Ok, I'll have a bunch of challenging questions from the AMC (Australian Mathematical Competition) I did today. I'll post one at a time so as not to confuse the posters and myself. The questions I suppose will be of ascending difficulty, those which I wasn't able to solve.

1. There's a given equation; . There was a sketch along, that of an inverted parabola, which had a positive y-intercept and the turning point was on the y-axis.

Which is true?
a) a + b + c = 0
b) a + b - c < 0
c) -a + b - c > 0
d) a + b + c < 0
e) There is not enough information.

I ruled out a) and d), since there is a solution other than 0 when putting x = 1.
The others, I'm at a lost.

Thanks for replying

Survivorboi, wanna make an attempt? I'm sure you'll be interested too to know how to solve the problems I'll post

morgaine300

Is this competition for fun, or for school? And how long before you know how you did?

As for the questions, if this is supposed to be the least difficult, I give up already.

Unknown008

I did it yesterday, and have no idea when the results will come. It's a competition among the students of the same age as me, in the country. I know, that was actually the eighth question. I managed with the first seven. Or, you want to take a look at them too?

morgaine300

I'm always willing to take a look at stuff. My math -- or at least my memory of it -- isn't near the level of the others here, so I just have to do stuff for my own fun.

I'm still trying to figure out what an "inverted" parabola is, as opposed to a, um, "normal" one. LOL. So is the curved part up or down?

Unknown008

LOL! Yup, my graphing program is not yet installed... I'll try get one online. It's a parabola with the coefficient of x^2 being negative (an inverted one, an upside down bowl )

morgaine300

Gotcha. If only I'd read this before the 2nd glass of wine. Now I'll have to think harder. :-)

So where's the first 7 questions?

Unknown008

Do you want me to post them too?

morgaine300

This took a lot of trial & error & thought since I don't know very many "rules" about parabolas. Like I had to spend quite a bit of time even finding an equation that would graph that way. (Can we do a linear one instead?) I did manage to figure out reasonably quickly that (a) can't be true.

I originally thought that C had to be greater than the absolute value of A, but finally figured out that isn't true. However, that I even had several equations where that was true eliminated (d). And (c) is kind of the reversal of that, cause as far as I can tell, B is 0 and makes no difference.

So I'm down to (b)... if I have everything else correct, this is it. We know A is negative and that C is positive. (We do know that, right? ) If I'm right that B is 0, then it's answer (b). If you start with negative and subtract C, you're more negative. Hence, less than zero.

Unknown008

Yup, I think you're correct...

c) would be correct also if -a was greater than c. But since b) does not put such conditions, it must be b).

I took some time to realise that b=0 (which was quite obvious).

Unknown008

Ok, if you want the first 7 questions, here they are:

1. Find the value of (2009+9)-(2009-9)
a. 4000
b. 2018
c. 3982
d. 0
e. 18

Obviously, the answer is e;18

2. In the diagram(first attachment sorry for the bad quality of the image ), x is equal to:
a. 140
b. 122
c. 80
d. 90
e. 98

Here the answer is e. 98

3. The graph of y=kx passes through (-2, -1). The value of k is
a. 2
b. -2
c. 4
d.
e.

Of course the answer is d=1/2

4. The value of is
a. -0.36
b. 0.036
c. 9/25
d. 25/9
e. 3.6

The answer is d. 25/9

5. (x - y) - 2(y - z) + 3(z - x) equals
a. - 2x - 3y + 5z
b. -2x - 3y - z
c. 4x + y - z
d. 4x + 3y - z
e. 2x + 3y - 5z

The answer is a. - 2x - 3y + 5z

6. On a string of beads, the largest bead is in the centre and the smallest beads are on the ends. he size of the beads increases from the ends to the centre as shown in the diagram (second attachment).

The smallest beads cost $1 each, the next smallest beads cost $2 each, the next smallest is $3 each and so on. How much change from $200 would there be for the beads on a string with 25 such beads?

a. $25
b. $31
c. $40
d. $52
e. $55

I got b. $31 I'm sure of it.

7. If for every pair a, b of positive numbers, the value of 1*(2*3) is
a. 10/3
b. 10/7
c. 11/6
d. 9/2
e. 3/10

I got b. 10/7

The answers are in white. Just highlight them to see them.

Attached Images