Mathematics Paper 1 Questions and Answers - Mathioya Mock Exams 2022

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INSTRUCTIONS TO CANDIDATES

  • The paper contains two sections: Section I and II
  • Answer ALL questions in section I and ONLY FIVE questions from section II.
  • All working and answers must be written on the question paper in the spaces provided below each question.
  • Marks may be awarded for correct working even if the answer is wrong.
  • Negligence and slovenly work will be penalized.
  • Non-programmable silent electronic calculators and mathematical tables are allowed for use.

SECTION 1 (50 Marks)
Answer ALL questions from this section

  1. Four wooden poles have lengths 280cm, 336cm, 476 and 420cm. The owner wishes to cut them into shorter pieces of equal length. Find the greatest possible length of each piece if no wood is left over. (3mks)
  2. simplify; 3/5 ÷ 2/3 -1/6 x 7/12 of (1/2 + 4/5) (4mks)
  3. A rectangular lawn measures 50m by 40m. There is a concrete path of width 100cm all round it. what is the area of the path (3mks)
  4. The length of the major and minor arcs of a circle are 154cm and 110cm respectively. Taking π =22/7, find the angle subtended at the centre by the minor arc. (4mks)
  5. Without using mathematical tables, evaluate (2mks)
    Log1020 + Log1040 - Log108
  6. A trader bought 360 trays of eggs at sh120 per tray. He later discovered that 10% of the eggs were bad and could not sell them. For how much must he sell the good eggs per dozen in order to make a profit of 50% (4 mks)
  7. Use matrices to solve the simultaneous equations (3 mks)
    4x -5y=13
    -2y+ 3x=8
  8. A form IV maths teacher originally worked out the mean mark of her 30 students to be 41. After the correction of the test, she added some marks to Amina, Nduku and Karimi in the ratio 2:3:4. If the new mean mark for the class is 42.5, determine how many more marks Karimi was added than Nduku (4mks)
  9. Find the value of K so that the expression 25x2 – 10x + 1/8 +k is a perfect square (3mks)
  10. Find all the integers satisfying the inequalities 3-2x < x-3 ≤4 (3mks)
  11. The cross-section of a head of a bolt is in the form of a regular hexagon as shown below. Determine the area of the cross-section to 2 dp (3 mks)
    Mathp1q11
  12. A truck left Kisumu to Mombasa a distance of 1000km at 11.30am on Monday at a speed of 60km/hr. On the way it stopped for 2 ⅓hrs. At what time did it reach Mombasa (3 mks
  13. Solve the equation (3 mks)
    4/x-1=    3 - x   
                 x+3
  14. The wheel of a bicycle has a diameter of 70cm and is rotating at the rate of 200 revolutions per minute. Calculate the speed of the bicycle in km/hr. (3mks)
  15. Given the column vectors Mathp1q15 and that p=2a - 1/3 b +c, express p as a column vector and hence calculate its magnitude to 3 significant figures. (3mks)
  16. A commission agent gets 15% commission on the total value of goods sold. In one month he earned sh11,250 as commission. Calculate the total value of goods sold (2mks)

SECTION 1I (50 Marks)
Answer ANY FIVE questions from this section

  1. A bus left Mombasa and travelled towards Machakos at an average speed of 60km/h. After 2 ½ hrs, a car left Mombasa and travelled along the same road at an average speed of 100km/hr. If the distance between Mombasa and Machakos is 500km, determine:
    1.        
      1. The distance of the bus from Mombasa when the car took off (2mks)
      2. The distance the car travelled to catch up with the bus (4mks)
    2. Immediately the car caught up with the bus, the car stopped for 25 minutes. Find the new average speed at which the car travelled in order to reach Machakos at the same time as the bus. (4mks)
  2.  
    1. On the graph paper provided, plot the triangle whose co-ordinates are A(1,2) B(5,4) and C(2,6). (1mk)
    2. On the same axes
      1. Draw the image A1B1C1 of ABC under a rotation of 90° clockwise about the origin (2mks)
      2. Draw the image A11B11C11 of A1B1C1 under a reflection in the line y=-x. State the coordinates of A11B11C11 (3mks)
    3. A111B111C111 is the image of A11B11C11 under the reflection in the line x=o. Draw the image A111B111C111 and state its coordinates (2mks)
    4. Describe a single transformation that maps A111B111C111 onto ABC (2mks)
  3.  
    1.      
      1. The gradient function for a certain curve is 3x2 -4x -15. Determine the gradient of the curve at x= -5 (2mks)
      2. Determine the coordinates of the turning points of the curve (3mks)
    2. A particle moving along a straight line covers a distance of S metres in t seconds from a fixed point O on the line where S=t3 _6t2 +8t – 4. Find;
      1. the acceleration when t=5 seconds (3mks)
      2. the time when the velocity of the particle is constant (2mks)
  4. Four towns P,Q,R and S are such that town Q is 120km due east of town P. Town R is 160km due north of town Q. Town S is on a bearing of 330° from P and on a bearing of 300° from R. Use a ruler and compasses only for all constructions in this question.
    1. Taking a scale of 1cm=50km, construct a scale drawing to show the positions of towns P,Q, R and S (6mks)
    2. Use the scale drawing to determine
      1. the distance SP (1mk)
      2. The distance SR (1mk)
      3. The bearing of town S from town Q (2mks)
  5. Three variables P,Q and R are such that P varies directly as Q and inversely as the square of R.
    1. When P=18, Q=24 and R=4 . Find P when Q=30 and R=10 (4mks)
    2. Express P in terms of Q and R (1mk)
    3. If Q is increased by 20% and R is decreased by 10% find:
      1. A simplified expression for the change in P in terms of Q & R (3 mks)
      2. The percentage change in P (2 mks)
  6. In the figure below OA=a and OB=b. M is the midpoint of OA and AN: NB= 2:1
    Mathp1q22
    1. Express in terms of a and b:
      1. BA (1 mk)
      2. NB (1mk)
      3. ON (2mks)
      4. Given that BX= hBM and OX=kON determine the values of h & k (6mks)
  7. The table below gives the marks obtained in a chemistry test by 47 students. By completing the table and using 55 as the assumed mean, calculate the measures that follow.
     Marks No. of students  Midpoint
    (x) 
    d = (X - A)  fd  fd2  CF 
     41 - 45          
     46 - 50          
     51 - 55 14           
     56 - 60 13           
     61 - 65          
     66 - 70          

    1. The actual mean mark (2mks)
    2. The median mark (2mks)
    3. The semi-interquartile range (3mks)
      1. The standard deviation (3mks)
  8. The diagram below represents a conical vessel which stands vertically. The vessel contains water to a depth of 30cm. The radius of the surface of water in the vessel is 21cm. (Take π =22/7)
    Mathp1q24
    1. Calculate the volume of the water in the vessels in cm3 (3mks)
    2. When a metal sphere is completely submerged in the water, the level of the water in the vessel rises by 6cm.
      Calculate:
      1. The radius of the new water surface in the vessel (2mks)
      2. The volume of the metal sphere in cm3 (3mks)
      3. The radius of the sphere (3mks)

Marking Scheme

  1. Find GCD of 280cm, 336cm, 476cm and 420 cm
     2 280  336  476  420 
     2 140  168  238  210 
     7 70  84  119  105 
       10 12  17  15 
    The greatest possible length is : 2 x 2 x 7 = 28cm
  2. 3/5 ÷ 2/3 - 1/6 x 7/12 of (1/2 + 4/5)
    = 3/5 ÷ 2/3 - 1/6 X 7/12 of 13/10 
    3/5 ÷ 2/3 - 1/6 X 7/12 x 13/10
    =  3/5 x 3/2 - 1/6 X 7/12 x 13/10
    = 9/10 - 91/720
    = 648 - 91  557   
           720           720
  3. Length of outer rectangle = 50 + 1+ 1= 52m
    Breath of outer rectangle = 40 + 1 + 1 = 42cm
    Area of lawn togther with path = 52 x 42 = 2184
    Area of lawn = 50 x 40 = 2000
    Area of patrh alone = 2184 - 2000 = 184m2
  4. Length of major arc + minor arc = circumference
    2πr = 154 + 110
    2 x 22/7 xr = 264
    r = 264 x 7/22 x 1/2 = 42cm
    If the angle is x , then:
    x/360 x 2πr = arc length
    x/360 x 2 x 22/7 x 42 = 110
    x = 110 x 7 x 360
            2 x 22 x 42
    = x = 150
  5. Log1020 + Log1040 - Log108 = Log10 (20 x 40/8)
    Log10100
    = 2
  6. Good eggs: 90/100 x 360 = 324
    1 tray = 30 eggs
    324 x 30  = 810 dozens
       12 
    cost of eggs = 360 x 120
    Selling    150    x 360 x 120
                   100           810
    = 80
  7.   
    Mathp1qa07
    coeff matrix M= (4 -5 3 -2)
    det M = (4 x -2) - (3x - 5)
    = (-8) - (-15) = 7
    Mathp1qa07b
  8.  Original total marks = 41 x 30 = 1230
    New total marks = 42.5 x 30 = 1270
    ∴ marks added = 1275 - 1230
    = 45
    Karimi got 4/9 x 45 = 20
    Nduku got 3/9 x 45 = 15
    Hence karimi got 5 more
  9.   b²   = c
      4a
    1/3 + k = 1
      (-10)²  = 1/3 + k 
    4 x 25
     100/100 = 1/3 + k
    1 = 1/3 + k 
    k = 2/3
  10. 3-2x < x-3     
    3 + 3 < 3x
    6< 3x
    2<x

    x-3 ≤4
    <  7
    2 < x < 7
    integers {3, 4, 5, 6, 7}

  11. Tan 60° = 5/x
    x = 50 = 2.88675
    Tan 60°
    Sin 60° = 5/L
    L = 5/Sin 60° = 5.7735
    Area = (5.7735 x 10) + (1/2 x 10 x 2.88675) x 2
    = 57.735 + 28.8675
    = 86.6025
    = 86.60
  12. Total time taken for journey:
    1000/60 + 21/3 = 19 hrs
    1130 + 1900
    = 3030 - 2400
    = 0630
    ∴ Tuesday at 6:30am
  13. 4/x - 1 =    3 - x   
                    x + 3
    4 - x  3 - x  
      x          x + 3
    Cross multiply:
    (x + 3)(4 - x) = x (3 - x)
    4x - x2 + 12 - 3x = 3x - x2
    x + 12 = 3x
    12 = 2x
    ∴ x = 6
  14. Distance moved in 1 revolution = πd
    Distance covered in 1 minute = 200πd
    = 200 x 22/7 x 70cm
    Distance moved in 1 hour = 200 x 22/7 x        70 x 60    
                                                                          100 x 100
    hence speed = 26.4km/hr
  15.   
    Mathp1qa15
    ∴ |P| = √((-3)² + (-1)² + (2)²)
    =√(9 + 1 + 4)
    = √14
    = 3.74
  16. 15% of goods sold = 11250
    100% of goods sold = ?
    11250 x 100
        15
    = 75, 000
  17.      
    1.          
      1. 2½ = 2.5 x 60 = 150km
      2. Time taken =        150        
                                  100 - 60
        = 3.75 - 3 hrs 45 minutes
        Distance = 3.75 x 100 = 375km
    2. Time taken by bus:
      125/60 
      = 21/12
      21/12 hrs of 2hrs 05 monutes
      25/12 - 25/60 = 12/3hrs
      Spd = 125/ 12/3 = 75km
  18.      
    Mathp1qa18
  19.         
    1.          
      1. 3 (25) + 20 - 15
        75 + 20 - 15 = 80
      2. 3x2 - 4x - 15 = 0
        3x2 - 9x + 5x - 15 = 0
        3x (x -3)+ 5(x -3)= 0
        (3x + 5)(x - 3)= 0
        3x + 5 = 0
        x =3, x = -5/3
        Coordinates of turning point 
        (-5/3 , 670/27) (3, -26)
    2.       
      1. ds/dt = 3t2 - 12t + 8
        dv/dt = a = 6t - 12
        When t = 5
        a = 30 -12
        a = 18m/s2
      2. At constant velocity, acceleration = 0
        6t - 12 = 0
        6t = 12
        t = 2 seconds
  20.           
    1. Scale: 1cm = 50km 
      Mathp1qa20a   
    2.        
      1. Distance SP = 7.9 x 50 = 395km (±5km)
      2. Distance SR = 7.4 x 50 = 370km (±5km)
      3. Bearing of S from R = 270° + 47° = 317° (±1)
  21.      
    1. P α Q/R2
      P = kQ/R2
      K 24/42 = k = 18 x 4²
                                24
      K= 12
      Q = 30 and r = 10
      P = 12Q/R2
      12 x 30/102 = 3.6 
    2. P = 12Q/R2
    3. P = K 120/1002
            _________   
              (90/100R)2
      P = K   1.2Q   ⇒ P = K1.48Q/R2
                  (0.9R)2
    4. P = 1.48 (KQ/R2)
      P increases by 48%
  22.       
    1.    
      1. BA
        →       →     →
        BA =  BO + OA
        = -b + a
        = a - b
      2. NB
                       →
        NB = 1/3AB
        = 1/3 b - 1/3a
      3. ON
         →      →     →
        ON = OA + AN
        = a + 2/3AB
        = 1/3a + 2/3b

    2. OX = k (1/3a + 2/3b)

      OX = 1/3ka + 2/3kb
       →       →   →
      OX = OB + BX
      = b + h (1/2a - b)
       →
      OX = (1 - h)b + 1/2ha
      1/3ka + 2/3kb = b - bh + 1/2ha
      2/3k = 1 - h
      1/2h = 1/3k
      h = 1/2
      k = 3/4
  23.      
     Marks No. of students  Midpoint
    (x) 
    d = (X - A)  fd  fd2  CF 
     41 - 45 43  -12  -36  432 
     46 - 50 48 -7  -49  343  10 
     51 - 55 14   53 -2  -28  56  24 
     56 - 60 13   58 39  117  37 
     61 - 65  63 56  448  44 
     66 - 70  68 13  39  507  47 
      47     21 1903  
     Mean mark

    X = 55 + 21/47 = 55.45
    Median mark
    50.5 + (24 - 10) x 5
                    14
    = 55.5
    Semi- interquartile range
    Q1 = 50.5 + (12 - 10) x 5 = 51.21
                            14
    Q3 = 55.5 + (36 - 24) x 5 = 60.12
                            13
    S.I.Q.R = (60.12 - 51.21)
                             2
    = 4.453
    Standard deviation
    S= √ ((1903/47) - (21/47)2)
    S= 6.347

  24.     
    1. Volume of water = 1/3 x 22/7 x 212 x 30
      = 13, 860cm3
    2.      
      1. h/H = r/R
        30/36 = 21/R ⇒ 21 x 36 = 25.2cm
                                       30
      2. New volume = 1/3 x 22/7 x 25.22 x 36
        = 23950.08cm3
        Volume of sphere = 23950.08 - 13860
        = 10090.08cm3
      3. 4/3 x 22/7 x r3 = 10090.08
        r3 = 10090.08 x 21
                   88
        r = ³√(10090.08 x 21/88)
        = 13.40cm

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