INSTRUCTIONS TO THE CANDIDATES
 Write your name and school and index number in the spaces provided above
 This paper contains two sections; Section 1 and Section 11.
 Answer all the questions in section 1 and only five questions from Section 11
 Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
 Marks may be given for correct working even if the answer is wrong.
 NonProgrammable silent calculators and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
FOR EXAMINERS’S USE ONLY
Section 1
Question  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  Total 
Marks 
Section 1I
Question  17  18  19  20  21  22  13  24  Total 
Marks 
QUESTIONS
SECTION I (50 marks):
Answer all questions in this section
 Use logarithm tables to evaluate ( 4 marks)
 Solve the equation 6^{2x+1} = 2^{3x+1} ( 3 marks)
 Kevin truncated 0.00627 to 3 decimals and 487.74 to 3 significant figures. Calculate his percentage error in calculating product of numbers in truncated values to 1 decimal places. ( 3marks)
 A new laptop depreciates at 8% per annum in the first year and 12% per year in the second year. If its value at the end of the second year was sh121,440. Calculate the original value of the laptop. (3marks)
 Rationalize the denominator and simplify ( 3 marks)
 In the figure below angle A=68º, B= 390, BC= 8.4cm and CN is the bisector of angle ACB. Calculate the length CN to 1decimal place. ( 3 marks)
 In the figure below R, T and S are points on a circle centre O. PQ is a tangent to the circle at T, POR is a straight line and angle QPR =20º. Find the size of angle RST (3marks)
 Use binomial expansion to find the value of (1.02)5 correct to 3 decimal place. ( 4 marks)
 Make x the subject of the equation (3 marks)
 The equation of the circle is given by x^{2} + y^{2} + 8x 2y 1 = 0 . Determine the radius and the centre of the circle. (3marks)
 Given that the minor arc of a circle subtends an angle of 140º at the centre of a circle of radii 3.5cm. Calculate the area of the major segment correct to 4 significant figures ( 3 marks)
 Given that the matrix is a singular matrix, find the values of x. (3marks)
 The mass of a mixture A of peas and millet is 72 kg. The ratio of peas to millet is 3:5 respectively;
 Find the mass of millet in the mixture. (1mark)
 A second mixture of B of peas and millet of mass 98 kg is mixed with A. The final ratio of peas to millet is 8:9 respectively. Find the ratio of peas to millet in B (2marks)
 Draw a line AB= 8cm long. Divide the line proportionally into5 equal parts. Locate a point Y on the line AB such that AY: YB = 3:2. ( 3 marks)
 A solid prism is made of a pentagonal cross section of sides 10cm. If the prism is 30cm long calculate area of the cross section hence the volume of the prism (3 marks)
 Given that X = 2i + j 2k, y = 3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures. (3marks)
SECTION II (50 Marks)
Answer any five questions in this section
 The masses in kilograms of patients who attended a clinic on a certain day were recorded as:
38 52 46 48 60 59 62 73 49 54 49 41 57 58 69 72 60 58 42 41
79 62 58 67 54 60 65 61 48 47 69 59 70 52 63 58 59 49 51 44
67 49 51 58 54 59 39 59 54 52 starting with class 3539, make a frequency distribution table for the data indicating the class and frequency. ( 3 marks)
 state the modal class ( 1 mark)
 Calculate the mean mass ( 3 marks)
 Calculate the median mass ( 3 marks)
 The income tax rates of a certain year were as shown below:
Monthly taxable income in Ksh
Tax rate in %
09680
10
968118800
15
880127920
20
27921 37040
25
37041 and above
30
In that year, Sayao monthly earnings were as follows; basic salary Ksh. 30 000, house allowance Ksh.15 000, and medical allowance of Ksh 3,500. He is entitled to a monthly tax relief of Ksh. 1056. Calculate Sayao’s taxable income ( 2 marks)
 Calculate his P.A.Y.E ( 5 marks)
 A part from P.A.Y.E, other deductions is education insurance policy Ksh. 1500 and Ksh 2500 as cooperative shares. Find his net income at end of the month. ( 3 marks)
 A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = 3, When p =  0.2, m = 3 and n= 2.
 Find
 The equation that connects p, m and n (4marks)
 The value of p when m = 10 and n = 4 (1mark)
 Express m in terms of p and n (2marks)
 If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)
 Find

 Complete the table below by filling in the blank spaces ( 2 marks)
x
0^{0}
30^{0}
60^{0}
90^{0}
120^{0}
150^{0}
180^{0}
210^{0}
240^{0}
270^{0}
300^{0}
330^{0}
360^{0}
y=Cos x
1.00
0.87
0.00
1.00
0.87
0.87
y=3sinx
0.00
3.00
1.50
0.00
 using the scale 1cm to represent 30^{0} on the xaxis and 2 cm to represent 1 unit on the vertical axis, draw on the graphs of cosx and 3sinx ( 5 marks)
 use your graph to solve the equation cosx =3sinx ( 2 marks)
 What is the difference in the values of y=cosx and y=3sinx at x=120^{0} ( 1 mark)
 Complete the table below by filling in the blank spaces ( 2 marks)
 The 5th term of an AP is 16 and the 12th term is 37.
 Find;
 The first term and the common difference ( 3 marks)
 The sum of the first 21 terms (2 marks)
 The second, fourth and the seventh term of an AP are the first 3 consecutive terms of a GP. If the common difference of the AP is 2.
Find: The common ratio of the GP ( 3 marks)
 The sum of the first 8 terms of the GP (2 marks)
 Find;
 In driving to work, John has to pass through three sets of traffic lights. The probability that he will have to stop at any of the lights is ¾
 Draw a tree diagram to represent the above information. (2 marks)
 Using the diagram, determine the probability that on any one journey, he will have to stop at:
 All the three sets. (2 marks)
 Only one of the sets (2 marks)
 Only two of the sets (2 marks)
 None of the sets. (2 marks)
 The figure below shows a lampshade in the form of a conical frustum
The top and bottom radii are 7cm and 14cm respectively. The slant height AB is 20cm. Calculate: The slant height of the original cone correct to two decimal places ( 2 marks)
 The height h, of the lampshade ( 2 marks)
 The curved surface area of the lampshade ( 3 marks)
 The volume of the lampshade correct to 4 significant figures ( 3 marks)
 Gary bought 5 tins of plums and 3 tins of peaches from a supermarket for Ksh.75, while Mike bought 3 tins of plums and 5 tins of peaches for Ksh.77
 Set up the simultaneous equations which represent the given information (2 marks)
 Write down the matrix equation ( 2 marks)
 Using the matrix method, find the cost of
 4 tins of plums ( 5 marks)
 2 tins of peaches ( 2 marks)
MARKING SCHEME
SECTION I (50 marks):
Answer all questions in this section
 Use logarithm tables to evaluate ( 4 marks)
No Standard form Log 0.4239
149.6
log6 = 0.7782
4.3324.239 × 10^{1}
1.496 × 10^{2}
7.782 × 10^{1}
4.335 × 10º1.6272 +
2.1750
1.9022 
1.8911
1.9111
3
0.6370M1
M1
M1
=4.335M  Solve the equation 6^{2x+1} = 2^{3x+1} ( 3 marks)
(2x + 10)log6 = (3x + 1)LOG2
2x + 1 = log2 M1
3x + 1 log6
2x + 1 = 0.3869
3x + 1
2x + 1 = (3x + 1)(0.3869) M1
2x + 1 = 1.1607x + 0.3869
2x  1.1607x = 0.3869  1
0.8393x = 0.6131
x = 0.7305 A1  Kevin truncated 0.00627 to 3 decimals and 487.74 to 3 significant figures. Calculate his percentage error in calculating product of numbers in truncated values to 1 decimal places. ( 3marks)
Actual = 0.00627 × 487.74
=3.0581298
Truncated = 0.006 × 488 M1
=2.928
% = Actual  Truncated
Actual
3.0581298  2.928
3.0581298
= 0.1301298 × 100
3.0581298
=4.255
=4.3% 1dp A1  A new laptop depreciates at 8% per annum in the first year and 12% per year in the second year. If its value at the end of the second year was sh121,440. Calculate the original value of the laptop. (3marks)
Let the original value be P
A = P(1  ^{r}/_{100})^{n}
P(1  ^{8}/_{100})^{1}
A = P(^{92}/_{100})
= 0.92P
2nd year
A = P(1  ^{r}/_{100})^{n}
0.92P(1  ^{12}/_{100})^{1}
0.92P(0.88)
A = 0.8096P
0.8096P = 121,440 M1
P = 121,440
0.8096
P = shs.150,000 A1  Rationalize the denominator and simplify ( 3 marks)
(√3 + 2√5)(√5 + √3)
(√5  √3)(√5 + √3)
Numerator
√3(√5 + √3) +2√5(√5 + √3)
√15 + 3 + 10 + 2√15
13 + 3√15 M1
Denominator
(√5  √3)(√5 + √3)
5  3 = 2 M1
13 + 3√15
2
6.5 + 1.5√15 A1  In the figure below angle A=68º, B= 390, BC= 8.4cm and CN is the bisector of angle ACB. Calculate the length CN to 1decimal place. ( 3 marks)
ACN = 180º  (68 + 39) = 365º
2
8.4 = x
sin68º sin39º
x = 8.4 × sin39º = 5.701
sin68º
CN = 5.701
sin68 sin75.5
cn = 5.701 × sin68
sin75.5
CN = 5.459
CN = 5.5(1dp) cm  In the figure below R, T and S are points on a circle centre O. PQ is a tangent to the circle at T, POR is a straight line and angle QPR =20º. Find the size of angle RST (3marks)
RST = 55º
Angle substended by a chord at the centre is thrice what is substends at the circumference  Use binomial expansion to find the value of (1.02)5 correct to 3 decimal place. ( 4 marks)
(1 + x)^{5} = 1^{5}(x)^{0} + 1^{4}(x)^{1} + 1^{3}(x)^{2} + 1^{2}(x)^{3} + 1^{1}(x)^{4} + 1^{0}(x)^{5}
x = 0.02
1 + x + x^{2} + x^{3} + x^{4} + x^{5}1 5 10 10 5 1
1 + 5x + 10x^{2} + 10x^{3} + 5x^{4 }+ x^{5 }M1
1 + 5(0.02) + 10(0.02)^{2} + 10(0.02)^{3} + 5(0.02)^{4 }+ (0.02)^{5 } M1
1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032 M1
= 1.1040808
= 1.104(3 dp) A1  Make x the subject of the equation (3 marks)
(^{t}/_{s})^{2} = (^{b}/_{√x  4})^{2}t^{2} = b^{2}
s^{2} (x  4)
t^{2}(x  4) = s^{2}b^{2}
t^{2}x  4t^{2} = s^{2}b^{2}
t^{2}x = s^{2}b^{2} + 4t^{2} M1
x = s^{2}b^{2} + 4t^{2} A1
t^{2 }  The equation of the circle is given by x^{2} + y^{2} + 8x 2y 1 = 0 . Determine the radius and the centre of the circle. (3marks)
x^{2} + 8x + 16 + y^{2}  2y + 1 = 1 + 16 + 1
(x + 4)^{2} + (y  1)^{2} = 18
Centre (4,1) radius = √18 = 4.243 units  Given that the minor arc of a circle subtends an angle of 140º at the centre of a circle of radii 3.5cm. Calculate the area of the major segment correct to 4 significant figures ( 3 marks)
360  140 = 220
220 x 22 x 3.5^{2} = 23.52778
360 7
= 23.53 cm^{2}  Given that the matrix is a singular matrix, find the values of x. (3marks)
x( x 1)  0 = 0
x^{2}  1 = 0
x^{2} = 1
x = √1
x = 1 or x = 1
x = ±1  The mass of a mixture A of peas and millet is 72 kg. The ratio of peas to millet is 3:5 respectively;
 Find the mass of millet in the mixture. (1mark)
Total mass = 72 kg
peas: millet
3:5
a) millet = ^{5}/_{8} x 72 = 45kg  A second mixture of B of peas and millet of mass 98 kg is mixed with A. The final ratio of peas to millet is 8:9 respectively. Find the ratio of peas to millet in B (2marks)
A + B mixture = 72 + 98 = 170 kg
mass of peas in A = ^{3}/_{8} x 72 = 27kg
A and B 8:9 of 170 kg
peas = 8/17 x 170 = 80kg
millet 9/17 x 170 = 90kg
In mixture
B:millet = 90  45
=45kg
In mixture
B :peas
=80  27
=53kg
Ratio = 53:45
 Find the mass of millet in the mixture. (1mark)
 Draw a line AB= 8cm long. Divide the line proportionally into5 equal parts. Locate a point Y on the line AB such that AY: YB = 3:2. ( 3 marks)
 A solid prism is made of a pentagonal cross section of sides 10cm. If the prism is 30cm long calculate area of the cross section hence the volume of the prism (3 marks)
tan36º = 5
h
h = 5
tan36º
h = 6.882
5(^{1}/_{2} x 6.882 x 10)
cross sectionesd are = 172.05cm^{2}  Given that X = 2i + j 2k, y = 3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures. (3marks)
= 19.7
SECTION II (50 Marks)
Answer any five questions in this section
 The masses in kilograms of patients who attended a clinic on a certain day were recorded as:
38 52 46 48 60 59 62 73 49 54 49 41 57 58 69 72 60 58 42 41
79 62 58 67 54 60 65 61 48 47 69 59 70 52 63 58 59 49 51 44
67 49 51 58 54 59 39 59 54 52 starting with class 3539, make a frequency distribution table for the data indicating the class and frequency. ( 3 marks)
class tally frequency x fx c.f 3539 ll 2 37 74 2 4044 llll 4 42 168 6 4549 llll lll 8 47 376 14 5054 llll llll 9 52 468 23 5559 llll llll l 11 57 627 34 6064 llll ll 7 62 434 41 6569 llll 5 67 335 46 7074 lll 3 72 216 49 7579 l 1 77 77 50 Σfx = 2775  state the modal class ( 1 mark)
55  59  Calculate the mean mass ( 3 marks)
mean = Σfx = 2775
50
= 55.5  Calculate the median mass ( 3 marks)
54.5 + 2 x 5 = 55.41
11
 starting with class 3539, make a frequency distribution table for the data indicating the class and frequency. ( 3 marks)
 The income tax rates of a certain year were as shown below:
Monthly taxable income in Ksh
Tax rate in %
09680
10
968118800
15
880127920
20
27921 37040
25
37041 and above
30
 Calculate Sayao’s taxable income ( 2 marks)
30,000 + 15,000 + 3,500 = 48,500  Calculate his P.A.Y.E ( 5 marks)
9680 × 10 = 968
100
9120 × 15= 1368
100
9120 × 20 = 1824
100
9120 × 25= 2280
100
11460 × 30 = 3438
100
Gross tax = 9878
P.A.Y.E = 9878  1056
= ksh. 8822 p.m  A part from P.A.Y.E, other deductions is education insurance policy Ksh. 1500 and Ksh 2500 as cooperative shares. Find his net income at end of the month. ( 3 marks)
Total deductions = 8822 + 2500 + 1500
= 12, 822
Net income = 48,500  12,822 = ksh. 35,678
 Calculate Sayao’s taxable income ( 2 marks)
 A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = 3, When p =  0.2, m = 3 and n= 2.
 Find
 The equation that connects p, m and n (4marks)
p = xm^{2} + yn
3.8 = 4x  3y
0.2 = 9x + 2y
7.6 = 8x  6y
0.6 = 27 + 6y +
7 = 35x
x = ^{7}/_{35} = ^{1}/_{5} = 0.2
3.8 = 0.8  3y
3 = =3y
y = 1
P = 0.2m^{2}  n  The value of p when m = 10 and n = 4 (1mark)
P= 0.2 x 100  4
=20  4
=16
 The equation that connects p, m and n (4marks)
 Express m in terms of p and n (2marks)
P = 0.2m^{2}  n
0.2m^{2} = P + n
m^{2} = p + n
0.2  If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)
 Find

 Complete the table below by filling in the blank spaces ( 2 marks)
x
0^{0}
30^{0}
60^{0}
90^{0}
120^{0}
150^{0}
180^{0}
210^{0}
240^{0}
270^{0}
300^{0}
330^{0}
360^{0}
y=Cos x
1.00
0.87
0.50 0.00
0.50 0.89
1.00 0.87
0.50 0.00 0.50 0.87
1.00 y=3sinx
0.00
1.50 2.60 3.00
2.60 1.50 0.00 1.50 2.60 3.00 2.60 1.50
0.00
 using the scale 1cm to represent 30^{0} on the xaxis and 2 cm to represent 1 unit on the vertical axis, draw on the graphs of cosx and 3sinx ( 5 marks)
 use your graph to solve the equation cosx =3sinx ( 2 marks)
x = 18º, 198º  What is the difference in the values of y=cosx and y=3sinx at x=120^{0} ( 1 mark)
2.60   0.5 = 3.1
 Complete the table below by filling in the blank spaces ( 2 marks)
 The 5th term of an AP is 16 and the 12th term is 37.
 Find;
 The first term and the common difference ( 3 marks)
T_{n} = a + (n  1)d
T_{5} = a + 4d = 16
T_{12} = a + 11d = 37
7d = 21
d = 3
a + 4 (3) = 16
a + 12 = 16
a = 4  The sum of the first 21 terms (2 marks)
S_{21} = 21 {2(4) + 20(3)}
2
= 714
 The first term and the common difference ( 3 marks)
 The second, fourth and the seventh term of an AP are the first 3 consecutive terms of a GP. If the common difference of the AP is 2.
Find: The common ratio of the GP ( 3 marks)
a + d, a + 3d, a + 6d
a + 2 a + 6 a + 12
a + 6 = a + 12
a + 2 a + 6
(a + b)^{2} = (a + 2)(a + 12)
a + 12a + 36 = a^{2} + 14a + 24
12= 2a
a = 6
comm ratio
r = 12 = 11
8 2  The sum of the first 8 terms of the GP (2 marks)
S_{8} = 8((^{3}/_{2})8  1 ) = 197.03 = 394.063
1^{1}/_{2} 1 ^{ 1}/_{2}
 The common ratio of the GP ( 3 marks)
 Find;
 In driving to work, John has to pass through three sets of traffic lights. The probability that he will have to stop at any of the lights is ¾
 Draw a tree diagram to represent the above information. (2 marks)
 Using the diagram, determine the probability that on any one journey, he will have to stop at:
 All the three sets. (2 marks)
P(sss) = ^{3}/_{4 }× ^{3}/_{4} × ^{3}/_{4} = ^{27}/_{64}  Only one of the sets (2 marks)
P(ss^{1}s^{1}) or P(s^{1}ss^{1}) or P(s^{1}s^{1}s)
(^{3}/_{4 }× ^{1}/_{4} × ^{1}/_{4}) + (^{1}/_{4 }× ^{3}/_{4} × ^{1}/_{4}) + (^{1}/_{4 }× ^{1}/_{4} × ^{3}/_{4}) = ^{3}/_{64} × 3 = ^{9}/_{64}  Only two of the sets (2 marks)
P(sss^{1}) or P(ss^{1}s) or P(s^{1}ss)
(^{3}/_{4 }× ^{3}/_{4} × ^{1}/_{4}) + (^{3}/_{4 }× ^{1}/_{4} × ^{3}/_{4}) + (^{1}/_{4 }× ^{3}/_{4} × ^{3}/_{4}) = (^{9}/_{64})3 = ^{27}/_{64}  None of the sets. (2 marks)
P(s^{1}s^{1}s^{1}) = ^{1}/_{4 }× ^{1}/_{4} × ^{1}/_{4} = ^{1}/_{64}
 All the three sets. (2 marks)
 Draw a tree diagram to represent the above information. (2 marks)
 The figure below shows a lampshade in the form of a conical frustum
The top and bottom radii are 7cm and 14cm respectively. The slant height AB is 20cm. Calculate: The slant height of the original cone correct to two decimal places ( 2 marks)
L + 20 = 14
L 7
14L = 7L + 140
7L = 140
L = 20
Slant height
= 20 + 20 = 40cm  The height h, of the lampshade ( 2 marks)
40^{2} = H^{2} + 14^{2}
1600 = H^{2} + 196
1600  196 = H^{2}
H^{2} = 1404
H = 37.47
20^{2} = X^{2} + 49
400  49 = X^{2 }
X = 18.73
37.44  18.73
h = 18.74cm  The curved surface area of the lampshade ( 3 marks)
T1R1  T1rL
^{22}/_{7} × 14 × 40  ^{22}/_{7} × 7 × 20
1760  440 = 1320cm^{2}  The volume of the lampshade correct to 4 significant figures ( 3 marks)
^{1}/_{3}πR^{2}H  ^{1}/_{3}πr^{2}h
^{1}/_{3 }× ^{22}/_{7 }× 14^{2} × 37.47  ^{1}/_{3 }× ^{22}/_{7 }× 7^{2} × 18.73
= 7693.84  961.47
= 6732.37cm^{3}
 The slant height of the original cone correct to two decimal places ( 2 marks)
 Gary bought 5 tins of plums and 3 tins of peaches from a supermarket for Ksh.75, while Mike bought 3 tins of plums and 5 tins of peaches for Ksh.77
 Set up the simultaneous equations which represent the given information (2 marks)
let plums be x and peaches be y
5x + 3y = 75
3x + 5y = ??  Write down the matrix equation ( 2 marks)
 Using the matrix method, find the cost of
 4 tins of plums ( 5 marks)
 2 tins of peaches ( 2 marks)
2 tins of peaches
10 × 2 = 20
IA = A
x = 9
y 10
x = 9
y = 10
4 tins of peaches = 9 × 4 = ksh. 36
 4 tins of plums ( 5 marks)
 Set up the simultaneous equations which represent the given information (2 marks)
Download Mathematics Paper 2 Pre Mock Questions and Answers  Mokasa I Joint Examination July 2021.
Tap Here to Download for 50/
Get on WhatsApp for 50/