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abhi758
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 00:13
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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If x^2 + 5y = 49, is y an integer?

(1) 1 < x < 4
(2) X^2 is an integer

Why is the OA not A??



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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) Updated on: 29 Aug 2009, 03:55
Originally posted by CasperMonday on 29 Aug 2009, 01:58.
Last edited by CasperMonday on 29 Aug 2009, 03:55, edited 1 time in total.
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\(x^2+5y=49\)
\(5y=49-x^2\)

(1) we have no restrictions on \(x\) except that it is \(1>x>4\)
a) if x is an integer, it can be either 2 or 3
\(x=2\), \(5y=49-4=45\), hence, \(y=9\), y is an integer
\(x=3\), \(5y=40\), y is an integer
b) if \(x\) is not an integer, if \(x=\frac{1}{3}\), \(5y=49-\frac{1}{9}=48\frac{8}{9}\), \(y\) is not an integer
Hence, (1) not sufficient
(2) \(x^2\) is an integer means that\(x\)is an integer. Together with (1) it is suff.
Hence, C
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 03:36
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OA is not C.. its E any explanations??
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) Updated on: 29 Aug 2009, 06:15
Originally posted by CasperMonday on 29 Aug 2009, 04:00.
Last edited by CasperMonday on 29 Aug 2009, 06:15, edited 1 time in total.
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Ah! I get it!

Condition 2 that \(x^2\) is an integer means that
a) \(x\) is integer and
b) \(-infinity<x<+infinity\) (how do we write infinity, by the way?)

Then, from combined statements, we only know that \(x\) is an integer and \(1>x>4\). Hence, \(x\) is either \(2\) or \(3\). However, we are not given with any directions as to how to choose between the two possible solutions, hence E!!!! Hooray!
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 04:12
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abhi758
OA is not C.. its E any explanations??
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Ans should be C

OA might be wrong.
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 06:17
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CasperMonday
Ah! I get it!

Condition 2 that \(x^2\) is an integer means that
a) \(x\) is integer and
b) \(-infinity<x<+infinity\) (how do we write infinity, by the way?)

Then, from combined statements, we only know that \(x\) is an integer and \(1>x>4\). Hence, \(x\) is either \(2\) or \(3\). However, we are not given with any directions as to how to choose between the two possible solutions, hence E!!!! Hooray!
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I was wrong, the question asks about \(y\), not \(x\). For either \(x\), \(y\) is positive. Hence, I return to my previous answer. C is right.
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 12:15
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It is clear !!..
X can be 2 or 3..
only when we take condition 2 into account.
C is OA :arrow:
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 16:37
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CasperMonday
\(x^2+5y=49\)
\(5y=49-x^2\)

(1) we have no restrictions on \(x\) except that it is \(1>x>4\)
a) if x is an integer, it can be either 2 or 3
\(x=2\), \(5y=49-4=45\), hence, \(y=9\), y is an integer
\(x=3\), \(5y=40\), y is an integer
b) if \(x\) is not an integer, if \(x=\frac{1}{3}\), \(5y=49-\frac{1}{9}=48\frac{8}{9}\), \(y\) is not an integer
Hence, (1) not sufficient
(2) \(x^2\) is an integer means that\(x\)is an integer. Together with (1) it is suff.
Hence, C

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If \(x^2\) is an ineger DOES NOT imply that x is an integer. Consider for example, \(x=\sqrt{2}\) and \(x^2=2\).
The answer for this question is E.

Statement 1: say that \(1<x<4\) => \(1<x^2<16\). There are only three values of \(x^2\) between 1 and 16, for which y will be an integer. These values are 4, 9 and 14. But there are other values that will satisfy the inequality but y will not be an integer (ex.: 5,8,11, etc)

Statement 2: states that x^2 is integer...x can be an integer but it can be NOT an integer as well...

Combined, obviously not enough since statement 2 does not add much info..the only thing we know is that \(1<x^2<16\) and x can be {2,3,4,5,6,7,8,....15}.
Hence E is an answer
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 17:03
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Agreed with LenaA..
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 29 Aug 2009, 17:25
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The OA should be E.

Case 1: Take x = 2. Then \(x^2\) = 4. Then y = 45/5 = 9 (integer)
Case 2: Take x = \(sqrt(2).\)Then \(x^2\) = 2. Then y = 47/5 = 9.4 (not integer)
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 30 Aug 2009, 01:07
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Agree with the last 3. Answer should be E. In no way X^2 being an integer also implies that X is also an Integer.
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If x^2 + 5y = 49, is y an integer? (1) 1 < x < 4 (2) 30 Aug 2009, 02:20
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CasperMonday
\(x^2+5y=49\)
\(5y=49-x^2\)

(1) we have no restrictions on \(x\) except that it is \(1>x>4\)
a) if x is an integer, it can be either 2 or 3
\(x=2\), \(5y=49-4=45\), hence, \(y=9\), y is an integer
\(x=3\), \(5y=40\), y is an integer
b) if \(x\) is not an integer, if \(x=\frac{1}{3}\), \(5y=49-\frac{1}{9}=48\frac{8}{9}\), \(y\) is not an integer
Hence, (1) not sufficient
(2) \(x^2\) is an integer means that\(x\)is an integer. Together with (1) it is suff.
Hence, C

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If \(x^2\) is an ineger DOES NOT imply that x is an integer. Consider for example, \(x=\sqrt{2}\) and \(x^2=2\).
The answer for this question is E.

Statement 1: say that \(1 [m]1<x^2<16\). There are only three values of \(x^2\) between 1 and 16, for which y will be an integer. These values are 4, 9 and 14. But there are other values that will satisfy the inequality but y will not be an integer (ex.: 5,8,11, etc)

Statement 2: states that x^2 is integer...x can be an integer but it can be NOT an integer as well...

Combined, obviously not enough since statement 2 does not add much info..the only thing we know is that \(1<x^2<16\) and x can be {2,3,4,5,6,7,8,....15}.
Hence E is an answer
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what a trap you trapped :-)



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