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2019 Edexcel GCSE Maths Higher : Predicted topics for paper 3

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ahh alrigtt thanks, we propbs also need the area of a pyramid which i think is 1/3 x base area x height. Also just to confirm volume of cylinder is pie x r^2 x height but im blanked about the surface area of a cylinder?
Original post by sqrt of 5
no, you only need to know the one for the cylinder. surface area + volume
Original post by DualGunZ
ahh alrigtt thanks, we propbs also need the area of a pyramid which i think is 1/3 x base area x height. Also just to confirm volume of cylinder is pie x r^2 x height but im blanked about the surface area of a cylinder?


2*pi*r2 + 2*pi*r*h
Reply 43

Yes it's not explained at all.

The "top 56" drivers were fined and you need to estimate the lowest speed of these top 56. Now looking at the table or the areas of the histogram bars you see that:

The top 5 lie between 50 and 60
The top 40 lie between 40 and 60
The top 80 lie between 30 and 60

So the 56th fastest will lie somewhere between 30 and 40. Can you continue from here? If you're still stuck let me know - this is one of the trickiest topics in GCSE.
hey, is there a markscheme?
Thank you. Yes I know that 56 lies in between 30 & 40 but I'm not sure how they got 36 in the end?
Original post by Notnek
Yes it's not explained at all.

The "top 56" drivers were fined and you need to estimate the lowest speed of these top 56. Now looking at the table or the areas of the histogram bars you see that:

The top 5 lie between 50 and 60
The top 40 lie between 40 and 60
The top 80 lie between 30 and 60

So the 56th fastest will lie somewhere between 30 and 40. Can you continue from here? If you're still stuck let me know - this is one of the trickiest topics in GCSE.
100% for real.
Original post by sqrt of 5
hell no i hate circle theorems. it takes me ages to spot them :frown: id rather a question on equation of a circle
Reply 47
Original post by liaente
Thank you. Yes I know that 56 lies in between 30 & 40 but I'm not sure how they got 36 in the end?

At this stage of the working it's best to look at the histogram bar.

The area of the whole bar is 40 but we need an area of 16 (to add to the areas of the bars to the right of it) to give 56. So you need to estimate how much of the bar you need to give you an area of 16.

The height of the bar is 4 so that means you need a width of 4 since 4 x 4 = 16.

So the area you need is the shaded area marked on the diagram and this leads to a value of 36.

Make sense? If not please let me know at which point you got lost.
:shock: Oh!!! Okay I understand! I was counting the squares, what I did was 400 squares = 40 for frequency, so I did 16 = 160 but then I got confused.
I understand now!
To be honest, in the exam I would count the other way, as I would do 140 x 40% which is 56, and then I'd count up - so I'd land between 20 & 30 and it would've been so inaccurate.
Thanks!
Original post by Notnek
At this stage of the working it's best to look at the histogram bar.

The area of the whole bar is 40 but we need an area of 16 (to add to the areas of the bars to the right of it) to give 56. So you need to estimate how much of the bar you need to give you an area of 16.

The height of the bar is 4 so that means you need a width of 4 since 4 x 4 = 16.

So the area you need is the shaded area marked on the diagram and this leads to a value of 36.

Make sense? If not please let me know at which point you got lost.
affirmative
Original post by G11T11
To be honest, I really hope quadratic simultaneous equations come up. They're very complicated, but at least it's a straight-forward question. It's much easier than a long-winded problem solving question which purposefully confuses you.
Well that was *******s
hahaha! Agreed! It could've been better 100% more.
Original post by Princeyk16
Well that was *******s
Reply 52
Grade boundary for a 6?
Last year it was 108. This year is gonna be similar or less. So roughly 100ish to around 118.
Original post by Aamilah12
Grade boundary for a 6?

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