INSTRUCTIONS TO THE CANDIDATES
 Write your name and index number in the spaces provided above
 This paper contains two sections; Section A and Section B
 All workings and answers must be written on the question paper in the spaces provided below each question.
 Marks may be given for correct working even if the answer is wrong.
 Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
 This paper consists of 15 printed pages.
 Candidates should check carefully to ascertain that all the pages are printed as indicated and no questions are missing.
Section A. (50mks)
Answer all the questions in this section in the spaces provided.
 Without using a calculator evaluate. (3mks)
(2^{1}/_{3}1^{1}/_{5} of 4)
¼  (^{1}/_{3})^{2}  Use logarithms to evaluate (4mks)
^{5}√(0.006628^{3}×193.9
2.2822  Without using a calculator or mathematical tables evaluate. (3mks)
4^{3⁄2} × (^{256}/_{2401})^{¾}  The base of a right pyramid is a rectangle of length 80cm and width 60cm. each slant edge of the pyramid is 150cm. Calculate the volume of the pyramid. (3mks)
 Simply. (3mks)
2x^{2} xy  6y ^{2}
x^{2} 4xy + 4y^{2}  A two digit number is such that, the sum of its digits is 13. When the digits are interchanged the original number is increased by 9. Find the original number. (4mks)
 The size of an interior angle of a regular polygon is 3x^{0} while its exterior is (x+20)^{2}. Find the number of triangles that makes the polygon. (3mks)
 In the triangle ABC below BC = 14cm <ABC= 70º and <ACB=40º. Calculate; correct to 4 significant figures the areas of triangle ABC. (3mks)
 A vector given by (^{2}_{4}) transforms a point A(3,6) to A^{1}. Find the distance A A^{1} (3mks)

 Using a rule and a pair of compasses only, construct a quadrilateral PQRS in which PQ = 6cm ,PS = 4cm QR = 5cm, <PQR=105º and <SPQ= 90º (2mks)
 The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in metres. (2mks)
 Without using mathematical tables or calculator evaluate. (3mks)
sin30º  sin60º
60º  Use matrix method to solve. (3mks)
5x+3y= 35
3x4y=  8  Use mathematical tables to find the reciprocal of 0.0247, hence evaluate. (4mks)
∛3.025 Correct to 2 decimal places.
0.0247  A Kenyan businessman intended to buy goods worth US dollar 20,000 from South Africa. Calculate the value of the goods to the nearest south Africa (S.A) Rand given that 1 US dollar = Ksh 101.9378 and 1 S.A Rand = Ksh 7.6326. (3mks)
 Find all integral values of x which satisfy the inequalities. (3mks)
53x ≤x7 <112x  Express 5184 and 2744 in terms of its prime factors hence determine the value of
√5184 (4mks)
∛2744
Section B (50mks)
Answer any five questions from this section on the spaces provided.  Two vertices of a rectangle ABCD are A(3,6) and B( 7,12)
 Find the equation of line AB. (3mks)
 Find the equation of the perpendicular bisector of line AB. (4mks)
 Given that BC is perpendicular to AB. Find the equation of BC. (3mks)
 Three business partners Abila, Bwire and Chirchir contributed Ksh 120,000, Ksh 180,000 and Ksh 240,000 respectively to boost their business. They agreed to put 20% of the profit accrued back into the business and to use 35% of the profits for running the business. The remainder was to be shared among the business partners in the ratio of their contribution. At the end of the year, a gross profit of Ksh 225,000 was realised.
 Calculate the amount.
 Put back into the business. (2mks)
 Used for official operations. (1mk)
 Calculate the amount of profit each partner got. (4mks)
 If the amount put back into the business was added to individual’s shares proportionately of their initial contributions, find the amount of Chirchir’s new shares. (3mks)
 Calculate the amount.
 Coast bus left Nairobi at 8.00Am and travelled towards Mombasa at an average speed of 80Km/hr. At 8.300am Lamu bus left Mombasa towards Nairobi at an average speed of 120Km/hr. given that the distance between Nairobi and Mombasa is 400Km, determine.
 The time Lamu bus arrived in Nairobi. (2mks)
 The time the two buses met. (4mks)
 The distance from Nairobi to the point where the buses met. (2mks)
 How far coast bus is from Mombasa when Lamu bus arrives in Nairobi. (2mks)
 A land is enclosed by four straight boundaries AB,BC,CD and DA. Point B is 25Km on bearing of 315º from A, C is directly south of B on a bearing of 260º from A and D is 30Km on a bearing of 210º from C
 Using a scale of 1cm to represents 5Km represent the above information on a scale drawing. (3mks)
 Using the scale drawing, determine the
 Distance in Kilometres of D from A. (2mks)
 Bearing of A from D. (1mk)
 Calculate the area, correct to 1 decimal place, of the land in square kilometres. (4mks)
 Complete the table below for the functions
 y= x^{2} 6x+7 (2mks)
x 0 1 2 3 4 5 6 x^{2} 0   9 16 25  6x  6 12    36 7 7 7 7 7 7 7 7 Y = x^{2}  6x + 7 7  1 2     On the grid provided below draw the graph of y= x^{2} 6x+7 for 0 ≤ x ≤6 and use it to estimate the roots of the equation.
x^{2} 6x+7=0 (4mks)  Use the graph above to solve the equation x^{2} 7x+9=0 (3mks)
 Determine the range of values of X for which x^{2} 6x+7 <x2 (1mk)
 y= x^{2} 6x+7 (2mks)
 In the figure below PQR and S are points on the circumference of the circle centre O. TP and TR are tangents to the circle at P and R respectively. POQ is a diameter of the circle and angle PQR = 64º (10mks)
Giving reasons on each case, find the size of <ROP
 <PSR
 <ORP
 <TRP
 <RTP
 The figure below shows two triangles ABC and BCD with a common base BC = 3,4cm. The area of triangle ABC = Area of triangle BCD and ABC =90º
 Calculate, correct to one decimal place.
 The area of triangle ABC. (3mks)
 The size of <BCD (3mks)
 The length of BD (2mks)
 The size of <BDC (2mks)
 Calculate, correct to one decimal place.
 The marks scored by 40 students in a mathematics test were as shown in the table below.
Marks 48 – 52 53 – 57 58 – 62 63 – 67 68 – 72 73 – 77 Number of students 3 4 10 12 8 3  State the modal class. (1mk)
 Using an assumed mean of 64, calculate the mean mark. (3mks)
 On the grid provided, draw the cumulative frequency curve for the data. (3mks)
 Use the graph to estimate the semi interquartile range (3mks)
MARKING SCHEME
 2^{1}/_{3} – 1^{1}/_{5}o of 4
^{6}/_{5} ×4= ^{24}/_{5}^{7}/_{3}= ^{24}/_{5}= ^{35}/_{15 } 72= ^{37}/_{15}
¼  (^{1}/_{3})2
¼  ^{1}/_{9}= 9  4 = 5
36 36
^{37}/_{15 }× ^{36}/_{5 }= 17^{19}/_{25}
M_{1} for ^{37}/_{15}
M_{1} for ^{5}/_{36}
A_{1} for 19^{19}/_{25}  No. std Log
0.006628 6.628×10^{3} 3 ̅ .8214
3
7 ̅ .4642
+
193.9 1.939×10^{2 } 2.2876
5 ̅ .7518
2.2822 2.2822×10º 0.3584
5 ̅ .3934
5
0.1198 1.198×10^{1 } 1 ̅.0.786
M_{1} for – correct logs
M_{1} for 5 ̅ .7518
M_{1} for 1 ̅.0.786
A_{1} for C.A  49^{ 3⁄2} = √49^{3}= 7^{3}=343
256 = ( ∜256 )^{3}=64
2401= ( ∜2401)^{3}=243
343 × ^{64}/_{343}=64
M_{1}for 343
M_{1} for 64, 343
A_{1} for C.A
Length of diagonal
√(80^{2 }+ 60^{2}=100
½ diagonal = 50
Height = 150^{2} 50^{2}
= 22500 – 2500
= 20000
=√20000=141.4
^{1}/_{3}×80 ×60 ×141.4
226. 2
M_{1} for height
M_{1} for expression of volume
A_{1} for C.A 2x^{2} – xy – 6y^{2}
2x^{2} – 4xy + 3xy – 6y^{2}
2x (x2y) + 3y( x2y)
(2x+3y) (x – 2y)
x^{2} 4xy + 4y^{2}x^{2}– 2xy – 2xy + 4y^{2}x(x2y) – 2y( x2y)
(x2y) (x2y)
(2x+3y)(x2y)
(x2y) (x2y)
2x+3y
x2y
M_{1} for (2x+3y) (x – 2y)
M_{1} for (x2y) (x2y)
A_{1} for C.A  x+y=13
(10y + x )  ( 10x + y) = 9
10y + x – 10x – y = 9
9y – 9x = 9
y – x = 1
 x+ y = 13
x + y = 1
2x = 12
y=6
x + y = 13
6 +y = 13
Y = 7
M_{1} for 9y – 9x = 9
M_{1 }for 2x = 12
A_{1} for both values of x and y.  int+ext=180
3x + x + 20 = 180
4x + 20 = 180
4x +160
X = 40
Ext = 40 + 20 = 60
Ext = ^{360}/_{n}= ^{360}/_{n}=60
n=^{360}/_{60}=6
Number of triangles = n – 2
= 6 2
= 4
M_{1} for expression
M_{1} for ^{360}/_{60}
A_{1} for C.A  <BAC=180110
= 70º
14 = AC
sin70 sin70
AC = 14 × sin70
sin 70
AC = 14
Area = ½ ×14 14 × sin40
=62.99
=63.0
M_{1} for expression of AC
M_{1} for for expression of area
A_{1} for C.A  O+T =1
(^{3}_{6})+ (^{2}_{4})= (^{5}_{2})
Distance  √(5^{2}+ 2^{2})
= 5.385
M_{1} for addition of vectors
M_{1} for for expression of distance
A_{1} for C.A
Rs = 7.cm
1: 4000
7.4: 7.4 ×4000
= 29600
B_{1} for PQR and SPQ
B_{1} for for complete figure
B_{1} for for RS
B_{1} for C.A sin 30º  sin 6º
tan60º
Sin 30º=½ tan60º = √3
sin60º= ^{√3}/_{2}
M_{1} for trigonometric ratio
M_{1} for for rationalising
A_{1} for C.A  5x + 3y =38
3x – 4y = 8
x=4 y=5
B_{1} for determinant
M_{1} for inverse
A_{1} for C.A  Reciprocal of
2.47 ×10^{2}
=0.4049 × 10^{2}=40.49
∛(3.025 ) =1.446
=40.49 ×1.446=58.55
M_{1} for reciprocal
M_{1} for ∛
A_{1} for C.A  1 US dollar = Ksh 101.9378
20,000 = 20,000 ×101.9378
=Ksh 2038756
1 S.A = 7.6326
2038756 =2038756
7.6326
= 267111.6
= 267112 S.A
M_{1} for expression
M_{1} for expression
A_{1} for C.A  53x ≤x7 <112x
5x – 3x ≤x 7 x7 <112x
5+ 7 ≤x+3x x+2x <11+7
12 ≤4x 3x <18
3≤x x <6
3≤x <6
3,4 ,5
M_{1} for separating inequality
M_{1} for 3≤x and x <6
A_{1} for C.A  5184 = 2×2×2×2×2×2×3×3×3×3
√5184=2×2×2×3×3
= 72
2744 = 2×2×2×7×7×7
∛2744=2×7=14
= 72/14 = 5^{1}/_{7}
M_{1} for factors of 5184
M_{1} for factros of 2744
M_{1} for √ and ∛
A_{1} for C.A 
 A( 3,6) B(7,12) (x ,y)
Gradient of AB = 126 = 6 = 3
73 4 2
y  6 = 3
x  3 2
2(y – 6) = 3( x 3)
2y – 12 = 3x – 9
2y = 3x – 9 + 12
2y = 3x + 3  mid point of AB
3+7 , 6+12
2 2
(5,9) (x,y)
Perpendicular bisector gradient =  ^{2}/_{3}
(y9 )/(x5 )= 2/(3 )
3( y9) = 2(x 5)
3y – 2y = 2x + 10
3y = 2x + 10 + 27
3y = 2x + 37
AB ⊥BC gradient of Bc=  ^{2}/_{3},B( 7,12)
(x,y ) x y12 = ^{2}/_{3} ) 3y= – 2x + 14 + 36
x7
3y = 2x + 50
3(y12)= 2(x7)
3y – 36 = 2x + 14
M_{1} for gradient
M_{1} for expression
A_{1} for equation
M_{1} for mid point
M_{1} for gradient
M_{1} for expression
A_{1} for equation
M_{1} for expression= ^{2}/_{3}M_{1} for xmultiplication
A_{1} for C.A
 A( 3,6) B(7,12) (x ,y)

 A ∶ B ∶ C
120000∶180000∶ 240000
12 ∶ 18 ∶ 24
2 ∶ 3 ∶ 4 ^{20}/_{100} ×225000=45000
 ^{35}/_{100} ×225000 = 78750
 Remainder = 2250004500078750
= 101 250
2: 3∶ 4 = 9
= 101250 ^{2}/_{9} ×101 250 = 22,500
 ^{3}/_{9} ×101 250 = 33,750
 ^{4}/_{9} ×101 250 = 45,000
 2:3:4=9
45000  ^{4}/_{9} ×45000 = 20,000
Chirchir new share = 240,000+20,000
= 260,000
M_{1} for multiplication
A_{1} for C.A
A_{1} for C.A
M_{1} for 101250
A_{1} for C.A
A_{1} for C.A
A_{1} for C.A
M_{1} for expression
M_{1} for addition
A_{1} for C.A
 A ∶ B ∶ C

 Nairobi Mombasa
8.00am 8.30am
80Km/h 120Km/h
Time = ^{D}/_{S}= ^{400̸}/_{120̸} =3^{1}/_{3} hrs
= 3: 20min
+8∶ 30
3∶ 20
= 11 ∶ 50 am  Distance by coast bus for ½ hr.
= 80 × ½=40Km
R.D = 400  40 = 360 Km
R.S = 80 + 120 = 200Km/h
R.T = ^{260}/_{200}=1^{4}/_{5}hr=1 hr:48min
Time met = 8∶ 30
1∶ 48
= 10: 18am  R.T = 1^{4}/_{5} hr
Coast by distance in 1^{4}/_{5}
= 80 × 1^{4}/_{5}
= 144 km
Distance = 40 + 144
= 184  In 3hr20min coast bus covers
= 80 × 3^{1}/_{3}
= 80 × ^{10}/_{3} =266 ^{2}/_{3} km
Distance from Mombasa = 400 – 266 ^{2}/_{3}
=133^{1}/_{3} Km
M_{1} for 3^{1}/_{3} hr
M_{1} for addition
A_{1} for C.A
M_{1} for 40km
M_{1} for RD and R.S
M_{1} for expression of R.T
A_{1} for C.A
M_{1} for expression of distance
A_{1} for C.A
M_{1} for 400 – 266 ^{2}/_{3}A_{1} for C.A
 Nairobi Mombasa
AD = 8.7 cm = 43.5 km ± 1
Bearing = 047º
B_{1} for point B
B_{1} for point C
B_{1} for Point D
B_{1} for 8.7 ± 0.1
A_{1} for 43.5 ± 1
A_{1 }for 047º ± 1 y = x6x+7
x 0 1 2 3 4 5 6 x^{2} 0 1 4 9 16 25 36 6x 0 6 12 18 24 30 36 7 7 7 7 7 7 7 7 Y = x^{2}  6x + 7 7 2 1 2 1 2 7 
 <ROP= 128º
Angles on a straight line  <PSR= 116º
Angle on cyclic quadrilateral  <ORP= 26º
Base angle of isosceles triangle  <TRP= 64º
Angle between a chard a tangent( Nagle in alternate segment)  <RTP=53º
Angles on a triangle.
B_{1 }for Angle
B_{1} for Reason
B_{1} for Angle
B_{1} for Reason
B_{1} for Angle
B_{1} for Reason
B_{1} for Angle
B_{1} for Reason
B_{1} for Angle
B_{1} for Reason
 <ROP= 128º
 AB height = ∛7.2^{2} 3.4^{2}
= 6.347 Area ABC
= ½ 3.4 ×6.347
= 10.78
= 10.8  Area of BCD = 10.8
A= ½ ×3.4 ×7.5 ×sinBCD
10.8
10.8 = 1.7 ×7.5 ×sinBCD=sinBCD
1.7 ×7.5
0.8471 sin BCD
sin^{1} 0.8471=BCD
= 57.8 = BCD  BD^{2}=3.4+7.52×3.4 ×7.5 cos57.8
BD = 6.37  <BDC = 34 3.4 = 6.37
sinBCD s inBDC sin57.8
Sin BDC – 3.4 × sin57.8
6.37
Sin BDC = 0.4517
BDC = sin^{1}0.4517
BDC = 26.9
M_{1} for expression of height
M_{1} for expression of Area
A_{1 }for C.A
M_{1} for expression of Area
M_{1} for 0.8471
A_{1} for C.A
M_{1} for expression
A_{1} for C.A
M_{1} for expression
A_{1} for C.A
 Area ABC
Marks x F Tx  A FT CF 48 – 52 50 3 14 42 3 53 – 57 55 4 9 36 7 58 – 62 60 10 4 40 17 63 – 67 65 12 1 12 29 68 – 72 70 8 6 48 37 73 – 77 75 ^{3}/_{40} 11 ^{33}/_{25} 40  Modal class 63 – 67
 Mean = A + ^{ƸFt}/_{ƸF}
64 + ^{25}/_{40}
= 63 ^{3}/_{8}
B_{1} for modal class
M_{1} for mid point
M_{1} for ^{25}/_{40}A_{1} for C.A
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