INSTRUCTIONS TO THE CANDIDATES
 This paper contains two sections; Section I and Section II.
 Answer all the questions in section I and only five questions from Section II.
 All workings and answers must be written on the question paper in the spaces provided below each question.
 Non programmable silent electronic calculators and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
 Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
SECTION I (50MKS)
 Simplify by rationalising the denominator
√ 2 + √3 (3mks)
√6  √3  Find the value of x in the equation log_{10} (2x1) + log_{10}3 = log_{10} (8x1). (3mks)
 Find the compound interest on sh. 200,000 for 2 years at 14% pa. Compounded semiannually. (3mks)
 The ratio of 12th to 10th term in a geometric series is 9:1. Find the common ratio. (3mks)

 Expand (2 – ¼x)^{5} (2mks)
 Use your expansion to find the value of (1.96)5 correct to 3 decimal places (2mks)
 Chord WX and YZ intersect externally at Q. The secant WQ = 11cm and QX = 6cm while ZQ = 4cm.
 Calculate the length of chord YZ. (2mks)
 Using the answer in (a) above, find the length of the tangent SQ. (2mks)
 Given that is a singular matrix, find the possible values of y. (3mks)
 The masses to the nearest kg of 50 adults were recorded as follows:
Mass (kg) Frequency (f) 45 – 50
51 – 56
57 – 62
63 – 68
69 – 74
75 – 802
10
11
20
6
1  P varies as the cube of Q and inversely as the square root of R. If Q is increased by 20% and R decreased by 36%, find the percentage change in P. (3mks)
 Solve 8 cos^{2} x – 2 cos x – 1 = 0 (3mks)
 Make χ the subject of the formula: (3mks)
A = √3 + 2χ
5  4χ  The position vectors of A and B are given as a= 2i3j+4k and b= 2ij+2k respectively. Find to 2decimal places, the length of the vector(AB). (3mks)
 Find the centre and the radius of a circle whose equation is x^{2}6x+y^{2}10y+30=0 (3mks)
 A point (x, y) is mapped onto (13, 13) by two transformations M followed by T where Find the point (x y) (3mks)
 Given that 2 ≤ A ≤ 4 and 0.1 ≤ B ≤ 0.2. Find the minimum value of AB (3mks)
A  B  In a transformation, an object with area 9cm^{2} is mapped onto an image whose area is 54cm^{2}. Given that the matrix of transformation is find the value of x (3mks)
SECTION II (50MKS)  The table below shows the rates of taxation in a certain year.
Income in K£ pa Rate in Ksh per K£ 1 – 3900
3901 – 7800
7801 – 11700
11701 – 15600
15601 – 19500
Above 195002
3
4
5
7
9 Calculate how much income tax Juma paid per month. (7mks)
 Juma’s other deductions per month were cooperative society contributions of sh. 2000 and a loan repayment of sh. 2500. Calculate his net salary per month. (3mks)
 Wainaina has two dairy farm A and B. Farm A produces milk with 3 ½ percent fat and farm B produces milk with 4 ¾ percent fat. Determine;
 The total mass of milk fat in 50kg of milk from farm A and 30kg from farm B. (3mks)
 The percentage of fat in a mixture of 50kg of milk from A and 30kg of milk from farm B. (2mks)
 Determine the range of values of mass of milk from farm B that must be used in a 50kg mixture so that the mixture may have at least 4 percent fat. (5mks)
 A cupboard has 7 white cups and 5 brown ones all identical in size and shape. There was a blackout in the town and Mrs. Kamau had to select three cups, one after the other without replacing the previous one.
 Draw a tree diagram for the information. (2mks)
 Calculate the probability that she chooses.
 Two white cups and one brown cup. (2mks)
 Two brown cups and one white cup. (2mks)
 At least one white cup. (2mks)
 Three cups of the same colour. (2mks)

 Complete the table below, giving the values correct to 2 decimal places (2mks)
Xº 0º 15º 30º 45º 60º 75º 90º 105º 120º 135º 150º 165º 180º Cos 2Xº 1.00 0.87 0.00 0.5 1.00 0.5 0.00 0.50 0.87 1.00 Sin(Xº+30º) 0.50 0.71 0.87 0.97 1.00 0.87 0.71 0.50 0.00 0.50  Using the grid provided draw on the same axes the graph of y=cos 2Xº and y=sin(Xº+30º) for 0º≤X≤180º. (4mks)
 Find the period of the curve y=cos 2x0 (1mk)
 Using the graph, estimate the solutions to the equations;
 sin(Xº+30º) = cos 2Xº (1mk)
 Cos 2Xº=0.5 (1mk)
 Complete the table below, giving the values correct to 2 decimal places (2mks)
 The For a sample of 100 bulbs, the time taken for each bulb to burn was recorded. The table below shows the result of the measurements.
Time(in hours) 1519 2024 2529 3034 3539 4044 4549 5054 5559 6064 6569 7074 Number of bulbs 6 10 9 5 7 11 15 13 8 7 5 4  Using an assumed mean of 42, calculate
 the actual mean of distribution (4mks)
 the standard deviation of the distribution (3mks)
 Calculate the quartile deviation (3mks)
 Using an assumed mean of 42, calculate

 Using a ruler and a pair of compasses only, construct a parallelogram ABCD such that AB=9 cm, AD=7 cm and angle BAD=60º. (3mks)
 On the same diagram, construct:
 The locus of a point P such that P is equidistant from AB and AD; (1mk)
 The locus of a point Q such that Q is equidistant from B and C; (1mk)
 The locus of a point T such that T is equidistant from AB and DC; (1mk).

 Shade the region R bounded by the locus of P, the locus of Q and the locus of T. (1mk)
 Find the area of the region shaded in (d)(i) above. (3mks)
 The points A (1,4), B(2,0) and C (4,2) of a triangle are mapped onto A1(7,4), B1(x,y) and C1 (10,16) by a transformation N =Find
 Matrix N of the transformation (4mks)
 Coordinates of B1 (2mks)
 A^{II}B^{II}C^{II} are the image of A1B1C1 under transformation represented by matrix M =Write down the coordinates of A^{II}B^{II}C^{II}(2mks)
 A transformation N followed by M can be represented by a single transformation K.
Determine K (2mks)
 The roof of a ware house is in the shape of a triangular prism as shown below
Calculate The angle between faces RSTU and PQRS (3mks)
 The space occupied by the roof (3mks)
 The angle between the plane QTR and PQRS (4mks)
MARKING SCHEME
 (√2 +√3) (√6 + √3)
(√6  √3) (√6 + √3)
√12 + √6 + √18 + 3
6 – 3
2√3 + √6 + 3√2 + 3
3  3(2x1) = 8x1
6x3 = 8x1
2x = 2
x = 1
= 200,000 (1.3107960)
= Sh. 262159.20
I = 262159.20 – 200000 = Sh.62,159 12th term = ar^{11}
10th term = ar^{9}ar^{11} = 9
ar^{9} 1
r^{119} = 9
r2 = 9
r = ± 3
r = 3 or 3 
 (2 – ¼ x)^{5} = 2^{5} + (2^{4})(5)(¼x)^{2} +
10(2^{2})(¼x)^{3}+ 5(2)(¼x)^{4} + (¼x)^{5}=32 – 20x + 5x^{2} – ^{5}/_{8}x^{3} + ^{5}/_{128}x^{4} – ^{1}/_{1024} x^{5}  1.96^{5} = 32  20(0.16) + 5(0.16)^{2} – ^{5}/_{8} (0.16)^{3} +^{5}/_{128}(0.16)^{4} – ^{1}/_{1024}(0.16)^{5}=28.925
 (2 – ¼ x)^{5} = 2^{5} + (2^{4})(5)(¼x)^{2} +

 QW x QX = QY x QZ
11 x 6 = 4(a+4)
4a +16 = 66
4a = 50
a = 25  QS^{2} = QY x QZ
= 4(4+12.5)
QS = √66
= 8.124
 QW x QX = QY x QZ
 x(x1) – 3x(x+1) = 0
x^{2} – x – 3x^{2} – 3x = 0
2x^{2} 4x = 0
2x (x + 2) = 0
x = 0 or x = 2  x f cf
45 – 50 2 2
51 – 56 10 12
57 – 62 11 23
63 – 68 20 33
69 – 74 6 39
75 – 80 1 40
¼ x 50 =12.5th = [56.5 + 12.5 – 12] 6
11
= 56.77kg
¾ x 50 = 37th ;
62.5 + [37.5 – 23] 6
20
= 66.85kg
Quartile deviation = ½ (66.85 – 56.77)
= 5.04  P = KQ^{3}
√R
P_{1} = K (1.2Q)^{3}
√0.64R
= 1.728KQ^{3}
0.8√R
= 2.16 KQ^{3}
√3
2.16 – 1 x 100
1
= 116%  Let cos x be y
8y^{2} – 2y – 1 = 0
(4y + 1) (2y – 1) = 0
y =  ¼ or ½
cos x = ¼ ⇒ x = 75.52
Angle in 2nd and 3rd quadrant
. : . x = 104.48, 255.52
Cosx = ½ ⇒ x = 60º
Angle in 1st and 4th quadrant.
x = 60º, 300º
.:. x = 104.48, 255.52º, 60º 300º  A² = 3 + 2χ A² (5  4χ) = 3 + 2χ
5  4χ
5A²  4A²χ = 3 + 2χ
5A²  3 = 2χ + 4A²χ
5A²  3 = χ(2 + 4A²)
χ= 5A²  3
4A² + 2
/AB/= √16 + 4 + 4
= √24
/AB/=4.90 x^{2}6x+9+y^{2}10y+25= 30+9+25
(x3)^{2}+(y5)^{2}=4
R=2
C (3,5)
3x+y=17
2x+4y=10
X=5.8
Y=0.4
(0.4,5.8) 2×0.1
40.1
=0.2
3.9
=^{2}/_{39}  Area scale factor =^{54}/_{9}= 6
4x – 2(x – 1) = 6
x = 2 
 Taxable income = 21,000 + 9000
p.a = sh. 30,000
30000 x 12 = K₤ 18,000 p.a
12
2 x 3900 = 7,800
3 x 3900 = 11,700
4 x 3900 = 15,600
5 x 3900 = 19,500
7 x 2400 = 16,800
71,400
^{15}/_{100} x 2000 = 300
Total relief p.a = (300 + 1056) 12
= sh. 16,272
Tax paid 71400 – 16272 = sh. 55, 128
P.A.Y.E 55128 = sh 4594
12  Total deductions = 4594 +2000 + 2000 + 2500 = sh. 11,094 per month
Net salary = 30,000 – 11,094
= sh. 18,906
 Taxable income = 21,000 + 9000
 ^{ }
 ^{7}/_{200} x 50 + ^{19}/_{400} x 30
1.75 + 1.425
= 3.175  3.175 x 100
80
= 3.96875%  let the masses be x
[^{19}/_{400} x + ^{7}/_{200}(50 – x)]100 = 4
50
[1.25x + 1.75]100= 4
50
1.25 x + 175 = 200
1.25x = 25
x = 25
1.25
x = 20
x > 20
 ^{7}/_{200} x 50 + ^{19}/_{400} x 30



 (^{7}/_{12} x ^{6}/_{11} x ^{5}/_{10}) + (^{7}/_{12} x ^{5}/_{11} x ^{6}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{6}/_{10})
= ^{21}/_{44}  (^{7}/_{12} x ^{5}/_{11} x ^{4}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{4}/_{10}) + (^{5}/_{12} x ^{4}/_{11} x ^{7}/_{10})
= ^{7}/_{22}  (^{5}/_{12} x ^{4}/_{11 }x ^{7}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{4}/_{10}) + (^{7}/_{12} x ^{5}/_{11} x ^{4}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{6}/_{10}) + (^{7}/_{12} x ^{5}/_{10} x ^{6}/_{10}) + (^{7}/_{12} x ^{6}/_{11} x ^{5}/_{10}) + (^{7}/_{12} x ^{5}/_{11} x ^{6}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{6}/_{10})
= 427/440
 (^{7}/_{12} x ^{6}/_{11} x ^{5}/_{10}) + (^{7}/_{12} x ^{5}/_{11} x ^{6}/_{10}) + (^{5}/_{12} x ^{7}/_{11} x ^{6}/_{10})


Class Mid point X f t =X  A
Cft t^{2} ft^{2} 1519
2024
2529
3034
3539
4044
4549
5054
5559
6064
6569
707417
22
27
32
37
42
47
52
57
62
67
746
10
9
5
7
11
15
13
8
7
5
45
4
3
2
1
0
1
2
3
4
5
630
40
27
10
7
0
15
26
24
28
25
2425
16
9
4
1
0
1
4
9
16
25
36150
160
81
20
7
0
15
52
72
112
125
144∑f=100 ∑ft=28 ∑ft^{2}=873 
 x = 42 + (^{28}/_{100} x 5) = 43.4
 ^{873}/_{100}  (^{28}/_{100})^{2} = 0.7856
 [49.5+ (7563) 5.29.5] x ½
13
= 12.31


A+4b = 7 ……….(i) x 4
4a + 16b ………..10 (ii)
4a + 16b = 28
4a  2b = 10
18b =18
b = 1
a = 94 = 3
c+4d =4 ………. (iii) x 4
4c – 2d = 16 ………..(iv)
4c + 16d = 16
4c2d=16
20d=0
d =0
c =4
Required angle is θ
82=102+1222x10x12cosθ
64=244240 cosθ
180= 240 cosθ
cosθ =180
240
θ= cos^{ 1 }180 = 41.41θ
240 Space =volume
= ½ x 6 x10sin 41.41ºx24
= 60sin41.41ºx24
=952.5m^{3}
TA= 10 sinθ= 10xsin41.41
Tan∞=10xsin41.41
24
tan∞=0.2756
= 15.41º
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