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".

(edited 8 years ago)

Reply 15

asunny1998

OP

2

Original post by SaadMuhammad

I got 108/5 or something similar not 31.

I think I got something similar to you

Reply 16

asunny1998

OP

2

I want someone to post an unofficial mark scheme 😝

Reply 17

asunny1998

OP

2

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 😝

Reply 18

gottapassitall

1

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