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

Instructions to candidates

• The paper contains TWO Sections: Section I and Section II.
• Answer ALL the questions in Section I and only five questions from Section II.
• All answers and working must be written on the question paper in the spaces provided below each question.
• Show all the steps in your calculations, giving your answers at each stage in the spaces 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.
• Answer all the questions in English.

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

1.  
   1. Evaluate 825 × 4 + 144 ÷ of (−36 + 30)                    (2 marks)
   2. Write the total value of the digit in the thousands place of the results obtained in (a) above  (1 mark)
2. Given the inequalities 2x − 3 ≤ 4x + 7 < x + 13 solve the inequalities and represent the solution on a number line. (3 marks)
3. Simplify completely     (3 marks)
         8x² − 2     
   2x³ −7x² + 3x
4. The figure below shows a sector of a circle centre O.
   
   Determine
   1. The radius OA of the circle (2 marks)
   2. The perimeter of the sector (2 marks)
5. Solve for in the equation (3 marks)
   (¹/₂₇)^x ÷ 81 = ³√729
6. Christmas bulbs made of different colours are set to light after 8 seconds, 10 seconds and 14 seconds. How many times will they light simultaneously in one hour if they start together?    (3 marks)
7. A bus left Nairobi at 7.00 a.m and travelled towards Mombasa at an average speed of 80 km/hr. At 8.00 a.m, a car left Mombasa towards Nairobi at an average speed of 120 km /hr. If the distance between Mombasa and Nairobi is 500 km , calculate the time of the day the two vehicles met (4 marks)
8.                                                                                                                                                                         →
   The position vectors of point  P and Q are 2i + 3j − k and 3i − 2j + 2k respectively. Find the magnitude of PQ to 4 significant figure. (3 marks)
9. Solve the equation 2 Cos 3x = √3 for 0° ≤ x ≤ 180°        (2 marks)
10. The diagram below represents a right rectangular based pyramid of 5 cm by 4 cm. The slant edge of the pyramid is 6 cm. Draw and label the net of the pyramid. (3 marks)
    
11. Two cylindrical containers are similar. The larger one has an internal cross-sectional area of 45 cm² and can hold 0.945 litres of liquid when full. The smaller container has internal cross-sectional area of 20 cm². Calculate the capacity of the smaller container. (3 marks)
12. Use logarithms correct to 4 decimal places to evaluate (3 marks)
    
13. One interior angle of a polygon is a right angle and each of the other interior angles is 126°. Calculate the number of sides of the polygon. (3 marks)
14. Use the exchange rates below to answer this question.
                              Buying                Selling
    1 US dollar        103.00                103.20
    1 UK £               125.30                125.95
    A tourist arrived in Kenya from Britain with 8400 UK Sterling pounds (£). He converted the whole amount to Kenya shillings. While in Kenya, he spent ¾ of this money and changed the balance to US dollars. Calculate the amount of money, to the nearest US dollars, that he received. (3 marks)
15. Joan and Gloria working together can do a piece of work in 12 hours. Joan, working alone takes 10 hours longer than Gloria. How many hours does it take Gloria to do the work alone? (3 marks)
16. The diagram below represents a solid of a conical frustrum. (use π = ²²/₇)
    
    Calculate the volume of the solid (4 marks)

SECTION II (50 marks)
Answer any five questions in this section in spaces provided

17. Given that a line passes through the points and , find;
    1. The equation of line L₁ in the form y = mx + c (2 marks)
    2. The equation of a line L₂ , which is perpendicular bisector of line L₁. Leave your answer in the form ax + by = c, where a, b and c are integers. (3 marks)
    3. Given that another line L₃ is parallel to L₂ and passes through point (−3, −5) and intersects line L₁ at point .
       (2 marks)
    4. The coordinates of the point of intersection of lines L₁ and L₃ (3 marks)
18.  
    1. Draw the graph of the function y = −x² + 4x − 1 for −1 ≤ x ≤ 5. (5 marks)
       

       x	−1	0	1	2	3	4	5
       y

    2. Use the graph above to solve the equations:
       1. x³ + 4x − 1 = 0        (2 marks)
       2. x² − 2x − 1 = 0(3 marks)
19.  
    1. Given that P =  , find (2 marks)
    2. Matawi bought 8 T-shirts and 5 pairs of shorts at a total cost of ksh. 4400. Had he bought 6 T-shirts and 9 pairs of shorts, he would have spent ksh. 1000 more.
       1. Form two equations to represent the above information (2 marks)
       2. Use the matrix method to determine the cost of a T-shirt and a pair of shorts (3 marks)
    3. Three months later the price of a pair of shorts went up. Matawi bought 5 T-shirts and 5 pairs of shorts at a total cost of ksh. 3650. Find the percentage increase in the price of a pair of shorts. (3 marks)
20. The marks of 50 students in a Mathematics test were taken from a form 3 class and recorded in the table below.
    

    Marks (%)	21-30	31-40	41-50	51-60	61-70	71-80	81-90	91-100
    Frequency (ƒ)	2	5	7	9	11	8	5	3

    1. On the grid provided, draw a cumulative frequency curve of the data above. Use a scale of 1 cm to represent 5 students on the vertical scale and 1 cm to represent 10 marks on the horizontal scale. (3 marks)
    2. From your curve in (a) above;
       1. Estimate the median mark. (1 mark)
       2. Determine the interquartile deviation. (2 marks)
       3. Determine the 10^th and 90^th percentile range. (2 marks)
    3. It is given that the students who scored over 45 marks pass the test. Use your graph in (a) above to estimate the percentage of students that pass the test. (2 marks)
21.  
    1. On the Cartesian plane below , draw triangle ABC with vertices A(0, −1), B(4,3) and C (2,2)       (1 mark)
    2. Draw triangle A'B'C', the image of ABC under a transformation defined by the translation vector T = . Write down the coordinates of A'B'C'.           (3 marks)
    3. A''B''C'' is the image of A'B'C' under an enlargement, scale factor −2, centre (3,1). On the same plane draw A''B''C'' and write down its coordinates. (4 marks)
    4. Draw A'''B'''C''', the image of A''B''C'', under reflection in the line x = 0 . (2 marks)
22. The figure below is of a right pyramid on a rectangle base of 8cm by 6cm and Its slant edge .
    TA = TB = TC = TD = 13cm
    
    Calculate:
    1. The length of AC (1 mark)
    2. The height of the pyramid (1 mark)
    3. The angle between TAB and plane TCD. (2 marks)
    4. The angle between TAB and plane TBC. (4 marks)
    5. The volume of the pyramid. (2 marks)
23. The figure below shows two circles, centres O₁ and O₂ intersecting at points A and B. The radius of the circle centre is O₁ 3cm. Further, and ∠ AO₁B = 90° and ∠ AO₂B = 63°
    
    Calculate correct to four significant figures
    1. The length of the common chord AB.    (2 marks)
    2. The radius of the circle center O₂        (2 marks)
    3. Hence, calculate the area of the shaded region. (use π = ²²/₇)                (6 marks)
24. During a surveying exercise to establish a housing estate a surveyor marked out four points W,X,Y and Z to represent an area to be left out for a shopping complex and social amenities. Point X is 240m on a bearing of 073° from point W. Point Y lies on a bearing of 145° at a distance of 300m from X. Z is directly south of W a distance of 320m.
    1. Draw a scale diagram to represent the relative positions of the area under survey
       Scale 1cm represents 40m. (4 marks)
    2. Using the scale diagram in (a) above, determine
       1. The distance and bearing of point Z from Y. (2 marks)
       2. The bearing of point X from point Z. (1 mark)
       3. A road is to be constructed directly South of X to meet another road from Y westwards at point P. Find the area enclosed by triangle PXY in hectares. (3 marks)

MARKING SCHEME

1.  
   1. 825 × 4 + 144 ÷ 4 of −6
      825 ×  4 + 144 ÷ −24
      3300 + −6
      =3294
   2. 3 ×  1000 = 3,000
2. 2x − 3 ≤ 4x + 7
   −3−7 ≤ 4x − 2x
   −10 ≤ 2x
   −5 ≤ x
   4x + 7 < x + 13
   4x − x < 13 − 7
   3x < 6
   x < 2
   
   −5 ≤ x < 2
3. 8x² − 2
   2(4x² − 1)
   2(2x −1)(2x +1)
   x(2x² − 7x + 3)
   2x² − 6x − x + 3
   2x (x −3) −1(x −3)
   x(2x −1) (x −3)
   2(2x −1)(2x +1)
   x(2x −1) (x −3)
   2(2x + 1)
    x(x − 3)
   4x + 2 
   x² − 3x\
4.  
   1. 1.57 × r = 5.5
      r = 5.5 ÷ 1.57
      r =] 3.5032 cm
   2. = (3.5032 + 3.5032 + 5.5) cm
      = 12.5064 cm
5. 3^−3x ÷ 3⁴ = 3² 
   −3x − 4 = 2
   −3x = 6
      x = −2
6. 8 = 2³ 
   10 = 2 × 5
   14 = 2 × 7
   L.C.M = 2³ × 5 × 7 = 280 seconds
   3600 ÷ 280 = 12.527
   =  12 times
7. Distance covered by bus by 8.00am
   1hr × 80 km/h = 80km
   Distance covered by both vehicles before meeting
   500 − 80 = 420km
   Relative speed 80 + 120 = 200km
   Time taken to meet ⁴²⁰/₂₀₀ = 2hrs 6 mins
   Time the two met 8:00am + 2hrs 6 mins = 10.06 am
8.  
   
   √(1² + (−5)² + 3²) = √35
   = 5.916 units
9.  cos3x = ^√3/₂ 
   3x = 30°, 330° ,390°, 690°
   x = 10°,110°,130°
10.  
    
11. A.S.F = ²⁰/₄₅ = ⁴/₉ 
    L.S.F = ²/₃ 
    V.S.F = (²/₃)³ = ⁸/₂₇ 
    ²⁷/₈ × 0.945 = 0.28 litres
12.  
    
13. 90 + 126(n − 1) = 180(n −2)
    90 + 126n − 126 = 180n −360
    90 + 360 − 126 = 180n −126n
    324 = 54n
    n = 6
14. 8400 × 125.3 = ksh 1,052,520
    balance = ¼ × 1,052,520 = Ksh 263,130
    amount of us dollars recieved
    263,130 = 2550
     103.2
15. ¹/₁₂ = ¹/_x + ¹/_x+10  
     1  =   2x + 10 
    12     x² + 10x
    24x + 120 = x² + 10x
    x² − 14x − 120 = 0
    x² − 20x + 6x − 120 = 0
    x(x −20) + 6(x − 20) = 0
    (x + 6) (x−20) = 0
    x = 20
    Gloria takes 20hrs
16.   h   = 2.1
    h+4    4.9
    h = 2.1h + 8.4
    h = 3
    Volume = ( ¹/₃ × ²²/₇ × 4.9² × 7) − (¹/₃ × ²²/₇ × 2.1² × 3)
                 = 176.07333 − 13.86
                 =162.2133
17.  
    1. gradient = −1−5 = −3
                        3−−1     2
       equation = y − 5 = −3
                         x + 1      2
       2(y − 5) = −3(x + 1)
       2y−10 = −3x −3
       2y = −3x + 7
    2. gradient of line 2 = ²/₃ 
       mid point of line L₁ =(1,2)
       equation = y − 2 = 2
                         x −1     3
       3(y − 2) = 2(x −1)
       3y − 6 = 2x − 2
       3y = 2x + 4
    3. gradient of L₃ = ²/₃ 
       equation = y + 5 = 2 
                         x + 3    3
       3(y+5) = 2(x +3)
       3y + 15 = 2x + 6
       3y = 2x − 9 
    4. 2(3y = 2x − 9)
       3(2y = −3x + 7)
       6y =   4x − 18}−
       6y = −9x + 21}
       0  =   13x −39
       x = 3
       3y = 2(3) − 9
       y = −1
       (3,−1)
18.  
    1.  
       

       x	−1	0	1	2	3	4	5
       y	−4	−1	4	11	20	31	44

        
    2.  
       1. x = − 4.2
          x = 0.2
       2. y = x² + 4x −1}
          0 = x² − 2x − 1} −
          y = 6x
          x = −0.4
          x = 2.4
19.  
    1.  
       det = (8 × 9) − (6 × 5) = 42
       P⁻¹ = 
    2.  
       1. 8x + 5y = 4400
          6x + 9y = 5400
       2.  
          
    3. 5 × 300 + 5y = 3650
       5y = 3650 − 1500
        y = 430
       % increase = ³⁰/₄₀₀ × 100% = 7.5%
20.  
    1.  
       

       U.C.L	30.5	40.5	50.5	60.5	70.5	80.5	90.5	100.5
       C.F	2	7	14	23	34	42	47	50

       
    2.  
       1. Q₂ = 62.5 marks
       2. Q₃ = 74.5, Q₁ = 48.5
          Q₃ − Q₁ = 74.5 − 48.5
                       = 26
          Q₃  − Q₁ = 26 = 13
                2          2
       3. 10^th percentile = 36.5 marks and 90^th percentile = 86.5 marks
          Range = 86.5 − 36.5 = 50 marks
    3. students who score 45 marks and below 11. 
       45 marks and above 50 − 11 = 39 students.
       ∴ % scoring over 45 marks = ³⁹/₅₀ × 100% = 78%
21.  
    1.  
       
    2. A'(1, −3), B'(5,1), C'(3,0)
    3. A''(7,9). B''(−1, 1), C''(3,3)
22.  
    1. = √(8² + 6²) = 10cm
    2. =√(13² − 5²) = 12cm
    3. tan θ = ³/₁₂ 
       θ = 14.04° × 2 = 28.08°
    4. √(13² − 4²) = √153
       √(13² − 3²) = √160
       √(4² + 3²) = 5
       5² = (√153)² + (√160)² − 2 × √153 × √160 cos x
       25 = 153 + 160 − 312.92 cos x
       25 − 313 = − 312.92 cos x
       − 288 = − 312.92 cos x
       cos x = 0.9204
       x = 23.02°
    5. ¹/₃ × 8 × 6 × 12 = 162cm³ 
23.  
    1. AB =  √(3² + 3²) 
             =  √26
            = 4.243 cm
    2. 1.121 = sin 31.5
          x
       x =   1.121   = 4.060cm
             sin 31.5
    3.  
       (⁹⁰/₃₆₀ × ²²/₇ × 3²) − (½ × 3 × 3) = 2.571
       (⁶³/₃₆₀ × ²²/₇ × 4.06²) − (½ × 4.06² sin 63°) = 1.723
       shaded area 
       2.571 + 1.723 = 4.294
24.  
    1.  
       
    2.  
       1. 420m, 248° ±1°
       2. 030° ±1°
       3. 2.083ha