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 log10 (2x-1) + log103 = log10 (8x-1). (3mks)
- Find the compound interest on sh. 200,000 for 2 years at 14% pa. Compounded semi-annually. (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 cos2 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= 2i-3j+4k and b= -2i-j+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 x2-6x+y2-10y+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 9cm2 is mapped onto an image whose area is 54cm2. 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) 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 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)
- AIIBIICII are the image of A1B1C1 under transformation represented by matrix M = Write down the co-ordinates of AIIBIICII(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(2x-1) = 8x-1
6x-3 = 8x-1
-2x = 2
x = -1
= 200,000 (1.3107960)
= Sh. 262159.20
I = 262159.20 – 200000 = Sh.62,159- 12th term = ar11
10th term = ar9
ar11 = 9
ar9 1
r11-9 = 9
r2 = 9
r = ± 3
r = 3 or -3 -
- (2 – ¼ x)5 = 25 + (24)(5)(-¼x)2 +
10(22)(-¼x)3+ 5(2)(-¼x)4 + (-¼x)5
=32 – 20x + 5x2 – 5/8x3 + 5/128x4 – 1/1024 x5 - 1.965 = 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 = 25 + (24)(5)(-¼x)2 +
-
- QW x QX = QY x QZ
11 x 6 = 4(a+4)
4a +16 = 66
4a = 50
a = 25 - QS2 = QY x QZ
= 4(4+12.5)
QS = √66
= 8.124
- QW x QX = QY x QZ
- x(x-1) – 3x(x+1) = 0
x2 – x – 3x2 – 3x = 0
-2x2 -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 = KQ3
√R
P1 = K (1.2Q)3
√0.64R
= 1.728KQ3
0.8√R
= 2.16 KQ3
√3
2.16 – 1 x 100
1
= 116% - Let cos x be y
8y2 – 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- x2-6x+9+y2-10y+25= -30+9+25
(x-3)2+(y-5)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
4-0.1
=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 t2 ft2 15-19
20-24
25-29
30-34
35-39
40-44
45-49
50-54
55-59
60-64
65-69
70-7417
22
27
32
37
42
47
52
57
62
67
746
10
9
5
7
11
15
13
8
7
5
4-5
-4
-3
-2
-1
0
1
2
3
4
5
6-30
-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 ∑ft2=873 -
- x = 42 + (28/100 x 5) = 43.4
- 873/100 - (28/100)2 = 0.7856
- [49.5+ (75-63) 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 = 9-4 = 3
c+4d =4 ………. (iii) x 4
4c – 2d = 16 ………..(iv)
4c + 16d = 16
4c-2d=16
20d=0
d =0
c =4-
Required angle is θ
82=102+122-2x10x12cosθ
64=244-240 cosθ
-180= -240 cosθ
cosθ =180
240
θ= cos -1 180 = 41.41θ
240- Space =volume
= ½ x 6 x10sin 41.41ºx24
= 60sin41.41ºx24
=952.5m3
TA= 10 sinθ= 10xsin41.41
Tan∞=10xsin41.41
24
tan∞=0.2756
= 15.41º
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