MAA-BIH ORG

I’ve been entered into the Wespac Maths competition in year 7 and i’ve been revising on the past Wespac Maths booklet 2005.

I need some help with these questions and what is the formula/ how did you figure it out.
———————————————————————————–
If S is the sum of the remainders when each of the numbers 30, 31, 32, 33, 34 and 35 is divided by , then what is the remainder when S is divided by 6?
a)0
b)1
c)2
d)3
e)5
———————————————————————————
If it is 9am now, what time will it be in 2003 hours from now?
a)11pm
b)8pm
c)8am
d)11am
e)12noon
———————————————————————————
At Utopia High each of the 1516 students voted either ‘YES’ or ‘NO’ on whether to change the school uniform.
There was 1162 more ‘YES’ votes than ‘NO’ votes. The number of students who voted ‘NO’ was
a)344
b)254
c)177
d)172
e)127
———————————————————————————
The side of each of the equilateral triangles in the figure is twice the side of the central hexagon. What fraction of the total area of the six triangles is the area of the hexagon?
a)1/6
b)1/12
c)3/4
d)1/4
e)2/3

Thank you ————————————-…
If S is the sum of the remainders when each of the numbers 30, 31, 32, 33, 34 and 35 is divided by 6 , then what is the remainder when S is divided by 6?
a)0
b)1
c)2
d)3
e)5
————————————–…

Awww, the Westpac Maths Competition. I remember doing that!

—————————————-
1. You have left out a word. "when each of the numbers 30, 31, 32, 33, 34 and 35 is divided by ?????"
—————————————-
24 hours in a day.
So 2003/24=83 with a remainder

If we add 24 hours to 9am today any number of times, the time will always equal 9am.

83*24=1992
2003-1992=11

That means, that 83 days from now, the time will be 9am. BUT that’s only 1992 hours away. We need to add another 11 hours onto it.

9am + 11 hours = 8pm
—————————————-
let y = number of students who voted yes
let n = number of students who voted no

There are 1516 students total so that means:
y + n = 1516

There were 1162 more yes votes than no votes, which means:
y = n + 1162

If we substitute the second equation into the first equation, we get:
n + 1162 + n = 1516
2n + 1162 = 1516
2n = 354
n = 177

So 177 students voted no.
—————————————-
I need the picture to work this one out!

MATHEMATICS is a subject which needs to be practised.it is fun to solve problems which are a little easy to solve but to solve difficult problems its seems to be more critical to think about it. what we do so that our mind may not disturbed while facing such problems? i have done some experiment on this topic and i found that the most simple trick to concentrate our mind while solving a critical problem and to make our mind relaxed while facing such problems is MUSIC. its very funny that music helps in studying mathematics but its quite true. while studying mathematics if some song is playing in your surrounding you feel more relaxed and tension free and also the problems are easily solved. So those who have problems in mathematics must try it……………….and please tell me about your experience…….

Mathematics is the fundamental subject for almost all the scientific knowledge. Music will not help u in solving problems of mathematics, it will only a means of relaxation and not the kit for finding solution to the problems. The solutions will be possible only when u know the subject otherwise not. Playing music without knowing mathematics will not give solutions to the problems of mathematics.

My sincere suggession to u is don’t ever try to mislead people with ur personal perceptions regarding great fundamentals of science like this.

Hello! I’m currently a freshman in high school, wishing to master math competitions. I have participated in the middle school Mathcounts competition, placing around 20th statewide in 7th and 8th grade.

Now, in high school, I struggle to comprehend many of the questions that are given in the various competitions. I’m looking for a strategy that will enable me to excel in these math competitions. I understand that I’ll have to invest many, many hours of studying, and more importantly, doing problems, but I don’t know where to begin. I’m asking for a starting point, or a program, or some books, etc. Here are two sample competitions:

http://www.math.duke.edu/dumu/mathmeet/
http://www.mandelbrot.org/

(sample questions are on the sites)

Thank you so much!

Here are a few books I’d recommend.

can someone tell me which site is the best for learning mathematics basics and get homeworkhelp?i need to learn the basics well i find it difficult to do mathematics problems as i have forgotten some basics and my foundation is not very good.

Alex,

Here are two great sites:

mathtv.com

and

mathbits.com

I have done relatively well on math contests like AMC (95th+ percentile since freshman year) in high school but gotten 70’s in high school math classes and getting 60’s in first year university calculus classes. Why is this? It’s opposite for people I know. I can’t get my math marks up.

I have seen many of my friends facing this problem. Generally the standard of the test conducted in school is lower than that of the Maths Competition. So, students are over confident in school test. They take those tests lightly.In COmpetition, students become very conscious about their ranks and hence solve problems very cautiously.(They take minimum risk as a single mistake can cost them a great loss !)

Why do some persons, although they are perfectly fine with many other academic disciplines, and even possess a high IQ, have an almost impossible time with understanding some mathematical applications? I am not actually speaking of simple addition and subtractions, but math applications that begin with algebra and higher.

I understand that Einstein had problems, but still did what he did. That fact alone is perplexing in itself.

Is it a brain function that is not functioning properly? Is it the idea of numbers themselves as being abstract? Is it a perception problem?

Thanks for any thoughts.

There is dyscalculia, which the the mathematical equivalent of dyslexia and has varying levels of severity. This is a problem partially perception and partially how the brain works, but tends to affect maths at an earlier level.

For the majority though, it’s more how they think, there’s nothing wrong with them. Much like an otherwise intelligent person with good hand to eye co-ordination cannot sketch a picture that looks almost life like, while someone else with the same IQ etc can. Or some people can be creative, inventing poems, or new machinary, or art, or so on, whereas other people with the same IQ don’t have the inventive spark. In the case of algebra, you need a form of completely abstract logical processing. It’s not that they don’t posess it, it is more how adept they are in using it. Everyone has weaknesses.

-Im 14 turning 15. I take part in competitions like AMC, ASMA and Mandelbrot, do any of you have any tips? Such as how to prepare, what to do, how to approach? Also anyone know any good sites where i could get practice questions of competition/olympiad level?

I’m going to just provide some links such as this to the Mathematical Association
of America with some sample questions

http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC1012/2009-10-AMC1012-bro.pdf

If you practice SAT questions, many of them require some math excellence to
get up towards 750 to 800

This link will help also

http://www.mathpropress.com/competitions.html

you could independently put math contests/math competitions into Yahoo! search
{or that search engine whose name begins with "G" } and find numerous helpful links

A circle ω is inscribed in a quadrilateral ABCD. Let i the centre of ω. Suppose that
(|Ai| + |Di|)^2 + (|Bi| + |Ci))^2 = (|AB| + |CD|)^2. Prove that ABCD is an isosceles trapezoid.

Here is your proof ( scroll down to Q. 6 )
with my best wishes :

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04USAMO_solutions.pdf#page=6
…………………………………………………………………………………………….

EDIT : Thanks to Mr. Daftary, the direct link to the question

. . .. . is now available.
……………………………………………………………………………….

Happy To Help !
………………………………………………………………………………………………………..

I have also tried searching for somewhere to download them, however, I haven’t been able to find anywhere with free download. I think the only option is to buy them from the AMC website. Here’s the link to buy them, if you are interested:

http://www.amtt.com.au/

Hope this helped!

Q Let z= x + iy be a complex number where x and y are integers. Then the area os the rectangle whose vertices are the roots of the equation
z(conj z)^3 + (conj z)(z)^3
is ? plz solve it step by step!!

For simplicity I’ll refer to the conjugate of z as z’.

z = x + yi
z’ = x - yi
zz’ = x^2 + y^2

The equation is: zz’^3 + z^3z’ = 0, and zz’ can be factored out.

zz’ (z^2 + z’^2)
(x^2 + y^2) (x^2 - y^2 + xyi + x^2 - y^2 - xyi) = 0
(x^2 + y^2) (2x^2 - 2y^2) = 0
2x^4 - 2y^4 = 0
x^4 = y^4

Since x and y are defined as integers, this solves to x = ±y. So the four corners form a perfect square, each side of which has a length of 2x. The area should be 4x^2, which should also equal 4y^2. If you want the answer in terms of z, 2zz’ also fits.