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STEP Maths I, II, III 1993 Solutions

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1. Re: STEP Maths I, II, III 1993 Solutions
STEP III Number 11 (not quite complete)
Can anyone finish this off please. Just the very last part is missing.

(i)

(i)

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2. Re: STEP Maths I, II, III 1993 Solutions
(Original post by brianeverit)
..
It's a very long while since I've done vector angular momentum, so this is from looking stuff up on Wiki.

I believe you can simply dot the vector moment (call it V_m) we previously calculated with the direction vector (1, 1, 1)/sqrt(3) to get the torque about the axis OO'. (I didn't find a direct statement that this is the case, but if the cube wasn't constrained to rotate on the OO' axis, then it's certainly true that . On the assumption that the constraint effectively acts as to remove any resultant torque NOT parallel to OO', then I think you can simply dot with the OO' axis to get the behaviour).

From which point it's simply a standard "acceleration = moment / MI" calculation.

[I don't consider this terribly satisfactory since I have no real idea if it's right. Sorry.]
3. Re: STEP Maths I, II, III 1993 Solutions
(Original post by SimonM)
(Updated as far as #213) SimonM - 11.05.2009
....
STEP 1 Q16
Taking the question's advice and letting we see that:
and so
as required.
Now, if X=x, consider the case when the pin has fallen in such a way that it's end only just touches the line. From a diagram it can be seen that a right angled triangle is formed by the pin, the line and the distance X such that:
. So, the probability tht the pin crosses the line, in this case, is the probability that which, since Y is uniformly distributed, is
Now, for the next part, the probability that the pin will cross the line for a general throw is the sum of the probabilities that X will take a particular value
less than a and that Y will be less than for this particular value of X, i.e:
P(crosses).
Taking a hint from the first part we let which is what we calculated in the first part so P(crosses) as required.
Last edited by ben-smith; 15-06-2011 at 18:13.
4. Re: STEP Maths I, II, III 1993 Solutions

5. Re: STEP Maths I, II, III 1993 Solutions
(Original post by Rabite)
I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

 Here it is anyway.

By the product/sum formulae that no one remembers.

But if m=±n, one of the fractions explodes. So in that case the question is:

If m=n=0, the integral turns to .

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt

Ignoring the +c for now

Which you can rewrite using the log form of arsinh.
Ignoring the +c for now

Last edited by klausw; 12-05-2012 at 21:34.
6. Re: STEP Maths I, II, III 1993 Solutions
(Original post by nota bene)
I'm on to Q1 STEP II (Further Pure A)

Is it just me or is it pure brute force?

Okay, if B knows A's numbers and they are of the form B can always chose so that he wins; the possible numbers are:
1+1+7
1+2+6
1+3+5
1+4+4
2+2+5
2+3+4
3+3+3 (and of course in other order as well...)
So here it is quite convincing B can always beat A if B knows A's numbers and the order of them.
edit: To make this clearer I'll list the possibilities explicitly like David told (so the person typing out these markschemes won't have to fill in half the answers themselves!)
A ------B
(117)(351)
(126)(351)
(135)(216)
(234)(441)
(333)(441)
(522)(144)
(144)(315)
Furthermore, if A choses 1+1+7 (or some other order of those numbers) it is impossible for A to win no matter what B choses (they can tie).

If now A gets to chose two triples and B are to find one triple that beats both it is quite clear from looking at the possible combinations that A shall not chose any of the combinations with few rearrangements (like 333 and 117 can easily be counteracted by 441). Many combinations leads to tied situations. From trial and error it is possible to eliminate all triples but some of the 531s. Testing these we can see 135/153 (by 171),135/531 (by 441), 135/513 (by 144), 135/351 (by 441), 135/315 (by 441), 153/531 (by 261), 153/513 (by 126), 153/351 (by 261), 153/315 (by 414), 531/513 (by 144), 531/351 (by 441), 531/315 (by NONE), 351/315 (by 414)
Conclusion: A shall chose 5+3+1 and 3+1+5 as his triples to always be sure of winning.

(okay and I am masochistic, here are all the stupid possible combinations:
117 (beats neither 531 nor 315)
711 (beats neither 531 nor 315)
171 (beats neither 531 nor 315)
144 (beats 531, not 315)
141 (beats neither 531 nor 315)
441 (beats 315, not 531)
135 (ties with both 315 and 531)
153 (beats 531, not 315)
531 (beats 315, ties with 531)
513 (ties with both 531 and 315)
351 (ties with both 531 and 315)
315 (beats neither 315 nor 531)
225 (beats neither 531 nor 315)
252 (beats 531, not 315)
522 (beats 315, ties with 531)
333 (ties with both 531 and 315)
234 (beats neither 531 nor 315)
243 (beats neither 531 nor 315)
432 (beats 315, ties with 531)
423 (beats 315, not 531)
342 (beats 531, ties with 315)
324 (beats neither 315 nor 531)
126 (beats 315, not 531)
162 (beats 531, not 315)
216 (beats neither 531 nor 315)
261 (beats neither 531 nor 315)
612 (beats 531, ties with 315)
621 (beats 315, ties with 531)
Firstly, as a1, a2, a3 are non negative numbers, so I think they can take the zero value as well.

For the first part, I first let A choose the numbers a1<a2<a3, then I let B choose b1=a1+1; b2=a2+1;b3= 7-a1-a2. Then I can prove that 7- a1-a2 is always non-negative so B can always find the valid b1,b2,b3 in this way.

For the second part I let A choose 0,0,9. Notice that A can never win the first two rounds and lose the last round. then it can be deduced that A can never win more rounds then the rounds he loses (if we say there are 3 rounds)

For the last bit, first condiser 5,3,1, then we can find the range of b1,b2,b3. Then we try to use b1,b2,b3 in these ranges to beat 3,1,5. But finally I proved that I can never beat it.

But I think brute force is quick indeed, at least for the first part.
7. Re: STEP Maths I, II, III 1993 Solutions
(Original post by generalebriety)
Done II/4, but I'm ****ed if I'm LaTeXing it. I swear TSR's LaTeX is messed up somehow. Written and scanned.
I agree with the first two parts but not the last one. The first part gives the perpendicular distance between the two lines, but this is not necessary the shortest distance between the two planes. This is because the position vector of the two planes is determined by the parameter t so they may not be able to just be at the two ends of the common perpendicular at the same tome.

My method is to express the distance in terms of v2 and treat all other things as constants. Then differentiate the expression with respect to v2 to get a minimum point. But my answer is ugly so I may have made some mistakes. Any one has an idea for this?
8. Re: STEP Maths I, II, III 1993 Solutions
(Original post by khaixiang)
I've done STEP II (paper A), question 2, 3, 8 and STEP III (paper B), question 1, 2, 4, 9 awhile ago. So I won't do them again and will try other questions. Here's question 9 of STEP II:

Sorry, I know this is very ugly, if it's too much of an eyesore I will remove it for conformity's sake.

This question seems way too easy compared to others, so perhaps I've overlooked something.
May be we need to notice that we only take 0 < =arg z < 2 pi, so arg(z1z2z3) = arg z1 + arg z2 + arg z3 > arg z3 is not necessary true. When arg z1 + arg z2 + arg z3 > 2 pi, we need to take 2 pi away from it to give a arg(z1z2z3) between 0 and 2 pi. And we need to prove that this value for arg(z1z2z3) is less than arg z1. But this is in fact no hard as well. I did it by some straightforward inequalities.
9. Re: STEP Maths I, II, III 1993 Solutions
From a brief scan for a couple of pages after the solution I haven't seen any answers to the final part of II, Q6. My solution is this:

Now, let

We also know that z lies on the circle with centre (0, 1) and radius 1:

so x=y

This gives solutions of x=0 or x=1, so our two solutions are 0+0i and 1+i.
Last edited by DJMayes; 16-05-2013 at 00:40.
10. Re: STEP Maths I, II, III 1993 Solutions
(Original post by brianeverit)
1993 STEP III number 13
...
(Original post by SimonM)
(Updated as far as #213) SimonM - 11.05.2009

If you see any mistakes please point them out.
There's nothing wrong with Brian's solution as such - he has got the required answer in exactly the way the examiners expected. But unfortunately, the required answer is wrong. It should be , not .

This is the only STEP question I've ever found with a mistake in it. It's particularly bad that it indicates that the examiner misunderstood Newton's Law.

I've got a document that proves the correct result four different ways, but I'll let people think about it a while before posting it.

EDIT This is being discussed at http://www.thestudentroom.co.uk/show...4#post43561514 . Unless that discussion is long finished, it's probably best to post there rather than here.
Last edited by MAD Phil; 17-07-2013 at 20:01.
11. Re: STEP Maths I, II, III 1993 Solutions
(Original post by SimonM)
..
There's nothing wrong with Brian's solution as such - he has got the required answer in exactly the way the examiners expected. But unfortunately, the required answer is wrong. It should be , not .

This is the only STEP question I've ever found with a mistake in it. It's particularly bad that it indicates that the examiner misunderstood Newton's Law.

I've got a document that proves the correct result four different ways, but I'll let people think about it a while before posting it.

EDIT This is being discussed at http://www.thestudentroom.co.uk/show...4#post43561514 . Unless that discussion is long finished, it's probably best to post there rather than here.

We've had a bit of a discussion over there, and the upshot is that the examiners used the wrong "Not Newton theorem" for the part of the chain hanging down. They used force = rate of change of momentum, which is not correct in this situation, as the parts of the chain joining the hanging section are not at rest just before joining the body. They should have used force = mass times acceleration, which is the correct theorem in cases where mass joins or leaves the body without changing its velocity. (I prove the theorems, in the appropriate situations, at http://www.thestudentroom.co.uk/show...4#post43576544 .)

Back in the 90's I wrote the attached document and sent it off to Cambridge; I can't remember what reply, if any, I got back. (That email system is long dead - I had to use a version of the document that I scanned in 2004.)

The document contains 4 proofs of the correct result. One uses the (correct) Not Newton Theorem and is very short. The next uses energy considerations; the rate of increase of the mechanical energy of the system is the rate at which (kinetic) energy is being fed into it by the mass joining it, minus the rate at which the system is doing work against the tension. The third finds a differential equation (with respect to position) satisfied by the tension in the hanging part of the chain, and solves it using boundary conditions derived from the lower end. The fourth uses rotational Newton for the pulley and the section of the chain in contact with it.

Those four methods all give the same answer, of course. I also give two "proofs" of the "as desired" value, but they both depend on using the incorrect Not Newton Theorem. I wanted to check that that was the full explanation of the value they asked for. I can't remember why I did it twice.
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Last updated: July 19, 2013
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