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
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 dunno for the rest...
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.)
Quote:
2. In the diagram(first attachment sorry for the bad quality of the image ), 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.
Quote:
4. The value of is
Well, that's easy. You pull out the calculator and put in (.6)(2nd)(y^x)(2)(+/-)(=).
Actually... I didn't cheat. I solved it manually. Really I did.
Quote:
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.
Quote:
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.
Quote:
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. ) 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.
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.
Lol, what I'm giving answers, I'm 100% 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.
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...
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...
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.
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.
) 
), x is equal to:
) 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.
Linear Mode
