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PROGRAM DIDIK CEMERLANG AKADEMIK
SPM
TTTT
SPM PAPER 2 MODEL & SOLUTIONS
ORGANISED BY:
JABATAN PELAJARAN NEGERI PULAU PINANG
ADDITIONAL MATHEMATICSFORM 5
MODULE 20
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SULIT3472/2MatematikTambahanKertas 22 2
1 hours
JABATAN PELAJARAN PULAU PINANG
SPM EXAMINATION MODEL PAPER
ADDITIONAL MATHEMATICS
Paper 2
Two hours and thirty minutes
DO NOT OPEN THE QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO
1. This question paper consists of three sections : Section A, Section B and Section C .
2. Answer all questions in Section A, four questions from Section B and two questions
from Section C.
3. Give only one answer/solution to each question.
4. Show your working. It may help you to get marks.
5. The diagrams in the questions provided are not drawn to scale unless otherwise stated.
6. The marks allocated for each question and sub-parts of a question are showns inbrackets.
7. A list of formulae is provided.
8. A booklet of four-figure mathematical tables is provided.
9. You allowed to use a non-programmable scientific calculator.
This question paper consists of 9 printed pages.
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2
Section A
[ 40 marks ]
Answer all questions in this section.
1. Solve the simultaneous equations 3x + 2y = 5 and 2x 2 + xy = – 3 .
[5 marks]
2. The equation of a curve is y = 3x 2 + 5x + c where c is a constant. The straight line
y + 7x + 5 = 0 is a tangent to the curve at a point P. Find
(a) the coordinates of P, [4 marks]
(b) the value of c. [2 marks]
3. Figure 1 shows the first few triangles of a series of triangles of equal heights drawn ona straight line, AB.
The lengths of the bases of the triangles form a geometric progression. The length ofAB is 16.38 m.
(a) Calculate the area of 8th triangle, [3 marks]
(b) If n triangles are drawn on the line AB, find the maximum value of n. [3 marks]
A B
3 cm
Figure 1
0.4 cm 0.8 cm 1.6 cm
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4. Table 1 shows the score distribution obtained by a group of contestants in a contest.
Table 1
(a) Calculate the varians of the distribution. [4 marks]
(b) If each score in the distribution is multiplied by 2 then subtracted by c, the mean
becomes 2.8. Calculate
(i) the value of c,
(ii) the new standard deviation. [3 marks]
5. (a) Given tan A =4
3, find the value of cos 2A. [3 marks]
(b) (i) Sketch the curve y = sin 2x for 0 x 2 .
(ii) Hence, by drawing a suitable straight line on the same axes, find the number
of solutions satisfying the equation sin x kos x =2
1
4
x
for
0 x 2 . [5 marks]
6. In figure 2, PQRS is a quadrilateral and PTR is straight line.
Given thatPQ = 6x,
QR = 24y,
SR = 15x + 12y dan PT =
3
1TR
(a) Express in terms of x dan y
(i)PR
(ii)QT . [3 marks]
(b) Show that PS is parallel to QT. [3 marks]
(c) If cm7x and o60TPQ , findPR [2 marks]
Score 1 2 3 4 5Number of contestants 4 6 12 5 3
S
T
R
P Q
Figure 2
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Section B
[40 marks]
Answer four questions from this section.
7. Use the graph paper provided to answer this question.
Table 2 shows the values of two variables, x dan y, obtained from an experiment.x dan y are related by the equation y – q = p( x + 1 )( x – 1 ), where p and q areconstants.
x 1.0 2.0 3.0 4.0 4.5 5.5
y 12.0 16.0 24.2 34.3 40.5 55.6
Table 2
(a) Plot y against x 2 using a scale of 2 cm to 5 unit on both axes.Hence, draw the straight line of best fit. [4 marks]
(b) Use your graph from (a) to find the value of
(i) p
(ii) q
(iii) y when x = 5. [6 marks]
8. In figure 3, the straight line PQ is a tangent to the curve y = x 3 + 4 at the point
T(2 , 12 ).
Find
(a) the value of k, [3 marks]
(b) area of shaded region, [4 marks]
(c) the volume generated when the shaded region is revolved 3600 about the y-axis,giving your answer in terms of . [3 mark]
y
xO
P
Q(k , 0 )
T(2 , 12)
y = x 3 + 4
Figure 3
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9. Solution by scaled drawing will not be accepted .
In figure 4, the straight line PR cuts y-axis at Q such that PQ : QR = 1 : 3. Theequation of PS is 2y = x + 3.
(a) Find
(i) the coordinates of R,
(ii) the equation of the straight line RS,
(iii) the area PRS.
[7 marks]
(b) A point T moves such that its locus is a circle which passes through the points
P, R dan S. Find the equation of the locus of T.
[3 marks]
P(- 3, 0 )
Q( 0, 4 )S
Ry
xO
Figure 4
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10. Figure 5 shows a semicircle APD with AD as a diameter, and PBQD is a sector whose
centre is D. AB = BC = 1 cm dan AD = 10 cm.
Calculate
(a) the value in radian, [4 marks]
(b) the length the arc AP in cm, [2 marks]
(c) the area of shaded the shaded region in cm 2. [4 marks]
11. For this question, give your answer correct to 3 significant figures.
(a) A survey done on a group teachers shows that 32 teachers of 40 teachers in a
district posses laptop computers. If 6 teachers from that district are selected at
random, calculate the probability that
(i) exactly 3 teachers posses laptop computers,
(ii) more than 2 teachers posses laptop computers.
[4 marks]
(b) The heights of teenagers in a residential area were found to have normaldistribution whose mean is 150 cm. It was found that a teenager whose height is156 cm has a standard score of 1.5 . Find(i) the value the standard deviation, ,
(ii) the probability that a teenager selected at random from that area has aheight between 148 cm and 158 cm,
(iii) the value h if 70% of the teenagers in that area are less than h cm tall.
[6 marks]
B DC
θ
A
P
Q
Figure 5
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Section C
[ 20 marks ]
Answer two questions from this section.
12. In Figure 6, BC is parallel to ED.
Find
(a) the length, in cm, of(i) EC,
(ii) EB .[6 marks]
(b) AEB,[2 marks]
(c) the area of ABE in cm2.[2 marks]
A
B C
DE
4 cm
7 cm
6.5 cm
3.5 cm
70 o
Figure 6
75 o
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13. Table 3 shows price indices of sport shoes of four brands sold in a shop. Figure 7 is a pichart which represents the number of pairs of shoes by brands sold in 1997.
Brands of
sport shoes
Price indices in 1994
(1991=100)
Price indices in 1997
(1994=100)
P 112 130
Q 108 125
R 123 X
S Y 110
Table 3
(a) (i) If the price of the P brand shoes in 1991 was RM50, calculate the price in
1994.
(ii) If the price of the Q brand shoes in 1997 was RM60.75, calculate the
price in 1991.
[3 marks]
(b) (i) Given that the price of the R brand shoes increased by 18% in 1997
compared to the year 1994. Find value x.
(ii) The price index for the S brand shoes in 1997 based in on the year 1991 is
165. Find the value of y.
[4 marks]
(c) Calculate the composite index number for the shoes of the four brands for the year
1997 based on the year 1994.
[3 marks]
110o
30o
RQ
SP
Figure 7
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14. Use the graph paper provided to answer this question.
The Mathematics Society of a school is selling x souvenirs of type A and ysouvenirs of type B in a charity project based on the following constraints :
I : The total number of souvenirs sold must not exceed 75.
II : The number of souvenirs of type A sold must not exceed twice the numberof souvenirs of type B sold.
III : The profit gained from the selling of a souvenir of type A is RM9 whilethe profit gained from the selling of a souvenir of type B is RM2. The totalprofit must not be less than RM200.
(a) Write down three inequilities other than x 0 dan y 0 which satisfy the aboveconstraints. [3 marks]
(b) Hence, by using a scale of 2 cm to 10 souvenirs on both axes, construct and shadethe region R which satisfies all the above constraints. [3 marks]
(c) By using your graph from (b), find
(i) the range of number of souvenirs of type A sold if 30 souvenirs of type B aresold.
(ii) the maksimum which may be gained. [4 marks]
15. An object, P, moves along a straight line which passes through a fixed point O.
Figure 8 shows the object passes the point O in its motion. t seconds after leaving the
point O , the velocity of P, v m s─1 is given by v = 3t2 – 18t + 24. The object P stops
momentarily for the first time at the point B.
P
O B
Figure 8(Assume right-is-positive)
Find:
(a) the velocity of P when its acceleration is 12 ms – 2 , [3 marks]
(b) the distance OB in meters, [4 marks]
(c) the total distance travelled during the first 5 seconds. [3 marks]
END OF QUESTION PAPER
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3472/2
AdditionalMathematics
Paper 2
JABATAN PELAJARAN PULAU PINANG
ADDITIONAL MATHEMATICS
Paper 2
Solutions and Marking Schemes
This document consists of 16 printed pages
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2
1. 3x + 2y = 5 ……………………………. (1)2x 2 + xy = – 3 ……………………………. (2)
From (1) : y =2
x35 ………………... …… (3) [1]
Substitute (3) into (2) :
2x 2 + x
2
x35= – 3 [1]
2 : 4x 2 + 5x – 3x 2 = – 6x 2 + 5x + 6 = 0
(x + 2)(x + 3) = 0 [1]x = – 2, – 3 [1]
When x = – 2 , y =2
11
2
)2(35
When x = – 3 , y = 72
)3(35
[1]
2. (a) Equation of curve : y = 3x 2 + 5x + c
dx
dy= 6x + 5 [1]
The given equation of tangent at P is y + 7x + 5 = 0,Rearranging, y = – 7x – 5
Gradient of tangent at P = – 7 6x + 5 = – 7 [1]
x = – 2 [1]and y = – 7(– 2) – 5 = 9
coordinates of P = ( – 2 , 9 ) [1]
(b) Point of contact P( – 2 , 9 ) lies on the curve y = 3x 2 + 5x + c
9 = 3(– 2 ) 2 + 5(– 2) + c [1] c = 7 [1]
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3
3. (a) Length of bases of triangles: 0.4 cm , 0.8 cm , 1.6 cm , ………..form a geometric progression with a = 0.4 and r = 2 [1]Length of base of 8th triangle = T 8
= ar 7
= 0.4 (2) 7 [1]= 51.2 cm
Height of triangle = 3 cm
area of 8th triangle =2
1 51.2 3 = 76.8 cm 2 [1]
(b) Length of AB = 16.38 m = 1638 cm and number of triangle drawn = nSum of lengths of bases of triangle 1638 cm
S n 1638
1r
)1r(a n
1638
12
)12(4.0 n
1638 [1]
2 n – 1 4.0
1638= 4095
2 n 4096 = 2 12
n 12 [1] the maximum value of n = 12. [1]
4. (a)Score, x Frequency, f Fx fx 2
1 4 4 42 6 12 243 12 36 1084 5 20 805 3 15 75
f = 30 fx = 87 fx 2 = 291[1] [1]
2 = 23087
30291
[1]
= 9.7 – 2.9 2 = 1.29 [1]
(b) (i) Old mean =30
87= 2.9
When every score is multiplied by 2 and then subtracted by c,new mean = 2.8 2(2.9) – c = 2.8 [1]
c = 3 [1]
(ii) new standard deviation = 2 ( 29.1 ) = 2.272 [1]
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5. (a) Given tan A =4
3
cos A = 5
4[1]
cos 2A = 2 cos 2 A – 1
= 2 1254
[1]
=25
7[1]
(b) (i)
(ii) The given equati
2 :
The required st
From the grap
y
xO
1
- 1
y = sin 2x
y = 12
x
3
– 3 4– 4
5
5
http:/
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4
on : sin xkosx
kxsin2
sin
raight line has th
x 0y – 1
h, the number of
21
/mathsmozac
4
x
xos
x2
e equati
2
xy
2 0
solutio
.blogs
2
1
12
x
12
x
on
1
n = 4.
23
pot.com
2Lengkung [2]
Garis lurus [1]
[1]
[1]
5
6.
(a) (i)PR =
PQ +
QR
= 6x + 24 y [1]
(ii)QT =
QP +
PT
=QP +
QR
4
1
= –6x +4
1( 6x + 24y) [1]
= –2
9x + 6y [1]
(b)PS =
PR +
RS
= )y12x15()y24x6( [1]
= – y12x9
= 2 ( – yx 629 )
= 2QT [1]
PS is parallel to QT. [1]
(c) SPT = 90 O and PS is parallel to QT PTQ = 90 O
PQ
PTcos 60 o =
2
1
PT =2
1 ( 6 7) = 21 [1]
PR = 4 PT
= 4 (21) = 84 cm [1]
60 o
24 y
6 x
15 x + 12 y
S
T
R
P Q
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7. (a)[1]
Refer to Appendix A
Correct axes and uniform scale [1]6 points marked correctly [1]Straight line of best fit [1]
(b) (i) y = p( x + 1 )( x – 1 ) + q= p ( x 2 – 1 ) + q= p x 2 – p + q [1]
p = gradient of graph
=125.20
125.40
[1]
= 1.48 [1]
(ii) – p + q = y intercept
–1.48 + q = 10.5 [1]q = 11.98 [1]
(iii) When x = 5, x 2 = 25y = 47.5 [1]
8. (a) Equation of curve : y = x 3 + 4
2x3dx
dy [1]
Gradient of tangent at T( 2 , 12 ) = 3 ( 2 2 ) = 12
12k2
012
[1]
2 – k = 1k = 1 [1]
x 2 1 4 9 16 20.25 30.25y 12.0 16.0 24.2 34.3 40.5 55.6
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(b)
Area under the curve RT = 2
0dxy
= dx4x2
0
3 [1]
=
2
0
4
44
x
x[1]
= 4 + 8 – (0) = 12
Area of QST = 612121
area shaded region = 12 – 6 [1]= 6 unit 2 [1]
(c) Volume generated = dyx21
4
2
= dy])4y([12
4
23
1
[1]
= dy)4y(12
4
3
2
=
35
)4(53
12
4
y [1]
= ]08[ 3
5
53
= 596 unit 3 [1]
y
xO
P
Q(k , 0 )
T(2 , 12)
y = x 3 + 4
R
12
12
S
y = 12
2
4
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9. (a) (i) Let the coordinates of R be (h , k)
04
9h
and 4
4
0k
[1]
h = 9 k = 16 coordinates of R = (9 , 16) [1]
(ii) The given equation of PS is 2y = x + 3
Rearranging, y = 21 x + 2
3
m PS = 21
From the diagram, RS PS.
m RS m PS = – 1
m RS = – 2 [1]
The straight line RS passes through the point R(9 , 16)
the equation of RS is y – 16 = – 2 (x – 9 )
Simplifying and rearranging, y = – 2x + 34 [1]
(iii) Equation of RS: y = – 2x + 34 …………….. (1)
Equation of PS: 2y = x + 3 ……………… (2)
Substitute (1) into (2) :
2(– 2x + 34) = x + 3 [1]
– 4x + 68 = x + 3
65 = 5x
x = 13
and y = – 2(13) + 34 = 8
coordinates of S = (13,8)
area of PRS =08160
31393
2
1
= )242080(072482
1 [1]
= 80 unit 2 [1]
(b) Centre of circle = mid-point of PR = )2
160,
2
93(
= (3 , 8) [1]
Radius of circle = 21 PR = 22
21 )160()93( = 10 [1]
Therefore, equation of locus of T is (x – 3) 2 + (y – 8 ) 2 = 10 2
Simplifying and rearranging, x 2 + y 2 – 6x – 16y – 27 = 0 [1]
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9
Other method:
Let coordinates of T be (x,y)
PR is a diameter. TP is perpendicular to TR.
gradient of TP X gradient of TR = ─1
,19
16
)3(
0
x
y
x
y
1)9)(3(
)16(
xx
yy
y(y – 16) = ─(x + 3)(x ─ 9)
y2 – 16y = ─(x2 – 9x +3x – 27)
= ─(x2 – 6x – 27)
= ─x2 + 6x + 27)
x2 + y2 – 6x – 16y – 27 = 0
10.
(a) Let O be the centre of semicircle APD.
Given AD = 10 cm.
Radius of semicircle APD = 21 10 = 5 cm
OP = 5 cm , CO = 5 – 1 – 1 = 3 cm [1]
kos POC =5
3[1]
POC = 0.9273 radian [1]
PO = OD POD is an isosceles triangle
ODPOPDPOC
0.9273 = = 2
0.9273 2 = 0.4637 radian. [1]
B DC
θ
A
P
Q
O1 cm1 cm 5 cm3 cm
5 cm
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10
(b) Length of arc AP = 5 0.9273 [1]
= 4.637 cm [1]
(c) Using Pythagoras’ theorem, PC 2 = 5 2 – 3 2 = 16
PC = 16 = 4 cm [1]
sinPD
PC
PD
4= sin 0.4637 [1]
PD = 8.943 cm
and PDQ = 2 = 0.9273
area of shaded region = 21 (8.493) 2 ( 0.9273 – sin 0.9273 ) [1]
= 4.591 cm 2 [1]
11. (a) (i) P(one teacher posseses laptop computers ) =40
32= 0.8
P(exactly 3 teachers posses laptop computers)
= 333
6 )2.0()8.0(C [1]
= 0.0819 (correct to 3 significant figures) [1]
(ii) P(more than 2 teachers posses laptop computers)
= P( X = 3, 4, 5, 6)
= 1 – P (X = 0, 1, 2)
= 1 – [ 0.2 6 + 16 C (0.8) 1(0.2) 5 + 2
6 C (0.8) 2 (0.2) 4 ] [1]
= 0.983 (correct to 3 significant figures) [1]
(b) (i) Given the standard score for 156 cm height = 1.5
5.1150156
6 = 1.5
=5.1
6= 4 [1]
(ii) P( 148 < X < 158 )
= P (4
150148 < Z <
4
150158 ) [1]
= P ( – 0.5 < Z < 2 )
= 1 – ( 0.3085 + 0.0228) [1]
= 0.669 (correct to 3 significant figures) [1]
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11
(iii) Given P( X < h ) = 70%
P( Z < )4
150h = 0.7
P( Z > )4
150h = 1 – 0.7 = 0.3
From tables, P(Z > 0.524) = 0.3
524.04
150
h[1]
h = 4(0.524) + 150
h = 152 (correct to 3 significant figures) [1]
12. (a) (i) Using the cosine rule in CDE :EC 2 = 6.5 2 + 3.5 2 – 2 (6.5)(3.5) cos 70 o [1]
EC = 94.38 = 6.24 cm [1]
(ii) Using the sine rule inCDE :
24.6
70sin
5.3
DECsin o
[1]
sin DEC = 0.5271
DEC = 31.81 O
Given ED is parallel to BC
BCE = DEC = 31.81 O [1]
From EBC , EBC = 180 o – 75 o – 31. 81 o = 73.19 o [1]
Using the sine rule in EBC :
oo
EB
19.73sin
24.6
81.31sin
EB = 3.436 cm [1]
(b) Using the cosine rule in ABE :
cos AEB =)436.3()4(2
7436.34 222 [1]
= – 0.7710 AEB = 140.44 o [1]
(c) Area of ABE
=2
1( 4 ) ( 3.436 ) sin 140.44 o [1]
= 4.377 cm 2 [1]
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12
13. (a) (i) Given the price of P brand shoes in 1991 = RM50 and
its price index in1994 based on the 1991 = 112
Its price in 1994 = 112 % RM50 = RM56 [1]
(ii) For the Q brand shoes,
price index in 1994 based on the year 1991 = 108
08.1100
108
H
H
91'
94' ……………………………….. (1)
price index in 1997 based on the year 1994 = 125
25.1100
125
H
H
94'
97' ………………………………. (2)
(1) (2) gives 91'
97'
H
H1.08 1.25 = 1.35 [1]
Given 97'H = RM60.75, 91'H =35.1
75.60= 45
price of Q brand shoes in 1991 = RM45 [1]
(b) (i) Given the price of R brand shoes increased by 18 % in 1997
compared to the price in 1994.
x = price index in 1997 based on the year 1994
= 100 + 18 = 118 [1]
(ii) Price index for S brand shoes in 1997 based on the year 1991
= 91'
97'
H
H100
= 100H
H
H
H
91'
94'
94'
97'
= 100100
y
100
110 = 1.1 y [1]
Given the price index = 165 1.1y = 165 [1]
y = 150 [1]
(d) Angle of sector for S brand shoes= 360 o – (30 o + 110 o + 90 o ) = 130 o [1]
composite price index in 1997 based on the year 1994
=360
)130(110)90(118)110(125)30(130 [1]
= 118.25 [1]
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13
14. (a) The three inequalities are
x + y 75 , x 2y dan 9x + 2y 200 [3]
(b) Refer to Appendix B :
All lines drawn correctly [2]
Region shaded correctly [1]
(c) (i) 16 number of A type souvenirs sold 45 [2]
(ii) Maximum profit
= RM [ 9(50) + 2(25) ] [1]
= RM500. [1]
15. (a) Velocity, v = 3t 2 – 18 t + 24
Acceleration, a =dt
dv= 6t – 18 [1]
When a = 12 ms – 2 ,
6t – 18 = 12
6t = 30
t = 5 [1]
v = 3( 5 2 ) – 18 (5) + 24 = 9 ms – 1 [1]
(b) When P stops momentarily, v = 0
3 t 2 – 18t + 24 = 0
:3 t 2 – 6t + 8 = 0
( t – 2 )( t – 4 ) = 0 [1]
t = 2 , 4
P stops momentrarily for the first time when t = 2 seconds [1]
v = 3 t2 – 18t + 24
Displacement, s = ( 3 t 2 – 18t + 24) dt
= t 3 – 9 t 2 + 24t + c [1]
When t = 0 , s = 0.
c = 0
s = t 3 – 9 t 2 + 24 t
When t = 2, s = 2 3 – 9 (2 2 ) + 24 (2 ) = 20
OB = 20 m [1]
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14
(c) When t = 5 , s = 5 3 – 9 ( 5 2 ) + 24( 5 ) = 20
When t = 4 , s = 4 3 – 9 ( 4 2 ) + 24( 4 ) = 16 [1]
P t=4 t=2, 5
O B
16 m 4 m
From the diagram, total distance travelled during the first 5 seconds
= 20 + 4 + 4 [1]
= 28 m [1]
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15
7 (a)
Appendix A
y
0x 2
10
10
20 30
20
30
40
50
x
x
x
x
x
x
10.5( 1 , 12 )
( 20.25 , 40.5 )
47.5
25
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16
14 (b)
Appendix B
100
40
90
y
80
70
60
50
30
20
10
O 10 20 30 40 50 60x
70 80
9x + 2y = 200
x = 2y
x + y = 75
y = 30
( 50 , 25 )
R
16 45
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