[Unknown008]

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, want to 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? :o) 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 :o), 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.

[Unknown008]

Ok, next questions are easier.

9. in a school of 1000 students, 570 are girls. One-quarter of the students travel to school by bus and 313 boys do not go by bus. How many girls travel to go to school by bus?
a. 7
b. 63
c. 153
d. 180
e. 133

The answer is e. 133

10. A box in the dressing shed of a sporting team contains 6 green and 3 red caps. The probability that the first 2 caps taken at random from the box will be the same colour is
a. 1/2
b. 5/12
c. 2/3
d. 3/4
e. 2/9

The answer is a. 1/2

11. QRST is a square with T at (0,1) and S at (2, 0). Which of the following is an equation of the line through the origin which bisects the area of the square? (attachment 1)
a. y = x/2
b. y = x/3
c. y = (2x)/3
d. y = 2x
e. y = 3x

The answer is y = x/3

12. [I'm stuck at this one too :()
A rectangle PQRS (attachment 2)has PQ = 2x cm and PS = x cm. The diagonals PR and QS meet at T. X lies on RS so that QX divides the pentagon PQRST into two sections of equal area. The length, in centimetres of RX is
a. x/2
b. x
c. (5x)/4
d. (3x)/2
e. (3x)/4

I ruled out a, b and e because they are too short. I don't know for the rest...

[galactus]

#12:

The area of the pentagon PQRST is

Half of that is

If we let RX have length y, then the area of the triangle formed by QRX has area half the area of the pentagon which is

Solving for y gives us

[morgaine300]

Interesting mix.

1. Find the value of (2009+9)-(2009-9)

They just wanted you to have a success before delving into the real test? :)
(I really shouldn't comment. I have no idea what this competition is, or the age groups involved, etc.)

2. In the diagram(first attachment sorry for the bad quality of the image :o), x is equal to:

I don't remember too many geometry rules cause I really don't work with it -- I remember some basics, and sometimes I can figure stuff out by logic. But I have no clue where your answer is coming from. I just made up a rule cause I didn't know and screwed it up. :eek:

4. The value of is

Well, that's easy. You pull out the calculator and put in (.6)(2nd)(y^x)(2)(+/-)(=). :D

Actually... I didn't cheat. I solved it manually. Really I did.

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).

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

You don't sound too sure. But it's right.

7. If for every pair a, b of positive numbers, the value of 1*(2*3) is

I don't even get what they're saying. 1*(2*3) is 6. How does that relate to a & b? Seriously, I don't get it.

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

Nifty idea.

Except for (7) and not remembering triangles properties, I thought this was all pretty easy. (I tutor most of this stuff.) The parabola one almost seems out of place with this stuff. (I only solved it cause I don't have a time limit. :p) But I haven't looked at the other set yet. I'm supposed to be getting work done in my yard while it's still light.

[morgaine300]

As for 9 - 12... well, 9 was easy.

10 I got but it was the long way. I knew there were 72 ways it could be done, and then had to manually figure out that 36 of those would be the same color. I don't know how to do that any shortcut way. Might be one of those things that I actually know, but don't recognize it as being such. Probabilities can be like that.

11 & 12 are both getting beyond my memory. I have a feeling I could get 11 if I tried hard enough, but my mind isn't up for the challenge at the moment. And 12 - no clue. I don't even know what galactus is talking about. Way too many years since I did that kind of stuff.

(Notice I'm having trouble with shapes? :D)

[Unknown008]

Lol, what I'm giving answers, I'm 100&#37; sure of them, morgaine! :) It's OK, I gave the paper to my mum too, and she too was rusty!

Thanks galactus! I had actually found the answer some time after I posted the question. I'm kind of a little frustrated it was as easy as that... but I can do nothing, I was thinking it was some kind of 'logical' work like some questions. I should've done the calculations... :-/

And, the level is actually 'Senior Division', with 'Australian School Years 11 and 12', 'Time allowed: 75 minutes'.

Those aussie do have a high level of maths! Even being 17, I'm stuck at some of their questions!

Ok, the next one, 13, which I had problems as well.

13. The solution to the equation is rational number of the form where b is not equal to 0 an a and b are positive and have no common factors. What is the value of a+b?
a. 3
b. 5
c. 7
d. 9
e. 11

I started like this:

Let y = 5^x

So, we have:









so a+b=9

Oh my... I have solved it... and saw my mistake... I'm weird, I'm really weird...

The mistake I did was 120*25=300, which of course is 3000. :o

14 How many points (x,y) on the circle [math]x^2=y^2=50[/math are such that at least on eof the coordinates x, y is an integer?
a. 16
b. 30
c. 48
d. 60
e. 100

I got d. 60

15. An eyebrowis an arrangement of the numbers 1, 2, 3, 4 and 5 such that the second and forth numbers are each bigger than both their immediate neighbours. For example, (1, 3, 2, 5, 4) is an eyebrow an (1, 3, 4, 5, 2) is not. The number of eyebrows is:
a. 16
b. 12
c. 15
d. 24
e. 18

My answer: a. 16

16. The sum of the positive solutions to the equation is :
a. 5
b. 7
c. 8
d. 9
e. 18

I didn't know at first, but after the competition, I figured it out (another frustration) The answer is b. 7

17 On a clock face, what is the size, in degrees, of the acute angle between the line joining the 5 and the 9 and the line joining the 3 and the 8?
a. 15
b. 22.5
c. 30
d. 45
e. 60

The answer is d. 45

18. A positive fraction is added to its reciprocal. The sum is x/60 in lowest terms, where x is an integer. The number of possible values of x is
a. 1
b. 2
c. 3
d. 4
e. 5

I don't know how to do this.

I started with:

Then, equating, ab = 60 , a^2+b^2 = x

What I think now, is that one of the two must be a fraction, for a^2+b^2 > ab for all positive integers. Or I'm overlooking something...

[galactus]

18. A positive fraction is added to its reciprocal. The sum is x/60 in lowest terms, where x is an integer. The number of possible values of x is
a. 1
b. 2
c. 3
d. 4
e. 5

I don't know how to do this.

I started with:

Then, equating, ab = 60 , a^2+b^2 = x

What I think now, is that one of the two must be a fraction, for a^2+b^2 > ab for all positive integers. Or I'm overlooking something...

The values of ab=60 that satisfy are:

15*4=60
20*3=60
10*6=20
30*2=60
12*5=60

15/4+4/15=241/60
10/6+6/10=34/15
30/2+2/30=226/15
12/5+5/12=169/60
20/3+3/20=409/60

3 of those have the required 60 as the denominator when in lowest terms.

[Unknown008]

Oh, so that's what I was overlooking... I was looking for proper fractions... Thanks galactus!

Ok, next (there are 30 in all):

19.

In triangle PQT, PQ = 10 cm, QT = 5 cm and angle PQT = 60 degrees. PW, PY and TQ are tangents to the circle with centre S at W, Y and V respectively. The radius of the crcle, in centimetres, is
a.

b.

c.

d.

e.

I managed to get PT = 75^{0.5) [from cosine rule] and angle PTQ = 90 degrees [from sine rule]. That makes angle TPQ = 30 degrees, and angle YSW = 150 degrees. I'm stuck here.

[galactus]

19.

In triangle PQT, PQ = 10 cm, QT = 5 cm and angle PQT = 60 degrees. PW, PY and TQ are tangents to the circle with centre S at W, Y and V respectively. The radius of the crcle, in centimetres, is
a.

b.

c.

d.

e.

I managed to get PT = 75^{0.5) [from cosine rule] and angle PTQ = 90 degrees [from sine rule]. That makes angle TPQ = 30 degrees, and angle YSW = 150 degrees. I'm stuck here.

See how I got that? I used some of the old surveyor trig. This problem is a lot how a curve in a road is laid out.

[Unknown008]

Is that some sort of formula? I can't get it. I see you took the 30 degrees of the place where the two longer tangents meet...

[galactus]

TV and TY are equal to the radius of the circle length.

If we look at angle VSW is equal to 60 degrees.

VQ=

And TV+VQ=5

and solve for R.

See, we can find the tangent length, VQ, from the formula

where I is a central angle VSW, which is 60 degrees.