# Thread: Need help on this one.

1. ## Need help on this one.

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.  Reply With Quote

2. ## Re: Need help on this one.

For example if x = 3/2 than 5y = 46, so y isnt an integer. Thats why the answer is E  Reply With Quote

3. ## Re: Need help on this one.

3/2 as in the fraction 3 over 2? But statement 2 satisfies the necessity for (x^2) of being an integer and not a fraction, no?  Reply With Quote

4. ## Re: Need help on this one. Originally Posted by ak78
3/2 as in the fraction 3 over 2? But statement 2 satisfies the necessity for (x^2) of being an integer and not a fraction, no?

yes.. I was thinking of 1.5... it's between 1 and 4 and 1.5 * 2 = 3 which is an integer... we know that x*2 is an integer and not necessary x is an integer... hope u understand what I want to say  Reply With Quote

5. ## Re: Need help on this one.

sorry --- when i typed x^2, i was using the format i usually used in my TI-82 or TI-85 calculator, i mean x squared.

sorry for the confusion. now do you see why I might be confused with the answer though?  Reply With Quote

6. ## Re: Need help on this one.

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.  Reply With Quote

7. ## Re: Need help on this one.

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 !!  Reply With Quote

8. ## Re: Need help on this one.

aHa.

geez. didn't think of sqrt2, sqrt3. thanks! i wish the frikkin test explanation itself would have said that.  Reply With Quote