INSTRUCTIONS TO CANDIDATES
- Write your name and admission number in the spaces provided at the top of this page.
- This paper consists of two sections: Section I and Section II.
- Answer all questions in section I and any five questions in Section II.
- 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.
- KNEC Mathematical tables may be used.
QUESTIONS
SECTION I (50 marks)
Answer all the questions in the spaces provided in this section.
- Make d the subject of the formulae;
(3 marks) - Use logarithm tables to evaluate (4 marks)
- The base and perpendicular height of a triangle are measured as 8.2cm and 6.3cm respectively. Calculate the percentage error in calculating the area correct to 3 d.p. (4 marks)
- A businesswoman borrowed Ksh. 240,000 from a financial institution that charges compound interest of 12% p.a for five years. She used the money to purchase a piece of land that appreciates at the rate of 15% p.a. If she sold the piece of land after five years, determine the amount of profit she made. (3 marks)
- Simplify (2 marks)
-
- Expand and simplify (1-3x)5 up to the term in x3 (1 mark)
- Hence use your expansion to estimate (0.97)5 correct to 4 d.p. (2 marks)
- A quantity P varies partly as Q and partly as the square root of Q. When P= -4, Q = 16 and when P = -42, Q = 36. State the equation connecting P and Q. (3 marks)
- Solve for x given that (3 marks)
Log (x-4) +2 =log 5 + log (2x+10) - Simplify
hence use the quadratic formula to find the value of x. (3 marks) - The equation of a circle is given as 4x2+4y2-16x+24y+3=0. Find the centre of the circle and its radius. (3 marks)
- The region in the figure below is defined by the inequalities L1, L2, L3 and L4
Form four inequalities that satisfy the unshaded region. (4 marks) - The three sides of a triangle ABC are AB¬¬¬=12cm, BC=10cm and AC=8cm. Using the Hero’s formula calculate to 2 decimal places the area of the triangle. (3 marks)
- Given the column vectors
and that p=2a-1/3 b+c.- Express p as a column vector. (2 marks)
- Determine the magnitude of p. (1 mark)
- In the figure below DC is tangent to the circle at D and O is centre of the circle. AOBC is a straight line. Given that DC=6cm and BC =4cm
Calculate the length AC (3 marks) - The figure below shows a triangle ABC in which AB = 6cm, BC = 11cm and angle ABC = 110°. Calculate to 2 decimal places the length of AC. (3 marks)
- Use tables of reciprocals only to work out
(3 marks) - In the figure below E is the midpoint of BC. AD: DC =3:2 and F is the meeting point of BD and AE. AC=c ̌ and AB=b
- Express the following vectors in terms of b and c
- BD (1 mark)
- AE (2 marks)
, Express AF in two different ways hence find the value of t and n (5 marks)- State the ratio in which F divides
- BD (1 mark)
- AE (1 mark)
- Express the following vectors in terms of b and c
- Use a pair of compass and a ruler only for all constructions in this question.
- Construct triangle PQR in which QR=5cm , PR=4cm and Angle PRQ=105° (3 marks)
- Measure line PQ. (1 mark)
- Calculate the area of the triangle PQR. (3 marks)
- Draw a circle passing through the vertices of the triangle. (3 marks)
- The table below shows the income tax rates for the year 2021.
In the year 2021, Omondi’s monthly income was as follows:Taxable income per annum per K£
Tax rate Sh per K£
0-5808 2
5809-11 280 3
11 281-16 752 4
16 753-22 224 5
22 224 and above 6
Basic salary Ksh. 22 600
House allowance Ksh. 12 000
Medical allowance Ksh. 2 880
Transport allowance Ksh. 340
Omondi was entitled to a monthly personal relief of Ksh. 1156. Every month the following deductions are made:
NHIF Ksh. 1 500
SACCO share contribution Ksh. 3 000
Calculate:- Omondi’s taxable income in Kenyan pounds. (2 marks)
- Total tax paid per year in shillings. (6 marks)
- Net salary for the month. (2 marks)
-
- Solve the equation (4 marks)
- The length of a floor of a rectangular hall is 9m more than its width .If the area of the floor is 136 m2,
- Calculate the perimeter of the floor (3 marks)
- A rectangular carpet is placed on the hall leaving an area of 64m2.If the length of the carpet is twice its width, determine the width of the carpet. (2 marks)
- Solve the equation (4 marks)
- The probability that our school will host soccer and rugby tournament this year is 0.8. If we host, the probability of winning soccer is 0.7. If we don’t host the probability of winning soccer is 0.4. If we win soccer the probability of winning rugby is 0.8, otherwise if we lose the probability of winning rugby is 0.3.
- Draw a tree diagram to represent this information. (2 marks)
- Use the tree diagram to find:-
- The probability that we lose both games (2 marks)
- The probability that we will win only one game (3 marks)
- The probability that we will host and lose both games (2 marks)
- The probability that we win at least one game, if we host (1 mark)
- The first, fifth and seventh terms of an Arithmetic Progression(A.P) form the first three terms of a decreasing Geometric Progression(G.P) respectively. Given that the first term of each progression is 64.
- Write two equations involving common ratio of Geometric Progression (G.P) and find the common difference of Arithmetic Progression (A.P). (4 marks)
- Find the sum of the first 24 terms of the Arithmetic Progression (AP) (3 marks)
- Find the number of terms for which the sum of the Geometric Progression (G.P) is 127 7/8 . (3 marks)
- The figure below shows a belt passing round two wheels of centres P and Q respectively. The radii of the pulleys are 15cm and 9cm respectively. RS and UV are tangents to the circles. ∠RPQ=72° and PQ = 20cm. (Use π=3.142)
Calculate- The length of RS (2 marks)
- Arc length RWV (3 marks)
- Arc length STU (3 marks)
- The total length of the belt (2 marks)
-
- Complete the table below for the equation y = x2 – 6x + 5 (2 marks)
0
1
2
3
4
5
6
0
4
9
25
0
5
5
5
5
5
5
5
5
5
0
- Draw the graph of y = x2 – 6x + 5 using values in the table (3 marks)
- Use the graph to solve the equation
- x2 – 6x + 5 = 0 (1 mark)
- x2= 6x – 7 (2 marks)
- x2– 6.5x + 5 = 0 (2 marks)
- Complete the table below for the equation y = x2 – 6x + 5 (2 marks)
MARKING SCHEME
- Make d the subject of the formulae; (3 marks)
- Use logarithm tables to evaluate (4 marks)
No
Std form
Log
45.3
0.00697
4.53 x 101
6.97 x 10-3
1.6561
3.8432 +
1.4993
0.534
5.34 x 101
1.7275
N
D1.4993
1.7275 -
1.7718 ÷ 30.8392
100.9239 x 101
1.9239
- The base and perpendicular height of a triangle are measured as 8.2cm and 6.3cm respectively. Calculate the percentage error in calculating the area correct to 3 d.p. (4 marks)
Actual area = 0.5 x 8.2 x 6.3 = 25.83cm2
maximum area = 0.5 x 8.25 x 6.35 = 26.19375cm2
minimum area = 0.5 x 8.15 x 6.25 = 25.46875cm2
absolute error = 26.193875 - 25.46875 = 0.3625
2
percentage error = 0.3625 x 100 = 1.403%
25.83 - A businesswoman borrowed Ksh. 240,000 from a financial institution that charges compound interest of 12% p.a for five years. She used the money to purchase a piece of land that appreciates at the rate of 15% p.a. If she sold the piece of land after five years, determine the amount of profit she made. (3 marks)
Amount as a result of borrowing = 240000(1 + 0.12)5 = sh. 422 962
Amount as a result of appreciation = 240 000(1 + 0.15)5 = sh. 482 725
profit = 482 725 - 422 962 = sh.59 763 - Simplify (2 marks)
-
- Expand and simplify (1-3x)5 up to the term in x3 (1 mark)
(1 - 3x)5 = 1 - 5(3x) + 10(9x2) - 10(27x3) + .......
=1 - 15x + 90x2 - 270x3 - Hence use your expansion to estimate (0.97)5 correct to 4 d.p. (2 marks)
0.97 = 1 - 3x
x = 0.01
0.975 = 1 - 15(0.01) + 90(0.01)2 - 270(0.01)3
=1 - 0.15 + 0.009 - 0.00027
= 0.858773 = 0.8587
- Expand and simplify (1-3x)5 up to the term in x3 (1 mark)
- A quantity P varies partly as Q and partly as the square root of Q. When P= -4, Q = 16 and when P = -42, Q = 36. State the equation connecting P and Q. (3 marks)
p = kQ + mQ2
-4 = 16k + 256m .................(i)
-42 = 36k + 1296m .............(ii)
Solving (i) and (ii) simultaneously
-42 = 36 (-4 - 256m) + 1296m
16
-672 = 36(-4 - 256m) + 20 736m
-672 = -144 - 9216m + 20 736m
-528 = 11520m
m = -11/240
k = -1/4 - 16(-11/240) = 29/60
p = 29/60Q - 11/240Q2 - Solve for x given that (3 marks)
Log (x-4) +2 =log 5 + log (2x+10)
100(x - 4) = 5(2x + 10)
100x - 400 = 10x + 50
90x = 450
x = 5 - Simplify hence use the quadratic formula to find the value of x. (3 marks)
- The equation of a circle is given as 4x2+4y2-16x+24y+3=0. Find the centre of the circle and its radius. (3 marks)
4x2 - 16x + 4y2 + 24y = -3
x2 - 4x + 4 + y2 + 6y + 9 = -0.75 + 4 + 9
(x - 2)2 + (y + 3)2 = (7/2)2
radius = 3.5 units
centre =>(2, -3) - The region in the figure below is defined by the inequalities L1, L2, L3 and L4
L1 ⇒ 2y + x ≤ 6
L2 ⇒ y ≤ x - 2
L3 ⇒ y > 0
L4 ⇒ x > 0
Form four inequalities that satisfy the unshaded region. (4 marks) - The three sides of a triangle ABC are AB¬¬¬=12cm, BC=10cm and AC=8cm. Using the Hero’s formula calculate to 2 decimal places the area of the triangle. (3 marks)
Area = √s(s - a)(s - b)(s - c)
s = 0.5(12 + 10 + 8) = 15
Area = √15 x 3 x 5 x 7
= √1575
= 39.69cm2 - Given the column vectors and that p=2a-1/3 b+c.
- Express p as a column vector. (2 marks)
- Determine the magnitude of p. (1 mark)
IpI = √9 + 1 + 4 = 3.742 units
- Express p as a column vector. (2 marks)
- In the figure below DC is tangent to the circle at D and O is centre of the circle. AOBC is a straight line. Given that DC=6cm and BC =4cm
Calculate the length AC (3 marks)
let AB = x
CD2 = AC x CB
62 = (4 + x)4
36 - 16 = 4x
x = 5
AC = 5 + 4 = 9 cm - The figure below shows a triangle ABC in which AB = 6cm, BC = 11cm and angle ABC = 110°. Calculate to 2 decimal places the length of AC. (3 marks)
Let AC = x
x2 = 62 + 112 - (2 x 6 x 11 x cos 110º)
x2 = 36 + 121 + 45.15 = 202.15
x = √202.15 = 14.22 cm - Use tables of reciprocals only to work out (3 marks)
1/0.0396 = 102 x 0.2525 = 25.25
1/0.593 = 101 x 0.1686 = 1.686
(5 x 25.25) + (12 x 1.686) = 126.25 + 3.372
= 129.622 - In the figure below E is the midpoint of BC. AD: DC =3:2 and F is the meeting point of BD and AE. AC=c ̌ and AB=b
- Express the following vectors in terms of b and c
- BD (1 mark)
- AE (2 marks)
- , Express AF in two different ways hence find the value of t and n (5 marks)
- State the ratio in which F divides
- BD (1 mark)
5:4 - AE (1 mark)
3:1
- BD (1 mark)
- Express the following vectors in terms of b and c
- Use a pair of compass and a ruler only for all constructions in this question.
- Construct triangle PQR in which QR=5cm , PR=4cm and Angle PRQ=105° (3 marks)
- Measure line PQ. (1 mark)
7.2 ± 0.1 cm - Calculate the area of the triangle PQR. (3 marks)
Area = 0.5 x 5 x 4 x sin 105º
= 9.66 cm2 - Draw a circle passing through the vertices of the triangle. (3 marks)
Line bisectors
Circle correctly drawn
- Construct triangle PQR in which QR=5cm , PR=4cm and Angle PRQ=105° (3 marks)
- The table below shows the income tax rates for the year 2021.
In the year 2021, Omondi’s monthly income was as follows:Taxable income per annum per K£
Tax rate Sh per K£
0-5808 2
5809-11 280 3
11 281-16 752 4
16 753-22 224 5
22 224 and above 6
Basic salary Ksh. 22 600
House allowance Ksh. 12 000
Medical allowance Ksh. 2 880
Transport allowance Ksh. 340
Omondi was entitled to a monthly personal relief of Ksh. 1156. Every month the following deductions are made:
NHIF Ksh. 1 500
SACCO share contribution Ksh. 3 000
Calculate:- Omondi’s taxable income in Kenyan pounds. (2 marks)
Taxable income = (22600 + 12000 + 2880 + 3340) x 12
20
= 22 692 pounds - Total tax paid per year in shillings. (6 marks)
1st band = 5808 x 2 = sh.11 616
2nd band = 5472 x 3 = sh. 16 416
3rd band = 5472 x 4 = sh.21 888
4th band = 5472 x 5 = sh. 27 360
5th band = 468 x 6 = sh. 2808
total tax payable = 11 616 + 16 416 + 21 888 + 27 360 + 2808
= sh. 80 088
Tax paid = sh.80 088 - 13 872
= sh. 66 216 - Net salary for the month. (2 marks)
net salary = 37 820 - (5518 + 1500 + 3000)
= sh. 27 802
- Omondi’s taxable income in Kenyan pounds. (2 marks)
-
- Solve the equation (4 marks)
- The length of a floor of a rectangular hall is 9m more than its width .If the area of the floor is 136 m2,
- Calculate the perimeter of the floor (3 marks)
- A rectangular carpet is placed on the hall leaving an area of 64m2.If the length of the carpet is twice its width, determine the width of the carpet. (2 marks)
=136 - 64 = 72m2
2x2 = 72
x = √36 = 6 cm
- Calculate the perimeter of the floor (3 marks)
- Solve the equation (4 marks)
- The probability that our school will host soccer and rugby tournament this year is 0.8. If we host, the probability of winning soccer is 0.7. If we don’t host the probability of winning soccer is 0.4. If we win soccer the probability of winning rugby is 0.8, otherwise if we lose the probability of winning rugby is 0.3.
- Draw a tree diagram to represent this information. (2 marks)
- Use the tree diagram to find:-
- The probability that we lose both games (2 marks)
P(HS'R') or P(H'S'R') = (0.8 x 0.3 x 0.7) + (0.2 x 0.6 x 0.7)
= 0.168 + 0.084
=0.252 - The probability that we will win only one game (3 marks)
P(HSR') or (HS'R) or P(H'SR') or P(H'S'R')
= (0.8 x 0.7 x 0.2) + (0.8 x 0.3 x 0.3) + (0.2 x 0.4 x 0.2) + (0.2 x 0.6 x 0.3)
= 0.112 + 0.072 + 0.016 + 0.036
= 0.236 - The probability that we will host and lose both games (2 marks)
P(HS'R') = (0.8 x 0.3 x 0.7) = 0./168 - The probability that we win at least one game, if we host (1 mark)
P(HSR) or P(HSR') or P(HS'R)
(0.8 x 0.7 x 0.2) + (0.8 x 0.3 x 0.3)
= 0.448 + 0.112 + 0.072
= 0.632
- The probability that we lose both games (2 marks)
- Draw a tree diagram to represent this information. (2 marks)
- The first, fifth and seventh terms of an Arithmetic Progression(A.P) form the first three terms of a decreasing Geometric Progression(G.P) respectively. Given that the first term of each progression is 64.
- Write two equations involving common ratio of Geometric Progression (G.P) and find the common difference of Arithmetic Progression (A.P). (4 marks)
64 = 64 + 4d
64 + 4d 64 + 6d
= 4096 + 512d + 16d2 = 4096 + 384d
16d = -128
d = -8 - Find the sum of the first 24 terms of the Arithmetic Progression (AP) (3 marks)
a = 64 & d = -8
s24 = 24/2(128 - 184)
= 12 x -56 = -672 - Find the number of terms for which the sum of the Geometric Progression (G.P) is 127 7/8 . (3 marks)
n = 10 terms
- Write two equations involving common ratio of Geometric Progression (G.P) and find the common difference of Arithmetic Progression (A.P). (4 marks)
- The figure below shows a belt passing round two wheels of centres P and Q respectively. The radii of the pulleys are 15cm and 9cm respectively. RS and UV are tangents to the circles. ∠RPQ=72° and PQ = 20cm. (Use π=3.142)
Calculate- The length of RS (2 marks)
sin 72º = RS/PQ
RS = 20 x sin 72 = 19.02 cm - Arc length RWV (3 marks)
∠RPV(Reflex) = 360 - (72 x 2) = 216
= 216/360 x 3.142 x 2 x 15
= 56.56 cm - Arc length STU (3 marks)
∠STU = 72 x 2 = 144
= 144/360 x 3.142 x 2 x 9
= 22.62 cm - The total length of the belt (2 marks)
total length = 2(19.02) + 56.56 + 22.62
= 117.22 cm
- The length of RS (2 marks)
-
- Complete the table below for the equation y = x2 – 6x + 5 (2 marks)
x 0
1
2
3
4
5
6
x2 0
1 4
9
16 25
36 -6x 0
-6 -12 -18 -24 -30 -36 5
5
5
5
5
5
5
5
y 5
0 -3 -4 -3 0
- Draw the graph of y = x2 – 6x + 5 using values in the table (3 marks)
- Use the graph to solve the equation
- x2 – 6x + 5 = 0 (1 mark)
x = 1 or 5 - x2= 6x – 7 (2 marks)
x = 1.6 or x = 4.4 - x2– 6.5x + 5 = 0 (2 marks)
x = 0.9 or x = 5.6
- x2 – 6x + 5 = 0 (1 mark)
- Complete the table below for the equation y = x2 – 6x + 5 (2 marks)
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