Mathematics Paper 1 Questions and Answers - KCSE 2020 Past Papers

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Instructions to Candidates

  • 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 calculations, giving your answers at each stage 
  • Marks may be given for correct working even if the answer is wrong.
  • 8 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)  

  • Answer all the questions in this section in the spaces provided.  
  1. Without using mathematical tables or a calculator, evaluate (3 marks)
    −3(6+−2)−−12÷4+5 
          −4x−6+−3x5  
  2. Express recurring decimal as a mixed number. (3 marks)
  3. Without using a calculator or mathematical tables, evaluate: (3 marks)  
    querry 3 kcse
  4. A rectangular floor of a room measures 5.4 m long and 4.2 m wide. The room is to be covered with square tiles. Calculate the minimum number of square tiles that can be used to cover the floor. (4 marks)  
  5. Simplify
     simplifying q 4   (3 marks)
  6. The line 2y+x= 1 is perpendicular to a line L. Line L passes through point (2,−1), Determine the equation of L in the form y = mx + c, where m and c are constants. (3 marks)  
  7. The sum of the interior angles of a regular polygon is 12600. Find the size of each exterior angle of the polygon. (3 marks)
  8. Using the grid provided below, solve the simultaneous equations.
    grid on simultaneous equation  
    X−4y = −5  (3 marks)  
    −x+2y =1
  9. Given that sin(θ+30°)=cos2θ, find the value of cos(θ+40°).(3 marks)
  10. In the figure below, AB and CD are arcs of two concentric circles, centre O. Angle AOB = 60° and AD=BC=7cm.  
    Concentric circle arcs
    Given that the perimeter of ABCD is 28 2/3 cm, find OA, the radius of the inner circle. (3 marks)  
  11. Using a ruler and a pair of compasses only, construct a parallelogram ABCD in which AB = 6cm, BC = 5cm and angle ADC = 150°.  (3 marks)  
  12. A Kenyan Non-Governmental Organization (NGO) received a donation of 200 000 US dollars. The money was converted into Kenyan shillings in a bank which buys and sells foreign currency as follows:  
                                                     Buying (Ksh)            Selling (Ksh)  
                      1 US Dollar                    102.40                   102.50  
                      100 Japanese Yen            92.80                     93.30  
    1. Calculate the amount of money, in Kenya Shillings, the NGO received. (1 mark )  
    2. The NGO used 90% of the donation to buy a machine from Japan. Calculate the cost of the machine to the nearest Japanese Yen.  (3 marks)
  13. A chord of a circle, 7 cm long subtends an angle of 60° at the centre of the circle. Determine the area correct to 2 decimal places, of the major segment of the circle. (4 marks)  
  14. A vertical electric pole was erected 6.4m from the foot of a vertical fencing pole on the same horizontal level. The fencing pole is 2m high. The angle of elevation of the top of the electric pole from the top of the fencing pole is 30°. Determine the height of the electric pole correct to 1 decimal place. (2 marks)  
  15. In a 4x400 m relay competition, a team of athletes each completed their round in 45 sec, 43 sec, 44 sec and 45 see respectively. If the race started at 1:35:31 p.m., find the time when the team completed the race.  
    (3 marks)  
  16. The figure below shows a quadrilateral ABCD and a mirror line M1M2.  
    Quadrilateral with mirror line
    1. Draw quadrilateral A'B'C'D', the image of ABCD, under a reflection in the mirror line M1M2.  (2 marks)  
    2. State the type of congruency between ABCD and A'B'C'D'. (1 mark)  

SECTION II (50 marks)  

  • Answer any five questions from this section in the spaces provided.  
  1. Three business partners, Kosgei, Kimani and Atieno contributed Ksh 1,750,000 towards an investment. Kosgei contributed 20% of the money. Kimani and Atieno contributed the remainder in the ratio 3:5 respectively. 
    1.  
      1. Calculate the amount of money Kimani contributed. (2 marks)  
      2. Find the ratio of the contributions by the three partners. (2 marks)   
    2. The money earned a compound interest at a rate of 8% per annum. After 3 years, the  partners withdrew the interest and donated 10% to a charitable organisation. The partners then shared the remainder in the ratio of their contributions.  
      Calculate the amount of money, to the nearest shilling, that each partner received. 
      (6 marks)  
  2. A solid consists of a conical part, a cylindrical part and a hemispherical part. All the parts have the same diameter of 12cm. The height of the cylindrical part is 15cm and the slanting height of the conical part is 10 cm. (Take π = 3.142).  
    Calculate the:  
    1. height of the solid; (2 marks) 
    2. surface area of the solid, correct to 1 decimal place,(4 marks)  
    3. volume of the solid, correct to 1 decimal place. (4 marks)   
  3. The average speed of a pick-up was 20km/h faster than the average speed of a lorry. The pick-up took 45 minutes less than the lorry to cover a distance of 180 km.  
    1. If the speed of the lorry was x km/h:  
      1. Write expressions in terms of x for the time taken by the lorry and the pick-up respectively to cover the distance of 180 km (2 marks)  
      2. Determine the speed of the lorry and that of the pick-up. (5 marks)  
    2. The distance between towns A and B is 240 km. On a certain day the pick-up started from town A at 8.30 a.m. and the lorry started from town B at the same time. Determine the time that the lorry and the pick-up met. (3 marks)   
  4. A forest is enclosed by four straight boundaries AB, BC, CD and DA. Point B is 25km on a bearing of 315° from A, C is directly south of B on a bearing of 260° from A and D is 30km on a bearing of 210° from C.  
    1. Using a scale of 1:500,000, represent the above information on a scale drawing
      (3 marks)  
    2. Using the scale drawing, determine the:  
      1. distance, in kilometres, of D from A: (2 marks)  
      2. bearing of A from D (1 mark) 
    3. Calculate the area, correct to 1 decimal place, of the forest in square kilometres.
      (4 marks)  
  5. The position vectors of points A and B are OA=q21 vector a  and OB = q21 vector b  .A point M is on AB such that AM = ½AB and a point N is on OB such that ON:NB= 1:3.  
    1. Find:  
      1. AB (2 marks) 
      2. NM; (3 marks)  
      3. lNMl, correct to 1 decimal place. (2 marks)  
    2. A translation vector maps A onto A'(5,−1). Find the coordinates of the image of B under the translation vector.  (3 marks)
  6. The number of students in 40 schools in a certain sub-county were recorded as follows:  
             238  100  173  115  398  597  572  299  
             255  750  587  403  250  217  797  386  
             362  486  453  279  436  248  780  316  
             529  305  372  286  842  235  420  492  
             226  245  695  720  267  617  198  830  
    1. Complete the frequency distribution table below for the data. (2 marks) 
       Number of students   100-199  200-249   250-399   400-599   600-799   800-849 
       Number of schools            
       
    2. On the grid provided, draw a histogram to represent the data. (4 marks)

      Data represented in a histogram     
    3. Use the histogram to estimate the:  
      1. median; (2 marks) 
      2. number of schools with more than 350 students. (2 marks)  
  7.  
    1. Given that matrix M =Matrix M is a singular matrix, 23, find the value of p. (3 marks)
    2. Kirai earned a basic salary of Ksh x per month and Ksh y for each extra hour worked. Mutua's basic salary is 20% more than Kirai's basic salary but also earns Ksh y for each extra hour worked. In a certain month, Kirai worked 30 extra hours and earned a total of Ksh 70,000 while Mutua worked an extra 40 hours and earned a total of Ksh 88,000.  
      1. Form two equations in x and y to represent Kirai's and Mutua's total earnings. ()  (2 marks)  
      2. Use matrix method to solve for x and y. (5 marks)  
  8. The displacement, s metres, of a moving particle after t seconds is given by
     displacement of a moving particle
    Determine the:  
    1. velocity of the particle when t = 5:  (3 marks)
    2. time, t seconds, when the particle is momentarily at rest; (3 marks)  
    3. displacement, s metres, when the particle is momentarily at rest;  (2 marks)
    4. acceleration of the particle when t = 4 seconds. (2 marks) 

Marking Scheme

  1. −3(6+−2)−−12÷4+5 
          −4x−6+−3x5
    −3(4) + 3 + 5 
         24 +−15
    =−4 
        9

  2. recurring numbers
     9r = 50
      r= 55/9

  3. Evaluate equation
     =26
     =64
  4. 42 =2×3×7
    54 =2×33
    G.C.D =2×3=6
    ∴ G.C.D of 5.4 and 4.2=0.6
    No of tiles = area room
                        area tile
    = 5.4 × 4.2 
        0.6×0.6
    =63 tiles
  5.  
    Simplifying equation
     simplyfying equation 2
    = (2x + 3y) 
        (x − 2y)
  6. m1= −½
    m2 = 2  (2,−1)
    y=m(x −x value) + y value
    y=2(x − 2)−1
    y=2x−5
  7. (n−2)180=1260
    n=9 sides
    Ext ∠ = 360 
                 9
            =40°
  8. X−4y = −5 
    −x+2y =1

    X−4y = −5
     x   −5   3 
     y   0  2

    −x+2y =1
     x   −1   7 
     y   0  4
    gradient graph
    x=3 and y=2
  9. sin(θ+30°)=cos2θ
    θ+30+2θ=90
    ∴ cos 60=0.5
  10. θ=60°=π/3radians
    Arc length =θr where θ is in radians
    π/× r + π/3×(r+7) +7+7 =282/3
    /3r =142/3− /3
            r=3.502
  11.  
    Parallelogram construction
  12.  
    1. 1dollar=Ksh 102.40
      200000dollars =?
      200000dollars   × Ksh102.40
            1dollar
      = Ksh 20,480,000
    2. 100 Jap yen= Ksh 93.30
           ?           = 0.9(20,480,000)
                         =0.9(20480000) × 100yen
                                  93.3
                         =19755627.01
                         ≈19,755,627
  13. θ =π/3                                         θ=360−60=300
    θ =2π−π/3                                   arc length, b = θ/360 . 2πr
       =12/3π                                     chord= radius = 7cm
    area=½θr2                                                 b = 300 × 2 × 22 × 7
            where θ is in radians                         360            7
    =½ × 5/3π × 7 × 7                                =362/3cm
    =128.28cm2                                    area=½b.r
                                                                =½ × 362/3 × 7
                                                                =1281/3cm2
                                                                ≈128.33cm2
  14. 2 + 6.4 tan 30
    = 5.695
    ≈ 5.7m
  15.   ℎr   min   s
      1     35   31
      0      0    45
      0      0    43
      0      0    44
    +0      0    45
      1     38    28
    = 1:38:28pm
  16.  

    1. quadrilateral image
    2. Indirect congruent or opposite congruency
  17.  
    1.  
      1. Ko :  Ki : At
        0.2: 0.3: 0.5
        Ki = 30 % of 1750000
            =Ksh. 525000
      2.  2:3:5
    2. A =PRnwhere R = 1+r
         =1750000(1.083)
         =2204496
      Shared amount=90% × 2204496
                            =1984046.4
      Kosgei=0.2 ×1984046.4
               =Ksh. 396809
      Kimani=0.3 ×1984046.4
                =Ksh. 595214
      Atieno=0.5 x1984046.4
                =Ksh. 992023
  18.  
    1.  
      Solid with conical cylindrical and hemispherical
      Total height =8+15+6
                       =29 cm
    2. SA= πrl + 2πr2 + 2πrh
          = 3.142× 6×10 + 2×3.142×62 + 2×3.142×6×15
          =188.52+565.56+226.224
          = 980.304
          ≈ 980.3 sq.cm
    3. V=1/3xAh + πr2h + 2/3πr3
        = 1/3 × 3.142×36×8 + 3.142×36×15 + 2/3 × 3.142 × 216
        =2450.76 cubic cm
        ≈2450.8cm3
  19.  
    1.  
      1. lorry time = 180km hrs
                             x
        Pick up time=180km hrs
                             x+20
      2.  180km  − 180km =  
            x            x+20     4
         x2 + 20x − 4800= 0
        (x − 60)(x + 80) = 0
                              x = 60km/h or −80km/h
                           ∴ x = 60km/h
                hence lorry = 60km/h &
                     pick up = 80km/h
    2. distance pick up +distance lorry = 240km
      80(t − 8.30am) + 60 (t − 8. 30am) = 240
      where t meeting time
                     140 t− 680 − 510 = 240
                                              t ≈ 1013hrs
  20.  
    1.  
      information in scale drawing
    2.  
      1. AD=8.7cm × 5km
                1cm
        = 43.5km ± 0.5
      2. bearing of A from D is 048° or N48°E
    3. A∆ABC =½ × 21 × 25 sin 45°
                = 185.62 km2
      A∆ACD = ½ ×30 × 43.5 sin 18°
                =201.63km2
      total area= 185.62 + 201.63
                    = 387.25km2
                    ≈ 387.3km2
  21.  

    1. Vector position and calculations 
      1. AB= (−2) + (12 ) = (10 )
               (−4)     (−4)    ( −8)

      2. Equation q21 ii

      3. position vector biii
                =√17
                =4.123
                ≈4.1
    2.  T=A−1−A
      answer 21b
      ∴ B1(15,−9)
  22.  
    1.  class  tally   f   fd   cf 
       100-199   llll  4  0.04  4
       200-249  llll l  6   0.12   10 
       250-399  llll llll ll   12   0.08  22
       400-599  llll llll  10  0.05  32
       600-799  llll l  6  0.03  38
       800-849  ll  2  0.04  40

    2. Histogram answer
    3.  
      1. Median psn=40/2 =20th value
           =10 + x × 0.08 = 20
        x = 125
        Median = 125+249.5
        =374.5
      2. 49.5 × 0.08 + 10 + 6 + 2
        = 21.96
        ≈ 22 schools
  23.  
    1. 3 −{p (p − ½)} = 0
           3 − p2 + ½p = 0
            2p2 − p − 6 = 0
       (p − 2)(2p + 3) = 0
                            p = 2 or − 1.5
    2.  
      1. Kirai        x + 30y = 70000 … (i)
        Mutua 1.2x + 40y = 88000 … (ii)
                     6x + 200y = 440000…(ii)

      2. matrix to answer question
        ∴ x = 40000 and y = 1000
  24.  
    1. V=2t2 − 7t − 6 iƒ t = 5
        =2t2−7t−6
        =9 m/s
    2. V=2t2−7t−6 =0
      q 24 b
      =4.212 s or -0.7122 s
      ∴ t = 4.212s
    3. S= 2/3t37/2t2 − 6t + 8 iƒ t = 4.212
      S= 2/3t3 − 7/2t2 − 6t + 8
        =−37.5487m
    4. a = dv/dt = 4t − 7 iƒ t = 4
                   =4 x 4 − 7
                    =9m/s2

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