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STEP maths I, II, III 1991 solutions

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1. Re: STEP maths I, II, III 1991 solutions

2. Re: STEP maths I, II, III 1991 solutions

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3. Re: STEP maths I, II, III 1991 solutions

Last edited by brianeverit; 20-07-2011 at 12:02.
4. Re: STEP maths I, II, III 1991 solutions

Last edited by brianeverit; 19-07-2011 at 17:16.
5. Re: STEP maths I, II, III 1991 solutions

Last edited by brianeverit; 20-07-2011 at 17:09.
6. Re: STEP maths I, II, III 1991 solutions
(Original post by toasted-lion)
On the last part of question 15 of STEP II I got instead of . Did anybody else get this? What are the chances they made a typo? I may type up my solution in a bit, but I just typed up 16 and I'm latexed out.
Sorry to resurrect an old thread, but came here looking for a solution so thought I might as well type mine up. I make no guarantees of correctness; feedback is more than welcome.

The integers are independent random variables with the discrete uniform distribution, between 1 and m.

So .

By symmetry, .

Now . By independence . Since this is . So , as was required.

In the case m = 2, note that either or , in which case , or or , in which case . Each of these have probability of , so . As n_2, n_3 are identically distributed, also, and summing the expectations (note ) gives an expectation of 1.

Letting m = 2 in gives the same result.

For general m:

Let be the position of the desk of the 'th candidate to call, so that is the position of the desk for the first candidate.

The invigilator starts at desk 1, so the distance travelled to the first candidate is and .

On all subsequent occasions, the distance travelled is .

Let S be the total distance walked, so and . This equals , which reduces to , as required.
7. Re: STEP maths I, II, III 1991 solutions
(Original post by DFranklin)
We find the equations for the asymptotes meet at (1,-1), and your guess that the axes are at +/-45 degrees seems a good one.

So we suspect that the axes have equations x+y=1, x-y =2. You could probably stop there, but partly to reassure myself, and partly for the hell of it, we can go further:

We look for such that resembles the given equation.

Easiest is to look first at the coefficients of x and y, which tells us that , and then the coefficient of x^2 tells us .

Expanding out, we find and so using the given equation for the curve, we obtain .

So we have explicitly obtained the equation of a hyperbola in terms of our new axes, confirming our guess.
Im thinking maybe this was the kind of answer they were looking for:-
Attached Files
8. Step1991Paper2Question2.pdf (79.8 KB, 26 views)
9. Re: STEP maths I, II, III 1991 solutions
(Original post by brianeverit)

what is your reasoning for the particle sliding up the rod?
10. Re: STEP maths I, II, III 1991 solutions
(Original post by Nick_)
what is your reasoning for the particle sliding up the rod?
On second thoughts I think the answer is probably that the ring does not move on the rod.
11. Re: STEP maths I, II, III 1991 solutions
The solution to II/8 is wrong (coshx -1)/sinhx ≠ tanhx. The final solution I get using e^x substitution is y = - x + 2ln(1+e^x) - ln4.
12. Re: STEP maths I, II, III 1991 solutions
(Original post by Snowdog94)
The solution to II/8 is wrong (coshx -1)/sinhx ≠ tanhx. The final solution I get using e^x substitution is y = - x + 2ln(1+e^x) - ln4.
...does the given solution work then?
13. Re: STEP maths I, II, III 1991 solutions
(Original post by TheMagicMan)
...does the given solution work then?
Yes, then they work out to be the same solution. Where does that hyperbolic result come from?
14. Re: STEP maths I, II, III 1991 solutions
(Original post by Snowdog94)
Yes, then they work out to be the same solution. Where does that hyperbolic result come from?
It's just writing them out in exponential form...the top becomes a perfect square and the bottom is a difference of two squares and they have a common factor

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