MOCKS 1 2023
121/1MATHSPAPER 1MARKING SCHEME
| 1. | Numerator: Denominator: | M1 M1
A1 | Numerator Denominator | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 2. | N;( D;
| M1
M1
A1 | Numerator
denominator | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 3. | UK = |
M1
M1
A1 |
Expression
Expression
CAO | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 4. | M1
M1
A1 |
Comparing powers
| |||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 5. | M1
M1 A1 | Equation
Expression
| |||||||||||||||||||||||||
| 03
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| 6. |
0.4428 | M1
M1
A1 | All logs correct
Attempt to divide | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 7. | B1
B1
B1 | 2y<x +4
4y ≥ – x- 4
x≤2 | |||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 8. | Midpoint ( | M1
M1
A1 |
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| 03 | |||||||||||||||||||||||||||
| 9. | LCM = 900 = 22 x 32 x 52 36 = 22 x 32 60 = 22 x 3 x 5 Least possible number = 2 x 3 x 52= 150
| B1
B1
B1 | GCD/LCM
36/60 | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 10. | M1
M1
A1 | ||||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 11. |
M1
M1
A1 |
Substitution
For both | |||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 12. | (a) 5 Tan θ = 4 3 (b) Cos (180 – θ) = – | B1
B1
B1 | |||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 13. |
|
B1
B1
B1 |
Complete net, well labelled | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 14. |
M1
M1
A1 | ||||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 15. | (i) <BOD = 2 <DAB = 2 x 87 = 1740
(ii) |
B1 B1
B1 B1 |
<AOB Property
<ADT Property | ||||||||||||||||||||||||
| 16. | (i)
(ii)
|
M1
M1
A1
B1 | |||||||||||||||||||||||||
| 04 | |||||||||||||||||||||||||||
| 17. | (a) Distance after 30mins Relative = 20km/hr
=
(b)
(c) |
M1
M1
M1
M1
A1
M1 A1
M1 M1 A1 |
For both distance
Relative distance
Relative speed
Relative time | ||||||||||||||||||||||||
| 10
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| 18. | (a) OP = a + (b – a) = a + b BQ = a – b
(b) (i)OC = h = OC = b + k = ka + (1 – k)b h = h = k h = 1 – k k = 1 – k 2k = 3 – 3k 5k = 3 k = h = (ii) OC = =
(iii)BC: CQ = : BC:CQ = 3:2 | B1
B1
M1
M1 M1
M1
A1
M1 A1
B1 | |||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||
| 19.
|
Total area = 2 + 22 + 50 + 24 + 2 = 100 50 – (2 + 22) = 26+26 = 2.5 x y, y = 10.4 Median = 34.5 + 10.4 = 44.9 | B1
M1 A1
B1
S1 B1 B1
B1
B1 A1 | fx
f.d
scale
For median line | ||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||
| 20. | a)
b) c) i) x = + 0.4 x = -1.7 + 0.1 ii) y = 3x2 + 4x – 2 0 = 3x2 + 7x + 2 y = -3x – 4 x = -2 or x = -0.4 + 0.1 | B2 B1
B1 B1
B1 L1
B1 | All ü at least 6 ü
For equation of line For ü line drawn | ||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||
| 21. | (b) A1(4,-4) B1(7,-3) C1(2,-1) (c) A11(4,4) B11(3,7) C11(1,2) (d) A111(4,-4) B111(3,-7) C111(1,-2)
|
B1 B1 B1 B1
B1
B1
B1
B1 B1 B1 |
For plotting For ∆ABC For ∆A1B1C1 For construction or otherwise For ∆A11B11C11
For construction or otherwise For ∆A111B111C111 | ||||||||||||||||||||||||
| 10
| |||||||||||||||||||||||||||
| 22. | a) b) 2.1 + 0.1cm 200km, 210km, 220km c) i) Bearing of M from N = 0100+ 10 ii) Bearing of N from M = 1900+ 10 | S1
B1
B1
B1
B2
M1 A1 B1 B1 | 1cm rep.100km
<300 at P
<450 at Q
üpositions of PQM and N
ülabelling 540km, 360km, 500km allü | ||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||
| 23. | (b)
(ii) Volume of frustum
(c) | M1 A1
M1
M1 A1
M1
M1
A1 | |||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||
| 24. | (i)
(ii) (c) Maximum speed,
|
M1
M1 A1
M1
A1 B1
M1 A1
M1 A1
M1 A1 |
For both
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| 10 |
NAME:……………………………………………….. INDEX NO………………………………
SCHOOL:……………………………………..……… STREAM:…………… ADM:………….
CANDIDATE’S SIGN …………………………….… DATE …………………………………..
121/1
MATHEMATICS
Paper 1
FORM 4
JULY 2023
Time: 2 ½ Hours
MOCKS 1 2023
Kenya Certificate of Secondary Education (K.C.S.E)
INSTRUCTIONS TO CANDIDATES
FOR EXAMINERS USE ONLY
Section I
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | Total |
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Section II Grand Total
| 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | Total |
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This paper consists of 16 printed pages. Candidates should check the question paper to ensure that all pages are printed as indicated and no questions are missing.
SECTION I (50 MARKS)
Answer all questions in this Section
| |
Buying Selling
1 US dollar 63.00 63.20
1 UK £ 125.30 125.95
Abwanja, a tourist arriving in Kenya from Britain had 9600 UK Sterling pounds (£). He converted the pounds to Kenya shillings at a commission of 5%. While in Kenya, he spent ¾ of this money. He changed the balance to US dollars after his stay. If he was not charged any commission for this last transaction, calculate to the nearest US dollars, the amount he received. (3 mks)
4x (8x – 1) = tan 45o
(3mks)
2y < x + 4; 4y ≥ –x – 4; x ≤ 2
y = mx + c (3mks)
(a) Tanq (2mks)
AB = 4cm and BC = 5cm. Draw a clearly labeled net of the prism. (3mks)
Giving reasons, Calculate the size of;
(i) Angle AOB. (2mks)
(ii) Angle ADT (2mks)
(i) The radius of the cone (3mks)
(ii) To one decimal place the vertical height of the cone (1mk)
SECTION II: 50 MARKS
Answer any FIVE questions in this section
(a) Express OP and BQ in terms of a and b (2 mks)
(b) If OC = hOP and BC = kBQ, Express OC in two different way and hence
(i) Deduce the value of h and k. (5 mks)
(ii) Express vector OC in terms of a and b only. (2 mks)
(iii) State the ratio in which C divides BQ (1 mk)
| Marks | Frequency | ||||
| 5 – 14 | 2 | ||||
| 15 – 34 | 22 | ||||
| 35 – 54 | 50 | ||||
| 55 – 84 | 24 | ||||
| 85 – 94 | 2 |
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y |
(a) On the grid provided plot the triangle ABC. (2 mks)
(b) A1B1C1 is the image of ABC under a translation . Plot A1B1C1 and state its coordinates. (2mks)
(c) Plot A11B11C11 the image of A1B1C1 under a rotation about the origin through a negative quarter turn. State its coordinates. (3 mks)
(d) A111B111C111 is the image of A11B11C11 under a reflection on the line y = 0. Plot A111B111C111 and state its coordinates. (3 mks)
Leave from P and Q respectively at the same time. Warplane M moves at 360km/h on a bearing of 0300. Warplane N moves at a speed of 240km/h on a bearing of 3150. The two warplanes landed at Police camps A and B respectively after 90 minutes. Using a scale of 1cm represent 100km
(b) Find the shortest distance between the police camps A and B. (2mks)
(c) Find the true bearing of;
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| |
| |
| |
| |
| |
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is given by h = -2t3 + 3/2 t2 + 3t
(a) Find the initial acceleration. (3mks)
(b) Calculate
(i) The time when the particle was momentarily at rest. (3mks)
(ii) Its displacement by the time it comes to rest momentarily. (2mks)
(c ) Calculate the maximum speed attained. (2mks)
MOCKS 1 2023
121/2 MATHEMATICS PAPER 2 MARKING SCHEME
| Qn | Workings | Marks | Comments | ||||||||||||||||||||||||
| 1. | b2 = 4ac 52 = c + 2 25 = c + 2 c = 23 | M1
A1 | Correct expression in C | ||||||||||||||||||||||||
| 02 | |||||||||||||||||||||||||||
| 2. | Truncated = 0.777 Rounded off = 0.778 A.E = 0.778 – 0.777 = 0.001 % E = x 100 = 0.12870012870012870012870012870013 | B1
M1 A1 | For both values correct
Expression for % Error Allow 0.1287 | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 3. | = P(-8.5, -20, -11) | M1 A1 B1 | Expression Correct matrix Co-ordinate form | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 4. | x
| M1
M1
M1
A1 | Correct substitution in sine rule
Surd form for sin 600
Correct attempt to rationalize CAO
| ||||||||||||||||||||||||
| 04 | |||||||||||||||||||||||||||
| 5. | (x3)6 -6 (x3)5 + 15 (x3)4 2 – 20 (x3)3 3 . . . – 20 (x3)3 3 – 20 x 8 = – 160 | M1
M1 A1 | Expansion up to the 4th term
Correct attempt to simplify Constant term stated | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 6. | Let log3x = y 2y2 – y – 3 = 0 (2y – 3)(y + 1) = 0 y = -1 or y = 1 ½ if log3x = -1, x = 3-1 = 1/3 if log3x = 1 ½ , x = 31.5 = 5.196 |
M1 A1
B1 |
Quadratic equation formed For both correct
For both correct | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 7. | P = cp – d = 13800 – 2280 = 11520 I = 11520 x 20 x 2/100= 4608 A = P + I = 11520 + 4608 = 16128 MI= 16128 ÷ 24 = 672 | M1
M1 A1 | Expression for simple interest
Expression for MI
| ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 8. | 2ax + x2 = 3v x2 + 2ax – 3v = 0 x2 + 2ax +a2 = 3v + a2 √(x + a)2 = √(3v +a2) x + a = ±√(3v +a2) x = -a ± √(3v +a2) | M1
M1
A1 | Formation of quadratic equation Completing the square Correct attempt to solve | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 9. |
A= 1(14.75 +26.75+77.75+68.75+98.75) = 253.75 square units |
B1 M1 A1 |
Correct values of mid-ordinates Expression for area
| ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 10. | OA = OP = 5 units AM = 5 – 2 = 3 units OM = √(52 – 32) = 4 units C(5,4) , r = 5 (x – 5)2 + (y – 4)2 = 52 x2 – 10x + 25 + y2 – 8y + 16 = 25 x2 + y2 -10x – 8y + 16 = 0 | M1
M1
M1
A1 | Expression for midpoint Radius, r
Expression for OM
Correct substitution
Correct expanded form | ||||||||||||||||||||||||
| 04 | |||||||||||||||||||||||||||
| 11. | = 1.736k % change = |
M1
M1 A1 |
Correct substitution
Expression for percentage change
| ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 12. | 3sin2x – sin x – 2 = 0 Let sin x = y 3y2 – y -2 = 0 (3y + 2)(y – 1 ) = 0 y = 1 or y = -2/3 sin-1(1) = 900 sin-1(-2/3) = 221.80, 317.80 x = 900, 221.80, 317.80 |
M1 M1 A1
B1 |
Quadratic equation formed Correct attempt to solve For both
All values correct | ||||||||||||||||||||||||
| 04 | |||||||||||||||||||||||||||
| 13. | i) k + 2k + 3k + 4k + 5k + 6k = 1 21k = 1 ii) P(5&6) 0r P(6&5) ( |
B1
M1
A1
|
Addition of probabilities (allow for any correct) Allow
| ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 14. | (a) Let VU = x 8(8 + x) = 122 8x = 144 – 64 =80 x = 10 b) VX = XU = XT = 6 + 8 = 14 SX = √(142 – 122) = 7.211 |
B1
M1
A1 |
x = 10
Expression for XT | ||||||||||||||||||||||||
| 03 | |||||||||||||||||||||||||||
| 15. |
Q1 = Q3 = Quartile deviation =
|
B1
M1
M1 A1 |
Cf
Q1 and Q3
Expression for quartile deviation Allow 16.47 | ||||||||||||||||||||||||
| 04 | |||||||||||||||||||||||||||
| 16. | = 4:5 |
M1
A1 |
Correct substitution
| ||||||||||||||||||||||||
| 02 | |||||||||||||||||||||||||||
| 17. | (a) (i) x-intercept x2(2x + 3) = 0 (ii) y-intercept When x =0, y = 0 (b) (i) Stationary points of curve = 0 6x(x + 1) = 0 x = 0 or x = -1 stationary points (0,0) and (-1,1) (ii)
maximum point (-1, 1), minimum point (0,0) iii)
| M1
A1
B1
M1
A1 B1 B1
B1
B1
B1
B1 | Factorized form
Both correct
Both correct
Derivative equated to zero
Attempt to solve For both
Checking points
For both
Points plotted (-1.5,0), (-1,1), (0,0)
Smooth curve | ||||||||||||||||||||||||
| 18. | a) (i)
ii)
r = 5.2cm ± 0.1
iii) h = 5cm ± 0.1
b) area of circle – area of triangle = 84.98 – 21.25 = 63.73cm2
| B1 B1 B1
B1 B1 B1
B1 B1
M1 A1 | Construction of 300 Construction of 1050 Complete triangle, well labeled
Line bisectors Complete Circle drawn radius
height dropped
follow through for r and h ± 0.1 | ||||||||||||||||||||||||
| 10 |
| 19 |
| B1 B1
S1 P1 C1 C1
B1 B1 B1 B1 |
Scale Plotting for both Smooth curve | ||||||||||||||||||||||||||||||||||||||||||
| 20. | a) i)
ii) b)i)
ii)
| B1
M1 A1
B1
M1 M1 M1
A1
M1 M1 A1 |
Tree diagram draw with probabilities indicates
ü1 probability Addition of the probability
ü probability
Addition
| ||||||||||||||||||||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||||||||||||||||||||
| 21. | (a)(i) Distance = =longitude difference =40+140=1800
=17,337.8Km b) =60Î2 =1200 Distance = = =13,346.7km (c) A(300N,400N)
B(300W,1400E) Difference in longitude=140+40 =1800 10=4min 180=? 180Î4=720minutes 8.00+12.00=20.00 =12.00hrs/8.00pm | 10
B1
M1 A1
M1 A1
M1
A1
M1
M1 A1 |
For 180o
| ||||||||||||||||||||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||||||||||||||||||||
| 22. |
c) i) Q3 = 19.25, Q1 = 17.15 ½ (Q3 – Q1) = ½ (19.25 – 17.15) = 1.05
ii) 13cm – – 15.2, 17cm – – 15.8 15.8 – 15.2 = 0.3
| B2 B1
S1 P1 C1
B1 M1 A1
B1 B1 | All values correct At least 4 values correct
Q3 & Q1 correct
Correct cf values
| ||||||||||||||||||||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||||||||||||||||||||
| 23. |
Error: 54.5-54=0.5
=
| B2
M1
A1
M1 M1 A1
B1
M1 A1 | |||||||||||||||||||||||||||||||||||||||||||
| 10 | |||||||||||||||||||||||||||||||||||||||||||||
| 24. | a)
b)
c) Objective function
| B1 B1 B1
B1
B1 B1
B1
B1
B1 B1
|
For each correct inequality
For each correct line drawn
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| 10 |
NAME:……………………………………………….. INDEX NO………………………………
SCHOOL:……………………………………..……… STREAM:…………… ADM:………….
CANDIDATE’S SIGN …………………………….… DATE …………………………………..
121/2
MATHEMATICS
Paper 2
FORM 4
Time: 2 ½ Hours
MOCKS 1 2023
Kenya Certificate of Secondary Education (K.C.S.E)
INSTRUCTIONS TO CANDIDATES
FOR EXAMINERS USE ONLY
Section I
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | Total |
|
|
Section II Grand Total
| 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | Total |
|
|
This paper consists of 18 printed pages. Candidates should check the question paper to ensure that all pages are printed as indicated and no questions are missing.
SECTION I
Answer all the questions in the spaces provided (50marks)
Find the co-ordinates of P (3mks)
n
2 2
2 Sin2x – 1 = Cos2x + Sin x, for 00 ≤ x ≤ 3600 (3mks)
|
T
|
U
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| Height | 10 – 19 | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 – 79 |
| No. of Seedlings | 9 | 16 | 19 | 26 | 20 | 10 | 4 |
Calculate the quartile deviation (4mks)
SECTION II
Answer ONLY five questions in this section (50marks)
(ii) the y-intercept of the curve (1mk)
(ii) For each point in b(i) above, determine if it is maximum or minimum (2mks)
(3mks)
| x | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 |
| (2cos x) -1 | 0 | -2 | -3 | -2 | -1 | 0 | 1 | ||||||
| Sin x | 0 | 1 | 0.50 | 0 | -1 | 0 |
Calculate
(a) (i) The distance from A to B along a parallel of latitude in kilometres. (3mks)
(ii) The shortest distance from A to B along a great circle in kilometre (4mks)
(Take =and radius of the earth =6370km)
(b) If the local time at B is 8.00am, calculate the local time at A (3mks)
| Length in cm | 9.5-12.5 | 12.5-15.5 | 15.5-18.5 | 18.5-21.5 | 21.5-24.5 |
| No. of Leaves | 3 | 16 | 36 | 31 | 14 |
| Cumulative frequency |
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