INSTRUCTIONS TO CANDIDATES
- This paper contains TWO sections: section I and section II
- Answer ALL the questions in Section I and only five questions from section II.
- Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
- Marks may be given for correct working even if the answer is wrong.
- Non-programmable silent electronic scientific calculators and KNEC mathematical tables may be used except where stated otherwise
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QUESTIONS
SECTION 1 (50 marks)
Answer all the questions in this section in the space provided
- Evaluate:
(3mks)
- Use logarithms to evaluate the following to 4 significant figures to:
(4mks)
- An electrician made a loss of 30% by selling a multi plug at sh.1400.what percentage profit would he has made if he sold the multi plug at sh. 2300. (3mks)
- In the triangle ABC below, BC=9cm, angle ABC=80°.and angle ACB=30°.
Calculate, correct to 4 significant figures, the area of the triangle. (3mks) - Given that the exterior angle of a regular hexagon is 2x. Find the value of x. Hence find the size of each interior angle of the hexagon. (3mks)
- Two numbers t and s are such that t4 x s2= 5625. Find t and s (3mks)
- Find the obtuse angle the line with equation 2y+5x+2=0 makes with the x-axis. (3mks)
- Simplify the expression (3mks)
- A plot is in the shape of a right angled triangle. The length of the shorter side is 15m and the area is 456.8m2. Calculate the length of the longest side of the garden. (3mks)
- The diagram below shows a region R bounded by three lines L1, L2 and L3. Form the three inequalities that satisfies the given region R (3mks)
- A bus travelled at an average speed of 63km/h left the station at 9.15 am. A car later left the same station at 10.00am and caught up with bus at 11.45 am. Find the average speed of the car. (3mks)
- A tourist came in Kenya and exchanged 1250 US dollars into Kenyan shillings at the rate shown below.
Buying (Kshs) selling (Kshs)
1US dollar 105.5 110.8
He spent Ksh. 85400 after which he converted the remaining balance to US dollars. How much US dollars did he get back to the nearest dollar. (3mks) -
- Complete the table below for y=x^2+5 (1mk)
X
0
1
2
3
4
5
6
Y=x2+5
5
- Use the trapezoidal rule with7 ordinates to estimate the area bounded by the curve y=x2+5, x –axis , y-axis and x=6 (2mks)
- Complete the table below for y=x^2+5 (1mk)
- Given that
, find c (3mks)
- A triangle T with vertices A(2,4), B(6,2)and C(4,8) is mapped onto triangle T’ with vertices A’(10,0) ,B’(8,-4) and C’(14,-2) by a rotation .
- on the grid provided draw triangle T and its image T’ (2mks)
- Determine the centre and the angle of rotation that maps T onto T’ (2mks)
- A small cone of height 8cm is cut off from a bigger cone leaving a frustum of height the 16cm. If the volume of the smaller cone is 160cm3, find the volume of the frustum. (3mks)
Section II (50MKS)
Answer any five questions in this section in the space provided - Three businessmen, Hassan, Mutua and Wanyonyi decided to start a business. The initial capital which was needed was Ksh. 4,000,000 of which they were able to raise 30% by making contributions in the ratio 3:3:2 respectively. The rest of the amount was obtained from a bank and was to be paid back within one year with an interest of 25% in the same ratio 3:3:2. The three men were to share the profit of the business in the ratio of their contribution. During the year, the business realized a profit of ksh.4, 800,000.
- How much of the initial amount did Wanyonyi raise? (3mks)
- How much did Hassan pay to the bank at the end of the year? (4mks)
- After paying the bank at the end of the year, how much was Mutua left with? (3mks)
- The marks scored by 50 students in a geography examination are as follows:
60
54
40
67
53
73
37
55
62
43
44
69
39
32
45
58
48
67
39
51
46
59
40
52
61
48
23
60
59
47
65
58
74
47
40
59
68
51
50
50
71
51
26
30
38
70
46
40
51
42
- Prepare a frequency distribution table using a class interval of 10 starting from 21-30 (3mks)
- Draw a histogram to represent the distribution (4mks)
- Use your histogram to estimate the modal class (1mk)
- Using the histogram estimate the median of the distribution of the data. (2mk
- A petrol tanker has a cross-section in the shape shown below. It is used to transport petrol. Its internal length is 7m while its internal radius is 3.5 m. Obtuse angle POQ = 144º.On one of its trips, it was filled to capacity. Taking π = 22/7
- Calculate the volume of petrol in the tanker in m3 (4 marks)
- In the parking lot at night, a third of the petrol was stolen.
- How many litres of petrol was the owner left with? (2 marks)
- What was the mass of the remaining petrol given that one cubic metre of petrol has a mass of 700kg? (2 marks)
- At the weigh bridge, any vehicle carrying excess of 50,000 Kg was charged Sh. 12.50 for every extra kilogram. How much fine did the owner of the tanker pay? (2marks)
- The figure below represents a quadrilateral piece of land ABCD divided into three triangular plots. The lengths BE and CD are 100m and 80m respectively. Angle ABE = 30º, ACE = 45º and ACD = 100º.
- Find to four significant figures:
- The length of AE. (2mks)
- The length of AD. (2mks)
- The perimeter of the piece of land. (3mks)
- The plots are to be fenced with five strands of barbed wire leaving an entrance of 2.8m wide to each plot. The type of barbed wire to be used is sold in rolls of lengths 480m. Calculate the number of rolls of barbed wire that must be bought to complete the fencing of the plots. (3mks)
- Find to four significant figures:
- A straight line L1 has a gradient -½ and passes through point P (-1, 3). Another line L2 passes through the points Q (1,-3) and R (4, 5).
Find:- The equation of L1 (2 marks)
- The equation of L2 in the form ax + by + c = 0 . (3 marks)
- The equation of a line passing through a point S (0, 5) and is perpendicular to L2. (3 marks)
- The equation of a line through R parallel to L1. (2 marks)
-
- A port B is on a bearing 080º from a port A and a distance of 95 km. A Submarine is stationed at a port D, which is on a bearing of 200º from A, and a distance of 124 km from B. A ship leaves B and moves directly Southwards to an Island P, which is on a bearing of 140º from A. The Submarine at D on realizing that the ship was heading to the Island P, decides to head straight for the Island to intercept the ship. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D and P. (4 marks)
Hence find:- The distance from A to D. (2 marks)
- The bearing of the Submarine from the ship when the ship was setting off from B. (1 mark)
- The bearing of the Island P from D. (1 mark)
- The distance the Submarine had to cover to reach the Island P. (2 marks)
- A port B is on a bearing 080º from a port A and a distance of 95 km. A Submarine is stationed at a port D, which is on a bearing of 200º from A, and a distance of 124 km from B. A ship leaves B and moves directly Southwards to an Island P, which is on a bearing of 140º from A. The Submarine at D on realizing that the ship was heading to the Island P, decides to head straight for the Island to intercept the ship. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D and P. (4 marks)
-
- Find the inverse of the matrix: (1mk)
- Amina bought 20 bags of oranges and 15 bags of mangoes for a total of sh. 9,500. Nafula bought 30 bags of oranges and 20 bags of mangoes for a total of sh. 13,500. If he price of a bag of oranges is X and that of mangoes is y:
- Form two equations to represent the information above. (2mks)
- Hence use the matrix A-1 above to find the price of one bag of each item. (4mks)
- The price of each bag of oranges was increased by 10% and that of mangoes reduced by 10%. The businesswomen (Amina and Nafula) bought as many oranges and as many mangoes as they bought earlier. Find the total cost of oranges and mangoes that each businesswomen bought after the percentage change. (3mks)
- Find the inverse of the matrix: (1mk)
- The displacement, s metres, of a moving particle from a point O, after t seconds is given by, s = t3-5t2+3t+10
- Find s when t=2 (2mks)
- Determine:
- the velocity of the particle when t=5 seconds; (3mks)
- the value of t when the particle is momentarily at rest (3mks)
- find the time , when the velocity of the particle is maximum (2mks)
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MARKING SCHEME
SECTION 1 (50 marks)
Answer all the questions in this section in the space provided
- Evaluate:
(3mks)
5/2 of 7/4 = 3.5/8
35/8 - 21/4
35 - 42 = -7
8 8
5/4 - 11/4 = -6/4
-6/4 x 2 = -12/4
7/5 + -12/4 = -13/5
-7/5 x 5/-8 = 35/64 - Use logarithms to evaluate the following to 4 significant figures to:
(4mks)
no std form log 95.75 9.575 x 101 1.9811 0.85 8.5 x 10-1 1.9294 + 5.758 5758 x 100 1.91105 0.7603 1.1502 x 2/3 = 0.7668 5.845 <--- 5845 x 10 < ---- 0.7668 - An electrician made a loss of 30% by selling a multi plug at sh.1400.what percentage profit would he has made if he sold the multi plug at sh. 2300. (3mks)
1400 --> 70/100
1400 x 200 = 2000
100
100% = 2000
2300 x 100 = 115%
2000
115 - 100 = 15% - In the triangle ABC below, BC=9cm, angle ABC=80°.and angle ACB=30°.
Calculate, correct to 4 significant figures, the area of the triangle. (3mks)
9/sin 70 = x/ sin 80
9 sin 80 = x sin 70
x = 9 sin 80
sin 90
= 9.4321 cm
A = 1/2 x 9 x 9.4331 sin 30
= 21.22 cm3 - Given that the exterior angle of a regular hexagon is 2x. Find the value of x. Hence find the size of each interior angle of the hexagon. (3mks)
n = 6 sides
360 = exterior angles
6
360 = 2x
6
12x = 360
12 12
x = 30
exterior = 2 x 30 = 60º
180º - 60º = 120º - Two numbers t and s are such that t4 x s2= 5625. Find t and s (3mks)
5 5625 t4 x s2 = 54 x 32 5 1125 t9 = 54 5 225 t = 5 5 45 s2 = 32 3 9 s = 3 3 3 1 - Find the obtuse angle the line with equation 2y+5x+2=0 makes with the x-axis. (3mks)
2y = -5x - 2
2 2 2
y = -25x - 1
m = -2.5
angle = tan-1(25) = 6 x 20
180 - 68.20
= 111.80º - Simplify the expression (3mks)
p = 12x - 6 = -72
s = -1
12xº + 90x - 86x - 64º
3x(4x + 3a) - 29(4x + 3a)
(3x - 2a)(4x + 3a)
(3x - 2a)(3x + 22)
= 4x + 3a
3x + 2a - A plot is in the shape of a right angled triangle. The length of the shorter side is 15m and the area is 456.8m2. Calculate the length of the longest side of the garden. (3mks)
1/2 x 15 x h = 456.8
7.5 h = 456.8
h - 60.91 m
x = √60.91º + 152
= 62.73 m - The diagram below shows a region R bounded by three lines L1, L2 and L3. Form the three inequalities that satisfies the given region R (3mks)
L1
x ≥ 0L2
x/5 + y/4 = 1
4x + 5y = 20
4x + 5y ≤ 20L3
x/4 + y/-3 = 1
-3x + 4y = -12
-3x + 4y > -12 - A bus travelled at an average speed of 63km/h left the station at 9.15 am. A car later left the same station at 10.00am and caught up with bus at 11.45 am. Find the average speed of the car. (3mks)
11.45
-9.15
2h 30 min
21/2 x 63 = 157.5 km
11.45
- 10.00
1 h 45 min
157.5 ÷ 13/4
= 157.5 x 4/7
= 90 km/h - A tourist came in Kenya and exchanged 1250 US dollars into Kenyan shillings at the rate shown below.
Buying (Kshs) selling (Kshs)
1US dollar 105.5 110.8
He spent Ksh. 85400 after which he converted the remaining balance to US dollars. How much US dollars did he get back to the nearest dollar. (3mks)
1 us dollar = sh 105.5
1250 dollars = ?
1250 x 105.5 = 131875
1
131875 - 85400
= 46475 = 419.449
110.8
= 419 -
- Complete the table below for y=x^2+5 (1mk)
X
0
1
2
3
4
5
6
Y=x2+5
5
6 9 14 21 30 41 - Use the trapezoidal rule with7 ordinates to estimate the area bounded by the curve y=x2+5, x –axis , y-axis and x=6 (2mks)
7 ordinates = 6 ships
A = 1/2 x 1 [(5 + 41) + 2(6 + 9 + 14 + 21 + 30)]
= 1/2 [46 + 2(80)]
= 1/2 x 206
= 103 sq units
- Complete the table below for y=x^2+5 (1mk)
- Given that
, find c (3mks)
- A triangle T with vertices A(2,4), B(6,2)and C(4,8) is mapped onto triangle T’ with vertices A’(10,0) ,B’(8,-4) and C’(14,-2) by a rotation .
- on the grid provided draw triangle T and its image T’ (2mks)
- Determine the centre and the angle of rotation that maps T onto T’ (2mks)
centre of rotation = 94, -2)
Angle of rotation = -90º
- on the grid provided draw triangle T and its image T’ (2mks)
- A small cone of height 8cm is cut off from a bigger cone leaving a frustum of height the 16cm. If the volume of the smaller cone is 160cm3, find the volume of the frustum. (3mks)
l.s.f = 24/8= 3/1
v.s.f = (3/1)2 = 27/1
27/1 = x/160
x = 27 x 160 = 4320cm3
4320 - 160 = 4160cm3
Section II (50MKS)
Answer any five questions in this section in the space provided - Three businessmen, Hassan, Mutua and Wanyonyi decided to start a business. The initial capital which was needed was Ksh. 4,000,000 of which they were able to raise 30% by making contributions in the ratio 3:3:2 respectively. The rest of the amount was obtained from a bank and was to be paid back within one year with an interest of 25% in the same ratio 3:3:2. The three men were to share the profit of the business in the ratio of their contribution. During the year, the business realized a profit of ksh.4, 800,000.
- How much of the initial amount did Wanyonyi raise? (3mks)
0.3 x 4000000 = 1200000
2/8 x 1200000 = 300000 - How much did Hassan pay to the bank at the end of the year? (4mks)
4000000 - 1200000
= 2800000 x 1.25
= 3500000
= 3/8 x 3500000
= 1, 312, 500 - After paying the bank at the end of the year, how much was Mutua left with? (3mks)
profit = 3/8 x 4800000 = 1800000
1800000 - 1312500
= sh 487,500
- How much of the initial amount did Wanyonyi raise? (3mks)
- The marks scored by 50 students in a geography examination are as follows:
60
54
40
67
53
73
37
55
62
43
44
69
39
32
45
58
48
67
39
51
46
59
40
52
61
48
23
60
59
47
65
58
74
47
40
59
68
51
50
50
71
51
26
30
38
70
46
40
51
42
- Prepare a frequency distribution table using a class interval of 10 starting from 21-30 (3mks)
mark f tally 21-30 3 III 31-40 9 IIII IIII 41-50 11 IIII IIII I 51-60 15 IIII IIII IIII 61-70 9 IIII IIII 71-80 3 III 50 - Draw a histogram to represent the distribution (4mks)
- Use your histogram to estimate the modal class (1mk)
51-60 - Using the histogram estimate the median of the distribution of the data. (2mk)
1/2 x 50 = 25th partition
25 - 23 = 2
x x 15 = 2
x = 2/15 = 0.1333
median = 50.5 + 0.1333
= 50.633
- Prepare a frequency distribution table using a class interval of 10 starting from 21-30 (3mks)
- A petrol tanker has a cross-section in the shape shown below. It is used to transport petrol. Its internal length is 7m while its internal radius is 3.5 m. Obtuse angle POQ = 144º.On one of its trips, it was filled to capacity. Taking π = 22/7
- Calculate the volume of petrol in the tanker in m3 (4 marks)
- [(22/7 x 3.5 x 3.5) - (22/7 x 144/360 x 3.5 x 3.5 - 1/2 x 3.5 x 3.5 x sin 144)] x 7 m
(26.7)7
= 186.9m3 - 186.9 x 1000 = 186900 litres
- [(22/7 x 3.5 x 3.5) - (22/7 x 144/360 x 3.5 x 3.5 - 1/2 x 3.5 x 3.5 x sin 144)] x 7 m
- In the parking lot at night, a third of the petrol was stolen.
- How many litres of petrol was the owner left with? (2 marks)
2/3 x 186900 = 124600 litres - What was the mass of the remaining petrol given that one cubic metre of petrol has a mass of 700kg? (2 marks)
1m3 ---> 700kg
124.6m3
124.6 x 700 = 87220 kg
1 - At the weigh bridge, any vehicle carrying excess of 50,000 Kg was charged Sh. 12.50 for every extra kilogram. How much fine did the owner of the tanker pay? (2marks)
87220 - 50 000 = 37 220
1 k = 1300
37220 = ?
= 465 250 shillings
- How many litres of petrol was the owner left with? (2 marks)
- Calculate the volume of petrol in the tanker in m3 (4 marks)
- The figure below represents a quadrilateral piece of land ABCD divided into three triangular plots. The lengths BE and CD are 100m and 80m respectively. Angle ABE = 30º, ACE = 45º and ACD = 100º.
- Find to four significant figures:
- The length of AE. (2mks)
tan 30 = AE
100
AE = 57.74 cm - The length of AD. (2mks)
AC2 = (57.74)2 + 957.74)2
AC = 81.66m
cosine rule
AD2 = 81.662 + 18012 - (2 x 81.66 x 80 Cos 100)
=123.83m - The perimeter of the piece of land. (3mks)
cos 30 = 100 = AB = 115.47
AB
perimeter = 115.47 - 2384 + 57.74 + 100 = 477.08 m
- The length of AE. (2mks)
- The plots are to be fenced with five strands of barbed wire leaving an entrance of 2.8m wide to each plot. The type of barbed wire to be used is sold in rolls of lengths 480m. Calculate the number of rolls of barbed wire that must be bought to complete the fencing of the plots. (3mks)
2.8 x 3 = 84
477.08 - 84 = 468.68
46868 x 5 = 1 x 5 = 5
480
= 5
- Find to four significant figures:
- A straight line L1 has a gradient -½ and passes through point P (-1, 3). Another line L2 passes through the points Q (1,-3) and R (4, 5).
Find:- The equation of L1 (2 marks)
-1/2 = y - 3
x + 1
2y - 6 = -x - 1
2y = -x + 5
y = -x/2 + 5/2
y = -x/2 + 21/2 - The equation of L2 in the form ax + by + c = 0 . (3 marks)
m = 5 - -3 = 8
4 - 1 3
8 = y - 5
3 x - 4
3y - 15 = 8x - 32
3y = 8x - 17
8x - 3y - 17 = 0 - The equation of a line passing through a point S (0, 5) and is perpendicular to L2. (3 marks)
m2 = 1/8/3 = -3/8
-3 = y - 5
8 x
= -3x = 8y - 40
3x + 8y - 40 = 0 - The equation of a line through R parallel to L1. (2 marks)
-1/2 = y - 5
x - 4
2y - 10 = -x + 4
2y = -x + 14
2y + x - 14 = 0
- The equation of L1 (2 marks)
-
- A port B is on a bearing 080º from a port A and a distance of 95 km. A Submarine is stationed at a port D, which is on a bearing of 200º from A, and a distance of 124 km from B. A ship leaves B and moves directly Southwards to an Island P, which is on a bearing of 140º from A. The Submarine at D on realizing that the ship was heading to the Island P, decides to head straight for the Island to intercept the ship. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D and P. (4 marks)
Hence find:- The distance from A to D. (2 marks)
- The bearing of the Submarine from the ship when the ship was setting off from B. (1 mark)
- The bearing of the Island P from D. (1 mark)
- The distance the Submarine had to cover to reach the Island P. (2 marks)
- A port B is on a bearing 080º from a port A and a distance of 95 km. A Submarine is stationed at a port D, which is on a bearing of 200º from A, and a distance of 124 km from B. A ship leaves B and moves directly Southwards to an Island P, which is on a bearing of 140º from A. The Submarine at D on realizing that the ship was heading to the Island P, decides to head straight for the Island to intercept the ship. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D and P. (4 marks)
-
- Find the inverse of the matrix: (1mk)
- Amina bought 20 bags of oranges and 15 bags of mangoes for a total of sh. 9,500. Nafula bought 30 bags of oranges and 20 bags of mangoes for a total of sh. 13,500. If he price of a bag of oranges is X and that of mangoes is y:
- Form two equations to represent the information above. (2mks)
20x + 15y = 9500 --> 4x + 3y = 1900
30x + 20y = 13500 --> 3x + 3y = 1350 - Hence use the matrix A-1 above to find the price of one bag of each item. (4mks)
- Form two equations to represent the information above. (2mks)
- The price of each bag of oranges was increased by 10% and that of mangoes reduced by 10%. The businesswomen (Amina and Nafula) bought as many oranges and as many mangoes as they bought earlier. Find the total cost of oranges and mangoes that each businesswomen bought after the percentage change. (3mks)
new price of oranges = 110/100 x 250 = 275
new price of oranges = 90/100 x 300 = 270
Amina = 20 x 275 + 15 x 270 = 9550
Nafula = 30 x 275 + 20 x 270 = 13650
- Find the inverse of the matrix: (1mk)
- The displacement, s metres, of a moving particle from a point O, after t seconds is given by, s = t3-5t2+3t+10
- Find s when t=2 (2mks)
s = (2)3 - s(2)2 + 3(2) + 10
= 4m - Determine:
- the velocity of the particle when t=5 seconds; (3mks)
v = ds/dt = 3t2 - 10t + 3
v = 3(5)2 - 10(5) + 3
= 28 m/s - the value of t when the particle is momentarily at rest (3mks)
- the velocity of the particle when t=5 seconds; (3mks)
- find the time , when the velocity of the particle is maximum (2mks)
l = 12/3
- Find s when t=2 (2mks)
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