<p>Does anyone remember having the problem</p>
<p>ax+by=c
y=mx+k</p>
<p>how do you find K in terms of the other variables?</p>
<p>This one stumped me and I could not take my mind off of it the rest of the test!</p>
<p>And is that problem that i mentioned in my most recent post, was that experimental, or no?</p>
<p>Hi all, to the question about the pentagons, here is how it was solved (for me):</p>
<p>They gave you that the angle from one point on Pentagon I to Pentagon II was 54 degrees. You know that the inner angle of a pentagon must equal 72, therefore the remaining angle distance between the point from Pentagon II to the nearest point of Pentagon I must be 18 degrees. If you solved correctly, the distance to the point they were asking for was 3 “18 degree angles” and 2 “54 degree angles”. </p>
<p>Therefore 3(18) + 2(54) = 162 degrees.</p>
<p>You could also tell that this answer is correct since 180 degrees from the original point on Pentagon II was not too far off from the designated point.</p>
<p>How did you find the area of the parallelogram if the area of the square was k? That one stumped me and I rushed the other ones.</p>
<p>@Darthpwner</p>
<p>The easiest way was to eyeball it, you know that the diamond consists of two isosceles triangles, and there were 2 on each side, and one to fill in the gap, therefore there are 4 of these “shaded” regions, and the answer was k/4.</p>
<p>The mathematical way was to assume one side equaled x.</p>
<p>Thus,
x^2 = k</p>
<p>Divide up the diamond in the center into two,
x/2 = base of triangle, x/2 = height of one triangle
A = 1/2(bh)
x^2/8 = area of triangle</p>
<p>Two triangles in the shaded region, x^2/4 = area of the shaded region; substitute k back in for x^2 and you get k/4, which I believe was (C).</p>
<p>the figure given did NOT say “not to scale”, so it’s ok to assume that it is to scale. each of the triangles (parallelogram=2 triangles) is 1/8th of the whole thing, since there are 8 of them (ones on the side are arranged differently, but same area). 8 portions, the parallelogram takes up 2, the parallelogram is 1/4th the whole thing, or k/4.</p>
<p>Was the one with the perpendicular line asking for the line that was bisected an experimental question? I just put down 5, I never realized at the time it had something to do with a triangle…</p>
<p>Someone explain the w,x,z question to me. It asked which of the pairs COULD have a mean of 4. Since we didn’t know the value of any of the numbers, except that they were less than 7, then couldn’t any or all of the pairs have a mean of 4?</p>
<p>Okay thanks xgoldy. 
I missed that one</p>
<p>Unseen it had to be positive integers.</p>
<p>@Lolol1</p>
<p>I had a writing experimental and I had that question, so unfortunately it was a real question.</p>
<p>The question stated there were two lines perpendicular to each other, forming a right triangle. It also said that the line connecting the two was twice as long as one of the lines. I believe it also said that one of the lengths was 10, or something of the sort, I remember getting 5^2 + x^2 = 10^2</p>
<p>25 + x^2 = 100
x^2 = 75
x = root(75) = root (25) * root (3) = 5 root(3)</p>
<p>@Unseen
The question said that given w < x < y < z < 7, where w, x, y, z were all POSITIVE INTEGERS (i.e. cannot be a decimal and must be positive), which of the following could have a mean of 4.</p>
<p>The possible combinations for a mean of 4 are only (2, 6) and (3, 5), therefore just by that you can eliminate the single or triple answers, which I believe were B and E.
Now let’s dissect this by Roman Numeral:
I. w & x
The only way we can get a mean of 4 is if we chose the largest possible numbers, meaning:
3 < 4 < 5 < 6 < 7
There is no other way to get a mean of 4, and if this is our only option, it is clear that the mean of 3 + 4 =/= 4, therefore I cannot be true.</p>
<p>II. x & y
Similar to I, we do the same process, however we attempt to choose one of the possibilities (2, 6 or 3, 5)
You will find that the two possibilities are
2 < 3 < 5 < 6 < 7
or
1 < 3 < 5 < 6 < 7
The mean of x & y or 3 and 5 = 4, therefore II is true.</p>
<p>III. y & z
Again, repeat the same process and we get:
1 < 2 < 3 < 5 < 7
The mean of y & z or 3 and 5 is 4.</p>
<p>NOTE: We cannot use -1 in place of w since the question specifically says positive, nor 0.5 or 0 since it says integer.</p>
<p>Thus the answer was (D) (I think), or II and III only.</p>
<p>Hope this clears some things up.</p>
<p>Parallelogram thing: Props to the people that were able to eyeball it / to scale it. If I were to guess, that was my original guess…</p>
<p>But then I took the long way of taking the two large right triangles bases being the sqrt(k) and sqrt(k/2) because the bottom bases of two congruent triangles make up a side of the square and likewise the little equilateral triangles make up half a side in both base and height… Then subtract the areas. k/4 is the end result.</p>
<p>What was the 2a+5b question? Shouldn’t it be 18?</p>
<p>missed the pentagon rotation one. *****…</p>
<p>iceblender yes it was 18</p>
<p>Wasn’t it saying sqrt(4a+10b)=6?</p>
<p>sqrt(4(4)+10(2))=6</p>
<p>2(4) + 5(2) = 18</p>
<p>Does anyone remember the problem with the sequence -2,3,-4,5 and it gave you the equation ((-1)^n)(n+1) = number in a sequence. It asked for the sum of 101 integers or something?</p>
<p>Was this an experimental problem? I put down -52 for it. Can someone verify?</p>
<p>Lolol1, I hope it is experimental!!! that one was hard.</p>
<p>@Lolol1</p>
<p>Basically everyone has agreed on here that it was indeed -52. That’s what I put as well</p>
<p>I made the stupidest error. I thought I got them all right until I read this thread. For the 1126 one I put 1026. Wow. And I was shooting for an 800, haha whoops</p>