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STEP Prep Thread 2015

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Since there seems a lot of discussion about combinations to get to 60. How would 3 x 20 do compared to 4 x 15 and 6 x 10?
(edited 9 years ago)
Reply 41
Avatar for Principia
Principia
OP
8
Original post by shamika

1) we can argue with what you've said because whilst doing the latter, harder parts of questions will yield more marks than doing fragments of lots of questions, I'm not aware of 4 x 15 marks is viewed as "better" as 6 x 10 marks. At the end of the day, both would receive the same grade (but also see below in my edit for more detail)..


I'm not sure this is necessarily true. The final mark is "based on" the best six, but it may not just be a summation of the marks of the individual questions - I think I saw several plausible systems which place additional weight on more complete solutions whilst still maintaining the range 0-120.
Original post by zetamcfc
Since there seems a lot of discussion about combinations to get to 60. How would 3 x 20 do compared to 4 x 15 and 6 x 20?


i think 6x20 *might* be considered better :tongue:

Joking aside, when someone says partial to me 5 or 6 marks comes to mind. I think someoene scoring about 10 would have done over half of a question and there probably arent 10 marks in the easier beginning parts of a questions so consistently getting 10 isnt exactly bad.

Im curious: what are people opinions on something like 2 fulls and 4 partials or as i found myself having to do in III (3 fulls and 3 halves/partials)?
Original post by newblood
i think 6x20 *might* be considered better :tongue:

Joking aside, when someone says partial to me 5 or 6 marks comes to mind. I think someoene scoring about 10 would have done over half of a question and there probably arent 10 marks in the easier beginning parts of a questions so consistently getting 10 isnt exactly bad.

Im curious: what are people opinions on something like 2 fulls and 4 partials or as i found myself having to do in III (3 fulls and 3 halves/partials)?


lol, sorry quite tired at the moment :colondollar:
Original post by jtSketchy
For the last part of 5, I believe we got to the point where you had just used the substitution I hinted at, and you had ended up with:

 ∫1α1β−1u(1u−α)(β−1u)⋅du\displaystyle\ \int^{\frac{1}{\beta}}_{\frac{1}{\alpha}}\frac{-1}{u\sqrt {(\frac{1}{u}-\alpha)(\beta - \frac{1}{u})}}\cdot du

Next, try and get rid of all the fractions within the denominator, and remember that both 1α \frac{1}{\alpha} and 1β \frac{1}{\beta} are just constants, and so you could call them c and d (this may make the next part easier to see). Once you have done that, look at what you've got, see if you can do a little more manipulation to get your integral into something which heavily resembles the first integral in the question (the αβ \sqrt {\alpha \beta} in the answer is a huge hint), and try and finish the question.

As for question 7, what have you tried so far?


I don't have a clue what is going on here at all.What the limits are doesn't matter until you have actually integrated the expression and surely we don't want it looking like the integral of the first part because that wouldn't lead to pi because we have already substituted for x unless we can get it looking like 2theta which I doubt somehow.Anyway I have put u into the square root as u^2 and multiplied things out but there is no way I can take a sqrt(alpha*beta) out.

For Question 7 I tried the formulas for sin a + sinb and sin(a+b) but it didn't lead to anything and I also considered similar sums like sin 30+sin90+sin150+sin210+sin270 but the sums with 2pi/23 don't work out nicely there is no way for it to cancel with like sin(44pi/23) to give 0 like sin30 and sin330 would so I don't get where to go there.
Original post by Dalek1099
I don't know though because the person who got 6 lots of 10 has shown a wider range of talent and hasn't just done questions the 4 that are easy for them.


The person who answered 4 questions has shown their ability to get further through a problem and not give up because it gets harder whereas the person who answered 6 has shown they can do the easy parts
Original post by DanTheMan358
The person who answered 4 questions has shown their ability to get further through a problem and not give up because it gets harder whereas the person who answered 6 has shown they can do the easy parts


I think people are underestimating a 10. While i agree 4*15 probably is better than 6*10, the easy parts of a question would in my mind and having seen examiner reports get you a lot less than 10. Id have thought you would get about 10 if you gave a 'good/decent' attempt to most of a question (i only know as much as you guys, just speculation)
Original post by newblood
I think people are underestimating a 10. While i agree 4*15 probably is better than 6*10, the easy parts of a question would in my mind and having seen examiner reports get you a lot less than 10. Id have thought you would get about 10 if you gave a 'good/decent' attempt to most of a question (i only know as much as you guys, just speculation)


You're probably right, from what i've read they're more interested in what you learn from the first (easy) part and how you apply it to the rest of the question. At least i think i read that somewhere :tongue:
Reply 48
Avatar for Principia
Principia
OP
8
Also to say I'm in the same situation as jtSketchy and Ben 385 - I did a lot of STEP prep this year and as such am happy to answer questions about either STEP or the Cambridge admissions process (specifically Christ's).
Original post by Dalek1099
I don't have a clue what is going on here at all.What the limits are doesn't matter until you have actually integrated the expression and surely we don't want it looking like the integral of the first part because that wouldn't lead to pi because we have already substituted for x unless we can get it looking like 2theta which I doubt somehow.Anyway I have put u into the square root as u^2 and multiplied things out but there is no way I can take a sqrt(alpha*beta) out.


Right, what I'm getting at is this:

Spoiler



For Question 7 I tried the formulas for sin a + sinb and sin(a+b) but it didn't lead to anything and I also considered similar sums like sin 30+sin90+sin150+sin210+sin270 but the sums with 2pi/23 don't work out nicely there is no way for it to cancel with like sin(44pi/23) to give 0 like sin30 and sin330 would so I don't get where to go there.


Right, I remember when I first tried this question I tried using the addition formulae (but for a more general term in the sequence). I didn't get very far at all. However I found a different method which worked. Have you seen how to express the sine function in terms of exponentials?
Original post by jtSketchy
Right, what I'm getting at is this:

Spoiler





Right, I remember when I first tried this question I tried using the addition formulae (but for a more general term in the sequence). I didn't get very far at all. However I found a different method which worked. Have you seen how to express the sine function in terms of exponentials?


I am not getting how the integral is equal to the one in the first part which equals pi because if u=1/x then x=1/u and you substitute this back in and change the limits back so they change from 1/alpha to alpha it isn't the same integral:confused:(EDIT:I have to divide by the dx/du bit to change from x to u so forget this).However, there is still a problem in that 1/alpha is where alpha should be so going back to the first question this would give the same integral but limits of 1/0 and 2/pi surely which is ridiculous?

You mean using sinx=(e^ix-e^-ix)/2i?
(edited 9 years ago)
Original post by Dalek1099
I am not getting how the integral is equal to the one in the first part which equals pi because if u=1/x then x=1/u and you substitute this back in and change the limits back so they change from 1/alpha to alpha it isn't the same integral:confused:(EDIT:I have to divide by the dx/du bit to change from x to u so forget this).However, there is still a problem in that 1/alpha is where alpha should be so going back to the first question this would give the same integral but limits of 1/0 and 2/pi surely which is ridiculous?

You mean using sinx=(e^ix-e^-ix)/2?


The integral in the first part simply tells us a result which is true for any values of beta and alpha so long as beta is greater than alpha. So we can relate it to this integral, since 1/alpha and 1/beta are just constants.
Not quite - we would use a slightly different substitution, not exactly the same one. So we wouldn't get those limits.

And yes I mean that. Try using that form of sine and see if you can get anywhere with it.


Posted from TSR Mobile
Does anyone know when we actually enter for this exam? I'm assuming we wait until we get an offer (IF we get one)?
Reply 53
Avatar for alpen
alpen
Moving over to this thread, sat STEP I and II this year so should be some help, when I can of course.

Original post by EXTREMEninja
Does anyone know when we actually enter for this exam? I'm assuming we wait until we get an offer (IF we get one)?

I believe it was around February/March time my school entered me for STEP upon request... you just have to ask your school/college, whether you have an offer based on step or not it should be available for most mathematicians
Original post by EXTREMEninja
Does anyone know when we actually enter for this exam? I'm assuming we wait until we get an offer (IF we get one)?


Entry opens around March 2015 - you should have offers by then to know what you're sitting.
Original post by jtSketchy
The integral in the first part simply tells us a result which is true for any values of beta and alpha so long as beta is greater than alpha. So we can relate it to this integral, since 1/alpha and 1/beta are just constants.
Not quite - we would use a slightly different substitution, not exactly the same one. So we wouldn't get those limits.

And yes I mean that. Try using that form of sine and see if you can get anywhere with it.


Posted from TSR Mobile


I substituted in for sine and took out e^(2pi/23) and e^(-2pi/23) respectively ended up with 2 geometric series and evaluated them but I have ended up with something that looks nothing like tanx=e^ix-e^-ix/i(e^ix+e^-ix)?
(edited 9 years ago)
I evaluated both integrals correctly in Question 8 but didn't correctly substitute x=1/t and didn't come to the conclusion that it didn't work because t can't equal 0 but the integral is between 1 and -1.How many marks would I lose?
Reply 57
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Pyoro
8
Original post by shamika
Suggested changes - you might want to get thoughts from others to ensure there's consensus before you make the changes, to avoid you having to change it back


I agree (mostly) with all of these proposed changes (well, except that you say you don't think #3 is good advice and then go on to justify it anyway :biggrin:).
I'm still a bit concerned about the information overload, so it might be worth looking at consolidating the three posts (e.g. I think mathematical logic should be emphasised beyond week-before revision).

Original post by Principia
[OP]

I have a list of changes that would clear things up a bit, which you could implement if they seem reasonable and you have the time/patience:

Updates:
– "Past papers, solutions, exam reports 1987-2012
(download separately for years 1998-2012 here)" → replace with "2014"
– "...a cross reference between the first Siklos booklet and" → replace with "Siklos booklets"​
– "Spreadsheet to print out and keep track of which questions you have done (download separately here)" [dead link]
– "TSR has a solution bank of its own if you are interested:" Add 2014 (links available here)
– "(note: 2008 and 2010 are incomplete, and 2009 seems to be non-existant)" → replace with "there are currently no solutions for 2009 on TSR" [2008 and 2010 were completed last year.]
–"For the moment, the links are there more for coherence than anything else." [We're aiming for a fully-fledged resource list.]

Restructuring:
– Make a new heading called "RESOURCES": put a star just before the Megapack link and then put this heading in underneath it. Remove the star before the TSR solution bank. Put the links to useful tricks and graph sketching in this section.
– Put the entire FAQ at the end (natural location) and adjoin "In the week before the exam..." to the bottom.
– Turn the second paragraph into a FAQ item ("Is STEP useful for interviews?") and reference it with "Note: STEP Prep may help with interview prep – see the FAQ for more information."
– Move the bottom half of "Where do I start?" to "When should I start preparing?")

Misc:
– Remove the extraneous line breaks to make it easier to read.
– Delete "Also, rumor has it that the applied questions tend to be slightly easier than the pure ones." (vague; dealt with better in following sentences)
– "Any fully-correct answer will be awarded maximum marks."
– "Most marks are method marks; according to DFranklin, small slip-ups will only cost you 1-2 points."
– Remove the bare URLs (e.g. link the word "here").

Don't feel you have to follow these word-for-word, or at all, if you don't agree with them. :smile:
Original post by shamika
Was there anything in that which tells us more about how STEP is marked, or any other info we don't know?


I don't know what you know! They said just doing the first bit of a question gets a disappointing amount of credit e.g. 5 marks out of 20.
Original post by Principia
I'm not sure this is necessarily true. The final mark is "based on" the best six, but it may not just be a summation of the marks of the individual questions - I think I saw several plausible systems which place additional weight on more complete solutions whilst still maintaining the range 0-120.


You're right that I'm making the assumption that marking is linear. This is based on the marks that people on TSR get relative to how they think they've done, and snippets from examiners' reports.

Once the 2014 set of mark schemes are out I'll send in a Freedom of Information request to find out once and for all.

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