MAth Questions (seemingly simple) that need to be explained from BB.

Test Preparation SAT Preparation
1

<p>In test 3 sec. 2 of the BB, there is a question that really confuses me.
It's #18.</p>

<ol>
<li>Any 2 points determine a line. If there are 6 points in
a plane, no 3 of which lie on the same line, how many
lines are determined by pairs of these 6 points?</li>
</ol>

<p>(A) 15
(B) 18
(C) 20
(D) 30
(E) 36</p>

<p>Can anyone tell me the shortcut to finding it?^^</p>

<hr>

<ol>
<li>Diagram on p. 519 of BB (please look)</li>
</ol>

<p>At a beach, a rectangular swimming area with dimensions
x and y meters and a total area of 4,000 square meters
is marked off on three sides with a rope, as shown above,
and bounded on the fourth side by the beach. Additionally,
rope is used to divide the area into three smaller rectangular
sections. In terms of y, what is the total length, in meters,
of the rope that is needed both to bound the three sides of
the are and to divide it into sections?</p>

<p>(A) y + (4,000/y)
(B) y + (16,000/y)
(C) y + (16,000/3y)
(D) 3y + (8,000/3y)
(E) 3y + (16,000/3y)</p>

<p>That is impossible.... almost^^</p>

<hr>

<p>---Also in sec. 2 16 and 17 and 19 were confusing. Got them right by guessing. Anyone care to explain them (esp. 17 and 19)??---</p>

<hr>

<p>SECTION 5</p>

<ol>
<li>If x is not equal to 0 and x is inversely proportional to y,
which of the following is directly proportional to 1/x^2 ?</li>
</ol>

<p>I don't get the language on this one, but got it right ^^^</p>

<hr>

<h2>Is there an easy way to do #7 w/out drawing anything??</h2>

<p>Omitted #8. I looked at it and was like ***?</p>

<ol>
<li>(x-8)(x-k) = x^2 - 5kx + m</li>
</ol>

<p>(A) 8
(B) 16
(C) 24
(D) 32
(E) 40</p>

<p>^^Probably the HARDEST one for me yet. I need an explanation, please ;D.</p>

<hr>

<p>Also, I don't feel like typing it all up, but can at least one person quicly explain to me 15-18 on this section 5??? Got them ALL wrong. </p>

<p>(I was really proud of myself for getting a 660 on the math... first time over 600 yet, so yay! I really appreciate all your help on CC.)</p>

<hr>

<p>SECTION 8</p>

<p>(Below) Is there an easy/quick way to do 11? I did it by plugging in variables and got it right. Can someone take a quick look at it... it looks simple. It's on p. 546 thanks.</p>

<ol>
<li>If k is a constant and 2x + 5 = 3kx + 5 for all values of x, what is the value of k?</li>
</ol>

<p>a 5
b 3
c 2
d 3/2
e 2/3</p>

<hr>

<ol>
<li><p>I was pretty sure I was right on this one. Not sure what I did wrong. I said 4, btw.</p></li>
<li><p>In a certain game, each token has one of three possble
values: 1 point, 5 points, or 10 points. How many different
combinations of these token values are worth a total of 17
points?</p></li>
</ol>

<p>a 2
b 3
c 4
d 5
e 6</p>

<p>^ I spend 2 mins on this one^. I was sure I was right. Any shortcuts
to doing it?</p>

<hr>

<ol>
<li>A graph question. Really easy. p. 547 please take a look.</li>
</ol>

<hr>

<ol>
<li>I omitted it... It was impossible. Can anyone be a superhero and solve it?</li>
</ol>

<p>A cube with volume 8 cubic cms is inscribed in a sphere so that each vertex
of the cube touches the sphere. What is the length of the diameter, in cm's,
of the sphere?</p>

<p>a 2
b sqrt(6)
c 2.5
d 2sqrt(3)
e 4</p>

<p>I thought it was (2)sqrt(2).</p>

<hr>

<p>I appreciate everyone who is helping me achieve a higher score. It's good practice
for you to help me, these are the harder questions.</p>

2

<p>Section 2
18. Since no 3 of the 6 pts are on the same line, you can get a unique line by connecting any combination of 2 points. There are 6*5/2 combinations of 2 points, and thus there are 15 lines.</p>

<ol>
<li><p>The total rope length is y+4x Since the area is 4000, x is 4000/y. Thus, the total length of rope is y+16000/y</p></li>
<li><p>g(2)=f(3*2+1)=f(7)=-5</p></li>
<li><p>There are 3 sizes of semicircles(AD, CD, and BD being diameters). Call the biggest semicircle A, the middle sized one B and the smallest one C. To find the area of half the shaded portions, do this: Area of semicircle A-Area of SC B+Area of SC C. Double this to bet the correct answer</p></li>
<li><p>f(a+b)=f(a+a)(since a=b)=f(a)+f(a)=2f(a)
This is not equal to f(a)^2 for obvious reasons
2f(b)=2f(a) since b=a</p></li>
</ol>

<p>Section 5
6. If x is inversely proportional to y, then x=k/y where k is a constant.
1/x^2 is thus y^2/k^2 Thus, it’s proportional to y^2, since they are related by a constant(1/k^2)</p>

<ol>
<li><p>Drawing it is the best way to solve it. You don’t have to draw all diagonals; just those from one point.</p></li>
<li><p>(x-8)(x-k)=x^2-5kx+m
Thus, expanding,
x^2-(8+k)x+8k=x^2-5kx+m
Thus,
-(8+k)=-5k and
8k=m
k is 2, so m is 16</p></li>
<li><p>Since x is 40, and OB bisects angle AOD, angle BOD is also 40. Thus, angle AOD is 80. Since OD bisects angle AOF, angle FOD is 80. Since y is 30, angle EOD is 50. Angle BOE is thus 40+50=90</p></li>
<li><p>There is 1 1, 2 2’s, 3 3’s, 4 4’s, etc. To get to the 1st 12, you need
1+2+3+4+5+6+7+8+9+10+11 terms, which is 66. The 1st 12 is thus the 67th term</p></li>
<li><p>There are 20 4 inch lengths in 80 inches. On each 4 inch length, there is a smooth 3 inch part and an equilateral triangle, whose sides are 1 inch. Thus, the bolded part of each notch is 2 inches, so each 4 inch length has 5 inches of bolded edge. 205=100, the answer.</p></li>
<li><p>The square’s side length is 8, to point R must be (4, 8). Plug it in, and you get 8=a4^2. Thus, a=1/2</p></li>
</ol>

<p>Section 8
11. Cancelling the 5’s, you get
2x=3kx
Cancelling x’s, you get
2=3k
Thus, k=2/3</p>

<ol>
<li><p>This of the possible numbers of 1’s: There can be 2 1’s, 7 1’s, 12 1’s or 17 1’s. Then, make a chart like this:
17 1’s–> 1 combo for the other 2 values(0 5’s, 0 10’s)
12 1’s—>1 combo(1 5)
7 1’s—>2 combos(1 10 or 2 5’s)
2 1’s—> 2 combos(3 5’s or 1 5, 1 10)
The total number of combos is thus 6</p></li>
<li><p>Every value is doubled, so the graph stretches vertically. The y-intercept, originally slightly less than 1, becomes greater than 1, eliminating B, C, and E. A shifts the graph up, so D is the answer</p></li>
<li><p>If every vertex touches the sphere, than the diagonal of the cube is equal to the diameter of the sphere. A volume 8 cube has an edge length of 2. Using Pythagorean theorem, you can derive that the diagonal of a cube is sqrt(3) times its side length. Thus, the diagonal is 2sqrt(3), which is thus the diameter of the sphere.</p></li>
</ol>

3

<p>Sorry, don’t wanna do all of the questions, but for the first one,
two points determine a line. Set up a combination.
6 cpr 2.</p>

4

<p>^
No worries, I already did all of them.</p>

<p>Remember to divide by 2, since AB and BA are the same line.</p>

5

<p>WOW!!! I’m impressed. I honestly didn’t think anyone was going to reply. However, I am still confused on a few problems an I’ll say why.</p>

<p>Section 2</p>

<ol>
<li><p>I have no idea what you mean. I don’t understand how you know that circles abc and bcd are half shaded in. </p></li>
<li><p>How did you get the 6*5/2. I can’t see your process. </p></li>
</ol>

<p>Section 5</p>

<ol>
<li><p>When you said something about ‘k’ I got confused. I just assumed inversely prop. meant that x = 1/y so x^2 = 1/y^2. Then x^2*y^2 = 1. Finally y^2 = 1/x^2.
Am I wrong?</p></li>
<li><p>How did you eliminate so many variables. I expanded it to:</p></li>
</ol>

<p>x^2 - kx - 8x + 8k = x^2 - 5kx + m </p>

<pre><code>4kx - 8x + 8k = m <What do I do from here?
</code></pre>

<p>4(kx - 2x + 2k) = m </p>

<p>lol i have no idea ^</p>

<p>(GREAT JOB on explained 15-18 and others. I actually understand. Just some were very tricky.)</p>

<p>last one…</p>

<p>Section 8</p>

<ol>
<li>The reason why I’m so confused on this one is strange. It seems like it’s so simple.
I thought that when you make a diagonal in A SQUARE, you bisect the right angles, and create a 45 45 90 triangle. I don’t understand. If this were true, it would have the length 2sqrt(2). But I don’t understand how you got 2sqrt(3)…</li>
</ol>

<p>Thanks again for all your help. I need it since I’m taking the SAT in 3 weeks for the 1st time as a sophomore.</p>

6

<p>oh yay you are taking it at the same time and as a sophomore. For section 2
18. Pairs of points determine a line. If you took algebra two then you might see that this is a combination problem. 6 choose 2. The formula is 6!/2!4! which equals 15.</p>

7

<p>Section 2
17. Think of it like this:
Divide the big circle in half along AD. Each side is the same. Look at the top semicircle formed by A and D. The area of that semicircle covers all the shaded parts in the top semicircle except the middle white section. The area of the middle white section is the area of the semicircle formed by B and D minus the area of the semicircle formed by C and D. Thus, the shaded area in the top half is Area of big semi-Area of White part, which is Area of big semi-(Area of Medium semi-Area of small semi), which simplifies to what I explained earlier.</p>

<ol>
<li>Okay. Each line is uniquely determined by 2 points. You have 6 choices for the 1st point and 5 choices for the second. Thus, the number of combinations is 6 times 5,or 30. However, since A and B make the same line as B and A, you have to divide this by 2 to get 15.</li>
</ol>

<p>Section 5
6. Your reasoning is almost correct. Proportional means that one thing is always a constant number times another thing. For example, the diagonal of a square is proportional to its side length because the diagonal is the side length times sqrt 2.</p>

<ol>
<li>You have (8+k)x+8k=5kx+m
Since this is true for all x, that means the coefficients for the x term( 8+k and 5k) have to be equal. Similarly, the constant terms(8k and m) also have to be equal.</li>
</ol>

<p>Section 8
16. The diagonal of a square is sqrt2 times its side. To find the diagonal of the cube, notice that the triangle formed by an edge of the cube, the diagonal of a face of the cube, and the diagonal of the cube is a right triangle, which the cube diagonal as the hypotenuse. For a cube with edge length 1, the diagonal is thus
sqrt(1^2+(sqrt 2)^2)=sqrt(1+2)=sqrt 3.
Thus, a cube with edge length 1 has a diagonal of sqrt 3. Sqrt 3 can then me multiplied by the edge length of any cube to get the cube’s diagonal.</p>

8

<p>

</p>

<p>x^2-8x-kx+8k=x^2-5kx+m (FOIL out the left side)
-8x-kx+8k = -5kx+m (subtract x^2)
-8x-kx = -5kx … and … 8k = m (break up the equation)
-x(8+k) = -x(5k) (Factor out a -x)
8+k = 5k
8 = 4k
k = 2</p>

<p>Plug into k = 2 … 8k = m.
8(2) = m
m = 16</p>

9

<p>I don’t know how big a pain in the a** I’m being right now, but I’d like to thank you for all your help. I understand what you’re.</p>

<p>I learned (and wrote on my whiteboard that 'the diagonal of a cube is the side’s length times sqrt(3)) so much today!! thanks a lot</p>

<p>Only 2 more questions. In observation of the pattern, I should have no more questions after this one so that’s good.</p>

<p>Section 2 </p>

<ol>
<li>I wish you were my tutor so you could show me. I began to understand what you we’re saying, then I got lost.</li>
</ol>

<p>Section 5</p>

<ol>
<li>Same as above^. I am soo lost.</li>
</ol>

10

<p>wait don’t answer that last one… I think I get it.</p>

11

<p>@AvidStudent I got lost at the ‘break-up the equation’ I srsly said to myself noooo! Probably cause I’ve never seen that before. How’d you get there?</p>

12

<p>Where exactly did you get lost? Did you get how there are 3 semicircles and how you can add/subtract their areas to find the area of the shaded regions?</p>

13

<p>wow. I see now. Correct me if I’m wrong, but you want me to tread -8x-kx as one thing? I still don’t see how that makes it somehow magically equal to -5kx.</p>

14

<p>@garfieldliker </p>

<p>I figured it out because of YOU! yay! I had to recall some 9th grade geometry to understand, and now I do. You just subtract the area of semicircles! Why did I think it was so hard? thanks a bunch</p>

15

<p>I think collegeboard’s answer is clearer than mine.
Go to SAT Study Guide – SAT Suite | College Board and you can see explanations for every answer.</p>

16

<p>Section 2 - #17:
In the case of #17 it’s really hard to explain the solution without showing you a picture.
Anyways here is another way to cross out some answers.
AD = 6
That means the radius of the circle = 3.
Area = pi(r^2)
Area = 9Pi</p>

<h2>Thus, you can then cross out D and E.</h2>

<p>I flipped the bottom part the 180 degrees so it mirrored the other part in white. Follow?
Radius of this circle is equal to 2.
Area of this circle is equal to 4Pi.</p>

<p>But, wait there is a smaller circle within the 4Pi circle with a radius of 1.
Subtract the 1Pi from the 4Pi and then subtract that from 9Pi.
9Pi - (4Pi - 1Pi) = 6Pi.</p>

17

<p>haha AvidStudent that’s what I did! I was like, no way D and E!, and I crossed them out. Then I was like, hmm looks like about 1/3 of the circle. so on the test I got it right, I just didn’t understand how.</p>

<p>oh and I didn’t know CB has explanations online. oops</p>