For example if x = 3/2 than 5y = 46, so y isn’t an integer…. That’s why the answer is E…
If (x^2) + 5y=49, is y an integer?
(1) 1 < x < 4
(2) (x^2) is an integer
I say combined together, they are sufficient, ....
....since the first statement, if x were either 2 or 3, y would be an integer, but we're not told that x is an integer so the fact that (x^2) could be fraction is possible. so INSUFFICIENT
... but the second statement tells us that x is an integer. alone, that's INSUFFICIENT..
.. but combined, we're basically told that x can either be 2 or 3 which makes 49-(x^2) a multiple of 5. So I chose C.
Please tell me why this way of solving the question is wrong since the answer is saying that it should be E.
OK, statement 1: x can be an integer, x can be afraction. really doesn't help.
Statement 2: Alone, this is insufficient, as x can equal sqrt2, sqrt3, 2, etc.
Combined: Still insufficient. If x were 2, yes, then y is an integer. If x = sqrt2, then y is not an integer.
Hi ak,
If (x^2) + 5y=49, is y an integer?
(1) 1 < x < 4
(2) (x^2) is an integer
Here's my explanation :
Q: x^2 + 5y=49
5y = 7^2 - x^2 => (7-x) (7+x)
y = (7-x) (7+x)/5
(1) 1 < x < 4 ,so x can be anything from 1.5, 2,2.5 ...so on upto 4
As you already know ,y is an integer when x is an integer 2 or 3 ,otherwise y is a fraction. So the data is insufficient.
(2) (x^2) is an integer :
Meaning x is sqrt "n" where n can be a number from 1,2,3,4....n. The answer is yes and no , that makes it insufficient too.
Combining statement (1) and (2) ,x is a sqrt and lies between 1 and 4. We are left with options like sqrt2 ,sqrt3 ,sqrt 4 ,sqrt 5...so on.
We know y = (7+sqrt2)(7-Sqrt2)/5 is not an integer.
We know y = (7+2)(7-2)/5 is an integer.
So the data is insufficient. There !!
Last edited by Salila; 11-15-2004 at 10:13 AM. Reason: smiley
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