M1 2014 lal paper need help.

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coconut64
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#1
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Hi, I did this paper from 2014 however, some of my answers don't match with the mark scheme's answer but it does match the other mark scheme posted by someone else (Arsey I think?).

So for question 8c) of this paper , I got 54s but the mark scheme is 126s ( if 126 is the right answer, please can someone explain?) https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

Also, my answer for question 4 doesn't match the mark scheme either as I got something like 0.8 and 0.6 as my final answer.Can anyone clarify ? Thanks,.
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Zacken
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(Original post by coconut64)
Hi, I did this paper from 2014 however, some of my answers don't match with the mark scheme's answer but it does match the other mark scheme posted by someone else (Arsey I think?).

So for question 8c) of this paper , I got 54s but the mark scheme is 126s ( if 126 is the right answer, please can someone explain?) https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf
The markscheme is not saying the the answer is 126. You are meant to note the symmetry of your graph and say that there are two points where the velocity is the same, at t =54 and by symmetry, at t = 126, this should come of no surprise to you as the "same speeds" are represented by the intersections of your velocity-time graphs for each train and there are two such points.
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Zacken
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(Original post by coconut64)

Also, my answer for question 4 doesn't match the mark scheme either as I got something like 0.8 and 0.6 as my final answer.Can anyone clarify ? Thanks,.
How can you get a number as your final answer when it says give your answer in terms of \mu?
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coconut64
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(Original post by Zacken)
How can you get a number as your final answer when it says give your answer in terms of \mu?
Hi, i mean 0.8 and 0.6 is part of my answer it is something like 0.8u-0.6/0.6mu-0.8 something like that. But the mark schme has 3 and 4 as part of their answer not o.6 or 0.8. Thanks
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coconut64
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(Original post by Zacken)
The markscheme is not saying the the answer is 126. You are meant to note the symmetry of your graph and say that there are two points where the velocity is the same, at t =54 and by symmetry, at t = 126, this should come of no surprise to you as the "same speeds" are represented by the intersections of your velocity-time graphs for each train and there are two such points.
I get it now thanks.
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Zacken
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(Original post by coconut64)
Hi, i mean 0.8 and 0.6 is part of my answer it is something like 0.8u-0.6/0.6mu-0.8 something like that. But the mark schme has 3 and 4 as part of their answer not o.6 or 0.8. Thanks
Ahhh, what happens when you simplify that by multiplying top and bottom of the fraction by 3 or 4?
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coconut64
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(Original post by Zacken)
Ahhh, what happens when you simplify that by multiplying top and bottom of the fraction by 3 or 4?
If I multiply the fraction by 4 it will just cancel the 4 in the fraction, this still wouldn't give me anything close to 0.8 or 0.6. Also the mark scheme specifies that this has to be the only answer. So I don't really get what's wrong with my answer.
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Zacken
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(Original post by coconut64)
If I multiply the fraction by 4 it will just cancel the 4 in the fraction, this still wouldn't give me anything close to 0.8 or 0.6. Also the mark scheme specifies that this has to be the only answer. So I don't really get what's wrong with my answer.
Can you show me a picture of your working?
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coconut64
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(Original post by Zacken)
Can you show me a picture of your working?
I did this a few days ago and can't actually find it now but I got the same answer as arsey I think that's his name. https://googledrive.com/host/0B1ZiqB...%20Edexcel.pdf It's question 4 I am on about, we both got a difrrent answer to the one in the mark scheme
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Zacken
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(Original post by coconut64)
I did this a few days ago and can't actually find it now but I got the same answer as arsey I think that's his name. https://googledrive.com/host/0B1ZiqB...%20Edexcel.pdf It's question 4 I am on about, we both got a difrrent answer to the one in the mark scheme
Surely it's the same thing as the markscheme?

\displaystyle k = \frac{0.8 \mu + 0.6}{0.8 + 0.6 \mu} \times \frac{5}{5} = \frac{4 \mu + 3}{4 + 3\mu}.
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coconut64
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#11
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(Original post by Zacken)
Surely it's the same thing as the markscheme?

\displaystyle k = \frac{0.8 \mu + 0.6}{0.8 + 0.6 \mu} \times \frac{5}{5} = \frac{4 \mu + 3}{4 + 3\mu}.
Oh okay. Why did they multiply it by 5 ?
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Zacken
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(Original post by coconut64)
Oh okay. Why did they multiply it by 5 ?
To make it look neater.
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coconut64
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(Original post by Zacken)
To make it look neater.
So does it mean I won't get the last mark because my answer is not exactly the same as the one in the mark schme cuz it said specifically. Thanks
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Zacken
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#14
(Original post by coconut64)
So does it mean I won't get the last mark because my answer is not exactly the same as the one in the mark schme cuz it said specifically. Thanks
It doesn't say specifically. It says the exact opposite. Can you see the "oe" at the end of the answer in the "notes" section of the markscheme? That means "or equivalent" and since your answer certainly is equivalent, then you'd get the mark.
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coconut64
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(Original post by Zacken)
It doesn't say specifically. It says the exact opposite. Can you see the "oe" at the end of the answer in the "notes" section of the markscheme? That means "or equivalent" and since your answer certainly is equivalent, then you'd get the mark.
Ok thx
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Zacken
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#16
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(Original post by coconut64)
Ok thx
Cheers.
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