<p>I transferred from a community college to UCLA. I graduated from my community college with a 3.56 grade point average, but, now, at UCLA, I believe I will be graduating with a 2.9. I understand that my grade point average doesn't quiet measure up to the competition, but how does my community college grade point average factor in? I'm not looking to apply to any ivy schools since I understand my grades are not what they look for. Therefore, I plan on applying to smaller, yet competitive, schools. For example, Loyola, St. Mary's, UCLA, USC, USD, UC Davis, & this one might be a huge long shot, but also UC Berkeley. I have high hopes, but I also want to know what kind of LSAT score should I be aiming for? I checked various score sites, but I would also like to hear it from people. Thank you in advance.</p>
<p>You say you “graduated” from your community college. Do you mean you got a degree, and if so was it a bachelors, or do you mean you had a 3.56 when you transferred?</p>
<p>When you graduate from a community college you receive an Associates (2- year) degree.</p>
<p>You gpa from your community college will be rolled in to your GPA from UCLA (hopefully raising your GPA to over 3.0)</p>
<p>Sybbie, that used to be the case, but these days many community colleges offer [bachelor’s</a> degrees](<a href=“Community Colleges Challenge Hierarchy With 4-Year Degrees - The New York Times”>Community Colleges Challenge Hierarchy With 4-Year Degrees - The New York Times). If it’s an associate’s it all gets rolled together, as you say. If it was a bachelor’s, however, it cuts off at that point.</p>
<p>Cal-Berkely is GPA-focused, and a 3.3 overall just won’t cut it without a big hook.</p>
<p>For the others, you’ll need a LSAT that exceeds their 75th %, because you will want them to give you money to attend.</p>
<p>Do not under any circumstances pay sticker to attend LS.</p>
<p>LOL, it’s clear that bluebayou has spent time on TLS :)</p>
<p>^^perhaps true, but unlike the TLS’ers, I understand UC admissions and what they look for, and more importantly, ‘why’…</p>