# AQA maths core 1, 13th May 2015

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#1
How did everyone find the paper?
0
6 years ago
#2
(Original post by asunny1998)
How did everyone find the paper?
horrible what was that cylinder question
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#3
How did everyone find the paper?
Also how did you work out the cylinder question?
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#4
I thought the last question was fine.
Hated the cylinder question
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6 years ago
#5
It was so much more difficult than anything else's before
1
#6
Can anyone remember what answers they got for questions?
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6 years ago
#7
(Original post by asunny1998)
How did everyone find the paper?
Also how did you work out the cylinder question?
Quite hard hoping the grade boundaries are low
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6 years ago
#8
(Original post by asunny1998)
Can anyone remember what answers they got for questions?
Got 31 something for the area of the shaded graph thingy.
2
6 years ago
#9
It was pretty hard! Hoping the grade boundaries are low.
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6 years ago
#10
Not gonna lie, did not even know there was a cylinder question
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6 years ago
#11
Cylinder question I messed up the first 5 marks. But the second part of it was okay.

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6 years ago
#12
Never mind wrong exam board
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6 years ago
#13
I tried the simultaneous equation so many times i got y=300 or sumthing it was the hardest paper i ever sat

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6 years ago
#14
I got 108/5 or something similar not 31.
6
6 years ago
#15
(Original post by asunny1998)
How did everyone find the paper?
Also how did you work out the cylinder question?
I ended up rushing the cylinder question in about 5 mins at the end, but somehow I got to the given answer. I worked out an equation for the surface area (Which i think was given as 49π?):

49π= πr^2 + h(2πr) where h is the height of the cylinder and 2πr if the circumference.

I then worked that through to make h the subject. For part b I worked out the volume to be V=2πr^2 x h and substituted in the equation for h from part a.

Hope that makes sense, I found the paper to be harder than most of the past papers I did so I think grade boundaries might be higher than usual.*

Hope everyone gets the grade they wanted!

EDIT: *Second to last sentence should say "I think grade boundaries will be lower than usual".
1
#16
I got 108/5 or something similar not 31.
I think I got something similar to you
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#17
I want someone to post an unofficial mark scheme 😝
5
#18
(Original post by ImLiterally12)
I ended up rushing the cylinder question in about 5 mins at the end, but somehow I got to the given answer. I worked out an equation for the surface area (Which i think was given as 49π?):

49π= πr^2 + h(2πr) where h is the height of the cylinder and 2πr if the circumference.

I then worked that through to make h the subject. For part b I worked out the volume to be V=2πr^2 x h and substituted in the equation for h from part a.

Hope that makes sense, I found the paper to be harder than most of the past papers I did so I think grade boundaries might be higher than usual.

Hope everyone gets the grade they wanted!
Left that question out 😝
0
6 years ago
#19
it was open top so the surface are was just pi r^2 + 2rh pi = 48 then you rearrange to get h=(48- pir^2) / 2r
then you workout the volume equation which was v=pihr^2 then sub what you got for h into the equation and simplify
2
#20
Hopefully someone posts an unofficial marks heme soon before I forget what my answers were x
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