ADUMAN SENIOR HIGH SCHOOL
MARKING SCHEME
CORE MATHEMATICS
MOCK EXAMINATION, 2023
1 D 11 A 21 D 31 B 41 A
2 B 12 D 22 C 32 B 42 B
3 C 13 B 23 B 33 C 43 B
4 D 14 D 24 B 34 A 44 B
5 D 15 B 25 C 35 D 45 C
6 A 16 A 26 A 36 A 46 B
7 B 17 C 27 C 37 A 47 A
8 B 18 C 28 A 38 B 48 D
9 D 19 A 29 D 39 B 49 C
10 B 20 B 30 A 40 C 50 C
Question Details Marks
Number
1. (a) 42−x × 16x+1 = 64
42−x × 42(x+1) = 43 M1
42−x+2(x+1) = 43
2 − x + 2(x + 1) = 3 M1
2 − x + 2x + 2 = 3
x+4=3
x=3−4
x = −1 A1
(b) a + x + 20 + 5 = 65
a + x = 65 − 25
a = 40 − x M1
b + 15 + 20 + 5 = 50
b = 50 − 40
b = 10 M1
c + x + 15 + 5 = 40
c + x = 40 − 20
c = 20 − x M1
a + b + c + x + 15 + 5 + 20 = 100
40 − x + 10 + 20 − x + x + 40 = 100 M1
110 − x = 100
110 − 100 = x
x = 10 A1
Total 8 marks
Question Details Marks
Number
2. (a) For himself = 100
20
× 250 = 50
Left for sale =250 − 50 = 200
sold for $6.50 = 115 × 6.50
= $747.50 M1
Left for sale = 200 − 115 = 85
sold for $5.00 = 85 × 5.00
= $425.00 M1
selling price (sp) = 747.50 + 425
= $1172.50
cost price (cp)= $1000.00
profit (p) =selling price−cost price
= 1172.50 − 1000.00
= $172.50
profit percent= costprof it
price
× 100
172.50
= 1000 × 100 M1
profit percent = 17.25% A1
(b) Let x = remaining interior angles
Number of size of polygon n = 3 + x M1
Sum of interior angle = (n − 2)180
3 × 160 + 120x = ((3 + x) − 2)180 M1
480 + 120x = 180 + 180x
480 − 180 = 180x − 120x M1
300 = 60x
x = 360
60
x=5
n=x+3=5+3
The polygon has 8 sides A1
Total 8 marks
Question Details Marks
Number
3. (a) ∠QSR = ∠QP R = 40o
∠T P S + ∠SP R + ∠QP R = 180o
74o + 40o + ∠SP R = 180o M1
∠SP R = 180o − 114o
∠SP R = 66o M1
∠SP R = ∠SQR = 66o
∠U QR + ∠SQR + ∠SQP = 180o
68o + 66o + ∠SQP = 180o M1
∠SQP = 180o − 134o
∠SQP = 46o
∠SQP = ∠P RS = 46o A1
(b) i. y = xy22 −y1
−x1
(x − x1 ) + y1
−6−3
y = 2−(−4) (x − (−4)) + 3 M1
y = −96
(x + 4) + 3
−3
y = 2 (x + 4) + 3
2y = −3(x + 4) + 6
2y = −3x − 12 + 6
2y + 3x + 6 = 0 A1
ii.
p
Q| = (y2 − y1 )2 + (x2 − x1 )2
|P p
= √ (−6 − 3)2 + (2 − (−4))2 M1
= 117 A1
Total 8 marks
Question Details Marks
Number
4(a)
B2
|BF |
tan 42o = |BP |
o |BF |
tan 42 = 180
180 tan 42o = |BF |
|BF | = 162.07m M1
|BF |
tan θ = |BQ|
tan θ = 162.07
45
θ = tan−1 162.07
45
θ = 74.48
The angle of elevation of the top of the A1
flagpole from Q is 74.50
(b) R S13
R = Sk3 M1
9 = 3k3
9×3=k
k = 243 M1
R = 243
S3
243
64
= 243
S3
M1
3 243×64
S = 243
S 3 = 64
S 3 = 43
S=4 A1
Total 8 marks
Question Details Marks
Number
5(a) Let x = elder sister’s age
Ama’s age = 23 x
4 years ago sister’s age = x − 4 M1
and Ama’s age = 23 x − 4
2
3
x − 4 = 12 (x − 4) M1
6 × 23 x − 6 × 4 = 6 × 21 (x − 4)
4x − 24 = 3x − 12 M1
4x − 3x = −12 + 24
x = 12
The sister is 12 years old A1
M1
(b) 4415 = 4(5)2 + 4(5)1 + 1(5)0
= 100 + 20 + 1
= 121 M1
B N R
4 121
4 30 1
B1
4 7 2
4 1 3
4 0 1
4415 = 13214 A1
Total 8 marks
Question Details Marks
Number
6. (a) Let x = radius
|OB|2 = |OA|2 + |AB|2
(x + 8)2 = x2 + 122 M1
(x + 8)(x + 8) = x2 + 144
x2 + 8x + 8x + 64 = x2 + 144
x2 − x2 + 16x = 144 − 64 M1
16x = 80
x = 80
16
x=5 A1
Radius of the circle is 5cm
(b) B1
2 2 2
|P Q| =√|QR| − |P R|
|P Q| = √132 − 52
|P Q| = 144
|P Q| = 12 M1
5
sin y = 13 , cos y = 12
13
, tan y = 5
12
3( 12 )−2( 5 )
3 cos y−2 sin y
6 tan y
= 136( 5 ) 13 M1
12
[ 36
13
− 10
13
] ÷ 5
2
= 26
13
÷ 25 = 2× 2
5
= 4
5
A1
(c) Un = ar n−1
U3 = ar U8 = ar7
2
ar2 = 4 · · · (1) M1
1
ar7 = 256 · · · (2) M1
7
(2) ÷ (1) : ar
ar2
1
= 256 ÷4
5 1
r =q 1024
r = 5 1024 1
= 14 A1
Put r = 14 in (1)
a( 14 )2 = 4
a = 4 × 16 = 64 A1
U2 = 64( 14 ) = 16
U1 + U2 = 64 + 16 = 80 A1
Total 12 marks
Question Details Marks
Number
7(a)(i)
B2
p
|XZ| = √ |XY |2 + |Y Z|2
|XZ| = √51202 + 4482 M1
|XZ| = 26415104
|XZ| = 5139.56
|XZ| ≈ 5140 km A1
|Y Z|
(ii) tan θ = |XY |
448
tan θ = 5120 M1
θ = tan−1 5120448
θ=5 M1
The bearing of Z from X is 345 + 5 = 350o A1
(iii) Speed= timetaken
distance
Distance it took to sail from X to Z through
Y = 5120 + 448 = 5568
120 = 5568t
M1
5140km
t = 120kmh −1
t = 46.4
Time taken to sail from X to Z through Y is A1
46 hours.
∗ 1 3 5 6
1 4 6 1 2
(b) 3 6 1 3 4 B3
5 1 3 5 6
6 2 4 6 0
Total 12 marks
Question Details Marks
Number
8 (a)
x 0o 30o 60o 90o 120o 150o 180o
y 1 0 -2 -3 -2 0 1 B2
x 0o 30o 60o 90o 120o 150o 180o
y -2 -1.8 -1.7 -1.5 -1.3 -1.2 -1 B2
(b)
Correct labeling of x− axes B 12
Correct labeling of y− axes B 12
Correct plotting and drawing of y = 2 cos 2x−1 B3
Correct plotting and drawing of y = 180
1
(x−360) B2
(d)(i) x = 30 ± 3 or x = 150 ±3
o o
A1
(ii) x = 57 ± 3 or x = 123 ±3 A1
Total 12 marks
Question Details Marks
Number
9.(a) P (K) = 10 9
P (K) = 10 1
M1
P (Y ) = 45 P (Y ) = 15 M1
P (A) = x P (A) = 1 − x M1
P (only one passing) =
P (K Y A) + (Y K A) + (A K Y )
9
50
9
= [ 10 × 15 × (1 − x)] + [ 10
1
× 54 × (1 − x)] + M1
1 1
[ 10 × 5 × x]
9 9 2 1
50
= [ 50 (1 − x)] + [ 25 (1 − x)] + [ 50 x]
9 = 9(1 − x) + 4(1 − x) + x M1
9 = 9 − 9x + 4 − 4x + x
12x = 4
4
x = 12
x= 3 1
A1
(b) (α) 2x + 3x + 4x − 1 + x + x + 2 + x − 3 = 42 M1
12x − 6 = 42
12x = 48
x = 4812
x=4 A1
Age(x) Freq (f) f(x)
7 8 56
8 12 96
9 15 135
(β) B2
10 4 40
11 2 22
12 P 1 P 12
P
x = 42 f (x) = 361
Mean x = P f f (x)
x = 361
42
M1
x = 8.595
The mean age is 9 years A1
Total 12 marks
Question Details Marks
Number
10 (a) p k + q12
p = k + kq21 M1
−1 = k + k121
k + k1 = −1 · · · (1) M1
2 = k + k221
4k + k1 = 8 · · · (2) M1
(2) − (1) : 3k = 9
k = 93 = 3 M1
Put k = 3 into (1)
3 + k1 = −1
k1 = −4 M1
p = 3 − q42
2 34 = 3 − q42 M1
11
4
= 3 − q42
11q 2 = 12q 2 − 16
16 =√12q 2 − 11q 2
q = 16
q=4 A1
(b)
p
|M 2 2
√N | = p (x2 − x1 ) + (y2 − y1 )
3 √10 = (−4 p − x)2 + (3 − 6)2 M1
(3 10)2 = ( (−4 − x)2 + (3 − 6)2 )2
90 = (−4 − x)(−4 − x) + 9
90 = 16 + 4x + 4x + x2 + 9 M1
x2 + 8x + 25 − 90 = 0
x2 + 8x − 65 = 0 M1
x2 − 5x + 13x − 65 = 0 M1
x(x − 5) + 13(x − 5) = 0
(x − 5)(x + 13) = 0
Eitherx − 5 = 0 or x + 13 = 0
x = 5 or x = −13 A1
Total 12 marks
Question Details Marks
Number
Amount to Rate of tax Tax (GH¢)
be taxed
(GH)¢
3000 300 Nil
11(a) 2700 5
100
× 240 12 M4
2460 7.5
100
× 480 36
1980 10
100
× 480 48
1500 12.5
100
× 960 120
540 Total Tax 216
(i) Tax paid by Mr Owusu is GH¢216.00 A1
(ii) % of income pay √ as tax= 3000
216
× 100 = 7.2% A1
(b) log5 (5 2x+1
−√20 5) = 2x
(52x+1
− 20 √ 5) = 52x M1
52x × 51 − 20 5 = √ 52x
5(52x ) − 52x = 20 5 √ M1
Let 52x √ = y: 5y − y = 20 5
4y = 20 √
5
y= √ 20 5
4
M1
y=5 √ 5
2x
5 =5 5 M1
1
52x = 51 × 5 2
1
52x = 51+ 2
2x = 1 + 12 M1
4x = 2 + 1
x = 34 A1
Total 12 marks
Question Details Marks
Number
12. y = ax2 + bx + c = 0
(a)
(0, 5); 5 = a(0)2 + b(0) + c
c=5 M1
(1, 0); 0 = a(1)2 + b(1) + c
0=a+b+c
Put c = 5 in a + b + c = 0
a+b+5=0
a + b = −5 · · · (1)
(3, −4); −4 = a(3)2 + b(3) + c
−4 = 9a + 3b + c
Put c = 5 in 9a + 3b + c = −4
9a + 3b + 5 = −4
9a + 3b = −9 · · · (2)
a + b = −5 · · · (1) M1
9a + 3b = −9 · · · (2)
Make a the subject in (1)
a = −5 − b
Put a = −5 − b
9(−5 − b) + 3b = −9
−45 − 9b + 3b = −9
−6b = −9 + 45
36
b = −6
b = −6 M1
Put b = −6 in (1)
a − 6 = −5
a=1 M1
y = x2 − 6x + 5
x -1 0 1 2 3 4 5 6 7
(b) B2
y 12 5 0 -3 -4 -3 0 5 12
Total 12 marks
Question Details Marks
Number
(c)
B3
(d) 3<x≤7 A1
(d) Minimum coordinates are (3, −4) A1
(e) 1<x<5 A1
Total 12 marks
Question Details Marks
Number
13. (a) )Let ∠M SN = x and ∠QN P = y
B2
132 + y + y = 180 M1
2y = 180 − 132
2y = 48
y = 48
2
= 24 M1
∠N P Q = 2(24) = 48o and ∠P N Q = 24o
∠QN P + ∠N P Q + ∠N QP = 180o
y + 2y + ∠N QP = 180o
∠N QP = 180o − 3y
∠N QP = 180o − 3(24)
∠N QP = 108o
∠SQN = 180o − 108o = 72o M1
∠SM N +∠M SQ+∠SQN +∠M N Q = 360
x + 2x + 72 + 132 = 360 M1
3x = 360 − 204
x = 156
3
x = 52o p A1
(b)(i) |AC| = √ |AB|2 + |BC|2
|AC| = √242 + 72 M1
|AC| = 625
|AC| = 25cm ∴ Radius = 12.5cm A1
(ii) Area = πr − 2 bh
2 1
= (3.142 × 12.52 ) − ( 12 × 7 × 24) M1
= 490.9375 − 84 M1
= 406.9375 − 84
Area of the shaded portion is = 407cm2 A1
Total 12 marks
