IGCSE MATHS Doubts

[astarmathsandphysics]

can some1 now do my Q....which i ve posted half an hour bfore.......if an1 knws how to do it...lol

Do d minus cii

[Dark Prince]

can some1 now do my Q....which i ve posted half an hour bfore.......if an1 knws how to do it...lol

Do d minus cii

thnkx...sir...can u xplain

[Adzel]

m/j 2007 p4 q7 + (explaination please) 

[Vin]

I'd solved this one earlier .. long time ago .. lemme check ..

Got it !! quite easy ..
 a) is quite direct ..
for such kinda ques wat i always do is look for the units .. for eq in this one..
they hav asked Calculate the number of cubic metres .. 3 measurements hav to be multiplied (this is jus a tip .. this is how i under the ques)

so for this vol = 0.3 m  x  1.2  x 3 m/min x 60 (for 1 hour)

= 64.8 cm3
-----------------------------------
b) same method as the previous ..
first
vol =  0.8m  x  1.2m  x  15 m/min  x  60
= 864 cm3

for % , difference is 864 - 64.8 = 799.2
per cent = diff/original  x 100
= 799.2
 ---------  x  100
   64.8
= 1230 %

-----------------------------------
c) firstly always remember to convert ur units to units asked in ques..
1.3 mm = 0.0013m

vol  =  pi rrh
2   =  pi r2 x 0.0013
on solvin = 22.1 m

-----------------------------------
d)  take the length as L ..

surface area =  50.4/0.12
= 420 m2

SA = (0.8 x L) +  (1.2 x L) + (0.8 x L)
420 = 0.8L + 1.2L + 0.8L
420 = 2.8 L
L = 150m .. done !! 

[mdwael]

I have a question I dont really think its maths but Im going to say it anyway. This is about my friend who did IT exam few days ago, well he seems to have forgotten to check the number of pages in his paper, only then after the exam had finished that he realized  2 pages worth 20 marks were missing from his paper which is out of (120) he got approximately 70 so will his mark be 70/100? or they wont forgive him and his mark will be 70/120?
Sorry for asking this here but Im new and I didnt find a place to post it..

[haris94]

lol.....there's nothing to worry about

the paper was out of 100 this year 
none of the pages were missing from your friend's paper

[Vin]

I have a question I dont really think its maths but Im going to say it anyway. This is about my friend who did IT exam few days ago, well he seems to have forgotten to check the number of pages in his paper, only then after the exam had finished that he realized  2 pages worth 20 marks were missing from his paper which is out of (120) he got approximately 70 so will his mark be 70/100? or they wont forgive him and his mark will be 70/120?
Sorry for asking this here but Im new and I didnt find a place to post it..

post here from now on ..

https://studentforums.biz/index.php/board,78.0.html

[the_grim_reaper]

Just one thing, in M/J 2008 paper 4 question part d (ii),  I first found i to be x+2n+2. I read the question again and it said find in terms of x and n. So I spent another half an hour trying to remove 2 from the equation. In the end, I gave up and checked in the mark scheme to find out that my initial answer was correct. So what exactly is meant by finding something in terms of something. I mean does it mean ONLY x and n here? If it does, I'm going to sue CIE.....

[Vin]

ur talking about Q2 right ?

see NOT necessarily it has to be ONLY  x and n .. if u hav solved 2 c)iii) the answer deduced is y=20x + 5 and not jus y=x u getting my point .. ?? be confidant while solving such sums and solve them twice so that u r sure of it .. By the way d ii) is jus factorisation right ? so r u talking abt c)iii) ??

[$H00t!N& $t@r]

hey hw did you get 20x+5  for some reason im getting 4x+5... can you explain??

[Vin]

y=(m-2n) (m+2n)
substitute

=[(2x + 3) -2(x - 1)] [(2x + 3) +2(x + 1)]
expand

= (2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
multiply each number and variable from the first bracket to the second ..

= (2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
(2x + 3 - 2x + 2) ( 2x + 3 + 2x - 2)
u get

= 4x² + 6x + 4x² - 4x + 6x + 9 + 6x - 6 - 4x² - 6x - 4x²  + 4x + 4x + 6 + 4x - 4
cancel out similar +ve and -ve numbers
 
4x² + 6x + 4x² - 4x + 6x + 9 + 6x - 6 - 4x² - 6x - 4x²  + 4x + 4x + 6 + 4x - 4
rearranging

= 6x + 6x + 4x + 4x + 9 - 4

= 20x + 5

[$H00t!N& $t@r]

Thanks for the answer. Actually what i did was that i substituted the m and n values into y = m2 ? 4n2. This is also correct but my mistake was the i worked it out wrongly. I tried again and it worked.

[Vin]

you can do it that way too .. ultimately u hav to factorise in the form of (a + b) ( a - b) .. jus one more step adds up .. finally u get the same ans .. safer method to do is the one i mentioned ..

[$H00t!N& $t@r]

you can do it that way too .. ultimately u hav to factorise in the form of (a + b) ( a - b) .. jus one more step adds up .. finally u get the same ans .. safer method to do is the one i mentioned ..

k ill keep that in mind 

[key04]

Hi Everyone!!!!!!!!!!

I have a question 4 maths paper 4.............
how to do ques 4 of may june 2006?
anyone have any idea to do so???
plz i need help urgently...........

Thanks