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201

<p>This is NOT a permuation. It is the combination 6C2 = 15 (as I think you understand). But you then need to multiply this by 2 because for each choice of 2 participants, 2 matches are played.</p>

<p>So the answer is 2(6C2) = 2*15 = 30.</p>

<p>Note that the “2” in the question is signifying that you need to multiply by 2 - it is NOT for the 2 in the combination. The 2 in the combination comes from the fact that for each match you are choosing 2 from the 6 participants (this 2 is not mentioned explicitly).</p>

<p>Compare this with the following slightly easier question:</p>

<p>In a tournament, each of the 6 participants played a match against each of the other participants. What was the total number of matches played during the tournament?</p>

<p>The answer to this question is 6C2 = 15.</p>

202

<p>@pckeller: thanks a bunch!! I did it 1/10 * 1*9 and not 2/10 * 1/9… I think I got mixed up in using the counting vs. probability method. Anyway, thanks again!! :D</p>

203

<p>folks, I have a probability question that’s been driving me crazy!! [Probability</a> (part 8) | Probability | Khan Academy](<a href=“http://www.khanacademy.org/math/probability/v/probability--part-8]Probability”>http://www.khanacademy.org/math/probability/v/probability--part-8)</p>

<p>The question says: You have 10 coins in a bag. 9 of them are fair, but 1 of them is unfair and has BOTH sides heads. What is the probability of getting the 2 sided coin, GIVEN you get 5/5 heads. </p>

<p>Sal solves it in the video, and he actually has a nice visualization at the end. The answer is 32/41.
Here’s how he did it: he first determined the probability of getting 5 out of 5 heads in 5 flips which is (1/10 *1) + (9/10 * .5^5) = 41/320. That is the total # of probabilities. He then said that the probability of choosing the 2 sided coin AND getting 5 heads is 1/10 * 1 = 1/10. Therefore, the prob. is 1/10 divided by 41/320 = 32/41.</p>

<p>The only thing I don’t get in this problem is why did we put he total no. of possiblities (41/320)? Shouldn’t that be 1 since we’ve already determined that we’re only talking about 5 heads? The way I think of it is: if we constrain the probability to getting 5 heads, then why do we count ITS probability? It’s the only event and therefore it should be 1? Thanks in advance!</p>

204

<p>This is a conditional probabilty. We are computing P(A|B) where the line is read “given.” There is a simple formula for this: P(A|B) = P(A intersect B)/P(B). That is the formula that Sal used to solve this.</p>

<p>Here is a bit of informal reasoning about why that formula should be correct (this is not a rigorous argument): We are assuming that B is true. So we are essentially restricting the sample space from S to B. This is why P(B) is in the denominator as opposed to P(S) (which would be 1 - I think this was your point of confusion). For the numerator, note that since the sample space is restricted to B we now only consider an outcome succesful if it is in BOTH A and B.</p>

<p>You might want to draw the usual Venn diagram consisting of 2 intersecting circles, one labelled A and one labelled B. Shade all of B lightly, and shade the intersection darkly. The dark shading represents the numerator, and the light shading the denominator.</p>

205

<p>@DrSteve: Thank you sooooooooo much!! I totally get it now, especially with the Venn Diagram thing at the end :D</p>

206

<p>Here’s another challenging question:</p>

<p>In a class of 30 students, the number of students that study German is 3 more than the number of students that study Italian. If the number of students that study both subjects is equal to the number of students that study one subject, find how many study Italian only.</p>

<p>Answer is 6, but I have no idea how to get to it. I think it’s solved using a venn diagram, but I’m not sure how. Thanks!</p>

207

<p>I’m assuming that every student studies at least one of these subjects - otherwise the problem is not well-defined.</p>

<p>Since there are 30 students, and the number that study both subjects equals the number that study only one subject, there are 15 students that study only one subject. So we want to split up 15 as a sum of two numbers such that one is 3 more than the other - we have 9 + 6 = 15 (we can get this using “informal” or “formal” algebra). So 6 students study Italian only.</p>

<p>Notes: (1) Here is the “formal” algebra in case you need it: Let x be the number of students that study Italian only. Then the number of students that study German only is x+3. So x + (x+3) = 15. Thus, 2x + 3 = 15. So 2x = 12, and x = 6.</p>

<p>(2) To do this with a Venn diagram, you would draw 2 overlapping circles labelling the first G and the second I. The intersection gets 15 right away, and then we need to put a total of 15 in the other two regions such that region G has 3 more than region I (by G, I actually mean “G only,” and similarly for I) . This should clearly be 9 and 6, respectively. (Remark: note that I am strongly assuming that there are no students that do not study either language - this should have been mentioned in the problem. Technically ther should be a recatangle surrounding the two overlapping circles. It is consistent with the given information that we could put 4 in the rectangle outside the two circles, 13 in the intersection, 8 in region G and 5 in region I.)</p>

<p>(3) To solve this algebraically, we can let x be the number of students that study German only, y the number of students that study both, and z the number of students that study Italian only. Then we get the system of equations:</p>

<p>x+y+z=30
x+z=y
x+y=y+z+3</p>

<p>This system simplifies to:</p>

<p>x+y+z=30
x-y+z=0
x-z=3</p>

<p>You can solve this system pretty quickly by hand using the elimination method, or you can do Gauss Jordan Reduction pretty quickly in your graphing calculator by inputting a matrix and using the rref( feature.</p>

208

<p>

<a href=“http://i48.■■■■■■■.com/oro6fk.jpg[/IMG]”>http://i48.■■■■■■■.com/oro6fk.jpg

</a></p>

<p>I understand the simplifying to [n/(n^2-1)] = 5/k, and I understand how to get 24 once you get the 5. But I don’t understand how you can assume that n is 5 from [n/(n^2-1)] = 5/k, and how you can narrow it down to that number. Can someone explain how to understand this problem and do similar problems on the SAT?</p>

209

<p>The way I would actually do this would simply be to guess n=5 because that is the most obvious guess (it makes the numerators equal). Here is a rigorous explanation:</p>

<p>Cross multiplying gives nk = 5(n^2-1). Since the right hand side is divisible by 5 and the equations are equal we need to make the left hand side divisible by 5. The easiest way to do that is to set n=5 (then k=n^2-1) which is clearly an integer (and of course we know n^2-1 = 5^2-1 = 24).</p>

<p>Proof that n=5 is the only solution: Note if we choose a different multiple of 5 for n, say n=5j, then we must ensure that n^2-1 is divisible by j. If n is chosen to be even, then j will be a multiple of 2. But n^2-1 is odd (even^2-1 = even - 1 = odd). So k wil not be an integer. If n is odd, and n=5j with j at least 3, then n-1 and n+1 are not divisible by j. So again k is not an integer. So j must be 1. </p>

<p>The above argument shows that n=5, k=24 is the unique solution to this problem.</p>

<p>Remark: I used that fact that remainders cycle. For example, note that 35 is divisible by 7. It follows that 36 is NOT divisible by 7 (it has a remainder of 1) and 34 is NOT divisible by 7 (it has a remainder of 6).</p>

<p>Last remark: For the average student all of this explanation is unnecesary. The standard SAT strategy of taking a guess is sufficient. Any student that could produce the reasoning I just gave can probably get an 800 in SAT math pretty easily.</p>

210

<p>Although what Dr. Steve said is impressive and absolutely correct, there is fortunately an easier way. Simplifying fractions we get n^2/(n(n^2-1))=5/k so 1/k=n^2/(5n(n^2-1)) and k=5n(n^2-1)/n^2=5n-5/n. The only integer positive integer values of n that make this expression an integer are 1 and 5 and only 5 makes it positive. When n=5, k=25-1=24.</p>

211

<p>[Image</a> - ■■■■■■■ - Free Image Hosting, Photo Sharing & Video Hosting](<a href=“http://■■■■■■■.com/view.php?pic=2pseow0&s=6]Image”>http://■■■■■■■.com/view.php?pic=2pseow0&s=6)</p>

<p>In the figure above, the smaller circles each have
radius 3. They are tangent to the larger circle at
points A and C, and are tangent to each other at
point B, which is the center of the larger circle.
What is the perimeter of the shaded region?</p>

212

<p>The diameter of a smaller circle is the radius of the larger circle. So the larger circle has radius 6. The circumference of the large circle is then 12pi, and the circumference of a small circle is 6pi. Now if we travel along the perimeter of the shaded region we trace out half the circumference of the large circle, then half the circuference of a small circle, then half the circuference of a small circle once more. So the answer is 12pi/2 +6pi/2 + 6pi/2 = 12pi.</p>

213

<p>@ Dr. Steve: thank you again!! Your explanation is freakin’ awesome :D</p>

214

<p>I’m having problems with these questions:</p>

<p><a href=“http://i.imgur.com/N3nB1.png[/url]”>http://i.imgur.com/N3nB1.png&lt;/a&gt;
<a href=“http://i.imgur.com/tIdno.png[/url]”>http://i.imgur.com/tIdno.png&lt;/a&gt;&lt;/p&gt;

<p>Thanks in advance!</p>

215

<p>16) (D) 3 and 1/2. Really doesn’t matter which path you follow. Either make a staircase or go down then to the right.</p>

<p>17) (A) Six. Two left, two down; two down, two left; one down, two left, one down; one left, two down, one left; one left, two down, one left; one down, two left, one down.</p>

<p>18) (D) Because otherwise you cannot have an integer distance.</p>

<p>15) (p+n)(p-n) = 12 where (p+n) > (p-n). Right off the bat you know it can’t be III, since that would mean p+n = 3. Now if p - n = 1, p + n = 12, so 2p = 13, p = 6.5, n = 5.5. That’s fine. If p - n = 2, p + n = 6, so 2p = 8 and p = 4, so n = 2. That works too. The answer is (C) I and II.</p>

<p>It’s always good to test things out for SAT math - better safe than sorry, right?</p>

<p>20) Ok, so if j, k, and n are consecutive, so if the units’ digit of jn is 9, j must end in 9 too… e.g. 9, 19, 29… etc, and n will be 1, 11, 21, etc. k, then, will be something like 10, 20, 30, etc. so its units digit is 0. SWAG. The answer is (A)!</p>

216

<p>I, too, am having trouble with a rather misleading problem.</p>

<p><any 2="" points="" determine="" a="" line.="" if="" there="" are="" 6="" in="" plane,="" no="" 3="" of="" which="" lie="" on="" the="" same="" line,="" how="" many="" lines="" determined="" by="" pairs="" these="" points?=""></any></p>

<p>My answer was 30. This is because every point can form a line with one of the five other points, meaning the possibilities are 6 x 5 (assuming one point can be part of multiple lines).
I feel the difficulty lies in deciphering the prompt, is my interpretation of “how many lines are ‘determined’” correct? What does the prompt mean, exactly?</p>

<p>Any help would be much appreciated! :slight_smile:
-euro</p>

217

<p>@Europeen you have to divide by two, because you are counting each line twice. For example, you counted P6P5, and P5P6, which are the same line. The answer is 6*5/2 = 15 (or 6C2).</p>

218

<p>A list consists of 1000 consecutive even integers. What is the difference between the greatest number in the list and the least number in the list?</p>

<p>Answer is…1998.
I got 2000. Because I had 0 as my first even integer. Then 2000 as my largest. I got 2000 as my largest even integer because I noticed that 20 was the 10th consecutive integer and 30 was the 15th consecutive integer so to get the 1000th even integer I multiplied 1000 by 2. Where did I go wrong? Thanks in advance!</p>

219

<p>There is probably a more simple way to do the above problem, but I used the arithmetic sequence formula: a(n) = a1 + (n-1)d. (d = number of increase, a1 = first number, n = term number</p>

<p>a(1) = a1 + (1-1)(2)
a(1) = a1 (first even integer and thus the least number in the list)</p>

<p>a(n) = a1 + (1000 - 1)(2) (1000 represents the final and greatest term of the list)
a(n) = a1 + (999)(2)
a(n) = a1 + 1998</p>

<p>a(n) - a(1) = difference of greatest number and least number
a1 + 1998 - a1
= 1998</p>

220

<p>@tranman: You know where you went wrong? If you take 0 as your first and 2000 as your last value you create a range of 1001 integers not 1000.
Imagine it this way: there are 2000 consecutive integeres from 1 to 2000. If you take half of them it makes 1000 even (or odd, whatever the prompt says) integers. Your smallest value would be 2 (or 1 if it’s about odd integers). Zero is not included! If you take zero as your smallest value your highest one is 1998, not 2000. 1998 - 0 = 1998 so it’s correct again.</p>