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# x^2 + 5y = 49. is y an integer? 1. 1 < x < 4 2. x^2 is

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Senior Manager
Joined: 18 Jun 2007
Posts: 274
x^2 + 5y = 49. is y an integer? 1. 1 < x < 4 2. x^2 is [#permalink]

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29 Sep 2008, 22:26
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x^2 + 5y = 49. is y an integer?

1. 1 < x < 4
2. x^2 is an integer

--== Message from GMAT Club Team ==--

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
VP
Joined: 17 Jun 2008
Posts: 1477
Re: DS: relation between x and x^2 [#permalink]

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29 Sep 2008, 22:39
E.

original equation can be translated to y = (49-x^2)/5.
In order for y to be an integer, numerator should either be 0 or a multiple of 5.

stmt1: insufficient. Take x = sqrt2, y is not an integer. Take x = 2, y is an integer.
stmt2: insufficient, for x^2 = 1, y is not an integer and for x^2 = 4, y is an integer.

Sombining 1 and 2, 1 < x^2 < 16.

Now, for x^2 = 4, y is an integer, for x^2 = 5, y is not an integer. Hence, insufficient.
Intern
Joined: 16 Jun 2008
Posts: 17
Location: Kiev, Ukraine
Re: DS: relation between x and x^2 [#permalink]

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30 Sep 2008, 02:37
I don't agree with E because from the first condition we know that 1<x<4 and from the second - that X^2 is an integer, which means that X is an integer as well. It means that X may be equal only to 2 or 3.

x^2+5y=49
5y=49-x^2
5y=(7-x)*(7+x)

If we substitute x with 2 or 3. we will get one multiplier 5 or 10, which means that (7-x)*(7+x) is divisible by 5, and in this case Y is an integer.
Senior Manager
Joined: 18 Jun 2007
Posts: 274
Re: DS: relation between x and x^2 [#permalink]

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30 Sep 2008, 03:03
1
nlutsenko wrote:
I don't agree with E because from the first condition we know that 1<x<4 and from the second - that X^2 is an integer, which means that X is an integer as well. It means that X may be equal only to 2 or 3.

x^2+5y=49
5y=49-x^2
5y=(7-x)*(7+x)

If we substitute x with 2 or 3. we will get one multiplier 5 or 10, which means that (7-x)*(7+x) is divisible by 5, and in this case Y is an integer.

nlutsenko, I made the same mistake cause i thought that if x^2 is an integer then x too has to be an integer. Reason: I could not think of a non integer number which can yield an integer when squared but that was a silly mistake on my part cause x can be 'root5' which will equal to '5' when squared.

This is a GMAT Prep2 question. Though I scored 740 with 9 mistakes in math section but 7 of them were such silly mistakes...this is the area of great concern for me...somehow my concentration is shakey.

btw OA is E. thx scthakur
Intern
Joined: 16 Jun 2008
Posts: 17
Location: Kiev, Ukraine
Re: DS: relation between x and x^2 [#permalink]

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30 Sep 2008, 04:12
Thank you for explanation. I really didn't take into consideration square root of an integer. Thank again
VP
Joined: 05 Jul 2008
Posts: 1333
Re: DS: relation between x and x^2 [#permalink]

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30 Sep 2008, 21:19
rishi2377 wrote:
nlutsenko wrote:
I don't agree with E because from the first condition we know that 1<x<4 and from the second - that X^2 is an integer, which means that X is an integer as well. It means that X may be equal only to 2 or 3.

x^2+5y=49
5y=49-x^2
5y=(7-x)*(7+x)

If we substitute x with 2 or 3. we will get one multiplier 5 or 10, which means that (7-x)*(7+x) is divisible by 5, and in this case Y is an integer.

nlutsenko, I made the same mistake cause i thought that if x^2 is an integer then x too has to be an integer. Reason: I could not think of a non integer number which can yield an integer when squared but that was a silly mistake on my part cause x can be 'root5' which will equal to '5' when squared.

This is a GMAT Prep2 question. Though I scored 740 with 9 mistakes in math section but 7 of them were such silly mistakes...this is the area of great concern for me...somehow my concentration is shakey.

btw OA is E. thx scthakur

Same mistake here. I was wondering why this its not C. The point we miss is x can be root 5 or root 8 and still be between 1 and 4.

I am slowly realizing one thing with all these number properties, inequalities. Bring all of them into one form such as x^2 here. I mean , the imp take away here is 1 < x < 4 means
1 < x^2 < 16 and the second one is x^2 is an integer. When we do the conversion we clearly see that x^2 can take the value of 2 through 15. other wise, we get trapped with 2 & 3.
VP
Joined: 17 Jun 2008
Posts: 1285
Re: DS: relation between x and x^2 [#permalink]

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01 Oct 2008, 21:56
rishi2377 wrote:
x^2 + 5y = 49. is y an integer?

1. 1 < x < 4
2. x^2 is an integer

Clearly E
1) x=2,3,or any fraction too hence answer is different
2)x^2 can be 2,3,4,1,5,6 ... answer is different in each case !!!

combine both we get x=2,3,4,5,etc hence different answers
IMO E

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Re: DS: relation between x and x^2   [#permalink] 01 Oct 2008, 21:56
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