Mathematics Paper 1 Questions and Answers - Momaliche Joint Mock Exams 2021/2022

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Mathematics Paper 2 Questions and Answers - Momaliche Joint Mock Exams 2021/2022

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INSTRUCTIONS TO CANDIDATES:

Write your name,admission and class in the spaces provided at the top of this page.
Sign and Write the date of examination in the spaces provided above.
This paper consists of TWO Sections; Section I and Section II.
Answer ALL the questions in Section I and only five questions from Section II.
Show all the steps in your calculation, giving your answer 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 maybe used except where stated otherwise.
This paper consist of 15 printed pages.
Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing
Candidates should answer the questions in English.

SECTION I (50 marks)

Use tables of reciprocal only to evaluate, 1 hence evaluate ; (4 marks)
0.325
³√0.000125
0.325
Solve the equation 3x² + 4x = 2 giving the roots correct to two decimal places. (3 marks)
The straight line through the points D (6, 3) and E (3, -2) meets the y-axis at the point F. Determine the coordinates of F. (3 marks)
Using the grid provided below, draw and shade the unwanted regions to show the region satisfied by R given the following inequalities; y + x < 5, y - x ≤ 1 and x + 5y > 5 (3 marks)
$1$
Given that a = -2, b = -1 and c= 3, evaluate 2(a + c)² - (a - b)(b - c) - 2c (3 marks)
3(a + b) -2(b - c)
Simplify: (3marks)
x–2 - 2x–4
x+2 x²-4
Two boys and a girl shared some money .The elder boy got ⁴/₉ of it, the younger boy got ²/₉ of the remeinder and the girl got the rest. Find the percentage share of the younger boy to the girl’s share. (3 marks)
Annette has some money in two denominations only. Fifty shilling notes and twenty shilling coins. She has three times as many fifty shilling notes as twenty shilling coins. If altogether she has sh. 3400, find the number of fifty shilling notes and 20 shilling coins.(3 marks)
The figure below shows a circle centre O and AOB is a sector of the circle and angle AOB = 72º as shown. Given that the area of a sector AOB is 5πcm², find the radius of the circle and hence calculate the area of the shaded part. (4mks)
$2$
A particle accelerates uniformly from rest and attains a maximum velocity of 30m/s after 16 seconds. It travels at this constant velocity for the next 20 seconds before decelerating to rest after another 8 seconds. Calculate the total distance covered by the car. (3 marks)
Find the value of x in the equation 5^x^/^₄= ¹/₂₅ (2 marks)
Given that x = 2.4, evaluate without use of tables and calculators, sin x - cos x in the form of ^a/_b where a and b are integers. (3
marks)
The difference between the interior and exterior angles at each vertex of a regular polygon is 162º. Find the number of sides of the polygon. (3 marks)
The surface area of two similar bottles is 12cm² and 108cm² respectively. If the larger one has a volume of 810cm³. Find the volume of the smaller one.(3mks)
A cylindrical iron pipe is 2.lm long and 12cm in external diameter, the metal is 1cm thick and its density is 7.8g,/cm³. Taking pie as 3 ½ find its mass. (4 Marks)
Draw the net of the solid below given that is a right pyramid and that AB = 4cm = BC = CD = AD and BE= 6cm (3mks)
$3$
SECTION II (50marks)
Answer only five questions in this section in the spaces provided.
Ruhu, Toru, and Lwamawa contributed a total of Kshs. 8041950.00 for their joint campaigns ahead of 2022 general elections. The ratios of their contributions were Ruhu to Toru 5:4 and Lwamawa to Toru 2:3.
1. How much did each contribute? (4 Marks)
2. Ruhu further contributed Kshs. 875,000.00 towards the campaigns kitty. in response, Toru and Lwamawa increased their contributions in the ratios 10:9 and 11:6 respectively. How much did Toru and Lwamawa further contribute altogether? (3 marks)
3. The three agreed that if they win elections they would share the 15 cabinet positions amongst them in the ratio of their contributions. How many cabinets positions did Lwamawa get? (3 Marks)
In the figure below, 0 is the centre of the circle. PQ and PR are tangents to the circle at P and R respectively Angle PQS = 40º and angle PRS 30° RTU is a straight line. ( 3mks)

Find with reasons the angles
1. QRS (2marks)
2. RTQ (2 marks)
3. RPQ (2 marks
4. Reflex angle QOR (2 marks)
5. TRO given that TR =TQ (2 marks)
Complete the table below for the function y=x² + 6x² + 8x for -5≤ x ≤1 (2 marks)

x -5 -4 -3 -2 -1 0 1

x² -125 -64 -1 0 1

6x² 54 6 0

8x -40 -24 -16 0 8

y 0 3 0 15
1. Draw the graph of the function (3 marks)
  (Use a scale of 1cm to represent 1 unit on the x axis. 1 cm to represent 5 units on the y- axis)
2. Hence use your graph to estimate the roots of the equation
  1. X³ + 6X² + 8X = 0 (1 mark)
  2. x³+ 5x² + 4x = -x² - 3x - 1 (4 marks)
Three islands P, Q, R and S are on at ocean such that island Q is 400kmon a bearing of 030º from island P. Island R is 520km and on a bearing of 120º from island Q. A port S is sighted 750km due south of island Q.
1. Taking a scale of 1cm to represent 100km, give a scale drawing showing the relative positions of P, Q, R and S. (4 marks)
  Use the scale drawing to
2. Find the bearing of:
  1. Island R from island P (1 mark)
  2. Port S from island R (1 mark)
3. Find the distance between island P and R (2 marks)
4. Find distance between S and R (2marks)
In the figure below, E is the midpoint of AB, OD : DB = 2 : 3 and F is the point of intersection of OE and AD

Given that OA = a and OB = b,
1. Express in terms of a and b
  1. AD (1 mark)
  2. OE (2 marks)
2. Given further that AF = sAD and OF = tOE, find the values of sand t (5 marks)
3. Show that E, F and O are collinear (2 marks)
A plastic water tank has a shape as shown below, with a frustrum of a cone on top, a cylindrical body and a hemispherical bottom.
1. Calculate
  1. The volume of the tank in m³.(5mks)
2. A filler pipe takes 3 hours to fill a third of the tank. If the tank is already ¼ full, at what time will the filler pipe fill the tank if the pipe is opened at 9.00a.m. (3mks)
3. A particle falls in the tank. If its chances of being in any part of the tank are equally likely, find the probability of it being in the hemispherical part (2mks)
Construct a parallelogram ABCD in which AB = 8.5cm, AD = 6cm and angle BAD = 75º. (Use a ruler and pair of compasses only in this question)
1. Measure the length of AC. (4mks)
2. On the same diagram, construct a perpendicular from D to line AB at M. Measure BM. Hence calculate the area of the parallelogram ABCD. (4mks)
3. Ex-scribe a circle to triangle BDA tangent to BD (2marks)
The figure below shows two circles intersecting at C and D. The centres are A and B with radii 8cm and 6cm respectively. AB = 10cm.
Determine to 2 decimal places
1. Size of angle DAC (2mks)
2. Size of angle DBC (2mks)
3. Area of sector ACMD (2mks)
4. Area of the shaded region (4mks)

MARKING SCHEME

SECTION I (50 marks)

Use tables of reciprocal only to evaluate, 1 hence evaluate ; (4 marks)
0.325
³√0.000125
0.325
1 = 1 x 1 = 0.307.7 x 10 = 3.077
0.325 3.25 10^-1
3√125 x 10^-6 = 5 x 10^-2 = 0.05
0.05 x 3.077 = 0.15385
Solve the equation 3x² + 4x = 2 giving the roots correct to two decimal places. (3 marks)
3x² + 4x - 2 = 0
x = -4 ± √4²- (4 x 3 x -2)
2 x 6
x = -4 ± √40
6
either x =-4 + 6.325 = 0.39
6
The straight line through the points D (6, 3) and E (3, -2) meets the y-axis at the point F. Determine the coordinates of F. (3 marks)
G = -2 -3 = ⁵/₃
3 - 6
(6,3) (x,y)
y - 3 = ⁵/₃
x - 6
y = ⁵/₃x - 7
y axis x = 0
y = -7
Using the grid provided below, draw and shade the unwanted regions to show the region satisfied by R given the following inequalities; y + x < 5, y - x ≤ 1 and x + 5y > 5 (3 marks)
$1$
Given that a = -2, b = -1 and c= 3, evaluate 2(a + c)² - (a - b)(b - c) - 2c (3 marks)
3(a + b) -2(b - c)
2[-2 + 3]² - [-2-(-1)][-1-3]-2(3)
3[(-2)+(-1)]-2[-1-3]
2(1)²-(-1)(-4)-6
3(-3)-2(-4)
-8 = 8
-1
Simplify: (3marks)
x–2 - 2x–4
x+2 x²-4
x - 2 - 2x - 4 = (x - 2)(x - 2) -1(2x - 4)
x + 2 x² - 4 x² - 4
= x² - 6x + 8
x² - 4
= x²- 4x - 2x + 8
(x - 2)(x + 2)
= x(x - 4) -2(x - 4)
(x - 2) (x + 2)
= x - 4
x + 2
Two boys and a girl shared some money .The elder boy got ⁴/₉ of it, the younger boy got ²/₉ of the remeinder and the girl got the rest. Find the percentage share of the younger boy to the girl’s share. (3 marks)
Let the amount of money shared be x
Elder boy = ⁴/₉ x
Younger boy = ²/₅ x ⁵/₉ = ²/₉x
Girl = x (⁴/₉x + ²/₉x) = ¹/₃x
= 66²/₃%
Annette has some money in two denominations only. Fifty shilling notes and twenty shilling coins. She has three times as many fifty shilling notes as twenty shilling coins. If altogether she has sh. 3400, find the number of fifty shilling notes and 20 shilling coins.(3 marks)
Let the sh20 coin be x
sh 50 note will be 3x
20x + 3x(50) = 3400
170x = 3400
x = 3400 = 20
170
sh20 coins = 20
The figure below shows a circle centre O and AOB is a sector of the circle and angle AOB = 72º as shown. Given that the area of a sector AOB is 5πcm², find the radius of the circle and hence calculate the area of the shaded part. (4mks)
$2$
A = θ x r²
360
5x = 72 x r²
360
r² =5x x 360
72x
r² = 25
r = 5cm
Area of shaded region = 5x - ½ x 5^-2sin7
= 15.71 - 11.89
= 3.82cm²
A particle accelerates uniformly from rest and attains a maximum velocity of 30m/s after 16 seconds. It travels at this constant velocity for the next 20 seconds before decelerating to rest after another 8 seconds. Calculate the total distance covered by the car. (3 marks)
$2$
Distance = Area under graph
A = ½ x 30(20 + 44)
= 15 x 64 = 960M
Find the value of x in the equation 5^x^/^₄= ¹/₂₅ (2 marks)
5^x^/^₄= 5^-2
^x/₄ = -2
x = -8
Given that x = 2.4, evaluate without use of tables and calculators, sin x - cos x in the form of ^a/_b where a and b are integers. (3
marks)
sin x = ²⁴/₂₆
cos x = ¹⁰/₂₆
²⁴/₂₆ - ¹⁰/₂₆ = ¹⁴/₂₆ = ⁷/₁₃
The difference between the interior and exterior angles at each vertex of a regular polygon is 162º. Find the number of sides of the polygon. (3 marks)
Let the exterior < be x
Let the interior < be y
x + y = 180
y - x = 162
(or any method)
2x = 18
x = 9
No of sides = 3
= 40sides
The surface area of two similar bottles is 12cm² and 108cm² respectively. If the larger one has a volume of 810cm³. Find the volume of the smaller one.(3mks)
ASF = ¹²/₁₀₈ = ¹/₉
LSF = ¹/₃
VSF = ¹/₂₇¹/₂₇ =^x/₈₁₀x = ⁸¹⁰/₂₇x = 30cm³
A cylindrical iron pipe is 2.lm long and 12cm in external diameter, the metal is 1cm thick and its density is 7.8g,/cm³. Taking pie as 3 ½ find its mass. (4 Marks)
V = πR²h - πr²h
= πh(R² - r²)
= 3.142 x 210 (6² - 5²) cm³
= 5938.38 cm³M = v x d
= 5938.38 x 7.8g
= 46319.364g
= 46.32kg
Draw the net of the solid below given that is a right pyramid and that AB = 4cm = BC = CD = AD and BE= 6cm (3mks)
$3$
SECTION II (50marks)
Answer only five questions in this section in the spaces provided.
Ruhu, Toru, and Lwamawa contributed a total of Kshs. 8041950.00 for their joint campaigns ahead of 2022 general elections. The ratios of their contributions were Ruhu to Toru 5:4 and Lwamawa to Toru 2:3.
1. How much did each contribute? (4 Marks)
  R T L
  5 4
  3 2
  Lwa = ⁸/₃₅ x 8041950 = sh183811
  15:12:8
  Ruhu = ¹⁵/₃₅ x 8041950 = sh3446550
  Toru = ¹²/₃₅ x 8041950 = sh2757240
2. Ruhu further contributed Kshs. 875,000.00 towards the campaigns kitty. in response, Toru and Lwamawa increased their contributions in the ratios 10:9 and 11:6 respectively. How much did Toru and Lwamawa further contribute altogether? (3 marks)
  Toru ¹⁰/₉ x 2757240 = 3063600
  Lwa ¹¹/₆ x 1838160 = sh 6.433560
3. The three agreed that if they win elections they would share the 15 cabinet positions amongst them in the ratio of their contributions. How many cabinets positions did Lwamawa get? (3 Marks)
  4321550 + 6433560 = 10755110
  3369960 x 15 = 4.7 ≈ 5 positions
  10755110
In the figure below, 0 is the centre of the circle. PQ and PR are tangents to the circle at P and R respectively Angle PQS = 40º and angle PRS 30° RTU is a straight line. ( 3mks)

Find with reasons the angles
1. QRS (2marks)
  40º - angles in the alternate segments are equal
2. RTQ (2 marks)
  70º sum of alternate angles
3. RPQ (2 marks)
  70º angles in quadrilateral add up to 360º
4. Reflex angle QOR (2 marks)
  140º isoceles triangle
5. TRO given that TR =TQ (2 marks)
  35º <s of isoceles
Complete the table below for the function y=x² + 6x² + 8x for -5≤ x ≤1 (2 marks)

x -5 -4 -3 -2 -1 0 1

x² -125 -64 -27 -8 -1 0 1

6x² 150 96 54 24 6 0 6

8x -40 -32 -24 -16 -8 0 8

y -15 0 3 0 -3 0 15
1. Draw the graph of the function (3 marks)
  (Use a scale of 1cm to represent 1 unit on the x axis. 1 cm to represent 5 units on the y- axis)
  $5$
2. Hence use your graph to estimate the roots of the equation
  1. X³ + 6X² + 8X = 0 (1 mark)
    x³ + 6x² + 7x + 1 = 0
    = x³ +6x² + 8x
    = x - 1
  2. x³+ 5x² + 4x = -x² - 3x - 1 (4 marks)
    x₁ = -4.4 ± 0.1
    x₂ = -1.3 ± 0.1
    x₃ = -0.3 ± 0.1
Three islands P, Q, R and S are on at ocean such that island Q is 400kmon a bearing of 030º from island P. Island R is 520km and on a bearing of 120º from island Q. A port S is sighted 750km due south of island Q.
1. Taking a scale of 1cm to represent 100km, give a scale drawing showing the relative positions of P, Q, R and S. (4 marks)
  $6$
  Use the scale drawing to
2. Find the bearing of:
  1. Island R from island P (1 mark)
  2. Port S from island R (1 mark)
3. Find the distance between island P and R (2 marks)
4. Find distance between S and R (2marks)
In the figure below, E is the midpoint of AB, OD : DB = 2 : 3 and F is the point of intersection of OE and AD

Given that OA = a and OB = b,
1. Express in terms of a and b
  1. AD (1 mark)
    ²/₅b - a
  2. OE (2 marks)
    a + ½(b - a)
    ½a + ½b
2. Given further that AF = sAD and OF = tOE, find the values of s and t (5 marks)
  OF = a + s(²/₅b - a)
  OF = (1 - s)a + ²/₅s b
  OF = t(½a + ½b)
  = ½ta + ½tb
  (1 -s)a + ²/₅s b = ½ta + ½tb
  1 - s = ½t
  2 - 2s = t
  t + 2s = 2
  ²/₅s b = ½tb
  s = ⁵/₂t
  t + 2(⁵/₂)t = 2
  6t = 2
  t = ¹/₂
  s = ⁵/₄t
  t + 2(⁵/₄)t = 2
  t + ⁵/₂t = 2
  t = ⁴/₇
  s = ⁵/₇
3. Show that E, F and O are collinear (2 marks)
  OF:OE
A plastic water tank has a shape as shown below, with a frustrum of a cone on top, a cylindrical body and a hemispherical bottom.
LSF =2.8 = 2
1.4
^H/_h = 2
h + 1.5 = 2
h
h = 1.5
1. Calculate
  1. The volume of the tank in m³.(5mks)
    A = ¹/₃x(1.4² x 3) -0.7² x 1.5) + ²²/₇ x 1.4 x 1.4 x 2 + ²/₃ x 1.4³ x ²²/₇
    = 5.39 + 12.32 + 5.749
    = 23.46cm³
2. A filler pipe takes 3 hours to fill a third of the tank. If the tank is already ¼ full, at what time will the filler pipe fill the tank if the pipe is opened at 9.00a.m. (3mks)
  = 9 hrs
  ³/₄ = ³/₄ x ⁹/₁ = 6 ³/₄ hrs
  9.00 + 6:45 = 3.45pm
3. A particle falls in the tank. If its chances of being in any part of the tank are equally likely, find the probability of it being in the hemispherical part (2mks)
  5.74y = 0.245
  23.46
Construct a parallelogram ABCD in which AB = 8.5cm, AD = 6cm and angle BAD = 75º. (Use a ruler and pair of compasses only in this question)
1. Measure the length of AC. (4mks)
  $8$
2. On the same diagram, construct a perpendicular from D to line AB at M. Measure BM. Hence calculate the area of the parallelogram ABCD. (4mks)
3. Ex-scribe a circle to triangle BDA tangent to BD (2marks)
  tangent to BD
The figure below shows two circles intersecting at C and D. The centres are A and B with radii 8cm and 6cm respectively. AB = 10cm.
Determine to 2 decimal places
1. Size of angle DAC (2mks)
  6² = 8² + 10² - 2 x 8 x 10 cosx
  x = cos 120 = 36.87º
  160
  <DAC = 36.87 x 2
  = 73.74º
2. Size of angle DBC (2mks)
  8² = 6² + 10² - 2 x 6 x 10Cosx
  y = cos y 72 = 53.13
  120
  <DBC = 53.13 x 2 = 106.25
3. Area of sector ACMD (2mks)
  =73.74 x 22 x 8 x 8
  360 7
  = 41.83cm²
4. Area of the shaded region (4mks)
  A = ABCD = ½ x 8 x 8sin73.74 + ½ x 6 x 6sin106.6
  30.72 + 17.28 = 48cm²= 48 - 41.83
  = 6.17cm²