OXFORD UNIVERSITY PRESS

MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Compulsory Part Paper 1

Solution Marks
1. 2(3a  11) = 3a  5b
6a  22 = 3a  5b 1M
3a = 22  5b 1M
22  5b
a= 1A
3
2(3a  11) = 3a  5b
3a 5b
3a  11 =  1M
2 2
3a 22  5b
= 1M
2 2
22  5b
a= 1A
3
----------(3)

m 6 n 3
2.
( m5 n  4 ) 2
m 6 n 3
= 1M
m10 n 8
n 3  ( 8)
= 1M
m10  6
n5
= 4 1A
m
----------(3)

5 4
3. 
3k  2 2k  7
5(2k  7)  4(3k  2)
= 1M
(3k  2)(2k  7)
10k  35  12k  8
= 1M
(3k  2)(2k  7)
 2k  27
= 1A
(3k  2)(2k  7)
----------(3)

4. (a) 25x2  4
= (5x + 2)(5x  2) 1A
(b) 5x2y  17xy + 6y
= y(5x2  17x + 6)
= y(5x  2)(x  3) 1A

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MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
2 2
(c) 5x y  17xy + 6y  25x + 4
= 5x2y  17xy + 6y  (25x2  4)
= y(5x  2)(x  3)  (5x + 2)(5x  2) 1M
= (5x  2)[y(x  3)  (5x + 2)]
= (5x  2)(xy  3y  5x  2) 1A
----------(4)

5x  3
5. (a) 3(x  4) 
6
18(x  4)  5x + 3
18x + 72  5x + 3
23x  69 1M
x3 1A

(b) 6x + 24 > 0
6x > 24
x > 4 1A
The integers which satisfy both inequalities are 3, 2, 1, 0, 1, 2 and 3.
∴ 7 integers satisfy both inequalities. 1A
----------(4)

6. (a) Let $x be the cost of the computer.

x(1 + 40%) = 7 000 1M
1.4x = 7 000
x = 5 000
∴ The cost of the computer is $5 000. 1A

(b) Selling price of the computer

= $7 000  (1  12%) 1M
= $7 000  0.88
= $6 160
∴ Percentage profit
6 160  5 000
=  100%
5 000
= 23.2% 1A
----------(4)

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
7. Let x and y be the present ages of Peter and Irene respectively.
x 4
 y  3 ............................................ (1)

 1A+1A
 x  7  3 ...................................... (2)
 y  7 2
4
From (1), x = y ........................... (3)
3
Substitute (3) into (2).
4
y 7
3 3
= 1M
y 7 2
8
y  14 = 3y  21
3
y
 = 7
3
y = 21
∴ The present age of Irene is 21. 1A
Let 4k and 3k be the present ages of Peter and Irene respectively, where k is a
positive constant. 1A
4k  7 3
= 1M+1A
3k  7 2
8k  14 = 9k  21
k = 7
k=7
Present age of Irene
=37
= 21 1A
----------(4)

41  47  49  50  2  (50  a)  55  2  62  70  (70  a)  2
8. (a) = 57 1M
12
669 + 3a = 684
3a = 15
a=5 1A

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
(b) Range = (75  41) kg
= 34 kg 1A
49  50
Q1 = kg = 49.5 kg
2
62  70
Q3 = kg = 66 kg
2
Inter-quartile range = (66  49.5) kg
= 16.5 kg 1A
Standard deviation = 10.7 kg, cor. to 3 sig. fig. 1A
----------(5)

9. (a) ∵ AO = AE
∴ AOE = AEO
AOE
ABD = 1M
2
In △BDE,
ABD + AEO = BDC
AOE
+ AOE = 48
2
3AOE
= 48
2
AOE = 32 1A
(b) Join OB.
BOC = 2BDC
= 2  48
= 96
BOC + AOB + AOE = 180
96 + AOB + 32 = 180 1M
AOB = 52

AB = 2    OA 
AOB
360
52
= 2    AE  1M
360
 0.907 571 211AE
< AE
∴ The claim is agreed. 1A

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks

Join OB.
∵ OB = OD
∴ OBD = ODB
= 48
In △OBD,
OBD + BOD + ODB = 180
48 + AOB + 32 + 48 = 180 1M
AOB = 52

AB = 2    OA 
AOB
360
52 1M
= 2    AE 
360
 0.907 571 211AE
< AE
∴ The claim is agreed. 1A
----------(5)

10. (a) From the question, f (x) = k1 + k2x2, where k1 and k2 are non-zero constants. 1A
f (1) = 206
k1 + k2(1)2 = 206
k1 + k2 = 206 ................. (1)
1M
f (3) = 254
for either
k1 + k2(3)2 = 254 substitution
k1 + 9k2 = 254 ................... (2)
(2)  (1): 8k2 = 48
k2 = 6
Substitute k2 = 6 into (1).
k1 + 6 = 206
k1 = 200
∴ f(x) = 200 + 6x2 1A
----------(3)

(b) f (x) = 80x

200 + 6x2 = 80x 1M
6x2  80x + 200 = 0
3x2  40x + 100 = 0
(x  10)(3x  10) = 0
10
x = 10 or 1A
3
----------(2)

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
11. (a) c+1+4+a=8+ba+c 1M
b = 2a  3
Note that a > 5 and b < 11.
When a = 6, b = 2(6)  3 = 9.
When a = 7, b = 2(7)  3 = 11 (rejected).
When a  8, b must be greater than 11.
∴ a = 6 and b = 9. 1A+1A
----------(3)

(b)(i) ∵ c > 0 and the mode is greater than 2.

∴ The least possible value of c is 1. 1A
(ii) ∵ The mode is greater than 2.
∴ c+1<8
c<7
∴ The greatest possible value of c is 6. 1A
----------(2)

(c) c=1
The required probability
(9  6)  1
= 1M
(1  1)  4  6  8  (9  6)  1
1
= 1A
6
----------(2)

12. (a) ∵ x + 1 is a factor of f (x).

∴ f (1) = 0
3 2
4(1) + (a + 2)(1) + 2(1)  3b = 0
4 + a + 2  2  3b = 0
1M
a  3b = 4 ........................ (1)
for either
∵ When f (x) is divided by x  2, the remainder is 9. one
∴ f (2) = 9
4(2)3 + (a + 2)(2)2 + 2(2)  3b = 9
32 + 4a + 8 + 4  3b = 9
4a  3b = 29 ................................ (2)
(2)  (1): 3a = 33
a = 11 1A
Substitute a = 11 into (1).
11  3b = 4
3b = 15
b=5 1A
----------(3)

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
3 2
(b) f (x) = 4x + (11 + 2)x + 2x  3(5)
= 4x3 + 13x2 + 2x  15 1M
Using long division,
4x  5
 x  2 x  3  4 x 3  13 x 2  2 x  15
2

 4 x 3  8 x 2  12 x
5 x 2  10 x  15
5 x 2  10 x  15
∴ f (x) = (x2 + 2x + 3)(4x  5)
g(x) = 4x  5
kx g(x) = f (x)
kx(4x  5) = (x2 + 2x + 3)(4x  5) 1M
(4x  5)[kx  (x2 + 2x + 3)] = 0
(4x  5)[x2 + (k  2)x  3] = 0 1M
5
x= or x2 + (k  2)x  3 = 0
4
Consider the equation x2 + (k  2)x  3 = 0.
 = (k  2)2  4(1)(3)
= (k  2)2 + 12
>0
∴ The equation x2 + (k  2)x  3 = 0 has two distinct real roots.
∴ The equation kx g(x) = f (x) has more than one real root for all real values
of k.
∴ The claim is agreed. 1A
----------(4)

13. (a) Let r cm be the radius of the metal sphere.

4r2 = 144 1M
r2 = 36
r = 36
=6
4
Volume of the metal sphere = (6)3 cm3
3
= 288 cm3 1A
----------(2)

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
(b) Let h cm be the original depth of water in the container.
Volume of the metal sphere +
volume of water in the container = capacity of the container
∴ 288 + (16)2h = (16)2(14) 1M+1M
256h = 3 296
103
h=
8
103
∴ The original depth of water in the container is cm. 1A
8
----------(3)

(c) Let R cm and  cm be the base radius and the slant height of the circular
conical vessel respectively.
2R = 48
R = 24
R  = 720
(24)  = 720
 = 30
Height of the vessel = 30 2  24 2 cm 1M
= 18 cm
1
Capacity of the vessel = (24)2(18) cm3 1M
3
= 3 456 cm3
Volume of water in the circular cylindrical container
 103  3
= (16)2   cm
 8 
= 3 296 cm3
∵ 3 456 cm3 > 3 296 cm3
∴ The water will not overflow. 1A
----------(3)

14. Marking Schemes for (a)(i) and (a)(ii):

Case 1 Any correct proof with correct reasons. 2
Case 2 Any correct proof without reasons. 1
(a)(i) In △BCE and △DCE,
BC = DC (by definition)
BCE = DCE = 45 (property of square)
CE = CE (common side)
∴ △BCE  △DCE (SAS)

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MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
(ii) In △BEG and △FEB,
BEG = FEB (common angle)
∵ △BCE  △DCE (proved in (a)(i))
∴ EBC = EDC (corr. s,  △s)
EDC = EFB (alt. s, AF // DC)
∴ EBC = EFB
i.e. EBG = EFB
BGE = 180  BEG  EBG ( sum of △)
= 180  FEB  EFB
= FBE ( sum of △)
∴ △BEG ~ △FEB (AAA)
----------(4)

(b) ∵ △BCE  △DCE

∴ BE = DE
∵ △BEG ~ △FEB
BE EG BG
∴ = = 1M
FE EB FB
BG
∵ = tan AFD and BE = DE.
FB
DE EG
∴ = = tan AFD
FE DE
i.e. DE = FE tan AFD and EG = DE tan AFD. 1M
3
Note that when 0 < AFD < 30, 0 < tan AFD < .
3
DE = FE tan AFD
= (EG + FG) tan AFD
= (DE tan AFD + FG) tan AFD
= DE tan2 AFD + FG tan AFD
2
 3  
< DE   + FG  3  1M
 3   3 
   
1 3
DE < DE + FG
3 3
2 3
DE < FG
3 3
3
DE < FG
2
∴ The claim is agreed. 1A
----------(4)

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MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
15. (a) Number of teams formed = C58  C 25 1M
= 560 1A
----------(2)

(b) Number of teams formed = C38  C 45 + C 28  C55 1M

= 308 1A
----------(2)

x2 x
16. (a) f (x) =  + 11
16 2
1 2
= (x  8x) + 11
16
1  2  8  8 
2 2
=  x  8 x       + 11 1M
16   2   2  
1 2
= (x  8x + 16)  1 + 11
16
1
= (x  4)2 + 10
16
∴ The coordinates of the vertex are (4 , 10). 1A
----------(2)

(b) g(x) = f (x) 1M

1
=  (x  4)2  10
16
Note that the graph of y = h(x) is obtained by translating the graph of y = g(x)
upward by [6  (10)] units, i.e. 16 units.
h(x) = g(x) + 16
1
=  (x  4)2 + 6 1M
16
1
h(0) =  (0  4)2 + 6 = 5
16
∴ The y-intercept of the graph of y = h(x) is 5. 1A
----------(3)

17. (a) (x + 2)(x  2) = 8(x  1)

x2  4 = 8x  8
x2  8x + 4 = 0
8
∴ p=  =8 1A
1
4
q= =4 1A
1
----------(2)

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MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
log 8
(b) Common ratio = 1M
log 4
log 2 3
=
log 2 2
3 log 2
=
2 log 2
= 1.5
∴ The general term Tn = (log 4)1.5n  1
T + 1 + T2 + 1 < log 22 020
(log 4)1.5( + 1)  1 + (log 4)1.5(2 + 1)  1 < log 22 020 1M
(2 log 2)1.5 + (2 log 2)1.52 < 2 020 log 2
(1.5)2 + 1.5  1 010 < 0
 1  12  4(1)(1 010)   1  12  4(1)(1 010)
∴ < 1.5 <
2(1) 2(1)

∵ 1.5 > 0 for all .
  1  12  4(1)(1 010)
∴ 1.5 <
2(1)
 1  4 041
log 1.5 < log 1M
2
 1  4 041
 log 1.5 < log
2
 1  4 041
log
< 2
log1.5
 < 8.49, cor. to 2 d.p.
∴ The greatest value of  is 8. 1A
----------(4)

18. (a) In △ABD, by the sine formula,

AB AD
=
sin ADB sin ABD
AB 15 cm
= 1M
sin 65 sin 58
15 sin 65
AB = cm
sin 58
 16.030 478 54 cm
= 16.0 cm, cor. to 3 sig. fig. 1A
In △ABC, by the cosine formula,
AC2 = AB2 + BC2  2  AB  BC  cos ABC
AC  16.030 478 54 2  17 2  2  16.030 478 54  17  cos116 cm 1M
 28.016 145 65 cm
= 28.0 cm, cor. to 3 sig. fig. 1A
----------(4)
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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
(b) In △ABD,
BAD + ABD + ADB = 180
BAD + 58 + 65 = 180
BAD = 57
In △ABK,
AK
cos BAK =
AB
AK
cos 57 
16.030 478 54 cm
AK  8.730 824 363 cm
In △ACD, by the cosine formula,
AC 2  AD 2  CD 2
cos CAD =
2  AC  AD
28.016 145 652  152  27 2
 1M
2  28.016 145 65  15
for any
CAD  70.475 050 57 one
In △ACK, by the cosine formula,
CK2 = AK2 + AC2  2  AK  AC  cos CAD
CK  (8.730 824 3632  28.016 145 652 
1
2  8.730 824 363  28.016 145 65  cos 70.475 050 57) 2 cm
 26.412 684 5 cm
CK 2  AK 2  AC 2
cos CKA =
2  CK  AK
26.412 684 52  8.730 824 3632  28.016 145 652

2  26.412 684 5  8.730 824 363
CKA  91.372 522 28
∴ CKA  90 1M
∴ BKC is not the angle between the face ABD and the face ACD.
∴ The claim is disagreed. 1A

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks

In △ABD,
BAD + ABD + ADB = 180
BAD + 58 + 65 = 180
BAD = 57
In △ABK,
AK
cos BAK =
AB
AK
cos 57 
16.030 478 54 cm
AK  8.730 824 363 cm
In △ACD, by the cosine formula,
AC 2  AD 2  CD 2
cos CAD =
2  AC  AD
28.016 145 652  152  27 2
 1M
2  28.016 145 65  15 for any
CAD  70.475 050 57 one
In △ACK, by the cosine formula,
CK2 = AK2 + AC2  2  AK  AC  cos CAD
 (8.730 824 3632 + 28.016 145 652 
2  8.730 824 363  28.016 145 65  cos 70.475 050 57) cm2
 697.629 902 6 cm2
CK2 + AK2  (697.629 902 6 + 8.730 824 3632) cm2  773.857 196 7 cm2
AC2  28.016 145 652 cm2  784.904 417 1 cm2
∵ AC2  CK2 + AK2
∴ CKA  90 1M
∴ BKC is not the angle between the face ABD and the face ACD.
∴ The claim is disagreed. 1A

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OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks

In △ABD,
BAD + ABD + ADB = 180
BAD + 58 + 65 = 180
BAD = 57
In △ABK,
AK
cos BAK =
AB
AK
cos 57 
16.030 478 54 cm
AK  8.730 824 363 cm
In △ACD, by the cosine formula,
AC 2  AD 2  CD 2
cos CAD =
2  AC  AD
28.016 145 652  152  27 2

2  28.016 145 65  15 1M
for any
CAD  70.475 050 57 one
Let N be a point on AD such that CN  AD.
In △ACN,
AN
cos CAN =
AC
AN
cos 70.475 050 57 
28.016 145 65 cm
AN  9.363 480 57 cm
∵ AN  AK
∴ CK is not perpendicular to AD. 1M
i.e. BKC is not the angle between the face ABD and the face ACD.
∴ The claim is disagreed. 1A
----------(3)

19. (a) ∵ G is the circumcentre of △PQR.

∴ GQ = GR
(6  h) 2  (9  3) 2 = (a  h) 2  (11  3) 2 1M
36  12h + h2 + 36 = a2  2ah + h2 + 64
2ah  12h = a2  8
(2a  12)h = a2  8
a2  8
h=
2a  12
 a2  8  1A
∴ The coordinates of G are  , 3  .
 2a  12  ----------(2)

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MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
4
(b)(i) Slope of RG =
3
3 11 4
=
ha 3
2 1
=
ah 3
6=ah
h=a6
a2  8
Substitute h = a  6 into h = .
2a  12
a2  8
a6= 1M
2a  12
(a  6)(2a  12) = a2  8
2a2  24a + 72 = a2  8
a2  24a + 80 = 0
(a  4)(a  20) = 0
a = 4 or 20
42  8
When a = 4, h = = 2 < 0
2(4)  12
202  8
When a = 20, h = = 14 > 0
2(20)  12
∴ a = 20 1A

(ii) Coordinates of G = (14 , 3)

Radius of C = (6  14) 2  (9  3) 2
= 10
The equation of C is
(x  14)2 + (y  3)2 = 102 1M
x2  28x + 196 + y2  6y + 9 = 100
x2 + y2  28x  6y + 105 = 0
Substitute y = kx into x2 + y2  28x  6y + 105 = 0.
x2 + (kx)2  28x  6kx + 105 = 0 1M
(1 + k2)x2  (28 + 6k)x + 105 = 0
 (28  6k )

x-coordinate of M = 1 k 2
2
14  3k
= 1
1 k2

© Oxford University Press 2019 P.15

OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
k (14  3k )
(iii) y-coordinate of M =
1 k 2
Note that OMG = 90.
OM = 2 41
2 2
 14  3k   k (14  3k ) 
 2
 0   2
 0 = 2 41 1M
 1 k   1 k 

(14  3k ) 2
2
= (2 41) 2
1 k
196 + 84k + 9k2 = 164 + 164k2
155k2  84k  32 = 0
(5k  4)(31k + 8) = 0
4 8
k= or  (rejected)
5 31
  
 14  3 4  4 14  3 4  
 5, 5   5  
Coordinates of M =  2 2  = (10 , 8) 1M
 1   4  4
1   
 5 5 
     
When P is farthest from M, MGA is a straight line.
When P is nearest to the y-axis,
coordinates of B = (14  10 , 3)
= (4 , 3) 1M
Note that the area of the circle passing through A and B is the least when AB is
a diameter of the circle. Hence, AUB = 90. 1M
Slope of AM = slope of GM
83
=
10  14
5
= 
4
83
Slope of BM =
10  4
5
=
6
Slope of AM  slope of BM
5 5
=   1M
4 6
25
= 
24
 1
∴ AMB  90
∵ AUB + AMB  180
∴ A, M, B and U are not concyclic. 1A

© Oxford University Press 2019 P.16

OXFORD UNIVERSITY PRESS
MOCK 19(I) COMPULSORY PART PAPER 1 SOLUTION

Solution Marks
k (14  3k )
y-coordinate of M =
1 k 2
Note that OMG = 90.
OM = 2 41
2 2
 14  3k   k (14  3k ) 
 2
 0   2
 0 = 2 41 1M
 1  k   1 k 

(14  3k ) 2
= (2 41) 2
1 k 2
196 + 84k + 9k2 = 164 + 164k2
155k2  84k  32 = 0
(5k  4)(31k + 8) = 0
4 8
k= or  (rejected)
5 31
When P is farthest from M, MGA is a straight line.
When P is nearest to the y-axis,
coordinates of B = (14  10 , 3)
= (4 , 3) 1M
Note that the area of the circle passing through A and B is the least when AB is
a diameter of the circle. Hence, AUB = 90. 1M
4
Substitute x = 4 and y = 3 into y = x.
5
L.H.S. = 3 1M
4 16
R.H.S. = (4) =
5 5
∵ L.H.S.  R.H.S.
i.e. The ordered pair (4 , 3) does not satisfy the equation.
∴ B(4 , 3) does not lie on OM. 1M
∴ AMB  90
∵ AUB + AMB  180
∴ A, M, B and U are not concyclic. 1A
----------(11)

© Oxford University Press 2019 P.17