Intermediate Maths Challenge - UKMT - 5th February 2015

21. shaded - X
One strip - Y

X = (Y + Y - X)/8
= 0.25Y - 0.125X
1.125X=0.25Y
9X = 2Y
X = 2/9 x Y

23

First Letter--------------------

C 1st to 24th
I 25th to 48th
K 49th to 72th
M 73th to 96th
U 97th to 120th

Second Letter---------------

C 97th to 102th
I 103th to 108th
K 109th to 114th
M 115th to 120th

Third Letter--------------------

C 109th to 110th
I 111th to 112th
M 113th to 114th

Fourth Letter--------------------

C 111th
M 112th

25.

non shaded area

1/4 x 3/4 x 2 =
3/8

1- 3/8 = 5/8

lol your working is far more succinct/ rigorous (i just counted the possibilities after it- still got 112th so oh well)
(sob)

Reply 241

Dizgurl

15

Doesn't 210/8 have a remainder of 2?

Reply 242

Yuuchin1

2

Original post by SpeedieABC

sorry if i sound stupid but the ant can only travel 4 times right?

yes

Reply 243

Dizgurl

15

Original post by SpeedieABC

sorry if i sound stupid but the ant can only travel 4 times right?

I agree with you speedie

Reply 244

Dizgurl

15

So the answer is 4 not 6

Reply 245

Yuuchin1

2

Original post by Dizgurl

Doesn't 210/8 have a remainder of 2?

i dont know how you got 210...
the q was
2^2*3^3*5^5*7^7 or something

since 2^2=4, and multiple of 4 divided by 8 only gives 4 or 0 as remainder, and as 0 wasnt an option, the answer was 4

im confuzed

Original post by Infaredz

My answers:
1. A
2. E
3. E
4. C
5. B
6. D
7. E
8. B
9. B
10. A (confirmed wrong, actually C)
11. D
12. A (confirmed wrong, actually B)
13. A
14. C
15. E (confirmed wrong, actually C)
16. D
17. C (confirmed wrong, actually D)
18. B (presumed wrong, probably A)
19. C
20. C
21. C (marked 2/9 first and changed my mind ;()
22. A
23. B
24. E
25. D

Tallying these up I get 99, unless question 18 was actually correct. This is exactly what I got last year (also best in year and best in school) and I sat the Hamilton paper and got a merit with an abysmal score. Hope you all did well and the answers will be on the UKMT website tomorrow probably.

I think you got 19 21 22 wrong too

Reply 248

Yuuchin1

2

Original post by Dizgurl

So the answer is 4 not 6

the question asks for the number of different possible black tiles the ant can find himself in after 4 moves- this could be more than the number of movements the ant can move. for example, if an ant was on a white tile and can move 1, it doesnt just have 1 black tile it can travel to, it has 4

(edited 9 years ago)

Reply 249

SpeedieABC

Original post by Dizgurl

Doesn't 210/8 have a remainder of 2?

where does 210 come from

(2x2x3x3x3x5x5x5x5x5x7x7x7x7x7x7x7)/8=

(3x3x3x5x5x5x5x5x7x7x7x7x7x7x7)/2=n...1

1 x (8/2)= 4

(edited 9 years ago)

Reply 250

Yuuchin1

2

Original post by SpeedieABC

where does 210 come from

she probably mistook 2^2*3^3*5^5*7^7 with 2*3*5*7- reading error

Reply 251

Infaredz

Question 19 was 37 I'm pretty sure, if not this will be an utter embarrassment.
Separating the non-shaded area into various right angles triangles and squares I discovered that of the total area,

14+15+15+12+16+9+3

Was unshaded. This adds up to 84. The total area being 121 (11*11), we know that 121-84=37, therefore the answer is C.

I spent too long on that damn question.

21 I did mention was wrong, I actually wrote the correct answer and changed my mind thinking It was not a trap answer.

22 is also wrong however it was the only one I couldn't understand how to work out, ran out of time and just risked it by putting down the first answer XD

(edited 9 years ago)

Reply 252

Yuuchin1

2

Original post by Infaredz

Question 19 was 37 I'm pretty sure, if not this will be an utter embarrassment.
Separating the non-shaded area into various right angles triangles and squares I discovered that of the total area,

14+15+15+12+16+9+3

Was unshaded. This adds up to 84. The total area being 121 (11*11), we know that 121-84=37, therefore the answer is C.

I spent too long on that damn question.

hmm i did the same and i got 35... where do you have your 2*2 square and 1*1 square in your calculation

Reply 253

SpeedieABC

Original post by Infaredz

Question 19 was 37 I'm pretty sure, if not this will be an utter embarrassment.
Separating the non-shaded area into various right angles triangles and squares I discovered that of the total area,

14+15+15+12+16+9+3

Was unshaded. This adds up to 84. The total area being 121 (11*11), we know that 121-84=37, therefore the answer is C.

I spent too long on that damn question.

sorry to break it to you but 2 squared plus 1 squared is 5 not 3