Post on 17-May-2018
transcript
Paper 2 - 2010 80
Solution
Paper 2 - 2005
Page
1
Paper 2 - 2006 15
Paper 2 - 2007 31
Paper 2 - 2008 47
Paper 2 - 2009 63
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MathematicsGeneral Proficiency
May 2005
1. a)
21 10= 5 3
63-50= 15
- 13/- 15
b) i) A: 12.50 x 3 = $37.50 'v
B:$33.90 =$li.25--
2
c: $31.00 =5--$6.20
D:15
- X 108.28 = $16.24100
ii) 6 x 0.75 = $4.506 x 0.40 = $2.40Total made = $6.90Total spent = $5.88
*NB:The exact profit was not
necessary in this case.
(Difference: $6.90 - $5.88 = $1.02)
Amanda made a Rmfit
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2 a) i) ab(5a + b)
ii) (3k-l) (3k+ 1)iii) 2yL4y-y+ 2
2y(y - 2) -l(y - 2)(2y -1) (y - 2)
b) 6xz - 8x + lSx -206xz + 7x - 20
c) i) 2(x - 3)= 2x-6
l ii) x + x - 3 + 2x - 6 = 394 x - 9 = 39
4x =48x = 12
·Solving the equation was not necessaryin this case.
3. a) i) 3x + x = 244x=24
x=6~ students take both Music and Drama.
ii) Student who take drama only = 7 + x:. 7 + 6 = 1.3.students take only Drama.
b) i) 2y- 5 = -(x + 3) y=mx+c3
2 2y='3x+2+S or 5=-(-3)+c
32
5 = -2 + cy='3x+7c=7
ii) 2 x - 3y = 02
Y = -x + 73
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-3y = -2x-2xy=- -32
Y = -X3
m =!. .:for the line 2x - 3y = 0 therefore it is parallel to y = !.x + 73 3
(Parallel lines have the same gradient)
4. a) Area of small pizza = rr(7.5)2= 176.79 em-
Area of medium pizza = rr(152)= 707.14em2
707.14 = 4176.79
The medium pizza is more than twice as large as the small pizza. It is fourtimes as large.
b) S· f1 f di . 707.14 23571 2Izeo -0 me IUmplzza=--= . em3 3
ratio of area to cost of medium pizza = 235.71 = 14.78 em2/$15.95
ratio of area to cost of small pizza = 176.79 = 13.65 em2/$12.95
Therefore, it is better to buy a slice of medium pizza, rather than a smallpizza.
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5. The graphbelow shows the answer for 5(a), 5(b) (i) 5(b) (ii) - lern: 1 unit on bothaxis.
_J __ ,
c. 1; ;;\
. , ;
.
F'. ,
;', .
" , '.
. ,
.; ....,
.' .
.' ,
~~"
..~,
" ,
" ' " ... ,
.. ".
...'..,.... . 'F';'
: .1.· .. · ~ I'.: ' I·
.'
! ; ..~. ; , :.'
" "
' .... ./. "
',: I' ,"
. "r:,I,·:'
.i . -: ,
, ', .. ':
s :
.. '\' .
",,:.. ',',y-,. ,;.:' ':~.:ii
'" ; " . ,:
,.., "Et','
,n
,·c.
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1 ,l
.' .~ .
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J' I: .;
.... "
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- ,,> "'" .• ,.
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:,'; i,':' 'I','i, ",:
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4\Page
.... "., 'cT,'., ',: .. ", ;t.
! ,:' "
Hi) Composite transformation
c)1.8
tan a=-2
a = tan"! 0.9a = 42°
6. a) i) EAD = 57° Reason: Alternate angles of parallel lines
Then x = 180 - (57 + 80)= 43°
ii) DAB=108-57= 51°
Since angle of a quadrilateral add up to 360°.Then y = 360 - (51 + 57 + 90)
= 360 -198= 162°
b) i) 32 + (-3)2 = 9 + 9=18
ii) Ifx = Yzy+ 5x-5=Yzyy=2(x-5)
r1(x) = 2x -10rl(6) = 2(6) -10
= 12 -10=~
iii) fg(x) = 1fzX2 + 5f g(2) = 1fz(22) + 5
=Yz(4)+5=2+5=Z
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7.
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a) Cumulative Frequency Table showing the height of applicants
Height/ern Cumulative Frequency
$155 10$160 65$165 170$170 280$175 360$180 390$185 400
400
3!lO
300
J 250'"~,.~ 200Q
150
100
50
Cumulative Frequency Curve of Heights of Applicantsy
!SS 16S 110 ,17,S; 1&0 lSS x
H.ipt (cmj
b) i) 275 applicants
ii) 167 em
iii) 162.5 em
iv) 95-400
8. a) i)(4 X 72) + (3 x 6) + 2 216
ii) I~------~------------------------
103 (8 X 112) + (3 x 10) + 2 1000
iii) 1,--_n_3 --l.1_C(_n_-2_)x_C_n_+1_)2_+_(3_X_n~_2_1,-- __ n3__
b) (a - b) (a - b) (a + b) + ab(a + b)= (a - b) (a- - b2) + a-b + ab-= a3 - ab- - a-b + b3 + a2b + ab-= a3 - ab- + ab- - a-b + a2b + b3= a3 + b3
Difference of squares• (a - b)(a + b) =aL b'
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9. a) 5xz + 2x-75(xZ + ~x) - 7
5
5 [Xl + ~X + (!)Z ~ - 7 - 5(!)25 5 5
1 1Sex +_)2- 7-5 5
b) i) 1- 7-5
ii) 1 1WhenX + 5 = 0,then x = - '5
c) S(x + !)2 = 365 5
(x + !)2 = 365 25
X + ~ = ../36/251 6x=--+ -5 - 5
E" h 1 6 1 6It er x = - - + - or x=----
5 5 5 5
=17-5
d)v
x
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10. a) i) 40-0_'J£.77/2- £".,J,LL m s15-0 --
Acceleration == gradientof first portion of graph
ii)Distance travelled = 22
0 (40 + 50)= 10(90)= 9.illlm
20
b) i) 12km = 6km/h2h
ii) He took a rest or stopped.
iii) ~=8km/h1.5
c) i) x=6
ii) x~6
5 +5y>-- x- 8
1y<-x+5-6
li. a) i) x
7.5
"evaluate ::-point in the region foreach line.
"Sine rule: "h ab sin C = Area oftriangle.
pL- ~ Q
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1h (7.5) (4.5)Sin 0' = 13.5S
. 13.5InO' =--
16.8750"' = sino! 0.8
0"' = 53°
ii) 1h (15) (9) sin 53 = 67.5 sin 53= 54 cm2
b) i) a) 180 - 136 = 44°SJM = 44-
2= 22° (base angles of an isosceles triangle)
b) SJM = JMK = 22° (alternate angles):. JKM = 180 - (124 + 22)
= 180-146= 34°
ii) a) MJ 50--sin 136 sin22
MJ =9.2..Zm
b) --=--JK 92.7
sin 22 sin 34JK =Qllm
12. a)
17°N
A-AntiguaB - BelizeG - Greenwich MeridianE - Equator
G
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b) C= 2 x 3.14 x 6400= 40192= 40192 cos17= 38423.6
1= 38423.6 (~)360::::2776km
c) 1= 40192 (~)360::::6140km
13. a) i) AB=DCTherefore. AB = 3K
ii) BD=BA+/iD= - 3K - 3y
Hi)---.-. 1 ---->DP=--BD
31:::: -"3 (-3x - 3y)
=x+y
b) AP=AD+DP= -3y + x + y=x 2y
c) PH=PD+DE3= -x - y +- X2
::::Yzx-y
and fiE = liP+PH1= X - 2y + -x - Y2
3::::-x - 3y
2
=> AP=2PHTherefore. A, P and E are collinear.
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d) DA == 3G)=G)
IDAI = .../32 + 32 = ..JI8
AB=AP+PE= X - 2y + lh X - Y
3= 2"X - 3y= ~(~)-3G)= G) - (D== (~3)
IABI = J 02+ (-3)2 = ~ = 3
DE =~ (~)= (~)
IDEI == .../32 + 02=~=3
A
o '-----t+----" E
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A
Therefore; triangle AED is isosceles.
14. a) i)
ii)
iii)
(2) (15) - (7) (5)= 30-35= -5
=> M is a non-singular matrix
M-I = 2. (15 -5)-5 -7 2
iv) (10 0) (X) 1 (151 Y = -5 -7
(X) = ~ (-45 - 85)Y -5 21 + 34
(X) = ~ (-130)Y -5 55
X = 26 y =:.11
b) i) (-1 1°) • The general matrix ofR= 0 rotation for iT' in theanuclockwlse directionabout the
ii) (-1-~)
origine:,:! -::::)N= 0
iii) T= (~3)
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