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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

Compulsory Part Paper 2

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

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

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

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

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

26. C
27. B
28. C
29. C
30. B

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

Solutions to Paper 2
1. B 6. A
6 648 (2  3) 648 ∵ a<0
= ∴ The graph opens downward.
16162 (2 4 )162
y-intercept of the graph = b < 0
2 648  3648 ∴ The answer is A.
=
2 648
= 3648 7. C
Alternative method: Rewrite the compound inequality as
6 648 6 648 2x  5  5x + 1 and 5x + 1 < 16.
=
16162 (2 4 )162 Solving 2x  5  5x + 1:
3x  6
6 648
= x  2 ................ (1)
2 648 Solving 5x + 1 < 16:
648
6 5x < 15
=  x < 3 .................. (2)
2
= 3648 ∵ x must satisfy (1) and (2).
∴ The solutions are 2  x < 3.
2. D
a b3 8. B
 =2 f(k) = 0
c b 2
(k + k)(k + 1) = 0
a 3
 1 = 2 k(k + 1)(k + 1) = 0
c b k = 0 (rejected) or 1
a 3 2
∴ f(x) = (x  1)(x + 1)
3=
c b Remainder = f(2)
a  3c 3 = [(2)2  1](2 + 1)
=
c b = 3
3c
b=
a  3c 9. C
Let x and y be the numbers of boys and girls in
the class respectively.
3. A x  75% + y  50% = (x + y)  65%
a2 + 8ab + 16b2  9a2b2 0.75x + 0.5y = 0.65x + 0.65y
= (a + 4b)2  (3ab)2 0.1x = 0.15y
= [(a + 4b) + 3ab][(a + 4b)  3ab] x 3
=
= (a + 3ab + 4b)(a  3ab + 4b) y 2
∴ The ratio of the number of boys to the
4. D number of girls is 3 : 2.

5. C 10. A
Rewrite the given equation as: Cost = $[70(1  40%) + 8]
 3r  5s  22  1 .......................... (1) = $50
 Profit = $(70  50)
4r  3s  1 ................................. (2)
= $20
(1)  4: 12r  20s + 88 = 4
12r  20s = 84 ...... (3)
(2)  3: 12r + 9s = 3 .................... (4)
(4)  (3): 29s = 87
s=3

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

11. D 13. A
a:b=3:1 Maximum error of the measurement
1 1
b= a =  1 cm
3 2
b:c =2:3 = 0.5 cm
3 Least possible volume of the sphere
c= b
2 4
= (5  0.5)3 cm3
31  3
=  a 243
23  = π cm 3
1 2
= a
2
a + 2b  3 c = 3 14. D
Let T(n) be the number of dots in the nth
1  1  pattern.
a + 2 a  3 a = 3
3  2  T(1) = 11
1 T(2) = T(1) + 6 = 11 + 6 = 17
a =3
6 T(3) = T(2) + 6 = 17 + 6 = 23
a = 18 T(4) = T(3) + 6 = 23 + 6 = 29
T(5) = T(4) + 6 = 29 + 6 = 35
12. A T(6) = T(5) + 6 = 35 + 6 = 41
k2 T(7) = T(6) + 6 = 41 + 6 = 47
From the question, f(x) = k1x + , where k1 T(8) = T(7) + 6 = 47 + 6 = 53
x2
∴ The number of dots in the 8th pattern is
and k2 are non-zero constants.
53.
f(2) = 5
k 15. B
k1(2) + 22 = 5
2 In △ABC, by Pythagoras’ theorem,
8k1 + k2 = 20 .................. (1) AC2 = AB2 + BC2
f(3) = 2
AC = 16 2  30 2 cm
k
k1(3) + 22 = 2 = 34 cm
3 ∵ △ABC ~ △CDE (AAA)
27k1 + k2 = 18 .................... (2)
AB AC
(2)  (1): 19k1 = 38 ∴ =
k1 = 2 CD CE
Substitute k1 = 2 into (1). 16 cm 34 cm
=
8(2) + k2 = 20 12 cm CE
k2 = 36 CE = 25.5 cm
36 Perimeter of ACEF
∴ f(x) = 2x  2 = 2(AC + CE)
x
= 2(34 + 25.5) cm
36 = 119 cm
f(1) = 2(1)  2
1
= 34

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

16. A Area of AEF  AF 

2

Join AE. = 
Area of CED  CD 
B E C 2
Area of AEF 2
2
= 
36 cm 3
10 cm Area of △ AEF = 16 cm2
Area of AEF AE
=
Area of CEF CE
A D
20 cm 16 cm 2 2
=
AE = AD = 20 cm Area of CEF 3
In △ ABE, Area of △ CEF = 24 cm2
AB Area of △ BCF
sin AEB =
AE = area of △ ACF
1 = (16 + 24) cm2
= = 40 cm2
2
∴ Area of ABCD
AEB = 30
By Pythagoras’ theorem, = area of △ ADE + area of △ CDE +
AE2 = AB2 + BE2 area of △ ACF + area of △ BCF
= (24 + 36 + 40 + 40) cm2
BE = 20 2  10 2 cm = 140 cm2
= 300 cm Alternative method:
DAE = AEB = 30 ∵ △AEF ~ △ CED (AAA)
Area of the shaded region 1
AB
= area of trapezium ADCE  AE AF 14 2
∴ = = 2 =   =
area of sector ADE CE CD CD 23 3
( EC  AD)CD DAE Area of ADE AE
=   (AD)2 =
2 360 Area of CDE CE
 (20  300  20)(10) 30  24 cm 2
2
=    π(20) 2  cm2 =
 2 360  Area of CDE 3
= 8.68 cm2, cor. to 3 sig. fig. Area of △ CDE = 36 cm2
Let G be a point on AB such that DG  AB and
17. B DG  DC.
∵ △AEF ~ △ CED (AAA) D C
1
AB
AE AF 14 2 E
∴ = = 2 =   =
CE CD CD 23 3
Area of ADE AE A B
= G F
Area of CDE CE
1
24 cm 2
2  AB  DG
Area of ABC
= = 2
Area of CDE 3 Area of ACD 1
Area of △ CDE = 36 cm2  DC  DG
2
Join CF. Area of ABC AB
D C =
Area of ACD DC
E Area of ABC 4
=
(24  36) cm 2 3
Area of △ ABC = 80 cm2
A B
F ∴ Area of ABCD
= area of △ ACD + area of △ ABC
= (24 + 36 + 80) cm2
= 140 cm2
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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

18. B 20. B
Total surface area of the solid I. Sum of the interior angles = 2 520
1  (n  2)  180 = 2 520
=  [4 π(2) 2 ]  2 π (2)(10  2)  π(2) 2  cm2
2  n  2 = 14
= 44 cm 2 n = 16
∴ The number of diagonals of the
19. D
  
∵ AB : BC : AC = 13 : 11 : 12
polygon is not 14.
∴ I is not true.
360
∴ ACB : BAC : ABC = 13 : 11 : 12 II. Each exterior angle = = 22.5
16
DEF = ABC and DFE = BAC. ∴ II is true.
In △DEF, 2 520
EDF + DEF + DFE = 180 III. Each interior angle =
16
EDF + ABC + BAC = 180 = 157.5
EDF = 180  ABC  BAC  140
= ACB ∴ III is not true.
13
= 180  ∴ Only II is true.
13  11  12
= 65 21. D
Alternative method: With the notation in the figure, F is a point on
DE and DF are produced to meet AC and BC at BC such that DF  BC.
G and H respectively. C
C

D
G F
H
E F
A B

D A E B
∵ DG // BC and DH // AC. In △ ADE,
∴ DGCH is a parallelogram. AE
cos  =
  
∴ EDF = ACB
∵ AB : BC : AC = 13 : 11 : 12
AD
AE = AD cos 
In △ CDF,
∴ ACB : BAC : ABC = 13 : 11 : 12
DF
13 sin  =
ACB = 180  CD
13  11  12 DF = CD sin 
= 65 BE = DF = CD sin 
∴ EDF = 65 AE AD cos
=
BE CD sin 
AD
=
sin 
CD 
cos
AD
=
CD tan 

22. B

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

23. B 25. A
Rectangular coordinates of the image of P  12
y-intercept of L1 =  =3
= ( 3 , 1) 4
With the notation in the figure, 6 6
y y-intercept of L2 =  =
b b
∵ L1 and L2 intersect at a point on the y-axis.
Q 6
x ∴ 3=
O  b
b=2
R( 3 , 1) a
Slope of L1 = 
4
OQ = 3 2
QR = 1 Slope of L2 = 
b
OR = OQ 2  QR 2 ∵ L1  L2
∴ Slope of L1  slope of L2 = 1
= ( 3 ) 2  12
 a  2 
=2      = 1
QR  4  b 
tan  = a = 2b
OQ
= 2(2)
1
= = 4
3
 = 30 26. C
The required polar coordinates Coordinates of the centre of the circle
= (OR , 360  )  k  12 
= (2 , 360  30) =  ,  
 2 2 
= (2 , 330)
 k 
=   , 6
24. A  2 
I. Let (x , y) be the coordinates of P. The equation of L is
∵ PA = PB y0 12  0
=
∴ [ x  (1)]2  ( y  4) 2 x 8 08
2y = 3x + 24
= [ x  ( 5)]2  ( y  2) 2 3
y =  x + 12
(x + 1)2 + (y  4)2 = (x + 5)2 + (y  2)2 2
x2 + 2x + 1 + Since L divides the circle into two equal parts,
y2  8y + 16 = x2 + 10x + 25 + L passes through the centre of the circle.
y2  4y + 4 k
Substitute x =  and y = 6 into the equation
8x  4y  12 = 0 2
2x + y + 3 = 0 of L.
∴ I must not be true. 3 k 
II must be true. 6 =     + 12
2 2
III. △PAB is an equilateral triangle only when
3k
PA = PB = AB. Otherwise, △PAB is an 6 =
isosceles triangle or APB is a straight line. 4
∴ III may not be true. k = 8
∴ Only II must be true.

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

27. B 30. B
The 1st number ∵ Mode = 8
Sum
1 2 3 4 5 6 7 8 ∴ x=8
1 3 4 5 6 7 8 9 Mean = 7
2 3 5 6 7 8 9 10 358888 y
The 2nd number

3 4 5 7 8 9 10 11 =7
7
4 5 6 7 9 10 11 12 40 + y = 49
5 6 7 8 9 11 12 13 y=9
6 7 8 9 10 11 13 14 Standard deviation
7 8 9 10 11 12 13 15
(3  7) 2  (5  7) 2  4(8  7) 2  (9  7) 2
8 9 10 11 12 13 14 15 =
The required probability 7
8 =2
=
56
31. A
1
= 1 x
7 2
 3
x  2x  3 x  27
1 x
28. C = 
Total number of banknotes in the cash box ( x  1)( x  3) 2
( x  3)( x  3x  9)
= 15 + 20 + 10 + 5 x 2  3x  9  x( x  1)
= 50 =
( x  1)( x  3)( x 2  3x  9)
Expected face value
2x  9
 15 20 10 5  =
= $ 10   20   50   100   ( x  1)( x  3)( x 2  3 x  9)
 50 50 50 50 
= $31
32. C
29. C Intercept on the vertical axis of the graph
Median = 27 =a<0
20  a  28 Slope of the graph = b > 0
= 27 log5 y = a + bx
2
y = 5a + bx
a + 48 = 54
y = 5a(5b)x
a=6
∵ a<0
Q1 = 21
Q3 = 30 + b ∴ 0 < 5a < 1 ........................ (1)
Inter-quartile range = 14 ∵ b>0
(30 + b)  21 = 14 ∴ 5b > 1 .............................. (2)
b=5 5x
Only y = satisfies (1) and (2).
3
∴ The answer is C.

33. C
250  1610 + 192  165 + 29
= (15  16 + 10)  1610 + (12  16)  165 +
16 + 13
= 15  1611 + 10  1610 + 12  166 + 1  16 +
13
= FA000C00001D16

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

34. B II. When z is a purely imaginary number,

Substitute y = 6 into x + y = 9. a2  2
x+6=9  0 and
a2  1
x=3
∴ The coordinates of P are ( , 6). a 3  2a
=0
Substitute x = 8 into x + y = 9. a2 1
8+y=9 a3 + 2a = 0
y=1 a(a2 + 2) = 0
∴ The coordinates of Q are ( , 1). a = 0 or a2 = 2 (rejected)
Let (4 , r) be the coordinates of R. When a = 0,
r 64 02  2
= z= 2 i
6 6 0 1
r=2 = 2i
∴ The coordinates of R are ( , 2). ∴ II must be true.
Let (s , 5) be the coordinates of S. a 3  2a
s 65 Real part of z 2
= III. = a2  1
6 6 Imaginary part of z a 2
s=1 a2 1
∴ The coordinates of S are ( , 5).
a ( a 2  2)
At P(3 , 6), 18 + x  4y = 18 + 3  4(6) = 3. =
At Q(8 , 1), 18 + x  4y = 18 + 8  4(1) = . a2  2
=a
At R(4 , 2), 18 + x  4y = 18 + 4  4(2) = .
∵ a can be a rational number or an
At S(1 , 5), 18 + x  4y = 18 + 1  4(5) = .
irrational number.
∴ The required minimum value = 3
∴ III is not necessarily true.
35. B ∴ Only II must be true.
1
I. z=a+i+ 36. D
a i
I. Let Sn = a1 + a2 + a3 + … + an.
ai
=a+i+ a1  a2  a3    an
(a  i )(a  i ) = bn
n
ai Sn = n(3n  8)
=a+i+ 2 2
a i = 3n2  8n
ai For n  2,
=a+i+ 2
a  (1) an = Sn  Sn  1
a 1 = 3n2  8n  [3(n  1)2  8(n  1)]
=a+i+ 2 + 2 i = 3n2  8n  (3n2  6n + 3  8n + 8)
a 1 a 1
= 6n  11
a (a 2  1)  a a2  1  1 a2 = 6(2)  11 = 1
= + i
a2 1 a2 1 a1 = S1 = 3(1)2  8(1) = 5
a 3  2a a2  2 a2  a1 = 1  (5) = 6
= 2 + 2 i
a 1 a 1 For n  3,
∵ a is a real number and a2  0. an  an  1 = 6n  11  [6(n  1)  11]
= 6n  11  (6n  6  11)
a2  2 =6
∴ 0
a2  1 ∵ a2  a1 = 6 and an  an  1 = 6 for n  3.
∴ z is not a real number. ∴ a1, a2, a3, …, an is an arithmetic
∴ I must not be true. sequence.
∴ I is true.

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

II. Let the kth term of the sequence 38. B

b1, b2, b3, … be 79. Let HM = CM = x cm.
bk = 79 Join AC.
3k  8 = 79 E
3k = 87 F H
k = 29 G
∵ k is a positive integer.
∴ 79 is a term of the sequence M
D
b1, b2, b3, ….
∴ II is true. A C
III. bn  bn  1 = 3n  8  [3(n  1)  8] 6 cm 3 cm
B
= 3n  8  (3n  3  8) In △ ABC, by Pythagoras’ theorem,
=3 AC2 = AB2 + BC2
∴ b1, b2, b3, … is an arithmetic
AC = 6 2  32 cm
sequence.
b1 + b2 + b3 + … + bn = 45 cm
n In △ ACM, by Pythagoras’ theorem,
= [3(1)  8 + (3n  8)] AM2 = AC2 + CM2
2
3n 2  13n AM = ( 45 ) 2  x 2 cm
=
2 = 45  x 2 cm
∴ III is true.
Similarly, FM = 45  x 2 cm.
∴ I, II and III are true.
AF = CH = CM + HM = (x + x) cm = 2x cm
In △ AFM, by the cosine formula,
37. C
sin2  = sin  cos  AM 2  FM 2  AF 2
cos AMF =
sin   sin  cos  = 0
2 2  AM  FM
sin  (sin   cos ) = 0 1 ( 45  x 2 ) 2  ( 45  x 2 ) 2  (2 x) 2
sin  = 0 or sin  = cos  =
9 2 45  x 2 45  x 2
When sin  = 0,
 = 0 or 180 1 45  x 2  45  x 2  4 x 2
=
When sin  = cos , 9 2(45  x 2 )
sin  1 45  x 2
= 1 (where cos   0) =
cos 9 45  x 2
tan  = 1 45 + x2 = 405  9x2
 = 45 or 225 10x2 = 360
∴ For 0   < 360, the equation has x2 = 36
4 roots. x = 6 or 6 (rejected)
CH = 2(6) cm = 12 cm
Volume of the cuboid
= AB  BC  CH
= 6  3  12 cm3
= 216 cm3

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MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

39. D 41. D
From the figure, the graph of y = h  sin kx is  x  2 y  k  0 .................................. (1)
obtained by reducing the graph of y = sin x  2 2
1  x  y  8 x  12 y  48  0 ............. (2)
to of the original along the x-axis and then From (1), x = k  2y .......................... (3)
2
Substitute (3) into (2).
translating upward by 1 unit.
(k  2y)2 + y2  8(k  2y) + 12y  48 = 0
∴ y = 1  sin 2x
k  4ky + 4y2 + y2  8k + 16y + 12y  48 = 0
2
∴ h = 1 and k = 2
5y2 + (28  4k)y + k2  8k  48 = 0
Let (x1 , y1) and (x2 , y2) be the coordinates of A
40. A and B respectively.
I. ∵ BD // PQ y1 and y2 are the roots of the equation
∴ BCP = CBD 5y2 + (28  4k)y + k2  8k  48 = 0.
BAC = BCP 28  4k
∴ BAC = CBD y1 + y2 = 
5
CAD = CBD ∴ y-coordinate of the mid-point of AB
∴ BAC = CAD y  y2
i.e. CA bisects BAD. = 1
2
∴ I must be true. 1  28  4k 
II. ∵ FB bisects ABD. =  
2 5 
∴ ABF = DBF
2k  14
In △ ABF, =
CFB = BAC + ABF 5
= CBD + DBF
= CBF 42. C
∴ FC = BC Coordinates of the mid-point of the line
∴ II must be true. segment joining (7 , 0) and (5 , 4)
III. With the notation in the figure,  7  5 0  (4) 
= , 
 2 2 
A
= (6 , 2)
Slope of the straight line passing through (7 , 0)
a a
and (5 , 4)
F 0  (4)
b =
75
b
B a E a D =2
2b Slope of the perpendicular bisector of the line
a a
P
C
Q segment joining (7 , 0) and (5 , 4)
FBC + BCD 1
= 
= b + a + BCA + 2b 2
= 3b + a + (180  2b  2a) Let (2 , a) be the coordinates of the
= 180 + b  a circumcentre.
FBC + BCD = 180 only when a = b. a  (2) 1
= 
∴ III is not necessarily true. 26 2
∴ Only I and II must be true. a+2=2
a=0
∴ The y-coordinate of the circumcentre is 0.

© Oxford University Press 2016 P.10

OXFORD UNIVERSITY PRESS
MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION

Alternative method: 45. A

Let (2 , a) be the coordinates of the I. Let m be the mean of {a, b, c, d, e}.
circumcentre. Mean of P = x + m
Distance between (2 , a) and (7 , 0) Mean of Q = y + m
= distance between (2 , a) and (5 , 4) ∵ x>y
(2  7) 2  (a  0) 2 = (2  5) 2  [a  (4)]2 ∴ x+m>y+m
i.e. Mean of P > mean of Q
25 + a2 = 9 + (a + 4)2
25 + a2 = 9 + a2 + 8a + 16 ∴ I is true.
8a = 0 II. Let r be the range of {a, b, c, d, e}.
a=0 Range of P = r
∴ The y-coordinate of the circumcentre is 0. Range of Q = r
∴ II is true.
43. C III. Let s be the standard deviation of
The required probability {a, b, c, d, e}.
Standard deviation of P = s
C 4C 5
= 1 93 Variance of P = s2
C4 Standard deviation of Q = s
20 Variance of Q = s2
= ∴ III is not true.
63
∴ Only I and II are true.

44. D
Consider the books of the same subject as one
unit.
Since there are 3 subjects, there are 3 units.
The number of ways of arranging these 3 units
is 3!.
Number of ways of arranging the
6 Mathematics books
= 6!
Number of ways of arranging the 2 Economics
books
= 2!
Number of ways of arranging the 3 Geography
books
= 3!
The required number of ways
= 3!  6!  2!  3!
= 51 840

© Oxford University Press 2016 P.11