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I struggled with the time though and made two really dumb mistakes :S

You'd think by now I'd know that 3/6 is not more than half...

You'd think by now I'd know that 3/6 is not more than half...

It was easy but i didnt do that well. I froze up in Q2 with the parallel lines. I wasted 15mins then skipped it. When i went back to it, I realized that it was very simple.

In the end i got probably 70-80 but i found it very easy compared to last year's.

In the end i got probably 70-80 but i found it very easy compared to last year's.

How the hell do you do the question with "14 boys in each row and 10 girls on each column. Prove that there are at least 567 chairs"

I think I did very well. Looking at the CEMC website, the date for testing is officially over so I believe discussion is allowed...

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

What's considered as a good score (for the Math Faculty)? Top 10% is around 70. Top 25% is around 60.

Did you guys get the clock one? 8a

Did you guys get the clock one? 8a

Aghhh do you remember the question? I will be able to do it up if you just post the numbers and stuff but I forgot what I got. I was being INSANELY stupid for the boys and girls problem I literally sat there and stared at it for 40 minutes!

Also, I think A good score is considered high 50s or higher, correct me if im wrong. I happen to know someone who got a 97 on Euclid last year and received a lot of scholarship.

A great score would be like 70s... which is pretty hard obviously.

Also, I think A good score is considered high 50s or higher, correct me if im wrong. I happen to know someone who got a 97 on Euclid last year and received a lot of scholarship.

A great score would be like 70s... which is pretty hard obviously.

Can you post the other questions as well? If you have them at least just so I can make sure i didnt misread anymore lol :(

@ahuang wrote

I think I did very well. Looking at the CEMC website, the date for testing is officially over so I believe discussion is allowed...

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

YAY ! that was my answer too !!! i think i did well. doubt i got over 80 though. hoping for low 70s :P

@cometmars wrote

Since it's been over 48 hours:

http://bit.ly/IiBhdB

8a

Also, when do we get our results?

I'm pretty sure I got another answer on top of that for 8a? I got -1 too I THINK. Not sure. I got a quadratic expression to find two roots, and then another was a simple isolating-for-x expression.

@ahuang wrote

I think I did very well. Looking at the CEMC website, the date for testing is officially over so I believe discussion is allowed...

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

Correct me if I'm wrong but I think you ignored the part of the question where it said that there are exactly 3 seats empty in the auditorium. If your array has a size of 24x24 it assumes that it's a full house.

@iliketurtles wrote

@cometmars wrote

Since it's been over 48 hours:

http://bit.ly/IiBhdB

8a

Also, when do we get our results?

I'm pretty sure I got another answer on top of that for 8a? I got -1 too I THINK. Not sure. I got a quadratic expression to find two roots, and then another was a simple isolating-for-x expression.

Yup there were 3 answers, -3/2, -1 and 0

@SevenFlow wrote

@iliketurtles wrote

@cometmars wrote

Since it's been over 48 hours:

http://bit.ly/IiBhdB

8a

Also, when do we get our results?

I'm pretty sure I got another answer on top of that for 8a? I got -1 too I THINK. Not sure. I got a quadratic expression to find two roots, and then another was a simple isolating-for-x expression.

Yup there were 3 answers, -3/2, -1 and 0

Good to know :)

What did people get for 8a? I did it a little rushed and got 11/13, but and my friends got answers like 10/13 and 12/13, so I'm hoping most of my process is right.

ALSO I just realized I gave another answer for 8B not 8a. My bad.

@ChronosKey wrote

Yeah 8b had 3 solutions: -3/2, -1, 0

8a's answer was 12/13

for 10a I got 12 how bout you guys?

I got 12 too, I just wrote them all out lol.

@ahuang wrote

Anyone remember what question numbers 7 was? I would like to do it again.

I think that was the Ferris wheel one? Or maybe the triangle or trapezoid? I don't have my sheet with me.

I didn't have time to review much before the Euclid, but still managed to get through to question 5 or 6 fairly easily. I looked at questions 7-10 and scribbled some random information I remembered down for a couple of them, but either didn't have time to finish them or forgot how to do them.

Same :P

I was thinking for 10b you probably have to find some pattern. E(n) = E(n-1) + x. and maybe there's some kinda pattern for x, you could find the answer with programming pretty easily probably.

I was thinking for 10b you probably have to find some pattern. E(n) = E(n-1) + x. and maybe there's some kinda pattern for x, you could find the answer with programming pretty easily probably.

@SevenFlow wrote

@ahuang wrote

I think I did very well. Looking at the CEMC website, the date for testing is officially over so I believe discussion is allowed...

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

Correct me if I'm wrong but I think you ignored the part of the question where it said that there are exactly 3 seats empty in the auditorium. If your array has a size of 24x24 it assumes that it's a full house.

Not only is he wrong in that regard, the dimensions of the auditorium have a product of 567, the only reasonable of which are 27 x 21. I can write up an entire proof if any of you are interested.

And the infamous 10b, that question is still bugging me. I wasn't able to solve it during the contest and im still working on a solution during my off times. I have a few relations and sums I worked out that seem to hold but I haven't found a "beautiful" way of expressing it.

I have my sheet. Question 7a was a triangle ABC with AB=AC and angle BAC<60 degrees. Point D is on AC with BC=BD. Point E is on AB with BE=ED. If angle BAC = theta, determine angle BED in terms of theta.

7b was a ferris wheel that has a diameter of 16m rotating at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he's 1m above the ground. At a point P he's 16m above the ground and is rising and takes 4 sec to reach the top, point T. 8 sec later he reaches point Q. What's his height at point Q.

For these questions I got 3theta and 9m respectively

7b was a ferris wheel that has a diameter of 16m rotating at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he's 1m above the ground. At a point P he's 16m above the ground and is rising and takes 4 sec to reach the top, point T. 8 sec later he reaches point Q. What's his height at point Q.

For these questions I got 3theta and 9m respectively

@AhFu wrote

@SevenFlow wrote

@ahuang wrote

I think I did very well. Looking at the CEMC website, the date for testing is officially over so I believe discussion is allowed...

Lets post up the questions I really want to look at 9 and 10! I forgot most of the other ones lol.

Oh the 14 boy and 10 girl problem. I just said that let us consider the columns first. There should be at least 10 rows since there are at least 10 girls in every column. Also,we must have 14 boys in every row. Also note that every row and every column have same amount of people. So the minimum case will be when the every row and column has 14+10 people so that gives (24)(24) people or 576 people. That is the minimum amount.

Correct me if I'm wrong but I think you ignored the part of the question where it said that there are exactly 3 seats empty in the auditorium. If your array has a size of 24x24 it assumes that it's a full house.

Not only is he wrong in that regard, the dimensions of the auditorium have a product of 567, the only reasonable of which are 27 x 21. I can write up an entire proof if any of you are interested.

And the infamous 10b, that question is still bugging me. I wasn't able to solve it during the contest and im still working on a solution during my off times. I have a few relations and sums I worked out that seem to hold but I haven't found a "beautiful" way of expressing it.

i got the same dimensions for the auditorium