Mathematics Paper 1 Questions and Answers - Kangundo Subcounty Pre Mock Exams 2021/2022

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INSTRUCTIONS TO THE CANDIDATES

  • Write your name and index number in the spaces provided above
  • This paper contains two sections; Section A and Section B
  • All workings and answers must be written on the question paper in the spaces provided below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
  • This paper consists of 15 printed pages.
  • Candidates should check carefully to ascertain that all the pages are printed as indicated and no questions are missing.

Section A. (50mks)
Answer all the questions in this section in the spaces provided.

  1. Without using a calculator evaluate. (3mks)
    (21/3-11/5 of 4)
       ¼ - (-1/3)2
  2. Use logarithms to evaluate (4mks)
    5(0.0066283×193.9
               2.2822
  3. Without using a calculator or mathematical tables evaluate. (3mks)
    43⁄2 × (256/2401)¾
  4. The base of a right pyramid is a rectangle of length 80cm and width 60cm. each slant edge of the pyramid is 150cm. Calculate the volume of the pyramid. (3mks)
  5. Simply. (3mks)
       2x2- xy - 6y 2
     x2- 4xy + 4y2
  6. A two digit number is such that, the sum of its digits is 13. When the digits are interchanged the original number is increased by 9. Find the original number. (4mks)
  7. The size of an interior angle of a regular polygon is 3x0 while its exterior is (x+20)2. Find the number of triangles that makes the polygon. (3mks)
  8. In the triangle ABC below BC = 14cm <ABC= 70º and <ACB=40º. Calculate; correct to 4 significant figures the areas of triangle ABC. (3mks)
    1
  9. A vector given by (2-4) transforms a point A(3,6) to A1. Find the distance A A1 (3mks)
  10.    
    1. Using a rule and a pair of compasses only, construct a quadrilateral PQRS in which PQ = 6cm ,PS = 4cm QR = 5cm, <PQR=105º and <SPQ= 90º (2mks) 
    2. The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in metres. (2mks)
  11. Without using mathematical tables or calculator evaluate. (3mks)
    sin⁡30º - sin⁡60º
           60º
  12. Use matrix method to solve. (3mks)
    5x+3y= 35
    3x-4y= - 8
  13. Use mathematical tables to find the reciprocal of 0.0247, hence evaluate. (4mks)
    ∛3.025 Correct to 2 decimal places.
    0.0247
  14. A Kenyan businessman intended to buy goods worth US dollar 20,000 from South Africa. Calculate the value of the goods to the nearest south Africa (S.A) Rand given that 1 US dollar = Ksh 101.9378 and 1 S.A Rand = Ksh 7.6326. (3mks)
  15. Find all integral values of x which satisfy the inequalities. (3mks)
    5-3x ≤x-7 <11-2x
  16. Express 5184 and 2744 in terms of its prime factors hence determine the value of
    √5184  (4mks)
    ∛2744  

    Section B (50mks)

    Answer any five questions from this section on the spaces provided.
  17. Two vertices of a rectangle ABCD are A(3,6) and B( 7,12)
    1. Find the equation of line AB. (3mks)
    2. Find the equation of the perpendicular bisector of line AB. (4mks)
    3. Given that BC is perpendicular to AB. Find the equation of BC. (3mks)
  18. Three business partners Abila, Bwire and Chirchir contributed Ksh 120,000, Ksh 180,000 and Ksh 240,000 respectively to boost their business. They agreed to put 20% of the profit accrued back into the business and to use 35% of the profits for running the business. The remainder was to be shared among the business partners in the ratio of their contribution. At the end of the year, a gross profit of Ksh 225,000 was realised.
    1. Calculate the amount.
      1. Put back into the business. (2mks)
      2. Used for official operations. (1mk)
    2. Calculate the amount of profit each partner got. (4mks)
    3. If the amount put back into the business was added to individual’s shares proportionately of their initial contributions, find the amount of Chirchir’s new shares. (3mks)
  19. Coast bus left Nairobi at 8.00Am and travelled towards Mombasa at an average speed of 80Km/hr. At 8.300am Lamu bus left Mombasa towards Nairobi at an average speed of 120Km/hr. given that the distance between Nairobi and Mombasa is 400Km, determine.
    1. The time Lamu bus arrived in Nairobi. (2mks)
    2. The time the two buses met. (4mks)
    3. The distance from Nairobi to the point where the buses met. (2mks)
    4. How far coast bus is from Mombasa when Lamu bus arrives in Nairobi. (2mks)
  20. A land is enclosed by four straight boundaries AB,BC,CD and DA. Point B is 25Km on bearing of 315º from A, C is directly south of B on a bearing of 260º from A and D is 30Km on a bearing of 210º from C
    1. Using a scale of 1cm to represents 5Km represent the above information on a scale drawing. (3mks) 
    2. Using the scale drawing, determine the
      1. Distance in Kilometres of D from A. (2mks)
      2. Bearing of A from D. (1mk)
      3. Calculate the area, correct to 1 decimal place, of the land in square kilometres. (4mks) 
  21. Complete the table below for the functions
    1. y= x2- 6x+7 (2mks)
      0 1 2 3 4 5 6
      -x2 0 - - 9 16 25 -
      -6x  - -6 -12 - - - -36
      7 7 7 7 7 7 7
      Y = x2 - 6x + 7 7 - -1 -2 - - -
    2. On the grid provided below draw the graph of y= x2- 6x+7 for 0 ≤ x ≤6 and use it to estimate the roots of the equation.
      x2- 6x+7=0 (4mks)
      2
    3. Use the graph above to solve the equation x2- 7x+9=0 (3mks)
    4. Determine the range of values of X for which x2- 6x+7 <x-2 (1mk)
  22. In the figure below PQR and S are points on the circumference of the circle centre O. TP and TR are tangents to the circle at P and R respectively. POQ is a diameter of the circle and angle PQR = 64º (10mks)
    3
    Giving reasons on each case, find the size of
    1. <ROP
    2. <PSR
    3. <ORP
    4. <TRP
    5. <RTP
  23. The figure below shows two triangles ABC and BCD with a common base BC = 3,4cm. The area of triangle ABC = Area of triangle BCD and ABC =90º
    4
    1. Calculate, correct to one decimal place.
      1. The area of triangle ABC. (3mks)
      2. The size of <BCD (3mks)
      3. The length of BD (2mks)
      4. The size of <BDC (2mks)
  24. The marks scored by 40 students in a mathematics test were as shown in the table below.
    Marks 48 – 52 53 – 57 58 – 62 63 – 67 68 – 72 73 – 77 
    Number of students 3 4 10 12 8 3
    1. State the modal class. (1mk)
    2. Using an assumed mean of 64, calculate the mean mark. (3mks)
    3. On the grid provided, draw the cumulative frequency curve for the data. (3mks) 
      2
    4. Use the graph to estimate the semi- interquartile range (3mks)


MARKING SCHEME 

  1. 21/3 – 11/5o of 4
    6/5 ×4= 24/5
    7/3-24/5= 35/15 - 72= -37/15
    ¼ - (-1/3)2
    ¼ - 1/9= 9 - 4 =  5  
                   36      36
    -37/15 × 36/= 1719/25
    M1 for -37/15
    M1 for 5/36
    A1 for 1919/25
  2. No.                  std                  Log
    0.006628    6.628×10-3       3 ̅ .8214
                                                          3
                                                7 ̅ .4642
                                                         +
    193.9          1.939×102            2.2876
                                               5 ̅ .7518
    2.2822        2.2822×10º       0.3584
                                               5 ̅ .3934
                                                    5
    0.1198         1.198×10-1       1 ̅.0.786
    M1 for – correct logs
    M1 for 5 ̅ .7518
    M1 for 1 ̅.0.786
    A1 for C.A
  3. 49 3⁄2 = √493= 73=343
    256 = ( ∜256 )3=64
    2401= ( ∜2401)3=243
    343 × 64/343=64
    M1for 343
    M1 for 64, 343
    A1 for C.A

  4. 5
    Length of diagonal
    √(80+ 602=100
    ½ diagonal = 50
    Height = 1502- 502
    = 22500 – 2500
    = 20000
    =√20000=141.4
    1/3×80 ×60 ×141.4
    226. 2
    M1 for height
    M1 for expression of volume
    A1 for C.A
  5. 2x2 – xy – 6y2
    2x2 – 4xy + 3xy – 6y2
    2x (x-2y) + 3y( x-2y)
    (2x+3y) (x – 2y)
    x2- 4xy + 4y2
    x2– 2xy – 2xy + 4y2
    x(x-2y) – 2y( x-2y)
    (x-2y) (x-2y)
    (2x+3y)(x-2y)
     (x-2y) (x-2y)
    2x+3y
     x-2y
    M1 for (2x+3y) (x – 2y)
    M1 for (x-2y) (x-2y)
    A1 for C.A
  6. x+y=13
    (10y + x ) - ( 10x + y) = 9
    10y + x – 10x – y = 9
    9y – 9x = 9
    y – x = 1
    - x+ y = 13
       -x + y = 1
    2x = 12
    y=6
    x + y = 13
    6 +y = 13
    Y = 7
    M1 for 9y – 9x = 9
    Mfor 2x = 12
    A1 for both values of x and y.
  7. int+ext=180
    3x + x + 20 = 180
    4x + 20 = 180
    4x +160
    X = 40
    Ext = 40 + 20 = 60
    Ext = 360/n= 360/n=60
    n=360/60=6
    Number of triangles = n – 2
    = 6- 2
    = 4
    M1 for expression
    M1 for 360/60
    A1 for C.A
  8. <BAC=180-110
    = 70º
      14   =   AC  
    sin⁡70   sin⁡70
    AC = 14 × sin⁡70
                 sin⁡ 70 
    AC = 14
    Area = ½ ×14 14 × sin⁡40
    =62.99
    =63.0
    M1 for expression of AC
    M1 for for expression of area
    A1 for C.A
  9. O+T =1
    (36)+ (2-4)= (52)
    Distance - √(52+ 22)
    = 5.385
    M1 for addition of vectors
    M1 for for expression of distance
    A1 for C.A

  10. 6
    Rs = 7.cm
    1: 4000
    7.4: 7.4 ×4000
    = 29600
    B1 for PQR and SPQ
    B1 for for complete figure
    B1 for for RS
    B1 for C.A
  11. sin⁡ 30º - sin⁡ 6º
        tan⁡60º
    Sin 30º=½ tan⁡60º = √3
    sin⁡60º= √3/2
    7
    M1 for trigonometric ratio
    M1 for for rationalising
    A1 for C.A
  12. 5x + 3y =38
    3x – 4y = -8
    8
    x=4 y=5
    B1 for determinant
    M1 for inverse
    A1 for C.A
  13. Reciprocal of
    2.47 ×102
    =0.4049 × 102=40.49
    ∛(3.025 ) =1.446
    =40.49 ×1.446=58.55
    M1 for reciprocal
    M1 for ∛
    A1 for C.A
  14. 1 US dollar = Ksh 101.9378
    20,000 = 20,000 ×101.9378
    =Ksh 2038756
    1 S.A = 7.6326
    2038756 =2038756
     7.6326
    = 267111.6
    = 267112 S.A
    M1 for expression
    M1 for expression
    A1 for C.A
  15. 5-3x ≤x-7 <11-2x
    5x – 3x ≤x- 7         x-7 <11-2x
    5+ 7 ≤x+3x            x+2x <11+7
    12 ≤4x                   3x <18
    3≤x                          x <6
    3≤x <6
    3,4 ,5
    M1 for separating inequality
    M1 for 3≤x and x <6
    A1 for C.A
  16. 5184 = 2×2×2×2×2×2×3×3×3×3
    √5184=2×2×2×3×3
    = 72
    2744 = 2×2×2×7×7×7
    ∛2744=2×7=14
    = 72/14 = 51/7
    M1 for factors of 5184
    M1 for factros of 2744
    M1 for √ and ∛
    A1 for C.A
  17.          
    1. A( 3,6) B(7,12) (x ,y)
      Gradient of AB = 12-6 ==
                                  7-3     4     2 
      y - 6 =
      x - 3    2
      2(y – 6) = 3( x- 3)
      2y – 12 = 3x – 9
      2y = 3x – 9 + 12
      2y = 3x + 3
    2. mid point of AB
      3+7 , 6+12
        2       2 
      (5,9) (x,y)
      Perpendicular bisector gradient = - 2/3
      (y-9 )/(x-5 )= -2/(3 )
      3( y-9) = -2(x- 5)
      3y – 2y = -2x + 10
      3y = -2x + 10 + 27
      3y = -2x + 37
      AB ⊥BC gradient of Bc= - 2/3,B( 7,12)
      (x,y ) x y-12 = -2/3 ) 3y= – 2x + 14 + 36
                   x-7
      3y = -2x + 50
      3(y-12)= -2(x-7)
      3y – 36 = -2x + 14
      M1 for gradient
      M1 for expression
      A1 for equation
      M1 for mid point
      M1 for gradient
      M1 for expression
      A1 for equation
      M1 for expression= -2/3
      M1 for x-multiplication
      A1 for C.A
  18.               
    1.        A   ∶     B     ∶      C
      120000∶180000∶ 240000
            12  ∶    18    ∶    24
             2   ∶     3     ∶      4
      1. 20/100 ×225000=45000
      2. 35/100 ×225000 = 78750
    2. Remainder = 225000-45000-78750
      = 101 250
      2: 3∶ 4 = 9
      = 101250
      1. 2/9 ×101 250 = 22,500
      2. 3/9 ×101 250 = 33,750
      3. 4/9 ×101 250 = 45,000
    3. 2:3:4=9
      45000
    4. 4/9 ×45000 = 20,000
      Chirchir new share = 240,000+20,000
      = 260,000
      M1 for multiplication
      A1 for C.A
      A1 for C.A
      M1 for 101250
      A1 for C.A
      A1 for C.A
      A1 for C.A
      M1 for expression
      M1 for addition
      A1 for C.A
  19.  
    9   
    1. Nairobi        Mombasa
      8.00am         8.30am
      80Km/h         120Km/h
      Time = D/S= 400̸/120̸ =31/3 hrs
      = 3: 20min
      +8∶ 30
        3∶ 20
      = 11 ∶ 50 am 
    2. Distance by coast bus for ½ hr.
      = 80 × ½=40Km
      R.D = 400 - 40 = 360 Km
      R.S = 80 + 120 = 200Km/h
      R.T = 260/200=14/5hr=1 hr:48min⁡
      Time met = 8∶ 30
                         1∶ 48
      = 10: 18am 
    3. R.T = 14/5 hr
      Coast by distance in 14/5
      = 80 × 14/5
      = 144 km
      Distance = 40 + 144
      = 184
    4. In 3hr20min coast bus covers
      = 80 × 31/3 
      = 80 × 10/3 =266 2/3 km
      Distance from Mombasa = 400 – 266 2/3
      =1331/3  Km
      M1 for 31/3  hr
      M1 for addition
      A1 for C.A
      M1 for 40km
      M1 for RD and R.S
      M1 for expression of R.T
      A1 for C.A
      M1 for expression of distance
      A1 for C.A
      M1 for 400 – 266 2/3
      A1 for C.A

  20. 10
    AD = 8.7 cm = 43.5 km ± 1
    Bearing = 047º
    B1 for point B
    B1 for point C
    B1 for Point D
    B1 for 8.7 ± 0.1
    A1 for 43.5 ± 1
    Afor 047º ± 1
  21. y = x-6x+7
    0 1 2 3 4 5 6
    -x2 0 1 4 9 16 25 36
    -6x  0 -6 -12 -18 -24 -30 -36
    7 7 7 7 7 7 7
    Y = x2 - 6x + 7 7 2 -1 -2 -1 2 7

    11
  22.    
    1. <ROP= 128º
      Angles on a straight line
    2. <PSR= 116º
      Angle on cyclic quadrilateral
    3. <ORP= 26º
      Base angle of isosceles triangle
    4. <TRP= 64º
      Angle between a chard a tangent( Nagle in alternate segment)
    5. <RTP=53º
      Angles on a triangle.
      Bfor Angle
      B1 for Reason
      B1 for Angle
      B1 for Reason
      B1 for Angle
      B1 for Reason
      B1 for Angle
      B1 for Reason
      B1 for Angle
      B1 for Reason
  23. AB height = ∛7.22- 3.42
    = 6.347
    1. Area ABC
      = ½ 3.4 ×6.347
      = 10.78
      = 10.8
    2. Area of BCD = 10.8
      A= ½ ×3.4 ×7.5 ×sin⁡BCD
      10.8
          10.8    = 1.7 ×7.5 ×sin⁡BCD=sin⁡BCD
      1.7 ×7.5
      0.8471 sin BCD
      sin-1 0.8471=BCD
      = 57.8 = BCD
    3. BD2=3.4+7.5-2×3.4 ×7.5 cos⁡57.8
      BD = 6.37
    4. <BDC =    34           3.4     =   6.37   
                   sin⁡BCD  s in⁡BDC    sin⁡57.8
      Sin BDC – 3.4 × sin⁡57.8
                              6.37
      Sin BDC = 0.4517
      BDC = sin-10.4517
      BDC = 26.9
      M1 for expression of height
      M1 for expression of Area
      Afor C.A
      M1 for expression of Area
      M1 for 0.8471
      A1 for C.A
      M1 for expression
      A1 for C.A
      M1 for expression
      A1 for C.A 

  24. Marks x F Tx - A FT CF
    48 – 52  50  3 -14 -42 3
    53 – 57  55 4 -9 -36 7
    58 – 62  60 10 -4 -40 17
    63 – 67  65 12 1 12 29
    68 – 72  70 8 6 48 37
    73 – 77  75 3/40 11 33/-25  40
    1. Modal class 63 – 67
    2. Mean = A + ƸFt/ƸF
      64 + -25/40
      = 63 3/8 
      12
      B1 for modal class
      M1 for mid point
      M1 for -25/40
      A1 for C.A

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