<p>I've taken about 4 practice tests and I can break out of getting between a 610-690 in Math. I am looking at top schools and I really need to get my math score up. The problem is that its not like I am bad at math. I do really well in math in school. But on the test I always leave 1-2 blank and make at least 2 stupid mistakes.</p>
<p>I always get the level 4 or 5 questions wrong, usually they involve relating two variables (what is x in terms of y) and combination probability type things (if there are 6 pants and 6 sweaters and in each group one sweater is red, one is yellow, ect., and one pant is red, one is yellow, etc. how many combinations are there is each pant and each sweater must be a different color). On the latter I spend way to long drawing out the scenario because I don't know how to effectively solve it.</p>
<p>Thanks for helping me. I am aiming for a 750+ score in math so I need help. I have until the January SAT to prepare.</p>
<p>taking consecutive tests in a short period of time really doesn’t help you. especially if you have no idea what is tested on the test. apparently you have timing issues, and that’s because you over think certain problems or you are not familiar with the concepts that are tested on the math section. keep one thing in mind, all the math questions on the SAT are EASY because they don’t expect you work on each question more than 1 minute(unrealistic math), it’s advised for you to get a book specifically on SAT math even if you are good at math in school . only the wording part sometimes confuses most people and takes time to figure out.</p>
<p>i.e if i had 10 majors to choose from, and i can only choose 2, what are possibility of 2 majors i would choose?</p>
<p>the formula should come right to your head instead of figuring out the real math behind it.</p>
<p>10!/2!(10-2)! and i would get 45 as my answer.</p>
<p>For combinatorics problems, it’s always just being able to recognize the simplest solution. Sometimes it helps to use “complementary counting.” On more complicated counting/combo problems (e.g. how many of the first 1000 positive integers are multiples of 5 but not 7?) you could draw a Venn diagram illustrating each of the subsets.</p>
<p>For probability and combinatorics, I suggest the book Algebra and Trigonometry by Paul A. Foerster (no relation). Chapter 12 of this book is a great source, and likely to be better than all but the top hs teachers in terms of the quality of explanation.</p>
<p>You can probably buy it on Amazon, or if that is not feasible, your local library may have it, or be able to get it for you through inter-library loan.</p>
<p>If you have questions about it after reading it, please post them on this thread, and I will notice & try to explain.</p>
<p>This book will probably be good for the other issues, too.</p>
<p>hey rspence, just wondering, for your problem in (), cant we just take multiples of 5 from 1-1000(Arth. Prog.) and multiply by (6/7) cause every 7th multiple of 5 is of 7 too?</p>
<p>I would suggest maybe considering the ACT if you are good at Math. The questions are a lot more straightforward. On the negative side, the questions do test some Trig, but if you are good at trig, they are a cinch.</p>
<p>Actually, the easiest solution is to note that there are 200 multiples of 5, 28 multiples of 35, so there are 200-28 = 172 multiples of 5 that are not multiples of 7 (sorry this wasn’t the best Venn diagram example).</p>
<p>You could multiply 200*6/7, which is 171.42, but do you round up or down?</p>
<p>If you have n objects, and you wish to choose k of them, the number of ways to do this is nCk (n choose k) = n!/((n-k)!k!).</p>
<p>This is because we have n choices for the first object, n-1 for the second, etc., n-k+1 choices for the k-th object. The product of all these terms is n!/(n-k)!. We divide by k! due to over-counting, since the order which objects is chosen is irrelevant. That’s where the formula comes from…</p>
<p>10! means 10<em>9</em>8<em>7</em>6<em>5</em>4<em>3</em>2<em>1
Likewise n! is n</em>(n-1)<em>(n-2)…</em>3<em>2</em>1.
10!/2!(10-2)! is the formula for a ‘combination’. It’s basically like selecting a few things from a larger group, in such a way that order doesn’t matter.
For example, drawing 4 cards from a pack of 52.</p>
<p>someone else already mentioned this, but definitely think about using an ACT score to apply if you are strong in math but getting tripped up on the SAT. There is a major shift of higher education institutions moving to ACT…some ivy-league schools don’t even take SAT scores anymore. The SAT Subject Tests is how the college-board is trying to overcome this, since the ACT is much more indicative of what “you really know” as opposed to how good of a test taker you are.</p>
<p>If you need the SAT score, I’d really suggest you get back to basics (yes, even though you said you are good at math). This is 99 times out of 100 where mistakes happen…you forgot a positive or negative sign, you get your order operations out of order, etc. Silly, simple, “how could I mess that up???” type stuff. Seems easy enough, but it’s amazing how many people just try doing more practice exams and never change anything and can’t understand why their score stays the same. I learned this from a book I used to study before taking the SAT a second time called Secrets to Conquering the SAT ([Accepted</a>, Inc. - SAT, SAT study guide, SAT review, SAT Secrets, SAT tips | Accepted, Inc.](<a href=“http://acceptedinc.com/collections/sat]Accepted”>http://acceptedinc.com/collections/sat)). Highly suggest it, it helped me.</p>
<p>Yeah… I’ve yet to hear of any school rejecting SAT scores and only opting for ACT scores. The Ivies accept applications from too diverse a crowd to reject SAT scores.</p>
<p>There are schools that are testing optional, but that’s about as far as ‘SAT rejection’ goes.</p>