Search results

Nah, I'm doing the intermediate. I don't think they vary too much in difficulty though as I've noticed that question 30 in the intermediate is around question 28/29 in senior. I've also noticed that their are quite a few common questions each year.

Here's a link to some past papers: Link removed due to copyright Good luck!

Nicely done, want to ask a new question?

Here:

Thanks, I should definitely draw a picture next time so I don't get it wrong. Can't make mistakes like this in the competition on Thursday.

NEW Q: . Small squares of side x cm have been removed from the corners, sides and centre of a square of side y cm to form the gasket shown. If x and y are prime numbers and the sum of the inside and outside perimeters of the gasket, in centimetres, is equal to the area of the gasket, in square...

Yeah this is right. Could you see what part of my working out is wrong please: Area of roll = pi*r^2 = 36pi (diameter = 12) Area of tube = pi*r^2 = 4pi (diameter = 4) Area of annulus = 32pi After half has been used: 16pi Which gives us a radius of 4cm and a diameter of 8cm. This is...

NEW Q: Thanom has a roll of paper consisting of a very long sheet of thin paper tightly rolled around a cylindrical tube. Initially, the diameter of the roll is 12 cm and the diameter of the tube is 4 cm. After Thanom uses half of the paper, the diameter of the remaining roll is closest to what...

Yep, I saw that x^2-8x = 1001y^2 x(x-8) = 1001y^2 So for x+y to be at a minimum, we had to have y as low as possible. So I substituted y values from 1 onwards until I got a result that worked, Y=3, X=99 (99*91 = 1001*3^2)

Around September iirc

Thought this was a good question: A regular octahedron has eight triangular faces and all sides the same length. A portion of a regular octahedron of volume 120 cm^3 consists of that part of it which is closer to the top vertex than to any other one. In the diagram, the outside part of this...

Great effort from the team.

Unfortuantely that's incorrect @si2136 The answer they've given is 40

Completed it for 1≤i≤2, 1≤i≤3 ... 1≤i≤5, and the answer was always a triangular number of n-1 For: 1≤i≤2, the sum of distances was 1 For: 1≤i≤3, the sum of distances was 3 For: 1≤i≤5, the sum of distances was 10 So for 1≤i≤42 the sum of distances will be 41*42/2 = 861 New Q: In a 3 × 3 grid...

The radius of the circle is 5, so the area is 25pi Area of shape = Area of Quarter circle + Area of Quarter circle + (25 - Area of Quarter circle) + (25 - Area of Quarter circle) 25 is the area of a 5*5 square The top left and bottom right form quarter circles the top right and bottom left...

Yep, the answer is 50. 25pi/4+25pi/4+(25-25pi/4)+(25-25pi/4)

Continuing the Marathon: 1. A high school marching band can be arranged in a rectangular formation with exactly three boys in each row and exactly five girls in each column. There are several sizes of marching band for which this is possible. What is the sum of all such possible sizes? 2...

Terry has invented a new way to extend lists of numbers. To Terryfy a list such as [1, 8] he creates two lists [2, 9] and [3, 10] where each term is one more than the corresponding term in the previous list, and then joins the three lists together to give [1, 8, 2, 9, 3, 10]. If he starts with a...

Would you have any tips for the amc (especially last 5) and aimo? Completing them both this year.