QUESTIONS
SECTION A (50 marks)
Answer all questions in this section in the spaces provided.
 Without using a calculator evaluate(3mks)
5x^{6} + (76)÷4+27÷3
(5)÷3 x (4) 
 Express 2268 in terms of its prime factors (1mk)
 Hence determine the smallest positive number x such that 2268x is a perfect square. (2mks)
 Elvis arrived in Kenya with 5000 sterling pound, he exchanged it to Kenya Shilling and spent sh. 267 100. Before jetting out of the country, he exchanged the balance into Euros. Using the exchange rates below, calculate the amount he obtained in Euros in Kenya shillings. (3mks)
Currency Buying Selling
1 Sterling pound 114.20 114.50
1Euro 101.20 101.30  Simplify the expression (3mks)
2x^{2}+3x2
x^{3} 4x  When two wires of length 179m and 234m are divided into pieces of equal lengths a remainder of 3m is left in each case. Find the least number of pieces that can be obtained. (3mks)
 Without using calculator, solve for n in the equation 1  (^{1}/_{3})^{n} = ^{242}/_{243} (3mks)
 Solve for y in the equation 7  y  9  2y = ½ (3mks)
4 3  Two similar solids have surface area of 48cm^{2} and 108cm^{2 }respectively. Find the volume of the smaller solid if the bigger solid has a volume of 162 cm^{3} (3 mks)
 Use reciprocal table only to evaluate ^{1}/_{0.325} (3mks)
Hence, evaluate √0.25 to 1.d.p
0.325  A plot measuring 1.2m by 19.1 m is surrounded by a path 0.5m wide. Find the area of the path in square metres. (3mks)
 The interior angle of a regular polygon is 600 more than its exterior angle, find the number of sides of the polygon. (3mks)
 Complete the following solid given that ABC is its crosssection (3mks)
 If tan x = ^{1}/_{√3} find without using tables or calculator the value of Sin (90x)+ cos (90x) leaving your answer in simplified surd form (3mks)
 A line perpendicular to the line 3y2x=2 passes through the point (3,2). Determine the equation of the line and write it in the form ax +by = c where a, b, and c are constant. (3mks)
 Given that A = (^{4}_{1} ^{3}_{2}) and C = (^{3}_{1} ^{7}_{2}) (4mks)
Find B such that A^{2} + B = C^{1}  Ali travelled a distance of 5km from village A to village B in direction of N60ºE. He then changed direction and travelled a distance of 4km in the direction of 135º to village C.
 Using a scale of 1cm to represent 1.0 km represent the information on an accurate diagram. (2mks)
 Using scale drawing (a) above determine
 distance between A and C (1mk
 bearing of A from C (1mk)
SECTION B (50 MARKS):
Answer all the questions in this section in the spaces provided.
 The figure below shows a frustrum. The top and bottom radii are 5cm and 10cm respectively, while the vertical height of the frustrum is 12cm.
Find the: Slant height of the frustum. (3marks)
 Curved area of the frustum. (3marks)
 Volume of the frustum. (4marks)
 Bumala is a market centre 600km from Kisumu town.A bus starts from Kisumu for Bumala at 7.00am at an average speed of 80 km/h. At 8.30 am a car started from Kisumu to Bumala and moved at an average speed of 120 km/hr. Calculate
 The distance bus covered before the car started moving. (3marks)
 The relative speed for the two vehicles. (2marks)
 The time the car overtook the bus. (1 mark)
 Distance covered by the car before overtaking the bus. (2marks)
 Distance from Bumala to the car at the time the car was overtaking the bus. (2marks)
 The height of 36 students in a class was recorded to the nearest centimeter as follows:
148 159 158 163 166 155 155 179 158
161 160 157 165 165 175 173 172 178
147 168 157 172 165 154 170 157 167
155 159 173 171 168 160 172 156 167 Make a frequency distribution table using a class interval of 5 and starting with the class 145 – 149. (2marks)
 From the table above
 Calculate the mean mark (3marks)
 Calculate the median (3marks)
 Draw a frequency polygon using the table in (a) above. (2 marks)
 Bujumba Boys Secondary School. Intends to buy a certain number of chairs For Ksh. 16,200. The supplier agreed to offer a discount of Ksh. 60 per chair which will enable the school to get 3 chairs more.
Taking y as the originally intended number of chairs: Write an expression in terms of y for
 Original price per chair. (1mark)
 Price per chair after discount. (1mark)
 Determine
 The number of chair the school originally intended to buy. (4marks)
 Price per chair after discount. (2marks)
 The amount of money the school would have saved per chair of it got the intended number of chairs at a discount of 15%. (2marks)
 Write an expression in terms of y for

 Without using a protractor, construct triangle ABC such that angle ABC = 60º, BC = 8cm and AC = 9cm.Measure AB. (3marks)
 Drop a perpendicular from A to BC and measure its length. (2marks)
 Hence calculate the area of triangle ABC. (2marks)
 Locate a point D on BC such that the area of triangle ABC is three times that of triangle ABD. (3marks)
 In triangle ABC, shown below, AB = a AC = b point M lies on AB such that AM: MB = 2:3 and point N lies on AC such that AN: NC = 5:1 line BN intersects line MC at X.
 Express the following in terms of a and b
 BN (1 mark)
 CM (1 mark)
 Given that BX = kBN and CX = rCM where k and r are scalars
 Write two different expressions for AX in term of a, b, k and r (4marks)
 Find the values of k and r (4 marks)
 Express the following in terms of a and b
 A triangle ABC has vertices A(2,1), B(5,2) and C(0,4).
 On the grid provided plot the triangle ABC. (2 marks)
 A^{1}B^{1}C^{1} is the image of ABC under a translation (^{2}_{5}). Plot A^{1}B^{1}C^{1} and state its coordinates. (2 marks)
 Plot A^{1}^{1}B^{1}^{1}C^{1}^{1} the image of A^{1}B^{1}C^{1 }after a rotation about the origin through a negative quarter turn. State its coordinates. (3 marks)
 A^{1}^{1}^{1}B^{1}^{1}^{1}C^{1}^{1}^{1} is the image of A^{1}^{1}B^{1}^{1}C^{1}^{1} after a reflection on the line y = 0.
Plot A^{1}^{1}^{1}B^{1}^{1}^{1}C^{1}^{1}^{1} and state its coordinates.(3 marks)
 On the grid provided plot the triangle ABC. (2 marks)
 The table below shows income tax for a certain year
Monthly Income in Kenya Shillings (Ksh) Tax Rate 0  10164 10% 10165  19740 15% 19741  29316 20% 29317  38892 25% Over 38892 30%  Calculate
 Employees total taxable income in that month (2 Marks)
 The tax payable by the employee in that month (5 Marks)
 The employee’s income include a house allowance of Ksh 15,000 per month. The employee contributed of the 5% basic salary to a cooperative society. Calculate the employee’s net pay for that month (3 Marks)
 Calculate
MARKING SCHEME
 5x^{6} +(76) ÷ 4 +27÷3
15÷ 3 x (4)
= 30 + (19) + 9
5 x4
= ^{20}/_{20} = 1  2268= 2^{2} x 3^{4} x 7
x = 2^{2} x 34 x 7^{2}= 7
2^{2} x 3^{4} x 7  Amt in Ksh = 5000 x 114.2
= 571000
Remainder = 571000276100
= 303900
Amt in Euros = 303900
101.30
= 3000 Euros M1  2x^{2} + 3x 2 = (2x 1)( x+ 2)
X^{3}4x x(x^{2} 4)
= (2x 1)( x+ 2)
x( x 2)( x + 2)
= 2x  1
x( x 2)  179  3 = 176
234 – 3 = 231
176 = 2^{4} x 11
234 = 3 x 7 x 11
GCD = 11
Number of pieces = ^{176}/_{11} + ^{231}/_{11}
= 16 + 21
= 37  1(^{1}/_{3})^{n} =^{ 242}/_{243}(^{1}/_{3})^{n}= 1  ^{242}/_{243}
= ^{1}/_{243}
(^{1}/_{3})^{n} = (^{1}/_{3})^{5}
∴ n =5  4  y  9  2y= ½
4 3
3(4y) – 4(92y) =6
123y 36 +8y = 6
5y =30
Y=6  For locus (i)
For locus (ii)
For locus (iii)
For the region  1 = 1
0.325 3.25×10^{1}
= 0.3077×10^{1}
= 3.077
^{√0.25}/_{0.325} = 0.5×3.077
= 1.5385
= 1.5  External area = 20.1 × 2.2 = 44.22
Internal area = 19.1 × 1.2 = 22.92
Area of path = 44.22 – 22.92
= 21.3 m^{2}  x + x +60 = 180°
2x = 120˚
x = 60˚
Exterior =60
No. of sides = ^{360}/_{60}
= 6 sides
Sides
Broken lines
Shape of solid Total = 42×24 = 1008
Total with abs = 100865
= 943
Average = ^{943}/_{23} =41  γ = ^{2x}/_{3} + ^{2}/_{3}
Gradient = ^{3}/_{2}
γ  2 = ^{3}/_{2}
x3
2(γ2) = 3(x+3)
2γ – 4 = 3x – 9
3x + 2γ = 5  B = A^{2} + C^{1}
= (^{4}_{1} ^{3}_{2})(^{4}_{1} ^{3}_{2}) + (^{2}_{1} ^{7}_{3})
= (^{13}_{6} ^{18}_{1}) + (^{2}_{1} ^{7}_{3})
= (^{11}_{5} ^{11}_{4}) 
 AC = 7.2km
 Bearing 273º

 X = 10078
= 22  Modal class 3544
x f cf d fd 1524 19.5 6 6 40 240 2534 29.5 14 20 30 420 3544 39.5 24 44 20 480 4554 49.5 14 58 10 140 5564 59.5 22 80 0 0 6574 69.5 10 90 10 100 7584 79.5 6 96 20 120 8594 89.5 4 100 30 120 100 940
14
= 44.5 + ^{6}/_{14} × 10
= 44.5 + 4.29
= 48.79 Mean = ^{∑fd}/_{Σf} + A
= 59.5  ^{940}/_{100}
= 50.1
 X = 10078

 x = ^{5}/_{10}
x + 12
10x = 5x + 60
5x = 60
X = 12
L = 24^{2} + 10^{2} = 26
^{1}/_{26} = ^{5}/_{10}
L = 13
Slant height 26 – 13
= 13  Curved area of the frustum
S.A = πRL  πrl
= π(260 – 65)
= 612.6  Volume of the frustrum
^{1}/_{3}π(R^{2}H – r^{2}h)
=^{1}/_{3}π(2400 – 300)
=^{1}/_{3}π(2100)
= 2200 cm^{2}
 x = ^{5}/_{10}

 1½ x 80
= 120km  120 – 50
= 40  Time taken
^{120}/_{40}= 3hrs
8.300 + 3
11.30A.m  3 x 120
= 360km  600 – 360 = 240km
For 240
Total marks 10
 1½ x 80

Class Tally Frequency Mid point Fx Cf 145 – 149 11 2 147 294 2 150 – 154 1 1 152 152 3 155 – 159 1 11 157 1727 14 160 – 164 5 162 810 19 165 – 169 11 7 167 1169 26 170 – 174 11 7 172 1204 33 175  179 111 3 177 531 36 Fx^{2 }5887 
 5887
36
= 163.53  159.5+(^{36}/_{2}14)5
11
= 159.5 + 1. 8182
= 161 . 32
 5887
 Graph
Plotting
Curve
(No hanging curve. Not drawn by use of free hand)
Total marks 10


 16200
Y  16 200 or 16 200 – 60
y + 3 y
 16200

 16 200 – 16 200 = 60
y y + 3
60y^{2} + 180y – 48 600 = 0
Y^{2} + 3y – 810 = 0
(y + 30) ( y – 27) = 0
= 27 or 30
= y = 27  16200 = Ksh 540
30  15 x 16 200
100 27
= Ksh 90
 16 200 – 16 200 = 60


Constructed locating point A
Dropping a
Length the 8.2 + 0.1
For a line at an angle to BC
For joining C to the last point
For locating point D
(following through)
Total marks 10 1⁄2×8×8.2=32.8Cm^{2}


 BN = BA + AN
= a + ^{5}/_{6} b
= ^{5}/_{6}b – a  CM = b + ^{2}/_{5}a
= b + r (^{2}/_{5}a  b)  AX = b + rCM
= b – rb + ^{2}/_{5}ra
= (1 – r) b + ^{2}/_{5}ar
Ax = a + k ( ^{5}/_{6}b – a)
(1 – k)a + ^{5}/_{6}kb  (1 – r) b + ^{2}/_{5}ar = ( 1 k) a + ^{5}/_{6}kb
^{5}/_{6}K = 1 – r – (i)
^{2}/_{5}r = 1 – k  (ii)
K = ^{6}/_{5} – ^{6}/_{5}r
^{2}/_{5} = 1 (^{6}/_{5} –^{6}/_{5}r) = 1 – ^{6}/_{5} +^{6}/_{5}r
4r = 1 = 7 r = ¼
^{5}/_{6}K = ¾
K = ^{9}/_{10}
 BN = BA + AN


 Equating two values of AX.
Two equations extracted  A^{1}(4,4) B^{1}(7,3) C^{1}(2,1)
 A^{1}^{1}(4,4) B^{1}^{1}(3,7) C^{1}^{1}(1,2)
 A^{1}^{1}^{1}(4,4) B^{1}^{1}^{1}(3,7) C^{1}^{1}^{1}(1,2)
For plotting
For ∆ABC
For ∆A^{1}B^{1}C^{1}
For construction or otherwise
For ∆A^{1}^{1}B^{1}^{1}C^{1}^{1}
For construction or otherwise
For ∆A^{1}^{1}^{1}B^{1}^{1}^{1}C^{1}^{1}^{1}
 Equating two values of AX.

 S = t (t^{2} – t – 2t + 2)
= t^{3} – 3t^{2} + 2t
^{ds}/_{dt} = 3t^{2} – 6 (2) + 2
V = 3 ( 4)  6 ( 2) + 2
V = 2m/s  ^{dv}/_{dt} = 0
^{dv}/_{dt} = 6t – 6
6t – 6 = 0
t = 1
= 3(1)2 – 6(1) + 2
= 1m/s
3t^{2} – 6t + 2 = 0  T = 6 + (6)2 – 4 ( 3) ( 2)
T = 6 + 3.464
T = 1.577 or 0.4227
a = 6 ( 3) – 6 = 12m/s^{2}
 S = t (t^{2} – t – 2t + 2)
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