Mathematics Paper 1 Questions and Answers - Lanjet Mock Exams 2021/2022

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INSTRUCTIONS TO THE CANDIDATES

  • This paper contains two sections; Section I and Section II.
  • Answer all the questions in section I and only five questions from Section II.
  • All workings and answers must be written on the question paper in the spaces provided below each question.
  • Non programmable silent electronic calculators and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.

Section I (50 marks)
Answer All the questions in this section

  1. Evaluate without using tables or calculators (2mks)
    ⁶/₇ of ¹⁴/₃ ÷ 80 x –²⁰/₃
    −2 x 5 ÷ (14 ÷ 7 ) x 3
  2. The line y = 3x + 3 meets the line L1 at the point (2, 9) and at right angles. Find the points at which the two lines intersect with the x- axis. (3mks)
  3. Given that p = 5a – 2b where a = Maths LJM PP1 Q3 2122and b = Maths LJM PP1 Q3b 2122, find :
    1. Column vector p. (2mks)
    2. P' the image of p under a translation vector Maths LJM PP1 Q3bb 2122 (1mk)
  4. Solve the following inequalities and represent the solution on a number line. (3mks)
    4x – 3 ≤ 6x -1 < 3x + 8
  5. The figure below shows a triangular prism of uniform cross section. AF=FB= 4cm. AB=7cm, BC= 12cm. given that angle BAF=30°, find the total surface area of the prism. (3mks)
    Maths LJM PP1 Q5 2122
  6. Use tables of cubes, cubes roots and reciprocals to evaluate; (4mks)
    Maths LJM PP1 Q6 2122
  7. The density of a substance A is given as 13.6g/cm3 and that of a substance B as 11.3g/cm3. Determine, correct to one decimal place, the volume of B that would have the same mass as 50cm3 of A. (3mks)
  8. Find x in 2(4x) – 10 (2x) + 8 = 0 (3mks)
  9. The sum of interior angles of a regular polygon is 24 times the size of the exterior angle.
    Find the number of sides of the polygon. (3mks)
  10. Triangle ABC has its vertices at A(3, 0), B(2,3) and C(5,1) if A΄(5, 0), B΄(3,6) and C΄(9,2) is the image of ABC under enlargement. On the same axes and grid provided below, determine the centre of enlargement and linear scale factor. (3mks)
  11. The following measurements were recorded in a field book of a farm using XY as the base line XY=400M.
        Y    
     C60
     
     
     
     B100
     A120 
     340
     300 
     240
     220
     140
       80
     
     120D 
     100E
     160E
     
     
        X  
    Using a suitable scale, draw an accurate map of the farm. (4mks)
  12. Waceke is a saleslady. She is paid Ksh15,375 per month. She is also paid a commission of 4½ % on the amount of money she makes from her sales. In a certain month, she earned a total of Ksh. 28,875. Calculate the value of her sales that month. (3mks)
  13. Two buses P and Q leave Kisumu at 7.30am and 9.30am respectively. If their speeds are 60km/h and 100km/h respectively,find when Q catches up with P. (3mks)
  14. The figure below shows a solid cone. It has a cylindrical hole drilled into it. The diameter of the hole is 7cm and its height 8cm. The radius of the cone is 10.5cm and its vertical height is 15cm.
    Maths LJM PP1 Q14 2122
    Calculate the volume of the solid. (3mks)
  15. Simplify completely: (3mks)
    (a +2b)² – ( 2a – b)²
           ab² – a²
  16. The figure below represents a speed time graph for a cheetah in 40 seconds.
    Maths LJM PP1 Q16 2122
    Calculate the total distance covered by the cheetah. (3mks)

SECTION II
Answer any five questions in this section.

  1. A rectangular tank whose internal dimensions are 2.4m by 2.5m by 3.7m is two thirds full of juice.
    1. Calculate the volume of the juice in litres. (3mks)
    2. The juice is parked in small packets in a shape of right pyramid with equilateral triangles of sides 20cm. the height of each packet is 15cm. a full packet is sold at kshs. 50 per packet.
      Calculate:
      1. The volume of the juice in cm3 of each packet to the nearest whole number. (3mks)
      2. The number of full packets of juice. (2mks)
      3. The amount of money realized from the sale of juice. (2mks)
  2. The masses in kilograms of patients who attended a clinic on a certain day were recorded as follows.
     67   49   57  58   69   58   39   61   51   47 
     38  59   46   52  60  72  59  49  54  52
     69  62  58  67  63  59  65  58  49  44
     49  41  70  58  54  60  60  59  42   41
     79  52  51  48  54  59  62  73  48  54
    1. Starting with the class 35-39, make a frequency distribution table for the data. (2mks)
    2.  
      1. Calculate the mean mass. (3mks)
      2. Calculate the median mass. (3mks)
    3. On the grid provided draw a frequency polygon to represent the data. (2mks)
  3. The position vectors of A, B and C are 3i – 2j, -6i + 4j and -9i – 3j respectively.
    1. State the column vectors.
      1. AB (2mks)
      2. CB (2mks)
    2. Find the distance from A to C. (2mks)
    3. Find the coordinates of the mid point of AC. (2mks)
    4. If point C' is the image of C under translation vector Maths LJM PP1 Q19d 2122 Find the co-ordinates of C'. (2mks)
  4. Complete the table below for the function y= x2– 3x + 6 in the range -2 ≤ x ≤ 8. (2mks)
     X   −2   −1   0   1   2   3   4   5   6   7   8 
     Y                      
    1. Use the trapezium rule with 10 strips to estimate the area bounded by the curve, y=x2− 3x + 6, the lines x = −2, x =8 and the x – axis. (3mks)
    2. Use the mid – ordinate rule with 5 strips to estimate the area bounded by the curve, y = x2 – 3x + 6, the lines x = −2, x = 8 and the x- axis. (3mks)
    3. By integration, determine the actual area bounded by the curve y = x2 – 3x + 6, the lines x = −2, x = 8, and the x – axis. (3mks)
  5. The corner points A, B, C and D of a ranch are such that B is 8km directly East of A and C is 6km from B on a bearing of 30°. D is 7km from C on a bearing of 30°.
    1. Using a scale of 1cm to represent 1km, draw a diagram to show the positions of A, B, C and D. (4mks)
    2. Use the scale drawing to determine:
      1. The bearing of A from D. (1mk)
      2. The distance BD in kilometers. (2mks)
      3. The perimeter of the ranch in kilometers. (3mks)
  6. Johana and MuchirucontributesKsh. 150,000 and Ksh. 180,000 every year respectively for a business, after one year Jacob joined the business and contributed Ksh. 135,000.
    1. Calculate the ratio of their investment after three years of business. (3mks)
    2. They agreed that 30% of the profits after 3 years be used to cater for the costs of running the businesses while the remaining would be shared proportionally. Calculate each person’s share if the profit after 3 years was shs. 240,000. (4mks)
    3. If each re-invested their shares back into the business, find the new individual investment at the beginning of the fourth year. (3mks)
  7. The figure below shows the end wall of a building with the axes shown and 1M as the unit of measurement. The roof line is given by y= a + bx2, where a and b are constants.
    Maths LJM PP1 Q23 2122
    1. State the value of a. (1mk)
    2. Calculate the value of b. (3mks)
    3. Calculate the values of y for x = 2, 4, 6, 8 and compete the table below. (2mks)
       X   −10   0   2   4   6   8   10 
       Y    9  15            9
    4. Calculate the area of the wall. (4mks)
  8. A ball is thrown upwards and its height after t seconds is 5 meters, where s = 20t – 5t2. Find
    1. The greatest height reached and the time when it is reached. (3mks)
    2. The time when it returns to the original level. (3mks)
    3. Its velocity after 3 seconds. (2mks)
    4. Its acceleration during the throw. (2mks)


MARKING SCHEME

  1. Evaluate without using tables or calculators (2mks)
    ⁶/₇ of ¹⁴/₃ ÷ 80 x –²⁰/₃
    −2 x 5 ÷ (14 ÷ 7 ) x 3
    6/7 x 14/3 = 4
    4 ÷ 80 = 4/80
    4/80 x = 20/3 =−1/3
    −2 x 5 ÷ 2 x 3
    −2 x 5/2 x 3
    −15
    −¹/₃
    −15
    1/3 ÷ −15/1 M1 – num/den
    1/3 x 1/45 M1 – single fraction
    =1/45
  2. The line y = 3x + 3 meets the line L1 at the point (2, 9) and at right angles. Find the points at which the two lines intersect with the x- axis. (3mks)
    G1=3 
    G2= −1/3 
    Equation L2 
    y-9 = -1 
    x-2     3 
    3y – 27 = −x +2
    3y= −x +29 
     y=−1/3 x + 29/3
    at x –axis, y=0
    4 → 0= 3x + 3
    3x = −3
    X = 11
    ( −1, 0) 
    for L2
    0= −1/3x + 29/3
    1/3x = 29/3
    x = 29
    (29,0) 
  3.  
    1.  
      Maths LJM PP1 Ans3a 2122
    2.  
      Maths LJM PP1 Ans3b 2122
  4.  
    1. Ux – 3 ≤6x – 1
      -2x ≤ 2
      x≥-1
    2. 6x -1<3x+8
      3x<9
      X<3
      -1≤x<3
      Maths LJM PP1 Ans4 2122
  5. The figure below shows a triangular prism of uniform cross section. AF=FB= 4cm. AB=7cm, BC= 12cm. given that angle BAF=30°, find the total surface area of the prism. (3mks)
    Area of Δ faces = ½ x 4x 4 x sin120 x 2 
                        = 13.856 
                        = 14cm2
    Area of rectangles = 12 x 4 x 2+12 x 6 
                                  = 96+72
                                  =168
    Total S.A =168 + 13.856
                  = 181.856 accept 182cm2
                   = 181.9cm2
    Alternative
    h=√(42 - 352)
       =√(16-12.25)
       =1.94cm
    Area of triangular
    Faces = ½ x 1.94 x 7x2
    =13.587cm2
    Total = 168+13.587
    =181.58 = 181.6cm2
  6. (2.35 x 101)3 - 3√(4.411 x 103)+       1        
                                                       7.1 x 10-3
    12.978 x 103 – 1.64 x 10 + 0.1408 x 103 
    12,978 – 16.4 + 140.8 = 13,102.4
  7. D = m/v
    Mass of A = 50cm3 x 13.6g/cm3 = 680g
    Mass of B = 680g 
    Density = 11.3g/cm
    Volume of B =     680g    
                            11.3g/cm3 
    = 60.2cm3
  8. 4x − 5(2x) + 4 = 0
    (22)x – 5 (2x)+4 = 0 
    Let 2x be u 
    U2 – 5u + 4 =0  
    (−4, −1) 
    u− 4u – u + 4 = 0
    u(u – 4) -1 (u – 4) = 0
    (u – 1 ) (u – 4) = 0
    u = 1 or 4
    2x = 1 = 20
    X = 0 
    also:
    2x = 4 = 22
    x = 2
    x = 0
    and 
    2
  9. The sum of interior angles of a regular polygon is 24 times the size of the exterior angle.
    Find the number of sides of the polygon. (3mks)
    (2n – 4) 90= 24 x 360/n
    180n – 360 = 8640
                             n
    180n2 – 360n = 8640
    n2 – 2n – 48 =0
    -48 = -8 x 6
    - 2 = -8+6
    (n – 8) (n + 6) = 0
    n=8
    Or
    -6
    8 sides 
  10. Triangle ABC has its vertices at A(3, 0), B(2,3) and C(5,1) if A΄(5, 0), B΄(3,6) and C΄(9,2) is the image of ABC under enlargement. On the same axes and grid provided below, determine the centre of enlargement and linear scale factor. (3mks)
    Maths PP1 LJM Ans 10 2122
  11. The following measurements were recorded in a field book of a farm using XY as the base line XY=400M.
        Y    
     C60 1.5cm 
     
     
     
     B100 2.5cm
     A120 3cm
     340 8.5cm
     300 7.5cm
     240 6cm
     220 5.5cm 
     140 3.5cm
       80 2cm
     
     120D 3cm 
     100E 2.5cm 
     160F 4cm 
     
     
        X  
    Maths LJM PP1 Ans11 2122
  12. Waceke is a saleslady. She is paid Ksh 15,375 per month. She is also paid a commission of 4 ½ % on the amount of money she makes from her sales. In a certain month, she earned a total of Ksh. 28,875. Calculate the value of her sales that month. (3mks)
    Earnings on sales = 28875 – 15375
                                  = sh 13,500
    4 ½ % of y = 13500
    Y= 100 x 13500 
           4.5
    Y = sh 300,000
  13. Two buses P and Q leave Kisumu at 7.30am and 9.30am respectively. If their speeds are 60km/h and 100km/h respectively, find when Q catches up with P. (3mks)
    Distance by p in 2hrs 
    60 x 2 = 120km 
    RS= 100-60 60
    = 40Km/hr
    Time to overtake 
    = 120 = 3hrs  
        40 
    9.30 + 3hrs = 12.30p.m
    alternative method
    Maths LJM PP1 Ans13 2122
    distance by P = y km
    time = (y/60)hrs
    distance by Q = 120 + y
    time = 120 + y
                  100
    100y = 7200 + 60y 
    40y = 7200 
    y= 180km
    time= 180 =3hrs
                60
    time to catch
    up = 9.30 + 3 = 12.30p.m
  14. The figure below shows a solid cone. It has a cylindrical hole drilled into it. The diameter of the hole is 7cm and its height 8cm. The radius of the cone is 10.5cm and its vertical height is 15cm.
    Calculate the volume of the solid. (3mks)
    Volume of the cone – volume of cylinder M1-values subst in formula
    1/3πR2H – πr2h M1-simplifying
    π(1/3 x 10.52 x 15 – 3.52 x 8) A1
    π(551.25 – 98) 
    453.25π
    =1424.5cm3
  15. let a + 2b =x 
    2a – b = y 
    x2−y2 = (x + y) (x−y) 
    (a+2b+2a-b) (a+2b)−(2a−b)) 
    (3a+b) (−a+3b)
    (3a+b) (3b-a) 
    (3b)2 –a2
    (3b+a)(3b-a) 
    (3b+a)(3b-a) 
    (3a+b)(3b–a)
    3a+b
    3b+a
  16. The figure below represents a speed time graph for a cheetah in 40 seconds.Calculate the total distance covered by the cheetah. (3mks)
    Maths LJM PP1 Ans16 2122
    Area of a = ½ x 10 (15+30) 
                    = 225m 
    Area of b = 30 x 10 
                    = 300m = 825m
    area of c = ½ x 30 x 20
                   = 300m
    total = 225 + 300 + 300
           = 825m
  17. A rectangular tank whose internal dimensions are 2.4m by 2.5m by 3.7m is two thirds full of juice.
    1. Calculate the volume of the juice in litres. (3mks)
      2/x 2.4 x 2.5 x 3.7
      14.8m3
      1m3     → 1000L 
      14.8m→ ? 
      14.8 x 1000 
      14800 litres 
    2. The juice is parked in small packets in a shape of right pyramid with equilateral triangles of sides 20cm. the height of each packet is 15cm. a full packet is sold at kshs. 50 per packet.
      Calculate:
      1. The volume of the juice in cm3 of each packet to the nearest whole number. (3mks)
        Base area
        =½ ab sin C
        = ½ x 20 x 20sin 60
        = 173.21
        Vol = 1/3 x 173.21 x 15
             = 866.025
              = 866 cm3 
      2. The number of full packets of juice. (2mks)
        1m3 = 1000000cm3
        14.8m3 = ?
        14.8 x 1000000cm3
        No of packets = 14.8x106
                                     866
        =17089.58 
        =17,089
      3. The amount of money realized from the sale of juice. (2mks)
        17089 x 50 = Sh. 854,450 
  18. The masses in kilograms of patients who attended a clinic on a certain day were recorded as follows.
    1. Starting with the class 35-39, make a frequency distribution table for the data. (2mks)
       Mass   F   Mid(x)   ƒx   cƒ 
       35-39   2    37   74   2
       40-44      42   168  6
       45-49      47   376  14
       50-54      52   468  23
       55-59      57   627  34
       60-64      62   434  41
       65-69      67   335  46
       70-74      72   216  49
       75-79      77    77  50
         Σƒ = 50      Σƒx = 2775   
    2.  
      1. Calculate the mean mass. (3mks)
        X = Σƒx
                Σƒ
        = 2775
             50
         =55.5kg 
      2. Calculate the median mass. (3mks)
        Maths LJM PP1 Ans18b 2122
        = 54.5 +0.91
        = 55.41
    3. On the grid provided draw a frequency polygon to represent the data. (2mks)
      Maths PP1 LJM Ans 18c 2122
  19.  
    1.  
      1. AB (2mks)
        Maths LJM PP1 Ans19ai 2122
      2. CB (2mks)
        CB = B-C
        Maths LJM PP1 Ans19aii 2122
    2.  Find the distance from A to C. (2mks)
      AC = C−A
      Maths LJM PP1 Ans19b 2122
      √145
      = 12.04 units A1
    3. Find the coordinates of the midpoint of AC. (2mks)
      Maths LJM PP1 Ans19c 2122
      = -3i, -2.5j
      Point (-3, -2.5) 
    4. If point C΄ is the image of C under translation vector 1
      Find the co-ordinates of C΄. (2mks)
      Maths LJM PP1 Ans19d 2122
      C΄ = (-8, -6) or -8i -6j
  20.  
     X   −2   −1   0   1   2   3   4   5   6   7   8 
     Y  16  10   6   4   4   6   10   16   24   34   46 
       Y0  Y1  Y2   Y3   Y4   Y5   Y6  Y7  Y8  Y9  Yn
    1. Use the trapezium rule with 10 strips to estimate the area bounded by the curve, y=x2 - 3x + 6, the lines x = -2, x =8 and the x – axis. (3mks)
      Let h=1
      = ½h (Y0 +Yn +2 (y1+y2+…)
      =0.5(16+46+2(10+6+4+4+6+10+16+24+34)
      =0.5 (62+228)
      =145 square units
    2. Use the mid – ordinate rule with 5 strips to estimate the area bounded by the curve, y = x2 – 3x + 6, the lines x = -2, x = and the x- axis. (3mks)
      A = h(y1+y2+……..yn)
      = 2 (10 + 4 + 6 + 16 + 34)
      = 140 sq units
    3. By integration, determine the actual area bounded by the curve y = x2 – 3x + 6, the lines x = -2, x = 8, and the x – axis. (3mks)
      Intergrate form -2 to 8 (x2 – 3x +6).dx
      Maths LJM PP1 Ans20c 2122
      (170.7 – 96 + 48) – (-4 -6-12)
      122.7 + 22 M1
      = 144.7 sq units 
  21. The corner points A, B, C and D of a ranch are such that B is 8km directly East of A and C is 6km from B on a bearing of 30°. D is 7km from C on a bearing of 300°.
    1. Using a scale of 1cm to represent 1km, draw a diagram to show the positions of A, B, C and D. (4mks)
      Maths LJM PP1 Ans21 2122
    2. Use the scale drawing to determine:
      1. The bearing of A from D. (1mk)
        = 180 + 30
        = 201°
        ±1°
      2. The distance BD in kilometers. (2mks)
        8.2cm
        8.2 x 1km
        =8.2km B1
        ±0.1
      3. The perimeter of the ranch in kilometers. (3mks)
        AD = 10cm = 10km
        P = 8km + 6km +7km + 10km 
        =31km 
  22. Johana and Muchiru contributes Ksh. 150,000 and Ksh. 180,000 every year respectively for a business, after one year Jacob joined the business and contributed Ksh. 135,000.
    1. Calculate the ratio of their investment after three years of business. (3mks)
      Muchiri’s contribution = 180000 x 3 = 540,000
      Johana’s contribution = 150000 x 3 = 450,000
      Jacob’s contribution = 135000 x 2 = 270,000
      Ratio Johana: Muchiri: Jacob 
      = 450,000: 540,000: 270,000
      = 5:6:3 
    2. They agreed that 30% of the profits after 3 years be used to cater for the costs of running the businesses while the remaining would be shared proportionally. Calculate each person’s share if the profit after 3 years was shs. 240,000. (4mks)
      30/100 x 240000 = 72000
      Remainder = 240000 - 72000
                         = 168000
      5:6:3
      5 + 6 + 3 =14
      Johana’s = 5/14 x 168000 = 60000
      Muchiri = 6/14 x 168000 = 72000
      Jacobs = 3/14 x 168000 = 36000 
    3. If each re-invested their shares back into the business, find the new individual investment at the beginning of the fourth year. (3mks)
      Johana’s = 450000 + 60000 = 510000 
      Muchiri’s = 540000 + 72000 = 612000
      Jacob’s = 270000 + 36000 = 306000
  23. The figure below shows the end wall of a building with the axes shown and 1M as the unit of measurement. The roof line is given by y= a + bx2, where a and b are constants.
    1. State the value of a. (1mk)
      A = y intercept
      A= 15
    2. Calculate the value of b. (3mks)
      Y = 15 + bx2
      At x = −10, y=9 (−10,9)
      9= 15 + (−102)b
      9= 15 + 100b
      100b = 9 − 15
      =−6 
      B=−6/100 = −3/50 
      Or = 0.06
    3. Calculate the values of y for x = 2, 4, 6, 8 and compete the table below. (2mks)
       X   −10   0   2   4   6   8   10 
       Y    9  15   14.76   14.04   12.84   11.16    9
      Y = 15 – 3x2
                     50
    4. Calculate the area of the wall. (4mks)
      10(15 – 3x2) dx
                   50
      Maths LJM PP1 Ans23d 2122
  24. A ball is thrown upwards and its height after t seconds is 5 meters, where s = 20t – 5t2. Find
    1. The greatest height reached and the time when it is reached. (3mks)
      Velocity (v) = ds
                            dt
      v = 20 – 10t
      at maximum, v=0
      20 – 10t =0
      T=2 seconds 
      S=20(2) – 5 (22)
      = 40 – 20 
      =20m 
    2. The time when it returns to the original level. (3mks)
      At original level1 s=0
      20t – 5t2=0
      5t (4 – t) =0 M1-eqn
      T=0 or 4 sec M1
      After 4 seconds A1
    3. Its velocity after 3 seconds. (2mks)
      V = 20 – 10t
      V = 20 – 10(3)
      = 20 – 30 M1-substuted
      = – 10m/s A1
    4. Its acceleration during the throw. (2mks)
      If V= 20 – 10t
      Acceleration = dv/At  
      =– 10ms– 2

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