theoneo

Whaaaat? Whatever, it was a good workout. I was using the red book, too.

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.

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.

25) The equation to find the area of a sector is (pi*r*r)*(angle/360). The first one is (pi*r*r)*(angle/360) = 3. The second one is (pi*(2*r)*(2*r))*(2*angle/360) = x. You want to find x. That simplifies to (4*pi*r*r)*(angle/180) = x --> (pi*r*r)*(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*r*r)*(angle/360) = x/8 --> substitute 3 for the left side --> 3 = x/8 --> x = 24. The answer is A.

quitejaded

Can you explain that?

gcf101

y1 = kx + 36
and
y2 = 6nx + n^2.

y1 = y2 for all values of x; that means y1 and y2 coincide.
==============================
In other words, straight line y1 is the same as straight line y2.

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:

k = 6n and 36 = n^2.
|n| = 6, n>0 => n=6.

k = 6*6 = 36.
k - n = 36 - 6 = 30

(y1 = y2 = 36x + 36)

ryukun

oh, nice. That's quite a question.

loganr

theoneo , just curius .. what was ur math score?

quitejaded

That's the part that throws me off. How'd you figure 36=n^2?

I don't know how I got higher math scores than verbal o_O

gcf101

"36" and "n^2" are the y-intercepts in
y1 = kx + 36
and
y2 = 6nx + n^2.

Now, your turn to enlighten: how do you do html thingies here (quote box, underlining, bold, cursive, whatnot)?

Cuong

You do not have to do excessive arithmetic, I hate terribly that part. Let's manipulate it a bit and we shall get it alright.

x=7+y (1)
4x = 6-2y (2)

deal with (2) first
4x = 2(3-y)
divide both sides you get:

2x = 3-y (3)

Compare (3) and (1) you can see that if you add the two equations together, you eliminate y:

2x = 3-y
x = 7+y
------------
3x = 10
x = 10/3

yumihooo

20. x=7+y ==> 2y=2x-14
4x=6-2y ==> 2y=6-4x

so, 2x-14=6-4x ==> x=10/3