Mathematics PP1 Questions and Answers - Joint Pre-Mock Exams 2021/2022

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INSTRUCTIONS TO CANDIDATES.

  • This paper consists of two sections; section I and section II
  • Answer ALL questions in sections I and only FIVE sections in section II
  • Show all the steps in your calculations; giving your answers 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.

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

  1. Evaluate −4{(−4+−15÷5)+−3−4÷2}      (3 marks)
                           84÷−7+3−−5
  2. Simplify completely the expression:    (3 marks)
    Math Joint PM PP1 Q2 2122
  3. Given that cos θ = 3/5, find sin θ − tan (90°− θ) without using tables or calculator. (2 marks)
  4. Under an enlargement, the images of points A(3,1) and B(1,2) are A1(3,7) and B1 (7,5). Without construction, find the centre and the scale factor of enlargement. (4 marks)
  5. List all the integral values of x that satisfy the inequalities; (3 marks)
    x − 3/2 ≤ 2x+1 < 5
  6. A bus travelling at an average speed of x km/h left station at 8.15 am. A car, travelling at an average speed of 80km/h left the same station at 9.00 am and caught up with the bus at 10.45 am. Find the value of x. (3 marks)
  7. The interior angle of a regular polygon with 3x sides exceeds the interior angle of another regular polygon having x sides by 40°. Determine the value of x. (3 marks)
  8. Use squares, cubes and reciprocals tables to evaluate, to 4 significant figures, the expression:   (3 marks)
           1     +      3        
    3√27.56     (0.0712)2 
  9. From a point 20m away on a level ground the angle of elevation to the bottom of the window is 27° and the angle of elevation of the top of the window is 32°. Calculate the height of the window. (3 marks)
  10. Solve for x in the equation: 53y+3 + 53y−1=125. 2    (4 marks)
  11. Mr. Kanja, Miss Kanene and Mrs. Nyaga have to mark a form three mathematics contest for 160 students. They take 5 minutes, 4 minutes and 12 minutes respectively to mark a script. If they all start to mark at 9.00 am non-stop, determine the earliest time they will complete the marking. (4 marks)
  12. Evaluate Math Joint PM PP1 Q12 2122 (2 marks)
  13. Two similar cylinders have diameter of 7cm and 21cm. If the larger cylinder has a volume of 6237cm3, find the heights of the two cylinders. (take π = 22/7)  (3 marks)
  14. The cost of providing a commodity consists of transport, labour and raw materials in the ratio 8:4:12 respectively. If the transport cost increases by 12%, labour cost by 18% and raw materials by 40%, find the percentage increase of producing the new commodity. (3 marks)
  15. Given that 4p − 3q = Math Joint PM PP1 Q15a 2122 and p+2q = Math Joint PM PP1 Q15b 2122 , find value of p and q (4 marks)
  16. In the figure below ABCDE is a cross-section of a solid. The solid has a uniform cross-section. Given that AP is an edge of the solid, complete the sketch showing the hidden edges with a broken lines. (3 marks) Math Joint PM PP1 Q16 2122

SECTION II (50 Marks)
Answer any five questions from this section in the spaces provided.

  1. The figure below represents a sector of a circle radius r units. The area of the sector is 61.6 cm2 and the length of the arc AB is one tenth of the circumference of the circle from which the sector was obtained. ( Take π= 22/7)
    Math Joint PM PP1 Q17 2122
    1. Calculate;
      1. the angle subtended by the sector at the centre. (2 marks)
      2. The radius r of the circle. (3 marks)
    2. If the sector above is folded to form a cone;
      1. Calculate the base radius of the cone. (2 marks)
      2. The volume of the cone. (3 marks)
  2. Two factories A and B produce both chocolate bars and eclairs. In factory A, it costs Kshs x and Kshs y to produce 1 kg of chocolate bars and 1 kg of eclares respectively. The cost of producing 1 kg of chocolate bars and 1 kg of eclairs in factory B increases by the ratio 6:5 and reduce by the ratio 4:5 respectively.
    1. Given that it costs Kshs 460 000 to produce 1 tonne of chocolate bars and 800kg of eclares in factory A and Kshs 534 000 to produce the same quantities in factory B, form two simplified simultaneous equations representing this information. (3 marks)
    2. Use matrix method to find the cost of producing 1 kg of chocolate bars and 1 kg of eclaires in factory A. (5 marks)
    3. Find the cost of producing 100 kg of chocolate bars and 50 kg of eclaires in factory B. (2 marks)
  3. The vertices of triangle ABC are A(6,2), B(8,2) and C(6,0).
    1. On the grid provided below, draw triangle ABC. (1 mark)
      Math JPM Q19 2122
    2. Triangle A’B’C’ is the image of triangle ABC under a reflection in the line y = x. On the same grid draw triangle A’B’C’ and state its coordinates (2 marks)
    3. Triangle A”B”C” is the image of triangle A’B’C’ under and enlargement scale factor 2 about the centre (−1,9). On the same grid, draw triangle A”B”C” and states its coordinates. (2 marks)
    4. By construction, find and write down the co-ordinates of the centre and angle of rotation which can be used to rotate triangle A”B”C” onto triangle A’’’B’’’C’’’ shown on the grid above. (3 marks)
    5. State any pair of triangles that are:\
      1. Oppositely congruent. (1 mark)
      2. Directly congruent. (1 mark)
  4. The figure below shows a velocity-time graph of an object a which accelerates from rest to a velocity of V ms−1 then decelerated to rest in a total time of 54 seconds.
    Math Joint PM PP1 Q20 2122
    1. If it covered a distance of 810 metres;
      1. Find the value of V. (2 marks)
      2. Calculate its deceleration, given that its initial acceleration was 12/3  ms−2 (2 marks)
    2. A bus left town X at 10.45 am and travelled toward town Y at an average speed of 60 km/h. A car left town X at 11.45 am on the same day and travelled along the same road toward Y at an average speed of 100km/h. The distance between town X and town Y is 500km.
      1. Determine the time of the day when the car overtook the bus. (3 marks)
      2. Both vehicles continued towards town Y at their original speeds. Find how long the car had to wait in town Y before the bus arrived. (3 marks)
  5. The masses to the nearest kilogram of some students were recorded in table below.
     Mass(kg)   41-50   51-55   56-65   66-70   71-85 
     Frequency      8     12     16     10     6
     Height of rectangle           
    1. Complete the table above to 1 decimal place. (2 marks)
    2. On the grid provided below, draw a histogram to represent the above information. (3 marks)
    3. Use the histogram to:
      1. State the class in which the median mark lies. (1 mark)
      2. Estimate the median mark. (2 marks)
      3. The percentage number of students with masses of at least 74kg. (2 marks)
  6.  
    1. a straight line L1 whose equation is 9y − 6x=- −6 meets the x-axis at Z. Determine the coordinates of Z. (2 marks)
    2. A second line L2 is perpendicular to L1 at Z. Find the equation of L2 in the form ax+by=c, where ,b and c are integers. (3 marks)
    3. a third line L3 passes through the point (2,5) and is parallel to L1. Find:
      1. The equation of L3 in the form ax+by=c, where a, b and c are integers. (2 marks)
      2. The coordinate of point R at which L2 intersects L3. (3 marks)
  7. In the diagram below, the coordinates of points O, P and Q are (0,0), (2,8) and (12,8) respectively. A is a point on OQ such that 4OA=3OQ. Line OP produced to R is such as OR=5OP.
    Math Joint PM PP1 Q23 2122
    1. Find vector RA. (3 marks)
    2. Given that point L is on PQ such that PL: LQ=12:5, find vector RL. (4 marks)
    3. Show that R, L and A are collinear. (2 marks)
    4. Find the ratio of RL:LA. (1 marks)
  8. Five points, P, Q, R, V and T lie on the same plane. Point Q is 53km on the bearing of 055° of P. Point R lies 162° of Q at a distance of 58km. Given that point T is west of P and 114km from R and V is directly south of P and S40°E from T.
    1. Using a scale of 1:1,000,000, show the above information in a scale drawing. (3 marks)
    2. From the scale drawing determine:
      1. The distance in km of point V from R. (2 marks)
      2. The bearing of V from Q. (2 marks)
      3. Calculate the area enclosed by the points PQRVT in squares kilometers. (3 marks)

MARKING SCHEME

 1 




 −4{(−4−5)−3−2}  ✓ 
      −12+8
 −4×−14   ✓
     −4 
 =−4 ✓

 M1 
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 A1


 2





Math Joint PM PP1 Ans2 2122

 2(3xy−4)(xy−2)   ✓
  2(xy+2)(xy−2) 
  (3xy−4)   ✓
   (xy+2) 

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 3

 4/5 − ¾ 
 = 1/20 
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 4









let centre be (x,y)

3−x  =  7−x ✓
3−x      1−x 
x=3
7−y = 5−y
1−y    2−y
y=3  ✓
∴centre(3,3)  ✓
S.F.=7−3 =  4  = −2  ✓
        1−3= −2 

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 5




 
x − 3/2 ≤ 2x+1

x ≥ −5/6 ✓
2x+1<5
x<2
5/6 ≤ x < 2 ✓
integral values=0,1 ✓

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A1
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 6



 time taken to catch up = 10.45 − 9.00 = 1.45=1¾ h ✓
  x × ¾  =  7  ✓
   x−80      4 
x=140 km/h ✓
M1
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 7






(3x−2)180 = (x−2)180 +40 ✓
     3x                  x
540x − 360 − 540x + 1080  = 40
                 3x
720 = 40✓
 3x
x=6✓

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 8







 =    1         3       =     1           3            ✓

    3.021    (0.071)2     3.021    50.41 ×10-2

=      1     + 3 ×       1                  ✓
    3.021           5.041×10-1 

=0.3310 + 3 × 0.1984 × 10

=6.283✓

 M1
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 9


=20 tan⁡ 32 − 20 tan⁡27✓
=12.50 − 10.19✓
=2.31m✓
 M1
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 10






 
 53y × 53 + 53y = 125.2✓
                 5
let 53y=x
125x + X/5 = 125.2
625x + x = 626✓
x=1✓
53y = 1 = 50 
3y = 0 ⇒ y = 0✓
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 11



 LCM of 5,4,12=60 minutes✓
no of scripts marked in 60 minutes = 60/560/460/12 = 32 scripts✓
time to mark 160 scripts=(160×60)/32=300 minutes.✓
time to complete marking=9.00+5.00=1400=2.00pm✓
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 12 

 

  
 437  −   21  ✓
  99        99

= 416 = 4 20/99 or Math Joint PM PP1 Ans12 2122
    99

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 13



height of larger cylinder =     6237       = 18 cm✓

                                         22/7 × 10.52
21/7 = 18/h  ✓
h=6 cm✓

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 14


%increase=(8/24 × 12/100 + 4/24 × 18/100 + 12/24 × 40/100) × 100✓

=27/100 ×100✓
=27%✓

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 15








 Math Joint PM PP1 Ans15 2122

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 16




 Math Joint PM PP1 Ans16 2122

 B1 - continuous lines 
 B1 - dotted lines
 B1 -  complete diagram 

 17











  (a) (i) 1/10 ×360✓

              =36° ✓
      (ii) 61.6 = 36/360 × 22/7 ×r2 ✓
             r = Math Joint PM PP1 Ans17a 2122
             r = 14 cm✓
(b) (i) 61.6 = 22/7 × r × 14✓
          r = 61.6 × 7/22 × 1/14 = 1.4 cm ✓
     (ii) h = √(142 − 1.42 )=13.93✓
           V = 1/3 × 22/7 × 1.42 × 13.93✓
              =28.60 cm3

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 18














 (a) 1000x + 800y = 460,000✓

      1000 × 6/5 x + 800 × 4/5y = 534,000✓
      5x + 4y = 2,300
     15x + 8y = 6675✓
(b) 
    Math Joint PM PP1 Ans18b 2122
     x = Ksh 415, y = Ksh 56.25 ✓
(c) 6/5 × 415 × 100 + 4/5 × 50 × 56.25✓
    = Kshs 52,050✓

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 19






















 (a)
 Maths JPM ans19a 2122
(b) A’(2,6), B’(2,8), C’(0,6) ✓

(c) A”(5,3), B”(5,7), C”(1,3) ✓
(d) Centre(4.5,1.5) angle = 90° ✓
(e)  (i) ABC and A’B’C’✓
      (ii) A”B”C” and A’’’B’’’C’’’✓

 B1 - Δ ABC
 B1 - Δ A'B'C'
 B1 - chord
 B1 - Δ A''B''C''
 B1 - chord
 B1 - bisectors
 B1 - centre identified
 B1 - centre & angle
 B1 - opp
 B1 - direct








 

 20 
















(a) (i) ½ × 54 × V=810✓

         V = 810 = 30 m/s✓
                 27
     (ii) 30 − 0 = 5 ✓
            t − 0     3
          t=18s
        acceleration =   0 −30  = − 8.8333
                                 54−18
        ∴deceleration=0.8333m/s2  ✓
(b) (i) distance apart = 1 × 60=60km✓
          time taken to overtake =  60  = 1.5✓
                                                   40
         time of the day = 11.45 + 1.30 = 1315 = 1.15pm✓
      (ii) time the bus arrived at Y = 10.45 + 600/60 =10.45 + 8.20=1905h✓
           time the car arrived at Y = 11.45 + 500/100 = 11.45 + 5.00=1645h✓
          waiting time=1905 −1645=2 h 20 mins✓

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 21





















 (a)
 Mass(kg)   41-50   51-55   56-65   66-70   71-85 
 Frequency      8     12     16     10     6
 Height of rectangle    0.4    1.2    0.8    1.0   0.2
(b)
Maths JPM ans 21b 2122
(c) (i) 56-65
      (ii) 55.5 +      6    
                      2 × 0.8
         = 55.5 + 37.5 = 59.25
     (iii) 5/26 × 100
         = 8.929%

 B1B1
 S1 - scale
 B1B1 - rectangles
 B1
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 22




















 
 (a) −6x = −6✓

        x=1
       ∴Z(1,0)✓
 (b) L1;y = 2/3 x − 6✓
       L2 ; y−0 = −  ✓
              x−1      2
       y = −3x + 3
               2      2
        3x + 2y = 3✓
 (c) (i) y−5  = 2/3  ✓
         x−2
        y−5 = 2/3x − 4/5
        2x−3y=−11✓
      (ii) (3x+2y=3)×2
          (2x−3y=−11)×3✓
          6x+4y=6
          6x−9y=−33−
              13y=39
       ∴y = 3,x = −1
        R(−1,3) ✓

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 23




















 Math Joint PM PP1 Ans 23 2122

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 24

















 (a)
 Math Joint PM PP1 Ans 24a 2122
 (b) (i) (7 ± 0.1) × 10km  ✓
            = 70km  ✓
      (ii) (180° + 25°) ± 1°  ✓
            =205°  ✓
     (iii) QV = 96km; PV = 56km; PT = 48km  ✓
           area of region = area of ∆QRV + area of ∆PQV + area of ∆TPV

          =½ × 58 × 96 sin⁡ 43 + ½ × 53 × 56 sin⁡ 125 + ½ × 48 × 56 ✓
          =2,559.8km≅2,560km. ✓

 B1 - Q
 B1 - R&T
 B1 - V
 B1
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