Math Help Plz

Test Preparation
1

<p>Level 5 difficulty
25) Sets X,Y and Z have 4 members tin common. Sets X and Y have a total of 15 members in common. Sets Y and Z have a total of 8 members in common. If each member of set Y is contained in aat least one of the other two sets, how many members are in set Y?</p>

<p>Answer: 19, care to explain your method of doing it PLEASE?</p>

<p>8) Four distinct lines in a plan are such that no two lines are parallel and no three lines intersect at the same point. What is the total number of points of intersection of these four lines? </p>

<p>Answr is six, but please explain how to be SURE that this is the maximum?</p>

2

<p>Okay, I'm kind of weird, but who knows - maybe this will work for you!</p>

<p>for #25:</p>

<p>STEP 1: "sets X,Y and Z have 4 members in common" indicates that X and Y have 4 members in common, Y and Z have 4 members in common, and X and Z have 4 members in common. These 4 members are all the same. Makes sense, right?</p>

<p>STEP 2: It says X and Y have a TOTAL of 15 members in common. TOTAL implies that at most - there are 11 more members from the 4 previously counted. Okay, so looking at that part, we have 11 members in Set Y. Onward!</p>

<p>STEP 3: Sets Y and Z have a TOTAL of 8 members in common. Same method as before: TOTAL members minus 4 = 4. This means that there are 4 more members.</p>

<p>Finally, step 1+step 2+ step3 = 19 numbers. Just in case, let me reiterate: 4 numbers from common members in all of the sets, plus 15 more members because of X and Y common (aka 15-4=11), plus 4 more members because of Y and Z common (aka 8-4=4).</p>

<p>Hope that makes sense! Kind of long when you read it, but superfast on paper! I'll have 8 in just a second!</p>

3

<p>The maximum number of intersections for two lines is 1. If you add another line, you can get another 2 intersections. If you add yet another line, you could get three intersections. It's just not possible to get more than that because they're straight lines so each additional line can only create a number of intersections equal to the number of pre-existing lines.</p>

4

<p>Okay, number 8 (again, let me remind you that I'm weird - but it works)! Here's what I did:</p>

<p>"4 distinct lines" - okay, I draw 4 straight parallel lines.
"no two lines are parallel" - oops can't do that, let's try 4 lines of different slopes
"no three lines intersect at the same point" - check! wait, but it's gotta intersect... and they can't all cross through the middle (like what a cardinal system on a map would look like) - so I draw a rectangle with sides of different slopes (like a lopsided 4 sided polygon)</p>

<p>"what is the total number of points of intersection?" - Okay here's the tough one. Well right now I have 4 points. Let us see if I can get any more - if not, then 4 is the maximum. Scooting one of the sides (making the polygon smaller or larger) gets me nowhere. Especially since these lines are infinitely long. What if I turned one of the sides at a different angle relative to the other side? hmmm nope, that's still 4 points. </p>

<p>Kind of like the riddle asking how you can cut a circular cake into 8 equal portions with only 3 cuts (you cut vertical and horizontal, and then cut it in 2 layers)... I just figured that if we made the polygon 3-sided, we could just "cut" the shape.</p>

<p>So I draw the triangular shape, and use the 4th line to cross 2 of the sides. Hmmm, so my total is 6 now. Pretty neat! </p>

<p>But just in case, let's see if we can get more. Can we draw a 2-sided shape, and then just use the other 2 lines to cross the lines? No, because the 2-sided shape doesn't really exist. Okay, how about if I had that extra 4th line cross all three of the sides, then it would be 7 points of intersection! Nope, because you can't a line cross all three sides of a triangle (unless it's 3D).</p>

<p>So, total = 6 maximum</p>

<p>I'm sure there is a formula for max points of intersection, but it's so subject pertaining to each problem :)</p>

5

<p>In addition to what I said and slipstream99, a line can only intersect 2 or 3 other lines. Three only happens if it is the middle (but we can't do that because of the problem saying it's not at the same point).</p>

6

<p>8) "Four distinct lines in a plan are such that no two lines are parallel and no three lines intersect at the same point. What is the total number of points of intersection of these four lines?"</p>

<pre><code>There is a general formula for all questions of this type.

                n(n-1)/2. Where N is the number of lines

</code></pre>

7

<p>Thanks a lot!</p>