<p>my friend said she did the whole math section with no calculator... and she got a 740. I am using a calculator and barely scraping 600's... THERE MUST BE A WAY FOR ME TO SCORE HIGHE RON MATH! Please please please please help me! Give me advice! Share your wealth of knowledge!!</p>
<p>I suggest you practice a lot.</p>
<p>Learn from your mistakes.</p>
<p>except for some i dont even know HOW to practice bcoz i flat out dont know the answers or how to even take a stab at the question!</p>
<p>Think about it this way: the math test questions are designed so that you don't have to use a calculator on them. Technically, you should be able to get an 800 without a calculator (though a calculator will definitely be useful).</p>
<p>If you're having trouble with a question, ask yourself, "how would I tackle this question if I didn't have a calculator with me?" Looking at it from a different viewpoint may help you get the answer.</p>
<p>Also, get a prep book and practice hard. The SAT tends to ask the same kinds of questions on every test (or so I've heard). If you don't know how to "take a stab at the question", look at the answer explanations and see how they are worked out. They will definitely help you with similar questions in the future. Good luck!</p>
<p>I suspect that the problem is you are doing what I once did - you are trying to use the calculator to solve the problems.</p>
<p>Calculators can not solve math problems.</p>
<p>Calculators can be used to solve equations, graph functions, give you the prime factorization of a number, factor a polynomial, and maybe a few other useful things.</p>
<p>But that's it. And notice that half of the things a calculator can do that I described above are not even seen on the SAT. What I'm trying to get at is, you really don't need a calculator. In fact, a calculator is often an impediment more than a help on the SAT, because instead of focusing on solving the problem, you will distract yourself by attempting to punch keys into a calculator. On the SAT, the only thing you should ever need a calculator for is:
finding the prime factorization a large number
multiplying or adding two big numbers (which rarely occurs)</p>
<p>SOMETIMES a calculator is useful for solving an equation, but this is RARELY so. Often the questions on the SAT that call for something as simple as solving an equation, are so simplistic that you can solve it in your head faster than you punch the keys into a calculator. Example:</p>
<p>if x+2 = 25, then what does 2x+2 equal?</p>
<p>Trying to do this problem on a calculator is a waste of time.</p>
<p>Post an example of the kind of problem you can't even take a stab at. We'll try to explain to you how you can do it without a calculator. It is not a joke when people say that you can do the whole SAT without one. The better you get at math, the less need you will have for a calculator.</p>
<p>A calculator is not necessary.</p>
<p>try doing this without a calc: 4^8*5/2</p>
<p>To celebrate a collegue's graduation, the {m} coworkers in an office agreed to contribute a catered lunch that costs a total of {y} dollars. If {p} of the coworkers fail to contribute, which of the following represents the additional amount, in dollars, that wach of the remaining coworkers must contribute?</p>
<p>a) y/m
b)y/(m-p)
c) py/(m-p)
d) y(m-p)/m
e)py/m(m-p)</p>
<p>"To celebrate a collegue's graduation, the {m} coworkers in an office agreed to contribute a catered lunch that costs a total of {y} dollars. If {p} of the coworkers fail to contribute, which of the following represents the additional amount, in dollars, that each of the remaining coworkers must contribute?"</p>
<p>This is the perfect example of an SAT problem for which a calculator is completely unhelpful, perhaps even absolutely useless.</p>
<p>This question, which at first seems very abstract and complicated, as though it requires some profound hereditary mastery of numerical properties, is, at its heart, little more than a 5th grade exercise in fractions.</p>
<p>Let's start with what we know. </p>
<p>We are asked to find the extra amount each member must pay as a result of p members not paying. We are given the total, y, that there are m members, p not paying.</p>
<p>how much do all members pay assuming everyone pays equally? Well, of course it would be the total cost, divided by the number of members.</p>
<p>which, using y, m and p, equals</p>
<p>y/m</p>
<p>Now, what is the cost for each member, assuming p members fail to pay?
Well, it's the total cost, divided by m members, but minus p who fail to pay, so that value is</p>
<p>y/(m-p)</p>
<p>So we easily can find the per-member cost of what m members must pay, and we can clearly find the total cost of what (m-p) members must pay. Now for the $1,000,000 question:</p>
<p>What is the DIFFERENCE between m members must pay, and what (m-p) members must pay?</p>
<p>Well, if you're millionaire material, and I'm sure you are, you are probably saying:</p>
<p>"Well gee, wouldn't it have to be what m members must pay minus what m-p members must pay?"</p>
<p>Real close! Keep in mind though that what (m-p) members must pay is actually MORE than what m members must pay. So it's actually the other way around: what (m-p) members would pay - what m members would pay. The other way would give you a negative number, although its absolute value would be correct, and you would still be able to get the correct answer if you did it either way.</p>
<p>So now that you've made that brilliant insight, you have the following equation:</p>
<p>difference in costs as a result of p members not paying = y/(m-p) - y/m</p>
<p>You're thinking 1 of 2 things right now: "Wow, I had no idea this was that easy" OR "I don't see how the hell that helps me." Don't feel guilty if you're thinking for the latter and not the former; I was once in your boat.</p>
<p>Notice we have two fractions to subtract. Whether we're working with variables or numbers, we do it the same way. Get a common denominator!</p>
<p>So we have:</p>
<p>y/(m-p) - y/m... getting a common denom gives us
y<em>m/(m-p)</em>m - y<em>(m-p)/(m-p)</em>m</p>
<p>That might look a little freaky in "netmath" but if you write it on paper you will definitely understand. All I did was multiply the first expression by m and the second expression by (m-p) so that they would have a denominator of m*(m-p). Now we have a common denominator. </p>
<p>So our numerator is:
y*m - y(m-p)</p>
<p>distributing the second y gives us</p>
<p>y<em>m - ym - yp
the y</em>m's cancel, giving us
- - yp</p>
<p>(notice how, even though the y*m's canceled out, the negative sign between them did not! we should maintain it, although like I said above, if you messed up here and kept it as negative yp, your answer would still be correct just with a sign difference)</p>
<p>since a double negative is a positive, our numerator has been reduced to simply: yp.</p>
<p>And our denominator is m*(m-p).</p>
<p>So our answer is yp/m(m-p).</p>
<p>Which is answer choice E.</p>
<p>And now that I've shown you the whole nine yards, there's actually a fast-track approach if you're not too comfortable with your algebra skills. </p>
<p>At the step where you got:</p>
<p>y/(m-p) - y/m</p>
<p>you could just assign arbitrary values to y, m and p, solved that expression, and then evaluated the answer choices using the same values for y, m and p, and find which one gives you the same answer</p>
<p>For instance, I did:
y = 10, m = 5, p = 3
so I got 3</p>
<p>checking the answers, when I got to E, I got 3<em>10/5</em>(2) = 3.</p>
<p>As you can see, no calculator is needed no matter how you do it. And while the second approach may seem faster, the first is actually just as quick for if you've practiced.</p>
<p>I sincerely believe that the SAT Math section is designed so that two types of people will get two types of scores: 1) people who basically do good at math in school but who do not "think" mathematically on a daily basis who and over-rely on their calculators will get AVERAGE scores (500's-600's) and 2) people who REALLY know how to THINK mathematically and don't need a calculator to reason through things will get 700-800. I think this kind of problem demonstrates that.</p>
<p>Feel free to post more questions.</p>
<p>The first way was better than the whole number thing; that's how I did it. Amazingly, the first one took less time than the second!!</p>
<p>"try doing this without a calc: 4^8*5/2"</p>
<p>163840
no calc ~</p>
<p>4^8 * 5 / 2 = 2^16 / 2 * 5 = 2^15 * 5</p>
<p>2^15 = 2^10 * 2^5 = 1024 * 32 (easy to get w/o calc) = 32768 </p>
<p>32768 * 5 = 32768 * (1/2) * 10 = 16384 * 10 = 163840</p>
<p>Sure, but that's basic arithmatic. Most people know how to do that, what about some of the harder questions??</p>
<p>the more difficult problems aren't straight calculation like "4^8 * 2.5"</p>
<p>that's what i said</p>