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# GCSE MATHS EdExcel watch

1. hi all i hope that it all went well, how did everyone do???, i think that it was much easier that the previous non calculator paper, i mean the first 16 questions were suprisingly easy, how did the rest of the nation find it
???
im trying to recall as many questions as i can whilst it is still fresh in my head

what ones can you remebers
2. there were lots of percentage ones! YAY lol, it was a hell of a lot better than the first maths one! I made stupid mistakes though and could've done with 10 more minutes...

There was a triangle with the area and you had to work out the perimeter, a triange and a circle thing with a quadratic equation and other stuff
3. yeah, i didnt know how to calculate the perimeter but the quadratic was farily straightforward, just bung in the formula. i didnt entirely understand the question about the prism and something about a square ABCD ets what did you do for that>
4. I supposed you lot did the higher exam, i did the intermediate paper and it was soooooooooooooo easy, athough i will admit that some bits were a bit tricky!!!!! i will def get a B!!!

WOOOOHHHHOOOOOO!!!!
5. The answer to the simultaneous equations were

x²+y²=25
y = 2x-2

(3, 4) and (-1.4, -4.8)
6. (Original post by manps)
The answer to the simultaneous equations were

x²+y²=25
y = 2x-2

(3, 4) and (-1.4, -4.8)

how do u get that? I completely made it up and got 13.5,25 and 0,5 lol at least i tried!
7. yer it was gr8 i got those answers 4 da quadratic question
8. for the simultaneous equations, because one is linear and one is quadratic you had to substitute one of the letters, (in this case i think it was Y). I only know that from Additional Maths though, lucky. I had to ask for extra paper, i must have big writing or something..
9. (Original post by vmong)
yeah, i didnt know how to calculate the perimeter but the quadratic was farily straightforward, just bung in the formula. i didnt entirely understand the question about the prism and something about a square ABCD ets what did you do for that>
They told you the area was ten.

You had to use 1/2abSinC = 10 to work out the middle angle.

then the cosine rule could be used to calculate the last side.

I know people who used pythagoras.... ....There was no right angle!
10. i think you could divide the area by the base an then double it to give the perpendicular height which meant u could use pythagoras.
11. These people were just using the values regardless of the fact that no right angle existed.
12. Could someone work out the x^2 + y^2 = 25 question on this thread so I can see how it's done.

I completely screwed it up I kept ending up trying to find the square root of -11 for some reason
13. i think the beginning of the paper was easy, although i felt the last questons were harder. i just couldn't work out the perimiter of the triangle one, i knew what you had to do but it just didn't work. i also couldn't do the last question or the question about the upper and lower bounds and the prism one. although i did do a lot ofworking out for them so hopefully i will get some marks.
14. what questions did other people find difficult. i think i must have left out about 20 marks worth. oh yeah for the stratified one, i made it so it was 50.
15. (Original post by starcrossed)
i think the beginning of the paper was easy, although i felt the last questons were harder. i just couldn't work out the perimiter of the triangle one, i knew what you had to do but it just didn't work. i also couldn't do the last question or the question about the upper and lower bounds and the prism one. although i did do a lot ofworking out for them so hopefully i will get some marks.
The prism one was easy.
Work out the area of the circle, then multiply it by the length of the prism. That's if I remember the question correctyl
16. oh no sorry didn't mean the prism one, i meant the triangle one with the line DE
17. (Original post by starcrossed)
oh no sorry didn't mean the prism one, i meant the triangle one with the line DE
Was that the one that was sort of two triangles in one? And there were two parts to the question?
18. (Original post by koldtoast)
Could someone work out the x^2 + y^2 = 25 question on this thread so I can see how it's done.

I completely screwed it up I kept ending up trying to find the square root of -11 for some reason
x²+y²=25
y = 2x-2 <--- subsitute this into the first equation for y

x²+(2x-2)² =25
x²+(4x²-8x+4) =25
5x²-8x+4=25
5x²-8x-21=0

a=5, b=-8, c=-21

x= --(8±(√(-8²-4(5)(-21)))/2(5)
x = (8 ± √(64--420))/10
x = (8+ √484)/10 OR x= (8-√484)/10
x = 30/10 OR x= -14/10
x = 3 OR x= -1.4

sub into initial equation to get y solutions.

y = 2x-2
y = 2(3) - 2
y = 8 when x=3

OR

y = 2x-2
y = 2(-1.4)-2
y= -4.8 when x=-1.4

x=3 y= 4
x= -1.4 y=-4.8
19. x²+y²=25
y = 2x-2

x²+ (2x-2)²=25
x²+ (2x-2)(2x-2)=25
x² + 4x² - 8x + 4 = 25
5x² - 8x – 21 = 0
Factorise and then substitute to get y.
20. (Original post by ThornsnRoses)
x²+y²=25
y = 2x-2

x²+ (2x-2)²=25
x²+ (2x-2)(2x-2)=25
x² + 4x² - 8x + 4 = 25
5x² - 4x – 21 = 0
Factorise and then substitute to get y.
Ironically, i got it wrong....oh the cleverness!

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