<p>good test score ( at least 500>) could show her effort?
try ACT, it is slightly better, but if you are from SAT dominated area but hide SAT score, they might get suspicious.
or
try this trick some wiz told me, thou my kid can’t do commonsense math, so it was too hard for him</p>
<p>THE SAT TRICK
This strategy only works if you are fast with the calculator - you can’t use it on every problem because this method can be time consuming.</p>
<p>Before you start, read the problem and see it it’s primarily an algebraic problem (i.e. it gives you an equation and wants a straight-up answer, or it’s disguised as a word problem but needs algebra to be solved). If it isn’t, don’t use this method. If it is, try to see if you can solve the problem WITHOUT using this method. If it’s an easy problem and you know the answer for sure, don’t waste valuable time. If you still can’t get it, proceed on.</p>
<p>First, for each algebraic coefficient, pick a different “test case” number. While a test case number can technically be any number not equal to 0 or 1, it is advantageous to pick small numbers to reduce the size of the problem. Don’t assign the same test case number to separate coefficients - this may occasionally cause errors.</p>
<p>Once you select a test case number (or numbers), plug ‘em into the problem, replacing each instance of the variable with its assigned test number. Once you have done that, there should be no more remaining variables. Then, do the arithmetic and simplify the problem down to a single number. (i.e. if you plug in 3 to 4x-5, simplify it down to 4(3)-5 which equals 7.</p>
<p>At this point, you know what the “test case” problem results in. So, the last step is finding the correct answer “test case” solution, and therefore, the solution to the problem. Go through the multiple choice answers one by one, starting with A, and do the same thing you did above - plug in the test case numbers where appropriate and simply. The correct answer to the entire problem is the test case answer that is equivalent is the test case problem.</p>
<p>Sample problem:</p>
<p>“ x?4 - x?9 =” (? is the square root sign)</p>
<p>A. -5x
B. -x?5
C. -x
D. x
E. 3x</p>
<p>This is actually a really, really simple problem. The square root of 4 is 2 and the square root of 9 is 3, so all this is asking is “2x-3x =” </p>
<p>But, some people might have trouble recognizing it and might not be sure how to do it. Simple manipulation of the algebra makes it easy.</p>
<p>Plug 3 into the original problem, such that each instance of x = 3.</p>
<p>3?4 - 3?9</p>
<p>Just do the multiplication on the calculator (you still need to be able to recognize arithmetic operations, and the correct order they go in).</p>
<p>Simplification yields 6 - 9. The answer is -3.</p>
<p>Then, plug in 3 into all the choices that are given. The one that is equal to the answer you just got (-3) is the correct solution to the problem.</p>
<p>So it’s C), because plugging 3 into -X results in -3.</p>
<p>Sample #2</p>
<p>“There are K gallons of gasoline available to fill a tank. After d gallons have been pumped, in terms of k and d, what percent of the gasoline has been pumped?”</p>
<p>A) 100d/k %
B) k/100d %
C) 100k/d %
D) k/100(k-d) %
E) 100 (k-d)/k %</p>
<p>When you plug in numbers, make sure to use different ones. Try k = 4 and d = 2.</p>
<p>The hard part is figuring out what the problem’s equation should be.</p>
<p>It’s asking you what percent of the gasoline has been pumped. Just reading the problem, 2 if half of 4, so you’ve pumped 50%. Now, you just have to plug in k and d into all the possible answers until you find the one that equals 50%.</p>
<p>A is [100(2)/4]% = [200/4]% = 50%. So A is the correct answer here.</p>
<p>Now, the caveat is that this method is slower than conventional problem solving. So don’t use it on every problem because you will otherwise run out of time. But you should still use it on problems you don’t know.</p>