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# STEP maths I, II, III 1991 solutions

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1. (Original post by insparato)
I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
Exactly.

When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.
Form a differential equation for R as a function of time. Solve it for both spheres. Add the answers to get a formula for height as a function of time. Etc...
2. (Original post by insparato)
I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
Ah, of course! Thanks, this should set me off
3. and

Separating variables
Separating variables on the other one

What do I do with the constants now? They're bugging me.

is the diameter (height) of the upper sphere and is the diameter of the lower sphere. So the height is

So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/
4. I'd do something like:

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

So

And

So , so for some constant k.

Subbing in for the initial values, we get .

Now find kt when R_u+R_l = 5R/2, etc...
5. (Original post by DFranklin)
I'd do something like:

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

So

And

So , so for some constant k.

Subbing in for the initial values, we get .

Now find kt when R_u+R_l = 5R/2, etc...
This question is number 7 in Advanced Problems in Mathematics
6. II/6:

The inequality is satisfied at x = 0. Differentiating: , which is true. So x increases faster than sin x which increases faster than 0, so the inequality will still be satisfied: so .

The inequality is satisfied at x = 0. Differentiating: , which has been shown previous. So cos x decreases slower than 1 - x^2/2, so the inequality will still be satisfied: so .

is satisfied at x = 0. Differentiating: , which has been previously shown. sin x increases faster than x - x^3/6, so the inequality holds.

Similarly:

7. STEP II Q16

Part 1a

Obviously it needs to rain on both days. The next few arguments assume it does. Note also it is necessary that . If this is not the case, then , and the most specific expression for u is . If then the following applies.

At midnight on the first day the volume will be u below the top.

We require that on the second day it rains at a time t such that ie .

The probability of this being the case is .

So and .

Part 1b

We require where R is a random variable representing the number of times it rains, .

can be approximated by .

where .

We require so we look for a value of k such that .

Now so .

The probability must be strictly less than 0.1 so he should set k=9.

Part 2

The variance in the total amount of rain will be lower, allowing for a lower value of k.
8. 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.
9. (Original post by justinsh)
I may have done mistakes but here it goes:

In my personal opinion, both solutions to STEP III Q1 part b on here fail to find a general term (although this second one is quite nice with two fairly simple general terms), so I am going to post the way I did it, which is quite similar to the above method except that I actually put a general term, rather than a set of general terms. I don't know whether they expect you to do this, but before looking at these I presumed you had to. Anyway, here goes:

It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the xn and x2n terms, and finding a fairly simple function that returns 1 for even n and 0 for odd n.
10. (Original post by Rabite)
I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.
I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
1. There is a typo early on, you put 11 instead of 1.
2. No minimum at (1, -4) of any sort.
3. But there is a minimum at (2, -5), which is also necessary for the graph.
4. In the last part, you are defining f-1(x) so the x we are talking about is a different x to before if that makes sense, ie it is what was on the y-axis before not the x-axis. So your table thing doesn't really make sense.
5. If we ignore point 4, the right-most box is wrong anyway, because the positive root is only valid to the right of 2.

So in the last part you need to consider . The first is undefined, all the others have two possibilities (one of which is the same for all four). It's seriously boring stuff. This question is not particularly challenging, but loads of cases to consider, and loads of room for little errors.
11. STEP III Q16
Part 1

Part 2

Part 3

for even y.

Part 4

for odd y.

Part 5

Woooooooo! This has taken me almost all day and I've finally done it! I'm almost certain the answer is right as it works for a=2 and a=3, and everything follows nicely.
12. This alternative to III/1 part (c) is quite cute.

from the binomial expansion.
13. (Original post by Rabite)
I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.
I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
Hi there just wanted to point out one final thing i noticed when doing this question about the inverse functions is you have to check the functions are still valid in the intervals you gave them:
For instance for -1<x<0

is undefined
Also, in the interval x<-1

is undefined once x<-5 so the interval should be -5<x<-1
and the inverse functions are undefined for the intervals -1<x<0 and x<-5
14. I think these solve 1991 Step 3 Q4 and Q9
Attached Files
15. Step1991Paper3Question4.pdf (56.5 KB, 137 views)
16. Step1991Paper3Question9.pdf (77.4 KB, 119 views)
17. STEP 1991 Paper II question 9

18. An absolute swine to Latex but here goes

19. Attached Thumbnails

20. Attached Thumbnails

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