/ .,.,,r.. Cambridge~gs'Rftffif papers.

com
::: International Education

Cambridge IGCSE™

CANDIDATE
NAME

CENTRE CANDIDATE
NUMBER
I I I I I I NUMBER
I I I I I
MATHEMATICS 0580/42
Paper 4 (Extended) October/November 2022

2 hours 30 minutes

You must answer on the question paper.

You will need: Geometrical instruments

INSTRUCTIONS
• Answer all questions.
• Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
• Write your name, centre number and candidate number in the boxes at the top of the page.
• Write your answer to each question in the space provided.
• Do not use an erasable pen or correction fluid.
• Do not write on any bar codes.
• You should use a calculator where appropriate.
• You may use tracing paper.
• You must show all necessary working clearly.
• Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.

;.,.:;.b~t
~~11-
• For n:, use either your calculator value or 3.142.

_: ·' I
INFORMATION
• The total mark for this paper is 130.
• The number of marks for each question or part question is shown in brackets [ ].
!
1f

This document has 20 pages. Any blank pages are indicated.

DC (RW/CGW) 302727/3
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2

(a) (i) At a football club, season tickets are sold for seated areas and for standing areas.
The cost of season tickets are in the ratio seated: standing= 5 : 3.
The cost of a season ticket for the standing area is $45.

Find the cost of a season ticket for the seated area.

3 YS
5 ~,t
J">t :: L/ ?X5
3}{ = :?:? 5
'vl ;, J~ S : :JS $ ....:r-5. ....................................... [2]
3
(ii) In 2021, the value of the team's players was $2.65 million.
In 2022 this value has decreased by 12%.
~0;}~-~Oii1 ~1
Find the value in 2022.
1
~- G5x (1- 101)
1£1 )
: ~.33~
$ .... .• 3J~...................... million [2]

(iii) The number of people at a football match is 1455.

This is 6.25% of the total number of people allowed in the stadium.

Find the total number of people allowed in the stadium.

(:.~S X ">I -= 145'3
1 00
3~ 80
rt -:.. 11.,155"' x 1DQ
,.~'3 =-

.......... J3..2.2U........................ [2]

(iv) The average attendance increased exponentially by 4% each year for the three years from
2016 to 2019.
In 2019 the average attendance was 1631. .:I OI 't - 0 I : 3

Find the average attendance for 2016.

3
IG 3 \ -.. (1-i~
JOO
l
l-< = \b.31 -; 144q. q5 ct 145 0
!:L ).3
( 4 + /()/)

············1.Y..5:0......................... [3]

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3

(b) Another club sells season tickets for individuals and for families.
In 2018, the number of season tickets sold is in the ratio family: individual= 2: 7.

(i) The number of family season tickets sold is x.

Write an expression, in terms of x, for the number of individual season tickets sold .
.:i -"'> ,{
=1- -
2 v :. =l,;,1
°'"clivid'-'~ (v)
-.:..) ...... tt . . . .... . . . . . . . . . . . . . [l]

(ii) In 2019, the number of family season tickets sold increases by 12 and the number ofindividual
season tickets sold decreases by 26.

Complete the table by writing expressions, in terms of x, for the number of tickets sold each
year.

Year Family tickets Individual tickets

2018 X -
~)1.

2019 ~-1-\~ -+i.t -:l6

;l
[2]

(iii) In 2019, the number of individual season tickets sold is 3 times the number of family season
tickets sold.
Write an equation in x and solve it to find the number of family tickets sold in 2018.

12:t - ~' : 3 (n -t I~)

f"M - llii = 3 ,1 ·l 3 '

::+-..t - 5.Q - 3">1-t 3 6

- -,,.><
1rt -'5~ "' G r1 + '+~
x = .. tH-:l....................................... [4]
1 Y(-bYl -=- =,~ -t;R
)1. -:: \~~

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2 All the lengths in this question are measured in centimetres.

NOTTO
9-x
SCALE

The diagram shows a solid cuboid with a square base.

(a) The volume, Vcm3,ofthecuboidis V=x\9-x).

The table shows some values of V for O x 9.

X 0 I 2 3 4 5 6 7 8 9
V 0 8 ~i 54 80 100 108 98 64 0

(i) Complete the table.

[I]
(ii) On the grid on the opposite page, draw the graph of V = x 2 (9- x) for O x 9. [4]

(iii) Find the values of x when the volume of the cuboid is 44cm 3 •

x = ...~.•.b ........ or x = i.i4............ [2]

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V
110

100

90

80

70

60

50

40

30

20

10

(b) (i) Show that the total surface area of the cuboid is (36x-2x 2)cm 2 .

~LIN --+ ~LH-+ ~~r\

~'>1)\',( i)t (0\- )() + (q.)t)
~ -¾- \1,-i~t t I ~ -J11,'
3(; '>1. -~'X'-

[2]
(ii) Find the surface area when the volume of the cuboid is a maximum.

o3
A !' vi - '<- i
= 2b( G ) - '2 ( C ) ~: 144

1uu
•· ····· · · .":'f.~ .. ... .. .. . . .. .. .... .. .. . cm 2 [ 3]

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3 Kai and Ann carry out a survey on the distances travelled, in kilometres, by 200 cars.

Kai completes this ~·gquency table ffR ~1e data collecteyi=, S 50 J SO

Distance (dkm) 80 < d~ 100

----<
150 d 200 200 <d 300
..,-- ......_
300 < d 400

Frequency 7 33 76 52 32

(a) (i) Calculate an estimate of the mean.

~00

......... ii.11:~.f5. ................... km [4]

(ii) Ann uses this frequency table for the same data.
There is a different interval for the final group.

Distance (dkm) 80 <d 100 100 <d 150 150 <d 200 200 <d 300 300 <d 360
Frequency 7 33 76 52 32

Without calculating an estimate of the mean for this data, find the difference between Ann's
and Kai's estimate of the mean.
You must show all your working.

3$01\(3~ ) - 3J0 )((3~~

l;L~O

......... 3 ~..tf\..
(') .. ... ...... .... ... .. .... km. [2]

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(iii) A histogram is drawn showing the information in Kai's frequency table.

The height of the block for the interval 200 < d 300 is 2.6cm. ~= _-r,:______ _
No, of c\ass ·,t\\e,v~
Calculate the height of the block for each of the following intervals.
a•G -: 5~

5o , No, o~ c\o.~ i"\-e.ruoJ '" ~oatc\ loo ,~ Ro 'i<. ~ s~ :!lo

J ,b
st-~f\c\CHd \ri~ervoJ =- -toe.> . _ S 5
«,()
For BoL o\ b-100 ; IDo-io -c. ¥ : lf
So
F-
q:- y -: 1·1-L/
?
80 <d I 00 .... .1.:.:1..Y.............................. cm

ro, IS o L cl ~~uo Q00 - I '7,Q. .., I 0 150 <d 200 ....}'.~ ................................ cm
1-t.. 5
Si> ..f::_ -
/1) - To -- '"l,T't)'"
300 <d 400 ......1!.€:i ................................ cm [3]
a::-°" f. 4 00 tmo - 300
fo -:
jooL d .... j_ o
(b) One car is picked at ?andom. So .. l• C':I
~o
Find the probability that the car has travelled more than 300 km .

p ,,, ~ -::L. -:. !L .. .... ...'J./.J 5............................. [ l]

;too So ,s
(c) Two of the 200 cars are picked at random.

Find the probability that

(i) both cars have travelled 150 km or less,

33 -+ =t = 40

.::J2 A J1__ 8'1

~oo 1q q q"qs
.... .-.?~/9.9.5......................... (2]

(ii) one car has travelled more than 200 km and the other car has travelled 100 km or less.

( J~Jt-J~
~oo
A±-)
l'JCf
~
- -
_ IY=l-
9'l15

1
.... ..j_!/1-:,_q1.5.,.................... [3)

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r - • - - - - ,- - - - - - - r - - • - - - -,- - - - - - - , - - - -- - - ;- - - - -
I I I I I I
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I
I
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t
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t
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, 1 : : : : :
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::::J:::::::::,i,::::::. . . ---'------'1!111'=--,f-- _____ : ------1--- 5 ------1·----~- ------i

1

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------~------~
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"'i-.---ioo-¥"~!, ------1.--- 3 ------,-------
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1 - - - - l - - ~-------r------,''
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I I I I I I I I I I I I I

1:·------.l.i : ___
!! l ! ! !
L------ .______ _t ______ J ______ J_ ____ _ : I : : I :
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I I 1 I I I I I L f I
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t.
I I I I I 1 : : I I : :

~8 ~7 ~6 ~5 -4 ~3 -1 0 I 2 3 5 6 X

---· --r-----r------r-----! ------r------1

I I I I I I I j j j j I I

-1
I I I I I I I I I I I I I
L - - - - - _J __ - - - - - I. - - - - - - -'-- - - - - -.l - - - - - - - '- - - - - - .J __
I I I I I L I

: l ! l l : :,
------1 ;------r-----1
I I I I I I

~-----+-----+-----+-----+-----+-----+ -- --:
I I I I I I I
1
I I I I 1 I I I
-2 _______ :
I I I I I I I I
I I I I

l ------i-------~------+--- -t - 3
I L I I I I I I
I I I I I I I I 1 I I I I

~------~-------t------ ------~-------
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I I I

:
f

: :
I I I I I I I I I I I

:
t
:I
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:
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: \
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4 -------1------- ------
i------r-----r------r------1-------t------r ----. ·-
t I I I
I I
ol - - -
I
----L----- -_.
I I
I I I I I
I I I I I
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1

r------r------r------T------1-------r------r------1
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------~-------~------+------4-------~------~
I I I I I I
I I I 1 I I
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: : : l : : : l
L______ J_______ t ______ _:_ ______ J_______ L_-____ J__ - - ___ l - 6 I
I
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------.l-------L-------'-------.l-------L------.
I
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I

(a) Describe fully the single transformation that maps

(i) shape A onto shape B,

········· t~t)··························································································································

············································································································································· [2]
(ii) shape A onto shape C,

......... .c.ea+r.t .. .C.a.,.1D. ........;, .... !io.~..... c.loc.J~JtO~St....................................................... .

............................................................................................................................................. [3]

(iii) shape A onto shape D.

...... Ef.\\cu.~~····~···c.entce ....L.~..1.J.~.~.)....k.j ....ScoJe .....6 .c.t.o.~.!.. ½. . .. . .

························ ·· ············································································ ······················ ················ · [3]
(b) On the grid, draw the image of shape A after a reflection in the line y = x + 8. [2]
j=1C.--t-i
')l:: 'o ( o, i)
")\.:. - 8 ~-010)

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9

(a) The diagram shows the speed-time I fi f · ·

grap 1 or part o a Journey for two vehicles, a car and a bus.

24 ................................................... -- - - - - - - - - ,
- I Car

Speed ----~---------JBm
(m/s)
NOTTO
10 SCALE

Time (seconds)

(i) Calculate the acceleration of the car during the first 18 seconds.
9<ctcl\f(\-\; :: !:)l - :h -;. ~L/ -10 ';;.-
'J(z_-?(, IS--O IZ
........... J~vl.j ................. mfs 2 [I]

(ii) In the first 40 seconds the car travelled 134m more than the bus.
v _ d\~\o.(\ce ~d)
$ t>ee c\ , -
Calculate the constant speed, v, of the bus. ~ Me C'S)

D coJ - 'D~\)~ "a. \3'-\ cli5~c..() c.. c.. -=- S£) ac.\ X-\\Me.

( Do...,\?> ~i....,. q o \ -D~v!> \'3\.\

\ i ( 0\ -\ lo)~ -\- \ X w) - (L-10 v J -; \_3 '-\
a 9
(J ( ~L\-\I O)~ -t (Q txQ4))-4D\J -= 13li

tiov c~ x3Y)-+ cnxJL-!) - 13LI

L-/o v -c- 1 00
V: a. I , .,5 rn /5 v = .... L:t:.S............................... mis [4]
1.-10
(b) A train takes 10 minutes 30 seconds to travel 16240111 .
...... , 00 0

Calculate the average speed of the train. M \c rv..,

Give your answer in kilometres per hour. s -..___?) (\"i '---7' ""'
"".""b O 7 1:; cJ

\0 :. Ll-._
eu Lio

Ave. Spud

C\~ .'....
............ 2 .......... .............. kin/h [3]

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10

6 (a) Solve.
4x+ 15 =9
lf'}( q- - \5
0

l\")1. ";... - b
?L ::. - ~ J
>1
'I\,-_ - 1 X = .. ::.~........................................ [2]
o(
(b) Factorise.
a 2 -9

CCl-+ J) C.P\. - 3)
.... ( ~-tJ.).(0.~.3)................... [1]
(c) Write as a single fraction in its simplest form.

4a...:... 3ad
5 · 10c t1

- 8 C.
3d
.•
i."
I
.... }.~/J.d............................... [3]

Find an expression for m in terms of n.

M
5 " ( I + \ -+\ t I -t \ ') -:. 5
'3,., xi,' =- 51¥>
11
5"' "' 5.-,..
I'(\ ~ f"l-t\ m = .... ~T.t..................... ................. [2]
(e) Solve by factorisation.
4x 2 +8x-5 =0
lj)t1. -t (0">1 - ~ J-t - 5 :c D

~>1. ( ~.-tt 5) - I (.R11 --i5 ) -=- 0

(~x 1S) ( 9t,,l - \.)i; O

~')\ - \ =- 0
a~ -- - 5 a")\ :. '
'l,\_:.. -S ?\ -_ .1
x = ....:..2 ........ or x = \ [3]
.;i '-:i'""""""

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11

(f) (i) y is directly proportional to (x + 3)3.

When x = 2, y = 13.5.

Findx when y = 108.

0( (11 t 3) 3 (,t-t3)3
3 ~50
'j::~(71.13)
"'IL ..... ::\O

,3,5 ::. \~(~13) /1,)'g ( I)( +3)3 1t .. \o -3 -::i:='

\<.:; 'H·
51 JSo !3_,U. ~so : ("Jt t3)3
a"t
X = ...3:-....................... .................... [3]

(ii) g is inversely proportional to the square of d.

When dis halved, the value of g is multiplied by a factor n.

r3
Find n.
I
j oc Ji
ri~ -- l cA~)~ ng 4 (~
}),2~4,Z
n<j c. k
d,-¾ 'C'\--'--\

n = ... ~\........................................... [2]

(g) Expand and simplify.

(2x+ 3)(x- l)(x+ 3)

( ,& 1( i 3 ) ( ')t l 3 '}(_ - ')( - '3 )

i

( Q,t 1 3) (. )t 2. ""t 9. -it. - 3)

~')t ?, -11,,,111 '2 -J:X( 4- 3">t 2 - 9
;{'1(.3 , -t1' 1- ~

........ -2.11 ~.:t.::b·t.~.::.~ ............. [3]

(h) Find the derivative, : , of y = 3x 2 +4x- l.
cl j ){ I\ l'i 'X. <'1-1
c\'1-l

.......c.~. :t.~............................ [2]

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NOTTO
SCALE

p --------.:~ Q
46.5cm

(i) Calculate angle QPR.

CoS~ QPR ,i ... ~ 1~- p~
~'!'}
Co!)~~~ = ('ac:i,L\)'t-4(.l\(;,$)~ - (Jt,~)~
~<J'1 4) tl..\6,S)
1

~?R ;: Cc~-\ \ ~·4)'l.-t(46•5)~-lJ1•a)~)

<.3q •4)(L\<i•5)

Angle QPR= .... .5.J....................................... [4]

(ii) Find the shortest distance from Q to PR. Q

S'(r'\5~-;
5
t\6,S P

\-\ -:. $\/'\ 1' l.j G,5

3G,~

(b) The diagram shows a cuboid.

--------.G

20cm NOTTO
C SCALE

A 29cm

(i) Calculate the length AG.

AG-~ " AC~- C.. G,~
f'it' :: AB 1. ~ ~c..A.
AC.:. j
Ae,-l ~c.~ f'\G. !.35.~ 2
-t !;le~
=- 35, -g <:..iv\ 41 c""

AG= .... ~.1.1..........................

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13

(ii) Calculate the angle between AG and the base ABCD.

/1 ~0
p.. ~ C
3$",1

......... d.1:.~............................... [3J

(c)
North

NOTTO
SCALE

I
The diagram shows the positions of a lighthouse, L, and two ships, Kand M .
The bearing of L from K is 155° and KL = 112 km.
The bearing of K from Mis O10° and angle KML = 96°.

Find the bearing and distance of ship M from the lighthouse, L.

Mlle. i.. /fQ- GJ6 - 35 :. l- /~

tJ I..
C,

St> l~I.~ ~S"

N LM _ J, o - 4 Cf - S- = 8~
0
"' \,eCI.-A f\5 ° J./i frotv\ L-

,,~ - '>t.
s\ r. <:\6 · ~?
Bearing .. J.8.~.~-·-············ ····················
?\. -- ,, 1, S \C\ &':I
$'1(\ qc; Distance .. ,G .':l.,.C,_..., ... ... ................... km [S]
';. 01,Sq -c. bY•' 't:-H\
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14

AB is a line with midpoint M.

A is the point (2, 3) and Mis the point ( 12, 7).

(a) Find the coordinates of B. r,c- .~

- - - - - - - - --
")(a_ ... ~Ii · ~ <'V?.I "'-l1i 1tt) i3
~l •• 'il l -I '){ ': '>, v. o:;.. 'j \ ""~ t
"ii'- ,J -
I~ a :~) g'-1 :- \l.t .. ~. t~
"1.~ ,~ a.+
( ······ .. ......... , ....\\.. .............. ) (2]
:c)

(b) Show th at th e equation

· . ~) ':h-= I\.\· 3 '"" I I
of the perpendicular bisector of AB is 2y+ Sx = 74.
lf\ ini,d.,~t o~ .L ~,~ec..~'" ,. - \ ~ --
tv\l'IQ. : "-~,
')t't.- )t l

- .. s
"•~<to; ol
j - 'j, ---.£ bc-1~)
i~ - II.\-=- -S')l.•H;'o

~j *:>-.-t ~'-\
-#,-

(4]

( c) The perpendicular bisector of AB passes through the point N.

The point Nhas coordinates (2, n).

Find the value of n.

~~ ..- ?"It. ....
atn) SLa.) "'" ":\-\\
~"" :. " "'\
n= ..... a.~ ...................................... (11
(\ 'C- 3~
(d) Points A, M and N form a triangle. ~ I

Find the area of the triangle.

('rrea -_ \~ "O~ e

=- _fl 1?.- • ~) "- + l 1" '3) i),

=ffi
~-:; ~ ~--3~)a [-:i~s
1 tm) )

" ~ \lo\'?
\1..\5
................................................. [21

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(a) On the diagram, sketch the graph of y = sinx for 0° x 360°.

[2]

(b) Solve the equation 5 sinx+4 =O for 0° x 360°.

;
/
;
s $\('\ 'X .. -
I
Sio "'>( ":- - ~

s--
')t -;. s,n-' {-.!) -- c;3,\~ 0

S'""

1IL fd Q1.10.c:l1'V'll. '111 Q l,,l~d '°"" t

() ~ lfOtcJ e"'3,o-cf
110 .. 53.i.3 ~""' JGo- ';SitJ
: ~21, 13 -:. 3CE.,'1

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fbe Jengtbs of the sides of a triangle are 11.4cm, 14.8 cm and 15.7 cm, all correct to 1 decimal
p/llce.
l
caJcU late the upper botmd of the perimeter of the triangle.
0
L, -= 1\,4 C..M ! ·5 P~',me.\::e., .. (L, 1-~ -t- L ) ';ppe., L;, Q\4"c.\
1
1,.. .. ltt,8 .t o,i; : 11,4 S x IL.j, '8, -t IS ,~s
l..3 ~I?,~ ±0•5 -:- 4~.05

(b)

15.6cm NOTTO
SCALE

The diagram shows a circle, r~dius 105.6cm.

The angle of the minor sector is 150 .

Calculate the area of the minor sector.

.. ....... 3.t.~ ........................ cm2 (2)

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NOTTO
SCALE

The diagram shows a circle, radiu~ rem and minor sector angle x 0 •
The perimeter of the major sector is three times the perimeter of the minor sector.
90(7t-2)
Show that x = 1t .

Perimeter ot- Meljc>r sec.tot :. 3 ( p e-i imete1 t'v\t tr\l f\ov .S~c_t,o,)

t-,-\o,~C)....- I\<~ ..,.. ~.r - 3 (~ \ l'\ CH o..c \ e"Cj""' )

1 . xG ..-¥"1, .. ll,, • 3 ~ , .. ~.,.. )

lfo

:. 3')\ Wr 1 ?t rrrt
I' O /iD

-
4...._.qrr
J'-go

(41
3,\Jr ( ~ · ~\ -- 4"1' r (.yd
'\O
)L -;: ~ / (~ . ;j)
){'Tf ;r

')t ~ o ('Tr - ~)
.,,.

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!(:a:)I= 2~5
1//
find the two possible values of m.

r ('lfY'I )~ -t&:-M)~
- -~'
<~+. t~;..,i~J .,_ l Blf f
~I rn + l~ou"' \ ( ~,,~

\(~ ! \ ('(\ ~ ~ (~')~

m = ......~:.5.......... or.::-...~.'.?. .......... [3]

(b)
A
, - - - - - - - -...... B

NOTTO
SCALE

0 C C

OABC is a parallelogram.
0A = a and OC = c.
Pis the point on CB such that CP: PB= 3: 1.

(i) Find, in terms of a and/or c, in their simplest form,

AC= .......... ~.::.~ ............... .............. [l]

OP= ...... C..t. .. l.. ..c.,............ .. ............. [ 11

\q

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