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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.
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.
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