STEP Prep Thread 2019 Watch

Zacken
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#61
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#61
(Original post by Narjoyvt)
Hey guys, just a quick question. When the folks responsible for STEP say that the exam is done in the morning, what time do they exactly mean (in British time)? Because I am worried that my STEP papers will clash with my A2 exams.
It’s usually 9AM, but there is a protocol for handling clashing exams, so there’s no need for undue worry.
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RuneFreeze
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#62
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#62
(Original post by Zacken)
It’s usually 9AM, but there is a protocol for handling clashing exams, so there’s no need for undue worry.
This is excellent I perform so much better in the morning when I'm awake and can concentrate properly
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_gcx
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#63
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#63
(Original post by Think2ice)
Probably the only student that does media and is also taking a STEP Exam
Perhaps but hey if you enjoy it and it's an easy grade for you.
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etothepiiplusone
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#64
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#64
(Original post by RuneFreeze)
This is excellent I perform so much better in the morning when I'm awake and can concentrate properly
Haha, I'd be the opposite - most of the practise STEP I and II papers I've done thus far have been between 7-10pm or later
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username3720230
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#65
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#65
(Original post by etothepiiplusone)
Haha, I'd be the opposite - most of the practise STEP I and II papers I've done thus far have been between 7-10pm or later
just ruin ur sleep schedule enough to where 9am feels like 7pm
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TheTroll73
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#66
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#66
good luck to all that will take STEP exams this year
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Drogo Baggins
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#67
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#67
There is now a "mapping document" available on the STEP support programme website, you can find it here. As always, let me know of any errors!
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Rohan77642
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#68
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#68
This is not a step question, but is similar to a question asked in Q5 STEP 1 2010.
I was looking at binomial identities and found this on Wikipedia.
Anyone got any proof for this? Or if anyone is bored, you might wanna try this question cause it looks interesting.
Attached files
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DFranklin
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#69
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#69
(Original post by Rohan77642)
This is not a step question, but is similar to a question asked in Q5 STEP 1 2010.
I was looking at binomial identities and found this on Wikipedia.
Anyone got any proof for this? Or if anyone is bored, you might wanna try this question cause it looks interesting.
I think this is hard. I think you *have* to know DeMoivre and complex roots of unity at a minimum to progress with this.

Sketch of approach:

Spoiler:
Show




Define
S0 = nC0 + nC3 + nC6 + ...,
S1 = nC1 + nC4 + nC7 + ...,
S2 = nC2 + nC5 + nC8 + ...

and let w = (-1 + i \sqrt{3})/2 be a complex cube root of unity. Note w^2= (-1 - i \sqrt{3})/2, so Re(w) = Re(w^2) = -1/2, and Im(w) = -Im(w^2).

We know S0+S1+S2 = (1+1)^n = 2^n

We can find an expression for (1+w)^n of the form R^n (\cos n \theta + i \sin n\theta) by DeMoivre.

But (1+w)^n = 1 + nC1 w + nC2 w^2 + nC3 + nC4 w + ...
= S0 + S1 w + S2 w^2

equating the real and imaginary parts gives us 2 linear equations involving S0, S1, and S2, while
S0+S1+S2 = 2^n gives us a 3rd linear equation.

Solving for S0, S1 anf S2 should give the desired result. ].





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Rohan77642
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#70
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#70
(Original post by DFranklin)
I think this is hard. I think you *have* to know DeMoivre and complex roots of unity at a minimum to progress with this.

Sketch of approach:

Spoiler:
Show





Define
S0 = nC0 + nC3 + nC6 + ...,
S1 = nC1 + nC4 + nC7 + ...,
S2 = nC2 + nC5 + nC8 + ...

and let w = (-1 + i \sqrt{3})/2 be a complex cube root of unity. Note w^2= (-1 - i \sqrt{3})/2, so Re(w) = Re(w^2) = -1/2, and Im(w) = -Im(w^2).

We know S0+S1+S2 = (1+1)^n = 2^n

We can find an expression for (1+w)^n of the form R^n (\cos n \theta + i \sin n\theta) by DeMoivre.

But (1+w)^n = 1 + nC1 w + nC2 w^2 + nC3 + nC4 w + ...
= S0 + S1 w + S2 w^2

equating the real and imaginary parts gives us 2 linear equations involving S0, S1, and S2, while
S0+S1+S2 = 2^n gives us a 3rd linear equation.

Solving for S0, S1 anf S2 should give the desired result. ].






Wow. Where did you get the intuition that complex numbers are involved in this, if you dont mind me asking ?
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DFranklin
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#71
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#71
(Original post by Rohan77642)
Wow. Where did you get the intuition that complex numbers are involved in this, if you dont mind me asking ?
TBH, that's pretty obvious to anyone who's done a fair bit of series summing; you're going to want something whose powers repeat every 3 goes and that basically says "cube root of unity". And the form of the RHS is going to be fairly familar as well (at least in the sense that "there's a 'cos' term, so it's probably going to involve DeMoivre").

But even knowing "it will involve a cube root of unity", it took me quite a while to find a plausible approach though.
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Rohan77642
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#72
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#72
(Original post by DFranklin)
TBH, that's pretty obvious to anyone who's done a fair bit of series summing; you're going to want something whose powers repeat every 3 goes and that basically says "cube root of unity". And the form of the RHS is going to be fairly familar as well (at least in the sense that "there's a 'cos' term, so it's probably going to involve DeMoivre".

But even knowing "it will involve a cube root of unity", it took me quite a while to find a plausible approach though.
Thank you. I will try and understand the solution you wrote now.
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I hate maths
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#73
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#73
DFranklin
Rohan77642


Look at STEP I question 2 1990.

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have
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#74
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#74
(Original post by I hate maths)
DFranklin
Rohan77642


Look at STEP I question 2 1990.

That's a nice question. I think that's the first complex number STEP Question that I've done (cos I've stuck to the new spec and haven't touched STEP III).
I wouldn't mind more Without using a calculator, calculate... questions
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I hate maths
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#75
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#75
(Original post by have)
That's a nice question. I think that's the first complex number STEP Question that I've done (cos I've stuck to the new spec and haven't touched STEP III).
I wouldn't mind more Without using a calculator, calculate... questions
You'll really like this one then.
Spoiler:
Show


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have
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#76
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#76
(Original post by I hate maths)
You'll really like this one then.
Spoiler:
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I'm.good thanks
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Rohan77642
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#77
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#77
For step 1 2005 question 12, part b
I used the correct method that the mark scheme used but they seem to be using the values of 0.6, 0.7 and 0.3 whereas in the question it is given values of 0.7, 0.4 and 0.8. I don't understand where they are getting these values from?
I have attached the question and the mark scheme below.
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username3720230
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#78
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#78
(Original post by Rohan77642)
For step 1 2005 question 12, part b
I used the correct method that the mark scheme used but they seem to be using the values of 0.6, 0.7 and 0.3 whereas in the question it is given values of 0.7, 0.4 and 0.8. I don't understand where they are getting these values from?
I have attached the question and the mark scheme below.
I attached an attempt at a venn diagram. But the reason it's 0.6, 0.7 and 0.3 and not the other values is because:
P(wizard wears only hat)=P(wizard wears a hat)-P(wizard wears a hat and a cloak)-P(wizard wears a hat and a ring)-P(wizard wears a hat and a ring and a cloak)=0.7-x-y-0.1=0.6-x-y
The other two results can be obtained in the same way. So the 0.1 difference comes from P(wizard wears a hat and a ring and a cloak) which is the very middle section of the venn diagram
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Rohan77642
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#79
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#79
(Original post by sameactuallylol)
I attached an attempt at a venn diagram. But the reason it's 0.6, 0.7 and 0.3 and not the other values is because:
P(wizard wears only hat)=P(wizard wears a hat)-P(wizard wears a hat and a cloak)-P(wizard wears a hat and a ring)-P(wizard wears a hat and a ring and a cloak)=0.7-x-y-0.1=0.6-x-y
The other two results can be obtained in the same way. So the 0.1 difference comes from P(wizard wears a hat and a ring and a cloak) which is the very middle section of the venn diagram
Thanks!
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Rohan77642
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#80
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#80
For STEP 3 2009, I did the part 7 ii proof without induction. Would I be awarded full marks?
What I basically did was I found out the derivative of Pn(x) and then rearranged the given formula to show that it equal 0.
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