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STEP I, II, III 1999 solutions

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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?

Reply 101

DFranklin

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

Reply 102

maltodextrin

DFranklin

Use calculus:

Spoiler

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]. 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

Reply 103

DFranklin

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).

Reply 104

maltodextrin

Great i understand, thanks again.

Reply 105

maltodextrin

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] = approx -(2n^2/pi^2 + n/2)
and then rewriting SUM k(sin(kpi/2n))^2 [k] = 1/2 (SUM k [k] - SUM kcos(kpi/n) [k])
Thus:
SUM k(sin(kpi/2n))^2 [k] = approx n(n + 1)/4 + 1/2(2n^2/pi^2 + n/2)

Reply 106

DFranklin

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.

Reply 107

maltodextrin

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.

Reply 108

brianeverit

1999 Paper 1 number 4

1999PAPER1.no.4.pdf24.3KB

Reply 109

brianeverit

1999 Paper 1 numbers 11,12,13

1999PAPER1.nos. 11,12,13.pdf43.2KB

Reply 110

brianeverit

Paper 2 number 1

1999PAPER2.no.1.pdf32.4KB

Reply 111

brianeverit

Paper 2 numbers 4,8,9

1999PAPER2.nos.4,8,9.pdf62.6KB

Reply 112

brianeverit

Paper 2 numbers 11 & 12

1999PAPER2.nos.11,12.pdf45.3KB

Reply 113

brianeverit

1999 Paper 3 numbers 9,11 and 14

(edited 9 years ago)

1999.PaperIII.questionb9.pdf86.0KB

1999PAPER3.no.11.pdf33.7KB

1999PAPER3.no.14.pdf31.6KB

Reply 114

brianeverit

lilman91

Step I Question 9

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

time=\frac{distance}{speed}

for both:

\frac{0.5}{v}=\frac{0.5}{V}+0.5 \Rightarrow \frac{1}{v}=\frac{1}{V}+1=\frac{V+1}{V}

Therefore,

v=\frac{V}{V+1}

...(A)

When the hare passes the tortoise on the way back, the hare has travelled

X+1.25

whilst the tortoise has travelled

X-1.25

. However, the tortoise has been racing for 1 hour longer.

Using

time=\frac{distance}{speed}

again:

\frac{X+1.25}{V}+1=\frac{X-1.25}{v} \Rightarrow \frac{X+1.25+V}{V}=\frac{X-1.25}{v}

Therefore,

v=\frac{V(X-1.25)}{X+1.25+V}

...(B)

From, (A) and (B)

\frac{V}{V+1}=\frac{V(X-1.25)}{X+1.25+V} \Rightarrow \frac{1}{V+1}=\frac{X-1.25}{X+1.25+V}

This gives:

X+1.25+V=V(X-1.25)+X-1.25 \Rightarrow V(X-2.25)=2.5 \Rightarrow V=\frac{2.5}{X-2.5}

. Multiplying top and bottom by 4 gives

V=\frac{10}{4X-9}

as required.

From A we know that

v=\frac{V}{V+1}

. Substituting expression for V gives:

v=\frac{\frac{10}{4X-9}}{\frac{10+4X-9}{4x-9}} = \frac{10}{4X+1}

.
So,

v=\frac{10}{4X+1}

.

The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
So,

\frac{2X}{V}=1.5 \Rightarrow 4X=3V

Substituting expression for V,

4X=\frac{30}{4X-9} \Rightarrow 16X^2-36X-30=0 \Rightarrow 8X^2-18x-15=0

.
Solving the quadratic but since

X>0

X=\frac{18+\sqrt{804}}{16}=\frac{18+2\sqrt{201}}{16}=\frac{9+\sqrt{201}}{8}

.
So, finally

X=\frac{9+\sqrt{201}}{8}

Have you overlooked the fact that the hare slept for an hour so was only moving for half an hour?

Reply 115

dtfan91

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? :woo:

Reply 116

jj193

Aurel-Aqua

STEP I, Question 5

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

Spoiler

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.

Reply 117

IDGAF

SimonM

STEP II, Question 14

Spoiler

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?

Reply 118

SimonM

IDGAF

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:

\displaystyle \frac{1^2+4^2+5^2+6^2}{4} - 16 = 3 + \frac{1}{2}

and "stopping >4" is:

\displaystyle \frac{1^2+5^2+[br]6^2}{3} - 16 = 4+\frac{2}{3}

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.

Reply 119

IDGAF

SimonM

\displaystyle \frac{1^2+4^2+5^2+6^2}{4} - 16 = 3 + \frac{1}{2}

and "stopping >4" is:

\displaystyle \frac{1^2+5^2+[br]6^2}{3} - 16 = 4+\frac{2}{3}

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.