KCSE 2010 Mathematics Paper 1 Questions with Marking Scheme

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QUESTIONS

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

  1. Without using a calculator evaluate,
    fractions kcse 2010
  2. Kutu withdrew some money from a bank. He spent 3/8 of the money to pay for Mutua’s school fees and 2/5 to pay for Tatu’s school fees. If he remained with 12 330, calculate the amount of money he paid for Tatu’s school fees. (4 marks)
  3. A straight line l passes through point (3, -2) and is perpendicular to a line whose equation is 2y-4= 1.
    Find the equation of l in form of y = mx + c, where m and c are constants. (3 marks)
  4. A bus left a petrol station at 9.20 a.m. and travelled at an average speed of 75 km/h to a town N. At 9.40 a.m. a taxi, travelling at a speed of 95 km/h, left the same petrol station and followed the route of the bus.
    Determine the distance, from the petrol station, covered by the taxi at the time it caught up with the bus. (3 marks)
  5. The sum of three consecutive odd integers is greater than 219. Determine the first three such integers. (3 marks)
  6. A Kenyan company received US Dollars 100 000. The money was converted into Kenya shillings in a bank which buys and sells foreign currencies as follows:
       Buying (in Kenya shillings)  Selling(in Kenya shillings)
     1 US Dollar  77.24  77.44
     1 Sterling Pound  121.93  122.27
    1. Calculate the amount of money, in Kenya shillings, the company received. (2 marks)
    2. The company exchanged the Kenya shillings calculated in (a) above, into sterling pounds to buy a car from Britain. Calculate the cost of the car to the nearest sterling pound. (2 marks)
  7. In the figure below, OPQR is a trapezium in which PQ is parallel to OR and M is the mid-point of QR. OP = p, OR = r and PQ = 1/3 OR.
    vectors kcse 2010
    Find OM in terms of p and r. (3 marks)
  8. Without using a mathematical table or a calculator, evaluate powers kcse 2010. (3 marks)
  9. The figure below is a net of cube with some dots on two faces.

    net of cube kcse 2010
    Given that the number of dots on pairs of opposite faces and add up to 7, fill in appropriate dots in each of the empty faces. (2 marks)
  10. Using a ruler and a pair of compasses only, construct a rhombus QRST in which angle TQR = 600 and QS = 10 cm. (3 marks)
  11. A fruit vendor bought 1948 oranges on a Thursday and sold 750 of them on the same day. On Friday, he sold 240 more oranges than on Thursday. On Saturday he bought 560 more oranges. Later that day he sold all the oranges he had at the price of Ksh 8 each. Calculate the amount of money obtained from the sales of Saturday. (4 marks)
  12. Simplify the expression simplify expression kcse 2010 . (3 marks)
  13. Given that 3θ is an acute angle sin 3θ = 2θ°, find the value of θ. (3 marks)
  14. A cylindrical solid whose radius and height are equal have a surface area of 154 cm2. Calculate the diameter, correct to 2 decimal places. (Take π = 3.142) (3 marks)
  15. The figure below shows two sectors in which CD and EF are arcs of concentric circles, centre O. Angle COD = 2/3 radians and CE = DF = 5 cm.
    concentric circles
    If the perimeter of the shape CDFE is 24 cm, calculate the length of OC. (3 marks)
  16. The histogram shown below represents the distribution of heights of seedlings of a certain plant.
    histogram kcse 2010
    The shaded area in the histogram represents 20 seedlings. Calculate the percentage number of seedlings with heights of at least 23 cm but less than 27 cm. (3 marks)

    SECTION II (50 marks)
    Answer only five questions in this section in the spaces provided.

  17. A sales woman is paid a commission of 2% on goods sold worth over 100 000. She is also paid a monthly salary of Ksh 12 000. In a certain month, she sold 360 handbags at Ksh 500 each.
    1. Calculate the saleswoman’s earning that month. (3 marks)
    2. The following month, the saleswoman monthly salary was increased by 10%. Her total earnings that month were Ksh 17 600.
      Calculate:
      1. The total amount of money received from the sales of handbags that month; (5 marks)
      2. The number of handbags sold that month. (2 marks)
  18. A carpenter constructed a closed wooden box with internal measurements 1.5 metres long, 0.8 metres wide and 0.4 metres high. The wood used in constructing the box was 1.0 cm thick and had a density of 0.6g/cm3.
    1. Determine the:
      1. Volume, in cm3, of the wood used in constructing the box; (4 marks)
      2. Mass of the box, in kilograms, correct to 1 decimal place. (2 marks)
    2. Identical cylindrical tins of diameter 10 cm, height 20 cm with a mass of 120 g each were packed in the box.
      Calculate the:
      1. Maximum number of tins that were packed; (2 marks)
      2. Mass of the box with the tins. (2 marks)
  19.  
    1. Find A-1, the inverse of matrix A = (569). (2 marks)
    2. Okello bought 5 Physics books and 6 Mathematics books for a total of Ksh 2 440. Ali bought 7 Physics books and 9 Mathematics books for a total of Ksh 3 560.
      1. Form a matrix equation to represent the above information. (1 mark)
      2. Use matrix method to find the price of Physics book and that of a Mathematics book.
    3. A school bought 36 Physics books and 50 Mathematics books. A discount of 5% was allowed in each Physics book whereas a discount of 8% was allowed on each Mathematics book. Calculate the percentage discount on the cost of all the books bought. (4 marks)
  20. The boundaries PQ, QR, RS and SP of a ranch are straight lines such that: Q is 16 km on a bearing of 0400 from P; R is directly south of Q and east of P and S is 12 km on a bearing of 120o from R.
    1. Using a scale of 1 cm to represent 2 km, show the information in a scale drawing. (3 marks)
    2. From the scale drawing determine:
      1. The distance in kilometers, of P from S; (2 marks)
      2. The bearing of P from S. (2 marks)
    3. Calculate the area of the ranch PQRS in square kilometers. (3 marks)
  21. Motorbike A travels at 10 km/h faster than motorbike B whose speed is x km/h. Motorbike A takes 1½ hours less than motorbike B to cover an 180 km journey.
    1. Write an expression in terms of x for the time taken to cover the 180 km journey by:
      1. Motorbike A; (1 mark)
      2. Motorbike B. (1 mark)
    2. Use the expression in (a) above to determine the speed in km/h , of motorbike A. (6 marks)
    3. For a journey of 48 km, motorbike b starts 10 minutes ahead of motorbike A. Calculate, in minutes, the difference in time of their arrival at the destination. (2 marks)
  22. In the figure below, ABCD is a square. Points P,Q, R and S are the midpoints AB, BC, CD and DA respectively.

    transformations kcse 2010
    1. Describe fully:
      1. A reflection that maps triangle QCE onto triangle SDE; (1 mark)
      2. An enlargement that maps triangle QCE onto triangle SAE; (2 mark)
      3. A rotation that maps triangle QCE onto triangle SED. (3 marks)
    2. The triangle ERC is reflected on the line BD. The image of ERC under the relection is rotated clockwise through an angle of 900 about P.
      Determine the images of R and C:
      1. Under the reflection; (2 marks)
      2. After the two successive transformations. (2 marks)
  23. The frequency distribution table below represents the number of kilograms of meat sold in a butchery.
    mode mean median kcse 2010
    1. State the modal frequency. (1 mark)
    2. Calculate the mean mass. (5 marks)
    3. Calculate the median mass. (4 marks)
  24. A rectangular box open at the top has a square base. The internal side of the base is x cm long and the internal surface area of the box is 432 cm2.
    1. Express in terms of x:
      1. The internal height h, of the box; (3 marks)
      2. The internal volume V, of the box. (1 mark)
    2. Find:
      1. The value of x for which the volume V is maximum; (4 marks)
      2. The maximum internal volume of the box. (2 marks)

Marking Scheme

  1. = -2(5 + 3) - 9 + 3 + 5 = -14
          -3 × -5 + (-2) × 4         7
    = -2
  2. Total fraction
    3 + ? = 31
    8    5    40
    remaining fraction = 1  - 31
                                         40
    =9
     40
    original amount = sh.12330 × 40
                                                 9
    =sh.54800
    Tatu's fees - sh 2 × 54800
                           5
    = sh.21920
  3. gradient(perpendicular) = -1/2
    y + 2 = -1
    x - 3      2
    y = -1x - 1
           2     2
  4. let the distance be d kn
     d   and   d 
    75          95 
    therefore:  d  -   d  = 20
                    75    95    60
    d = 118.75km
  5. let odd integers be:
    x, (x + 2), (x + 2 + 2)
    x + (x + 2) + (x + 2 + 2) > 219
    3x > 213
    x = 71
    the numbers are 73, 75 , 77
  6.                        
    1. sh 77.24 × 100,000
      =sh.7,724,000
    2. sh 77.24 × 100000
              122.27
      =63,172
  7.                             
    q7 uatfgytad

    q7 ans khagyudga
  8.                   
    q8 auy g ydta
  9.                    
    q9 u atfytda
  10.                        
    q10 uyafgytda
    <TQR = 60º, QS = 10cm and bisects < TQR
    Mediator(⊥ or bisector) of QS drawn or
    <RSQ=<QST=<RQS=30º
    Rhombus completed
  11. No of oranges for friday
    1948 - (750 + 750 + 240)
    = 208
    no of oranges for saturday
    208 + 560 = 768
    therefore: amount = sh 8 × 768
    =sh.6144
  12. x2 + x - 4xy - 4y = x(x + 1) - 4y(x + 1)
    (x + 1)(4y2 - xy)     (x + 1)(y)(4y - x)
    = (x - 4y)(x + 1)
    (x + 1)(-y)(x - 4y)
    =-1
       y
  13. sin 3θ = cos 2θ°
    therefore: sin 3θ = sin(90° - 2θ)
    therefore: 3θ = 90° - 2θ
    5θ = 90
    θ = 18°
  14. 2πr2 + 2πrh = 154
    r = h
    2πr2 + 2πr2 = 154
    4πr2 = 154
    r = √      154      
             4 × 3.142
    r = 3.500
    therefore didameter = 2r = 3.500 × 2
    =7.00 (2 d.p)
  15.                         
    q15 yatfytfad
    let OC = r
    therefore: CD = 2r and EF = 2(r + 5)
                            3                3
    2r + 2(r + 5) + 5 + 5 = 24
    3     3
    4r = 102
    3         3
    r = 8
  16. total number of seedlings
    (5 × 1) + (10 × 3) + (15 × 1) + (20 × 4) + (30 × 1) + (10 × 2)
    = 5 + 30 + 15 + 80 + 30 + 20 = 180
    % of height (h) : 23 ≤ h < 27
    = 30 + 15 ×  100
         180
    =25%
  17.                
    1. total sales = sh.360 × 500
      =sh.180,000
      commission = sh.(180,000 - 100,000) × 2/3
      =sh.1600
      total earnings = sh.(12,000 + 1600)
      =13600
    2.                 
      1. new salary = sh.(12000 + 12000 × 10/100)
        =sh.13200
        commission paid = sh(17,600 - 13,200)
        =sh.4400
        commission is paid on sh 4400 × 100/2
        =220,000
        total sales = sh.220,000 + 100,000
        =320,000/=
      2. no of handbags sold = 320000
                                            500
        =640
  18.              
    1.                           
      1. internal volume of box  = 150 × 80 ×40cm3
        =480,000cm3
        external volume of box = 152 × 82 × 42cm3
        =523488cm3
        therefore volume of wood = (523488 - 480000)cm3
        =43488cm3
      2. mass of box = 43488 × 0.6
                                   1000
        =26092.8
        =26.1kg
    2.              
      1. no of tins  = 150 × 80 × 40
                            10     10    10
        =240
      2. total mass = 26.1 + 240 × 120
                                          1000
        =54.9kg
  19.            
    1. Det 45 - 42 = 3
      q19 a hfaytdfa
    2.                             
      1.                   
        q19 b jgfytafd
      2.                         
        q19 c jhaftrda
        q19 d jygautfgda
        therefore: x = 200; y = 240

    3. total cost of books = (36 × 200) + (50 × 240)
      =19200
      total cost of discount 
      36 × 200 × 95 + 50 × 240 × 92
              100                    100
      =17880
      %discount
      =19200 - 17880 × 100
                 19200
      =6.875%
  20.                        
    1. Given scsale: 1cm to 2km
      q20 jagfytdfaytd
    2.                      
      1. Distance of P from S =10.8 ± 0.1cm
        =21.6km
      2. <PSN = 74 ± 1°
        Bearing of P from S=286 ± 1°
    3. Area of triangle PQR = 1/2 x 10.2 x 12.2
      = 62.22km2
      Area of triangle PRS = 1/2 x10.2 x 12 sin 150°
      =30.6km3
      Area of ranch PQRS
      =62.22 + 30.6
      =92.82km2
  21.                      
    1.                          
      1. A takes  180
                    x +10
      2. B takes    180
                         x
    2. 180 - 180 = 3
        x    x+10   2
      180(x+10) – 180x = 3x(x+10)
                                     2
      360x+3600 - 360x=3x3 + 30x
      x2 + 10x - 1200 = 0
      (x-30) (x +40) = 0
      X= 30 or x = -40
      Speed of A = 30+ 10 = 40
    3. Time taken by A = 48 x 60 = 72min
                                  40
      Time taken by B = 48 x 60 = 96min
                                  30
      Time for B = 96-10 = 86 min
      86 - 72 = 14min
  22.                  
    1.                        
      1. Reflection in the line PR or ER Or PER
      2. Enlargement centre E
        Scale factor - 1
      3. Rotation about pt R
        Through 90°
        Clockwise
    2.                    
      1. R → S
        C → A
      2. R →Q
        C → E
  23.                    
    1. Modal frequency = 8
    2.              
      no of kg of meat  fre.(f)  mid pts(x)  fx  cf 
      1-5 2 3 6 2
      6-10 3 8 24 5
      11-15 6 13 78  11
      16-20 8 18 144 19
      21-25 3 23 69 22 
      26-30 2 28 56 24
      31-35 1 33 33 25
        Σf = 25   Σfx = 410  
      mean = 410 = 16.4
                   25        
    3.  2, 5, 11, 19, 22, 24,25
      Median - 15.5 + 12.5 - 11 x 5
                                    8
      =15.5 + 1.5 x 5
                    8
      =16.4375
  24.                              
    1.                          
      1. Area of base x2
        Or Area of sides = 4xh
        X2 + 4xh = 432
        h = 432 - x2
                   4x
      2. Volume = x2h
        =x2(432 - x2)
    2.                  
      1. Volume (v) = 108x – 1/4x3
        dv = 108 - 3x2 = 0
                        4
        x=12
      2. Vol = 108x - 1/4x3
        =(108 x2) - 1/4 x 123
        =864cm3

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