IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: Q80BOY on May 05, 2009, 06:19:33 pm

Title: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 06:19:33 pm
IGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001

(i've attached the question paper)

Page 3, Question 5 (Done)

Page 4, Question 9 (Done)

Page 5, Question 12 (Done)

Page 8, Question 19, Part (d) (Done)

Page 9, Question 20, Part (a) and (b) (Done)

Page 10, Question 22, Part (a) and (b) (Done)

Page 11, Question 24, Part (c) (Done)


Special thanks to Sanity_Master, Amber, Twilight and Sweetsh for their gr8 help !!
 
keep up the good work !!  ;)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sanity_master on May 05, 2009, 06:36:51 pm
Page 3, Q 5::

M.F (multiplying factor) = 99>>>>100%-1%

(99*4)รท100= 3.96 hours
3 hours......0.96 hours *60= 57.6 minutes
3 hours, 57 minutes.....0.6 min*60= 36 seconds
3 hours, 57 minutes, 36 seconds

_____________________________________________________________________________________________________________________
Page 4, Q 9::

X= 70*2 ==> 140 (degrees)

OSC = 90 (degrees)
so, Y= 90-40 ==> 50 (degrees)

OAC...X+2a=180...........the a is a virtual letter for the angle at OCA which is equal for OAC
a=20 (degrees)

SO, BCA = 40+20 ==> 60
AND, BAC = 180-60-70
BAC= Z+20
SO, Z+20=50
Z=30 (degrees)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 06:58:37 pm
Thanks very much Sanity_master !!! + rep !!  ;D
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Christy on May 05, 2009, 07:03:12 pm
Q9) x=140 degress (70*2)
      y= 50 ( 90 - 40)
      z= 30 ( 180 - 140 /2 = 20 and 40+20 = 60
                                                            60+70+20+x = 180
                                                                           x = 30 )  
Q12) a) 60 and 300
b) i think its is 270 <x<359

Q19) d) because BC is double of AQ

Q20) a) 70
b) 15.3cm ( 25/sin70 = x/sin35)

Q22) a) d= 42 ( 72-30)
           e= 74 ( 180-30 = 150+136= 360-286 = 74)
           f= 64 (180- [74+42])
b) 252 ( 180-136 = 44+64 = 180-108 = 72+180 = 252)

Q24) no .. the area under the graph is the distance so u use the formula :

1/2 * b * h
1/2 * 4 * 20 = 40 :D

couldnt do the first one sorry .. hope i helped    
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sanity_master on May 05, 2009, 07:06:26 pm
Page 5, Q 12::

(a) Cos(-) 1/2 = X

X=60
OR
X=300

(b) on the paper draw lines to show the part after 90 and below the half on Y and above 0.....

u will have the range......270<X<300
^^^^
i got it by calculation (trying, beside the drawing)

____________________________________________________________________________________________________________________
 sry, i g2g.......i hope i helped u :)

Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: twilight on May 05, 2009, 07:08:11 pm
Page 5, Question 12

(a)       cos x = 1/2
        cosine is positive in first and fourth quadrant
            x   =  cos-1 0.5    =   60o
     and  x   =           360 - 60            = 300o             =>     therefore   x = 60 or 300

(b)       cos x   is greater than 0 in these    0 < x < 90    and    270 < x < 360
             and it is less than 1/2  in this      60 < x < 300
  so combining both the required vales of x    are    270 < x < 300
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sweetsh on May 05, 2009, 07:08:58 pm
Quote from: Q80BOY on May 05, 2009, 06:19:33 pm
IGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001

(i've attached the question paper)

Page 3, Question 5 (Done)

Page 4, Question 9 (Done)

Page 5, Question 12

Page 8, Question 19, Part (d)

Page 9, Question 20, Part (a) and (b)

Page 10, Question 22, Part (a) and (b)

Page 11, Question 24, Part (c) < -- when i use the formula distance = speed * time i get 20 (ans in m/s is 40)

Total Number of Questions: 7

I really hope u can help me out here  :D


Did they answer everything? If no tell me so I can answer the rest
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sanity_master on May 05, 2009, 07:09:31 pm
Amber......u got 12 (b) wrong......the answer of cosX should be below 1/2

and 359 is 0.99984...
it is 270<x<300

and for 19 (d)......i guess they wont accept that answer as u didnt use part (b) and (c).......i would've solved it if i had time


By the way.....u better start canceling Qs or ppl will just keep answering the same questions :D
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: twilight on May 05, 2009, 07:12:39 pm
sanity master .. look at my answer up there  .. i xplained y
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Christy on May 05, 2009, 07:14:39 pm
Quote from: sanity_master on May 05, 2009, 07:09:31 pm
Amber......u got 12 (b) wrong......the answer of cosX should be below 1/2

and 359 is 0.99984...
it is 270<x<300

sorry my bad ... but the rest is correct right ??
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sanity_master on May 05, 2009, 07:20:37 pm
well....i should've been gone now

so if u r not sure about ur answers...let twilight check it for u :)

By the way twilight.......Thanks......i answered the Q with the correct answer without knowing the exact way of explaining (gd way  ;) )



CYA L8r GUYs!!..........BYE
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 07:22:50 pm
Quote from: Amber on May 05, 2009, 07:03:12 pm
Q9) x=140 degress (70*2)
      y= 50 ( 90 - 40)
      z= 30 ( 180 - 140 /2 = 20 and 40+20 = 60
                                                            60+70+20+x = 180
                                                                           x = 30 )  
Q12) a) 60 and 300
b) i think its is 270 <x<359

Q19) d) because BC is double of AQ

Q20) a) 70
b) 15.3cm ( 25/sin70 = x/sin35)

Q22) a) d= 42 ( 72-30)
           e= 74 ( 180-30 = 150+136= 360-286 = 74)
           f= 64 (180- [74+42])
b) 252 ( 180-136 = 44+64 = 180-108 = 72+180 = 252)

Q24) no .. the area under the graph is the distance so u use the formula :

1/2 * b * h
1/2 * 4 * 20 = 40 :D

couldnt do the first one sorry .. hope i helped    

Thanks for the help amber, appreciate it !!  ;D + rep
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 07:23:42 pm
Quote from: twilight on May 05, 2009, 07:08:11 pm
Page 5, Question 12

(a)       cos x = 1/2
        cosine is positive in first and fourth quadrant
            x   =  cos-1 0.5    =   60o
     and  x   =           360 - 60            = 300o             =>     therefore   x = 60 or 300

(b)       cos x   is greater than 0 in these    0 < x < 90    and    270 < x < 360
             and it is less than 1/2  in this      60 < x < 300
  so combining both the required vales of x    are    270 < x < 300

Thanks, that's what i was looking for, +rep !!!  ;)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: twilight on May 05, 2009, 07:25:01 pm
Quote from: sanity_master on May 05, 2009, 07:20:37 pm
well....i should've been gone now

so if u r not sure about ur answers...let twilight check it for u :)

By the way twilight.......Thanks......i answered the Q with the correct answer without knowing the exact way of explaining (gd way  ;) )



CYA L8r GUYs!!..........BYE

Thanks sanity master  ;)  and Q80BOY   ;)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: sweetsh on May 05, 2009, 07:25:23 pm
Quote from: Q80BOY on May 05, 2009, 06:19:33 pm
IGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001

(i've attached the question paper)

Page 3, Question 5 (Done)

Page 4, Question 9 (Done)

Page 5, Question 12 (Done)

Page 8, Question 19, Part (d)

Page 9, Question 20, Part (a) and (b)

Page 10, Question 22, Part (a) and (b)

Page 11, Question 24, Part (c) < -- when i use the formula distance = speed * time i get 20 (ans in m/s is 40)

Total Number of Questions: 7

I really hope u can help me out here  :D

Page 8 Q19
d)triangle AOQ and traingle BOC are similar so OB/OA=OC/OQ
proprotion so AQis parallel to BC

Page 9 Q2a
a) angle KTP=70 as the small triangle has (20 +90+70) and the 7o angle is opposite to KTP so its 70
b)you use the sine rule
KT/sin35 =25/sin70
(sin70)(KT)=25sin35
KT=15.23m

Page11 q24
c)you dont have to use this formula! o my god dont you take physics?! You calculate the distance under the graph to calculate the distance..
distance=0.5*4*20=40m
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: twilight on May 05, 2009, 07:30:53 pm
if u need further explaination of any particular question tell me  ;)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Christy on May 05, 2009, 07:33:10 pm
you welcome :)
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 07:34:15 pm
Quote from: sweetsh on May 05, 2009, 07:25:23 pm
Quote from: Q80BOY on May 05, 2009, 06:19:33 pm
IGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001

(i've attached the question paper)

Page 3, Question 5 (Done)

Page 4, Question 9 (Done)

Page 5, Question 12 (Done)


Page 8, Question 19, Part (d)

Page 9, Question 20, Part (a) and (b)

Page 10, Question 22, Part (a) and (b)

Page 11, Question 24, Part (c) < -- when i use the formula distance = speed * time i get 20 (ans in m/s is 40)

Total Number of Questions: 7

I really hope u can help me out here  :D

Page 8 Q19
d)triangle AOQ and traingle BOC are similar so OB/OA=OC/OQ
proprotion so AQis parallel to BC

Page 9 Q2a
a) angle KTP=70 as the small triangle has (20 +90+70) and the 7o angle is opposite to KTP so its 70
b)you use the sine rule
KT/sin35 =25/sin70
(sin70)(KT)=25sin35
KT=15.23m

Page11 q24
c)you dont have to use this formula! o my god dont you take physics?! You calculate the distance under the graph to calculate the distance..
distance=0.5*4*20=40m

Thanks a lot !!!! ;D
Title: Re: Math P2 Oct/Nov 2001 HELP !!!
Post by: Q80BOY on May 05, 2009, 07:35:02 pm
Quote from: twilight on May 05, 2009, 07:30:53 pm
if u need further explaination of any particular question tell me  ;)

Thanks a lot for ur time and help, ur the best !!  ;D