Author Topic: All doubts here!  (Read 56115 times)

Offline adorabledaughter_07

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Re: ANY HELP!
« Reply #105 on: October 29, 2009, 04:36:49 pm »
yea yea sure..i havnt uploaded bt hav de link ...here it iz/... :)
http://www.scribd.com/doc/3182857/19982003-Paper-4-Maths
"Never say a problem is difficult, because if it were not difficult,it wudnt be a problem!"

Offline Ghost Of Highbury

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Re: ANY HELP!
« Reply #106 on: October 29, 2009, 04:42:52 pm »
i hope you have already found the volume of the cylinder, the hemisphere and the cone in the previous question
        taking the radius as 5cm, height of the cylinder as 5cm, and  the perpendicular height of the cone also 5 cm..
you get the volume of the hemisphere as - 523.5987
           the volume of the cylinder as      - 392.699
           the volume of the cone    as       - 130.899

as we are unaware of the mases..v take their volumes previously found out to get the ratio
first..relate the volume of the hemisphere, the cylinder and the cone to the volume of the cone as it is the least ...

the ratio of the volume of the hemisphere to the volume of the cone
i.e 523.5987/130.899 = 4
the ratio of the volume of the cylinder to the volume of the cone
i.e 392.699/130.899 = 3
the ratio of the volume of the cone to the volume of the cone is obviously 1

therefore the answer is 4:3:1


check the markscheme too...just check if the answer is corrcect
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Offline astarmathsandphysics

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Re: ANY HELP!
« Reply #107 on: October 29, 2009, 04:43:51 pm »
Phew that was quick Iam!

Offline Ghost Of Highbury

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Re: ANY HELP!
« Reply #108 on: October 29, 2009, 04:55:12 pm »
Phew that was quick Iam!
lol..actually i had already answered this question somewhere.

so i just copy pasted it here.
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Offline adorabledaughter_07

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Re: ANY HELP!
« Reply #109 on: October 29, 2009, 05:00:44 pm »
ohk...bt thank 4 de ans anywayz!...hpe its rite..don hav de ms wid me!>,..
bt in such cases its ok to compare the volumes instead??..
nd ya..i had 1mor q..
may/jun02..q4 a part ...how u find angle "e"?? ???
"Never say a problem is difficult, because if it were not difficult,it wudnt be a problem!"

Offline astarmathsandphysics

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Re: ANY HELP!
« Reply #110 on: October 29, 2009, 05:02:30 pm »
ohk...bt thank 4 de ans anywayz!...hpe its rite..don hav de ms wid me!>,..
bt in such cases its ok to compare the volumes instead??..
nd ya..i had 1mor q..
may/jun02..q4 a part ...how u find angle "e"?? Huh?


which papers?

Offline adorabledaughter_07

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Re: ANY HELP!
« Reply #111 on: October 29, 2009, 05:04:36 pm »
paper 4 sir.. :)
"Never say a problem is difficult, because if it were not difficult,it wudnt be a problem!"

Offline Ghost Of Highbury

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Re: ANY HELP!
« Reply #112 on: October 29, 2009, 05:04:53 pm »
ohk...bt thank 4 de ans anywayz!...hpe its rite..don hav de ms wid me!>,..
bt in such cases its ok to compare the volumes instead??..
nd ya..i had 1mor q..
may/jun02..q4 a part ...how u find angle "e"?? ???

yea .. its rite..i saw the ms..

http://www.scribd.com/doc/3182918/Math-Answers-P2-P4-1993-2003

and yea..for the next question

triangle ADX is an isosceles triangle because tangents are equal

angle DAX = angle XDA = 450

angle CAX = 900

therefore, angle CAD = 90 - 45 = 450
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Offline astarmathsandphysics

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Re: ANY HELP!
« Reply #113 on: October 29, 2009, 05:08:18 pm »
a and b both 90 cos they are between dadii and tangents
AXD=45 since AX=XD hence R=180-90-45=45

Offline astarmathsandphysics

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Re: ANY HELP!
« Reply #114 on: October 29, 2009, 05:09:01 pm »
You are real quick to fire Iam. I am no competition

Offline ~~~~shreyapril~~~~

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Re: ANY HELP!
« Reply #115 on: October 29, 2009, 05:09:45 pm »
i guess he has copy pasted them again!!  ;) ;)
Friendship is like peeing on yourself. Everyone can see it but only you get the warm feeling that it gives :D :P :)

Offline Ghost Of Highbury

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Re: ANY HELP!
« Reply #116 on: October 29, 2009, 05:10:41 pm »
SIR, ur just occupied wid ur work, i am no comparison to u sir!..ur a genius

@shrey - not this time bro
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Offline adorabledaughter_07

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Re: ANY HELP!
« Reply #117 on: October 29, 2009, 05:16:46 pm »
now dats a real complement to u !am!...(thanks 4 de ans By the way..)
"Never say a problem is difficult, because if it were not difficult,it wudnt be a problem!"

Offline Ghost Of Highbury

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Re: ANY HELP!
« Reply #118 on: October 29, 2009, 05:17:44 pm »
now dats a real complement to u !am!...(thanks 4 de ans By the way..)

ur welcome..
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Offline ~~~~shreyapril~~~~

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Re: ANY HELP!
« Reply #119 on: October 29, 2009, 05:22:01 pm »
ohh!!

great!!

i guess aadi is right!!

sir occuoied by work!!

but how can you ans that fast??

have u solved it somewhere else??
then u just worte it like hard copy se soft copy??
odr u did it just now??!??
Friendship is like peeing on yourself. Everyone can see it but only you get the warm feeling that it gives :D :P :)