MATHEMATICS
PAPER 1
INSTRUCTION TO CANDIDATES:
 This paper consists of two Sections; Section I and Section II.
 Answer ALLthe questions in Section I and any five questions from Section II.
 Show all the steps in your calculation, giving your answer at each stage in the spaces provided below each question.
 Marks may be given for correct working even if the answer is wrong.
 Nonprogrammable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
 andidates should check the question paper to ascertain that all the pages are printed as indicated
and that no questions are missing.  Candidates should answer the questions in English.
Questions
SECTION I:(50 MARKS)
Answer all the questions in this section.
 Evaluate without using mathematical tables. (3mks)
1.9 x 0.032
20 x 0.0038  Use tables of reciprocals only to find the value of (3mks)
5  14
0.0829 0.581  You are given that cos θ= 8/10 . Without using mathematical tables express in fraction form the value of
 Sin θ (2mks)
 tan (90  θ) (1mks)
 An open right circular cone has radius of 5cm and a perpendicular height of 12cm .
Calculate the surface area of the cone correct to 1 decimal place.(Take π to be 3.142) (3mks)  Nyongesa spends a total of sh.970 on buying three text books and five pens. If he had bought two text books and five pens he would have saved sh.90. Find the cost of one text book. (3mks)
 A Kenyan tourist left Germany for Kenya through Switzerland. While in Switzerland he bought a watch worth 52 Deutche marts. 1 swiss Franc = 1.28 DM and 1 Swiss Franc = 45.21 Kenya shillings.
Find the value of the watch in; Swiss Franc (2mks)
 Kenya shillings using the exchange rates above. (2mks)
 The figure below is a velocity –time graph for a car that was in motion for 24 seconds.
 Find the total distance traveled by the car? (2mks)
 Calculate the deceleration of the car. (2mks)
 Form the quadratic equation whose roots are x=  ^{5}/_{3} and x=1 in the form ax^{2} +bx+c=0. (2mks)
 Three towns are situated in such a way that town B is 40km due south of town A and town C is 30 km due East of town B.
 Draw a sketch diagram showing the position of town A.B and C. (1mk)
 From your sketch, calculate:
 Distance AC (1mk)
 To the nearest degree the bearing of town A from town C. (2mks)
 Find the inequalities that defines the region R shown in the figure below (3mks)
 The interior angle of a regular polygon is 20º more than three times the exterior angle. Determine the number of sides of the polygon. (3 marks)
 The circle below whose area is 18.05cm^{2} circumscribes a triangle ABC where AB=6.3cm, BC =5.7cm and AC =4.8cm. Find the area of the shaded part. (3mks)
 Solve for x in the equation. (3mks)
9^{(x1)} x 3^{(2x+1)} = 243  The graph below shows frequency densities for the masses of some 200 students selected from a class. Use it to answer the questions that follow:
 Complete the frequency distribution table below (2mks)
Mass in (Kg) Frequency  State the modal frequency (1mk)
 Complete the frequency distribution table below (2mks)
 The volumes of two similar solid cylinders are 4752 cm^{3} and 1408 cm^{3} .If the area of the curved surface of the smaller cylinder is 352 cm^{2} , find the area of the curved surface of the larger cylinder (3mks)
 The line which joins the point A (3, K) and B (2, 5) is parallel to the line whose equation is 5y+2x 7=0. Find the value of K. (3mks)
SECTION II (50 MARKS)
Answer only five questions in this section in the spaces provided.
 John bought 3 brands of tea, A B and C. The cost price of the three brands were sh 25, sh 30, sh 45 per kg respectively. He mixed the three brands in the ratio 5:2:1 respectively: After selling the mixture he made a profit of 20%.
 How much profit did he make per kilogram of the mixture? (4 mks)
 After one year the cost price of each brand was increased by 10%
 For how much did he sell one kilogram of the mixture to make a profit of 15%? (Give your answer to the nearest 5 cents) (3 mks)
 What would have been his percentage profit if he sold one kilogram of the mixture at sh. 45. (3 mks)
 The equation of a curve is given as y=2x^{3}  ^{9}/_{2}x^{2} 15x+3
 Find:
 the value of y when x=2 (2 marks)
 the equation of the tangent to the curve at x=2 (4 marks
 determine the turning points of the curve (4 marks)
 Find:

 Using the trapezium rule with 7 ordinates, estimate the area of the region bounded by the curve y= x^{2} +6x+1, the lines x=0, y=0 and x=6. (5mks)
 Calculate :
 The area of the region in (a) above by the integration; (3mks)
 The percentage error of the estimated area to the actual area of the region, correct to 2 decimal places. (2mks)
 A country bus left town A at 11.45 am and traveled towards town B at an average speed of 60km/h.Amatatu left town B at 1.15 pm on the same day and traveled towards town A along the same road at an average speed of 90km/h. the distance between the two towns is 540km. Determine
 The time of day when the two vehicles met (4mks)
 How far from town A did they meet ? (2mks)
 How far outside town B the bus was when the matatu reached town A (4mks)
 A straight line passing through the points (8,2) and (4,4) has its equation in the form ax + by + c = 0, where a, b and c are integers.
 Determine the numerical values of a, b and c. (3mks)
 If the line in (a) above cuts the xaxis at point P, determine the coordinates of P. (2mks)
 Another line, which is perpendicular to the line in (a) above passes through point P and cuts the y axis at Q. determine the coordinates of point Q. (3mks)
 Find the length of QP (2mks)

 The figure below shows a metal solid consisting of a right cone mounted onto a hemisphere. The height h of the cone is twice the radius r. if the volume of the solid is 36cm^{3} , find the radius of the hemisphere. (4mks)
 The solid is totally immersed in water contained in a cylindrical tin of radius 9cm. Through what height does the water level in the tin rise? (2mks)
 The solid is melted and recast into a right pyramid of vertical height 4.2cm. Find the base area of the pyramid. (2mks)
 If the solid is of mass 14.4g. Find its density in kg/m^{3}. (2mks)
 The figure below shows a metal solid consisting of a right cone mounted onto a hemisphere. The height h of the cone is twice the radius r. if the volume of the solid is 36cm^{3} , find the radius of the hemisphere. (4mks)
 In the diagram below, AOB is a triangle such that OA = a, OB = b and <AOB is obtuse.
If P is a point on AB such that AP = 3PB and Q is the midpoint of OA Express in terms of a and b
 AB (1mk)
 BQ (2mks)
 If X is a point on BQ such that BX =hBQ, express OX in terms of a b and h, where h is a scalar. (3mks)
 Given further that OX = kOP where k is another scalar, obtain the values of h and k. (4mks)
 Express in terms of a and b
 The table below shows the length of 40 seedlings.
Length in (mm) Frequency 118126 3 127 – 135 4 136 – 144 10 145 – 153 12 154 – 162 5 163 – 171 4 172180 2 
 The modal class (1 Mark)
 The median class (2 mks)

 The mean of the seedlings (4 mks)
 The median of the seedlings (3 mks)

Marking Scheme
 1.9 x 0.032 x 10000
20 x 0.0038 x 10000
19 x 32
20 x 38
^{16}/_{20}
^{8}/_{10 }  5 x 1  14 x 1
8.29 x 10^{2} 5.81 x 10^{1}5 x 0.1206 x 10^{2} 14 x 0.1721 x 10
=60.324.094
=36.206  Cos = Ads = 8
hps 10
QR=8 and PQ=10
PR^{2} =PQ^{2} QR^{2} =102 82
PR=√36 = 6 sin θ = 6/10 or 3/5
 tan (90θ) =^{QR}/_{PR} = ^{8}/_{6} or ^{4}/_{3}

SA= πrl
1= √5^{2} +12^{2}= √169
=13cm
=3.142x13x5
=188.5  Let cost of text bk be t and pen p.
3t +5p = 970.......(i)
2t + 8p = 880.......(ii)
Multiply e.g(i) by 8 and (ii0 by 5
24t + 40p = 7760
10t+40p=4400
14t=240
Cost of 1 text bk=sh.240 
 1.28 Deutche marks=1 Swiss franc
52 Deutche marks =?
52 x swiss franc
1.28 Deutche marks
=40.635
≈40.63 Swiss francs  I swiss franc =42.21Ksh.
40.63 Swiss franc=?
42 .21 x 40.63
1
=1,714.99 Kenya shillings
 1.28 Deutche marks=1 Swiss franc

 Dist= 1⁄2 x 4 x 80 + 80 x 16 + 1⁄2 x 4 x 80
=160+1280+160
=1600m  Deceleration= ^{080}/_{4} = 20m/s^{2}
 Dist= 1⁄2 x 4 x 80 + 80 x 16 + 1⁄2 x 4 x 80
 Soln:x= ^{5}/_{3} and x=1
= 5 and x=1
∴ (3x+5)(x1)=0
3x(x 1)+5(x1)=0
3x^{2} 3x+5x 5=0
3x^{2} +2x 5=0 
 Diag,
Bearing of A from C
=270+ θ 
 AC^{2} =AB^{2} +BC^{2} =40^{2}+30^{2}
=√ 2500
=50km  sin θ = 40/50 = 0.8
θ =53.13
≈ 53º
 AC^{2} =AB^{2} +BC^{2} =40^{2}+30^{2}
 Diag,
 L_{1} = x>0
L_{2} = ^{x}/_{5} + ^{y}/_{4} =1
4x + 5y = 20
4x + 5y < 20
L_{3}: ^{x}/_{4} + ^{y}/_{3} = 1
3x  4y=12
∴3x 4y <12
 Let ext. < xº
Interior < = 3x + 20º
4x = 160º
4x = 160º
x = 40
360/40= 9 sides 
s = √s(sa)(sb)(sc)
s = 1/2(4.8 + 6.8 + 5.7)= 8.4
Area = √8.4(2.7)(3.6)(2.1)
18.05  13.0943
= √171.4608
= 13.0943
=4.9557
≈4.96  Change to base 3
3^{2(x1)} x 3^{(2x+1)} =3^{5}
2(x1)+2x+1=5
2x2=2x+1=5
4x1=5
4x=6
x=1.5 
Mean in kg 4150 5155 5665 6670 7180 Frequency 20 70 50 50 10  Modal frequency is 70.
 L.S.F=V.S.F
lsf= √^{4752}/_{1408}=1.5
l.s.f^{2} =A.S.F
1.5^{2 }=^{ }Area of larger cylinder
352
Area of larger cylinder 792cm^{2}  Reduce 5y+2x=7 in the form y=mx+c
∴ y= 2/5x + 7/5
But gradient =
k5 = k5
32 5
k5 = 2
5 3
k=3 
 5 x 25 + 2 x 30 + 45 x 1 = 28.75
8
Profit 28.75 x 20/100 = sh5.75 
 New price = 28.75/100 x 110 x 115/100 = 36.40
 ^{28.75 x 110}/_{100} = 31.60
New profit 45  31.60 x 100
31.60
= 13.4 x 100
31.60
=42.41%
 5 x 25 + 2 x 30 + 45 x 1 = 28.75

X 3 2 1 0 1 2 3 4 X2 9 4 1 0 1 4 9 16 2X2 18 8 2 0 2 8 18 32 3X 9 6 3 0 3 6 9 12 5 5 5 5 5 5 5 5 5 Y 22 9 0 5 6 3 4 15
minimum at point(1,6)
y= 2 x 2  5x 5
0= 2 x 2  3x  5
Y=0
X=1 or x=2.5
y= 2x2  3x 5
0= 2 x 2 5x 3
Y= 2x 2
X= 0.5 or x=3 
 XV=MO2 = 300sin 70 = 281.9cm
 YBW = 140/360 x 2πr
MO1 = 300 Cos 70 = 102.6cm
Radius of small circle = 180  102.6 = 77.4cm
VBW = 140/360 x 2 x 3.142 x 77.4 = 189.1cm  XAY = 220/360 x 2π x 180
= 220/360 x 2 x 3.142 x 180 = 691.2cm  Length of conveyor belt
= 281.9 x 2 + 189.1 + 691.2
=1444.1cm^{2}
 XV=MO2 = 300sin 70 = 281.9cm

 11.45 = 1^{1}/_{2}h
Distance moved by bus = 1.5 x 609 = 90km
Distance between 2 vehicles
540  90=450
R. speed = 60 + 90 = 150km/h
T.taken to meet = 450 ÷ 150 = 3h
= 1.15 + 3 = 4:15pm  Total time moved by bus when they meet
16:15  11:45 = 4^{1}/_{2}Distance from A = 4.5 x 60 = 270km  Time taken by matatu to travel from B to A = 540 ÷ 90 =6
∴ Time matatu reaches A
13:15 + 6 = 19:15hrs
∴Time travelled= 19:1511:45 = 7^{1}/_{2} hrs
distance = A= 7.5 x 60
Hence:
540  450 = 90km
 11.45 = 1^{1}/_{2}h

 y + 2 = 4 + 2 = 1
x8 48 2
2y + 4 = x 8
⇒ x  2y  12 = 0
∴ u =1, b= 2 and c= 12  y = ^{x}/_{2} 6 , when y=0
^{x}/_{2}  6= 0 ⇒ x=12
∴ coordinates of pane (12,0)  y = 2
x 12
⇒ y = 24  2x
but when x = 0, y=24  ∴ coordinates of q are(0,24)
QP = √(120)^{2} + (024)^{2}=√144 + 576 = 26.83 units
 y + 2 = 4 + 2 = 1

 Volume of solid =volume of cone + vol of hemisphere
=^{1}/_{3}πr^{2}h + ^{1}/_{2}x^{4}/_{3}πr^{3} = 36
= ^{1}/_{3}πr^{2} + ^{2}/_{3}πr^{3}=36
=^{4}/_{3}πr^{3}= 36
r^{3} = 36 x 3
4 x 3.142
r^{3} = 8.593
r= 2.048cm  Let the level rise be x
πr2x = 36
x = 36
3.142 x 81
36 = 0.14cm
254.502  Volume = ^{1}/_{2} Ah
= ^{1}/_{3} x A x 4.2 = 36
A = 36 x 3
4.2
= 25.71cm^{2}  Density = mass/ volume
= ^{14.4}/_{36} x 1000kg/m^{3}
= 400kg/m^{3}
 Volume of solid =volume of cone + vol of hemisphere


 AB = b  a
BQ = BO = OQ = 1/2 a  b  BX = h(^{1}/_{2} a  b)
OX = b + h (^{1}/_{2} a  b)
^{1}/_{2}ha + (1h)b
 AB = b  a
 OX = ROP
but OP = 1/4a + 3/4b= 1/4(a +3b)
OX = R/4 (a + 3b)
OX = OX2
1/2ha + (1h)b = 1/4Ra = 3/4kb
1/2h = 1/4⇒(43h= 4)
4h2k=0
4h3k=4
5k=4
k=4/5
h=2/5


Length
(cm)Mid pt
(x)F xf CF 118135 122 3 366 3 127135 131 4 524 7 136144 140 10 1400 17 145153 149 12 1788 29 154162 158 5 790 34 163171 167 4 668 38 172180 176 2 352 40 40 5888 
 Modal class ⇒145153
 Median class 145 153

 Mean of feeding
X= Σxf/Σf
= 5888/40
= 147.2  Median = L( ^{(ⁿ ⁺ ¹)/₂  c}/_{f}) i
= 144.5 + (^{(⁴¹/₂  17)}/_{12})9
=144.5 + 20.5  17 x 9
12
144.5 + 2.625
=147.125
 Mean of feeding

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