USA June 2011 SAT: MATHEMATICS Discussion Thread

Test Preparation SAT Preparation
281

<p>ok -6 2 omit is how much?</p>

282

<p>rdpgn, 640-ish</p>

283

<p>i think i got 5pi/2
15 degrees cause it was a isoscelese right triangle inside an equilateral triangle, so
60-45 = 15</p>

<p>answer to the RS question?</p>

284

<p>any chance of 650?</p>

285

<p>good curve for us?</p>

<p>i took the march one, this one was definitely harder, and march was -1=790.</p>

<p>keeping my thumbs crossed for -1=800.</p>

286

<p>For the 10% question, was it A,B,C,D, or E?</p>

287

<p>@Skorpius: Each kid had 3 writing utensils, either pencils or pens. The average number of pencils was 1.4. Since the average number of total utensils was 3, the average pens would be 3 - 1.4 = 1.6
Think about it as if it were one kid. If one kid has a total of 3 whole utensils, and has 1.4 pencils, it follows that the remainder would have to be 1.6. :slight_smile: (even though that doesn’t work in reality - the concept is the same.)</p>

<p>@rd: Read back a couple of pages and see the explanations. The question said ANY two sides = twice third side. Well if I take your triangle and I pick ANY sides I want, 2+3 is not 8. Therefore that triangle doesn’t work.</p>

<p>x+y=2z –> x=2z-y substitute
x+z=2y –> 2z-y+z=2y –> 3z=3y –> z=y subsitute
y+z=2x –> z+z=2x –> 2z=2x –> z=x AND
–> y+y=2x –> 2y=2x –> y=x
I JUST PROVED MATHEMATICALLY IT MUST BE AN EQUILATERAL TRIANGLE.</p>

<p>For the cube, I got 5 because in the first two pictures you could see the 1 next to the 0, 1, 2, 3, and 4; therefore, since those numbers are next to and not across from 1, 5 must be across from it. (Numbers might be off, but that’s the correct reasoning.)</p>

288

<p>Rdpgn: around 650-690.
According to previous curves:</p>

<p><a href=“http://www.erikthered.com/tutor/SAT-Released-Test-Curves.pdf[/url]”>http://www.erikthered.com/tutor/SAT-Released-Test-Curves.pdf&lt;/a&gt;&lt;/p&gt;

289

<p>For the question about the 63 integers remaining after subtracting multiples of 3 and n from 100, would 20 have worked as an answer?</p>

290

<p>@ shyams: Yes, 20 works. You needed 4 to be taken out. 20, 40, 60, 80, 100, but 60 was already taken out by the three. Therefore 20 works.</p>

291

<p>Confirming 5 on opposite cube. it was pretty obvious from the picture. The overlapping circles with the Triangle the answer was 4. </p>

<p>One had a radius of 8 I believe and the other of 6. The hypotenuse or RS was 10 because the triangle in radius 8 was a 6,8,10 triangle which simplifies to 345. Then you just do the differences in the radii to get 4.</p>

292

<p>two questions:

  1. does 14 work for the grid-in and
  2. was the point for the absolute-value graph (4,-6)?</p>
293

<p>What do you think the curve for math will be? Do you think you could miss one and still get an 800?</p>

294

<p>@nadi on page 19, the y had to be negative. Absolute value means there are no negative y values.</p>

295

<p>4, -6 for one of the last questions is correct.</p>

296

<p>How do you guys remember the specific page numbers and everything hahaha that’s so amazing. I mean, I remember the problems and stuff, but there’s no way I could remember specific page numbers</p>

297

<p>Triangle</p>

<p>Jesus, the triangle question is not that hard. ANY two sides of the triangle add up to be twice the third side. ANY. ANY ANY ANY ANY.</p>

<p>Stop throwing out theses bs numbers, 2,4,6 this 3,4,5 that.</p>

<p>You have to be able to add ANY and ALL two sides and have that be twice the third side. The only way this is possible if is every side is the same length, hence II. equilateral. </p>

<p>Can we stop wasting space over this question now?</p>

<p>Quadrilateral</p>

<p>It is a quadrilateral, so two sides must be parallel. Just ignore the whole quadrilateral thing. You have a line intersection two parralel lines. So on the top left corner, you have (given) 75 inside the corner, and (180-75) 105 outside. Math your angles up, therefore x has to be 105.</p>

298

<p>16 works for the 63 integers one, correct?</p>

299

<p>I highly doubt a -1 will be 800, since the math section seemed pretty easy. I’m crossing my fingers, though. I only have -1 so far.</p>

<p>Did anyone get experimental math grid-in? The question with Crazy Hat Day? I wasn’t sure how to do that one so I just multiplied the numbers and got some absurdly large number (1050?). How do you do it?</p>

300

<p>Wait I put 20 for that multiples one. How is that wrong? #s can be used more than once.</p>