[kashagupta]

(Original post by boing360)
heyy can anyone help me i got 2 As so far - 104 and 98 ive woeked out i need 118 UMS for an A but does anyone know roughly what sorta mark out of 80 i need ? thankss

Around 59?

[Herschel]

I need 118 ums
what mark would it roughly be?
what would the grade boundaries be any idea?

[v.007]

Hey guys can someone help me solve this question from the edexcel unit 3 book ??, its under the section of setting up and solving quadratics , ive been stuck for 30 minutes on this question !

thanks
BTW its question 9 about circles ...

[24hoursmiles]

Does anyone know what roughly the grade boundaries will be for the unit 3 mock paper?

[Hi, How are you ?]

Hey guys, if you work something out and the answer is correct, but the mark scheme uses a different method and your method is not on the mark scheme, would you get all the mark for that question? Thanks

[amyAcurran]

(Original post by kashagupta)
All the papers I have are attached I also have the mark scheme for the mock paper, but I don't have a digital copy of the actual paper :/

do the practise papers have mark schemes?

[kashagupta]

(Original post by amyAcurran)
do the practise papers have mark schemes?

Yepp, there are 4 papers with their mark schemes

[v.007]

When working out the area of a triangle using 1/2abSinC , how do you know angle to take as sin C ???

[Hayley W]

Sorry If this has already been asked, but if you get the correct answer using a different method to the one shown on the mark scheme, would you still get the marks?

[Hayley W]

(Original post by v.007)
When working out the area of a triangle using 1/2abSinC , how do you know angle to take as sin C ???

It doesn't matter as long as you have the two side lengths on either side of the angle and use them as a and b

[QnRtt]

(Original post by Hayley W)
Sorry If this has already been asked, but if you get the correct answer using a different method to the one shown on the mark scheme, would you still get the marks?

if it makes sense then yes

[Laureeenn159]

I'm confused, I have just done the paper by Churchill Maths on top of the thread, but the mark scheme isn't the right one for the paper? Does anyone know where the answers are?

[nievans1]

Well spotted Welbeck - posted the wrong file, apologies. Here is the correct one - apparently sample assessment material for this module, with mark scheme. It's too big to post here, so link added below. Alternatively search for file "GCSE-2010-Spec-B-Unit-3H.pdf" for other d/l sources.


http://ebookbrowse.com/gdoc.php?id=1...f77eaa1c45230b

[Dilzo999]

(Original post by v.007)
Hey guys can someone help me solve this question from the edexcel unit 3 book ??, its under the section of setting up and solving quadratics , ive been stuck for 30 minutes on this question !

thanks
BTW its question 9 about circles ...

It's easy once you grasp the entire way on how to do it, here's how ^2 = Squared:
Lets call the small circle A and the bigger circle B
Area of circle A =
0.5*pi(x+3)^2 (Area of circle = pi * r^2)
expand it out gives us
0.5*pi(x^2+6x+9)
Area of circle B =
pi*(2x)^2
Expand it out = pi*4x^2
4pi*x^2-0.5*pi(x^2+6x+9)
divide the whole thing by pi so it now =
4x^2-0.5*pi(x^2+6x+9) which if divide by 0.5 it goes into the equation so
1/2*pi(8x^2-(x^2+6x-9) now simplify it becomes
1/2*pi(7x^2-6x-9)=84pi
now divide by 1/2pi so you get
(7x^2-6x-9)=168
now make it so 0 is on one side so -168 which =
7x^2-6x-177=0
now just use completing the square or the quadratic formula, whichever you prefer and you've got your answer so, the answer is 5.475291923615913 however it doesn't state it to 1 decimal place so this is still the right answer.
Sorry if it's a little complicated didn't know an other way to explain it.

Also I've pretty much revised everything but the only thing I still can't fully grasp is how I'm going to remember the equations to the volume of 3d shapes the only thing I can do is pray it doesn't come up

[v.007]

(Original post by Dilzo999)
It's easy once you grasp the entire way on how to do it, here's how ^2 = Squared:
Lets call the small circle A and the bigger circle B
Area of circle A =
0.5*pi(x+3)^2 (Area of circle = pi * r^2)
expand it out gives us
0.5*pi(x^2+6x+9)
Area of circle B =
pi*(2x)^2
Expand it out = pi*4x^2
4pi*x^2-0.5*pi(x^2+6x+9)
divide the whole thing by pi so it now =
4x^2-0.5*pi(x^2+6x+9) which if divide by 0.5 it goes into the equation so
1/2*pi(8x^2-(x^2+6x-9) now simplify it becomes
1/2*pi(7x^2-6x-9)=84pi
now divide by 1/2pi so you get
(7x^2-6x-9)=168
now make it so 0 is on one side so -168 which =
7x^2-6x-177=0
now just use completing the square or the quadratic formula, whichever you prefer and you've got your answer so, the answer is 5.475291923615913 however it doesn't state it to 1 decimal place so this is still the right answer.
Sorry if it's a little complicated didn't know an other way to explain it.

Also I've pretty much revised everything but the only thing I still can't fully grasp is how I'm going to remember the equations to the volume of 3d shapes the only thing I can do is pray it doesn't come up

Thank you so much !!!! I got it now ! I. Realised I went wrong when diving by 0.5 . Thanks


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[Kayannk]

(Original post by A193)
I have th same problem as yours, but you are much luckier. I am at 200 UMS, i need 60 marks out of 80 to get my A, but this exam is much harder. I need an A even though my target grade is A*. LOL

I know right! I'm not saying the last two units were easy... but compared to this one!

Good luck friend!

[Kayannk]

(Original post by gmseahorse)
You should remember that ums is your friend, the a* mark might be considerably lower than you think, fingers crossed anyway.

Hopefully! Good luck if you're taking this also

[Me123456789]

I just got 42/80 made such stupid mistakes
Does anyone know roughly what grade that would be?

[kashagupta]

(Original post by Dilzo999)
It's easy once you grasp the entire way on how to do it, here's how ^2 = Squared:
Lets call the small circle A and the bigger circle B
Area of circle A =
0.5*pi(x+3)^2 (Area of circle = pi * r^2)
expand it out gives us
0.5*pi(x^2+6x+9)
Area of circle B =
pi*(2x)^2
Expand it out = pi*4x^2
4pi*x^2-0.5*pi(x^2+6x+9)
divide the whole thing by pi so it now =
4x^2-0.5*pi(x^2+6x+9) which if divide by 0.5 it goes into the equation so
1/2*pi(8x^2-(x^2+6x-9) now simplify it becomes
1/2*pi(7x^2-6x-9)=84pi
now divide by 1/2pi so you get
(7x^2-6x-9)=168
now make it so 0 is on one side so -168 which =
7x^2-6x-177=0
now just use completing the square or the quadratic formula, whichever you prefer and you've got your answer so, the answer is 5.475291923615913 however it doesn't state it to 1 decimal place so this is still the right answer.
Sorry if it's a little complicated didn't know an other way to explain it.

Also I've pretty much revised everything but the only thing I still can't fully grasp is how I'm going to remember the equations to the volume of 3d shapes the only thing I can do is pray it doesn't come up

Most of the volumes of 3D shapes are at the front of the paper though, aren't they? It's only really the cylinder, right?


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[KatiexZoe]

(Original post by Me123456789)
I just got 42/80 made such stupid mistakes
Does anyone know roughly what grade that would be?

My teacher said that just over half marks would be a B.


This was posted from The Student Room's iPhone/iPad App