2

This post received KUDOS

If x+\frac{2}{y}= 3,is xy an integer?

I. x is an integer less than 3 x=-1 y=0.5 -1+4=3 xy = -0.5 no integer x=2 y=2 2+1=3 xy=4 integer Not sufficient

II. x >-1 Clearly not sufficient. x can be any decimal value >-1

I and II) x is an integer between -1 and 3 ( not included) so x is 0,1 or 2 Case x=0 xy is an integer Case x=1 so y=1 xy integer Case x=2 so y=2 xy integer. Sufficient C

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If x+2/y = 3, is xy an integer? [#permalink]
07 May 2013, 01:22

1

This post received KUDOS

vinaymimani wrote:

If x+\frac{2}{y}= 3,is xy an integer?

(1) x is an integer less than 3 (2) x >-1

Kudos for correct solution

The answer is C for me as well......

as you can find out the value of 1 variable x or y if you have the value of one variable on hand. ..& then you can find out whether xy is an integer or not.

stmt 1... is clearly insufficient as it says that x<3.. & from this the value of xy can be non-integer or integer. Therefore, Insufficient stmt 2... is clearly insufficient as it says that x>-13..& from this as well the value of xy can be non-integer or integer. Therefore, Insufficient.

From 1+2 ...... we have a a definite range I,e.; -1<x<3. & for all the value of x in this range the solution of xy must be integer as well. Therefore, C is sufficient.

_________________

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

Last edited by manishuol on 07 May 2013, 01:46, edited 1 time in total.

Re: If x+2/y = 3, is xy an integer? [#permalink]
07 May 2013, 01:30

manishuol wrote:

The answer is C for me as well......

as you can find out the value of 1 variable x or y if you have the value of one variable on hand. ..& then you can find out whether xy is an integer or not.

stmt 1... is clearly insufficient as it says that x<3.. & from this the value of xy can be non-integer or integer. Therefore, Insufficient stmt 2... is clearly insufficient as it says that x>-13..& from this as well the value of xy can be non-integer or integer. Therefore, Insufficient.

From 1+2 ...... we have a a definite range I,e.; -1<x<3. & for all the value of x in this range the value of y must be integer as well. & as we know Integer * Integer = Integer. Therefore, C is sufficient.

Hi manishuol, your answer is correct. But not your solution

Statement 1 says that x is an integer, read carefully

From 1+2 it not true that we have int*int. If x=0 then y is not an integer. xy is an integer because x is 0.

Hope this clarifies

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If x+2/y = 3, is xy an integer? [#permalink]
07 May 2013, 01:33

Expert's post

manishuol wrote:

vinaymimani wrote:

If x+\frac{2}{y}= 3,is xy an integer?

From 1+2 ...... we have a a definite range I,e.; -1<x<3. & for all the value of x in this range the value of y must be integer as well. & as we know Integer * Integer = Integer. Therefore, C is sufficient.

Was about to post the same thing as Zarrolou has already mentioned above.No need now.

Re: If x+2/y = 3, is xy an integer? [#permalink]
07 May 2013, 01:44

Zarrolou wrote:

manishuol wrote:

The answer is C for me as well......

as you can find out the value of 1 variable x or y if you have the value of one variable on hand. ..& then you can find out whether xy is an integer or not.

stmt 1... is clearly insufficient as it says that x<3.. & from this the value of xy can be non-integer or integer. Therefore, Insufficient stmt 2... is clearly insufficient as it says that x>-13..& from this as well the value of xy can be non-integer or integer. Therefore, Insufficient.

From 1+2 ...... we have a a definite range I,e.; -1<x<3. & for all the value of x in this range the value of y must be integer as well. & as we know Integer * Integer = Integer. Therefore, C is sufficient.

Hi manishuol, your answer is correct. But not your solution

Statement 1 says that x is an integer, read carefully

From 1+2 it not true that we have int*int. If x=0 then y is not an integer. xy is an integer because x is 0.

Hope this clarifies

Thanks !! but I know this ... & that's how I reached the solution .... oderwise I would have not get the answer .... Anyways thanks for pointing out my typo...... Appreciate that. Thanks !!

_________________

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

Re: If x+2/y = 3, is xy an integer? [#permalink]
07 May 2013, 01:46

Zarrolou wrote:

manishuol wrote:

The answer is C for me as well......

as you can find out the value of 1 variable x or y if you have the value of one variable on hand. ..& then you can find out whether xy is an integer or not.

stmt 1... is clearly insufficient as it says that x<3.. & from this the value of xy can be non-integer or integer. Therefore, Insufficient stmt 2... is clearly insufficient as it says that x>-13..& from this as well the value of xy can be non-integer or integer. Therefore, Insufficient.

From 1+2 ...... we have a a definite range I,e.; -1<x<3. & for all the value of x in this range the value of y must be integer as well. & as we know Integer * Integer = Integer. Therefore, C is sufficient.

Hi manishuol, your answer is correct. But not your solution

Statement 1 says that x is an integer, read carefully

From 1+2 it not true that we have int*int. If x=0 then y is not an integer. xy is an integer because x is 0.

Hope this clarifies

Thanks !! Typo Edited !!..............

_________________

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

Re: If x+2/y = 3, is xy an integer? [#permalink]
11 Mar 2014, 04:42

we can re-write the equation as: x = 3 - (2/y)

1) x is an integer less than 3: So (2/y)>0. The answer is Yes when y = 2, 1/3 and answer is no when y = 1/4, 1/8. Satisfies both Yes & No. Insufficient. 2) x > -1: (2/y) < 4. Since nothing has been mentioned about x, it can be either integer or non integer. Answer is yes y<(-1/2), y>(1/2) and answer is no for all other values of (-1/2)<y<(1/2) except zero. Insufficient.

combining: Answer is yes when it satisfies both[y = 2, 1/3 & y<(-1/2), y>(1/2) ] Only value satisfying is y = 2 when x = 2. Sufficient.

Answer: C

Tried to attempt somewhat different method. Kindly comment.

Re: If x+2/y = 3, is xy an integer? [#permalink]
17 Apr 2014, 07:51

I really understand this, why my method doesn't solve this problem

x + 2/y = 3 so, xy + 2 =3y so, xy= 3y - 2

We need to know if xy is integer.

(1) x is an integer less than 3 xy = 3y -2 So, this depends upon value of y, y=0.1, xy= -1.7 ( Not an integer) y=1, xy=1 (Integer) Not sufficient (2) x >-1 xy = 3y -2 So, this depends upon value of y, y=0.1, xy= -1.7 ( Not an integer) y=1, xy=1 (Integer) Not sufficient

Re: If x+2/y = 3, is xy an integer? [#permalink]
17 Apr 2014, 07:59

Expert's post

umeshpatil wrote:

If x+\frac{2}{y}= 3,is xy an integer?

(1) x is an integer less than 3 (2) x >-1

I really understand this, why my method doesn't solve this problem

x + 2/y = 3 so, xy + 2 =3y so, xy= 3y - 2

We need to know if xy is integer.

(1) x is an integer less than 3 xy = 3y -2 So, this depends upon value of y, y=0.1, xy= -1.7 ( Not an integer) y=1, xy=1 (Integer) Not sufficient (2) x >-1 xy = 3y -2 So, this depends upon value of y, y=0.1, xy= -1.7 ( Not an integer) y=1, xy=1 (Integer) Not sufficient

(1)+(2) give the same solution. E is my answer.

Please let me know what is wrong with my method.

The point is that the first statement says that x is an integer. So, you cannot randomly pick a value for y substitute into xy= 3y - 2 and check. You must ensure that x will be an integer for y you choose.

Also, notice that when we consider both statements together x can only be 0, 1, or 2. For each of those values xy will be an integer.