I agree. The ans ahould be C.
If x^2 + 5y = 49, is y an integer?
(1) 1 < x < 4
(2) x^2 is an integer
It's C because if you have both 1 & 2, you get 1 < x < 4 where x is an integer, so x must either be 2 or 3, and so y is an integer. However, the Kaplan explanation says if we choose C, then x can be either 2 or 3 OR 1. Wouldn't x only be 1 if we have 1 <= x <= 4, instead of 1 < x < 4 ??
Is this an error on Kaplan's part?
Actually Kaplan is right because from 1) you get that y=(49-x^2)/5 which may or may not be an integer
well, 2 says that x^2 is an integer, but what if x^2 = 3, when x = sqrt(3), 1< sqrt(3) < 4 which also satisfies 1) hence E
Even a stopped clock is right twice a day.
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