Mathematics Paper 2 Questions and Answers - Mokasa II Joint Mock Exams 2023

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Instructions to candidates

This paper consists of two sections: Section I and Section II.
Answer all questions in section I and only five questions from section II.
Show all the steps in your calculations, giving the 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 KNEC mathematical tables may be used, except where stated otherwise.
Candidates should answer the questions in English.

SECTION I (50 Marks)

Solve the following equations (3mks)
2x + y = 5
− x² + y + 2x = 1
Find the percentage error in the volume of a cone whose radius is 7.0 cm and has an exact vertical height of 18cm. (3 mks)
Solve the equation, 2 cos 2x = 3 sin x for 0 ≤ x ≤ 360 (3mks)
Simplify without using tables or calculator (3mks)
Sin 480° − Cos 765°
tan 225° − cos (−330°)
Without using mathematical tables or a calculator, solve the equation . (3 marks)
2log₁₀x - 3log₁₀2 = 1 − log₁₀5
A businessman invested ksh 1,000,000 in a fixed deposit account that pays 12% per annum compound interest, every 2 month. Calculate the number of years his amount will be Ksh 1,126,162.42 (3marks)
Using the properties of chords and tangents, construct a tangent touching point P on the circumference of the circle below showing clearly the centre of the circle. (3marks)
In a soccer competition, the number of goals (G) scored in penalty kicks is partly constant and partly varies as the skill (S) of the player. Given that G = 8 when S = 2 and G = 12 when S = 4, find the value of G when S = 6. (3 marks)
The position vector of A is OA = i − 3j − 3k and that of B is OB= 3i − j + 2k. A point N divides AB externally in the ratio 3:1. Find the magnitude of ON. (3marks)
1. Expand (3x − y)⁵ up to the fourth term. (2marks)
2. Use the expansion to evaluate (2marks)
The figure below shows a circle Centre O, radius 7cm. Angle AOB= . If a point is selected at random inside the circle, find the probability that it lies in the shaded region. (3marks)
The mass of a mixture P of beans and maize is 72kgs. The ratio of beans to maize is 3:5. A second mixture R of maize and beans of mass 98kg is mixed with P. The final ratio of beans to maize is 8:9 respectively. Find the ratio of beans to maize in R. (3marks)
A curve has a turning point at the point (1,1). Given that the gradient function of the curve is 6x² + ax − 12, find the value of a and the equation of the curve. (3 marks)
Find the area bounded by the curve y = x²− x − 2 , x =− 3, x = 2 and the x-axis (4marks)
Study the figure below hence describe transformation mapping the wave y = sin x to y = ¹/₂ sin 2x (2mks)
In the figure below, ST is a tangent to the circle at S. AXBT and CXS are straight lines.

Find;
1. The length of XB. (2 marks)
2. The size of angle STX. (2 marks)

SECTION B – 50MARKS

X and Yare two points on the earth’s surface and on latitude 30°N. The two points are on the longitude
40 °W and 140°E respectively (Take π = ²²/₇ and radius of the earth R = 6370km) Calculate:
1. The distance from X to Y along a parallel of latitude in kilometres. (3marks)
2. The shortest distance from X to Y along a great circle in kilometres (4mks)
3. If the local time at Y is 8.00am on Wednesday, What is the day and the local time at X in 24 hours system. (3mks)
The table below shows the masses measured to the nearest Kg of 200 people.

Mass kg 40-49 50-59 60-69 70-79 80-89 90-99 100-109

No. of people 9 27 70 50 26 12 6
1. Draw a cumulative frequency curve for the data above. (4 marks
2. Use your graph to estimate
  1. The median mass. (1mark)
  2. The number of people whose mass lies between 70.5 kg and 75.5 kg (1 mark)
3. From your graph find
  1. The lower quartile (1 mark)
  2. The upper quartile (1 mark)
  3. The interquartile range (2 marks)
The figure above shows the model of a roof with rectangular base PQRS. PQ= 40 cm and PS= 24 cm. The ridge XY = 30 cm and is centrally placed. The faces PSX and QRY are equilateral triangles. M is the midpoint of QR.
Calculate correct to 2 decimal places the:
1. Perpendicular distance of XY from the plane PQRS (3 marks)
2. Angle between SX and PQRS (4 marks)
3. Angle between planes RSXY and QPXY (4 marks)
The principal of mazuri high school intends to spend not more than ksh 18,000 to transport minimum of 70 students to the county games using two matatus A and B. Matatu A has a passenger capacity of 10 and B a capacity of 30.The cost per trip for matatu A is ksh 2000 and that of matatu B is ksh 3000. Given that A makes less than 5 trips.Taking trips made by matatu A be x while trips made by matatu B be y
1. Write down all the inequalities to represent the above information. (4marks)
2. Use the grid below to represent the above information. (4mks)
3. Find the number of trips that each matatu should make to minimize the amount of money the school will spend. (2mks)
1. A quantity p varies directly as the square of q and inversely as the square root of r. If q increases by 20% and r decreases by 36%, find the percentage change in p. (3 marks)
2. The velocity of water flowing through a pipe is inversely proportional to the square of the radius of the pipe. If the velocity of the water is 30cm/s when the radius of the pipe is 2cm. Find the velocity of water when the radius of the pipe is 4cm. (3 marks)
3. Three quantities X and Y and Z are such that X varies partly as Y and partly as the inverse of the square of Z. When X= 6, Y= 3 and Z = 2.When X = 8,Y=5 and Z = 1.Find the value of X when Y = 10 and Z = 8 (4mks)
1. Complete the table below for the function y = x³ − 3x² − 5x + 7 . (2 marks)
  
  x −3 −2 −1 0 1 2 3 4 5
  
  y
2. On the grid provided, draw the graph of y = x³ − 3x² − 5x + 7 for −3 ≤ x ≤ 5 . (3 marks)
3. Use your graph to solve the equation x³ − 3x² − 5x + 7 = 0 (2 mks)
4. By drawing a suitable line, use the graph in (b) to solve the equation x³ − 3x² − 10x + 17 = 0
  (3 marks)
A triangle with vertices A(1,1),B(3,1) and C(2,3) undergoes a transformation represented by the matrix to A’B’C’.
1. Draw the triangle ABC on the grid provided. (1mk)
2. Calculate the coordinates of A’B’C’ and draw it on the grid hence describe the transformation that maps ABC to A’B’C’. (3mks)
3. The triangle A’’B’’C’’ A’’(2,0) ,B’’(6,−2) and C’’(4,1) is the image of triangle ABC.
  1. Draw triangle A’’B’’C’’ on the grid provided. (1mark)
  2. Find the matrix of transformation that maps A’’B’’C’’ to ABC. (3marks)
4. Triangle DEF area 60cm² undergoes transformation represented by the transformation matrix . Find area of the image of triangle DEF. (2marks)
1. Using a ruler and a pair of compass only, construct a triangle ABC in which AB=6cm, BC = 7cm and angle ABC = 75° (3mks)
  Measure:
  1. Length of AC (1mark)
  2. Angle ACB (1mark)
2. Locus of P is such that BP = PC. Construct P (2marks)
3. Construct the locus of Q such that Q is on one side of BC, opposite A and angle BQC = 60° (3mks)

MARKING SCHEME

Y = 5 − 2X
− x² + 5 − 2x + 2x = 1
−x² = −4
x = ± 2
x = −2
y = 5 − 2x − 2
= 9
x = 2
y = 5 − 2 × 2
= 1
x = 2, y = 1, x = −2, y = 9
Actual = 7.0 × 7.0 × 18
= 882
Max = 7.05 × 7.05 × 18
= 894.645
Min = 6.95 × 6.95 × 18
=869.445
Absolute Error = 894.645 − 869.445
2
= 12.6
%Error = 12.6 × 100
882
= 1.429% or 1³/₇%
2(1 − Sin²x) = 3 sin x
2y² + 3y − 2 = 0
2y(y+2) − 1(y+2) = 0
(y+2) (2y−1) = 0
y =¹/₂ or −2
⇒ Sin x = ¹/₂
x = 30,150
log10 (5x²) = 1
8
5x² = 10
8
80 = 5x²
16 = x²
x = ±4
1, 126, 162.42 = 1,000,000 (1 + ²/₁₀₀)ⁿ
1.12616242 = (1.02)ⁿ
n = log 1.12616242
log 1.02
n = 6
= 1 year
8 = k + 2m .....(i)
12 = k + 4m ....(ii)
−4 = −2m
m = 2
k = 8 − 2m
= 8 − 2 × 2
k = 4
G = 4 + 25
S = 6
G = 4 + 2 × 6
= 4 + 12
G = 16
1. (3x)⁵ − (3x)⁴(y) + (3x)³(y)² − (3x)²(y)³
  243x⁵ − 405x⁴y + 270x³y² − 90x²y³
2. (3 − 0.02)⁵ = 3x − y)⁵
  3 = 3x
  x = 1
  − 0.02 = − y
  y = 0.02
  243 × 1 − 405 × 0.02 + 270 × (0.02)² + 90(0.02)³
  243 − 8.1 + 0.108 − 0.00072
  = 235.000728 At least 4s.f
AC = ²²/₇ × 7² = 154
¹/₄ × ²²/₇ × 7² = 38.5
¹/₂ × 7 × 7 = 24.5
14.0
p(s) = ¹⁴/₁₅₄ = ¹/₁₁
In P
maize = ⁵/₉ × 72 = 45kg
beans = ³/₈ × 72 = 27kg
72 + 98 = 170 − P & R
beans = ⁸/₁₇ × 170 = 80kg
maize = ⁹/₁₇ × 170 = 90kg
80 − 27 = 53
90 − 45 = 45
53 : 45
6x² + ax − 12 = 0
x = 1
6 + a − 12 = 0
a = 6
y = 2x³ + 3x² − 12x + C
1 = 2 + 3 − 12 + C
C = 8
y = 2x³ + 3x² − 12x + 8
A stretch parallel to y-axis x-axis invariant, stretch factor ¹/₂, followed by a stretch parallel to x-axis, y-axis invariant scale factorv(stretch factor)¹/₂
1. XB × 6 = 5 × 12
  XB = 5 × 12
  6
  = 10cm
2. 24 × 8 = 57²
  ST = √192
  12² = 8² + (√192)² − 2 × 8 × √192 Cos T
  144 = 64 + 192 − 2 × 8 × √192 Cos T
  −112 = −16√192 Cos T
  Cos T = −112
  −16√192
  T = 59.66°
1. 40 + 140 = 180
  ¹⁸⁰/₃₆₀ × ²²/₇ × 2 × 6370 Cos 30
  = 17,337.82858km
  At least 1d.p - 17337.8kkm
2. 180 − (30×2) = 120
  ¹²⁰/₃₆₀ × 2 × ²²/₇ × 6370
  13346²/₃ or 13346.66667
  At least 1d.p or 13346²/₃
3. 1° = 4min
  180
  180 × 4 = 720
  1 60
  = 12 hours
  8 + 12 = 2000hrs tuesday

Mass kg	40-49	50-59	60-69	70-79	80-89	90-99	100-109
No. of people	9	27	70	50	26	12	6
	9	36	106	156	182	`194	200

1. 68.5
2. 70.5kg -112
  70.5kg - 140
  140 − 112 = 28
1. 62.5kg
2. 77.5kg
3. 77.5 - 62.5 = 15

1. √(24² − 12²) = 20.7846
  H = √ 20.7849² − 5²
  = 20.17cm
2. Sin Q = √407
  24
  G = Sin⁻¹(√407)
  24
  = 57.20°
  
  Cos Q = ¹³/₂₄
  Q = COs⁻¹(¹³/₂₄)
  = 57.20°
3. tan Q = 12
  (√407)
  Q = tan⁻¹ (12)
  (√407)
  = 30.74494004
  <btn RSXY & QPXY = 2(30.74494004)
  = 61.49°
1. x + 3y ≥ 7......(i)
  2x + 3y ≤ 18 ......(ii)
  x < 5 ...........(iii)
  x < 0 .....(iv)
  y < 0 ''
3. 2000x + 3000y = k
  2000x + 3000y = 6000
  A - 1 trip
  B - 2 trips
1. P = kq²
  √r
  P₁ = (1.2q)²k = 1.44q²k
  √0.648 0.85r
  P₁ = 1.8kq²
  √r
  %Δ = (1.8 − 1) 100
  1
  = 80%
2. V= k
  r²
  30 = ^k/₄
  k = 120
  V = 120
  r²
  V = 120
  4²
  = 7.5cm/s
3. x = kY + m
  Z²
  6 = 3k + ^m/₄ ......(i)
  8 = 5k + m ........(ii)
  24 = 12k + m
  8 = 5k + m
  16 = 7k
  k = ¹⁶/₇
  m = 24 − 12k
  = 24 − 12 × ¹⁶/₇ = −3³/₇
  = −²⁴/₇
  x = ¹⁶/₇ × 10 − 24
  7×64
  = 22⁴⁵/₅₆ or 22.80
1. x −3 −2 −1 0 1 2 3 4 5
  
  y −32 −3 8 7 0 −7 −8 3 32
3. x = −1.8 or 1 or 4 ± 0.1
4. x = − 2.8 or 1.4 or 4.2 ± 0.1
  drawing line y = 5x − 10
4. Def = A.S.F
  l3.8l = 5
  5 = A.I
  90
  A.I = 300cm²
1. 1. 8cm ± 0.1
  2. 47°