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# STEP I, II, III 1999 solutions Tweet

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IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Re: STEP I, II, III 1999 solutions
(Original post by speedy_s)
I was wondering if someone could help me with the last part of STEP II Q5

basically what i tried to do:

and we have to show a < 3pi/4

so

putting the two together I got

which is obviously true since is just less than 192. Taking as just more than 35 the number is > 0. Hence a is <3pi/4.

But, I am not sure if this is a valid argument to secure all the marks.

I am pretty sure there is a better method and if some one can shed some light on it, it would be great!
I used the fact that 120 < 121 thus 120^1/2 < 11, to show that

(12(120^1/2) - 35)/169 > 97/169

Then proved by contradiction that 97/169 < (1/2)^1/2 and with the aid of a sketch it follows that a < 3pi/4. You end up having to calculate 169^2 but it avoids making approximations. It put me off that asked you to proove this 'carefully', makes me feel like i haven't been thorough enough.
2. Re: STEP I, II, III 1999 solutions
Could anyone help me with II question 4? I am stuck on the last part. The RHS of the equality you need to proove suggests n(1 + x)^2n might be used. But this doesn't seem to help with getting the sum on the LHS.

I can generate the sum on the LHS using 2nx(1 + x)^(n-1)*((1 + x)^n - x^n) but then I cant see how I could turn this into a single bracket that generates the expression on the RHS. Any ideas?
3. Re: STEP I, II, III 1999 solutions
(Original post by maltodextrin)
I can generate the sum on the LHS using 2nx(1 + x)^(n-1)*((1 + x)^n - x^n) but then I cant see how I could turn this into a single bracket that generates the expression on the RHS. Any ideas?
Use calculus:

Spoiler:
Show
Try differentiating the original identity in an appropriate manner.
4. Re: STEP I, II, III 1999 solutions
(Original post by DFranklin)
Use calculus:

Spoiler:
Show
Try differentiating the original identity in an appropriate manner.
Thanks, i think i'm on the right track but still a bit stuck....

If i differentiate both sides, the RHS comes out ok but for the LHS (after some fiddling) i get SUM 2(n - r)*(nCr)^2 [r = 0, n - 1]. I'm not sure if this is correct, and if it is i can't see how to turn it into the sum they're asking for
5. Re: STEP I, II, III 1999 solutions
nCr = nC(n-r).

So if you make the "substitution" t = n - r, you will get the sum in the question (note that as r goes from 0 to n-1, t goes from 1 to n).
6. Re: STEP I, II, III 1999 solutions
Great i understand, thanks again.
7. Re: STEP I, II, III 1999 solutions
In Question 8 i either keep making the same numerical slip or i'm doing something wrong as i get the approximation in part ii to be (1/4 + 1/pi^2)n^2 + n/2 any help would be great (sorry this is going to look ugly).

This was done by differentiating the original equality to get SUM kcos(kpi/n) [k = 0, n] = approx -(2n^2/pi^2 + n/2)
and then rewriting SUM k(sin(kpi/2n))^2 [k = 0, n] = 1/2 (SUM k [k = 0, n] - SUM kcos(kpi/n) [k = 0, n])
Thus:
SUM k(sin(kpi/2n))^2 [k = 0, n] = approx n(n + 1)/4 + 1/2(2n^2/pi^2 + n/2)
8. Re: STEP I, II, III 1999 solutions
I haven't done the actual calculations, but it may well be your answer is "good enough" - that is, you're supposed to simply discard the n/2 term on the grounds that it is small compared with the n^2 term.
9. Re: STEP I, II, III 1999 solutions
That's pretty harsh of them if that's the case, knowing how important a 'full answer' is i'd probably have spent 20 minutes going over my working and trying other methods in the assumption that my answer was wrong, but i guess that's STEP for you.
10. Re: STEP I, II, III 1999 solutions
1999 Paper 1 number 4
Attached Files
11. 1999PAPER1.no.4.pdf (23.7 KB, 301 views)
12. Re: STEP I, II, III 1999 solutions
1999 Paper 1 numbers 11,12,13
Attached Files
13. 1999PAPER1.nos. 11,12,13.pdf (42.2 KB, 218 views)
14. Re: STEP I, II, III 1999 solutions
Paper 2 number 1
Attached Files
15. 1999PAPER2.no.1.pdf (31.6 KB, 513 views)
16. Re: STEP I, II, III 1999 solutions
Paper 2 numbers 4,8,9
Attached Files
17. 1999PAPER2.nos.4,8,9.pdf (61.2 KB, 701 views)
18. Re: STEP I, II, III 1999 solutions
Paper 2 numbers 11 & 12
Attached Files
19. 1999PAPER2.nos.11,12.pdf (44.3 KB, 182 views)
20. Re: STEP I, II, III 1999 solutions
1999 Paper 3 numbers 9,11 and 14
Attached Files
21. 1999PAPER3.no.9.pdf (34.7 KB, 216 views)
22. 1999PAPER3.no.11.pdf (32.9 KB, 86 views)
23. 1999PAPER3.no.14.pdf (30.9 KB, 86 views)
24. Re: STEP I, II, III 1999 solutions
(Original post by lilman91)
Step I Question 9

When the hare overtakes, both have travelled 0.5km, but the tortoise has travelled for 0.5hours longer. Using for both:

Therefore, ...(A)

When the hare passes the tortoise on the way back, the hare has travelled whilst the tortoise has travelled . However, the tortoise has been racing for 1 hour longer.

Using again:

Therefore, ...(B)

From, (A) and (B)

This gives: . Multiplying top and bottom by 4 gives as required.

From A we know that . Substituting expression for V gives:

.
So, .

The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
So,
Substituting expression for V,
.
Solving the quadratic but since , .
So, finally .
Have you overlooked the fact that the hare slept for an hour so was only moving for half an hour?
25. Re: STEP I, II, III 1999 solutions
(Original post by brianeverit)
1999 Paper 3 numbers 9,11 and 14
Theres a mistake in question 9. The distance from P to the sphere isnt constant for all theta... and it definitely isnt a. I cant get it though so could someone please solve this one?
26. Re: STEP I, II, III 1999 solutions
(Original post by Aurel-Aqua)
STEP I, Question 5

(Please someone check this, I'm not very confident, because I did last year)

Spoiler:
Show

(i) Triangle with x on the adjacent, x on the opposite, and h+R on the hypotenuse, implies , so .

(ii) h is small relative to h, so distance tends to , but (for obviously small ), so , so .

(iii) We draw some strange diagram... and then... !

, where is an angle ZOK, where K is a side of the earth, as well as that angle being the angle XZO. So two triangles are congruent, and the opposite is R, hypotenuse h+R and x the adjacent.

, and since h gets large compared to R, . Then, , so and .
you say sintheta=x/R
the hypotenuse is r+h in you're working since you found out the opposite,x using pythagoras and did (r+h)^2 as the hypotenuse.

/i think
edit: i geuss having done the question you've not bothered to add in the small h.
Last edited by jj193; 28-10-2009 at 14:24.
27. Re: STEP I, II, III 1999 solutions
(Original post by SimonM)
STEP II, Question 14

Spoiler:
Show
Your tactic will essentially be "stop rolling if greater than a given number", this is because each result is independent on the previous ones, so you must take a decision based on what you have rolled. The probability of obtaining any of the scores in the "stop" range is the same (symmetry, relabelling, etc). Therefore, considering cases we have:

Stop after first:

Stop if larger than 2

Stop if larger than 3

Stop if larger than 4

Stop if larger than 5

Therefore we should stop if we get greater than 3. (or 4, I guess we should consider variances)
My reasoning for this question was mainly focused on E(X) = 3.5

I said that if X was 2 or 3 then we would roll again as the expected value is larger than this, and if X was 4, 5 or 6 then we should stop rolling because it was larger than the expected value already.

Your reasoning makes more sense having read it, I was just wondering where the flaws in my method were?

Is it just that I haven't considered the throws after each possible outcome, or is there more to it? In particular how would I have used variances as you suggested?
28. Re: STEP I, II, III 1999 solutions
(Original post by IDGAF)
My reasoning for this question was mainly focused on E(X) = 3.5

I said that if X was 2 or 3 then we would roll again as the expected value is larger than this, and if X was 4, 5 or 6 then we should stop rolling because it was larger than the expected value already.

Your reasoning makes more sense having read it, I was just wondering where the flaws in my method were?

Is it just that I haven't considered the throws after each possible outcome, or is there more to it? In particular how would I have used variances as you suggested?
What is X? The expected value of one role?

What if there were no rules, you could stop when you like? How would your argument scale to only stopping when you role a 6?

Ok, the argument using variances would go like this:

The variance for "stopping >3" is:

and "stopping >4" is:

With stopping at >4 there is a bigger spread, so depending on what properties you want your system to have, you might pick 3 more than 4 or vice-versa.
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