<p>im thinking its 10 now...cause i remember an example from barrons 2400 with students who took statistics, programming and both. It was like, out of 200 students 90 took statistics, 20 took both programming and statistics and 50 took programming. The rest did not take statistics or programming. The venn diagram had 70 in statistics, 20 in the middle, and 30 in programming. So I think you would add the middle sections into the total.</p>
<p>Sorry for the terrible example lol...</p>
<p>ok. think of y, z, and x as conditions. so if an object fits all 3 conditions (x,y,z) then it must therefore fit y and z as well. so it is a part of the intersection of y and z, not exclusive. so the 3 is included in 7.</p>
<p>answer is 7.</p>
<p>my teacher taught us this too and said the middle counted</p>
<p>to bad i was sleeping when she said that =(</p>
<p>no wendy its 10</p>
<p>ok. x = even numbers, y = multiples of 6, z = multiples of 10. those that are in the intersection of x, y, and z must be even, multiple of 6 AND multiple of 10. those that are only in x and y must only be multiples of 6. therefore, the elements in x/y/z are INCLUDED in x/y.</p>
<p>thats what i think anyways. not gonna argue.</p>
<p>except that's not the condition of the question...</p>
<p>ya me neither.... but they wouldnt put that question on the sat if it would be 7.... Princeton Reviews 1300 dollar class taught me that.... lol but hey well find out in 19 days</p>
<p>For the last time, its 10. I put 10 as many others on this forum did, my friend who made USAMO last year also says its 10.</p>
<p>i kno i kno i kno</p>
<p>its 10...10....10</p>
<p>but Wendy wont give up</p>
<p>this is standard notation. x/y is written only in that region. you don't need to include any other regions.</p>
<p>i made USAMO in 2006 too.. i got 7. There are many hints but here is one.</p>
<ol>
<li><p>Difficulty is in order. Usually obscure questions come last. The question was in early-middle, which is full of easy problems.</p></li>
<li><p>No SAT problems are that obscure. If it was AMC maybe but this is SAT.
I got 800 on Math the two times previous to this test. I didn't even need to think on the problem. </p></li>
</ol>
<p>It asked what is the intersection between y and z.</p>
<pre><code>x is an element of A ∩ B if and only if
* x is an element of A and
* x is an element of B.
</code></pre>
<p>if it asked for X ∩ Y ∩ Z it should be 3.</p>
<p>THE QUESTION ASKED FOR "Y ∩ Z" and it was 7 shown in the diagram.</p>
<p>you can't word questions and interpret them differently to your self.
This is a very obvious rule. X ∩ Y ∩ Z is not equal to Y ∩ Z alone.</p>
<p>now that the 7 vs. 10 dispute is settled, can someone clarify on the "rectangle inside circle" question. did it ask "how many rectangles you can inscribe using any point of the circle" or "how many can be drawn using the given 8 points".</p>
<p>ETA- lol apparently, its not over yet.</p>
<p>nope it was 10........ and no i didnt get 800s on 15 previous tests/...... buts thats still nt gonna change my opinion on the fact that the answer is 10</p>
<p>
[quote]
ok. x = even numbers, y = multiples of 6, z = multiples of 10. those that are in the intersection of x, y, and z must be even, multiple of 6 AND multiple of 10. those that are only in x and y must only be multiples of 6. therefore, the elements in x/y/z are INCLUDED in x/y.
[/quote]
</p>
<p>Right, elements in x/y/z are a subset of x/y, but they are NOT already counted. That is why there is that special category in the middle - to set apart those elements in the intersection of x/y that also happen to fit into z. </p>
<p>Also, think: y is a circle and z is a circle. Their INTERSECTION should be formed by two arcs... so just count up everything inside...</p>
<p>I will give an easy example</p>
<p>x={4,5,8,11}
y={4,5,9,11}
z=(4,5,6,11}</p>
<p>Y ∩ Z = 4,5
X ∩ Y ∩ Z= 11</p>
<p>there are two intersections between y and z and
one between x y and z
does that mean the intersection of y and z is 2+1=3???</p>
<p>yea snowia makes sense. its a toss.
if they set x/y/z apart, as in they took it out of set y/z, then you'd have to add it back into y/z. but, in standard notation, they don't take it out, the number written in y/z is inclusive of x/y/z.
all depends on notation.</p>
<p>Just because "11" is in set x does not mean it stops being in both set y and set z - thus it is part of the intersection of y and z. You're making this out to be a question testing vocabulary, when it is just to test a basic understanding of a venn diagram.</p>
<p>EDIT: Actually I think you're right about standard notation... so maybe it would be 7? When I read the problem I didn't even think about technicalities in notation... that seems too hard for the SAT!</p>
<p>I guess my example wasn't large enough to show anything.</p>
<p>so what your saying is</p>
<p>10 - set in x = Y ∩ Z
which makes set in x=0?</p>
<p>Maybe this question could possibly be thrown out?</p>
<p>Elements common to X and Y and Z are still common between any two of the three of them. It X, Y, and Z have 1, 2, and 3 in common, Y and Z still have 1, 2, and 3 in common.</p>
<p>And for the rectangle in the circle question, the answer is infinite. If you rotate one of the circles around it's center (on top of the circl'e center) it should be able to spin freely. The rectangles could be placed in ANY postion (any = infinite).</p>
<p>And I have my own question. On the question with the cylinder and the smallest 3-D rectangle that could contain the whole cylander, the area of the recangle would be (d^2)(h), right?</p>