SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.
 Use a calculator to evaluate (3.84)^{2}  √110.592
0.03885 (2 marks)  In a certain year, Ondege's annual salary was Ksh 120600. For the next seven years, his annual increment was Ksh 2 880.
Determine: Ondege's annual salary in the 7th year; (2 marks)
 the total salary that Ondege earned during the first six years. (2 marks)
 A curve crosses the xaxis at x = 1^{1}/_{3} and x = ½. Determine the equation of the curve in the form ax^{2} + bx + c = 0 where a, b and e are integers. (2 marks)
 Sifa invested an amount of money in a financial institution which paid a simple interest rate of 5% per annum. After 3½ years, the total amount of money in Sifa's account was Ksh 37 600. Calculate the amount of money that Sifa invested. (3 marks)
 A county allocated funds for various projects as shown in the piechart below.
The allocation for Agriculture was twice that of Peace Initiative. Calculate the size of the angle of the sector that represents the Peace Initiative.  Given that matrix (3 marks)
 The table below shows the amount of milk, in litres, delivered to a milk processing plant, on a certain day, by 80 farmers.
Milk in litres 1014 1519 2024 2529 3034 3539 4044 4549 No. of farmers 2 6 14 24 13 10 8 3  A point A is mapped onto A'(4, 6) by a transformation whose matrix is
Find the coordinates of A. (3 marks)  In a children's home, the amount of water in litres (L) used per month is partly constant and partly varies with the number n, of the children. In a certain month, there were 50 children and the amount of water used was 78 000 L. In another month, there were 70 children and 85 200 L of water was used.
 Form an equation connecting L and n. (1 mark)
 Determine the amount of water used in a month when the number of children was 100. (3marks)
 The figure below shows two circles, centres O_{1} and O_{2}. PQ is a common tangent to the circles. The radius of the smaller circle is 0.12m while the radius of the larger circle is 0.8 m. The distance between O_{1} and O_{2} is 2.5 m.
Calculate the length PQ, correct to 2 decimal places. (3 marks)  T and R are two towns on the equator. The longitude of T is 12°E and that of R is 5°W. If the local time at T is 2245 h, find the time at R in the 12hour clock system. (4 marks)
 Given that 4 tan x  5 = 0, find the value of x correct to 2 decimal places for 0° ≤ x ≤ 360°. (3 marks)
 A box contains 13 balls which are identical except for the colour. Three of the balls are red while the rest are white. Two balls are picked at random from the box, one at a time, without replacement.
 Using a tree diagram, show all the possible outcomes. (2 marks)
 Find the probability that a red and a white ball are picked. (2 marks)
 The x and y values of points on a curve are as shown in the table below.
x 0.5 1 1.5 2 2.5 3 y 1 3 6 10 15 21  Points P, Q and R lie on a straight line, such that PQ = iPR. Given P(2,5) and R(6,1), express in terms of i and j:
 the position vector of P; (1 mark)
 PQ. (2 marks)
 In an experiment, water was heated and its temperature changes recorded at intervals of 2 minutes as shown in the table below.
Time (min) 0 2 4 6 8 10 12 14 16 Temperature (°C) 25 35 42.5 50 60 67.5 77.5 85 92.5  On the grid provided, plot the points and draw the line of best fit. (2 marks)
 Use the line of best fit to estimate the time taken for the temperature of the water to reach 75°C. (1 mark)
SECTION II (50 marks)
Answer any five questions from this section in the spaces provided.
 A factory blends three types of juice in the ratios A:B = 3:4 and B:C = 1:2.
 Determine:
 the ratio A:B:C; (1 mark)
 the amount of type A juice in a 20litre mixture. (2 marks)
 The cost of producing one litre of A is Ksh 80, one litre of B is Ksh 84 and one litre of C is Ksh 90.
 Find the cost of producing one litre of the mixture. (2 marks)
 Calculate the selling price of one litre of the mixture if the factory makes a profit of 25%. (2 marks)
 The factory uses two types of machines P and Q to blend the juices. Machine P takes 7 hours to blend 14 000 litres and Q takes 5 hours to blend 12 000 litres. Determine the time it would take the factory to blend 550 000 litres. (3 marks)
 Determine:
 The diagram below represents a map of a settlement scheme. The map is drawn on a one centimetre square grid. The scale of the map is 1:50000
 Estimate:
 the area of the map in square centimetres; (2 marks)
 the area of the settlement scheme in square kilometres. (3 marks)
 The settlement scheme was subdivided into parcels of land each of 5 hectares.
 Find the number of the 5 hectare parcels of land obtained. (3 marks)
 Determine the area in hectares of the settlement scheme that remained after the subdivision. (2 marks)
 Estimate:
 In the figure below, TD is a tangent to the circle at T. Chord AC produced intersects TD at D. CD = 4.1 cm, TD = 7 cm and angle TAC = 33°.
 Giving a reason, find the size of angle CTD. (2 marks)
 Calculate the length of AC correct to one decimal place. (3 marks)
 Calculate to the nearest degree the value of:
 the obtuse angle TCD; (3 marks)
 angle ABC. (2 marks)

 Complete the table below for values of y = x^{3} + 3x^{2} + 5 for 4 ≤ x ≤ 2, correct to 1 decimal place. (2 marks)
x 4 3.5 3 2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 y 1.1 5 9 5.6 5.9 15.1  On the grid provided, draw the graph of y = x^{3} + 3^{2} + 5 for 4 ≤ x ≤ 2. Use the scale: 2cm for 1 unit on the xaxis and 2cm for 5 units on the y axis. (3 marks)
 Using the graph, determine:
 the average rate of change between x = 0.5 and x = 1.8; (2 marks)
 the instantaneous rate of change of the curve at x = —3, correct to one decimal place. (3 marks)
 Complete the table below for values of y = x^{3} + 3x^{2} + 5 for 4 ≤ x ≤ 2, correct to 1 decimal place. (2 marks)
 A pavement is of length (x  1)m and width (x  8)m. The area of the pavement is 4.56 m^{2}.

 Write a quadratic equation for the area of the pavement in the form ax^{2} + bx + c = 0 where a, b and c are constants. (2 marks)
 Using the method of completing the square, find the actual length and width of the pavement. (6 marks)
 The pavement is covered with rectangular tiles measuring 0.4 m by 0.3 m. Determine the number of tiles used to cover the pavement completely. (2 marks)

 Omari bought a house valued for Ksh 4 000 000. The value of the house appreciated at 20% per annum for the first three years and then at 15% per annum for the next two years.
 Calculate the value of the house after:
 three years, (2 marks)
 five years. (2 marks)
 After the five years, the value of the house depreciated for the next two years. At the end of the two years, Omari sold the house through an agent. Omari received Ksh 7 125 000 after paying a 5% commission to the agent.
Calculate: the value of the house after the two years; (2 marks)
 the annual rate of depreciation in the two years. (4 marks)
 Calculate the value of the house after:
 A quadrilateral ABCD has vertices A(3, 1), B(2, 3), C(3, 4) and D(4, 3).

 Find the coordinates of A'B'C'D', the image of ABCD, under a transformation whose matrix is (2 marks)
 On the grid provided, draw the quadrilateral ABCD and its image A'B'C'D'. (2 marks)
 A"B"C"D" is the image of A'B'C'D' under a transformation whose matrix is followed by an enlargement scale factor 2, centre (0,0).
 Determine the coordinates of A"B"C"D". (2 marks)
 On the same grid as in (a) (ii) above, draw A"B"C"D". (1 mark)
 Find a single transformation matrix that maps A"B"C"D" onto ABCD. (3 marks)

 The masses, to the nearest kg, of 65 patients who attended a medical clinic are as shown in the table below:
Mass (kg) 2630 3135 3640 4145 4650 5155 Frequency (f) 9 13 20 15 6 2  the mean mass of the patients; (4 marks)

 the variance mass; (5 marks)
 the standard deviation. (1 mark)
MARKING SCHEME
 3.84^{2}  ^{3}√110.592
0.03885
14.7456  4.8
0.03885
= 256
B1 Squaring and cube roo 
 Annual salary in the 7th year:
= 120 600 + 2880 x 6
= Ksh137 880  Total earnings for 6 years
S_{6} = ^{6}/_{2} {2 x 120 600 + 5(2880)}
= Ksh 766 800
 Annual salary in the 7th year:
 (x  ^{4}/_{3})(x+½) = 0
x^{3}  ^{5}/_{6}x  ^{2}/_{3} =0
6x^{2}  5x 4 = 0  P + PRT = 37600
100
P + P x 5 x 3.5 = 37600
100
P(1.175) = 37600
P = 37600
1.175
= Ksh 32 000  3x = 100  (17.5 + 15 + 30)
x = 12.5%
Angle for peace initiative
= 12.5 x 360
100
= 45°
Multiplication by scalar
Classes 1014 1519 2024 2529 3034 3539 4044 4549 frequency (f) 2 6 14 24 13 10 8 3 c.f 2 8 22 46 59 69 77 80
Median = 24.5+ 40  22 x 5
24
= 24.5 +3.75
= 28.25
2a = 4 = a = 2
2b = 6 b = 3
Coordinates of A=(2,3)
 L=C+ an
 78 000 = C + 50a
85 200 = C +70a
20a = 7200
a = 360
A1 For both C and a
C = 78 000  50 X 360
= 60 000
L = 60 000 + 100 x 360
= 96 000
 O_{2}R = 0.8 0.12 = 0.68
PQ = √2.5^{2} 0.68^{2}
= 2.41  Difference in longitude = 12° +5 = 17°
Time difference = 17 x 4
= 68 minutes
Time at R:
= 2245 h  1 hr 08 min = 2137 h = 9.37 pm  4 tan x = 5
tan x = ^{5}/_{4} = 1.25
x = tan ^{1} (1.25)
= 51.34°
and 231.34° 
B1 for 1st branch correct
B1 for 2nd branch correct P(R+W)
= 3 x 10 + 10 x 3
13 12 13 12
= 5
13
 Area
½ x ½[1+21+2(3+6+10+15)]
= ¼[22 +2 x 34] = 225 
 OP = 2i + 5j
 PQ = ^{3}/_{4} {(6i+j)(21 – 5j)}
= ^{3}/_{4} (4i  4j)
=3i  3j

√Scale
B1 Vplotting and line of best fit Time taken to reach = 750 C = 11.6 min


 A:B = 3:4
B:C = 1:2 B:C = 4:8
A:B:C = 3:4:8  ^{3}/_{15} x 20
= 4 litres
 A:B = 3:4

 Cost of production per litre
= 80 x 3 + 84 x 4+90 x 8
15
= Ksh 86.4  Selling price of 1 litre for a 25% profit:
= 86.4 x125
100
= Ksh 108
 Cost of production per litre
 In 1h machines P and Q blend
14000 + 12000
7 5
= 2000 + 2400  4400 litres
Time taken to blend 550000 litres
550000
4400
= 125 hours



 Full squares = 37
Part squares = 27
Total area = 37+27
2
= 50.5 cm^{2}  area in km^{2}
= 50.5 x 50000 x 50000
1000 x 1000 x 100 x 100
= 12.625 km^{2}
M1 conversion
 Full squares = 37

 Number of 5 ha parcels:
= 12.625 x 1000 x 1000
5 x 100 x 100
= 252.5
Number of equal parcels = 252  Remainder in hectares:
= 0.5 x 5
= 2.5 ha
 Number of 5 ha parcels:


 Size of angle CTD = 33°
 Angle in alternate segment  (x + 4.1) 4.1 = 7^{2}
4.1x + 4.1^{2} = 49
4.1 x = 32.19
x = 7.851219512
≈ 7.9 
 sin θ =sin 33
7 4.1
sinθ = 7 sin 33
4.1
= 0.9299
θ = 68°
obtuse angle = 180°  68° = 112°  ACT = 180°  112° = 68°
ATC = 180°  (68 +33) = 79°
ABC  180°  79°  101°
 sin θ =sin 33
 Size of angle CTD = 33°


x 4 3.5 3 2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 y 11 8.1 8.4 7 5 9 25 
 20.5  5.9
1.8  0.5
= 11.23  tangent line at x = 3
gradient : 10  5
24  3
= 8.3
 20.5  5.9



 (x  1)(x  8) = 4.56
x^{2}  9x + 8 = 4.56
x^{2}  9x + 3.44 = 0  x^{2} – 9 x +( 9 )^{2}=( 9 )^{2}  3.44
2 2 2
(x^{2} – 9 )^{2} = 81 = 3.44
2 4
(x^{2} – 9 ) = √16.81
2
x = 4.5 ± 4.1
= 8.6 or 0.4
x = 8.6 m
Length of pavement = 8.6  l = 7.6 m
Width of pavement = 8.6  8 = 0.6 m
 (x  1)(x  8) = 4.56
 No. of 0.4 m x 0.3 tiles
7.6 x 0.6
0.4 x 0.3
= 38



 Ksh. 4 000 000 x (1.2)^{3}
= Ksh 6912 000  4 000 000 x (1.2)^{3} x (1.15)^{2}
= Ksh 9 141 120
 Ksh. 4 000 000 x (1.2)^{3}

 7125 000 x 100
95
= Ksh 7 500 000  9141120 × (1  r )^{2} = 7500000
100
(1 r )^{2} = 7500000 = 0.82
100 9141120
1 r = 0.91
100
r = 1  0.91 = 0.09
100
r = 9%
 7125 000 x 100



Coordinates: A'(1,3), B'(3,2), C(4,3), D'(3,4)
Object ABCD drawn
Image A'B'C'D' drawn

Coordinates: A"(6,2), B"(4,6), C"(6,8), D'(8,6) image A"B"C"D" √drawn


Masses (kg) f Midpoint fx d=x38 fd fd^{2} 26  30 9 28 252 10 90 900 31  35 13 33 429 5 65 325 36 40 20 38 760 0 0 0 41  45 15 43 645 5 75 375 46  50 6 48 288 10 60 600 51  5 2 53 106 15 30 450 ∑=65 ∑fx=2480 ∑fd=10 ∑fd^{2} =2650
fx column
Mean = ∑fx = 2480
∑f 65
= 38.15 
 variance:
∑fd ^{2} (∑fd)^{2}
∑d ∑f
= 2650  (10)^{2}
65 65
= 40.769 0.02367
= 40.75  Standard deviation
= √40.75 = 6.38
 variance:

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