<ol>
<li>The plane whose equation is 5x + 6y + 10z = 30 forms a pyramid in the first octant with the coordinate planes. Its volume is
a) 15
b) 21
c) 30
d) 36
e) 45</li>
</ol>
<p>The answer is A. The book says the plane cuts the x-axis at 5, y-axis at 6, and z-axis at 3. Why does the plane cut the axes at those points??? Is there a formula I should be cognizant of?</p>
<ol>
<li>If five coins are flipped and all the different ways they could fall are listed, how many elements of this list will contain more than three heads.
a) 5
b) 6
c) 10
d) 16
e) 32</li>
</ol>
<p>The answers is 5C5 + 5C4 = 6. Why do we have to use combination for this problem?</p>
<ol>
<li><p>The plane cuts the x-axis at x=6 and y-axis at y=5 (set y=z=0 or x=z=0). The vertices of the pyramid are (0,0,0), (6,0,0), (0,5,0) and (0,0,3). From this it can be determined that the volume is 15.</p></li>
<li><p>How many strings of length 5 have 5 H’s? 4 H’s?</p></li>
</ol>
<ol>
<li><p>How do you know that the origin is one of the vertices?</p></li>
<li><p>What do you mean? This is how I approached the problem:
HHHHH
HTHHH
HHTHT
HHHTH
HHHHT
THHHH</p></li>
</ol>
<p>^ If I opted to list all the outcomes, I would no doubt get the answer, but I can’t do this for every problem (where the answer is 32, for example). But what’s the logic behind using combination to solve the problem? “Choose 4 heads from 5 coins” doesn’t make sense…</p>
<p>The pyramid lies in the first octant. Draw a picture and you should see.</p>
<p>For 2, you are choosing four “positions” (out of 5) to assign heads to. For example, you have five coins in a line and you are choosing four of them to assign heads to. The number of ways to do this is 5C4 = 5!/(4!1!) = 5.</p>
<p>I agree, it is a waste of time to list all the possibilities. If you find yourself checking 10+ cases on an SAT problem, you’re likely doing it the slow way.</p>
<p>See, 5c4 means you choose any 4 coins (which will show heads), irrespective of their position. That is because in the question, it isn’t written that head should come up at 2nd, 3rd or 5th position etc. </p>
<p>Moreover, its written that more than 3 heads so that means 4 or 5 heads. </p>
<p>So answer is 5C4 +5C5</p>
<p>Try solving this problem to get accustomed to PnC. (question with condition)</p>
<p>Q. In how many ways a grandfather along with two of his grandsons & 4 grand daughters can be seated in line for a photograph such that grandfather is always in middle and the two grandsons never together ?</p>
<p>Also, I have another Math IIC question from Barron’s:</p>
<p>A line, m, is parallel to a plane, X, and is 6 inches from X. The set of points that are 6 inches from m and 1 inch from X form:
a. a line parallel to m
b. two lines parallel to m
c. four lines parallel to m
d. one point
e. the empty set</p>
<p>How the heck do I visualize this problem…I suck at spatial concepts.</p>
<p>answer is (b). just draw a plane and a line in your textbook. now( answer is according to me, maybe right,maybe wrong.) </p>
<p>Stand in front of a wall and raise your hand, let your hand be line m and the wall, plane. Now turn your head 45 degrees to left and imagine a line near 1 inch unit of the wall (plane), do the same for the right side. now place the 2 lines 6 inches from your hand and 1 inch from wall and voila Tu as ton r</p>
<p>and answer to PnC is not correct, hint: break the question into 3 parts.
1- both boys sit on right
2 - both on left
and 3 - either side of grandparent</p>
<p>Your answer does not take into the account, the changes in position of daughters when you calculate the ways in number of positions when both boys sit together. </p>
<p>Don’t worry too much. You got plenty of time to practice</p>
<p>What is complementary counting in layman’s language :D?</p>
<p>And for that plane question, should I draw it like this:</p>
<hr>
<p>. .</p>
<pre><code> .
</code></pre>
<p>I have trouble representing a plane on a piece of paper…because the sheet is more or less “2D” I can’t effectively visualize a 2D/3D problem in a 2D context lol</p>
<p>Number of ways the grandchildren may be seated around him: 6! = 720
Number of ways two grandsons are seated together: 4 x 2 x (1 x 4 x 3 x 2) = 192</p>
<p>In the figure above, ABCD is a square. M is the point 1/3 of the way from B to C. N is the point one-half of the way from B to C. N is the point one half of the way from D to C. Then angle <MAN =
a) 50.8
b) 45
c) 36.9
d) 36.1
e) 30.0</p>
<p>The answer is B, and the way the book does it is assign an arbitrary value to the side of the square (6 in this case because it’s divisible by both 3 and 2). Then it uses sincostan to solve and whatnot. However, I initially approached the problem in a simpler, yet somehow inaccurate way. What’s wrong with this method:</p>
<ol>
<li>Draw a diagonal from A to C. This splits the square in two halves, 45 degrees each.</li>
<li>Since M is 1/3 of the way from B to C, angle <BAM = 45/3 = 15.</li>
<li>Since N is one half of the way from D to C, angle <NAD = 45/2 = 22.5.
Answer: angle <MAN is 90 - 15* - 22.5 = 52.5 = wrong, which cost me a crapload of time re-doing it the way the book does.</li>
</ol>
<p>“N is the point one-half of the way from B to C. N is the point one half of the way from D to C.”</p>
<p>Uhh what?</p>
<p>I’m going to assume N is 1/2 of the way from D to C. Your statement 2 is wrong (<BAM = tan^(-1) 1/3, not 15 deg). Similarly, statement 3 is wrong. You should use trig.</p>
<p>Let the length of side be 3x. Then one side is divided into 2x and x.
The Other side 3x/2 and 3x/2.
Now tan(angle NAD) = ND/AD - 1.5x/3x = .5
therefore, arctan(.5) = agnle NAD</p>
<p>Similarly, In triangle BMA,
tan(MAB) = x/3x = 1/3
this implies - angle MAB = arctan(1/3) </p>
your way is fine too (shorter)</p>