[edu.joshuatly.com] kedah spm trial 2011 maths t (w ans)
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CONFIDENTIAL*/SULIT*
954/1 STPM TRIAL 2011
REMINDER:
This marking scheme is specially for the use of examiners and cannot be given to unauthorized persons.
This scheme consists of 14 printed pages.
MARKING SCHEME FOR
MATHEMATICS T PAPER 1 (954/1)
STPM TRIAL EXAMINATION 2011
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CONFIDENTIAL*/SULIT* 2
954/1 STPM TRIAL 2011 [Please turn over
1. Determine the value of m if i
miz524
5
is a real number and find this real number
[ 4 marks ] Solution:
iix
imiz
524524
5245
M1
=2016
5241054
mmii
= 36
)104(36
5254 imm
M1
Z is a real number, 036
104
m
4m + 10 = 0
m = -25 A1
z = 45 A1
[4]
2. Given that 53x . Express x
x 2 in the form rqp .
State the values of randqp , . [4 marks] Solution:
53
253
=535612
M1
=5353x
535612
M1
= 523
23 M1
5 ,23 ,
23
rqp , All correct A1
[4]
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CONFIDENTIAL*/SULIT* 3
954/1 STPM TRIAL 2011 [Please turn over
3. A cylinder open at one end is constructed from thin metal. The total surface area of the cylinder is 192 cm 2 . The cylinder has radius of r cm and height of
h cm. Show that the volume V cm 3 of the cylinder is, V = 319221 rr .
Given that the radius of the cylinder can vary, find the maximum value of V.
[6 marks] Solution: 19222 rhr
r
rh2
192 2 B1
rrrhrV
2192 2
22 M1
319221 rrV A1
2
2396 r
drdV
M1
02396,0 2 r
drdV
8,642 rr . B1
.512)8(2
)8(96 33max cmV
A1
[6]
4. In a geometric progression , the third term is 6 and the sixth term is 92
.
Find the first term and the common ratio of the series. [2 marks] Hence, find (a) nS , the sum of the first n terms of the series. [2 marks]
(b) S , the sum to infinity of the series. [2 marks]
Solution: ar2 = 6 , a = 54 either one correct M1
r = -31 , a = 54 both correct A1
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CONFIDENTIAL*/SULIT* 4
954/1 STPM TRIAL 2011 [Please turn over
(a) nS =
n
311
311
54 M1
=
n
311
281 A1
(b)
311
54S M1
= 281 A1
[6] 5. The function f is defined by
3, 53, 53
)( 2 xxxxx
xf
(a) Without sketching the graph, determine whether f is continuous at x = 3. [4 marks] (b) Sketch the graph of f in the domain [0,5] and state the range of f. [3 marks] Solution: (a) 6)(lim
3
xxf B1
6)(lim
3
xxf B1
6)( xf B1
3
)(limx
xf = 3
)(limx
xf = )3(f =6, f is continuous at x = 3. A1
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CONFIDENTIAL*/SULIT* 5
954/1 STPM TRIAL 2011 [Please turn over
(b)
Straight line and curve meet at (3,6), all points correct. D2 Range = [0,15] B1 [7]
6. Given that .431)4)(31(
135822
2
xCBx
xA
xxxx
Show that C = 0 and determine the values of A and B. [3 marks]
Hence, evaluate
0
1 2
2
)4)(31(1358 dx
xxxx correct to 3 decimal places. [3 marks]
Solution:.
)31)(()4(1358 22 xCBxxAxx
Let x = 2491
9113
358,
31
AA B1
Let x = 0, 8 = 4A + C 0C B1 Let x = 1, 8 + 5 – 13 = 5A – 2B – 2C 5 B B1
dxx
xx
dxxx
xx
0
1 2
0
1 2
2
45
312
)4)(31(1358
= 0
1
2 4ln2531ln
32
xx M1
=
5ln
254ln
324ln
251ln
32 M1
= 0.366 A1 [6]
6
15
3 5
0 x
y
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CONFIDENTIAL*/SULIT* 6
954/1 STPM TRIAL 2011 [Please turn over
7. Find the equation of the tangent at the point P
tt 2,2 to the rectangular hyperbola xy = 4.
[3 marks]
The tangent meets the y-axis at point A. The straight line that passes through point A and parallel
to the x-axis meets the hyperbola at point Q. Find the coordinates of point Q. [3 marks] If M is the midpoint of chord PQ , find the equation of the locus of point M. [3 marks]
Solution:
y = x4
24xdx
dy
At point P, 244tdx
dy
= 21t
M1
Equation of tangent at point P is
)2(12
2 txtt
y M1
t2y -2t =-x+2t x + t2y = 4t A1 on y –axis , x=0 t2y = 4t
y = t4 M1
The coordinates of point A are ( 0, t4 )
When y = t4 , xy = 4
x(t4 ) = 4 M1
x = t
the coordinates of point Q are (t, t4 ) A1
the coordinates of point M =
2
42
,2
2 tttt
B1
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CONFIDENTIAL*/SULIT* 7
954/1 STPM TRIAL 2011 [Please turn over
=
tt 3,
23
Let x = 23t and y =
t3
xy =
tt 3
23 M1
xy = 29
A1 [9] 8 (a) Find the coefficient of 5x in the binomial expansion of 8)3( x . [2 marks]
(b) Expand 21
)21(
x as a series in ascending powers of x up to and including the term in 3x expressing the coefficients in their simplest form. State the range of values of x for which the expansion
is valid. By substituting 100
1x , estimate the value of 2 correct to four decimal places.
[7 marks] Solution:
(a) 6T = 53358
x
M1
Coefficient of 5x = 1512 A1
(b) ...26
)25)(
23)(
21(
22
)23)(
21(
)2(211)21( 322
1
xxxx M1
= ...25
231 32
xxx A1
Range of x for which the expansion is valid : .21
21:
xx B1
Substituting 100
1x
LHS = 21
)100
1(21
= 7
25 M1
RHS = ...100
125
1001
23)
1001(1
32
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CONFIDENTIAL*/SULIT* 8
954/1 STPM TRIAL 2011 [Please turn over
1.01015 M1
7
25 1.01015 M1
4142.12 A1 [9]
9. Given that f(x) = x4 – 3x3 + ax2 + 15x + 50, where a is a constant, and x + 2 is a factor of f(x), find (a) the value of a, [2 marks] (b) f(5) and hence factorise f(x) completely into linear factors, [3 marks] (c) the set of values of x for which f(x) > 0, [2 marks] (d) the set of values of x for which f(|x|) > 0. [3 marks] Solution:
(a) (x + 2 ) is a factor f(-2) = 0 (-2)4-3(-2)3 + a(-2)2 + 50 = 0 M1 a = -15 A1 (b) f(5) = 54 – 3 (5)3 + (-15)(5)2 +15 (5) +50 = 625 – 375 – 375 + 75 +50 = 0 B1 f(x) = ( x + 2 )( x - 5 )( x2 – 5 ) M1 = ( x + 2 )( x - 5 )( x - 5 )( x + 5 ) A1 (c) f(x) > 0, use graphical or table method, M1 (d) Solution set = 5525,: orxxorxxx A1
For f(|x|) > 0 , |x| < - 5 ( reject) or -2<|x|< 5 or |x| > 5 M1 0 |x|< 5 or |x| > 5 M1 Solution set = 5555,: xororxxxx A1 [10]
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CONFIDENTIAL*/SULIT* 9
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10. Given matrices A and B where
682121440
123122121
BandA . Show that matrices A and B abide by the
commutative law of multiplication. [3 marks] Find adjoint A. [3 marks] A hawker offers three type of noodles, Hokkien Noodle, Hailam Noodle and Tomyam Noodle. The price for each bowl of noodle is fixed and the price of 2 bowls of Hailam Noodle is equal to a bowl Hokkien Noodle and a bowl of Tomyam Noodle. A family pays RM23.00 for 3 bowls of Hokkien Noodle, 2 bowls of Hailam Noodle and a bowl of Tomyam Noodle. Another family pays RM19.50 for 2 bowls of Hokkien Noodle, 2 bowls of Hailam Noodleand a bowl of Tomyam Noodle. By using x, y and z to represent the price of a bowl of Hokkein Noodle, Hailam Noodle and Tomyam Noodle respectively, obtain a matrix equation to represent the information. Hence, find the price of a bowl of each kind of noodle. [6 marks]
10. AB =
400040004
682121440
123122121
B1
BA =
400040004
682121121
682121440
B1
As AB = BA, matrices a and B obey the communication law of multiplication. A1
Minors of A =
614824210
B1
Cofactors of A =
614824210
B1
Adjoint A =
682121440
A1
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CONFIDENTIAL*/SULIT* 10
954/1 STPM TRIAL 2011 [Please turn over
AB = - 4I
682121440
411A B1
x – 2y + z = 0, 2x + 2y + z = 19.50, 3x + 2y + z = 23 B1
2350.19
0
123122121
zyx
B1
235.19
0
682121440
41
zyx
M1
=
50.400.450.3
A1
x = RM3.50, y = RM4.00, z = RM4.50 A1 [12]
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CONFIDENTIAL*/SULIT* 11
954/1 STPM TRIAL 2011 [Please turn over
11. Sketch the graphs of x
xy 3 and y = 4 on the same coordinate axes. Mark the region
bounded by the two graphs as A. [4 marks] Find the area of A. [4 marks] Find, in terms of , the volume of the solid formed when region A is rotated 360o about the x- axis. [4 marks] Solution:
Sketch graph: Curve G1
Line y = 4, x = 1, x = 9 G2
Area, dxx
xA 39
132
=9
1
23
63232
xx M1
=
16)1(
3296)9(
3232 2
323
M2
= 2units 312932 A1
= 2units 322
y
xxy 3
y=4 4
1 0 9 x
A
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CONFIDENTIAL*/SULIT* 12
954/1 STPM TRIAL 2011 [Please turn over
Volume,
9
1
22 dx 3)8()4(
xxV M1
=
9
1
dx 69128x
x M1
= 9
1
2
6ln92
xxx M1
=
)9(69ln9
292
)1(6)1ln(9
212
M1
= 78.107 unit3 A1 [12]
12. Given a curve 22
x
x
eey .
(a) Show that .12 2ydxdy
[5 marks]
(b) Show that 0dxdy for all values of x. [3 marks]
(c) Find y
x lim and y
x lim . [2 marks]
(d) Sketch the graph of 22
x
x
eey . [3 marks]
Explain how the number of roots of the equation 22)12(
x
x
eexk ,depends on k.
[2 marks]
Solution:
(a) 2)2( xx eye B1
xxx edxdyeye )2( M1
121
dxdy
ey x
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CONFIDENTIAL*/SULIT* 13
954/1 STPM TRIAL 2011 [Please turn over
1
122
21
dxdy
yyy M1
1111
dxdy
yyy
11
2
dxdy
yy M1
ydxdy
y
1
12
212 ydxdy
A1
(b) 22 2
22
eeeee
dxdy xxxx
M1
= 22
4
x
x
ee A1
0dxdy for all values of x. A1
(c) 122lim 2
ee x
x A1
122lim 2
ee x
x A1
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CONFIDENTIAL*/SULIT* 14
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(d)
curve G1 x = ln 2 G1 point (0,-3) and label G1 Sketch y = k(2x – 1) on the same axes. M1 2 roots when k > 0 and no root when k 0. A1 [15]
y
x x=ln2
y =1
y=-1 O
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RESTRICTED* PERCUBAAN STPM 2011
954/2 1 RESTRICTED
This document is specially for the use of authorized examiner only.
954/2 PERCUBAAN
STPM 2011
MARKING SCHEME
MATHEMATICS T ( MATEMATIK T)
954/2
PAPER 2 ( KERTAS 2)
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1. The variables x and y are connected by
ye
x
dx
dy
where x ≥ 0. Find y in terms of x given y = 0 when x = 0. [5 marks]
Solution:
√ dy = √ dx
∫
dy = ∫
dx B1
2
=
+ c M1
x = 0, y = 0 c = 2 M1
2
=
+ 2
= ln (
+ 1 ) M1
y = 2 ln (
+ 1 ) A1 (5 marks)
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2. Show that tan x + cot x = 2 cosec 2x [ 3 marks] Hence, solve the equation 2 cosec 2x = 3 tan x + 1 for 0 ≤ x ≤ 2π giving your answers in terms of π. [ 4 marks]
Solution: LHS = tan x + cot x
= x
x
cossin +
x
x
sincos
= xx
xx
cossincossin 22 M1
= xx cossin
1
= xxcossin2
2 M1
= 2 cosec 2x A1 (3 marks) 2 cosec 2x = 3 tan x + 1
tan x + cot x = 3 tan x + 1 B1
tan 2 x + 1 = 3 tan 2 x + tan x
2 tan 2 x+ tan x – 1 = 0 M1
( 2 tan x – 1) ( tan x +1)= 0
tan x = 21 or tan x = –1 M1
x =0.148 π , 1.148 π or x = 4
3 , 4
7
Solution : x = 0.148 π , 43 ,1.148 π ,
47 A1 ( 4 marks)
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3. The diagram below shows ABCD is a cyclic quadrilateral. The tangent to the circle at A is parallel to the line BC. Prove that triangle ABC is an
isosceles triangle. [3 marks]
The lines AD and BC extended meet at the point E. Prove that AB is a tangent to the
circle passing through the points B, D and E. [4 marks]
Solution :
ACB = XAC (alternate angles, AX // BC) B1 ABC = XAC (alternate segment theorem) B1 ACB = ABC triangle ABC is an isosceles triangle. (Base angles equal) B1 (3 marks) DEB = XAD (alternate angles, AX // BC) B1 ABD = XAD (alternate segment theorem) B1 ABD = DEB Since any three points can determine a unique circle passing through them there exist a circle passing through the points B, D and E B1 The line AD is tangent to the circle passing through the points B, D and E. ( because angle between the tangent and the chord equals to the angle opposite the chord) B1 (4 marks)
E
A
C
D
B
E
A
C
D
B
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4. Using the substitution v = x + y, show that the equation dy
dx =
x + y + 2x + y + 1 can be reduced to
dv
dx =
2v + 3v + 1 . [3 marks]
Hence, show that 2y–2x – ln(2x + 2y + 3) + 10 = 0 is the particular solution that satisfies
x = 2 and y = – 3. [5 marks]
Solution: v = x + y
dvdx = 1 +
dydx
dydx =
x + y + 2x + y + 1 reduced to
dvdx – 1 =
v + 2v + 1
dvdx =
2v + 3v + 1 A1
v + 12v + 3 dv =
dx
12
1 – 1
2v + 3 dv =
dx
v – 12 ln(2v + 3) = 2x + c
when x = 2, y = –3, c = – 5
the particular solution is:
2y –2x – ln(2x + 2y + 3) + 10 = 0.
B1
M1
B1
M1
A1
M1
M1
(3 marks)
(5 marks)
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5. The diagram below shows that ABT is a straight line parallel to DC. CT is a tangent to the circle, centre O. If AOD = 2BTC, prove that
(a) AC = CT, [5 marks]
(b) BT = BC, [2 marks]
(c) BCT and DAC are congruent, [2 marks]
(d) DB parallel to CT. [2 marks]
Solution:
(a) Let BTC = AOD = 2 (Given)
Let ACD= AOD = 2 ( at centre = 2× at circumference)
=
Let CAT= = (alt. s , AB // DC)
=
ACT is isosceles (Base s are equal)
AC = CT
B1
B1
B1
B1
B1 (5 marks)
C
D
T B
A
O
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(b) Let BCT= = ( between chord and tangent = in alt segment)
= since = from (a)
BC = BT (BCT is isosceles)
(c) Let TBC = b and ADC= d.
then b = d (ext of cyclic quadrilateral = int. opp. )
= and AC = CT (from (a))
BCT and DAC are congruent (ASA)
(d) AOD = 2×ABD ( at centre = 2× at circumference)
ABD = = BTC
DB // CT (corresponding s are equal)
B1
B1
B1
B1
B1
B1
(2 marks)
(2 marks)
(2 marks)
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6. An aircraft leaves P to fly to Q which is 200 km due North of P. The pilot sets a course due North but after 20 minutes he realizes that, owing to a wind blowing from East, the plane is at a point R, where R is 80 km from P and 26 km west of the line joining P and Q. Find
(a) the speed, in km h-1, of the wind. [2 marks] (b) the speed, in km h-1, of the aircraft in still air. [2 marks] (c) the course the pilot should have set from P in order to have arrived directly at Q.
[2 marks]
Find also the course the pilot should set from R in order to fly directly to Q and, in this case, the time, to the nearest minute, taken for the journey from P to Q via R. [7 marks]
Solution :
(a) sin =
=
M1
u = 78 kmj-1 A1 (2 marks)
(b) V = √ M1 = 226.97 kmj-1 A1 (2 marks)
(c)
sin =
M1
= 020.1 A1 (2 marks)
tan =
= 78.19
v 240
u
75.66
124.34 26
80
d
α
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=
M1
= 19.660
course = 19.660 + 11.810 M1
= 031.5o A1
= 58.53o
=
M1
w = 197.77 km/h
d = √ = 127.03 km/h M1
t =
= 38.5 min M1
total time = 20 + 38.5 59 minutes A1 (7 marks)
226.97
101.81
11.81
78
W
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7. 10 students are seated at random in a straight line. Find the probability that
(a) three specific students sit next to each other. [ 2 marks]
(b) two brothers in the group do not sit next to each other. [ 2 marks]
Solution:
(a) P( 3 specific students sit next to each other) = !
!!
1038 M1
= 151 A1 ( 2 marks)
(b) P ( 2 brothers do not sit next to each other)
= 1 –P ( the 2 brothers sit next to each other)
= 1– !10!2!9 M1
= 54 A1 (2 marks)
8. The random variable X is normally distributed with mean and standard deviation . It
is known that P( X > 30 ) = 0.6 and P( X > 38 ) = 0.2. Find the value of and .
[5 marks]
Solution: XN(, 2)
P(X > 30) = 0.6 – 0.253 = 30 –
P(X > 38) = 0.2 0.842 = 38 –
= 7.306
= 38 – 0.842(7.306)
= 7.31, = 31.8 (3 s. f.)
A1
B1M1
(5 marks)
M1
A1
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9. A discrete random variable X takes the integer value x with probability P(x) defined by
otherwise,06,5,4),9(3,2,1),2(
)( xxk
xxk
xP
(a) Find the value of k, [2 marks]
(b) Determine the mean and variance of the probability distribution and hence state the
mean and variance of the variable Y where Y = 4X – 2. [6 marks]
Solution: (a) ( ) 1x
P x
24 1
124
k
k
(b) E(X) = ( )x
xP x
= 84k = 3.5
Var(X) = E(X2) – [E(X)]2
= 352k – 3.52
= 2.42.
Y = 4X – 2
E(Y) = 4E(X) – 2 = 12
Var(Y) = 42Var(X) = 38.72
M1
M1A1
M1
B1
B1
A1 ( 2 marks)
( 6 marks)
A1
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10. The continuous random variable X has the cumulative distribution function
F(x) =
1110
0,0
x
xxk
x
,
,
(a) Find the value of k if the median of X is 91 . [2 marks]
(b) Write down an expression for the probability density function of X, f(x). [1 mark]
Hence,
(c) show that E(X) = 21 . [3 marks]
(d) find Var (X). [3 marks]
Solution :
(a) k√ =
M1
k = A1 (2 marks)
(b) f(x) =
√ , 0 < x < 1 B1 (1 mark)
(c) E(X) = ∫ (
√ )
M1
= [
]
M1
= A1 (3 marks)
(d) E(X2) = ∫ (
√ )
= [
]
=
M1
Var(X) =
– ( )2 M1
=
A1 (3 marks)
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11. During each working day in a certain factory a number of accidents occur independently
according to a Poisson distribution with mean 0.5.
Calculate the probability that
(a) during any one day , there are two or more accidents, [3 marks]
(b) during two consecutive days there are exactly three accidents altogether. [3 marks]
By using a suitable approximation, calculate the probability that there are 3 weeks that
are accident –free, out of 50 consecutive five –day weeks . [5 marks]
Solution:
(a) X Po( 0.5) : P ( X ≥2) = 1 – 5.0e – 5.0
21 e
M1
= 1 – 5.0
23 e
M1
= 0.0902 A1 (3 marks)
(b) X Po() where = 0.5 x 2 = 1 B1
P ( X = 3) = !31 31e = 0.0613 M1A1 (3 marks)
X Po() where = 5 x 0.5 = 2.5
P ( no accident in a specific 5- day working week)
= P ( X = 0 )
= 5.2e or 0.0821 B1
Y = number of 5-day working weeks that are accident– free , in 50 consecutive weeks.
Y B ( 50 , 5.2e ) B1( can be implied)
Approximately , Y Po( 4.11) B1
P ( Y = 3 ) =
!311.4 3
11.4e M1
= 0.190 A1 ( 5 marks)
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12. The following table shows the height ( in cm) of 400 students chosen at random.
Draw a cumulative frequency curve for this distribution. [ 3 marks]
Hence, use your cumulative frequency curve to estimate the median and semi interquartile range for the distribution. [3 marks]
Estimate the mean and standard deviation for the data in this grouped frequency distribution. [6 marks]
Solution:
median = 129 cm, Q1 = 121 cm, Q3 = 138 cm B1(any one)
Semi interquartile range = 5.812113821
cm M1A1 (3 marks)
Height ( cm) Cumulative frequency
< 100 0 < 110 27 < 120 85 < 130 215 < 140 320 < 150 370 < 160 395 < 170 400
R1 ( points correct and
uniform scale)
Cumulative frequency
( No. of students)
Height/cm
100
300
200
400
121 100 129 138 170
R1 ( smooth curve passes
through all points)
R1 ( all correct and
label on axes)
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B1( midpoints) B1(frequency)
7.129
40051880
mean M1A1
1.13400
51880400
6797600..2
devstd M1A1 ( 6 marks)
End of Marking Scheme
Midpoint,x / cm 105 115 125 135 145 155 165 frequency 27 58 130 105 50 25 5
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