ak78 Posted November 14, 2004 Share Posted November 14, 2004 If (x^2) + 5y=49, is y an integer? (1) 1 (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. Quote Link to comment Share on other sites More sharing options...
maria1_03 Posted November 14, 2004 Share Posted November 14, 2004 For example if x = 3/2 than 5y = 46, so y isn’t an integer…. That’s why the answer is E… Quote Link to comment Share on other sites More sharing options...
ak78 Posted November 14, 2004 Author Share Posted November 14, 2004 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? Quote Link to comment Share on other sites More sharing options...
maria1_03 Posted November 14, 2004 Share Posted November 14, 2004 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 Quote Link to comment Share on other sites More sharing options...
ak78 Posted November 14, 2004 Author Share Posted November 14, 2004 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? Quote Link to comment Share on other sites More sharing options...
gmat168 Posted November 14, 2004 Share Posted November 14, 2004 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. Quote Link to comment Share on other sites More sharing options...
Salila Posted November 15, 2004 Share Posted November 15, 2004 Hi ak, If (x^2) + 5y=49, is y an integer? (1) 1 (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 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 !! Quote Link to comment Share on other sites More sharing options...
ak78 Posted November 15, 2004 Author Share Posted November 15, 2004 aHa. geez. didn't think of sqrt2, sqrt3. thanks! i wish the frikkin test explanation itself would have said that. Quote Link to comment Share on other sites More sharing options...
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