QUESTIONS
SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.
 Without using a calculator evaluate,
 Kutu withdrew some money from a bank. He spent ^{3}/_{8} of the money to pay for Mutua’s school fees and ^{2}/_{5} to pay for Tatu’s school fees. If he remained with 12 330, calculate the amount of money he paid for Tatu’s school fees. (4 marks)
 A straight line l passes through point (3, 2) and is perpendicular to a line whose equation is 2y4x = 1.
Find the equation of l in form of y = mx + c, where m and c are constants. (3 marks)  A bus left a petrol station at 9.20 a.m. and travelled at an average speed of 75 km/h to a town N. At 9.40 a.m. a taxi, travelling at a speed of 95 km/h, left the same petrol station and followed the route of the bus.
Determine the distance, from the petrol station, covered by the taxi at the time it caught up with the bus. (3 marks)  The sum of three consecutive odd integers is greater than 219. Determine the first three such integers. (3 marks)
 A Kenyan company received US Dollars 100 000. The money was converted into Kenya shillings in a bank which buys and sells foreign currencies as follows:
Buying (in Kenya shillings) Selling(in Kenya shillings) 1 US Dollar 77.24 77.44 1 Sterling Pound 121.93 122.27  Calculate the amount of money, in Kenya shillings, the company received. (2 marks)
 The company exchanged the Kenya shillings calculated in (a) above, into sterling pounds to buy a car from Britain. Calculate the cost of the car to the nearest sterling pound. (2 marks)
 In the figure below, OPQR is a trapezium in which PQ is parallel to OR and M is the midpoint of QR. OP = p, OR = r and PQ = ^{1}/_{3} OR.
Find OM in terms of p and r. (3 marks)  Without using a mathematical table or a calculator, evaluate . (3 marks)
 The figure below is a net of cube with some dots on two faces.
Given that the number of dots on pairs of opposite faces and add up to 7, fill in appropriate dots in each of the empty faces. (2 marks)  Using a ruler and a pair of compasses only, construct a rhombus QRST in which angle TQR = 60^{0} and QS = 10 cm. (3 marks)
 A fruit vendor bought 1948 oranges on a Thursday and sold 750 of them on the same day. On Friday, he sold 240 more oranges than on Thursday. On Saturday he bought 560 more oranges. Later that day he sold all the oranges he had at the price of Ksh 8 each. Calculate the amount of money obtained from the sales of Saturday. (4 marks)
 Simplify the expression . (3 marks)
 Given that 3θ is an acute angle sin 3θ = 2θ°, find the value of θ. (3 marks)
 A cylindrical solid whose radius and height are equal have a surface area of 154 cm^{2}. Calculate the diameter, correct to 2 decimal places. (Take π = 3.142) (3 marks)
 The figure below shows two sectors in which CD and EF are arcs of concentric circles, centre O. Angle COD = ^{2}/_{3} radians and CE = DF = 5 cm.
If the perimeter of the shape CDFE is 24 cm, calculate the length of OC. (3 marks)  The histogram shown below represents the distribution of heights of seedlings of a certain plant.
The shaded area in the histogram represents 20 seedlings. Calculate the percentage number of seedlings with heights of at least 23 cm but less than 27 cm. (3 marks)SECTION II (50 marks)
Answer only five questions in this section in the spaces provided.  A sales woman is paid a commission of 2% on goods sold worth over 100 000. She is also paid a monthly salary of Ksh 12 000. In a certain month, she sold 360 handbags at Ksh 500 each.
 Calculate the saleswoman’s earning that month. (3 marks)
 The following month, the saleswoman monthly salary was increased by 10%. Her total earnings that month were Ksh 17 600.
Calculate: The total amount of money received from the sales of handbags that month; (5 marks)
 The number of handbags sold that month. (2 marks)
 A carpenter constructed a closed wooden box with internal measurements 1.5 metres long, 0.8 metres wide and 0.4 metres high. The wood used in constructing the box was 1.0 cm thick and had a density of 0.6g/cm^{3}.
 Determine the:
 Volume, in cm^{3}, of the wood used in constructing the box; (4 marks)
 Mass of the box, in kilograms, correct to 1 decimal place. (2 marks)
 Identical cylindrical tins of diameter 10 cm, height 20 cm with a mass of 120 g each were packed in the box.
Calculate the: Maximum number of tins that were packed; (2 marks)
 Mass of the box with the tins. (2 marks)
 Determine the:

 Find A^{1}, the inverse of matrix A = (^{5}_{7 }^{6}_{9}). (2 marks)
 Okello bought 5 Physics books and 6 Mathematics books for a total of Ksh 2 440. Ali bought 7 Physics books and 9 Mathematics books for a total of Ksh 3 560.
 Form a matrix equation to represent the above information. (1 mark)
 Use matrix method to find the price of Physics book and that of a Mathematics book.
 A school bought 36 Physics books and 50 Mathematics books. A discount of 5% was allowed in each Physics book whereas a discount of 8% was allowed on each Mathematics book. Calculate the percentage discount on the cost of all the books bought. (4 marks)
 The boundaries PQ, QR, RS and SP of a ranch are straight lines such that: Q is 16 km on a bearing of 040^{0} from P; R is directly south of Q and east of P and S is 12 km on a bearing of 120^{o} from R.
 Using a scale of 1 cm to represent 2 km, show the information in a scale drawing. (3 marks)
 From the scale drawing determine:
 The distance in kilometers, of P from S; (2 marks)
 The bearing of P from S. (2 marks)
 Calculate the area of the ranch PQRS in square kilometers. (3 marks)
 Motorbike A travels at 10 km/h faster than motorbike B whose speed is x km/h. Motorbike A takes 1½ hours less than motorbike B to cover an 180 km journey.
 Write an expression in terms of x for the time taken to cover the 180 km journey by:
 Motorbike A; (1 mark)
 Motorbike B. (1 mark)
 Use the expression in (a) above to determine the speed in km/h , of motorbike A. (6 marks)
 For a journey of 48 km, motorbike b starts 10 minutes ahead of motorbike A. Calculate, in minutes, the difference in time of their arrival at the destination. (2 marks)
 Write an expression in terms of x for the time taken to cover the 180 km journey by:
 In the figure below, ABCD is a square. Points P,Q, R and S are the midpoints AB, BC, CD and DA respectively.
 Describe fully:
 A reflection that maps triangle QCE onto triangle SDE; (1 mark)
 An enlargement that maps triangle QCE onto triangle SAE; (2 mark)
 A rotation that maps triangle QCE onto triangle SED. (3 marks)
 The triangle ERC is reflected on the line BD. The image of ERC under the relection is rotated clockwise through an angle of 90^{0} about P.
Determine the images of R and C: Under the reflection; (2 marks)
 After the two successive transformations. (2 marks)
 Describe fully:
 The frequency distribution table below represents the number of kilograms of meat sold in a butchery.
 State the modal frequency. (1 mark)
 Calculate the mean mass. (5 marks)
 Calculate the median mass. (4 marks)
 A rectangular box open at the top has a square base. The internal side of the base is x cm long and the internal surface area of the box is 432 cm^{2}.
 Express in terms of x:
 The internal height h, of the box; (3 marks)
 The internal volume V, of the box. (1 mark)
 Find:
 The value of x for which the volume V is maximum; (4 marks)
 The maximum internal volume of the box. (2 marks)
 Express in terms of x:
Marking Scheme
 = 2(5 + 3)  9 + 3 + 5 = 14
3 × 5 + (2) × 4 7
= 2  Total fraction
3 + ? = 31
8 5 40
remaining fraction = 1  31
40
=9
40
original amount = sh.12330 × 40
9
=sh.54800
Tatu's fees  sh 2 × 54800
5
= sh.21920  gradient(perpendicular) = ^{1}/_{2}
y + 2 = 1
x  3 2
y = 1x  1
2 2  let the distance be d kn
d and d
75 95
therefore: d  d = 20
75 95 60
d = 118.75km  let odd integers be:
x, (x + 2), (x + 2 + 2)
x + (x + 2) + (x + 2 + 2) > 219
3x > 213
x = 71
the numbers are 73, 75 , 77 
 sh 77.24 × 100,000
=sh.7,724,000  sh 77.24 × 100000
122.27
=63,172
 sh 77.24 × 100,000




<TQR = 60º, QS = 10cm and bisects < TQR
Mediator(⊥ or bisector) of QS drawn or
<RSQ=<QST=<RQS=30º
Rhombus completed  No of oranges for friday
1948  (750 + 750 + 240)
= 208
no of oranges for saturday
208 + 560 = 768
therefore: amount = sh 8 × 768
=sh.6144  x^{2} + x  4xy  4y = x(x + 1)  4y(x + 1)
(x + 1)(4y^{2}  xy) (x + 1)(y)(4y  x)
= (x  4y)(x + 1)
(x + 1)(y)(x  4y)
=1
y  sin 3θ = cos 2θ°
therefore: sin 3θ = sin(90°  2θ)
therefore: 3θ = 90°  2θ
5θ = 90
θ = 18°  2πr^{2} + 2πrh = 154
r = h
2πr^{2} + 2πr^{2} = 154
4πr^{2} = 154
r = √ 154
4 × 3.142
r = 3.500
therefore didameter = 2r = 3.500 × 2
=7.00 (2 d.p) 
let OC = r
therefore: CD = 2r and EF = 2(r + 5)
3 3
2r + 2(r + 5) + 5 + 5 = 24
3 3
4r = 102
3 3
r = 8  total number of seedlings
(5 × 1) + (10 × 3) + (15 × 1) + (20 × 4) + (30 × 1) + (10 × 2)
= 5 + 30 + 15 + 80 + 30 + 20 = 180
% of height (h) : 23 ≤ h < 27
= 30 + 15 × 100
180
=25% 
 total sales = sh.360 × 500
=sh.180,000
commission = sh.(180,000  100,000) × 2/3
=sh.1600
total earnings = sh.(12,000 + 1600)
=13600 
 new salary = sh.(12000 + 12000 × 10/100)
=sh.13200
commission paid = sh(17,600  13,200)
=sh.4400
commission is paid on sh 4400 × 100/2
=220,000
total sales = sh.220,000 + 100,000
=320,000/=  no of handbags sold = 320000
500
=640
 new salary = sh.(12000 + 12000 × 10/100)
 total sales = sh.360 × 500


 internal volume of box = 150 × 80 ×40cm^{3}
=480,000cm^{3}
external volume of box = 152 × 82 × 42cm^{3}
=523488cm^{3}
therefore volume of wood = (523488  480000)cm^{3}
=43488cm^{3}  mass of box = 43488 × 0.6
1000
=26092.8
=26.1kg
 internal volume of box = 150 × 80 ×40cm^{3}

 no of tins = 150 × 80 × 40
10 10 10
=240  total mass = 26.1 + 240 × 120
1000
=54.9kg
 no of tins = 150 × 80 × 40


 Det 45  42 = 3



therefore: x = 200; y = 240

 total cost of books = (36 × 200) + (50 × 240)
=19200
total cost of discount
36 × 200 × 95 + 50 × 240 × 92
100 100
=17880
%discount
=19200  17880 × 100
19200
=6.875%
 Det 45  42 = 3

 Given scsale: 1cm to 2km

 Distance of P from S =10.8 ± 0.1cm
=21.6km  <PSN = 74 ± 1°
Bearing of P from S=286 ± 1°
 Distance of P from S =10.8 ± 0.1cm
 Area of triangle PQR = 1/2 x 10.2 x 12.2
= 62.22km^{2}
Area of triangle PRS = 1/2 x10.2 x 12 sin 150°
=30.6km^{3}
Area of ranch PQRS
=62.22 + 30.6
=92.82km^{2}
 Given scsale: 1cm to 2km


 A takes 180
x +10  B takes 180
x
 A takes 180
 180  180 = 3
x x+10 2
180(x+10) – 180x = 3x(x+10)
2
360x+3600  360x=3x^{3} + 30x
x^{2} + 10x  1200 = 0
(x30) (x +40) = 0
X= 30 or x = 40
Speed of A = 30+ 10 = 40  Time taken by A = 48 x 60 = 72min
40
Time taken by B = 48 x 60 = 96min
30
Time for B = 9610 = 86 min
86  72 = 14min



 Reflection in the line PR or ER Or PER
 Enlargement centre E
Scale factor  1  Rotation about pt R
Through 90°
Clockwise

 R → S
C → A  R →Q
C → E
 R → S


 Modal frequency = 8

no of kg of meat fre.(f) mid pts(x) fx cf 15 2 3 6 2 610 3 8 24 5 1115 6 13 78 11 1620 8 18 144 19 2125 3 23 69 22 2630 2 28 56 24 3135 1 33 33 25 Σf = 25 Σfx = 410
25  2, 5, 11, 19, 22, 24,25
Median  15.5 + 12.5  11 x 5
8
=15.5 + 1.5 x 5
8
=16.4375


 Area of base x^{2}
Or Area of sides = 4xh
X^{2} + 4xh = 432
h = 432  x^{2}
4x  Volume = x^{2}h
=x^{2}(432  x^{2})
 Area of base x^{2}

 Volume (v) = 108x – ^{1}/_{4}x^{3}
dv = 108  3x^{2} = 0
4
x=12  Vol = 108x  ^{1}/_{4}x^{3}=(108 x2)  ^{1}/_{4} x 12^{3}
=864cm^{3}
 Volume (v) = 108x – ^{1}/_{4}x^{3}

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