Mathematics Paper 1 Questions and Answers - Royal Exam Series Post Mock Trial Exams 2022

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MATHEMATICS
PAPER 1

INSTRUCTION TO CANDIDATES:

  • This paper consists of two Sections; Section I and Section II.
  • Answer ALLthe questions in Section I and any 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 may be used, except where stated otherwise.
  • andidates 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.


Questions

SECTION I:(50 MARKS)
Answer all the questions in this section.

  1. Evaluate without using mathematical tables. (3mks)
       1.9  x  0.032   
       20  x  0.0038
  2. Use tables of reciprocals only to find the value of (3mks)
         5         14     
    0.0829      0.581
  3. You are given that cos θ= 8/10 . Without using mathematical tables express in fraction form the value of
    1. Sin θ (2mks)
    2. tan (90 - θ) (1mks)
  4. An open right circular cone has radius of 5cm and a perpendicular height of 12cm .
    Calculate the surface area of the cone correct to 1 decimal place.(Take π to be 3.142) (3mks)
  5. Nyongesa spends a total of sh.970 on buying three text books and five pens. If he had bought two text books and five pens he would have saved sh.90. Find the cost of one text book. (3mks)
  6. A Kenyan tourist left Germany for Kenya through Switzerland. While in Switzerland he bought a watch worth 52 Deutche marts. 1 swiss Franc = 1.28 DM and 1 Swiss Franc = 45.21 Kenya shillings.
    Find the value of the watch in;
    1. Swiss Franc (2mks)
    2. Kenya shillings using the exchange rates above. (2mks)
  7. The figure below is a velocity –time graph for a car that was in motion for 24 seconds.
    MathPstMp1q7
    1. Find the total distance traveled by the car? (2mks)
    2. Calculate the deceleration of the car. (2mks)
  8. Form the quadratic equation whose roots are x= - 5/3 and x=1 in the form ax2 +bx+c=0. (2mks)
  9. Three towns are situated in such a way that town B is 40km due south of town A and town C is 30 km due East of town B.
    1. Draw a sketch diagram showing the position of town A.B and C. (1mk)
    2. From your sketch, calculate:
      1. Distance AC (1mk)
      2. To the nearest degree the bearing of town A from town C. (2mks)
  10. Find the inequalities that defines the region R shown in the figure below (3mks)
    MathPstMp1q10
  11. The interior angle of a regular polygon is 20º more than three times the exterior angle. Determine the number of sides of the polygon. (3 marks)
  12. The circle below whose area is 18.05cm2 circumscribes a triangle ABC where AB=6.3cm, BC =5.7cm and AC =4.8cm. Find the area of the shaded part. (3mks)
    MathPstMp1q12
  13. Solve for x in the equation. (3mks)
    9(x-1) x 3(2x+1) = 243   
  14. The graph below shows frequency densities for the masses of some 200 students selected from a class. Use it to answer the questions that follow:
    MathPstMp1q14
    1. Complete the frequency distribution table below (2mks)
      Mass in (Kg)                                      
      Frequency          
    2. State the modal frequency (1mk)
  15. The volumes of two similar solid cylinders are 4752 cm3 and 1408 cm3 .If the area of the curved surface of the smaller cylinder is 352 cm2 , find the area of the curved surface of the larger cylinder (3mks)
  16. The line which joins the point A (3, K) and B (-2, 5) is parallel to the line whose equation is 5y+2x- 7=0. Find the value of K. (3mks)

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

  1. John bought 3 brands of tea, A B and C. The cost price of the three brands were sh 25, sh 30, sh 45 per kg respectively. He mixed the three brands in the ratio 5:2:1 respectively: After selling the mixture he made a profit of 20%.
    1. How much profit did he make per kilogram of the mixture? (4 mks)
    2. After one year the cost price of each brand was increased by 10%
      1. For how much did he sell one kilogram of the mixture to make a profit of 15%? (Give your answer to the nearest 5 cents) (3 mks)
      2. What would have been his percentage profit if he sold one kilogram of the mixture at sh. 45. (3 mks)
  2. The equation of a curve is given as y=2x3 - 9/2x2 -15x+3
    1. Find:
      1. the value of y when x=2 (2 marks)
      2. the equation of the tangent to the curve at x=2 (4 marks
    2. determine the turning points of the curve (4 marks)
  3.        
    1. Using the trapezium rule with 7 ordinates, estimate the area of the region bounded by the curve y=- x2 +6x+1, the lines x=0, y=0 and x=6. (5mks)
    2. Calculate :
      1. The area of the region in (a) above by the integration; (3mks)
      2. The percentage error of the estimated area to the actual area of the region, correct to 2 decimal places. (2mks)
  4. A country bus left town A at 11.45 am and traveled towards town B at an average speed of 60km/h.Amatatu left town B at 1.15 pm on the same day and traveled towards town A along the same road at an average speed of 90km/h. the distance between the two towns is 540km. Determine
    1. The time of day when the two vehicles met (4mks)
    2. How far from town A did they meet ? (2mks)
    3. How far outside town B the bus was when the matatu reached town A (4mks)
  5. A straight line passing through the points (8,-2) and (4,-4) has its equation in the form ax + by + c = 0, where a, b and c are integers.
    1. Determine the numerical values of a, b and c. (3mks)
    2. If the line in (a) above cuts the x-axis at point P, determine the coordinates of P. (2mks)
    3. Another line, which is perpendicular to the line in (a) above passes through point P and cuts the y axis at Q. determine the coordinates of point Q. (3mks)
    4. Find the length of QP (2mks)
  6.       
    1. The figure below shows a metal solid consisting of a right cone mounted onto a hemisphere. The height h of the cone is twice the radius r. if the volume of the solid is 36cm3 , find the radius of the hemisphere. (4mks)
      MathPstMp1q22
    2. The solid is totally immersed in water contained in a cylindrical tin of radius 9cm. Through what height does the water level in the tin rise? (2mks)
    3. The solid is melted and recast into a right pyramid of vertical height 4.2cm. Find the base area of the pyramid. (2mks)
    4. If the solid is of mass 14.4g. Find its density in kg/m3. (2mks)
  7. In the diagram below, AOB is a triangle such that OAa, OB = b and <AOB is obtuse.
    MathPstMp1q23
    If P is a point on AB such that AP = 3PB and Q is the mid-point of OA
    1. Express in terms of a and b
      1. AB (1mk)
      2. BQ (2mks)
    2. If X is a point on BQ such that BX =hBQ, express OX in terms of a b and h, where h is a scalar. (3mks)
    3. Given further that OX = kOP where k is another scalar, obtain the values of h and k. (4mks)
  8. The table below shows the length of 40 seedlings.
    Length in (mm) Frequency
    118-126 3
    127 – 135 4
    136 – 144 10
    145 – 153 12
    154 – 162 5
    163 – 171 4
    172-180 2
    Determine
    1.        
      1. The modal class (1 Mark)
      2. The median class (2 mks)
    2.       
      1. The mean of the seedlings (4 mks)
      2. The median of the seedlings (3 mks)


Marking Scheme

  1. 1.9 x 0.032 x 10000
    20 x 0.0038 x 10000
    19 x 32
    20 x 38
    16/20
    8/10
     
  2.    5 x          1            - 14 x           1             
               8.29 x 10-2                5.81 x 10-1
    5 x 0.1206 x 102 -14 x 0.1721 x 10
    =60.3-24.094
    =36.206

  3. Cos =    Ads     8   
                 hps          10
    QR=8 and PQ=10
    PR2 =PQ2 -QR2 =102 -82
    PR=√36 = 6

    1. sin θ = 6/10 or 3/5
    2. tan (90-θ) =QR/PR = 8/6 or 4/3
  4.            
    MathPstMp1qa4
    SA= πrl
    1= √52 +122
    = √169
    =13cm
    =3.142x13x5
    =188.5
  5. Let cost of text bk be t and pen p.
    3t  +5p = 970.......(i)
    2t + 8p = 880.......(ii)
    Multiply e.g(i) by 8 and (ii0 by 5
    24t + 40p = 7760
    10t+40p=4400
    14t=240
    Cost of 1 text bk=sh.240
  6.          
    1. 1.28 Deutche marks=1 Swiss franc
      52 Deutche marks =?
         52 x swiss franc
      1.28 Deutche marks
      =40.635
      ≈40.63 Swiss francs

    2. I swiss franc =42.21Ksh.
      40.63 Swiss franc=?
      42 .21 x 40.63
               1
      =1,714.99 Kenya shillings

  7.         
    1. Dist= 1⁄2   x 4 x 80 + 80 x 16 + 1⁄2 x 4 x 80
      =160+1280+160
      =1600m

    2. Deceleration= 0-80/4 = 20m/s2

  8. Soln:x= -5/3 and x=1
    = -5 and x=1
    ∴ (3x+5)(x-1)=0
    3x(x -1)+5(x-1)=0
    3x2 -3x+5x -5=0
    3x2 +2x -5=0

  9.        
    1. Diag,
      Bearing of A from C
      =270+ θ
    2.      
      1. AC2 =AB2 +BC2 =402+302
        =√ 2500
        =50km
      2. sin θ = 40/50 = 0.8
         θ =53.13
        ≈ 53º
  10.  L1 = x>0
    L2 = x/5 + y/4 =1
    4x + 5y = 20
    4x + 5y < 20   
    L3: x/4 + y/-3 = 1
    3x - 4y=12
     ∴3x -4y <12   
       
  11. Let ext. < xº
    Interior < = 3x + 20º
    4x = 160º
    4x = 160º
    x = 40
    360/40= 9 sides

  12.  
    s = √s(s-a)(s-b)(s-c)
    s = 1/2(4.8 + 6.8 + 5.7)= 8.4
    Area = √8.4(2.7)(3.6)(2.1)
    18.05 - 13.0943 
    = √171.4608 
    = 13.0943
    =4.9557
    ≈4.96

  13. Change to base 3
    32(x-1) x 3(2x+1) =35
    2(x-1)+2x+1=5
    2x-2=2x+1=5
    4x-1=5
    4x=6
    x=1.5

  14.          

    1. Mean in kg  41-50 51-55 56-65 66-70 71-80
      Frequency 20 70 50 50 10
    2. Modal frequency is 70.

  15. L.S.F=V.S.F
    lsf= √4752/1408
    =1.5
    l.s.f2 =A.S.F
    1.52 = Area of larger cylinder
                         352
    Area of larger cylinder 792cm2

  16. Reduce 5y+2x=7 in the form y=mx+c
    ∴ y= -2/5x + 7/5
    But gradient =
       k-5      k-5   
       3-2         5
      k-5    2 
      5          3
    k=3

  17.        
    1.    5 x 25 + 2 x 30 + 45 x 1 = 28.75
                       8
      Profit 28.75 x 20/100 = sh5.75

    2.        
      1. New price = 28.75/100 x 110 x 115/100 = 36.40
      2. 28.75 x 110/100 = 31.60
        New profit 45 - 31.60 x 100
                             31.60
         13.4   x 100
            31.60
        =42.41%
  18.             
    X -3 -2 -1 0 1 2 3 4
    X2  9 4 1 0 1 4 9 16
    2X2 18 8  2  0  2  8  18 32
    -3X 9  6  3  0  -3  -6  -9 -12
    -5  -5  -5  -5  -5  -5 -5   -5 -5
     Y  22  9  0  -5  -6  -3  4 15

    minimum at point(1,6)
    y= 2 x 2 - 5x -5
    0=  2 x 2 - 3x - 5
    Y=0
    X=-1 or x=2.5
    y= 2x2 - 3x -5
    0= 2 x 2 -5x -3
    Y= 2x -2
    X= 0.5 or x=3
    MathPstMp1qa18
  19.           
    MathPstMp1qa19
    1. XV=MO2 = 300sin 70 = 281.9cm

    2. YBW = 140/360 x 2πr
      MO1 = 300 Cos 70 = 102.6cm
      Radius of small circle = 180 - 102.6 = 77.4cm
      VBW = 140/360 x 2 x 3.142 x 77.4 = 189.1cm

    3. XAY = 220/360 x 2π x 180
      = 220/360 x 2 x 3.142 x 180 = 691.2cm

    4. Length of conveyor belt 
      = 281.9 x 2 + 189.1 + 691.2
      =1444.1cm2
  20.                      
    1. 11.45 = 11/2h
      Distance moved by bus = 1.5 x 609 = 90km
      Distance between 2 vehicles
      540 - 90=450
      R. speed = 60 + 90 = 150km/h
      T.taken to meet = 450 ÷ 150 = 3h
      = 1.15 + 3 = 4:15pm

    2. Total time moved by bus when they meet
      16:15 - 11:45 = 41/2
      Distance from A = 4.5 x 60 = 270km

    3. Time taken by matatu to travel from B to A = 540 ÷ 90 =6
      ∴ Time matatu reaches A 
      13:15 + 6 = 19:15hrs
      ∴Time travelled= 19:15-11:45 = 71/2 hrs
      distance = A= 7.5 x 60
      Hence:
      540 - 450 = 90km
  21.                   
    1. y + 2 = -4 + 2  1   
        x-8      4-8          2
      2y + 4 = x -8
      ⇒ x - 2y - 12 = 0
      ∴ u =1, b= -2 and c= -12

    2. y = x/2 -6 , when y=0
      x/2 - 6= 0 ⇒ x=12
      ∴ coordinates of pane (12,0)

    3.       y      = -2
        x -12
      ⇒ y = 24 - 2x
      but when x = 0, y=24

    4. ∴ coordinates of q are(0,24)
      QP = √(12-0)2 + (0-24)2
      =√144 + 576 = 26.83 units
  22.                 
    1. Volume of solid =volume of cone + vol of hemisphere
      =1/3πr2h + 1/2x4/3πr3 = 36
      = 1/3πr2 + 2/3πr3=36
      =4/3πr3= 36
      r3      36 x 3      
                 4 x 3.142
      r3 = 8.593
      r= 2.048cm 

    2. Let the level rise be x
      πr2x = 36
      x =      36         
             3.142 x 81
          36        = 0.14cm
      254.502

    3. Volume = 1/2 Ah
      = 1/3 x A x 4.2 = 36
      A = 36 x 3
              4.2
      = 25.71cm2

    4. Density = mass/ volume
      = 14.4/36 x 1000kg/m3
      = 400kg/m3
  23.                   
    1.          
      1. AB = b - a
        BQ = BO = OQ = 1/2 a - b

      2. BX = h(1/2 a - b)
        OX = b + h (1/2 ab)
        1/2ha + (1-h)b

    2. OX = ROP
      but OP = 1/4a + 3/4b= 1/4(a +3b)
      OX = R/4 (a + 3b)
      OX = OX2
      1/2ha + (1-h)b = 1/4Ra = 3/4kb
      1/2h = 1/4⇒(-4-3h= -4)
      4h-2k=0
      4h-3k=-4
      -5k=-4
      k=4/5
      h=2/5
  24.    
    Length
    (cm)
    Mid pt 
    (x)
    F xf CF
    118-135 122 3 366 3
    127-135 131 4 524 7
    136-144 140 10 1400 17
    145-153 149 12 1788 29
    154-162 158 5 790 34
    163-171 167 4 668 38
    172-180 176 2 352 40
        40  5888  

    1.          
      1.   Modal class ⇒145-153
      2. Median class 145 -153
    2.           
      1. Mean of feeding 
        X= Σxf/Σf
        = 5888/40
        = 147.2

      2. Median = L( (ⁿ ⁺ ¹)/₂ - c/f) i 
        = 144.5 + ((⁴¹/₂ - 17)/12)9
        =144.5 + 20.5 - 17 x 9
                          12
        144.5 + 2.625
        =147.125

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