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OCR FSMQ Additional Maths 6th June 2016 Official Thread

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Original post by Lelanor
I'm doing this exam, unfortunately it clashes with another exam (General Studies, which is also an extra subject I'm doing). I have no plans to revise for either GCSE maths or fsmq, except learning the kinematics equations - anyone have some good ways to remember them? I can only remember v=u+at :/


I'm not sure if this will help but the way that I memorise it su su and vu vu.
Su su is a Thai phrase.
Then I know that the displacement equations start with s=u and s=u. Then I remember it from there.

This is very weird but it works for me. You could try to link it to something more familiar to you.
Does anybody know how hard the 2015 paper was. Does anybody have it?
Original post by gs41299
Does anybody know how hard the 2015 paper was. Does anybody have it?


I did it as a mock and found it harder than a lot of the past papers but scraped an A (which was miraculous considering I missed an easy five marks and forgot the cosine rule...)
2015 paper wasn't too bad - I did it as my mock as well.
I want to find it so I can redo it because I've done all the papers - so again, anyone know where I can find it? My teacher is too "busy" to print it out for me.
Does anyone know when the results day for this is? GCSE results or A levels?
Original post by Swimetti
Does anyone know when the results day for this is? GCSE results or A levels?


gcse
Original post by richpanda
gcse

thank you
Did the 2015 paper yesterday. It wasn't too bad, but to me it was still the hardest paper I've done and I've done every one since 2003. Bearing in mind I've averaged 85%, and even managed 98% in the 2011 paper, I was aiming for around 90. Don't think I got that, so I'd genuinely be happy with an A, even if it's a low one, because it was quite difficult.
Original post by kennethdcharles
Did the 2015 paper yesterday. It wasn't too bad, but to me it was still the hardest paper I've done and I've done every one since 2003. Bearing in mind I've averaged 85%, and even managed 98% in the 2011 paper, I was aiming for around 90. Don't think I got that, so I'd genuinely be happy with an A, even if it's a low one, because it was quite difficult.

Do you still have it, if so could you send pics of the questions here
Original post by gs41299
Do you still have it, if so could you send pics of the questions here
Sorry, I don't have it on m as it was a mock we did at school and so it's yet to be marked. If I get hold of it I will definitely send pics of it.
Original post by gs41299
Do you still have it, if so could you send pics of the questions here


I have a copy of it... but I can't upload things on my phone to here and I don't know how to do it on the computer...
But I'll try and see if I can find a way to upload it.
The quality of it was really bad, it was illegible and blurry, so I'm not going to upload it - but if I find a way where it's easier to read, then I'll upload it.
Cheers guys
Original post by gs41299
Cheers guys


We were given questions 1 to 10 to do during our lesson today. I can see if I can scan it?
I meant to post this the other day, i have the 2015 paper but I'm having trouble uploading the pictures of it here, so i put them on google drive: https://drive.google.com/folderview?id=0B-r2fgcRzAs_YWRNaFFidGpKb2M
(They haven't finished uploading as of when I'm posting this)
@gs41299
Original post by Lelanor
I meant to post this the other day, i have the 2015 paper but I'm having trouble uploading the pictures of it here, so i put them on google drive: https://drive.google.com/folderview?id=0B-r2fgcRzAs_YWRNaFFidGpKb2M
(They haven't finished uploading as of when I'm posting this)
@gs41299


Thank you so much
'The angleθis greater than 90◦ and less than 360◦ and cosθ= 2/3. Find the exact value of tanθ . '
By 'SOHCATOA', Cos=Adjacent / Hypoteneuse,
therefore Adjacent / Hypoteneuse = 2/3
so adjacent = 2
and hypoteneuse = 3
So then I used Pythagoras to find the Opposite angle (2^2 + b^2 = 3^2)
So b = square root of 5

By SOHCATOA, Tan= Opposite / Adjacent
so, I took the square root of 5 and divided it by 2 to get the value of Tanθ

But on the mark scheme it says that I should have divided -(square root of 5 / 2)
Why is the negative needed?
Original post by Tasha_140
'The angleθis greater than 90◦ and less than 360◦ and cosθ= 2/3. Find the exact value of tanθ . '
By 'SOHCATOA', Cos=Adjacent / Hypoteneuse,
therefore Adjacent / Hypoteneuse = 2/3
so adjacent = 2
and hypoteneuse = 3
So then I used Pythagoras to find the Opposite angle (2^2 + b^2 = 3^2)
So b = square root of 5

By SOHCATOA, Tan= Opposite / Adjacent
so, I took the square root of 5 and divided it by 2 to get the value of Tanθ

But on the mark scheme it says that I should have divided -(square root of 5 / 2)
Why is the negative needed?


Remember your quadrants.
Original post by Zacken
Remember your quadrants.


Sorry, still not following..?
Original post by Tasha_140
Sorry, still not following..?


If the angle is greater than 90, then what quadrant are you in? What sign does tangent have in that quadrant?

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