Mathematics Paper 1 Questions and Answers - Catholic Diocese of Kakamega Mock Exams 2023

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Mathematics Paper 2 Questions and Answers - Catholic Diocese of Kakamega Mock Exams 2023

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INSTRUCTIONS

This paper consists of two sections: Section I and Section II.
Answer ALL questions from section 1 and ONLY 5 questions from Section II All answers and workings must be written on the question paper in the spaces provided below each question.
Show all the steps in your calculation, giving your answer at each stage in the spaces below each question.
Non-programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.

SECTION I (50 MARKS)
Attempt ALL questions in this section

Work out the following, giving your answer as a mixed number in its simplest form; (3mks)
²/₅ ÷ ¹/₂ of ⁴/₉ − 1¹/₁₀
¹/₈ − ¹/₆ × ³/₈
In Shamshinjiri Primary School, each lesson in the lower primary and upper primary classes lasts 30 minutes and 40 minutes respectively. The first time the two bells ring at the same time, the pupils go out for tea break. If the lessons for the whole school start at 8.10a.m.
1. What time does tea break start? (2mks)
2. How many lower primary lessons and upper primary lessons take place before tea break? (2mks)
A certain amount of money was shared out among three men X, Y and Z in the ratio 5: 3: 1. If Y got shs. 700 more than Z, calculate how much X got. (3mks)
Given that sin(x + 60)° = cos(2x−30)°, find tan(2x)° to 4 significant figures. (3mks)
A retailer bought an IPAD for shs. 57500. The marked price at the retailer's shop was 12% more than the buying price. After allowing a discount, the retailer sold the IPAD for shs. 61,180. Calculate the percentage discount. (3mks)
List all the integral values of x which satisfy the inequalities below: (3mks)
2x + 21 > 15 − 2x ≥ x + 6
Simplify the expression (3mks)
4c² − d²
2c² − 7cd + 3d²
Solve for x in the equation 9^x+1 + 3^2x+1 = 36. (3mks)
A solid block in the shape of a cylinder has a height of 21cm and weighs 22kg. If it is made of a material of density 5g/cm³, find the radius of the cylinder. Give your answer to I decimal place. (take n = 3.142). (3mks)
The exterior angle of a regular polygon is equal to one-third of the interior angle. Calculate the number of sides of the polygon and give its name.
(3mks)
Use reciprocal tables to evaluate to 4 significant figures; (3mks)
28 + 4
0.3 0.0046
A clothes dealer sold three shirts and two trousers for Kshs. 840 and four shirts and five trousers for Kshs. 1680. Find the total cost of one shirt and one trouser. (3mks)
The diagram below shows a triangle ACD in which angle DAC = 31°, BC= 24cm and BD = xcm. Calculate the value of x to 2 decimal places.
(3mks)
A curve has the equation y = 2x² − 5x + 3. Find the co-ordinates of its turning point. (3mks)
The image of P(0, 2) under an enlargement with scale factor 3 is P' (4, 6). By calculation, find the centre of enlargement. (3mks)
Use a ruler and a pair of compasses only to construct triangle ABC with sides AB=5cm, BC = 5cm and AC = 6cm. Construct an escribed circle on side BC. (4mks)

SECTION II (50 MARKS)
Attempt ONLY FIVE questions in this section

Complete the following table.

Class	Mid-point(X)	Frequency (F)	FX	CF
50 - 54	52	5
55 - 59	57	3
	62		434
65 - 69			335
	72	5
	77		231
		4
		Σf =	Σfx =

State the modal frequency. (5mks)
Calculate the mean to 2 decimal places. (2mks)
Calculate the median. (2mks)

A lorry left town A at 8.30a.m for town B at an average speed of 70km/h. A Matatu left town B at 9.00a.m for town A at an average speed of 80km/h. The two towns are 180km apart.
1. Calculate the distance from town A when the two vehicles met. (5mks)
2. A car left town A at 9.00a.m at an average speed of 100km/h for town B. Calculate
  1. The time the car overtook the lorry. (2mks)
  2. The distance from town B when the car met the Matatu. (3mks)
A transport company has two types of vehicles for hire, Lorries and buses. The vehicles are hired per day. The cost of hiring 2 Lorries and 5 buses is kshs. 156,000. The cost of hiring 4 Lorries and 3 buses is kshs. 137,000.
1. Form two equations to represent the above information. (2mks)
2. Use matrix method to determine the cost of hiring one lorry and that of hiring a bus. (5mks)
3. The company makes a profit of 55% on a lorry hired and 60% on each bus hired. Find the profit made by the company on a day when 5 Lorries and 4 buses were hired. (3mks)
A village Q is 7 km from village P on a bearing of 045°. Village R is 5 km from Village Q on a bearing of 120° and village S is 4 km from village R on a bearing of 270°.
1. Taking a scale of 1 cm to represent 1 km, locate the four villages. (3mks)
2. Use the scale drawing to find the:
  1. Distance and bearing of the village R from village P. (2 mks)
  2. Distance and bearing of village P from village S. (2 mks)
3. Calculate the area enclosed by the four villages PQRS (3 mks)
The vertices of triangle ABC are A(2, 0), B(5, 3) and C(5, 1).
1. Find the coordinates of triangle A^l B^lC^l the image of triangle ABC after transformation by the matrix T (2mks)
2. Find the coordinates of triangle A^llB^llC^ll, the image of triangle A^lB^lC^l after a transformation by the matrix S = (2mks)
3. On the grid provided, draw triangle ABC and its images. (3mks)
4. Determine the single matrix that can map triangle A^l^l B^l^lC^l^lon to triangle ABC. (3mks)
A straight line passes through the points (8,-2) and (4,-4).
1. Write its equation in the form ax + by + c = 0 where a, b and c are integers. (3mks)
2. If the line in (a) above cuts the x - axis at point P, determine the coordinates of P. (2mks)
3. Another line which is perpendicular to the line in (a) above passes through point P and cuts the y-axis at the point Q. Determine the coordinates of point Q. (3mks)
4. Find the length of QP to 3 decimal places. (2mks)
The figure below shows triangle OAB in which M divides OA in the ratio 2: 3 and N divides OB in the ratio 4: 1. AN and BM intersect at X.
1. Given that OA = a, OB = b, express in terms of a and b ;
  1. AN (1mk)
  2. BM (1mk)
2. If AX = SAN and BX = tBM, where s and t are constants.
  1. Write two expressions for OX in terms of a, b, s and t. (2mks)
  2. Find the values of s and t (5mks)
  3. Find the ratio of AX: XN (1mk)
The displacement S metres of a particle moving after t seconds is given by
S = 2t³ − 5t² + 4t + 2
Determine
1. The velocity of the particle when t = 3seconds. (3mks)
2. The value of t when the particle is momentarily at rest. (3mks)
3. The displacement when the particle is momentarily at rest. (2mks)
4. The acceleration of the particle when t = 3seconds. (2mks)

MARKING SCHEME

Num
²/₅ ÷ ¹/₂ of ⁴/₉ − ¹¹/₁₀
²/₅ ÷ ²/₉ − ¹¹/₁₀
²/₅ × ⁹/₂ − ¹¹/₁₀
⁹/₅ − ¹¹/₁₀
= ⁷/₁₀
Den
= ¹/₈ − ¹/₆ × ³/₈
= ¹/₈ − ¹/₁₆
= ¹/₁₆
= ⁷/₁₀ ÷ ¹/₁₆
= ⁷/₁₀ × ¹⁶/₁
= 11¹/₅
1. L.C.M of 30 &40
  
  ⇒ LCM = 2⁴ × 3 × 5
  = 120 mins
  = 2hrs
  Break starts at
  8.10
  + 2.00
  10.10 a.m
2. 1 Lower primary Lesson = 30mins
  ? = 120 mins
  ⇒ 120 × 1
  30
  = 4 lessons
  1 Upper primary lesson = 40 mins
  ? = 120 mins
  ⇒ 120 × 1
  40
  = 3 lessons
x:y:z = 5:3: 1
y−z = 3 − 1 = 700
2 = 700
∴ 5 = ?
5 × 700
2
∴ x got Kshs. 1750
(x+60) + (2x-30)=90°
3x+30 = 90°
3x = 60°
x = 20°
Tan (2x20) = Tan 40°
= 0.839099
≅ 0.8391 (4s.f).
Mr P = ¹¹²/₁₀₀ × 57,500
= 64,400/=
Discount = 64,400 − 61,180 M1
= 3,220/=
% discount = disc x100%
m.p
= 3220 x 100%
64,400
= 5%. A1
2x + 21 > 15 − 2x
4x > −6
x > − 6/4
x > − 1.5
15 − 2x ≥ x + 6
−3x ≥ −9
x ≥ 3

Integral values;
−1, 0, 1, 2 and 3
Num: 4c² − d²4(2c − d)(2c + d)
Denom: 2c² − 7cd +3d²
P = 6, S = −7 −1, −6
2c² − cd − 6cd + 3d²
c(2c − d) − 3d(2c − d)
(c − 3d)(2c − d)
⇒ (2c − d)(2c + d)
(c − 3d)(2c − d)
= 2c + d
c − 3d
3^2(x+1) + 3^2x+1 = 36
3^2x+2 + 3^2x+1 = 36
3^2x = ³⁶/₁₂
3^2x = 3¹
⇒ 2x = 1
x = ½
From f = m/v
⇒ v = m/f
= (22 × 1000)g
5g/cm³
= 4,400cm³
But V = πr²h
⇒ 3.142 x r² x 21 = 4400
r² = 4400
3.142 × 21
= 66.68
r = √ 66.68° = 8.166 = 8.2
x + ¹/₃x = 180°
⁴/₃x = 180
x = 180 × ³/₄
= 135°
Each ext ∠ = ¹/₃ x 135
= 45°
No.of sides = 360
45
= 8 sides
Name ; Octagon
28(Rec.0.3)+4(Rec: 0.0046).
28(Rec. 3.0x10⁻¹) + 4(Rec. 4.6x10⁻³)
(28 x 0.3333 x 10) + (4 x 0.2174 × 10³)
= 93·324 + 869.6
= 962.92
≅ 962.9 (4sf) A1
3s + 2t = 840 x 4
4s + 5t = 1680 × 3
⇒ 12s + 8t = 3360
12s + 15t = 5040 −
−7t = − 1680
t = 240/=
3s + 2(240) = 840
3s + 480 = 840
3s = 360
s = 120
Total cost of 1 trouser & 1 shirt
= 120 + 240
= 360/=
Tan 31° = ²⁴/_x
x = 24 = 24
tan 31° 0.6009
= 39.94008986
≅ 39.94 (2dp)
dy/dx = 4x − 5
At turning point, dy/dx = 0
⇒ 4x − 5 = 0
4x = 5
x =⁵/
= 1.25
When x = ⁵/₄, y = 2(⁵/₄)² − 5(⁵/₄) + 3
= 3.125 − 6.25 + 3
= − 0.125
∴ Turning pt is (1.25, − 0.125)
OR (1¹/₄, − ¹/₈)

Class	Mid-point(X)	Frequency (F)	FX	CF
50 - 54	52	5	260	5
55 - 59	57	3	171	8
60 - 64	62	7	434	15
65 - 69	67	5	335	20
70 - 74	72	5	360	25
75 - 79	77	3	231	28
80 - 84	82	4	328	32
		Σf = 32	Σfx = 2119

(5mks)

modal frequency = 7
x̄ = Σfx = 2119
Σf 32
= 66.21875
= 66.22 (2dp)
Median position = ^N/₂ = ³²/₂ = 16^th
∴ median = 64.5 + (16 − 15) × 5
5
= 64.5 + 1
= 65.5

1. Distance travelled by Lorry in ¹/₂h.
  = ST = (70 × ¹/₂)km
  = 35km
  Relative distance = 180 − 35
  = 145km
  Relative speed = (70 + 80)km/h
  = 150km/h
  T.T to meet = R.d = 145km
  R.s 15km/h
  = ²⁹/₃₀ hrs
  Distance from A to meetup point;
  = 35 + (70 × ²⁹/₃₀)
  = 35 + 67²/₃
  = 102²/₃km or 102.67km
2. 1. R.d = 35km
    R.s =(100 − 70)km/h
    = 30km/hr
    T.T to overtake lorry
    = ^R.d/_R.s = ^35km/_30km/hr
    = 1¹/₆hrs = 1hr 10 min
    Car overtook lorry at;
    9.00
    +1.10
    10.10 a.m
  2. R.d = 180kmk
    R.s = 100 + 80
    = 180km/h
    T.T to meet car = ^R.d/_R.s
    = 180km
    180km/hr
    = 1hr
    ∴ Distance from B
    = (80 × 1)km
    = 80km
1. 2L + 56 = 156,000 .....(i)
  4L + 36 = 137,000 .....(ii)
3. 153% = 15,500/-
  100% = 100X15,500
  155
  = 10,000/=
  160% = 25,000/=
  100% = 100 × 25000
  160
  = 15,625/=
  Profit on each horry = 15,500 − 10,000
  = 5,500/=
  Profit on each bus = 25,000 − 15,625
  = 9,375/=
  Total profit = (5500x5) + (9375x4)
  = 27,500+37,500
  = 65,000/=
2. 1. Distance and bearing of the village R from village P. (2 mks)
    9.6 km:
    75°
  2. Distance and bearing of village P from village S. (2 mks)
    5.9 Km.
    245°
3. Calculate the area enclosed by the four villages PORS.
  = ¹/₂ x 7 x 5 Sin 105° − ¹/₂ x 4 x 5.8 Sin 155°
  = 16.90370196 − 5.085105303
  = 11.81859666
  = 11·82 cm² (4 s.f)
1. Gradient = −4+2 = ⁻²/₄
  8−4
  = − ½
  2y + 4 = −x + 8
  2y + x − 4 = 0
  x + 2y − 4 = 0
2. P(x, 0), at x-Intercept, y=0
  x + 2(0) − 4 = 0
  x − 4 = 0
  x=4
  ∴ P(4,0)
3. Gradient = 2, (4,0)
  y−0 = 2
  x−4
  y = 2x − 8
  At, y-intercept, x=0
  Q(0,y)
  and y = 2(0) − 8
  y = −8
  ∴ Q(0, −8)
1. V = ^ds/_dt = 6t² − 10t + 4
  At t = 3, V = 6(3²) − 10(3) + 4
  = 54 − 30 + 4
  = 28m/s
2. At rest, V = 0
  ⇒ 6t² − 10t + 4 = 0
  {P = 24 S =−10 } −4, −6
  6t² − 6t − 4t + 4 = 0
  6t (t−1) −4(t−1) = 0
  (6t − 4) (t −1) = 0
  6t − 4 = 0
  6t = 4
  t = ⁴/₆
  = ²/₃s
  OR
  t − 1 = 0
  t = 1s
3. At rest , t = ²/₃s or 1s
  When t = ²/₃,
  S = 2(²/₃)³ − 5(²/₃)² + 4(²/₃) + 2
  =¹⁶/₂₇ − ²⁰/₉ + ⁸/₃ + 2
  = 3¹/₂₇m or 3.037037m
  When t = 1,
  S = 2(1)³ − 5(1)² + 4(1) + 2
  = 2 − 5 + 4 + 2
  = 3m
4. a = dv/dt = 12t − 10
  At t = 3, a = 12(3) − 10
  = 36 − 10
  = 26m/s²