<p>since the exam's in little over a week, figured we shud have a thread to discuss questions and review material.</p>
<p>do u have barron’s
if yes pls answer question 50 of model test 2, the one wid da graphs</p>
<p>i think you are getting confused because they have named the axis y^2 and x. Know that this is same as marking y^2 vs x graph and naming co -ordinates x and y</p>
<p>so basically u will get y^2 if u square the (2x+1) term
so plot the graph of y = (2x+1)^2 and u will see that the figure is a parabola with vertex -.5 which corresponds to answer c</p>
<p>NOTE - HERE Y IS ACTUALLY Y^2 AND CORRESPONDS TO Y AXIS.</p>
<p>I’ve tried practice tests in McGraw Hill’s book. They seemed easier than other practice tests that I previously tried. Are they close to the actual test ? :)</p>
<p>I dont know i will give maths IIC on 7th this December. I prefer Barron’s. Its tougher and over preps you, if you manage 760 on barron’s, it means you can manage a 800 on actual</p>
<p>If any1 wants to review and check whether they are ready, here are some questions -:</p>
<p>PnC - How many triangles can you form from joining points if there are 15 points out of which 5 are co-linear?</p>
<p>Probability - A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is ?</p>
<p>Sequence and Series - Sum of first 20 terms of sequence - .7, .77, .777, .7777 …</p>
<p>Sets - Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of A </p>
<p>MaximeP I am finding that the McGrawHill tests are easier than the Barron’s tests, but I didn’t have the sense that they were too easy either. Perhaps the questions in the 40s could have been a little more advanced on some of them, but some of those seem hit or miss on the real thing too from the actual exams that I’ve seen.</p>
<p>Its better to be actually over-prep than '‘thinking’ that you are prepared. Barron’s may be tougher (i myself am getting 740,760 and a 740) on it but then again at least i know that a 740 here means a 780+ on actual SAT. (well that’s the hope but who knows).</p>
<p>I don’t have McGrawHill so i can’t comment on its difficulty but general opinion tends towards Barron’s if you are looking for a hard and challenging paper.</p>
<p>@Shaurya Your first problem is ambiguous, but I will proceed assuming that you meant is 1 line that contains 5 of the points, and no other points are on that or any other line that contains another point.</p>
<p>Problem 1: 10<em>9</em>8/(3!) triangles with no points on the line, 5<em>10</em>9/(2!<em>1!) with only one vertice on the line, and 5</em>4<em>10/(2!</em>1!) with two vertices on the line. So we have 120+225+100= 345 triangles.</p>
<p>Problem 2: There are 5 choose 4 = 5 ways to get 4 correct, and 5 choose 5 = 1 way to get 5 correct. There are 3^5=243 possible outcomes. Thus the probability is: (5+1)/(243)=2/81.</p>
<p>Problem 3: I can’t think of a non-tedious way to do this one.</p>
<p>Problem 4: There are 8 elements in A</p>
<p>Answer 1 : correct but your way is quite long and tedious. I have a much much shorter method. Calculation Error - 445 - 10 - </p>
<p>Answer 2: Wrong - (-2.5) - its a good question</p>
<p>Answer 3: Not answered - </p>
<p>Answer 4 : Correct - 10 -</p>
<p>Answer 5: Not answered - 0 -</p>
<p>Answer 6 : this ones a Poor Joke which nerds might enjoy. - not marked -</p>
<p>Overall - 17.5/50 Not Bad 
considering the ones you have not answered are little tougher than normal.</p>
<p>What’s the best way to study for this? Just taking a crapload of practice tests, and straightening out the ones you miss?</p>
<p>I used to know all of this stuff after I took PreCal but have since gone blank and can’t do much…</p>
<p>Q. Sum of series - .7 + .77 + .777 … upto 20 terms</p>
<p>Ans. </p>
<p>7 ( .1 + .11 + .111 + …)</p>
<p>(7/9) * ( .9 + .99 + .999 …)</p>
<p>(7/9)* ( (1 - .1) + (1. .01) + (1 -.001) …)</p>
<p>Separating 1s and .1s</p>
<p>(7/9)* ( 1 (20 times) - ( .1 + .01 + .001 +.0001 …) )</p>
<p>Use GP to solve the addition portion of decimal numbers. a = .1 , common multiple = .1
:. S (20) = (7/9)*( 20 - ( .1(1-.1^20)/1-.1)</p>
<p>Use your calculator to solve the product and get your answer.
good question ain’t it?</p>
<p>All those are just as tedious as plugging .7+.77… into your calculator, you’re still hitting just as many buttons. Except for your sequence way of solving it. I don’t understand how you got that at all.</p>
<p>What do you mean “All those are just as tedious as plugging .7+.77… into your calculator” What do you suppose by “all those”? All other questions or this one?</p>
<p>.7 + .77 + .777 + … = .7(20) + .07(19) + .007(18) + … + 7e(-20)(1)
The summation is pretty easy to enter into your calculator.</p>
<p>If I were doing the problem without a calculator, I would likely use shauryagupta’s method. Also note .1 + .01 + .001 + … = .11111…(20 1’s) = (1 - 10^(-20))/9.</p>
<p>I meant that all your solutions involved plugging an absurd amount of digits and numbers into your calculator except for the last one, where you identified a geometric sequence, which is pretty fast to put into your calculator.</p>
<p>EDIT: Replying to shauryagupta. Why does the “Quote message in reply?” box never work?</p>
<p>Question from Sparknotes:</p>
<ol>
<li> A figure skater competing at a local competition needs a mean of 5.8 (out of a total 6.0) to win the first-place medal. If there are three judges and one of them gives her a 5.5, what is the lowest score she can get from the remaining two judges if she still wants to finish first?
(A) 4.8
(B) 5.2
(C) 5.5
(D) 5.9
(E) 6.0</li>
</ol>
<p>5.8 = (5.5 + 2s) / 3
17.4 = 5.5 + 2s
11.9 = 2s
s = 5.95
Which I would be rounded up to 6 since a 5.9 would not get her a 5.8 mean.</p>
<p>Yet, the answer is D. I don’t understand what’s wrong with my equation?</p>
<p>your method of attempting is wrong.</p>
<p>you can’t assume both will give same marks.</p>
<p>(x+y+z)/3 = 5.8</p>
<p>x+y+z = 17.4
X+Y = 11.9</p>
<p>now for 1 to give min, other has to give max.</p>
<p>therefore, x(min) = 11.9 - 6</p>
<p>x(min)= 5.9</p>
<p>and bass guitar, all of my questions can be done by hand. want me to tell you all the answers?</p>
<p>Ans : 1 : 15C3 - 5C3 - easily doable by hand</p>
<p>Ans 2 : choose any 4 questions out of 5, then probability of correct - 1/3 and incorrect - 2/3 and then choose 5 out of 5 queestions
= 5C4<em>(1/3)(1/3)(1/3)(1/3)(2/3) + 5C5</em>(1/3)(1/3)(1/3)(1/3)(1/3)
= 11/(3^5) = 11/243</p>
<p>Ans 4 : AXB will have 8 elements. :. = 8C3+8C4…8C8 - very easy to calculate without a calc.</p>
<p>Ans 5 : we know that, z=1/z(bar)</p>
<p>replacing in given equation</p>
<p>therefore, arg(1+z/(1+zbar)) ----> arg(z)---->X</p>