<p>sunday may test 1996, on page 433</p>
<p>section 1. number 17,18,19,21,22</p>
<p>i know I missed alot, but i was timing myself, btw can u mention shortcuts to these if u find any</p>
<p>i think it would help if u posted the actual problems</p>
<p>The page # is 327.</p>
<ol>
<li><p>if x=7+y and 4x=6-2y what is the value of x
a.-4
b.-4/3
c.-1/6
d.10/3
e.10</p></li>
<li><p>if n>0 and 9x^2+kx+36=(3x+n^2)for all x what is the value
of k-n
a.0
b.6
c.12
d.30
e.36</p></li>
</ol>
<p>thats all I feel like posting, (really need to know those) ill try 2 figure the others out later.</p>
<ol>
<li>substitution
4(7+y)=6-2y
28+4y+2y=6
6y=-22
y=-22/6=-11/3</li>
</ol>
<p>x=7-(11/3)=(22/3)-(11/3)
x=10/3
Answer: D</p>
<p>I don't know how to do the second one..btw is this SAT I MATH?</p>
<p>I just did the second one..</p>
<p>24) it says for all x, so substitute the x for 1 and 0 respectively
when you substitute x for 1:
9+k+36=3+n^2
42+k=n^2
k-n^2=42</p>
<p>when you substitute x for 0:
36=n^2
n=6</p>
<p>and now you substitute the second equation for first
k-n^2=42
k-6^2=42
k+36=42
k=6</p>
<p>k-n=?
6-6=0</p>
<p>iwantfood ---</p>
<p>I was curious too about those questions, and followed along with your example. However, I'm not sure I understand it:</p>
<p>Math's never been my strongest point, so maybe I'm wrong, but:</p>
<p>You said:
42+k=n^2 -- makes sense
k-n^2=42 - How? If 42+k = n^2 then doesn't 42+k-n^2 = 0 and k-n^2 = -42?
so if n = 6, k = -6?
and you said:
k-n^2=42
k+36=42 -- How does k-6^2 = 42 become k + 36 = 42? I thought that according to PEMDAS, k-6^2 is equivalent to k-(6^2) or k-36 = 42, in which case k is 78.
And when I did the substitution of x with 1 ended up w/... = k + 42 = n^2
n^2 = 36
so k + 42 = 36 or k = - 6?</p>
<p>Now I know this can't be the right answer, because -6 - 6 is -12, which is not an option. What am I not getting here?</p>
<p>Thanks to anyone who can help
Andre</p>
<p>obssesed andre everything u said is correct until the very end. its k-n
which like u said both are -6 so its -6--6 which is 0. just a little mistake but makes a huge difference so answer should be a. 0 i think</p>
<p>ijm8710, I thought exactly the same thing you did, I thought maybe that was the problem, but the problem asked for k-n
and I got k = -6, and n = 6
so k-n = (-6) - 6 = -12
I thought maybe n could equal -6 too, which would be (-6) - (-6) = 0, but we were told in the problem that n > 0 so n must be the positive square root of 36, not the negative one, so n must be 6.</p>
<p>Man, that's the November 1995 test, not May 1996. I was searching my book cause I knew I'd seen that second problem somewhere. I just did this test a couple days ago, so here it goes:</p>
<p>17) What is the slope of a line that passes through the origin and the point (-2, -1)?
slope = (y2 - y1)/(x2 - x1), (x1, y1) = (0, 0), (x2, y2) = (-2, -1)
slope = (-1 - 0)/(-2 - 0) = (-1)/(-2) = 1/2 --> B</p>
<p>18) Julie has cats, fish, and frogs for pets. The number of frogs she has is 1 more than the number of cats, and the number of fish is 3 times the number of frogs. Of the following, which could be the total number of these pets?
frogs = cats + 1, fish = 3frogs
total = fish + cats + frogs
You want to find the total, so you can just start adding them together and simplifying through substitution
= (3 x frogs) + cats + (cats + 1) <-- substituted values of fish and frogs
= (3 x (cats + 1) ) + cats + (cats + 1) <-- substitutded value of frogs again
= 3 x cats + 3 + cats + cats + 1
= 3 x cats + 2 x cats + 4
= 5 x cats + 4
Since the number of cats must be an integer greater than 0, the total can be 5(1)+4 = 9, 5(2)+4 = 14, 5(3)+4 = 19, etc. Notice how the only answer choice that fits is E)19.</p>
<p>19) If x is an integer, which of the following could NOT equal x^3?
Don't be fooled by A, since you can take the odd root (3, 5, etc) of a negative number, but not an even one (2, 4, etc). In this case, you could just start from 0 and work your way up until the cube of the number is greater than 27. so 0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64. Logically, the cube root of 16 should be between 2 and 3, but no INTEGER value works, so D) 16 is the answer.</p>
<p>21) The incomplete table above categorizes the members of a ches club according to their age and status. During a tournament, each member of the club plays exactly one game with each of the other members. How many games of chess are played between amateurs 20 years or older and professional under 20 years old during the tournament?</p>
<p>First you need to finish the chart. 5 goes in the top right box and 3 goes in the bottom left box in order to satisfy the totals. So you need to figure out how many games were between the 3 young professionals and the 5 old amateurs. If you don't know this already, memorize it: the number of games between a group of x people and a group of y people in which each person competes with every single person in the other group once but never with someone in his/her own group (in other words, exactly what this problem is) is equal to x times y. In this case, that's 3 x 5 = 15 games. To check that, you could write A, B, C, D, and E in order to represent the five amateurs and draw three lines from each to represent one game with each professional, or vice versa, and count the number of lines. The answer is C.</p>
<p>22) A bag contains a number of pieces of candy of which 78 are red, 24 are brown, and the remainder are yellow. If the probability of selecting a yellow piece of candy from ths bag at random is 1/3, how many yellow pieces of candy are in the bag?
probability = chance of happening / all possibilities, which is the same as p(x) = x/total. You can use thiss as a formula. You know the probability p(Y) is equal to 1/3. 1/3 = Y/(R + B + Y). How can we simplify this? By getting rid of the R and B variables. Just substitute their actual values. Now 1/3 = Y/(78 + 24 + Y) = Y/(102 + Y). Now you can just solve for Y: 1/3 = Y/(102 + Y) --> (1/3)(102 + Y) = Y --> 34 + Y/3 = Y --> 34 = Y - Y/3 --> 34 = 2Y/3 --> 102 = 2Y --> Y = 51, which is B.</p>
<p>I hope this helped!</p>
<p>MWUHAHAH I saved the best for last! loganr typed problem 24 wrong. Here it is:</p>
<p>24) If n > 0 and 9x^2 + kx + 36 = (3x+2)<b>^2</b> for all values of x, what is the value of k - n?</p>
<p>Hahahahah this was awesome. Since I found the mistake, I get to solve it, woohoo!</p>
<p>You know that (3x + n) = 9x^2 + 6xn + n^2, right? Doesn't that look oddly familiar? Ah, yes! The left side of the equation! But we'll get back to that in a minute.</p>
<p>On the left side, aren't the 9 and 36 perfect squares? Hm.. perhaps we'll be able to use that. Can we factor? Hey, since the right side is a perfect square, can the left side be one, too? Using just the 9 and the 36, the only possible square would be (3x + 6)^2. When expanded, that equals 9x^2 + 12x + 36. IF that works, then k = 36. Now, over to the right side. Hey, that (3x + n) looks similar to the (3x + 6) we squared a minute ago. Since both sides are equal, n would have to equal 6. Now, expand both sides, plugging in 36 for k and 6 for n. My word, it works! k - n = 36 - 6 = 30 --> D.</p>
<p>I'm not sure why I narrate in math problems. You'll just have to suck it up and deal :P.</p>
<p>The 9x^2 easily drop out; we don't need to look at them anymore. Now, the problem says FOR ALL VALUES OF X, so there must be more than one possible value of x. Therefore, we cannot expect it to have only one value. To accomodate this</p>
<p>Sorry, theoneo,
seems your sleepy hands could not hold a pen any longer... I mean, keyboard.</p>
<p>If you don't mind, squeezing your narrative:
++++++++++++++++++++++++</p>
<h1>If n > 0 and 9x^2 + kx + 36 = (3x+n)^2 for all values of x, what is the value of k - n?</h1>
<p>9x^2 + kx + 36 = 9x^2 + 6xn + n^2
kx + 36 = 6nx + n^2</p>
<p>Left and right sides of this equation are equations of straight lines for any set pair of values of k and n:
y1 = kx + 36
and
y2 = 6nx + n^2.</p>
<p>y1 = y2 for all values of x; that means y1 and y2 coincide:
k=6n and 36=n^2,
n=6 (since n>0),
k-n = 30.</p>
<p>BTW, this question is from Saturday, November 1994 test, white
"10 Real SATs" p.316.</p>
<p>guys thank for the contribution but thats not the test im on, I already got those correct btw theoneo, </p>
<p>ok its my fault for not giving enough info,
im working from the RED 10 real SAT, SECOND EDITION "the only source of real sat question from cover to cover"</p>
<p>on page 408,
I need help on 23,24,25</p>
<p>heres how they begin </p>
<ol>
<li>if n >1 and each...
24.if j and K are integers</li>
<li>in the figure above</li>
</ol>
<p>o true my bad forgot bout the greater than 0 part but it seems like that problem was impossible if the one is right lol</p>
<p>Whaaaat? Whatever, it was a good workout. I was using the red book, too.</p>
<p>23) n, n+2, and n+4. It's easier to think of them as every other number, like 1, 3, and 5. To begin with, the only even prime is 2, and every "every other" number after that will be even as well, so 2 cannot be in the answer. So now we know that the three numbers must all be odd. Well, right after 2 is 3, so we might as well try that. 3, 5, 7 - it works. Myself, I just started counting every other number up to like 39 when I realized that I wasn't getting anything, and there was probably some kind of strategy to this. So, IF 3-5-7 is the only prime triple, then there must be a reason for it that has to do with either the 3, 5, or 7. Is there something with one of those numbers that occurs every odd (n, n+2, n+4), every other odd (n, n+4, n+6) , or every third odd (n, n+6, n+12) that made it composite? Then it hit me. Every third odd. The number three is in that. Look at the multiples of 3: 3, 6, 9, 12, 15... How often are odd numbers found in the set of all multiples of 3? Every other number. How often are multiples of three found in the set of odd numbers? Every third number. That's why there can never be a prime triple beyond 3-5-7: each triple would contain exactly one multiple of 3. So the answer is B.</p>
<p>24) Subtract 2j from both sides and you end up with k - j = 4. That's an even number. III is in. Are there any other restrictions in terms of being odd/even in the equation k - j = 4? Nope. In fact, you could even graph it as k = j + 4 or j = k - 4. The graphs wouldn't have points only when j or k were even, so I and II don't make sense. The answer is D.</p>
<p>25) The equation to find the area of a sector is (pi<em>r</em>r)<em>(angle/360). The first one is (pi</em>r<em>r)</em>(angle/360) = 3. The second one is (pi<em>(2</em>r)<em>(2</em>r))<em>(2</em>angle/360) = x. You want to find x. That simplifies to (4<em>pi</em>r<em>r)</em>(angle/180) = x --> (pi<em>r</em>r)<em>(angle/45) = x. Maybe we can get the left side to equal the equation for the first circle's sector? You know you're going to have to use the first circle somehow to solve the problem. Divide both sides by 8: (pi</em>r<em>r)</em>(angle/360) = x/8 --> substitute 3 for the left side --> 3 = x/8 --> x = 24. The answer is A.</p>
<p>
[quote]
y1 = y2 for all values of x; that means y1 and y2 coincide:
k=6n and 36=n^2,
n=6 (since n>0),
k-n = 30.
[/quote]
Can you explain that?</p>
<p>y1 = kx + 36
and
y2 = 6nx + n^2.</p>
<h1>y1 = y2 for all values of x; that means y1 and y2 coincide.</h1>
<p>In other words, straight line y1 is the same as straight line y2.</p>
<p>If two equations represent one line, then the slope and y-intercept of one is respectively equal to the slope and y-intercept of another:</p>
<p>k = 6n and 36 = n^2.
|n| = 6, n>0 => n=6.</p>
<p>k = 6*6 = 36.
k - n = 36 - 6 = 30</p>
<p>(y1 = y2 = 36x + 36)</p>
<p>oh, nice. That's quite a question.</p>
<p>theoneo , just curius .. what was ur math score?</p>