The Ultimate SAT Math Thread

Test Preparation SAT Preparation
1

<p>Hey everyone,</p>

<p>I've gone through every BB math section and have exhausted almost all of my QAS's. I have started seeing a pattern with almost every type of question and have improved tremendously. I used to get 2-4 wrong per section and now I'm down to 0-2 wrong per section (sometimes 3...but rarely. I've mostly been getting 1 wrong per section).</p>

<p>After completing all of these math sections, I have printed out each question I have gotten wrong and pasted them in my math notebook and have been slowly solving them on my own and with the help of satquantum.com (an invaluable resource). However, I now need someone to help me with the questions that I need more time and guidance in understanding, and this is where I'm hoping you guys can help me out.</p>

<p>Some/Most of these questions will be easy for you SAT math gurus, but for those of us not at that level yet, I'm hoping this thread can help.</p>

<p>So, everyday, before 1 PM, I will have the questions for that day posted. You guys have about 24 hours to help solve, ask questions, or post your own questions if you want. Detailed responses to certain questions will be MUCH appreciated.</p>

<p>*Note: I did not get all of these questions wrong. Some questions I will be posting are interesting questions that I think may help others or I would feel better having a little more knowledge about the topic.</p>

<p>So, here are the first set of questions:
Question 1

<a href="http://i.imgur.com/PtURMTU.png%5B/IMG%5D"&gt;http://i.imgur.com/PtURMTU.png

</a></p>

<p>Question 2

<a href="http://i.imgur.com/KfB17EE.png%5B/IMG%5D"&gt;http://i.imgur.com/KfB17EE.png

</a></p>

<p>Question 3

<a href="http://i.imgur.com/sxuSdaJ.png%5B/IMG%5D"&gt;http://i.imgur.com/sxuSdaJ.png

</a></p>

<p>Question 4

<a href="http://i.imgur.com/NdKGMk1.png%5B/IMG%5D"&gt;http://i.imgur.com/NdKGMk1.png

</a></p>

<p>Question 5

<a href="http://i.imgur.com/1WDnNoQ.png%5B/IMG%5D"&gt;http://i.imgur.com/1WDnNoQ.png

</a></p>

<p>This will be a looong journey, since I have many questions in my notebook, so thanks to everyone who participates.</p>

<p>If all of the questions are answered before the 24 hours, I'll happily post more. :)</p>

2

<p>Q1. This one looks really complex at first. Calm yourself down. For this one, I would pretty much guess and check the answers. </p>

<p>It can’t be (A. 1) because t and k are positive numbers, but k has to be less than 1…so how does that work?</p>

<p>It can’t be (B.2). k would be equal to 1. 2+1 does not equal to more than 4.</p>

<p>C works! If t=3, k can equal 2. 3^2 - 2^2 is 5, which is less than 6.</p>

<p>It can’t be D. If t=4, k can equal 1,2,or 3. Try plugging those values in the first mathematical sentence. It doesn’t work. Do the same for E.</p>

<p>The lesson from #1 is that sometimes you don’t need to try and figure it out all in your head. Try running the answers in and see which work and which do not.</p>

<p>I’ll leave someone else to look at the others and try their hand at them. If they are still unanswered, I’ll have a look.</p>

3

<p>WOW I just posted my reply and it didn’t go through! All my hard work disappeared :frowning: </p>

<p>I’ll start from 2 I guess. </p>

<ol>
<li><p>If x is inversely proportional to y, then 1/x is directly proportional to y. I think you should just memorize that. Square both sides and you’ll get that 1/x^2 = y^2. </p></li>
<li><p>The distance from P to Q is equal to 3. Now X is 2/3 the distance of P to Q FROM Point P. So we find 2/3 of 3, which is 2, and add that to point P, which makes X = 3.5</p></li>
<li><p>Since the parabola opens upwards, the a value has to be positive. It can’t be C because the parabola looks too narrow and when the a value is a fraction, it is wider.
I don’t think it can be D either; just using logic here. The parabola looks too narrow, so I’ll go with E. Not the best explanation sorry about that. </p></li>
<li><p>You’ll see that the line from X to Y is actually the radius of both circles, so the circles are congruent. Also, all the lines of the quadrilateral are radii; you should be able to visualize these types of questions. You’ll see that they are at one end of the circle and connect to a center. So all the sides are equal. Therefore, 24/4 = 6 = radius. Circumference of each circle is 12pi (2pi*radius).
Since all the sides of the quadrilateral are equal, the angles are equal as well, which makes each angle 90 degrees, which is a fourth of the circle. So subtract a fourth from the circumference, which gives 9pi. 9pi + 9pi = 18pi which is the length of the line. </p></li>
</ol>

<p>Sorry I didn’t explain it that well, I had written so much but it didn’t post and I lost all the info! Hope it helps though! And I certainly could’ve made some silly mistakes here, correct me if I’m wrong</p>

4

<p>@Day, Thank you. Totally forgot about plugging in for this problem.</p>

<p>Curious: How would you do it algebraically?
I see that t^2 - k^2 can be broken down into (t+k)(t-k) and so maybe I can plug in t+k>4 there?</p>

<p>So, we have (5)(t-k) < 6.
t-k = 6/5
and now we subtract
t-k = 6/5
t+k = 5
equals
2t = 6.2
t=3.1</p>

<p>Hence, answer choice C (3) ? </p>

<p>But if I plugged in t -k = 10, I wouldn’t have gotten an answer around 3.</p>

<p>Where am I going wrong?/</p>

5

<p>Yeah, once I find a way, I’ll just stick to that. It works pretty well for me! </p>

<p>Anyways, that’s the case where you think too hard :slight_smile: </p>

<p>I believe that you can’t plug in inequalities (I think, not sure). You would have 2 different inequalities, and now I’m lost lol.</p>

6

<p>@5am,
Thanks for the help!</p>

<p>For number 5, the answer is B(16pi) though.</p>

<p>SATQuantum has an explanation for it here: [SECTION</a> 3 QUESTION 20 MATH SAT QAS JAN 2013 - satqasjan2013journal - SATQuantum](<a href=“http://www.satquantum.com/satqasjan2013journal/section-3-question-20-math-sat-qas-jan-2013.html]SECTION”>http://www.satquantum.com/satqasjan2013journal/section-3-question-20-math-sat-qas-jan-2013.html)</p>

<p>but I don’t understand his reasoning right after the 1:55 mark.</p>

7

<p>Sorry I made a mistake on the last part of 5.
Here’s a picture to help explain it, I think this is right.
<a href=“http://s23.postimg.org/6i3y0515n/image.jpg[/url]”>http://s23.postimg.org/6i3y0515n/image.jpg&lt;/a&gt;&lt;/p&gt;

<p>Excuse the terrible handwriting!</p>

8

<p>The figure is a rhombus, therefore the diagonals and perpendicular lines (creating the 90 degree angle). The diagonals also bisect the vertex angles, which is why both the angles are 60 degrees</p>

9

<p>^Ahh, I think I get it, although my understanding may be a little too rudementary. Is the reasoning behind that part being 1/3 of the circle because of this: </p>

<p>

<a href=“http://i.imgur.com/3WTCNG6.png[/IMG]”>http://i.imgur.com/3WTCNG6.png

</a></p>

<p>(or in other words, central angle?..Except the central angle isn’t shown so explicitly in the question)</p>

10

<p>Yeah the angle is equal to 120 degrees because the diagonal bisects the vertex angle, so if one side is 60 then the other is 60 as well (as show in my picture). 60+60 = 120
You know it is the central angle because X is the center and the lines extend all the way to the edge of the circle.
It’s a really tough problem to explain lol but I hope you get the idea. This is definitely one of the toughest problems I’ve ever seen on the sats!</p>

11

<p>^Okay lol, thanks.</p>

<p>Any ideas about solving #1 algebraically?</p>

12

<p><a href=“http://s14.postimg.org/52b6l3fg1/image.jpg[/url]”>http://s14.postimg.org/52b6l3fg1/image.jpg&lt;/a&gt;&lt;/p&gt;

<p>Maybe this will help. You really shouldn’t use this approach though</p>

<p>EDIT: t^2 - k^2 cannot equal 2 or 1, sorry about that but that isn’t important for the problem anyway.</p>

13

<p>Here is a batch of new questions :slight_smile:
Question 6

<a href=“http://i.imgur.com/a6VU8tB.png[/IMG]”>http://i.imgur.com/a6VU8tB.png

</a></p>

<p>Question 7

<a href=“http://i.imgur.com/7mHJjgp.png[/IMG]”>http://i.imgur.com/7mHJjgp.png

</a></p>

<p>Question 8

<a href=“http://i.imgur.com/JRbnFca.png[/IMG]”>http://i.imgur.com/JRbnFca.png

</a></p>

<p>Question 9

<a href=“http://i.imgur.com/eTQkYhP.png[/IMG]”>http://i.imgur.com/eTQkYhP.png

</a></p>

<p>I hate questions like question 6…I try and draw a venn diagram and fill in all of the information, but I can’t for the life of me, decide what to do next. </p>

<p>Question 7’s wording is tripping me up. I know that when you reflect a point across the y=x, you just switch the original points, but I can’t decide what the coordinates for P are.</p>

<p>Question 8 - I’ve seen 3 surface area questions out of all of the QAS tests I’ve done (there are no SA questions in the BB), and I’ve gotten all three wrong. I know that the formula for SA of a cube is 6*(area (of each surface)). If anyone is willing to go in depth as to what exactly I need to know for surface area on on the SAT’s, I’d appreciate that greatly.</p>

<p>Question 9 - This question reminds me of question 6, which means I suck at them. I know that I suck at these types of questions, but I just don’t know how to fix the issue and start getting them all correct. Any ideas?</p>

14

<p>Did anyone solve Question 4 yet? I’m just having trouble with that one.</p>

15

<ol>
<li><p>[View</a> image: image](<a href=“http://postimg.org/image/icei5iqz5/]View”>http://postimg.org/image/icei5iqz5/)
You just have to set up the venn diagram like I did and it should be fine. We are looking for the z value </p></li>
<li><p>Point P = (1, 3), just switch the x and the y
Point T = (-1, 3), just negate the x </p></li>
<li><p>9*6 = 54 is the surface area before one cube is removed.
54-3 = 51. You subtract 3 because 3 of the faces on the surface are removed.
51+3 = 54. You add 3 because now the surface of the 3 cubes surrounding the one that has been removed show one more face. So the answer is 54. I think that’s how you work it out, not entirely sure. </p></li>
<li><p>[View</a> image: image](<a href=“http://postimg.org/image/59ivm90r5/]View”>http://postimg.org/image/59ivm90r5/)</p></li>
</ol>

<p>Again, I could be wrong so please correct me if I am.</p>

16

<p>@5am6996 you’re correct for number 9. Another way of thinking about it is overcounting and then correcting. Adding all of the numbers gives you 90, and 8 students are counted for twice, and 1 student is counted for thrice. 90-8-1-1 = 80, and 80/0.05 = 1600.</p>