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If x+2/y = 3, is xy an integer? [#permalink]
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Updated on: 19 May 2013, 07:50
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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.Question made by me.
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Originally posted by mau5 on 07 May 2013, 00:45.
Last edited by mau5 on 19 May 2013, 07:50, edited 3 times in total.



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Re: If x+1/y = 3, is xy an integer? [#permalink]
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07 May 2013, 00:55
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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
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Re: If x+2/y = 3, is xy an integer? [#permalink]
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Updated on: 07 May 2013, 02:46
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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 noninteger 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 noninteger 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.
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Originally posted by manishuol on 07 May 2013, 02:22.
Last edited by manishuol on 07 May 2013, 02:46, edited 1 time in total.



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Re: If x+2/y = 3, is xy an integer? [#permalink]
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07 May 2013, 02:30
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 noninteger 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 noninteger 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
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Re: If x+2/y = 3, is xy an integer? [#permalink]
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07 May 2013, 02:33
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.
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Re: If x+2/y = 3, is xy an integer? [#permalink]
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07 May 2013, 02:44
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 noninteger 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 noninteger 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 !!
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Re: If x+2/y = 3, is xy an integer? [#permalink]
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07 May 2013, 02:46
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 noninteger 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 noninteger 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 !!..............
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Re: If x+2/y = 3, is xy an integer? [#permalink]
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11 Mar 2014, 05:42
we can rewrite 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.



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Re: If x+2/y = 3, is xy an integer? [#permalink]
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17 Apr 2014, 08:51
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
(1)+(2) give the same solution. E is my answer.
Please let me know what is wrong with my method.



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Re: If x+2/y = 3, is xy an integer? [#permalink]
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17 Apr 2014, 08:59
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. Hope it's clear.
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