One might reasonably conclude that with the addition of a methyl group, it would unequally affect the deshielding of the ortho, meta, and para protons. However, Toluene is one of those special cases in which the addition of substituent does not significantly affect the proton nmr of the benzene protons because the R group does not contribute well to any deshielding on the ring. In this case, all 5 protons on the benzene carbons register as one signal. It's one of the more esoteric exceptions to h nmr theory, and I wouldn't dwell too much on it, so don't read this question and spaz out about how you'll know when you get a question like this. Very likely you won't.