INSTRUCTIONS TO THE CANDIDATES
- This paper contains two sections; Section A and Section B
- 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
FOR EXAMINER’S USE ONLY
Section A
Question |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
Total |
Marks |
Section B
Question |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
Total |
Marks |
QUESTIONS
Section A. (50mks)
Answer all the questions in this section in the spaces provided.
- Without using a calculator evaluate. (3mks)
- Use logarithms to evaluate (4mks)
- Without using a calculator or mathematical tables evaluate. (3mks)
- 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)
- 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 3xº 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 transforms a point A(3,6) to A1. Find the distance A A1 (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)
- Use matrix method to solve. (3mks)
5x+3y=35
3x-4y= -8 - Use mathematical tables to find the reciprocal of 0.0247, hence evaluate. (4mks)
Correct to 2 decimal places. - 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)
5 - 3x ≤ x - 7 < 11 - 2x - Express 5184 and 2744 in terms of its prime factors hence determine the value of (4mks)
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= x2- 6x+7 (2mks)
x
0
1
2
3
4
5
6
-x2
0
-
-
9
16
25
-
-6x
-
-6
-12
-
-
-
-36
7
7
7
7
7
7
7
7
Y = x2 - 6x + 7
7
-
-1
-2
-
-
-
- On the grid provided below draw the graph of y= x2- 6x+7 for 0 ≤ x ≤6 and use it to estimate the roots of the equation.
x2- 6x+7=0 (4mks) - Use the graph above to solve the equation x2- 7x+9=0 (3mks)
- Determine the range of values of X for which x2- 6x+7 <x-2 (1mk)
- Complete the table below for the functions
- 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
-
No. Std Log 0.006628 6.628 x 10-3 3.8214 3 + 7.4642 193.9 1.939 x 102 2.2876 5.7518 2.2822 2.2822 x 10º 0.3584 5.3934 5 - 0.1198 1.198 x 10-1 1.0786 - 493/2 = (√49)3 = 73 = 343
256 = (4√256)3 = 64
2401 = (4√2401)3 = 243
343 x 64/343 = 64
Length of diagonal
√802 + 602 = 100
1/2 diagonal = 50
Height = 1502 - 502
= 22500 - 2500
= 20000
= √20000 = 141.4
1/3 x 80 x 60 x 141.4
= 226.2- 2x2 - xy - 6y2
2x2 - 4xy + 3xy - 6y2
2x (x - 2y) + 3y(x - 2y)
(2x + 3y)(x - 2y)
x2 - 4xy + 4y2
x2 - 2xy - 2xy + 4y2
x(x - 2y) - 2y(x - 2y)
(x - 2y) (x - 2y)
(2x + 3y)(x - 2y)
x - 2y ) ( x - 2y)
2x + 3y
x - 2y - 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 - 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 traingles = n - 2
= 6 - 2
= 4 - <BAC = 180 - 110
= 70º
14 = AC
sin 70 sin 70
AC = 14 x Sin 70
sin 70
Ac = 14
Area = 1/2 x 1414 x sin 40
= 62.99
63.0 - Rs = 7 cm
1 : 4000
7:4 : &.4 x 4000
= 29600 - reciprical of
2.47 x 102
= 0.4049 x 102 = 40.49
3√3.025 = 1.446
= 40.49 x 1.446 = 58.55 - 1US dollar = ksh 101.9378
20000 = 20000 x 101.9378
= Ksh 2038756
1 S.A = 7.6326
2038756 = 2038756
7.6326
= 267111.6
= 267112 S.A - 5 - 3x ≤ x - 7 < 11 - 2x
5x - 3x ≤ x - 7
5 + 7 ≤ x + 3x
12 ≤ 4x
3 ≤ x
3 ≤ x < 6
3 , 4 , 5
x - 7 < 11 - 2x
x + 2x < 11 + 7
3x < 18
x < 6 - 5184 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3
√5184 = 2 x 2 x 2 x 3 x 3
= 72
2744 = 2 x 2 x 2 x 7 x 7 x 7
3√2744 = 2 x 7 = 14
= 72/14 = 51/7 -
- A(3,6) B(7,12) (x,y)
Gradient of AB = 12 - 6 = 6 = 3
7 - 3 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
y - 9 = -2
x - 5 3
3(y - 9) = -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 y - 12 = -2
x - 7 3
3y = -2 + 14 + 36
3y = -2x + 50
3(y - 12) = -2(x - 7)
3y - 36 = -2x + 14
- A(3,6) B(7,12) (x,y)
-
-
A B C 120000 180000 240000 12 18 24 2 3 4 - 20/100 x 225000 = 45000
- 35/100 x 225000 = 78750
- reminder = 225000 - 45000 - 78750
= 101250
2:3:4 = 9
= 101250- 2/9 x 101250 = 22500
- 3/9 x 101250 = 33750
- 4/9 x 101250 = 45000
- 2:3:4 = 9
45000 - 4/9 x 45000 = 20000
Chirchir new share = 240 000 + 20 000
= 260 000
-
-
- Time = D/s = 400/120 = 31/3 hrs
= 3:20 min
8:30
+3:20
11:50 am - Distance by coast bus for 1/2 hr
= 80 x 1/2 = 40km
R.D = 400 - 40 = 360 km
R.S = 80 + 120 = 200 km/h
R.T = 260/200 = 14/5 hr = 1hr : 48 min
Time met =
8 30
1 48
10 18 am - R.T = 14/5 hr
coast by distance in 14/5 hr
= 80 x 14/5 hr
= 144 km
Distance = 40 + 144
= 184 - In 3hr 20 min coast bus covers
= 80 x 31/3
= 80 x 10/3 = 2662/3 km
Distance from mombasa = 400 - 2662/3
= 1331/3 km
- Time = D/s = 400/120 = 31/3 hrs
AD = 8.7 cm = 43.5 km ± 1
Bearing = 047º- y = x - 6x + 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 (angle in alternate segment) - <RTP = 53º
Angles on a traingle
- <ROP = 128º
-
- 10.8
- =57.8 = BCD
- BD = 6.37
- BDC = 26.9
-
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 + ξFt / ξ F
64 + -25/40
= 633/8
- Modal class
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