<p>I got a 90 on the AMC 10 and I need at least a 90 on the AMC 12 (which is harder). What do you guys suggest I do?</p>
<p>AMC is more of an aptitude test (the first step to IMO team). This means if you aren’t already naturally talented at math then you can only improve so much. But 90 is pretty close if you want to qualify for 10B so if you do some practice problems on the art of problems website you should be able to get those last few points. However AMC 12 is much more difficult, a student who gets 90 on AMC 10 will probably get around a 70 on AMC 12 so you will need ALOT of practice.</p>
<p>I’m in a similar situation as you. I got a 90 on the 10, and now I’m trying to get 100 on the 12. I actually got back up to a 90 on practice tests on the 12, but I’m having trouble getting to the 100.</p>
<p>Art of Problem Solving dot com has online forums and classes that are incredibly helpful. I’m a mom of a 3 time USAMO taker who is now at MIT. Best!</p>
<p>The AMC is not an aptitude test. It’s all hard work. Sure talented people can get off to a head start, but it is mostly hard work. I would recommend if you have some extra money to go one the aops website and order their volume 1.</p>
<p>@ appmaster</p>
<p>AMC does have problems that require techniques seen in previous tests but for the most part you need creativity to solve the problems. You can study for years and still be unable to complete even 80% of the test. This is why AMC is an aptitude test and studying does not always ensure a high score.</p>
<p>^ He’s right, it’s more of an aptitude test. The toughest problems aren’t solved by 1 or 2 formulas (although they might be PART of the solution). They are solved by realizing what to do and how to apply what you know to get the answer. Look at #14 from last year’s AMC 12 A. #14 is the last problem needed to score to guarantee a 100.5, and AIME qualification (assuming you get 1-14 right and omit 15-25).</p>
<p>"Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx? "</p>
<p>I solved this problem because I realized that b had to be greater than y(a), then plugged in the necessary values, 1-3, to find the probability. No amount of work from a textbook could tell you how to solve this problem.</p>