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# UKMT Senior Individual Maths Challenge 2016 - Preparation and Tips watch

1. (Original post by Mathspro2580)
For 9 i don't see how y= -X^2 +1 passes through two. I thought it would only pass through (0,1) .
less chance for a gold cuz I'm wrong
If x=1, you have -(1^2)+1=-1+1=0 (1,0)
If x=0, you have -(0^2)+1=-0+1=1 (0,1)
2. (Original post by LaurenLovesMaths)
For some reason I really struggled to do Q8 - I think it was difficult to draw an accurate diagram and I struggled with counting haha

Other than that I think the rest of my answers were correct, and that I got 100 marks pretty happy with that!
Yeah for some reason I did (2+4+6+8+10+12)*2 but not really sure why

100 is insane though!! Well done
3. How do you do question 24?
4. (Original post by maths00)
The numbers in the boxes go:

10, 13, 16, 19, 22, 25
How did you get that? It works, but I just want to know how, for the future.
5. (Original post by 123Master321)
I accidentally did 2017 digits rip, think I got 115 in the end, didnt read scalene on Q4 wbu
I did exactly the same. Damn.

(Original post by sqrt(e/m)=c)
Can people explain how they got their answers for:
11, 20, 23, 13, 15, 21 and 24.
Thanks
11. I have a series of long, awkward and confusing simultaneous equations I used to solve this, but I don't think they'd help you. The numbers in the boxes were 10, 13, 16, 19, 22 and 25, increasing by 3 each time, if that helps.

13. If you've come across this spiral before you should know the side lengths of each square, but if not hopefully you can see that (going from the start of the spiral) they are 1, 1, 2, 3 and 5. Each spiral is the circumference of a quarter circle - a circle has circumference 2πr, so the part of the spiral will have length πr/2. Sub in 1, 1, 2, 3, 5 and add them up; you get π(1+1+2+3+5)/2 = 12π/2 = 6π.

And the others you mentioned were almost an exhaustive list of the ones I couldn't answer, so I'm afraid I can't help you there.

EDIT:
(Original post by surina16)
Yeah for some reason I did (2+4+6+8+10+12)*2 but not really sure why
This is how I did it! Everyone else seemed to be drawing it out, which I thought was too tedious to be the intended method. So I drew n = 2 and n = 3, and used the given diagram of n = 4 to work out a pattern. For instance, for n = 4, there were 2, 4, 6, 6, 4 and 2 squares (2+4+6+6+4+2 = (2+4+6)*2) on each 'row' going diagonally upwards. Comparing with n = 3 and n = 2, I found you just kept adding on the next even number so with n = 7, there are (2+4+6+8+10+12)*2 = 84 squares.
6. (Original post by surina16)
Yeah for some reason I did (2+4+6+8+10+12)*2 but not really sure why

100 is insane though!! Well done
Thank you!!
7. (Original post by sqrt(e/m)=c)
Thanks, but damn you
I misread the question too and did the same thing as you
8. (Original post by maths00)
The numbers in the boxes go:

10, 13, 16, 19, 22, 25
How did you work that out? I did some long complicated simultaneous equations ahah
9. (Original post by sqrt(e/m)=c)
Can people explain how they got their answers for:
11, 20, 23, 13, 15, 21 and 24.
Thanks
11: I got the numbers 10, 13, 16, 19, 22, 25
23: I used Pythagoras to find the distance between opposite corners of the cuboid (this is the diameter of the sphere) then I used Pythagoras to find the length of a cube whose distance between opposite corners is equal to this length.
13: The smallest squares contain a half circle, with arc length (2*pi*r)/2 = (2*pi*1)/2 = 2*pi/2 = pi
The next smallest square contains a quarter circle with radius 2 so its arc length is 2*pi*r/4 = pi
The next has radius 2 + 1 = 3, with a quarter circle arc length again, and so on.
Then you add them all up: pi + pi + 3/2*pi + 5/2*pi = 6*pi
15: I continued the side of each rectangle so that it met the adjacent line, creating three tiny triangles. Then I played with the angles a bit and it's not too hard from there
10. q 11. enjoy
Attached Images

11. Thanks.
I did really similar thing, but figure that I forgot to subtract a C, thereby giving me X=17.
12. (Original post by physicsgirl98)
How did you work that out? I did some long complicated simultaneous equations ahah
There was probably a better way of doing it, but I just tested each answer in the 'X' box, then found the average of X and 25 to find the box second from the end. Then I called the box to the left of X 'Y', then tested whether each one worked or not.
13. (Original post by maths00)
11: I got the numbers 10, 13, 16, 19, 22, 25
23: I used Pythagoras to find the distance between opposite corners of the cuboid (this is the diameter of the sphere) then I used Pythagoras to find the length of a cube whose distance between opposite corners is equal to this length.
13: The smallest squares contain a half circle, with arc length (2*pi*r)/2 = (2*pi*1)/2 = 2*pi/2 = pi
The next smallest square contains a quarter circle with radius 2 so its arc length is 2*pi*r/4 = pi
The next has radius 2 + 1 = 3, with a quarter circle arc length again, and so on.
Then you add them all up: pi + pi + 3/2*pi + 5/2*pi = 6*pi
15: I continued the side of each rectangle so that it met the adjacent line, creating three tiny triangles. Then I played with the angles a bit and it's not too hard from there
Thanks a lot.
14. (Original post by surina16)
That's what I thought but apparently it's 7! = 5040
No it's not....

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15. Guys,
I answered 14 questions and now I'm worried. Can I at least get a Bronze??
16. (Original post by d010534)
Guys,
I answered 14 questions and now I'm worried. Can I at least get a Bronze??
don't worry bronze is like 8 or 9 right. u will probably get silver. Maybe gold ( but unlikely)
17. (Original post by qeyoo)
No it's not....

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Ah great haha I was really happy with my answer of 84 but everyone was telling me it was 7 factorial
18. (Original post by Mathspro2580)
don't worry bronze is like 8 or 9 right. u will probably get silver. Maybe gold ( but unlikely)
really?! thank you!! I'm so worried (I'm in year 13 so I can't ever do this again) so I really want to get a certificate at least!!
19. (Original post by d010534)
Guys,
I answered 14 questions and now I'm worried. Can I at least get a Bronze??
Yes, you can get max 25+14x4= 81, last year it was 50 for a bronze

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20. (Original post by surina16)
Ah great haha I was really happy with my answer of 84 but everyone was telling me it was 7 factorial
I legit drew out the thing and counted the squares like but oh well, I still got the answer right!!

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