# 2018 MATHS P2

```N
12/26/2018
15° N
A
40°
70° E
E
35 °S
B
C
S
๐ฆ
1
๐ฆ = cos ๐
90°
270°
180°
360°
-1
MATHEMATICS
PAPER 2
SUGGESTED SOLUTIONS
b (i)
Z
b (ii)
5.5 cm
T
7 cm
P (c)
38 °
X
9 cm
Y
FULL SOLUTIONS | 2018
QUESTION ONE
Question
Number
Solution
suggest
|83 ๐ฆ5| = |41
1 (a) (i)
|
โ5
2
(40) - (3๐ฆ) = 8 โ ( โ 5)
40 โ 3๐ฆ = 13
โ 3๐ฆ = 13 โ 40
โ 3๐ฆ = โ 27
๐ฆ=9
(a)(ii)
๐ฉ=
(๐๐ ๐๐)
|๐ฉ| = ๐๐
๐ด
1 (b) (i)
โ๐
(
๐
๐
= ๐๐ โ ๐
โ๐
๐
)
P(both black) = P(ww )
๐
๐
= ๐๐ × ๐๐
๐
=๐
P(different colours) = P(WG or GW )
๐
๐
๐
๐
= ๐๐ × ๐๐ + ๐๐ × ๐๐
=
๐๐ + ๐๐
๐๐๐
๐๐
= ๐๐
2
By T.H Musondela
QUESTION TWO
Question
Number
Solution
2 (i)
๐
(๐)
๐=
๐=
๐๐ โ ๐ โ ๐ = ๐
โ ( โ ๐) ± ( โ ๐)๐ โ ๐(๐)( โ ๐)
๐(๐)
๐ ± ๐๐
๐
๐ =1.47 or -1.14
๐๐๐ญ๐ก๐๐ฆ๐๐ญ๐ข๐๐ฌ
๐๐ก๐๐ฆ๐ข๐ฌ๐ญ๐ซ๐ฒ
๐
๐
๐
3
๐
0
๐
๐๐ก๐ฒ๐ฌ๐ข๐๐ฌ
or
M
P
C
4
5
3
2
6
(b) (ii)
(a) 6
(๐) ๐ + ๐ + ๐ = ๐
(c) ๐
3
By T.H Musondela
QUESTION THREE
Question
Number
3 (a)
Solution
โซ๐ (๐ + ๐ โ ๐๐)๐๐
โ๐
=
[
๐ ๐ ๐ ๐
๐๐ + ๐๐ โ ๐๐
[
๐
๐
๐
]
โ๐
๐
= ๐(๐) + ๐(๐) โ ๐(๐)
= (๐ + ๐ โ
๐
] โ [๐( โ ๐) + ๐๐( โ ๐)๐ โ ๐๐( โ ๐)๐]
๐
๐
๐
)
โ
(
โ
๐
+
+
)
๐
๐
๐
=
๐๐
๐
โ
(
โ
)
๐
๐
=
๐๐
๐
+
๐
๐
๐
=๐
๐
= 4.5 or ๐๐
(a) ((ii)
(b) ๐ = ๐ + ๐๐
๐๐
=
๐๐
1-
โ๐
๐
๐
๐
๐
๐
๐ = ๐ โ ๐๐ = ๐
๐๐๐ =โ ๐
๐
โด ๐๐ =โ ๐
๐๐ + ๐ = ๐ at (๐,๐)๐ = ๐ and y= 5
๐
โ ๐(๐) + ๐ = ๐
โ ๐๐ + ๐๐ = ๐๐
๐=
๐๐
๐
๐
โด ๐ =โ ๐ +
๐๐
๐
or 3y + ๐๐ = ๐๐
4
By T.H Musondela
QUESTION FOUR
Question
Number
Solution
b (i)
Z
b (ii)
5.5 cm
T
7 cm
P (c)
38 °
9 cm
X
Y
QUESTION FIVE
Question
Number
Solution
5 (a)
๐โ๐
๐
๐(๐ โ ๐)
๐
๐ โ๐
= (๐ โ ๐)(๐ + ๐)
โ ๐(๐ โ ๐)
= (๐ โ ๐)(๐ + ๐)
โ๐
= ๐+๐
5(b)(i)
๐
(๐๐ โ ๐๐)(๐ + ๐) = ๐ค
๐
๐
๐๐ + ๐๐ โ ๐๐๐ โ ๐๐ = ๐
๐
๐ โ ๐๐ + ๐๐ = ๐
๐
๐ โ ๐๐๐ + ๐๐ โ ๐๐ = ๐
๐(๐ โ ๐๐) + ๐(๐ โ ๐๐) = ๐
(๐ โ ๐๐)(๐ + ๐) = ๐
๐ = ๐๐
5
By T.H Musondela
5(b) (ii)
๐ป๐ = ๐ + ๐ = ๐๐ + ๐ = ๐๐
๐ป๐ = ๐ = ๐๐
๐ป๐ = (๐๐ โ ๐๐) = ๐(๐๐) โ ๐๐ = ๐
โด 16, 12, 9
5(b) (ii)
๐
๐บโ = ๐ โ ๐
๐บโ =
๐๐
๐
๐โ๐
= 64
QUESTION SIX
Q no
Solution
6 (i) (a)
(a)
๐
๐จ๐ฌ = ๐ ๐จ๐ช
๐
= ๐(๐ + ๐๐)
(b) ๐ฉ๐ฌ = ๐ฉ๐จ + ๐จ๐ฌ
๐
๐
= โ ๐ + (๐ + ๐๐)
๐
= ๐ ( โ ๐๐ + ๐ + ๐๐)
๐
= ๐ ( โ ๐๐ + ๐๐)
๐
๐
= ๐ ( โ ๐ + ๐) or = ๐ (๐ โ ๐)
(c) ๐ฉ๐ซ = ๐ฉ๐จ + ๐จ๐ซ
= โ๐+๐
๐
(b) ๐ฉ๐ฌ= ๐ ( โ ๐ + ๐) and ๐ฉ๐ซ =
โ๐+๐
๐
๐ฉ๐ฌ= ๐ ๐ฉ๐ซ โด the points B, E and D are colinear
6
By T.H Musondela
6 (b)
Start
Enter x, y
M= sqrt (x^2+y^2)
Is M < ๐?
Yes
Error โMโ must be positive
no
Display M
Stop
7
By T.H Musondela
QUESTION SEVEN
Question
Number
7 (a) (i)
Solution
3
2
(a) ๐ฆ = 2๐ฅ โ 3๐ฅ + 5
When ๐ฅ =โ 2,
3
2
๐ = 2( โ 2) โ 3( โ 2) + 5
๐ =โ 23
7 (a) (ii)
7 (a) (iii)
๐ฆ=๐ฅ
X= -1.1
8
By T.H Musondela
(a) (1.5,5) and (2,7.3)
7 (a) (iv)
๐ฆ2 โ ๐ฆ1
M =๐ฅ
7.3 โ 5
= 2 โ 1.5
2 โ ๐ฅ1
=
2.25
0.5
= 4.5
7 (b)
3
2
โ
x+1 xโ1
=
3(x โ 1) โ 2(x + 1)
(x + 1)(x โ 1)
=
3x โ 3 โ 2๐ฅ โ 2
(x + 1)(x โ 1)
xโ5
= (x + 1)(x โ 1)
QUESTION EIGHT
Question
Number
Solution
๐พ๐
8 (i)
80
(a) sin 60 = sin 52
๐พ๐ =
80sin 60
sin 52
= 87.92015
= 87.9°
(b) A=
1
(80)(50)๐ ๐๐ 60
2
= 1732.0508
2
= 1700 ๐ (3sf)
8 (ii)
๐
(iii) ๐ (sd)(KN) = Area of ฮ KNR
๐ฌ๐ =
๐ × 3260
๐๐
= 81.5 (m)
9
By T.H Musondela
8 (B)
๐ฆ
1
๐ฆ = cos ๐
90°
180°
270°
360°
๐ฅ
-1
8 (c)
3
12๐๐
3
15๐๐
3
÷
9๐ ๐
2 2
10๐ ๐
12๐๐๐๐
10๐๐๐๐
×
15๐๐๐๐
9๐๐๐๐
๐
=
๐๐
๐
๐๐
QUESTION NINE
Question
Number
8 (a)
Solution
๐
2
10
15
23
30
10
๐
15
25
35
45
55
65
โ๐ = 90
๐๐๐๐ง (๐) =
โ๐๐ฅ
๐
๐๐
30
250
525
1035
1650
650
๐๐
450
6250
18375
46575
90750
42250
โ๐๐ฅ = 4140
โ๐๐ฅ = 204650
2
4140
โ๐ = 90
= 46
SD =
=
๐ฎ๐๐
โ๐
๐
- (๐)
๐๐๐๐๐๐
๐๐
๐
- (46)
๐
= ๐๐๐.๐๐๐๐๐๐๐
= ๐๐.๐๐๐๐๐
= 12.6
(3sf)
10
By T.H Musondela
(b)(i)
Marks(x )
โค10
โค 20
โค 30
โค 40
โค 50
โค 60
โค 70
CF
RCF
0
0
2
0.02
12
0.13
27
0.3
50
0.56
80
0.89
90
1
๐
0.๐
0.๐
0.๐
0.๐
0.๐
0.๐
0.๐
0. ๐
0. ๐
๐
๐๐
(iii)
65
× 1 = 0.65
100
๐๐
๐๐
๐๐
,
โด 65 percentile 53
11
By T.H Musondela
๐๐
๐๐
๐๐
QUESTION 10
Question
Number
Solution
(ii) Un anticlockwise Rotation of 270 about (0,0)
10 (a)
Or
clockwise Rotation of 90 about (0,0)
10 (b)
it is enlargment, centre (0,0) and scale factor 2
10 (c)
This is a stretch in which y axis is invariant and k = -2
โด the required matrix is
( โ02 01)
Or
(
)( 21
๐ ๐
๐ ๐
4
1
)
=
( โ14
โ8
1
)
Solving the simultaneous equation
๐ =โ 2 , b = 0 , c = 0 and d = 1
โด the required matrix is
10 (d)
(i)
(
)( 21
1 0
โ2 1
( โ02 01)
2 4
4 1
)
(
2 2
= โ3 0
โด the coordinates are
(2,-3) ,
(2,0) ,
( 4,-7)
12
By T.H Musondela
4
โ7
)
QUESTION ELEVEN
Question
Number
11 (a)
Solution
}
๐ฅ + ๐ฆ โค 60
๐ฅ + ๐ฆ โฅ 35
๐๐ ๐๐๐ฆ ๐๐๐๐๐
๐ฆโฅ6
๐ฆ โค 14
11 (b)
70
๐ โ ๐๐๐๐
60
50
๐ + ๐ = ๐๐
40
๐ + ๐ = ๐๐
30
20
๐ = ๐๐
Solution
set
10
๐=๐
0
11 (c)
10
20
30
40
(i) 10 teachers
(ii) 52 learners
d
K 30.00
13
By T.H Musondela
50
60
70
QUESTION 12
Question
Number
Solution
1
3 × 11.4 × [14 × 10 + 8 × 4 +
14 × 10 × 8 × 4 ]
1
(3 possible answers do exist here)
= 3 × 11.4 × [140 + 32 + 4480 ]
1
= 3 × 11.4 × [2389328 ]
= 907.94484
N
12 (b)
15° N
A
40°
70° E
E
C
B
35 °S
S
(i)
Differ in latitude = 15+35 = 50°
XZ =
=
50
× 2 × 3.142 × 6370
360
2,001,454
360
= 5,559.5944444 km
= 5,560 km (3sf)
(ii) (a)
๐
× 2 × 3.142 × 6370cos 35
360
= 900
900 × 360
๐= 2 × 3.142 × 6370cos 35
= 9.881088 (3sf)
= 9.9° (1dp)
(ii) (b)
Longitude of Q is (70-9.881088) E = 60.1
โด Q (35 °S, 60.1 °E)
14
By T.H Musondela
This not a frustum. Me I donโt know
teacher
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