Help rewrite TSR's STEP solutions! Watch

Quirky Object
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Hi everyone

We've noticed some comments that there are incorrect answers in TSR's collection of worked STEP solutions, so here is the thread to correct them! If you find a mistake in a solution, post the year, question number and your alternative solution here so that it can be checked and if there is agreement that it is correct, one of the ST or CAs will edit it into the relevant solutions thread (tag us if we're being slow).

Here are the solutions threads:
1987 1988 1989 1990 1991
1992 1993 1994 1995 1996
1997 1998 1999 2000 2001
2002 2003 2004 2005 2006
2007 2008 2009 2010 2011
2012 2013 2014 2015 2016
2017

Good luck to everyone preparing!
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Quirky Object
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Looks like I messed up the tag list :getmecoat:
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have
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(Original post by Sonechka)
If you find a mistake in a solution
It's not if, but when and how many.

What springs to mind is the very first new spec STEP I question.

STEP 1 1994, Q1
The poster does everything right (or at least what I did), until at the end he writes
Total volume = Volume of Prism + Volume of Pyramid
It should be imo
Total volume = Volume of Prism + 4 * Volume of Pyramid,
since there are 4 of his pyramids in each corner of the roof.
So then the volume comes up differently.

I'm revising for exams right now so don't have time to write it up nicely, but it's not hard.
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etothepiiplusone
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(Original post by Sonechka)
Hi everyone

We've noticed some comments that there are incorrect answers in TSR's collection of worked STEP solutions, so here is the thread to correct them! If you find a mistake in a solution, post the year, question number and your alternative solution here so that it can be checked and if there is agreement that it is correct, one of the ST or CAs will edit it into the relevant solutions thread (tag us if we're being slow).
Not the worst culprit, but STEP I 2000 Q6's solution, here, is incomplete.

First part:
Spoiler:
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Just expand (x-y+2)(x+y-1) and you get x^2-y^2+x+3y-2.


Sketch 1:
Spoiler:
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We're sketching where

x^2-y^2+x+3y>2 \Leftrightarrow (x-y+2)(x+y-1)> 0

This occurs when and only when those brackets have the same sign; our work comes down to sketching y = x + 2 and y = 1 - x on the same axis, and then shading the regions above one of the curves, and below the other of the curves: here's a desmos.


Sketch 2:
Spoiler:
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We first factorise the expression in a similar manner to before (by guessing coefficients, then checking):

x^2-4y^2+3x-2y+2 = (x+2y+2) (x-2y+1) < 0

And then sketch y=-\frac12 x - 1 and  y = \frac12 x + \frac12; we again want regions between the two curves, as such. Again, a desmos.


Final part:
Spoiler:
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(x,y)=(10,10) for example (admittedly, I don't see the significance of this part).
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DFranklin
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(Original post by Sonechka)
Hi everyone

We've noticed some comments that there are incorrect answers in TSR's collection of worked STEP solutions, so here is the thread to correct them!
In the older threads, what I notice a lot more is broken LaTeX rather than bad solutions (tied, I suspect, to TSR changing the LaTeX implementation several years ago).

I have to be honest, there's a certain "dog returning to its vomit" feeling about revisiting these threads for me, (particularly to fix other people's solutions). But if there's a particular case people are unsure about, etc. I'm happy to be tagged and take a look.
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have
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ignore this, I was being stupid with what it meant by average

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Another one that's wrong. I think it would be better for admins to just flag solutions that are wrong than waiting for a new one to be made

STEP I 1999 Q1
For the averages in both parts I got an answer different to the post. Lo and behold, java agrees with me


I doubt STEP examiner would give many marks for a wrong solution consisting soley of: By symmetry, the answer is x.

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DFranklin
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(Original post by have)
Another one that's wrong. I think it would be better for admins to just flag solutions that are wrong than waiting for a new one to be made

STEP I 1999 Q1
For the averages in both parts I got an answer different to the post. Lo and behold, java agrees with me


I doubt STEP examiner would give many marks for a wrong solution consisting soley of: By symmetry, the answer is x.
You've divided by 1000000 (or 4179) when you should have been dividing by the number of integers satisfying the conditions (400000 or 2388 I believe).
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have
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(Original post by DFranklin)
You've divided by 1000000 (or 4179) when you should have been dividing by the number of integers satisfying the conditions (400000 or 2388 I believe).
Oh alright thanks. Hmm. I read it to mean average as in divide by the million total numbers. or whatever.
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DFranklin
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(Original post by have)
Oh alright thanks. Hmm. I read it to mean average as in divide by the million total numbers. or whatever.
I think we'd discussed a similar question not long before Simon answered that one, so the method was in his mind. Basically if x is not a multiple of 2 or 5, then neither is 1000000-x. So you can pair each value x < 500000 with a value 1000000 - x that's > 500000 with a sum of 1000000, and so the average is going to be 1000000/2 (and similarly for the divisible by 3 and 7 problem).

FWIW: SimonM was a *very* able mathematician; his solutions may be overly brief (and I'm sure he would write more in an exam), but I don't think he made many mistakes.
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have
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(Original post by DFranklin)
I think we'd discussed a similar question not long before Simon answered that one, so the method was in his mind. Basically if x is not a multiple of 2 or 5, then neither is 1000000-x. So you can pair each value x < 500000 with a value 1000000 - x that's > 500000 with a sum of 1000000, and so the average is going to be 1000000/2 (and similarly for the divisible by 3 and 7 problem).

FWIW: SimonM was a *very* able mathematician; his solutions may be overly brief (and I'm sure he would write more in an exam), but I don't think he made many mistakes.
Yeah that's what I thought he meant by symmetry, and I was confused at how he got 500000(from my perspective of dividing by a million).

(I have no doubts about his ability, but we all make mistakes, like i just did now :^)))))

Anyway thanks for your help.
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Quirky Object
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Thanks everyone I'll edit out the two errors.
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etothepiiplusone
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(Original post by DFranklin)
I have to be honest, there's a certain "dog returning to its vomit" feeling about revisiting these threads for me, (particularly to fix other people's solutions). But if there's a particular case people are unsure about, etc. I'm happy to be tagged and take a look.
(Original post by Sonechka)
Thanks everyone I'll edit out the two errors.
I wouldn't want my solution edited in because of the quoted post form DFranklin for the time being. If anything, just those two desmos links could be edited in, since the solution wasn't wrong, just brief. In addition, that problem (00 S1 Q6) is in Siklos' Advanced Problems in Mathematics so is by far one of the least urgent solutions needing updating.

However, DFranklin I mentioned in passing in my post on the STEP prep thread that I felt that because TSR had become the number one google search result for 'step 19xx solutions' (and other old papers) certain solutions should be updated - users could as you mention above tag you in such posts but with so many STEP questions in existence I think they'd be more likely to just mark their answers as wrong and move on, which I thought should be rectified.
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Quirky Object
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(Original post by etothepiiplusone)
I wouldn't want my solution edited in because of the quoted post form DFranklin for the time being. If anything, just those two desmos links could be edited in, since the solution wasn't wrong, just brief. In addition, that problem (00 S1 Q6) is in Siklos' Advanced Problems in Mathematics so is by far one of the least urgent solutions needing updating.

However, DFranklin I mentioned in passing in my post on the STEP prep thread that I felt that because TSR had become the number one google search result for 'step 19xx solutions' (and other old papers) certain solutions should be updated - users could as you mention above tag you in such posts but with so many STEP questions in existence I think they'd be more likely to just mark their answers as wrong and move on, which I thought should be rectified.
It might not be urgent but that doesn't mean the solution shouldn't be edited anyway; the point of this whole thing is to improve as many solutions as possible so that TSR can be a more reliable STEP prep resource, since so many people look at it.
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I hate maths
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There seems to be a lot of broken links for solutions by Mathemagicien, I'm guessing he removed his posts. It's not great as he did STEP 3 2016 questions 11 to 13 so there is quite a large hole there.

Solution for Q12:

i) Consider X~B(100n,0.2). Then \displaystyle \alpha = P(16n\leqslant X \leqslant 24n)=1-P(|X-20n|&gt;4\sqrt{n}\sqrt{n}) \geqslant 1-\frac{1}{(\sqrt{n})^2} by Chebyshev's inequality, as required.

ii) Consider X~\text{Po}(n). Note \displaystyle 1-P(|X-n|&gt;n) \geqslant 1-\frac{1}{n} by Chebyshev's inequality.

Thus,
\displaystyle



\begin{align*} 1-\frac{1}{n}&\leqslant P(X \leqslant 2n) \\ &= \sum_{r=0}^{2n}\frac{e^{-n}n^{r}}{r!} \end{align*}

so \displaystyle (1-\frac{1}{n})e^n \leqslant \sum_{r=0}^{2n}\frac{n^r}{r!} as required.
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I hate maths
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Sonechka please could you consider attaching some of my answers to the step 3 2016 thread if DFranklin gives the green light? If we include questions that are straight up missing alongside Mathemagicien's posts there are five unanswered questions, the paper is actually in somewhat of a terrible situation on TSR.

Solution for Q11:

i)
Spoiler:
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I shall use the variable x to denote displacement of car. Given that resistive force is R=kv for some constant k, and driving force is \displaystyle F=\frac{P}{v} for constant power P, where v is the velocity of the car, we model the situation as following by Newton's second law:

\displaystyle



\begin{align*} \frac{P}{v}-kv=ma \ \text{(*)}\end{align*}

Using \displaystyle a=v\frac{dv}{dx} and that acceleration is zero when v=4U to find that \displaystyle k=\frac{P}{16U^2}, we have:

\displaystyle



\begin{align*} X_1 &= m\int_{U}^{2U}\frac{v^2}{P-\frac{P}{16U^2}v^2} \ \mathrm{d}v \\ &= \frac{16mU^2}{P}\int_{U}^{2U} \frac{v^2-16U^2+16U^2}{16U^2-v^2} \ \mathrm{d}v \\ &= \frac{16mU^2}{P} \Big[-v+2U \text{ln}\Big|\frac{4U+v}{4U-v}\Big| \Big]_{U}^{2U} \\ &= \frac{16mU^3}{P} \Big(-1+2\text{ln}\frac{9}{5} \Big) \end{align*}

and hence \displaystyle \lambda X_1 = 2\text{ln}\frac{9}{5}-1 as required.

ii)
Spoiler:
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If instead we are given that resistive force is R=kv^2, we find that \displaystyle k=\frac{P}{64U^3} instead, and thus:

\displaystyle



\begin{align*}X_2 &= \int_{U}^{2U} \frac{mv^2}{P-\frac{P}{64U^3}v^3} \ \mathrm{d}v \\ &= \frac{64mU^3}{P} \int_{U}^{2U} \frac{v^2}{64U^3-v^3} \ \mathrm{d}v \\ &=\frac{64mU^3}{3P} \Big[-\text{ln}|64U^3-v^3| \Big]_{U}^{2U} \\ &= \frac{64mU^3}{3P}\Big(\text{ln} \frac{9}{8}\Big) \end{align*}

hence \displaystyle \lambda X_2=\frac{4}{3}\text{ln}\frac{9}  {8} as required.


iii)
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Notice that X_1 &gt;X_2 \iff \lambda X_1 &gt; \lambda X_2 \iff 2\text{ln}\frac{9}{5}-1 &gt; \frac{4}{3}\text{ln}\frac{9}{8} \iff 4\text{ln}24 &gt; 3+6\text{ln}5

Using the given inequalities, we find that 12.68&lt;4\text{ln}24&lt;12.72 and 12.6&lt;3+6\text{ln}5&lt;12.66, thus X_1&gt;X_2.


New edit: minor corrections regarding capital letters and part iii.
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Quirky Object
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#16
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(Original post by I hate maths)
Sonechka please could you consider attaching some of my answers to the step 3 2016 thread if DFranklin gives the green light? If we include questions that are straight up missing alongside Mathemagicien's posts there are five unanswered questions, the paper is actually in somewhat of a terrible situation on TSR.

Solution for Q11:

i)
Spoiler:
Show


I shall use the variable x to denote displacement of car. Given that resistive force is R=kv for some constant k, and driving force is \displaystyle F=\frac{P}{v} for constant power P, where v is the velocity of the car, we model the situation as following by Newton's second law:

\displaystyle



\begin{align*} \frac{P}{v}-kv=ma \ \text{(*)}\end{align*}

Using \displaystyle a=v\frac{dv}{dx} and that acceleration is zero when v=4U to find that \displaystyle k=\frac{P}{16U^2}, we have:

\displaystyle



\begin{align*} X_1 &= m\int_{U}^{2U}\frac{v^2}{P-\frac{P}{16U^2}v^2} \ \mathrm{d}v \\ &= \frac{16mU^2}{P}\int_{U}^{2U} \frac{v^2-16U^2+16U^2}{16U^2-v^2} \ \mathrm{d}v \\ &= \frac{16mU^2}{P} \Big[-v+2u \text{ln}\Big|\frac{4u+v}{4u-v}\Big| \Big]_{U}^{2U} \\ &= \frac{16mU^3}{P} \Big(-1+2\text{ln}\frac{9}{5} \Big) \end{align*}

and hence \displaystyle \lambda X_1 = 2\text{ln}\frac{9}{5}-1 as required.




ii)
Spoiler:
Show


If instead we are given that resistive force is R=kv^2, we find that \displaystyle k=\frac{P}{64U^3} instead, and thus:

\displaystyle



\begin{align*}X_2 &= \int_{U}^{2U} \frac{mv^2}{P-\frac{P}{64U^3}v^3} \ \mathrm{d}v \\ &= \frac{64mU^3}{P} \int_{U}^{2U} \frac{v^2}{64U^3-v^3} \ \mathrm{d}v \\ &=\frac{64mU^3}{3P} \Big[-\text{ln}|64U^3-v^3| \Big]_{U}^{2U} \\ &= \frac{64mU^3}{3P}\Big(\text{ln} \frac{9}{8}\Big) \end{align*}

Hence \displaystyle \lambda X_2=\frac{4}{3}\text{ln}\frac{9}  {8} as required.



iii)
Spoiler:
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Suppose that X_1 &gt; X_2.

Notice that X_1 &gt;X_2 \iff \lambda X_1 &gt; \lambda X_2 \iff 2\text{ln}\frac{9}{5}-1 &gt; \frac{4}{3}\text{ln}\frac{9}{8} \iff 4\text{ln}24 &gt; 3+6\text{ln}5

Using the given inequalities, we find that 12.68&lt;4\text{ln}24&lt;12.72 and 12.6&lt;3+6\text{ln}5&lt;12.66, thus X_1&gt;X_2 indeed.

Will do; they do need to be checked first so that they're as reliable as possible though (not doubting your ability at all ).
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Zacken
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(Original post by I hate maths)
Solution for Q11:
I did this in my exam because of how easy it was, despite having done no mechanics preparation.

(Original post by Sonechka)
Will do; they do need to be checked first so that they're as reliable as possible though (not doubting your ability at all ).
All good, last part is confusingly written, in my opinion (It starts off by assuming the result). All that's needed is:

\displaystyle

\begin{align*} \lambda(X_1 - X_2) &= 2 \log \frac{9}{5} - 1 - \frac{4}{3}\ln \frac{9}{8} \\ & = \frac{4}{3} \log 3 - 2 \log 5 - 1 + 4 \log 2 \\ & &gt; \frac{1}{3} (4 \times 3.17  - 6 \times 1.61 - 3)  &gt; 0 \end{align*}

and \lambda \geqslant 0 so X_1 &gt; X_2.
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Quirky Object
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Thanks I hate maths and Zacken, I'll put up the solution now Mathemagicien's deleted posts just say "Edit: Sorry, wrong thread" so I'm not sure what happened there.

Since STEP III 2016 is so incomplete. write-ups for questions 7, 10 and 13 would be great too, whenever anyone has time to do those or uses the paper for practice.
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I hate maths
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(Original post by Sonechka)
Thanks I hate maths and Zacken, I'll put up the solution now Mathemagicien's deleted posts just say "Edit: Sorry, wrong thread" so I'm not sure what happened there.

Since STEP III 2016 is so incomplete. write-ups for questions 7, 9 and 13 would be great too, whenever anyone has time to do those or uses the paper for practice.
Mikelbird actually put up a solution for 7 on the latest page. However, he does not explain why he can let X_1,X_2,...X_{n-1} be \omega,\omega^2,...\omega^{n-1} without loss of generality, but it's a minor thing really, the solution is fine in my opinion.
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Quirky Object
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(Original post by I hate maths)
Mikelbird actually put up a solution for 7 on the latest page. However, he does not explain why he can let X_1,X_2,...X_{n-1} be \omega,\omega^2,...\omega^{n-1} without loss of generality, but it's a minor thing really, the solution is fine in my opinion.
Do you think it would get 20? The solutions should be pretty much flawless (I'm not sure how signficant an error/omission needs to be for a mark to be dropped).
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