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If x^2+5y=49, is y an integer? a) 1<x<4 b) x^2 is an

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If x^2+5y=49, is y an integer? a) 1<x<4 b) x^2 is an [#permalink] New post 01 Feb 2004, 08:04
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C
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E

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If x^2+5y=49, is y an integer?
a) 1<x<4
b) x^2 is an integer
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Ans E [#permalink] New post 01 Feb 2004, 09:47
(1) X = 1.5 & X = SQRT(6) NOT SUFF
(2) X^2 = 4 & X^2 = 6 NOT SUFF
(1) + (2) NOT SUFF

Ans : E
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 [#permalink] New post 01 Feb 2004, 09:47
E ?

A) if x = 3 y is integer if x!=3 then y is not integer : so insufficient
B) if X^2 is integer then Y may or manot be integer
if X^2 is 9 then Y is integer ifX^2 != 9 not then Y is not an integer.
Combining both we cannot tell as well because X could be sqrt(3) or 3
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 [#permalink] New post 01 Feb 2004, 10:05
Answer is E.

But, if x is an integer, then C wud be correct. Wudnt it?

Anand- was x! a typo or u refering to x(x-1)(x-2) etc.
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 [#permalink] New post 01 Feb 2004, 11:01
Sorry mate. I come from computer background. x != means x not equal to. I will keep that in mind.
Yeah you are right if X were integer then C would be correct.
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 [#permalink] New post 02 Feb 2004, 11:48
How would that be correct.

If X in integer then C would be the answer. i think it still will be E

5y = 49-x^2

1. 1<x<4
2. X is an intger

X can be between 1 and 4.Just substitite the values and see

5y = 49 - (1^2) = 48 - Not a multiple of 5
5y = 49-2^2 = 45 - Multiple of 5

Please correct me where am i workng
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 [#permalink] New post 03 Feb 2004, 08:35
For 1<x<4, x cannot be equal to 1
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What about A? [#permalink] New post 09 Feb 2004, 08:23
A in itself is enough:

1< X < 4

X = 2 => y =9 integer!

X = 3 => y = 8 integer!

B is not enough

X*X is integer


what if X = 6 then X*X = 36 => 5y = 13 y is NOT integer!
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 [#permalink] New post 09 Feb 2004, 08:46
Answer is C.

1 .
If x is 2, then y is 9
if x is 3 then y is 8

2 x2 is integer so.
if x2 is integer then we have to take the value of x=2 and x=3

Hence Answer is C.
We need both together to answer the question
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 [#permalink] New post 10 Feb 2004, 03:40
If x is not an integer, y will not be an integer.
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 [#permalink] New post 11 Feb 2004, 00:12
Moma please consider sqrt(2) and sqrt(3) also. these are not integers but the square of these is integer. So you have 4 valuse for X within the limit 1<X<4. For X = 2 and 3 the eq. is integer but for X = sqrt(2) and sqrt(3) it is not an integer.
Hence the answer is E.
  [#permalink] 11 Feb 2004, 00:12
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