# Mathematics Paper 2 Questions and Answers - Form 3 End Term 3 Exams 2023

Instructions

• This paper consists of two sections; Section I and Section II.
• Answer all the questions in Section I and any 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 calculators and Kenya National Examinations Council Mathematical tables may be used, except where stated otherwise.

SECTION I (50 Marks)

Answer all the questions in this section

1. Use tables of squares, square roots and reciprocals only to evaluate: (4 marks)
2. Three computer key boards and 2 charger cost Kshs. 5,100 while 4 similar keyboards and 3 similar chargers cost Kshs. 7,050. Use matrices to find the cost of each item. (4 marks)
3. Given that the log 2 =0.30103 and log 3 = 0.47712 , without using a calculator or mathematical table, find the value of log 1080 . (3 marks)
4. Simplify by rationalizing the denominator, (3 marks)
7    −     4
3 − 5     3+5
5. Calculate the percentage error in the calculation of volume of a sphere of radius 7.0 cm (3 marks)
6. Nadia a branding machine on hire purchase. The cash price of the branding machine is Kshs. 56,000. She pays a deposit of Kshs. 14,000 and followed by 24monthly installments of Kshs. 3,500 each. Calculate the monthly rate at which compound interest was charged. (3 marks)
7. The figure below shows a circle centre A.

Find the equation of the circle in below in the form x2 + y2 + ax + by + c = 0, where a, b and c are integers. (3 marks)
Make x the subject of the formula (3 marks)
8. A quantity Z varies directly as the square of X and inversely as the square root of Y. If X increases by 20% and Y decreases by 36%, find the percentage change in Z. (3 marks)
9. On the line PQ below,
1. Use a ruler and a pair of compasses only to construct ∠PQR = 37½° such that R is on the upper side of PQ. (1 mark)
2. By construction and using line QR, locate a point T on PQ such that PT: TQ = 4: 3 (2 marks)
10.
1. Expand the expression below in increasing powers of x, leaving coefficients as fractions in their simplest forms
(1 − 3/10x)5
2.  Use the first three terms of the expansion in part (a) above to estimate the value of (0.97)5 (4 marks)
11. John bought 3 brands of tea A, B and C. The cost price of the three brands were Kshs. 250, Kshs. 300 and Kshs. 450 per kilogram respectively. He mixed the three brands in the ratio 5:2:1 respectively. After selling the mixture he made a profit of 20%. How much profit did he make per kilogram of the mixture? (4 marks)
12. A farmer has two tractors P and Q. Tractor P, working alone, can plough a piece of land in 5 hours while tractor Q would take 12/3 hours less than tractor P. Determine the time the two tractors ploughing together, would take to complete the work. (3 marks)
13. In the figure below, O is the centre of a circle of radius 5 cm. TA and TB are tangents to the circle at A and B respectively. OT=13 cm

Using π = 3.142, calculate the shaded area, correct to 3 decimal places. (4 marks)
14. During an inter-schools games competition, hockey and basketball teams from Zeraki School took part. The probability that rugby team would win the first match was 5/8 while that of the hockey team losing was 1/7. Determine the probability that at least one team won the first match. (3 marks)
15. Given that cos θ = 9/41 and 0°≤90°, find tan (90 − θ) +sin θ without using a mathematical table or a calculator. (3 marks)

SECTION II (50 Marks)
Answer any five questions in this
16.
1. By taking integral values of x from x = −2 to x = 6, make a table of values for the function y = 3x(4−x) (2 marks)
2. On the same pair of axes and using the scales: 1 cm for 1 unit on the x-axis and 1 cm for 5 units on the y-axis, draw the graphs of y = 3x(4−x) and
y = 5(x − 2) (4 marks)
3. From the graphs;
1. Find the roots of the equation 3x (4−x) = 0 (1 mark)
2. Write the maximum value of y = 3x(4−x) (1 mark)
3. Deduce the roots to the equation 3x(4 − x) = 5( x − 2) (2 marks)
17. Three ships P, Q and R are approaching a harbour T. P is 16 km due east of T. Q is 14 km from T on a bearing of 1300 and R is 26.31 km to the west of Q and on a bearing of 2400 from T.
1. Make a sketch drawing showing the relative positions of the ships and the harbour (1 mark)
2. By calculation, find
1. The distance between P and Q
2. The distance from R to T
3. The bearing of
4. The bearing of
18. The table below shows how income tax was charged in a certain year.
 Taxable Income (Kenya pounds p.a.) Tax Rate (Kshs. per Kenya pound) 1 – 3,630 2 3,631 – 7,260 3 7,261 – 10,890 4 10,891 – 14,520 5 14,521 – 18,150 6 18,151 – 21,780 7 21,781 and above 7.5
During the year, Agnetta earned a basic salary of Kshs. 25,200 and a house allowance of Kshs. 12,600 per month. She was entitled to a personal tax relief of Kshs. 1,162 per month.
Calculate:
1. Agnetta’s taxable income in Kenya pounds per annum (2 marks)
2. The net tax she pays per month (6 marks)
3. Apart from income tax, she also contributes monthly NHIF Kshs. 1,600, WCPS Kshs. 1,000. Calculate her net monthly pay (2 marks)
19.
1. The first term of an arithmetic progression is 3 and the sum of its first 8 terms is 164.
1. Find the common difference of the arithmetic progression. (2 marks)
2. Given that the sum of the first n terms of this arithmetic progression is 570, find n. (3 marks)
2. The first, fifth and seventh terms of another arithmetic sequence forms a decreasing geometric progression. If the first term of the geometric progression is 64, find
1. The values of the common difference of the arithmetic sequence (3 marks)
2. The sum of the first 12 terms of the geometric progression. (2 marks)
20. The figure below is a cyclic quadrilateral PQRS. Given that TRX is a tangent at R and O is the centre of the circle and that PSX is a straight line, angle PRS = 50° and angle QPR = 30° and cord RS = PS.

1. Giving reason in each case, find the sizes of the following angles,
1. ∠XRS (2 marks)
2. ∠SXR (2 marks)
3. ∠PQR (2 marks)
4. Reflex ∠QOR (2 marks)
2. Given that RX = 12cm, SX = 8cm, Find the length of chord PS. (2 marks)
21. In triangle OAB below OA = a and OB = b. point M lies on ON such that OM : MA= 2:3 and point N lies on OB such that ON: NB = 5:1. Lines AN and MB intersect at X.

1. Express in terms of a and b;
1. AN (1 mark)
2. BM (1 mark)
2. Given that AX = hAN and BX = kBM, where h and k are constants
1. Express OX in two different ways in terms of a, b, h and k (2 marks)
2. Find the values of h and k (4 marks)
3. Determine the ratio MX:MB (2 marks)
22. The table below shows the wages of workers in a construction firm in Kenya Shillings deposited by selected students in a Zeraki School.
 Wages (Kshs.) Number of Workers 510 – 519 2 520 – 529 4 530 – 539 9 540 – 549 11 550 – 559 10 560 – 569 6 570 – 579 5 580 – 589 3 590 – 599 1

1. Calculate the mean wage (4 marks)
2. Calculate the median wage (3 marks)
3. On the grid provided, draw a frequency polygon to represent the information above. (3 marks)
23.
1. On the grid provided, draw triangle ABC with coordinates A(-2,6), B(2, 3) and C(-2, 3) (1 mark)
2. ∆ABC is reflected along the line x+3=0 to obtain ∆A'B'C'. On the same pair of axes, draw ∆A'B'C'. (2 marks)
3. Translate ∆A'B'C' through 10 2 and label the image as ∆A''B''C'' . Write down the coordinates of ∆A''B''C'' (3 marks)
4. A'''6, -6, B'''2,-3 and C(6,-3)is the image of ∆A''B''C''after a transformation P. Plot ∆A'''B'''C''' and describe P fully. (2 marks)
5. What single transformation would have mapped ∆ABC onto ∆A'''B'''C'''? (2 marks

MARKING SCHEME

NO.  WORKING  MARKS  REMARKS
1.  45.92 = (4.59 × 10)2 = 4.592 x102 = 21.068×100
= 2106.8
√12.08×10-3 = √1.208×10-4 = 1.0990 ×10-2
= 0.010990
1/1.0990×10−2 = 1/1.099 x 10= 0.9136 × 100 = 91.36
4 × 91.36 = 365.44
2106.8 − 365.44 = 1,741.36

M1

M1

M1

A1

Square of 45.9 (2106.8 seen)

Square root i.e. 0.010990 seen

Reciprocal i.e. 91.36 seen

1,741.36

2.  3k+2c=5100

4k+3c=7050

3 2 4 3 k c =5100 7050

Inverse=13×3-4×23 -2 -4 3 =3 -2 -4 3
3 -2 -4 3 3 2 4 3 k c =3 -2 -4 3 5100 7050
1 0 0 1 k c =1200 750 k c =1200 750

Hence keyboard costs Kshs. 1200 and charger Kshs. 750

4

Forming matrix equation

Premultiplying by inverse matrix

Both values correct

3.   1080=23 x 33 × 5
1080=22 x 33 × 10
log 1080 = 2 log  2 +3 log 3 + log 10
log 1080 = 2 × 0.30103 + 3 × 0.47712 + 1
log 1080 = 0.60206 + 1.43136 + 1
log 1080 = 3.03342

M1

M1

A1

Expressing in logs
Substituting given values of logs
3.03342 seen

3
4.   7(3 + √5)− 4(3 −√5)
(3 − √5)(3 + √5)
21 + 7√5−12 + 4√5
3− (√5)2
9+11√5
9 − 5
9 + 11√5 = 9 +11√5
4           4    4

M1

M1

A1

Rationalizing the denominator

Simplification

9/4 + 11/4√5 seen

3
5.  Max V = 4/3 × π × 7.05= 467.2035π
Min V = 4/3 × π × 6.953 = 447.6031667π
Actual V = 4/× π × 7.03 = 1372/3
A.E. = 467.2035π − 447.6031667π
2
A.E.= 9.80016665π
% age error = 9.80016665π×100%
1372/3π
% age error = 2.143%

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A1

0.05 + 0.05 + 0.05
7.0      7.0      7.0

3/140 × 100%

21/7%

3
6.  Total Amount of instalments = 24×3,500=84,000
Total payment = 84,000 + 14,000 = 98,000
Amount borrowed = 98,000 − 56,000 = 42,000
Let the rate of interest be r% per month
84,000 = 42,000(1 + r/100)24
84,000 = 1 + r/100)24
42,000
24√2 = 1 + r/100
1.0293 = 1 + r/100 → 102.93 = 100 + r
r = 102.93 − 100 = 2.93%

M1

M1

A1

Finding amount borrowed
3
7.  Centre 3, −2, radius = 4

(x − 3)2 + (y + 2)2 = 42
x− 6x + 9 + y2 + 4y + 4 = 16
x+ y2 − 6x + 4y + 9 + 4 −16 = 0

x+ y− 6x + 4y − 3 = 0

B1

M1

A1

Substituting centre and radius into general equation of a   circle

x2 + y2 − 6x + 4y – 3 = 0 seen

3
8.  P= kx + 1
1 − kx
P(1− kx) = kx + 1
P− kxP= kx + 1
P− 1 = kx + kxP2 or (− kxP− kx = 1 − P2)
P2 − 1 = x(k + kP2) or x(− kP2 − k) = 1− P2
x =  P− 1    or x =   1− P2
K + KP2           −kP2 − k

M1

M1

A1

Removal of square root

Grouping of terms in x

3
9.  Z = kX2
√Y
X= 120X = 1.2X1
100
Y1 = 64Y = 0.64Y
100
Z1 = kX12
√Y1
Z1 = k(1.2X)2
√0.64Y
Z1 = 1.44kX2 = 1.44Z
0.8Y         0.8
%age change = 1.44/0.8Z− Z  X 100%
z
%age change = 144/80 − 1 × 100% = 80%

M1

M1

A1

Value of Z1

Expression for % change

3
10.

B1

B1

Dividing PQ into 7 equal parts

Point R correctly located

2
11.   1 − 5(3/10x) + 10(3/10x)2 − 10(3/10x)3 + 5(3/10x)4 – (3/10x)5

1 − 3/2x + 910x227/100x3 + 81/2000x4243/100000x5
1 − 3/10x = 0.97
10 − 3x = 9.7
10 − 9.7 = 3x→3x = 0.3→x = 0.1
1 − 3/2(0.1) + 9/100 0.12

= 0.859

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A1

M1

A1

Expansion using the binomial coefficient
Simplified as fractions

Substituting x = 0.1 into the expansion

4
12.  5 × 250 + 2 × 300 + 1 × 450
5+2+1
= 287.50
120 × 287.50
100
= 345

M1

M1
A1

3
13.  P→5 hours→1/5→1 hour
Q→5 −12/= 31/= 10/→ 3/10→1 hour
1 hour both→1/+ 3/10 = 1/2
1/→1hour
1→1×1×2
= 2 hours

M1

M1
A1

Amount of work done in 1 hour by both
3
14.  Let AOT = θ

cos = 513
θ = cos−1(5/13) = 67.38°

Area of ∆ = 1/2×13×5×sin 67.38°
= 29.99997
Area of sector = 67.38/360×3.142×5×5
= 14.70194167
Shaded area = 29.99997 − 14.70194167 = 15.298 cm2

M1

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M1

A1

Angle subtended by the radii at the centre

Area of ΔTOA

Area of sector

4
15.  Pat least 1 team won = P(RH') + P(R'H) + P(RH)
= (5/8 x 1/7) + (3/8 x 6/7) + (5/8 x 6/7)
= 5/56 + 18/56 + 30/56
= 53/56

M1

M1

A1

P(R'H') = 3/8 x 1/= 3/56
1 – 3/56
3
16.     height = √41− 92 = 40
tan (90 − θ) = 9/40
sin θ = 40/41
tan (90 − θ) + sin θ = 9/40 + 40/41
= 1969
1640

B1

B1

B1

Getting height of the triangle

Values of tan (90 − θ ) and sin θ

1969
1640 seen

3
17.
1.
 x − 2 − 1 0 1 2 3 4 5 6 y − 36 − 15 0 9 12 9 0 −15 −36

2.

3.  From graph
1. x = 0 and x = 4
2. 12
3. x = −1 ± 0.2 and x = 3.4 ± 0.2

B2

S1

P1

C1

L1

B1

B1

B1, B1

All values correct (B1 for at least 6 values correct)

Given scales used

Plotting the points

Smooth curve drawn and labeled

Line y = 5(x − 2) drawn

10
18.
1. Sketch
2.  Calculations
1. Consider ΔTQP
t2 = p2 + q2 − 2pqcos T
t2 = 14+ 16− 2×14×16×cos 50°
t2 = 452 − 287.96885
t = √164.03115
t = 12.82 km
2. Consider ΔRTQ
q     t
sin Q    sin T
q       =   26.31
sin 50°   sin 100°
q = 26.31sin 50°
sin 100°
q = 20.47 km
3. Consider ΔTQP
14    =  12.82
sin P      sin 50°
sin P = 14sin 50°
12.82
P = sin−1(14sin 50°)
12.82
P = 56.78°
900 − 56.78° = 33.22°
Bearing = 1800 + 33.22° = 213.22°

B1

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M1
A1

M1

A1

M1

A1
B1
B1

Sketch correctly drawn

Cosine rule applied

Taking square root
12.82 seen

Sine rule applied

20.47 seen

10
19.
1. Taxable income in K£ per annum
23,400 + 10,000 × 12
20
K£ 20,040 per annum
2.  Net tax
3,630×2 = 7,260
3,630×3 = 10,890
3,630×4 = 14,520
3,630×5 = 18,150
3,630×6 = 21,780
1890×7 = 13,230
Gross tax p.m.= 7,260 + 10,890 + 14,520 + 18,150 + 21,780 + 1323012
= 85,830 = Kshs. 7,152.50
12
Net tax = 7.152.50 − 1,162 = Kshs. 5,990.50
3. Net pay
= 23,400 + 10,000 − 5,990.50 +1 ,600 + 1000
= 33,400 − 8,590.5
= Kshs.24,809.5

M1

A1

M1

M1

M1

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M1, A1

M1

A1

First three bands

Next two bands

Last band

Gross tax either p.a. or p.a.

10
20.
1.
1. Common difference
a = 3, S8 = 164
164 = 8/2{2×3 + (8 − 1)d}
164 = 4{6 + 7d}
41 = 6 + 7d
7d = 41 − 6 = 35
d = 35/7 = 5
2. finding n
570 = n22×3 + n −1×5
570 = n26 + 5n − 5)
1140 = n(5n + 1)
5n+ n −1140 = 0
2.
1. 64, 64 + 4d, 64 + 6d
64 + 4d = 64+6d
64        64+4d
642 + 384d = 642 + 512d + 16d2
16d2 + 128d = 0
d(d + 8) = 0
d = − 8

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10
21.
1.
1. RPS = PRS = 50° − base angles of isosceles triangle are equal
PSR = 180° − 2×50°= 80° – sum of angles in a triangle = 1800
PQR = 180° − 80° = 100°
2.  RPS = SRX = 50°
Angle subtended by a chord and a tangent is equal to the angle subtended by same chord in alternate segment to the circle

3.  RSX = PQR = 100° – interior angle equals to exterior opposite
SXR = 180° − 100° − 50° = 30°
Sum of angles in a triangle equals 180°

4.  QPR = QSR = 300

Angles subtended by same arc on the circumference
QOR acute = 2×300 = 600
Angle at the centre twice angle at the circumference
Reflex ∠QOR = 360°− 60° = 300°
Sum of angles at a point is 360°

2.  Let PS = x
x(x + 8) = 12
x2 + 8x − 12 = 0

B1
B1

B1

B1

B1
B1

B1
B1

M1

A1

10
22
1.
1. AN = AO + ON
AN = −a + 5/6b

2. BM = BO + OM
BM = −b + 2/5a
2.
1. OX = OA + AX
OX = a + (h − a + 5/6b)
OX = a − ha + 5/6hb
OX = (1− h)a + 5/6hb
OX = OB + BX
OX = b + k−b + 25a
OX = b − kb + 2/5ka
OX = (1− k)b + 2/5ka

Comparing the coefficients
1 − h = 25k
5 − 5h = 2k
2k + 5h=5…i
56h = 1−k
5h = 6 − 6k
6k + 5h = 6…ii

2k + 5h = 5
6k + 5h = 6
−4k = −1→k =14
1/4 + 5h = 5
5h = 5 −1/2 = 9/2
h = 9/2 x 1/5 = 9/10

2. BX = 14BM
4BX = BM→BX:BM = 1:4
MX:MB = 3:4

B1

B1

B1

B1

M1

M1

M1

A1

B1

Value of AN

Value of BM

First expression of OX

Second expression of OX

Comparing the coefficients

Forming the two equations

Solving the two equations

Values of h and k

10
23
1.  Mean
 Wages (K£) f x fx cf 20 – 29 2 23.5 49 2 30 – 39 5 34.5 172.5 7 40 – 49 9 44.5 400.5 16 50 – 59 11 54.5 599.5 27 60 – 69 10 64,5 645 37 70 – 79 6 74.5 447 43 80 – 89 4 84.5 338 47 90 - 99 3 94.5 283.5 50 Ʃ = 50 Ʃ = 2935

Mean = 2935 = K£ 58.7
50
2. Median
Median = 49.5 + (25 − 16) ×10
11
Median = 57.68
3. Histogram

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M1, A1

M1

A1

S1

P1

L1

Median

All mid points (x) correct

All fx correct

cf column correct

Substitution into median   formula

Appropriate linear scales   used on both axes

10
24.
1. P'(2, −1), Q'4, -1, R'(6, -3) and S'(2, -3)
2. P''(−1, −2), Q''−1, −4, R''(−3, −6) and S''(−3, −2)
3. Transformation – Enlargement
Centre (0, − 6),
Scale Factor – 3

B1

B1

B1

B1

B1

B1

B1

B1

B1
B1
B1

PQRS drawn

Reflection along y=0

P'Q'R'S' drawn

Coordinates P'Q'R'S'

Rotation implied

P''Q''R''S'' drawn

Coordinates P''Q''R''S''

P'''Q'''R'''S''' drawn

10

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