<p><a href="http://i475.photobucket.com/albums/rr116/watex/helpdude.png%5B/url%5D">http://i475.photobucket.com/albums/rr116/watex/helpdude.png</a>
HELP!</p>
<p>another 1 if interested : <a href=“http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/quick.png[/url]”>http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/quick.png</a></p>
![http://i475.■■■■■■■■■■■■■■■/albums/rr116/watex/quick.png[/url]](http://i475.photobucket.com/albums/rr116/watex/quick.png%5B/url%5D){kind=link}
<p>Is that first one B? because the ticks are equally spaced, and it goes from one side of B to the other and it goes from X to X+Y, so B represents a change of Y. I’m not suer but it made sense to me-</p>
<p>1st: E
2nd: A
So no.</p>
<ol>
<li><p>I think it’s E. (X+Y)/2 represents an average. So, (x+y)/2 is halfway between x and y, so y is at point E. Now, you may say: Wait, how can X+Y be where it is, then? Doesn’t an average have to be less than the simple sum? Not if one of the numbers is negative, which in this case it is. The zero mark is placed at point D, because the average has to be half of the sum, and the sum is two ticks away from the average, so to make the average the halfway mark, zero must go two points to the right of the average, or point D. Once you see this, you can see how everything fits. Confusing at first (seemingly impossible) so you kind of have to think outside the box and realize that this numberline includes negative numbers.</p></li>
<li><p>A. If this were in one dimension, the only possible answers are 1 and 11, neither of which are choices. So, you have to extend this into two dimensions and make a triangle (the two segments never had to be on the same line). By the triangle inequality (A+B > or = C, where A and B are the shorter legs and C is the longest leg) 10 is the only possible answer choice that works. Everything over 5+6=11 is too large.</p></li>
</ol>
<p>Two problems, seemingly typical at first, in which you have to think outside the box to solve. Me gusta estos problemas!</p>
<p>I spent a while on the first, but I used algebra (took me like 5 minutes because I realized that the points are arbitrary).</p>
<p>If you are as clueless as me, just use the trial and error method and put stuff together useing algebra. Note, I don’t get the above explanation explained by Gamma, so this is how I went about it…</p>
<p>t = a tick mark, it specifies that the ticks are all equal, then</p>
<p>x + 2t = x + y (2 tick marks away)
and
x+y+2t = x+y\2</p>
<p>so y= 2t, then x+2y = (x+y)/2
simplify x+3y = 0, x = -3y now test any point, I did y = 4, it doesn’t matter what you put in, as it’s arbritatry.</p>
<p>So that means x = -12, and the next point is -8 and the (x+y)\2 = -4 by plugging in the numbers. Each 2 spaces is 4 dfference, so continuing, -12,-8,-4,0,4 <- That’s on E.</p>
<h1>2.</h1>
<p>This question is all trial and error too. SAT only tests trial and error, not really “out side the box”. It’s whoever can reach a conclusion first from random trying.</p>
<p>The only two distances are 11 and 1 if A, B, and C lie on 1 plane, and none of those are listed. So make a triangle and use the pythagorean property A+B > C, so 6+5 > C, so only 10 works of the choices.</p>
<p>You might be interested about why a triangle. It’s because the 2 segments are connected by a point, I call it a hinge. If there’s ever something where 2 segments are connected to 1 point, it’s probably one of those pythagorean questions where the angle varies between the 2 segments.</p>
<ol>
<li><p>The way Gamma did it is most efficient</p></li>
<li><p>The longest possible length AC could be is 11 (when AC is a straight line composed of segments AB and BC). The only choice less than or equal to 11 is 10.</p></li>
</ol>
<p>All of them are good explanations, but here are mine. I’m an SAT tutor, so I hope I can explain it more friendly and easier to understand (not that other solutions weren’t good).</p>
<ol>
<li><p>Since (x+y)/2 > (x+y) [since it’s more to the right), we know that (x+y)/2 and (x+y), and x are all negative numbers. Keep that in mind.
Regardless of negativity, however, there is something that does not change: the distance.
The distance from (x+y) to (x+y)/2 is same as that of x to (x+y). So, we can establish this equation:
(x+y)/2 - (x+y) = (x+y) - x
Simplifying gives -3y = x. From here, cjgone, who arrived at the same result, decided to substitute. That’s fine. But, we don’t have to. Here’s why.
If we put -3y in place of x, then we get (x+y) = (-3y+y) = -2y. So, B becomes -2.5y. In other words, each tick mark is 1/2*y. So, using this simple logic, we see:
A = -4y
B = -2.5y
C = -1.5y
D = 0y (or 0)
E = y
E is the correct answer, and we didn’t have to substitute any value.</p></li>
<li><p>I disagree that this is a question of “trial and error.” It’s a clear problem with a clear method.</p></li>
</ol>
<p>WLOG (Without loss of generality), let’s establish segment AB such that AB = 5. Then, the locus of C with the distance of 6 is a circle with all possibilities for C. If C lies in the same segment as A, then we get AC = 1. Not the answer choice. If C lies in the diametrically opposite position, we get AC = 11. Another option is to consider if CAB = 90 to apply Pythagorean Theorem. Unfortunately, this does not give an integer value. We know that by the numbers given, AC is not 5, which would’ve formed an isosceles triangle.
So, what does this leave us? We are ended with the array of points scattered in the circle, and we can apply Triangle Inequality. Simple triangle inequality application gives one restriction of 11 > AC, so AC = 10.</p>
<p>You do have to try to find the values, but I wouldn’t call Triangle Inequality trial-and-error. And, in future, think about using locus in problems like this. It makes life a lot easier.</p>
<p>Regards,</p>
<p>For the first one…</p>
<p>I try to avoid algebra and use numbers any time I can. So I assumed that the tick marks were one unit apart and tried to make up numbers that fit. </p>
<p>It falls into place like this: </p>
<p>y has to be 2 (bc x+y is 2 units to the right of x)</p>
<p>x+y has to be a negative number that increases by 2 when you cut it in half…so it is -4</p>
<p>So x=-6 and y =2. Then, just find 2 on the number line counting up from the point where you knew x+y=-4</p>
<p>ok i understand the last 2 posts and i kinda understand gammas post for the 1st problem. im just gonna stick with gray falcons. is that the best/fastest one or is there a better 1 and i should try harder to understand that 1</p>
<p>The fastest way is to just know that the average between x and y is (x+y)/2. So, the point that is (x+y)/2, regardless of being negative or positive, will always be the midpoint of the segment from x to y. There, you’re done…</p>
<p>Granted, this doesn’t seem correct, due to the position of x+y, especially since we generally think of number lines without marks as being positive. But a way to confirm the negativity is to find zero. This is simple because (x+y)/2 will have to be halfway between zero and x+y, whether positive or negative. I think Grayfalcon’s is the most comprehensive, but not the fastest method. No algebra is really needed to solve this problem. Nor is guess and check or substitution needed either. </p>
<p>But I would, if I had time, just plug in numbers once I get the answer to see if things work out (which I did).</p>
<p>I hate that I had such a terrible geometry teacher… I am completely unfamiliar with what a locus is or with what the properties of the sides of a triangle are :. Does anyone know where I can review this stuff? Does Gruber’s have it?</p>
<p>Locus? I’ve never heard of the term before now, on either the math for the SAT I or the Math II subject test. Definitely don’t need it, far as I’m concerned. Math is about concepts, not formulas and tricks. If you know your math concepts (which you can get a good review with either Grubers or Barrons), you should be fine with anything that collegeboard throws at you. The math they give is extremely simple, simple enough so that no obscure terms like “locus” can mess up your score.</p>
<p>^Locus is a concept. If I recall, it is what you call a group of points. Example: what is the equation for the locus of all points 4 units away from (0,0)? (answer would be x^2 + y^2=16). I don’t recall if this terminology is used on the standardized tests, but now you know in case it is. I don’t know any good comprehensive math review sites. I generally recommend hippocampus for review stuff, but they do not have geometry. Just do a google search and I’m sure you can find some decent online reviews. I think a major advantage of reviewing geometry online is that there will (hopefully) be animated simulations, which are quite nice for more visual subjects.</p>
<p>You don’t need to know the term “locus.” At the rudimentary level, it’s an obvious observation.</p>
<p>If you put a point O and start to pick points that are x distances away from O, you will see that it will form a circle. So, this circle with center O and radius of x refers to all the points, or locus, that are x units away from O.</p>
<p>The term “locus,” in Latin, means “location.” It’s mentioned in many geometry textbooks, but you don’t really learn to use it much… It’s either in Circles or Construction pages. I learned it outside the classroom for competition-issues, so no worries if your high school teacher did not cover it.</p>
<p>Now, as far as to answer anyone’s question on finding geometry review (along with other math review), consider this website.</p>
<p>[Math</a> Study Guides - SparkNotes](<a href=“http://www.sparknotes.com/math/]Math”>http://www.sparknotes.com/math/)</p>
<p>Hope this helps,</p>
<p>Regards,</p>
<p>
</p>
<p>It’s trial and error because it doesn’t mention anything specific like, in a 2d plane etc… I really don’t find questions that leave out information for the solver to test out to show mathmatical skill. </p>
<p>It’s whoever notices meaningless things. Like when I did the question I first tested both 1 and 11, then I moved to the next possible choice. I consider that trial and error. I hate it-- the SAT should tell you that it’s a 2d plane. Some people consider this a test of intellect, but I beg to differ. I call it poorly worded, subpar english questions.</p>