ICAS Mathematics Competition (1 Viewer)

[Web Addict]

It was definitely the radius, 4.

Yeah, I know. I managed to get the diameter, but I didn't half it to get the radius because I misread the question.

 

[Resource]

Is the J Paper much harder than H?

 

[Web Addict]

Is the J Paper much harder than H?

To be honest, Paper J is much harder than Paper H. They have a lot of common questions. In fact, I believe that you'll be able to do most of the questions if you have the time.

Does anyone have a copy of the paper to upload? I need to check some of my answers. I think I made a few other silly mistakes.

 

[Resource]

What was the answer to that question with a,b,c that had a sum of 117 and something else..?

 

[Recondit]

What was the answer to that question with a,b,c that had a sum of 117 and something else..?

I think it was c) 35XXX

 

[jihad1234]

I think 7hr and 30 min is correct. A teacher explained to me that 60% faster means 1.6x faster so you would go 12/1.6= 7.5 which converted to hours and minutes becomes 7 hours and 30minutes. Thanks for all your help this question baffled most of our class!

 

[obliviousninja]

Moral of the ICAS Comp: Guess & Check.
I seriously do not see how to create any equations to solve some of them.
ie. Sum of 3 numbers is 117. One number is a prime. Other two are multiples of this prime. One of these numbers is larger than the other.
What is the greatest product of the 3 numbers?

My numbers were 13, 39, 65. But I did not get any of the numbers on the multiple choice, so logically there must have been a greater number than what I got, and hence I choose that as my answer

Hopefully I can continue my distinction streak from yr 7
Lucky it wasn't some bs Olympiad paper or BHP one, if any of you are familiar with that one?

 

[Web Addict]

Moral of the ICAS Comp: Guess & Check.
I seriously do not see how to create any equations to solve some of them.
ie. Sum of 3 numbers is 117. One number is a prime. Other two are multiples of this prime. One of these numbers is larger than the other.
What is the greatest product of the 3 numbers?

My numbers were 13, 39, 65. But I did not get any of the numbers on the multiple choice, so logically there must have been a greater number than what I got, and hence I choose that as my answer

Hopefully I can continue my distinction streak from yr 7
Lucky it wasn't some bs Olympiad paper or BHP one, if any of you are familiar with that one?

I agree. The ICAS Mathematics Competition has a lot of trial and errors.

By the way, what do you guys think the cut-offs for Credit, Distinction and High Distinction are this year?

 

[obliviousninja]

Do you by any chance know the cut-offs?

I agree. The ICAS Mathematics Competition has a lot of trial and errors.

By the way, what do you guys think the cut-offs for Credit, Distinction and High Distinction are this year?

 

[RealiseNothing]

I think it was c) 35XXX

Yep I got this as well.

 

[RealiseNothing]

I agree. The ICAS Mathematics Competition has a lot of trial and errors.

By the way, what do you guys think the cut-offs for Credit, Distinction and High Distinction are this year?

I barely used trial and error in the test.

 

[obliviousninja]

dafuq? pls explain how you mathematically solved the a+b+c=117 question, im just curious, don't think it will be any use for extension though.

I barely used trial and error in the test.

 

[Resource]

I made a=13. Then I decided that the two multiples of 13 that would add to 117 and give the highest product were 52 and 52.

13+52+52=117.
13x52x52=35152.

Was this question in paper J? Cause it was in Paper H.

 

[Web Addict]

I barely used trial and error in the test.

Wow. How did you manage without trial and error?

 

[RealiseNothing]

dafuq? pls explain how you mathematically solved the a+b+c=117 question, im just curious, don't think it will be any use for extension though.

B and C are multiples of A, so we can say:





A however is prime, so we find the prime factorisation of 117:



So A=13 or A=3

Now we want to find the maximum value of , which is

Since the A is cubed, we want to make A as large as possible, so since we can choose from 3 or 13, we let A=13 since by a lot. Also consider the product , it will be maximised when

by letting

Substitute in A=13:







So the product of A, B, and C is maximised when in the form where A=13, x=4, and y=4.

Thus we get:

The maximum product is

 

[obliviousninja]

Did they teach you this in like year 7 at nsg? ie, all these random FYI things such as prime factorisation.

B and C are multiples of A, so we can say:





A however is prime, so we find the prime factorisation of 117:



So A=13 or A=3

Now we want to find the maximum value of , which is

Since the A is cubed, we want to make A as large as possible, so since we can choose from 3 or 13, we let A=13 since by a lot. Also consider the product , it will be maximised when

by letting

Substitute in A=13:







So the product of A, B, and C is maximised when in the form where A=13, x=4, and y=4.

Thus we get:

The maximum product is

 

[Resource]

Didn't you learn factor trees in year 7? And if not it isn't too difficult to know that 13 is a factor of 117 and then (117-13)/2 =52.

 

[MineTurtle]

I think 7hr and 30 min is correct. A teacher explained to me that 60% faster means 1.6x faster so you would go 12/1.6= 7.5 which converted to hours and minutes becomes 7 hours and 30minutes. Thanks for all your help this question baffled most of our class!

YES! That's what I got. I'm so happy!!

 

[Web Addict]

B and C are multiples of A, so we can say:





A however is prime, so we find the prime factorisation of 117:



So A=13 or A=3

Now we want to find the maximum value of , which is

Since the A is cubed, we want to make A as large as possible, so since we can choose from 3 or 13, we let A=13 since by a lot. Also consider the product , it will be maximised when

by letting

Substitute in A=13:







So the product of A, B, and C is maximised when in the form where A=13, x=4, and y=4.

Thus we get:

The maximum product is

Wow. Nice working out. So, there is no need for trial and error after all.

I think we have a medallist here.

 

[nerdasdasd]

Did anyone ever get the CD with the old papers ICAS papers on them ?