<p>I really doubt i'll get any but if I could get around 5000 (loans) to 7500 that would just help me. Alot. (or any amount actually)</p>
<p>Are there any scholarships for students who already attend USC and want to get it next yr? Last year, i didn't apply soon enough for the december deadline because i just read that the deadline was in january from collegeboard.com (sniff)</p>
<p>Does anybody know what i have to find more info and what forms i'll have to fill out (gah i have to go through the application stuff all over again don't I???)</p>
<p>Town and Gown! I'm getting 3500 this year, but I didn't demonstrate need-based financial aid so I might have gotten less than other applicants. It requires an application, due sometime around December, and an interview, which is February/March, if I remember correctly.</p>
<p>Here's the link to the official alumni scholarship page:</p>
<p>You can also ask your school/department. There IS merit-based aid for continuing students, especially if you get good grades & are active in school activities. Good luck! I imagine the Financial Aid Office might also be able to let you know of other opportuntiies for continuing students or refer you to how to get more info.</p>
<p>Above is the website with info about scholarships/grants for continuing USC undergrads. As I said, there MAY be money in the department/school you're studying in as well. I know there is in the School of Engineering & suspect it is true in other departments/schools as well.</p>
<p>Thanks XD any more links by any chance? I just hope i can get 5000 so i can cover my loans and 2500 would cover my work study (and i would still work so it would add on) i can't go over 10000 (i doubt i would ever get this much but) or my grant's gonna stat to go down. XP</p>
<p>If you're an architecture major by any chance, make sure to fill out the departmental scholarship application. They were very generous to me throughout my last three years, and it seems that they've added more scholarships for second years than there used to be.</p>