<p>I was doing some more practice tests and got stumped on this problem:</p>
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<p>Anyone care to help explain? Thanks!</p>
<p>i’m thinking it’s E:
D = 0
and each tick mark is one unit, so x = -6
x + y = -4
(x + y) / 2 = -2
and y = 2
and E is at 2</p>
<p>This problem seems familiar. Did you get it from the PSATs 2008?</p>
<p>smeagz your answer is correct but there should be some pure algebraic solution.
In my opinion x and y are in arithmetic sequence and x+y = (x + x+y/2)/2</p>
<p>Well we know x, x+y, and (x+y)/2 are all negative because if they were positive, (x+y)/2 would be less than x+y (and hence to the left of it). If you experiment with the four rightmost points and try to figure out which one is zero, you find that D is 0. y must be positive since x+y is greater than x, so y is point E.</p>
<p>@Ivan: your reasoning supports y being equal to point E. x = -3y, and x = -6 and y = 2. -6 = -3(2).</p>
<p>Ok, I get it now…just a bit of guessing and checking I see…Thanks!</p>
<p>No, it was some online SAT released by CB I think :)</p>
<p>Right jamesford that’s how I did it, once you figure out D=0 you realize it has to be E because y is positive. :)</p>
<p>Here’s how I solved it.
Let 1 increment on the line = z.</p>
<p>x+2z = x+y
2z = y.</p>
<p>Next, looking at (x+y)/2,
we have x+y+2z = (x+y)/2
Rearranging,
2x+2y+4z = x+y
x+y+2y = 0
x+3y = 0.</p>
<p>Remember that y = 2z.
x+6z = 0.
Six increments to the right of x = 0 (point D).
Then assuming that z = 1, you have x = -6.
Because x+y = -4, then y = 2.</p>
<p>y is the point two increments to the right of D, which is
point E.</p>
<p>(x+y) - x = (x+y)/2 - (x+y)
y = -(x+y)/2</p>
<p>D is 0 because (x+y)/2 is a midpoint between (x+y) and D
E is opposite of (x+y)/2, so E is -(x+y)/2, that is E is y.</p>
<p>Here
x+4units=(x+y)/2
Solving this u get y=x+8units
which means y is at E</p>
<p>^wow! This is the ultimate shortcut. Kudos, ccprofile!</p>
<p>^wow! This is the ultimate shortcut. Kudos, ccprofile!</p>
<p>o_o</p>
<p>you can tell that
(x+y)/2 is a midpoint thingy.</p>
<p>since x is 0, and ((x+y)/2) is 4 over, then y must be 8 over.</p>
<p>no algebra needed :D</p>
<p>Thanks gcf :D</p>
<p>@anhtimmy
Your way is fantastic too.
What I like about ccprofile’s solution is that it gives a generic approach to this type of questions: find the offset, and you’re set.
Some people may have difficulty unferdstanding what you mean by “x is O”. I’ll take a liberty of expanding on your telegraphic style. lol</p>
<p>(x+y)/2 is a midpoint between x and y.
Since x is 4 units to the left of (x+y)/2, y is 4 units to the right of (x+y)/2, thus y is at E. </p>
<p>It’s amazing how some SAT questions look so simple in retrospect - but not in the middle of taking the test.</p>
<p>When you’ve never seen it before, it’s a problem. When you know how to do it, it’s an exercise. SAT math is supposed to be problem solving, thus it SHOULD feel new to you when you take the exam :)</p>
<p>i am so gonna fail</p>
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