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HARDEST MATH PAPER IN THE HISTORY OF IGCSE 0580 MATH 42 DISCUSSION

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I will agree with ramie on this it was tricky and challenging but not impossible.
I would say A* should be around 175/200.
 
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i am losing about 16 marks in both the exams , do you guys think i will get an A* or no
please respond

in both as in 16 in p2 and 16 in p4, or is it 16 in total?

losing 16 in total then you are most likely getting an A*.
If 16 each then you wont.

We cant tell for sure since we dont know the grade boundaries i could be wrong.
 
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guys u niticed that many made a mistake in finding the major arc length.... u had to first multiply the angle 39*2 and then minus it from 360 some people ddnt do that getting the wrong answer...
its 39 * 4 not 2 then minus from 360 because they want the arc AC thus u do mob * 4 u get AOC and minus that from 360..
 
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Guys when they asked us to do rotation 90 degrees anti clockwise in the traslation, did the image we form get formed half on top of the other one or right next to it?
 
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Guys when they asked us to do rotation 90 degrees anti clockwise in the traslation, did the image we form get formed half on top of the other one or right next to it?
It was not on top of it, but it was next to it. It was to the right of it, if i remember correctly.
 
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Paper was challenging So i had to use my brain
first i had to go find where my brain was
secondly i had to figure out if it was my brain
finally i had to use it
but then found out it was empty :sick:
this took me a lot of time so time was wasted and limited
and then I did the whole paper without my brain
Yay i did all the question other than the last "n" thingy which i gave a guess and worte 12 lol Guessing without a brain makes no sense And my replay doesn't either i couldn't find my brain but did you see it?? oh gtg saw it passing by having to catch up with it before Bio
 
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If you guys say that A* will be 170/200 then how much will A be in your opinion?
 
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why 4 if one triangle is half
it had to be multiplied by 2.

You multiply by 4. The arc AC has 2 triangles there: AOB and BOC. The angle MOB = 39, which makes up for a half of ONE OF THE TRIANGLES. Which means there are 3 other halves left to account for. Therefore the angle AOC = 39*4.

(360-39*4 = 204 ---- (204/360)*2(8.5)*pi = 30.264 = 30.3 (3 s.f.))

[That's what I got anyway]
 
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