The Student Room Group

STEP I, II, III 1999 solutions

Scroll to see replies

SimonM
STEP I, Question 1

By symmetry, their average value will be 5×105\boxed{5 \times 10^5}

What symmetry? :s-smilie: I think I want to die, I can never beat STEP. :frown:
Reply 121
Physics Enemy
What symmetry? :s-smilie: I think I want to die, I can never beat STEP. :frown:


Well, I was probably doing it to show off, there are much more "standard" ways to approach that question. However, consider the numbers 0-10

0,1,2,3,4,5,6,7,8,9,10

Striking out multiples of 2 and 5 leaves things evenly balanced on both sides: Symmetry
SimonM
Well, I was probably doing it to show off, there are much more "standard" ways to approach that question. However, consider the numbers 0-10

0,1,2,3,4,5,6,7,8,9,10

Striking out multiples of 2 and 5 leaves things evenly balanced on both sides: Symmetry

Yes I was gonna add and subtract sums of various arithmetic series. :smile:
SimonM
STEP I, Question 6

Spoiler


LOL @ this Q, what a gift. I wish my exam on Mon has Q's like these. This isn't even A-Level standard surely.
SimonM
STEP I, Question 8

Spoiler


Haha yes the last part is basically a fiddle, though I noticed they didn't use the word prove, just 'show' ...

I said N < 10^30 => lnN < ln(10^30) = 30ln10
But using our inequality before: ln10 <= (1 + 1/2 + 1/3 + ... + 1/10) < 3
So ln10 < 3 thus lnN < 90

I then said the summation approximates well to lnN for large N, thus is only slightly more than lnN. In addition, we used ln10 < 3 in our deduction of lnN < 90 thus providing even more margin of error (albeit uneeded).

Thus Summation < 90 + (0.05*90) < 101 (5% more than adequate margin for error)

Dunno if that would do, they may say it's not rigorous enough, decent stab though, would expect something for my efforts. :p:
How do you know that 1+1/2+1/3+...+1/10 < 3? (It is, but it's close enough that I would want some fairly careful justification).

The rest is blatantly unjustified but I assume you already knew that.

To be honest, I wouldn't give you much for any of the waffle.
DFranklin
How do you know that 1+1/2+1/3+...+1/10 < 3? (It is, but it's close enough that I would want some fairly careful justification).

First 6 terms give it away.

DFranklin
The rest is blatantly unjustified but I assume you already knew that.

I haven't proved it approximates well for large N, true. But based on that, I don't think it's unjustified. If you know it approximates well, how could the error be that large?

DFranklin
To be honest, I wouldn't give you much for any of the waffle.

Apart from not proving the approximation, I don't think it's too bad. I think you're being a little harsh, though I would say that naturally!

TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.
Physics Enemy
First 6 terms give it away.
Not to me they don't. I honestly can't see any examiner accepting that as an explanation.

I haven't proved it approximates well for large N, true. But based on that, I don't think it's unjustified. If you know it approximates well, how could the error be that large?
You can't expect to get marks based on "It would be nice if this was true, so I'll say that it is true, and then I can answer the question".

Apart from not proving the approximation, I don't think it's too bad. I think you're being a little harsh, though I would say that naturally!
It's possible I'm being harsh - but if so, it's basically because you'd be getting "pity marks". That is: "this is all nonsense, but he's only an A-level student, we can't really expect any better".

TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.
I don't see anything wrong with what Simon did, other than maybe a tiny bit more explanation about why ln 2 < 1.
Reply 128
DFranklin
...

if he just said e^3>10 => 3>ln10 =>91>1+30ln10, then surely that would do, everyone knows e^3>10, but 125/8 does it quite easily.
Reply 129
DFranklin
I don't see anything wrong with what Simon did, other than maybe a tiny bit more explanation about why ln 2 < 1.


e>5/2>2 => ln e > ln 2 => 1 > ln 2
Physics Enemy

TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.

The 2^10 thing was so you can use that 2^10 > 10^3
SimonM
e>5/2>2 => ln e > ln 2 => 1 > ln 2
Sure. What you'd written before was fine as an exam answer - I would not in a million years expect to see an examiner complain about it. But it was perhaps a little abrupt for people stuck and trying to understand how to do the question. :dontknow:
DFranklin
Sure. What you'd written before was fine as an exam answer - I would not in a million years expect to see an examiner complain about it. But it was perhaps a little abrupt for people stuck and trying to understand how to do the question. :dontknow:

I still don't understand how to do the Q lol. No offence to Simon, but when I read his solutions I don't understand what's going on at all. That's probably because he's very bright and answers it so well. That's why he's at Cambridge doing Maths.
Reply 133
Basically he says
10^30=1000^10<1024^10
Follow from there, (using the result given)
the inequality is preserved by ln'ing both sides since ln is a stricly increasing function (if your concerned)
Original post by toasted-lion
STEP III Question 13

Now taking a proportion X of the cake gives probability X^4 that all 4 currents will be in that portion.

So taking a portion X and getting all 4 currents in that portion has pdf:

f(x)=2x.x4=2x5 f(x) = 2x.x^4 = 2x^5

Hence, P(all4inportion)=012x5dx=1/3 P(all4inportion) = \int_0^1 2x^5 dx = 1/3

and, P(all4inportionbiggerthanhalf)=1/212x5dx=1/31/192=63/192 P(all4inportionbiggerthanhalf) = \int_{1/2}^1 2x^5 dx = 1/3 - 1/192 = 63/192

So finally P(portionbiggerthanhalfall4inportion)=(63/192)/(1/3)=189/192 P(portionbiggerthanhalf \mid all4inportion) = (63/192)/(1/3) = 189/192



The number of currants X in a portion x is actually Po(4x)\sim Po(4x)

So the integrals are actually (2x)[e4x(4x)44!]dx\displaystyle \int (2x)[e^{-4x}\frac{(4x)^{4}}{4!}] \, dx
Original post by jasperleeabc
The number of currants X in a portion x is actually Po(4x)\sim Po(4x)




but this gives P(X>4)>0.
Original post by Get me off the £\?%!^@ computer
but this gives P(X>4)>0.
Agreed. The original approach looks fine to me. (Poisson might make sense if the question said the average number of currents was exactly 4).
I don't think anything in the question is Poisson distributed. (The integral you have is not something you could reasonably be expected to evaluate, either).
Sorry my bad, I know what's actually wrong now.
**Ignore me**
(edited 12 years ago)

Quick Reply

Latest