Senior Four Mathematics Paper One Solutions. Section A
Senior Four Mathematics Paper One Solutions. Section A
Senior Four Mathematics Paper One Solutions. Section A
SECTION A.
1 4X=Y+7
6X+5Y=17
Rearrange the equation
4X-Y=7
6X+5Y=17
4 -1 X = 7
6 5 Y 17
5 6 4 -1 X = 5 1 7
-6 4 6 5 Y -6 4 17
26 0 X = 52
0 26 Y 26
26X = 52
X=2
and 26Y = 26
Y=1
2. 5 5 - 2 10 + 125
5 5 - 2 10 + 5 5
5 5 - 40 + 5 5
10 5 - 40
10 5 - 2 10
Page 1
3. 8: 10: 14
Sum = 32
Angle sum of a triangle is 1800
8 0
32 X 180 = 45
10 0
32 X 180 = 56.25
14 0
32 X 180 = 78.75
a b = 2 3
c d 1 2
a = 2, b = 3, c = 1, d = 2
Therefore the image of (7 , -4 )
= 2 3 7 = 2
1 2 -4 -1
T’ (2 , -1 )
5. Cos x = 0.2
0 x 90 x is Acute
10 b= 4 6
2
b = 10 2 2 2
b = 96
b =4 6
Page 2
4
Sin x = 6
10
2
Sin x = 6
5
4
tan x = 6
2
tan x = 2 6
6. Area = 5cm 2 R
P Q
4cm
1
Area = PR X PQSin30
2
1
5= PR X 4 X Sin30
2
5 X 4 = PR X 4
5cm = PR
7. (a) Mean =
fx
f
5X1) + (2X3) + (3X6) + (4X2) + (5X4)
=
20
57
=
20
= 2.85
x
n
(b) +1) th
Median (
2
For n – even
= 11 th Position
The median is 3
8. 8 2 n X 3 m = 36
2 6 n X 3 m = (6) 2
2 6 n X 3 m = (2 2 X 3 2 )
Page 3
6n = 2 and m = 2
1
n=
3
9. 1 : 30,000
This means that 1cm on the map represents 30,000cm on the ground
1cm → 30,000cm
3cm →90,000cm
4cm →120,000cm
W= 4cm
L=3cm
= 2 (L + W)
= 2 (90,000 + 120,000)
= 2 (210,000)
=420,000cm
But 1m → 100cm
? →420,000cm
=420,000X10 2 m
= 4,200m
10 n(A) = 10
n(B) = 6
Lets look at the possibilities
n(AnB) = 0, n(AuB)=16
Max of n(AnB) is 6
Max of n(AuB) is 16
Min of n(AnB) is 0
Min of n(AuB) is 10
11.
Y = x 2 -4x
X -2 -1 0 1 2 3 4 5
X2 4 1 0 1 4 9 16 25
Y 12 5 0 -3 -4 -3 0 +5
Page 5
Solving the Equations
(i) Y = x 2 - 4x and x 2 - 4x-3 = 0
0 = x 2 -4x – 3
Y=0+3
Y=3
So x 1 = -0.6 , x 2 = 4.6
(ii) x 2 -5x + 2 = 0
Subtracting equations
y = x 2 - 4x
0 = x 2 -5x +2
y=x–2
x 1 =0.5
x 2 = 0.6
Page 6
c. Range X 2 -4x<0
Simply draw the line Y=0 and note the values where the curve is below the line y=o
0≤ x ≤ 4 As required
P
12.
6cm
U Q
4cm
T R
6cm 10cm
P P
U Q T R
Since UQ is parallel to TR, this means that angle PRT is equal to angle PQU (corresponding angles)
Since Q = R and U = T then it follows that the triangle PUQ is similar to PTR .
b. (i) UV
10 16
=
6 UQ
UQ = 9.6cm
(ii)
Page 7
10 10
=
6 VQ
VQ = 6cm
UV = UQ-VQ
=9.6 – 6 = 3.6cm
5
L.S.F = ( )
3
5 2
A.S.F = ( )
3
25 40
=
9 PUQ
40x9
D PUQ =
25
= 14.4cm 2
(ii)Angle PRS
1
A x10 x16 Sin
2
160 Sin
40 =
2
80
Sin
160
1
Sin
2
1
Sin 1
2
30 0 .
Page 8
13. Allowances
120,000 /
1- Marriage = 10,000
12
10
2 – House & Transport x 600,000
100
= 60,000/=
240,000
_ Medical care =
12
= 20,000
= 90,000/=
Taxable income =
600,000 - 90,000
= 510,000/=
1 – 100,000 % x 100,000 0
5
100,000 – 200,000 x100,000 5000
100
10
200,000 - 300,000 x100,000 10,000/=
100
20
300,000 – 450,000 x150,000 30,000/=
100
Page 9
30
450,000 – 550,000 x 60,000 18,000/=
100
= 510,000 – 63,000/=
= 447,000/=
14. P 1Q 1 R 1
(a) 3 0 2 2 4 6 6 12
1 2 5 2 3 = 12 6 10
P 1Q 1 R 1 P 11Q 11 R 11
1
(b) - 0 6 6 12 3 3 6
2
1
0 12 6 10 =
6 3 5
2
Page 10
Single Matrix
1 3
0 3 0 = 0
2 2
1 1
0 1 2 -1
2 2
3
2 x 1 0 x3units
3 9
x 3 =
2 2
15.
1 2 3
10 18 24
5 7 10 3x3
100
200
80 3x1
Monthly requirements;
Page 11
5 7 10 80 500 + 1400 + 800
= 740
6520
3700
T R C
S 1 10 5 T 10,000 S 160,000
N 2 18 7 R 5,000 =N 250,000
H 3 24 10 C 20,000 H 350,000
250,000
Page 12