STEP Prep Thread 2015

Thats what I thought, but he was told this when he went to a STEP day at Fitzwilliam. It is possible the tutors there are misinformed?

Hmm, when he said college I assumed he meant college as in for A-levels.
It still seems unlikely to me, simply because the classing system has a fixed number of firsts produced, whereas in STEP, there is no such thing.
The tripos has longs and shorts which differs from STEP again, whilst I can see that in the spirit of things more credit is given to fuller solutions, and of course the marking process differs from A-levels I can't see that they use the exact same system.

Do the tripos exams use a root mean square method of marking (quadratic means)? Maybe that's what they meant, the later parts of questions have a higher allocation of marks where as it is fairly linear at A-Levels.

Do the tripos exams use a root mean square method of marking (quadratic means)? Maybe that's what they meant, the later parts of questions have a higher allocation of marks where as it is fairly linear at A-Levels.

On STEP I 2005 Q8 (i) the question specified that y=2 when x =1. Which I interpreted as implying that $y = -(x+1)$ was not a solution, and also that the solution to the differential equation represented a straight line.

However in the official solutions they state that $y = \pm (x+1)$ and also that the solution represents a pair of straight lines. I have two questions - 1) What is wrong with my interpretation of the question? 2) If my interpretation is correct, would I still lose marks? (and roughly how many 2-3?)

Ok so, albeit it was STEP 1, and they were all really easy questions in comparison to others, I managed to get 3 full solutions in a Step paper today! Great feeling

Hmm, when he said college I assumed he meant college as in for A-levels.
It still seems unlikely to me, simply because the classing system has a fixed number of firsts produced, whereas in STEP, there is no such thing.
The tripos has longs and shorts which differs from STEP again, whilst I can see that in the spirit of things more credit is given to fuller solutions, and of course the marking process differs from A-levels I can't see that they use the exact same system.

Obviously I don't know for a fact that this is the case - I don't know how it's really broken down, but then very few people do.

Do the tutors have a reason for not explaining the STEP Marking system? It's fairly obvious that the later stages of questions are worth more marks, so I don't think it would be detrimental to precisely explain it.

Do the tripos exams use a root mean square method of marking (quadratic means)? Maybe that's what they meant, the later parts of questions have a higher allocation of marks where as it is fairly linear at A-Levels.

At least that would make the marks greater than or equal to the marks using arithmetic means!




I'll show myself out.

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Any hints with the last part of STEP III 2004 Q4?

I got the area of AOB to be:
$\frac{r_0^2}{sin(\alpha)cos(\alpha)}$

Then I set up the following inequality:
$\frac{4r_0^2}{5sin(\alpha)cos(\alpha)}<\frac{\pi r_0^2(1+sin(\alpha)^2)}{4sin(\alpha)}$

Given our range for then angle, sin and cos are always positive so we can rearrange to get:

$0<5\pi cos(\alpha)(2-cos(\alpha)^2)-16$

Since I couldn't solve that cubic and didn't want to use numerical methods I
considered the function:
$f(x)=5\pi cos(\alpha)(2-cos(\alpha)^2)-16$

We get local extrema at:
$cos(\alpha)^2=0, \frac23$

The 0 point is negative so I checked the other one to see if the function was greater than 0 at that point.

This leads to the inequality :
$25\pi^2>25 \times 9=225>216$

I'm not sure if I've made any mistakes but there must have been a neater way of approaching this question.

Any time I hear I'm getting free stuff I get soo excited

Both lines satisfy the differential equation for all values, but you are right, only the positive one satisfies the initial conditions.

To be really safe I would note the pair of solutions to the differential equations and then say that one violates the boundary conditions.

I sincerely doubt you would lose any marks at all, the wording of the question isn't ambiguous at all. The solution should satisfy the conditions stated.

I agree with you. But if you look here at the official solutions they haven't excluded a solution - they only used the boundary conditions to find the constant of integration.

What is the hardest STEP question you guys have done?