Math P2 Oct/Nov 2001 HELP !!!

[Q80BOY]

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 !! 

[sanity_master]

Page 3, Q 5::

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

(99*4)÷100= 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

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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)

[Q80BOY]

Thanks very much Sanity_master !!! + rep !! 

[Christy]

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

couldnt do the first one sorry .. hope i helped    

[sanity_master]

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

[twilight]

Page 5, Question 12

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

[sweetsh]

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 

Did they answer everything? If no tell me so I can answer the rest

[sanity_master]

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

[twilight]

sanity master .. look at my answer up there  .. i xplained y

[Christy]

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 ??

[sanity_master]

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

[Q80BOY]

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

couldnt do the first one sorry .. hope i helped    

Thanks for the help amber, appreciate it !!  + rep

[Q80BOY]

Page 5, Question 12

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

[twilight]

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   

[sweetsh]

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 

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