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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

Compulsory Part Paper 2

Question No. Key Question No. Key
1. A 31. A
2. B 32. C
3. A 33. A
4. C 34. D
5. D 35. B

6. C 36. A
7. C 37. D
8. A 38. A
9. D 39. D
10. C 40. B

11. A 41. B
12. B 42. C
13. B 43. D
14. C 44. B
15. A 45. D

16. B
17. B
18. C
19. A
20. C

21. B
22. D
23. C
24. C
25. D

26. B
27. D
28. A
29. B
30. D

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

Solutions to Paper 2
1. A 6. C
m2 – 4m + 12n – 9n2 A. y-intercept = 25
= m2 – 9n2 – 4m + 12n ∴ A is not true.
= (m + 3n)(m  3n)  4(m  3n) B. Substitute x = 3 and y = 22 into
= (m  3n)(m + 3n  4) y = 25 + 4x  x2.
R.H.S. = 25 + 4(3)  (3)2
2. B =4
 1   1   L.H.S.
85 n  3  3n  1  = (23 )5 n  3  5 3n  1 
 32   (2 )  ∴ The graph does not pass through the
 1  point (3 , 22).
= 215 n  9  15 n  5  ∴ B is not true.
2 
=2 15n + 9  (15n + 5) C.  = 42  4(1)(25)
= 24 = 116
= 16 >0
∴ The graph cuts the x-axis at two
3. A distinct points.
p q ∴ C is true.
=
a  b b 1 D. Coefficient of x2 = 1 < 0
p(b + 1) = q(a  b) ∴ The graph opens downward.
pb + p = qa  qb ∴ D is not true.
pb + qb = qa  p ∴ The answer is C.
b(p + q) = qa  p
qa  p 7. C
b=
pq ∵ f(x) is divisible by x + k.
∴ f(k) = 0
4. C (k)3 + k(k)2  4(k)  8 = 0
4k = 8
5. D k=2
3 2
f(2n – 1) + f(1) ∴ f(x) = x + 2x  4x  8
= 5(2n – 1)2 – 4(2n – 1) + 1 + 5(1)2 – 4(1) + 1 Remainder when f(x) is divided by x  1
= 5(4n2  4n + 1)  4(2n – 1) + 3 = f(1)
= 20n2  20n + 5  8n + 4 + 3 = (1)3 + 2(1)2  4(1)  8
= 20n2  28n + 12 = 9

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

8. A 11. A
x  10 x2 kx 2
Solving 3x  2 > : Since z  , we have z = , where k  0.
2 y y
6x  4 > x + 10
k[ x(1  10%)]2
7x > 14 New value of z =
y (1  20%)
x < 2 ........................ (1)
k (0.81x 2 )
Solving 9  2x > 1: =
1.2 y
2x > 8
0.675kx 2
x < 4 ........................ (2) =
y
∵ x must satisfy (1) or (2).
∴ The solution is x < 2. Percentage change in z
0.675kx 2 kx 2

9. D y y
= 2
 100%
Interest rate per quarter kx
6% y
=
4 kx 2
(0.675  1)
= 1.5% =
y
 100%
Total number of quarters in 4 years kx 2
=44 y
= 16 = 0.325  100%
Interest = 32.5%
 16
 ∴ z is decreased by 32.5%.
 1.5 
= $ 30 0001    30 000
  100  
12. B
= $8 070, cor. to the nearest dollar
Let T(n) be the number of dots in the nth
pattern.
10. C
T(1) = 2
Let x cm2 be the area of the farm on the map.
2
T(2) = T(1) + 4 = 2 + 4 = 6
x cm 2  1  T(3) = T(2) + 4 = 6 + 4 = 10
2
= 
50 000 m  25 000  T(4) = T(3) + 4 = 10 + 4 = 14
2
x cm 2  1  T(5) = T(4) + 4 = 14 + 4 = 18
2
= 
50 000  100  100 cm  25 000  T(6) = T(5) + 4 = 18 + 4 = 22
x = 50 000  100  100  T(7) = T(6) + 4 = 22 + 4 = 26
2
 1  T(8) = T(7) + 4 = 26 + 4 = 30
 
 25 000  ∴ The number of dots in the 8th pattern is
= 0.8 30.
= 8  101
∴ The required area is 8  101 cm2.

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

13. B 14. C
Maximum absolute error 18 cm
1
=  1 cm
2 15 cm d cm
= 0.5 cm
Lower limit of the length of BC
27 cm
= (12  0.5) cm
Let d cm be the height of the trapezium.
= 11.5 cm
d2 + (27  18)2 = 152
Lower limit of the length of CD
= (6  0.5) cm d = 152  9 2
= 5.5 cm = 12
(18  27)  12
Upper limit of the length of AF 30  30  30   h = 21 600
2
= (4 + 0.5) cm
27 000  270h = 21 600
= 4.5 cm
270h = 5 400
Upper limit of the length of EF
h = 20
= (3 + 0.5) cm
= 3.5 cm
15. A
Actual area of the hexagon
5
 1  FC = BC
> 11.5  5.5   4.5  3.5  cm2 1 5
 2  5
= 55.375 cm 2 =  24 cm
1 5
Upper limit of the length of BC = 20 cm
= (12 + 0.5) cm OC = AB = 32 cm
= 12.5 cm A B
Upper limit of the length of CD
= (6 + 0.5) cm E
F
= 6.5 cm
Lower limit of the length of AF
= (4  0.5) cm
= 3.5 cm O C G D
Lower limit of the length of EF Let G be a point on CD such that EG  CD.
= (3  0.5) cm Join OB and EG.
= 2.5 cm In △OBC,
Actual area of the hexagon OB = OC 2  BC 2
 1 
< 12.5  6.5   3.5  2.5  cm2 = 322  242 cm
 2 
2
= 40 cm
= 76.875 cm
OE = OB = 40 cm
∴ The range of values of x is
EG = FC = 20 cm
55.375 < x < 76.875.
In △OEG,
EG
sin EOG =
OE
20
=
40
EOG = 30
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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

The required area Alternative method:

30 A B
= (40)2  cm2
360
= 419 cm2, cor. to the nearest cm2

16. B F
∵ AB : DC = 4 : 5
4 D
∴ AB = DC E C
5
Join CF.
∵ DE : EC = 3 : 2
∵ AB : DC = 4 : 5
3 3
∴ DE = DC = DC 4
3 2 5 ∴ AB = DC
5
4
AB
DC
4 ∵ DE : EC = 3 : 2
∴ = 5 = 3 3
DE 3 3 ∴ DE = DC = DC
DC
5 3 2 5
∵ △ABF ~ △EDF (AAA) 4
DC
AB 4
AF AB 4 ∴ = 5 =
∴ = = DE 3 3
EF ED 3 DC
5
Area of  AFD AF
= ∵ △ABF ~ △EDF (AAA)
Area of  DFE EF
AF BF AB 4
108 cm 2 4 ∴ = = =
= EF DF ED 3
Area of DFE 3 Area of  AFD AF
=
Area of △DFE = 81 cm2 Area of  DFE EF
3 108 cm 2 4
∵ DE = DC =
5 Area of DFE 3
DE 3
∴ = Area of △DFE = 81 cm2
DC 5
Area of  ADE Area of  DFE DE
=
DE =
Area of  BDC DC Area of CFE EC

( 108  81) cm 2 3 81 cm 2 3
= =
Area of BDC 5 Area of CFE 2
Area of △BDC = 315 cm2 Area of △CFE = 54 cm2
∴ Area of BCEF Area of FBC BF
=
Area of DFC DF
= area of △BDC  area of △DFE
Area of  FBC 4
= (315  81) cm2 2
=
(81  54) cm 3
= 234 cm2
Area of △FBC = 180 cm2
∴ Area of BCEF
= area of △CFE + area of △FBC
= (54 + 180) cm2
= 234 cm2

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

17. B 20. C
∵ AE = DE ACD = 
∴ DAE = ADE In △ACD,
In △ADE, CD
cos  =
DAE + ADE = 90 AC
2ADE = 90 x
AC =
ADE = 45 cos 
CAB = 
CBD = ADE = 45
ABC + ADC = 180
ABE + CBD + BCD = 180
ABC + 90 = 180
36 + 45 + BCD = 180
ABC = 90
BCD = 99
In △ABC,
BC
18. C sin  =
AC
∵ △ABC is an equilateral triangle.
BC = AC sin 
∴ AB = BC = CA and
x sin 
ABE = BCD = BAD = 60. =
cos 
In △ABD and △CAE,
AB = CA
21. B
AD = CE
∵ BC is the diameter of the semi-circle.
BAD = ACE
∴ BDC = 90
∴ △ABD  △CAE (SAS)
In △BCD,
∴ ABD = CAE = 10
BCD + BDC + CBD = 180
In △ABD,
54 + 90 + CBD = 180
BDC = ABD + BAD
CBD = 36
= 10 + 60
Let DFO = x.
= 70 In △BFO,
BOF + CBD = DFO
19. A
BOF + 36 = x
BC = AD = 12 cm
BOF = x  36
In △ADE and △ABC,
∵ AE = EO
DAE = BAC
∴ DAB = BOF = x  36
ADE = ADC 1 1
= ABC ADB = BOF = (x  36)
2 2
AED = 180  DAE  ADE In △ABD,
= 180  BAC  ABC DAB + ADB = CBD
= ACB 1
x  36 + (x  36) = 36
∴ △ADE ~ △ABC (AAA) 2
AD DE 3
∴ = (x  36) = 36
AB BC 2
12 cm DE x  36 = 24
=
18 cm 12 cm x = 60
12  12 ∴ DFO = 60
DE = cm
18
= 8 cm

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

22. D 24. C
I. Let x be each exterior angle of the I. L1: 5x + py = q
polygon. py = 5x + q
Then each interior angle of the polygon is 5 q
y=  x+
2x. p p
x + 2x = 180 L2: rx + 2y = s
3x = 180 2y = rx + s
x = 60 r s
y=  x+
∴ Each exterior angle of the polygon is 2 2
60. From the figure,
∴ I is true. slope of L1 > 0
II. Number of sides of the polygon 5
 >0
360 p
=
60 p<0
=6 Slope of L2 < slope of L1
∴ The polygon is a regular hexagon. r 5
 <
∴ The number of folds of rotational 2 p
symmetry of the polygon is 6. r 5
>
∴ II is true. 2 p
III. Sum of the interior angles of the polygon pr < 10
= (6  2)  180 (or 2  60  6) ∴ I is true.
= 720 II. Substitute y = 0 into 5x + py = q.
∴ III is true. 5x + p(0) = q
∴ I, II and III are true. q
x=
5
23. C q
∴ x-intercept of L1 =
Let O be the pole. 5
Substitute y = 0 into rx + 2y = s.
In △OPQ,
rx + 2(0) = s
OP2 + OQ2 = 92 + 122 = 225
s
PQ2 = 152 = 225 x=
r
∴ POQ = 90
s
∴  = 100  90 or 100 + 90 ∴ x-intercept of L2 =
r
= 10 or 190 From the figure,
slope of L2 > 0
r
 >0
2
r<0
x-intercept of L2 > x-intercept of L1
s q
>
r 5
5s < qr
∴ II is not true.

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

III. From the figure, 27. D

y-intercept of L1 < 1 I. 3x2 + 3y2 – 24x + 18y + 26 = 0
q 26
< 1 x2 + y2  8x + 6y + =0
p 3
q > p Coordinates of the centre
p+q>0  8 6
=  , 
∴ III is true.  2 2
∴ Only I and III are true. = (4 , 3)
∴ I is true.
25. D II. Radius
Substitute x = 0 and y = 8 into hx + ky + 24 = 0. 26
= 42  (3) 2 
h(0) + k(8) + 24 = 0 3
8k = 24 49
=
k=3 3
∵ L1  L2 Area of the circle
2
∴ Slope of L1  slope of L2 = 1  49 
=  
h  3   3 
   = 1  
k  4 49 π
h 3 =
  = 1 3
3 4 > 16
h=4 ∴ II is true.
Substitute y = 0 and h = 4 into III. Distance between the origin and the centre
3x  4y  15h = 0.
= (4  0) 2  ( 3  0) 2
3x  4(0)  15(4) = 0
=5
3x = 60
x = 20 49
>
∴ The x -intercept of L2 is 20. 3
∴ The origin lies outside the circle.
∴ III is true.
26. B ∴ I, II and III are true.
Let (x , y) be the coordinates of P.
∵ PA = 9
28. A
∴ [ x  (4)]2  ( y  2) 2 = 9 The required probability
(x + 4)2 + (y  2)2 = 81 15  5
=
x2 + 8x + 16 + y2  4y + 4 = 81 8  10  12  15  5
2
x2 + y2 + 8x  4y  61 = 0 =
5
∴ The equation of the locus of P is
x2 + y2 + 8x  4y  61 = 0.

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

29. B 31. A
abc 20B00CE0000016
=9
3 = 2  1611 + 11  169 + 12  166 + 14  165
a + b + c = 27 = 512  169 + 11  169 + 192  165 + 14  165
∵ a, b and c are positive numbers. = 523  169 + 206  165
∴ a, b and c are less than 27.
For the nine numbers 6, 27, 28, 29, 31, 32, a, b 32. C
and c, when they are arranged in ascending I. ∵ a>b>0
order, the 5th number is 27. ∴ ak > bk
∴ h = 27 ∴ I must be true.
6  27  28  29  31  32  a  b  c II. Since k > 1, log k > 0.
k=
9 When a > b > 1,
6  27  28  29  31  32  27 log a > log b > 0
=
9 1 1
<
= 20 log a log b
log k log k
<
30. D log a log b
(20  x)  (20  x) loga k < logb k
Q1 = = 20 + x
2 When a > 1 and 0 < b < 1,
33  (30  y ) 63  y
Q3 = = log a > log b
2 2
1 1
∵ Inter-quartile range > 14 > (∵ log a > 0 and log b < 0.)
log a log b
63  y
∴  (20 + x) > 14 log k log k
2 >
log a log b
63 + y  40  2x > 28
y  2x > 5 loga k > logb k
∵ x and y are non-negative integers. ∴ II may not be true.
∴ When y = 6, x = 0. III. ∵ a > b > 0
When y = 7, x = 0. a
∴ >1
When y = 8, x = 0 or 1. b
∵ k>1
When y = 9, x = 0 or 1.
a
∴ 0  x  1 and 6  y  9. log
a b >0
∴ I is not true and II is true. ∴ logk =
b log k
The range of the distribution is the greatest
∴ III must be true.
when x = 0 and y = 9.
∴ Only I and III must be true.
∴ Range
 (40 + 9)  (10 + 0)
= 39
∴ III is true.
∴ Only II and III are true.

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

33. A 35. B
log8 y  1 0 1 The shaded region in the figure represents the
=
log8 x  (1) 2  (1) solutions of the given system of inequalities.
log8 y  1  1 y
=
log8 x  1 3
3 log8 y  3 =  log8 x  1 A B
y=8
3 log8 y =  log8 x + 2
log8 y3 = log8 x1 + log8 82 x  3y = 4
log8 y3 = log8 64x1 D C
∴ y3 = 64x1
1 x
 O
y = 4x 3
2x + y = 20
x=2
∴ k=4
The coordinates of A are (2 , 8).
Alternative method:
Substitute y = 8 into 2x + y = 20.
y = kxa
2x + 8 = 20
log8 y = log8 (kxa)
2x = 12
log8 y = log8 k + log8 xa
x=6
log8 y = log8 k + a log8 x ............. (1)
∴ The coordinates of B are (6 , 8).
Substitute log8 x = 2 and log8 y = 0 into (1).
2 x  y  20 ................ (1)
0 = log8 k + 2a ............................. (2) 
Substitute log8 x = 1 and log8 y = 1 into (1).  x  3 y  4 ................ (2)
(1)  (2)  2: 7y = 28
1 = log8 k  a ............................... (3)
y=4
(3)  2 + (2): 2 = 3 log8 k
2 Substitute y = 4 into (2).
log8 k = x  3(4) = 4
3
2 x=8
k = 83 ∴ The coordinates of C are (8 , 4).
=4 Substitute x = 2 into x  3y = 4.
2  3y = 4
34. D
3y = 6
ki10  2ki11  4ki12 k (1)  2k (i )  4k (1) y=2
=
1 i 1 i ∴ The coordinates of D are (2 , 2).
3k  2ki 1  i
=  At A(2 , 8), 3y  7x + 1 = 3(8)  7(2) + 1 = 11.
1 i 1 i
At B(6 , 8), 3y  7x + 1 = 3(8)  7(6) + 1 = 17.
3k  2ki  3ki  2ki 2
= At C(8 , 4), 3y  7x + 1 = 3(4)  7(8) + 1 = 43.
12  i 2
3k  5ki  2k (1) At D(2 , 2), 3y  7x + 1 = 3(2)  7(2) + 1 = 7.
= ∴ The least value of 3y  7x + 1 is 43.
1  ( 1)
k 5k
=  i
2 2
5k
∴ Imaginary part = 
2

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MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

36. A 37. D
I. For n  2, The graph of y = f(x) is reflected in the x-axis
an = 3n2 + 8n  [3(n  1)2 + 8(n  1)] to become the graph of y = f(x). The
= 3n2 + 8n  (3n2  6n + 3 + 8n  8) y-intercept of the graph of y = f(x) is 6. The
= 6n + 5 y-intercept of the graph of y = g(x) is 2. The
Let ak = 95. graph of y = g(x) can be obtained by translating
6k + 5 = 95 the graph of y = f(x) downward by
6k = 90 4 (= 6  2) units.
k = 15 ∴ g(x) = f(x)  4
∵ k is a positive integer.
∴ 95 is a term of the sequence. 38. A
∴ I is true. In △ABC, by the sine formula,
II. a1 = 3(1)2 + 8(1) a b
=
= 11 sin A sin B
= 6(1) + 5 a sin B
b=
∴ For all positive integers n, sin A
5a
an = 6n + 5. =
4
For n  2,
a c
an  an  1 = 6n + 5  [6(n  1) + 5] =
sin A sin C
= 6n + 5  6n + 1 a sin C
c=
=6 sin A
∴ The sequence is an arithmetic 6a
=
sequence. 4
∴ II is true. 3a
=
III. a1, a3, a5, …, a2n  1, … form an arithmetic 2
sequence with first term 11 and common By the cosine formula,
difference 12. b2  c2  a 2
cos A =
Note that 2 019 = 2(1 010)  1. 2bc
2 2
i.e. a2 019 is the 1 010th term of the  5a   3a  2
    a
4 2
sequence a1, a3, a5, …. =    
∴ a1 + a3 + a5 + … + a2 019   3a 
5 a
2  
1 010  4  2 
= [2(11)  (1 010  1)(12)] 45 2
2 a
= 6 125 650 = 16
15 2
= 6.125 65  106 a
4
> 7  105
3
∴ III is not true. =
4
∴ Only I and II are true.

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OXFORD UNIVERSITY PRESS
MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

39. D
∵
 
BCD : DE = BAD : DAE
E M H 45 72
∴ =
6 DAE
3 72
F G =
2 DAE

14 cm ∵
 
DAE = 48
ABCD : BCD = AED : BAD
3  4  5 AED
∴ =
D C 45 72
4 AED
6 cm =
3 72
A 8 cm B AED = 96
Join FM. In △ADE,
The required angle is AMF. ADE + AED + DAE = 180
EF = BC = 6 cm ADE + 96 + 48 = 180
1 1 ADE = 36
EM = AB =  8 cm = 4 cm
2 2 TAE = ADE
In △FEM, = 36
FM2 = EF2 + EM2
FM = EF 2  EM 2 41. B
= 2 2
6  4 cm y
= 52 cm
AF B(0 , 40)
tan AMF =
FM
14
=
52 E
AMF = 63, cor. to the nearest degree F
∴ The required angle is 63. x
O D A(30 , 0)
40. B Suppose OA, AB and OB touch the inscribed
D circle of △OAB at D, E and F respectively.
Let r be the radius of the circle.
Note that OD = OF = r and the coordinates of
C P are (r , r).
In △OAB,
E
AB2 = OA2 + OB2
AB = OA2  OB 2
B = 302  40 2
T = 50
A
Join AD.
BCD + BAD = 180
108 + BAD = 180
BAD = 72

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OXFORD UNIVERSITY PRESS
MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

AD = OA  OD = 30  r 43. D
AE = AD = 30  r The multiples of 4 between 1 and 52 inclusive
BF = OB  OF = 40  r are 4, 8, 12, …, 52. There are 13 such
BE = BF = 40  r numbers.
AE + BE = AB The multiples of 13 between 1 and 52 inclusive
30  r + 40  r = 50 are 13, 26, 39, 52. There are 4 such numbers.
2r = 20 The number that is a multiple of both 4 and 13
r = 10 between 1 and 52 inclusive is 52. There is
∴ Coordinates of P = (10 , 10) 1 such number.
Substitute x = 10 and y = 10 into x + y  30 = 0. P(a multiple of 4 or a multiple of 13)
L.H.S. = 10 + 10  30 = P(a multiple of 4) + P(a multiple of 13) –
= 10 P(a multiple of both 4 and 13)
 R.H.S. 13 4 1
=  
∴ P does not lie on x + y  30 = 0. 52 52 52
Substitute x = 10 and y = 10 into 3x  y  20 = 0. 4
=
L.H.S. = 3(10)  10  20 13
The required probability
=0
= P(Daisy gets a multiple of 4 in the 1st draw) +
= R.H.S.
P(both Daisy and Elaine get neither a multiple of 4
∴ P lies on 3x  y  20 = 0.
nor a multiple of 13 in their 1st draw) 
Substitute x = 10 and y = 10 into 4x  y  40 = 0.
P(Daisy gets a multiple of 4 in the 2nd draw) +
L.H.S. = 4(10)  10  40
P(both Daisy and Elaine get neither a multiple of 4
= 10
nor a multiple of 13 in their first 2 draws) 
 R.H.S.
P(Daisy gets a multiple of 4 in the 3rd draw) + …
∴ P does not lie on 4x  y  40 = 0.
1  4  4  1 
Substitute x = 10 and y = 10 into 5x + 4y  160 = 0. =  1  1    
4  13  13  4 
L.H.S. = 5(10) + 4(10)  160
 4  4  4  4  1 
= 70 1  1  1  1     
 R.H.S.  13  13  13  13  4 
2 4
∴ P does not lie on 5x + 4y  160 = 0. 1  9  1  9  1
=           
∴ The answer is B. 4  13   4   13   4 
1
42. C = 4
2
9
Number of ways of selecting 2 boys and 3 girls 1  
 13 
= C216  C312
169
Number of ways of selecting 1 boy and 4 girls =
352
= C116  C 412
Number of ways of selecting 5 girls
= C512
∴ The required number of teams
= C 216  C312 + C116  C 412 + C512
= 35 112

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OXFORD UNIVERSITY PRESS
MOCK 18(I) COMPULSORY PART PAPER 2 SOLUTION

44. B
Let x and  be the mean and the standard
deviation of the scores of the test respectively.
54  x
= 1.5

54  x = 1.5 .............................. (1)
65  x
= 1.25

65  x = 1.25 .............................. (2)
(2)  (1): 11 = 2.75
=4
∴ The standard deviation of the scores of the
test is 4.

45. D
Let d be the common difference of the
sequence, where d  0.
2x3 = 2(x1 + 2d)
2x6 = 2(x4 + 2d)
2x9 = 2(x7 + 2d)
∴ 2d is added to each of the three numbers
x1, x4 and x7, and each resulting number is
then multiplied by 2 to obtain 2x3, 2x6 and
2x9.
∴ v2 = 22  v1
v1 1
=
v2 4

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