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AQA Unit 3 Mathematics 13 June 2012 - Unofficial Mark Scheme [43603H] [GCSE]

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The arc was from two points on an equilateral triangle, meaning the angle was 60 degrees.
(60/360)*16pi=8.377...
Answer = 8.38 (3sf)



Nopeeee, i dont think you use 8cm as the length, being an A* style question, you had to work out the radius for the circle using the sin rule.

Picture 24.jpg


If you cant read my working:

Use sine rule to work out x, x is the radius.
90 + 30 = 120. 180-120=60

x/sine 90 = 8/sine 60

x= 9.236
9.238 x 2 = diameter Diameter=18.48
To find arc length
60/360 x( 3.14x18.48) = 9.68
Reply 21
Original post by jacqueline789
can you include how many marks each questions is worth on a mark scheme?
and grade boundaries?


Sorry, I can't remember how much each question was worth, if anyone is sure of how many marks please post...

Also I have no idea on the grade boundaries, totally depends on how everyone found it... quite a few people on here are saying it was 'hard' though, meaning the boundaries will be lowered.
Reply 22
Original post by Xx1CALUM1xX
...

I have seen others agree with the posted answer and have not seen another post with your answer, so I shall keep it as it is for now. Also, the arc is just a section on the circumference, meaning that 8cm would be the radius, as the line from the centre to anywhere on the circumference is the radius.
Edit:But I have said disputed answer
(edited 11 years ago)
Reply 23
I put 20 for the first question oh FFS... I only lose 1 mark?
Reply 24
hey many marks will you need for a very high A? 60 enough?
Ok the arc length question is WRONG.

the triangle was equilateral. So all sides were 8cm!
So the SEMI-circles DIAMETER was 8cm.
Therefore its radius must be 4cm.

The arc length is pi x 2r or pi x D
Now the next bit is where everyone messed up and involved the triangles angles in their answers.
The angle you should have used was 180 degrees as it infact a semi circle.
180/360 x pi x D = answer.
Original post by PrinceUpsb
I have seen others agree with the posted answer and have not seen another post with your answer, so I shall keep it as it is for now. Also, the arc is just a section on the circumference, meaning that 8cm would be the radius, as the line from the centre to anywhere on the circumference is the radius.
Edit:But I have said disputed answer


My below answer may be incorrect. However I can assure you 100% the radius was not 8cm.
Reply 27
Original post by IAmYourdog
Ok the arc length question is WRONG.

the triangle was equilateral. So all sides were 8cm!
So the SEMI-circles DIAMETER was 8cm.
Therefore its radius must be 4cm.

The arc length is pi x 2r or pi x D
Now the next bit is where everyone messed up and involved the triangles angles in their answers.
The angle you should have used was 180 degrees as it infact a semi circle.
180/360 x pi x D = answer.

Take a look at the diagram, that is not a semi-circle.
Original post by IAmYourdog
Ok the arc length question is WRONG.

the triangle was equilateral. So all sides were 8cm!
So the SEMI-circles DIAMETER was 8cm.
Therefore its radius must be 4cm.

The arc length is pi x 2r or pi x D
Now the next bit is where everyone messed up and involved the triangles angles in their answers.
The angle you should have used was 180 degrees as it infact a semi circle.
180/360 x pi x D = answer.


The radius was 18 as is said 0 marks the centre meaning centre of the circle so the radius was 8


This was posted from The Student Room's iPhone/iPad App
Doc1.docx sorry i don't know any other way to upload it like the original. :redface:
Reply 30
That was a fun exam counting the questions I knew were right I have got at least 55/80. Answered every question but don't know if some where right. Is it too much to hope thats an A.

As for the initial graph question where you have to fill in the table and then draw the graph, for the second part where did people put that they intersected? For some reason I believe I have messed up on that one.
Has anybody got an unofficial mark scheme for Foundation tier for Unit 3 13/06/2012?
Reply 32
some more answers... final vector question was t+1/2s.

the question with x and y:

2.5y=x

2x +2y

2.8x

graph points of intersection, -2.5 and 2.5
Reply 33
Original post by Xx1CALUM1xX
The arc was from two points on an equilateral triangle, meaning the angle was 60 degrees.
(60/360)*16pi=8.377...
Answer = 8.38 (3sf)



Nopeeee, i dont think you use 8cm as the length, being an A* style question, you had to work out the radius for the circle using the sin rule.

Picture 24.jpg


If you cant read my working:

Use sine rule to work out x, x is the radius.
90 + 30 = 120. 180-120=60

x/sine 90 = 8/sine 60

x= 9.236
9.238 x 2 = diameter Diameter=18.48
To find arc length
60/360 x( 3.14x18.48) = 9.68

dreadfully sorry to have to prove you wrong however it was an equilateral triangle thus meaning 8cm was the length
Original post by Will2012
some more answers... final vector question was t+1/2s.



are you sure, cause i took quite a bit of time on that question and i got 0.5(s+2t).
:confused:
can you explain how you got it? please
Reply 35
Original post by Mathsismything
The radius was 18 as is said 0 marks the centre meaning centre of the circle so the radius was 8


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


not sure if this is what you have just said but this is what i did...

1) worked out what percentage of the circle the sector was.. 360/60= 6

so therefore the length of the arc is: radius =8
2 x pi x r = circumference of whole circle
2 x pi x 8 = 50.something
50.something :P /6= 8.38

anywayy this is what i did as i swear it asked for the length of the arc not the area like a large porportion of people have incorrectly done but not sure compltely
(edited 11 years ago)
Reply 36
^ radius was 8cm, so diameter is 16. Equilateral triangle therefore 60 degrees. 60/360*16pie
Reply 37
full unofficial markscheme anyone please??
Reply 38
Original post by mooooo.nisah
are you sure, cause i took quite a bit of time on that question and i got 0.5(s+2t).
:confused:
can you explain how you got it? please


you had s going ----> and t was going bottom left. X was on the top right. So -s +t + the bottom bit which was 1.5s, as the ratio was 3:1 so 1.5:1. so -s+t+1.5s =t+1/2s
(edited 11 years ago)
omg how could i not see this, i just factorized it and you didn't. we did the exact same thing. :smile:

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