Instructions to candidates
 The paper contains TWO sections: Section I and Section II.
 Answer ALL the questions in section I and strictly any five questions from Section II.
 All answers and working must be written on the question paper in the spaces provided below each question.
 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 electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
This paper consists of 17 printed pages.
For Examiner’s Use Only
Section I
1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  Total 
Section II
17  18  19  20  21  22  23  24  Total 









QUESTIONS
 The length and width of a rectangle were measured as 12.4cm and 5.0cm respectively. Find to 4 significant figures, the percentage error in the area of the rectangle (3mks)
 Simplify (2mks)
 A chord AB whose length is 8cm subtends an angle APB =60° at the circumference of a circle. Calculate to 4 significant figures;
 The perpendicular distance from the centre of the circle to the chord (2mks)
 The radius of the circle (2mks)
 Make h the subject of the formula (3mks)
 Tap A takes 4 hours to fill a tank when empty, tap B takes 3 hours to fill the same tank when empty. Tap C takes 6 hours to empty the same tank when full. Tap A is opened then one hour later tap B and tap C are opened simultaneously. Calculate the total time it takes to fill the tank (3mks)
 In the figure below, AB is a diameter of the circle and AB=8cm, BC=(x+2)cm and AC=2x cm. Calculate the length of AC to 2 decimal places (4mks)
 Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (4mks)

 Expand(3+x)^{6} upto the terms in x^{3} (2 marks)
 Use the expansion in (a) above to estimate (2.97)^{6} correct to 4 decimal places. (2 marks)
 The equation of a circle is x^{2} + y^{2} +4x − 2y − 20 = 0 . On the grid provided below, draw the circle.(4 mark)
 The weights of six boys in kilograms are 10,11, 12, 13, 14, and 15 while those of six girls are 8, 9, 10, 11,12 and 13. A boy and a girl are picked at random and the sum of their weight is recorded.
 Draw a probability space to show all the possible outcomes. (2 marks)
 Find the probability that the sum of their weights is at most 22 kilograms. (1 mark)
 Use completing the square method to solve: 3x^{2} + x −10 = 0 (3 marks)
 The value of a piece of land was Ksh. 400000 five years ago. Currently the piece of land is valued at Ksh. 587731.20 . Find the annual rate of appreciation of the piece of land. (3 marks)
 Find the length of an arc of a circle which subtends an angle of 0⋅8 radians at the centre of the circle. The radius of the circle is 15 cm. (3 marks)
 Lisa, a retailer buys two grades of rice. Grade A costing sh.90 per kilogram and grade B costing sh.120 per kilogram. She mixes the two grades of rice and sells the mixture at a cost of sh.127.5 per kilogram, making a profit of 25%. Find the ratio at which she mixed the two grades of rice. (3 marks)
 The figure below O is the centre of the circle. PQ is parallel to ST and angle PTS = 35º. Find the size of reflex angle QOT (2 marks)
 Given that OP = −4 i ̃+10(j ̃), OQ = 2 i ̃+j ̃ and OR = 6 i ̃−5(j ̃). Show that the points P, Q and R are collinear. (3 marks)
Section II(50 marks)
Answer any five questions from this section in the spaces provided  The income tax rate of a certain year was as shown in the table below;
Monthly taxable income in Kenya shillings (Ksh) Tax rate percentage (%) in each shilling. 0 to 9680
9681 to 18800
18801 to 27920
27921 to 37040
37041 and above10
15
20
25
30
Mwaniki was entitled to a monthly tax relief of Ksh 1162. Calculate Mwaniki’s;
 Taxable income (2 marks)
 Net tax (5 marks)
 Apart from income tax, the following monthly deductions were made; NHIF of Ksh 600, Sacco contributions of Ksh 1500 and 2% of his basic salary for widow and children pension scheme. Calculate Mwaniki’s monthly net income from his employment. (3mks)
 Calculate Mwaniki’s;

 A quantity y varies directly as the square of x and inversely as the square root of z. Given that y = 16 when x = 4 and z = 25,
 find the equation connecting y, x and z. (3 marks)
 if x is increased by 20% and z decreased by 36%, find the percentage change in y. (3 marks)
 The cost (C) per day of feeding examiners in a marking centre partly varies as the number of senior examiners (S) present and partly inversely as the number of ordinary examiners (P) present. It costs sh.32000 to feed 60 senior examiners and 400 ordinary examiners. The cost of feeding 80 senior examiners and 640 ordinary examiners is sh.41250. Find the cost of feeding 100 senior examiners and 500 ordinary examiners. (4 marks)
 A quantity y varies directly as the square of x and inversely as the square root of z. Given that y = 16 when x = 4 and z = 25,
 The 8th and the 15th terms of an Arithmetic sequence are 31 and 59 respectively.
 Find the first term and the common difference (3 marks)
 List the first 4 terms of the sequence (1 mark)
 Find the sum of the first 20 terms of the sequence (3 marks)
 Jeremy’s salary is Ksh980000 per annum. His salary increases by 10% annually. Calculate the total amount to the nearest Ksh he will have earned in 7 years. (3 marks)
 In the figure below O is the centre of the circle. MT is the tangent to the circle at point P. Angle PQS = 56°. PQRS are points on the circumference of the circle. SQT is a straight line.
 Find:
 Angle PRQ (2 marks)
 Angle QPT (2 marks)
 Angle PTQ (2 marks)
 Given that QT = 5 cm and PT = 9.4 cm, calculate to 1 decimal place the area of the circle.
(Use π= 3.142) (4 marks)
 Find:
 Find the area enclosed by the curve y=x^{2} – 2x + 5 with the xaxis between x=2 and x=5 by using;
 Trapezium rule using 6 trapezia (4mks)
 Midordinate rule using 3 midordinate (4mks)
 Find the percentage error in using the midordinate rule as compared to the trapezium rule (2mks)
 The marks scored by 40 students in mathematics test were as shown in the table below.
Mark
4852
5357
5862
6367
6872
7377
No of students
3
4
10
12
8
3
 Using an assumed mean of 64, calculate the mean mark (4mks)
 On the grid provided, draw the cumulative frequency curve for the data (3mks)
 Use the graph to estimate the semi interquartile range (3mks)
 A ship sails from A(0°,70°W) due North to B (25°N, 70°W) then due east to C(25°N,12°E) and finally a further 1800 nautical miles due East to D
 Calculate the total distance covered in nautical miles (4mks)
 If the whole journey took a total time of 300 hours, find its average speed in knots correct to 1 decimal place (2mks)
 Find to the nearest degree the final position of the ship (4mks)
 A quadrilateral ABCD has vertices A(4,−4),B(2,−4),C(2,−2) and D(4,−2)
 On the grid provided, draw the quadrilateral ABCD. (1 mark)
 AʹBʹCʹDʹ is the image of ABCD under a transformation represented by the matrix
Find the coordinates of Aʹ, Bʹ, Cʹ and Dʹ,hence draw the quadrilateral AʹBʹCʹDʹ on the same grid provided. (3 marks)  Given that A’(4,4) is mapped onto A’’(−4,4) by a shear with the x −axis invariant, draw the quadrilateral A’’B’’C’’D’’ ,the image of A’B’C’D’ under the shear. (3 marks)
 Determine a single matrix that maps ABCD onto A’’B’’C’’D’’ (3 marks)
 On the grid provided, draw the quadrilateral ABCD. (1 mark)
MARKING SCHEME
 The length and width of a rectangle were measured as 12.4cm and 5.0cm respectively. Find to 4 significant figures, the percentage error in the area of the rectangle (3mks)
max area = 12.45 x 5.05 = 62.8725
min area = 12.35 x 4.95 = 61.1325
actual area = 12.4 x 5.0 = 62.0
absolute error = 62.8725  61.1325
2
% error = 0.87 x 100%
62
= 1.403%  Simplify (2mks)
 A chord AB whose length is 8cm subtends an angle APB =60° at the circumference of a circle. Calculate to 4 significant figures;
 The perpendicular distance from the centre of the circle to the chord (2mks)
Tan 60° = 4/h
h = 4 = 2.309
tan 60°  The radius of the circle (2mks)
Sin 60° = 4/r
r = 4/sin60°
= 4.619 cm
 The perpendicular distance from the centre of the circle to the chord (2mks)
 Make h the subject of the formula (3mks)
hs^{2} = 2wdh  wd^{2}
2
2hs^{2} = 2wdh  wd^{2}
wd^{2} = 2wdh  2hs^{2}
wd^{2} = h(2wd  2s^{2})
h = wd^{2} or wd^{2}
2wd  2s^{2} 2s^{2}  2wd  Tap A takes 4 hours to fill a tank when empty, tap B takes 3 hours to fill the same tank when empty. Tap C takes 6 hours to empty the same tank when full. Tap A is opened then one hour later tap B and tap C are opened simultaneously. Calculate the total time it takes to fill the tank (3mks)
Fraction of tank remaining after
1 hr = 1  1/4 = 3/4
3 taps after 1 hr full => 1/4 + 1/3  1/6 = 5/12
1 hr  5/12
?  3/4
Time to fill 3/4 = 3/4 x 1 x12/5 = 1^{4}/_{5}
Total time => 1 + 1^{4}/_{5 }= 2^{4}/_{5} hrs  In the figure below, AB is a diameter of the circle and AB=8cm, BC=(x+2)cm and AC=2x cm. Calculate the length of AC to 2 decimal places (4mks)
angle ACB = 90º (subtended by D)
(2x)^{2} + (x + 2)^{2} = 64
4x^{2} + x^{2} + 4x + 4 = 64
5x^{2} + 4x = 60
x^{2} + 0.8x = 12
x^{2} + 0.8x + (0.4)^{2} = 12 + 0.4^{2}
x = 3.087 or 3.887
AC = 2 x 3.087 = 6.17 cm  Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (4mks)
cos^{1}0.8070 = 36.196 = 36.2º
2x = 36.2 x = 18.1
2x = 323.8 x = 161.9
2x = 396.7 x = 198.1
2x = 683.8 x = 341.9 
 Expand(3+x)^{6} upto the terms in x^{3} (2 marks)
1 6 15 20 3^{6} 3^{5} 3^{4} 3^{3} x^{0} x1 x^{2} x^{3} 729 + 1458x + 1215x^{2} + 540x^{3}  Use the expansion in (a) above to estimate (2.97)^{6} correct to 4 decimal places. (2 marks)
3 + x = 2.97
x = 2.97  3 = 0.03
729 + 1458(0.03) + 1215(0.03)^{2} + 540(0.03)^{3}
= 686.3389
 Expand(3+x)^{6} upto the terms in x^{3} (2 marks)
 The equation of a circle is x^{2} + y^{2} +4x − 2y − 20 = 0 . On the grid provided below, draw the circle.(4 mark)
 The weights of six boys in kilograms are 10,11, 12, 13, 14, and 15 while those of six girls are 8, 9, 10, 11,12 and 13. A boy and a girl are picked at random and the sum of their weight is recorded.
 Draw a probability space to show all the possible outcomes. (2 marks)
10 11 12 13 14 15 8 18 19 20 21 22 23 9 19 20 21 22 23 24 10 20 21 22 24 24 25 11 21 22 23 25 25 26 12 22 23 24 26 26 27 13 23 24 25 27 27 28  Find the probability that the sum of their weights is at most 22 kilograms. (1 mark)
15/36
 Draw a probability space to show all the possible outcomes. (2 marks)
 Use completing the square method to solve: 3x^{2} + x −10 = 0 (3 marks)
x^{2} + x/3 = 10/3
x^{2} + x/3 + (1/6)^{2} = 10/13 + (1/6)^{2}
(x + 1/6)^{2} = 3.3611
x + 1/6 = ± 1.8333
x = 1.8333  0.1667
= 1.6667 = 1^{2}/_{3}
x = 1.8333  0.1667
= 2  The value of a piece of land was Ksh. 400000 five years ago. Currently the piece of land is valued at Ksh. 587731.20 . Find the annual rate of appreciation of the piece of land. (3 marks)
A = P (1 + r/100)^{s}
587731.20 = 400,000 (1 + r/100)5
1.469328 = (1 + r/100)5
1 + r/100 = 1.08
r/100 = 0.08
r = 8%  Find the length of an arc of a circle which subtends an angle of 0⋅8 radians at the centre of the circle. The radius of the circle is 15 cm. (3 marks)
c/15 = 0.8
c = 15 x 0.8
= 12cm  Lisa, a retailer buys two grades of rice. Grade A costing sh.90 per kilogram and grade B costing sh.120 per kilogram. She mixes the two grades of rice and sells the mixture at a cost of sh.127.5 per kilogram, making a profit of 25%. Find the ratio at which she mixed the two grades of rice. (3 marks)
125% = 127.5
100% = 100 x 127.5 = 102
125
18 : 12
3 : 2  The figure below O is the centre of the circle. PQ is parallel to ST and angle PTS = 35º. Find the size of reflex angle QOT (2 marks)
TPQ = 35º
QOT = 70º
Reflex = 360  70
= 290 º  Given that OP = −4 i ̃+10(j ̃), OQ = 2 i ̃+j ̃ and OR = 6 i ̃−5(j ̃). Show that the points P, Q and R are collinear. (3 marks)
4 = 6k
k = 2/3
QR = 2/3 PQ hence parallel
Q is common hence collinear
Section II(50 marks)
Answer any five questions from this section in the spaces provided  The income tax rate of a certain year was as shown in the table below;
Monthly taxable income in Kenya shillings (Ksh) Tax rate percentage (%) in each shilling. 0 to 9680
9681 to 18800
18801 to 27920
27921 to 37040
37041 and above10
15
20
25
30
Mwaniki was entitled to a monthly tax relief of Ksh 1162. Calculate Mwaniki’s;
 Taxable income (2 marks)
=35400 + 7000 + 5000 + 4520 + 10400
=sh 62 320  Net tax (5 marks)
1st tax = 9680 x 0.1 = sh 968
2nd tax = 9120 x 0.15 = sh 1368
3rd tax = 9120 x 0.2 = sh 1824
4th tax = 9120 x 0.25 = sh 2280
5th tax = 25280 x 0.3 = sh 7584
gross tax = sh 14024
net tax = 14024  1162
= sh 12862
 Taxable income (2 marks)
 Apart from income tax, the following monthly deductions were made; NHIF of Ksh 600, Sacco contributions of Ksh 1500 and 2% of his basic salary for widow and children pension scheme. Calculate Mwaniki’s monthly net income from his employment. (3mks)
deductions = 12862 + 600 + 1500 + 2/100 x 35400
= 15670
62320  15670 = sh 46650
 Calculate Mwaniki’s;

 A quantity y varies directly as the square of x and inversely as the square root of z. Given that y = 16 when x = 4 and z = 25,
 find the equation connecting y, x and z. (3 marks)
 if x is increased by 20% and z decreased by 36%, find the percentage change in y. (3 marks)
 find the equation connecting y, x and z. (3 marks)
 The cost (C) per day of feeding examiners in a marking centre partly varies as the number of senior examiners (S) present and partly inversely as the number of ordinary examiners (P) present. It costs sh.32000 to feed 60 senior examiners and 400 ordinary examiners. The cost of feeding 80 senior examiners and 640 ordinary examiners is sh.41250. Find the cost of feeding 100 senior examiners and 500 ordinary examiners. (4 marks)
c = ms + n/p
3200 = 60m + n/400
12800000 = 24000m + n .....(i)
41250 = 80m + n/640
36400000 = 51200m + n .....(ii)
13600000 = 27200m
27200 27200
m = 500
n = 800,000
c = 500s + 800,000
p
c = 500(100) + 800,000
500
c = sh 51, 600
 A quantity y varies directly as the square of x and inversely as the square root of z. Given that y = 16 when x = 4 and z = 25,
 The 8th and the 15th terms of an Arithmetic sequence are 31 and 59 respectively.
 Find the first term and the common difference (3 marks)
a + 7d = 31
a + 14d = 59
7d = 28
d = 4
a + 28 = 31
a = 3  List the first 4 terms of the sequence (1 mark)
3, 7, 11, 15  Find the sum of the first 20 terms of the sequence (3 marks)
20^{th} term = 3 + (19 x 4) = 79
s_{20} = ^{20}/_{2} (3 + 79)
= 820  Jeremy’s salary is Ksh980000 per annum. His salary increases by 10% annually. Calculate the total amount to the nearest Ksh he will have earned in 7 years. (3 marks)
110 x 980,000 = 1078000
100
r = 1078000 = 1.1
98000
sn = a(r^{n}  1)
r  1
= 98000 (1.1^{7}  1)
0.1
= sh 9297428
 Find the first term and the common difference (3 marks)
 In the figure below O is the centre of the circle. MT is the tangent to the circle at point P. Angle PQS = 56°. PQRS are points on the circumference of the circle. SQT is a straight line.
 Find:
 Angle PRQ (2 marks)
QPS = 90º
QPS = PRQ = 34º  Angle QPT (2 marks)
QPT = 34º (alternate angle theorem)  Angle PTQ (2 marks)
PQT = 124º
PTQ = 180º  (124 + 34)
= 22º
 Angle PRQ (2 marks)
 Given that QT = 5 cm and PT = 9.4 cm, calculate to 1 decimal place the area of the circle.
(Use π= 3.142) (4 marks)
let the diametre SQ = x
5(x + 5) = 9.4^{2}
5x + 25 = 88.36
5x = 63.36
x = 12.672
area = 3.142 x ( 12.672/2)^{2}
= 1261 cm^{2}
 Find:
 Find the area enclosed by the curve y=x^{2} – 2x + 5 with the xaxis between x=2 and x=5 by using;
x 2.0 2.5 3.0 3.5 4.0 4.5 5.0 y 5 6.25 8 10.25 13.0 16.25 20  Trapezium rule using 6 trapezia (4mks)
A = 1/2 x 0.5 [{5 + 20) + 2(6.25 + 8 + 10.25 + 13 + 16.25)]
= 0.25 (25 + 2(53.75))
= 0.25 (25 + 107.5)
= 33.125 sq units  Midordinate rule using 3 midordinate (4mks)
y1 = 6.25 y2 = 10.25 y3 = 16.25
A = 1(6.25 + 10.25 + 16.25)
= 32.75  Find the percentage error in using the midordinate rule as compared to the trapezium rule (2mks)
% error = (33.125  32.75) x 100%
33.125
= 0.375 x 100%
33.125
= 1.132%
 Trapezium rule using 6 trapezia (4mks)
 The marks scored by 40 students in mathematics test were as shown in the table below.
Mark
4852
5357
5862
6367
6872
7377
No of students
3
4
10
12
8
3
 Using an assumed mean of 64, calculate the mean mark (4mks)
x f t ft 50 3 14 42 55 4 9 36 60 10 4 40 65 12 1 12 70 8 6 48 75 3 11 33 40 25
40
= 63.375  On the grid provided, draw the cumulative frequency curve for the data (3mks)
 Use the graph to estimate the semi interquartile range (3mks)
Q1 = 59 mk
Q3 = 68 mk
semi interquatile range = 68  59
2
= 4.5
 Using an assumed mean of 64, calculate the mean mark (4mks)
 A ship sails from A(0°,70°W) due North to B (25°N, 70°W) then due east to C(25°N,12°E) and finally a further 1800 nautical miles due East to D
 Calculate the total distance covered in nautical miles (4mks)
AB = 60 x 25
= 1500 nm
BC = 82 x 60cos25
= 4459.034 nm
Total distance = 1500 + 4459.034 + 1800
= 7759.034 nm  If the whole journey took a total time of 300 hours, find its average speed in knots correct to 1 decimal place (2mks)
Average speed = 7759.034
300
= 25.863
= 25.9 knots  Find to the nearest degree the final position of the ship (4mks)
1800 = α x 60 cos 25
α = 1800 = 33º
60 cos 25
longitude = 33 + 12 = 45ºE
Position = (25ºN, 45ºE)
 Calculate the total distance covered in nautical miles (4mks)
 A quadrilateral ABCD has vertices A(4,−4),B(2,−4),C(2,−2) and D(4,−2)
 On the grid provided, draw the quadrilateral ABCD. (1 mark)
 AʹBʹCʹDʹ is the image of ABCD under a transformation represented by the matrix
Find the coordinates of Aʹ, Bʹ, Cʹ and Dʹ,hence draw the quadrilateral AʹBʹCʹDʹ on the same grid provided. (3 marks)
A^{1}(4, 4)
B^{1}(4, 2)
C^{1}(2, 2)
D^{1}(2, 4)  Given that A’(4,4) is mapped onto A’’(−4,4) by a shear with the x −axis invariant, draw the quadrilateral A’’B’’C’’D’’ ,the image of A’B’C’D’ under the shear. (3 marks)
d = 2A  Determine a single matrix that maps ABCD onto A’’B’’C’’D’’ (3 marks)
 On the grid provided, draw the quadrilateral ABCD. (1 mark)
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