Mathematics PP2 Questions and Answers - Joint Pre-Mock Exams 2021/2022

INSTRUCTIONS TO CANDIDATES

• This paper consists of TWO sections: Section I and Section II.
• Answer ALL the questions in Section I only five questions from Section II.
• Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.

SECTION 1 (50 MARKS)

1. Evaluate using squares, cubes and reciprocal tables (4 marks)
   
2. Make x the subject in  (3 marks)
3. Ali deposited Ksh.100,000 in a financial institution that paid simple interest at the rate of 12.5% p.a. Mohamed deposited the same amount of money as Ali in another financial institution that paid compound interest. After 4 years, they had equal amounts of money. Determine the compound interest rate per annum to 1 decimal place. (3 marks)
4. Simplify (3 marks)
   
5. Expand (1 - 2x)⁴, hence find the value of (1.02)⁴ correct to 3 significant figures.   (3 marks)
6. If sin⁡ x =2b  and cos⁡ x=2b√3, find the value of b (3 marks)
7. Find the relative error in (a + b) given that a=77ml, b=23ml, c=36ml, and d=16ml. (3 marks)
                                           (c − d)
8. Without using a calculator or mathematical tables, express      √3          in surd form and simplify. (3 marks)
                                                                                                  1− cos30°
9. The equation 3x² − 8px + 12 = 0 has real roots. Find the value of P. (2 marks)
10. A construction company employs 200 artisans and craftsmen in the ratio 1:3 every week. An artisan is paid 2 ½ times as much as a crafts man. At the end of 3 weeks the company paid ksh 1485000 to those employees. Find how much each artisan and each craftsman is paid. (a working week has six days) (3 marks)
11. A dam containing 4158m³ of water is to be drained. A pump is connected to a pipe of radius 3.5cm and the machine operates for 8 hours per day. Water flows through the pipe at the rate of 1.5m per second. Find the number of days it takes to drain the dam. (4 marks)
12. Two brands of coffee Arabica and Robusta costs sh.4,700 and sh.4,200 per kilogram respectively. They are mixed to produce a blend that costs shs.4,600 per kilogram. Find the ratio of the mixture. (3 marks)
13. Under a transformation represented by a matrix  , a triangle of area 10cm² is mapped onto a triangle whose area is 110cm². Find x (3 marks)
14. Find the distance between the centre 0 of a circle whose equation is 2x² + 2y² + 6x + 10y + 7 = 0 and a point B(−4,1). (3 marks)
15. Solve for x in the equation: (log_2⁡x)²+ log₂⁡8 = log₂⁡x⁴ (4 marks)
16. The figure below shows a circle inscribed in an isosceles triangle ABC. If Q, P and R are the points of contact between the triangle and the circle, O is the centre of the circle,
    BO=19.5cm and BQ=18cm. Find the radius of the circle and hence the length of the minor arc PQ. (3 marks)
    

SECTION II (50 MARKS)
ANSWER ONLY FIVE QUESTIONS

17.  
    1. Mr. Mackey pays a tax of Kshs.5,800 per month according to the income tax
       table given below. He is married and entitled to a family relief of   K€ 420p.a.
       

       Taxable income	Rate (Ksh per K€ )
       (K€ p.a.)  1 – 9,600  9,600 - 19,200  19,201 - 29,800  29,801 - 38,400  38,401 - 47,200  Over 47,200	2        3        5        7        9       10

       Calculate Mackey’s gross annual salary in K€ (6marks)
    2. The difference between compound interest and simple interest on Kshs.P over a duration of 36 months at the rate of 15% p.a. is Kshs.52,477.50. Calculate the value of P. (4 marks)
18.  
    1. Complete the table below for  y = x³ + 4x² − 5x − 5 (2 marks)
       

       X	−5	−4	−3	−2	−1	0	1	2
       Y			19			−5

    2. On the grid provided, draw the graph of y = x³ + 4x² − 5x − 5 for −5 ≤ × ≤ 2 (3 marks)
    3.  
       1. Use the graph to solve the equation x³ + 4x² − 5x − 5 = 0     (2 marks)
       2. By drawing a suitable straight line on the graph, solve the equation x³ + 4x² − 5x − 5 = − 4x − 1 (3 marks)
19. OPQ is a triangle in which OP=P and OQ=q. x is a point on OP such that OP:XP=5:2 and y is another point on PQ such that PY:YQ=1:2. Lines OY and XQ intersect at T.
    1. Express the following vectors in terms of P and q
       1. PQ (1 mark)
       2. OY (1 mark)
       3. OX (1 mark)
    2. If OT=kOY and QT=hQX express OT in two different ways. Hence or otherwise find the values of h and k. (6 marks)
    3. Determine the ratio OT:TY (1 mark)
20. If (x-1 1⁄8), x and (x+3⁄2) are the first three consecutive terms of a geometric progression;
    1. Determine the values of x and the common ratio. (4 marks)
    2. Calculate the sum of the first 6 terms of this progression. (3 marks)
    3. Another sequence has the terms;
       − 13,− 16,− 19, ……………………………− 310. Find the sum of this sequence. (3 marks)
21. The figure below shows a belt passing round two pulleys of centres A and B. The radius of the pulleys is 4cm and 6cm respectively and the distance between the centres is 25cm.
    
    Calculate the length of the belt used for the pulley system. (10 marks)
22. The points P(2,1), Q(4,1) R(4,3) and S(3, 3) are coordinates of a quadrilateral.
    1. Plot the quadrilateral PQRS on the grid provided. (1 mark)
    2. Find the coordinates of P¹Q¹R¹S¹ the image of PQRS under the transformation represented by the matrix
       M = (2 marks)
    3. Draw and label P¹Q¹R¹S¹ on the same grid.
    4. Find the coordinates of P¹¹Q¹¹R¹¹S¹¹ on the image of P¹Q¹R¹S¹ under the transformation represented by the matrix N =  (2 marks)
    5. Draw and label P¹¹Q¹¹R¹¹S¹¹ on the same grid. (1 mark)
    6. Determine the matrix that maps PQRS directly onto P¹¹Q¹¹R¹¹S¹¹ . (3 marks)
23. The table below shows the ages of people in years who attended a wedding ceremony.
    

    Age in years	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79
    Frequency	2	4	4	8	6	3	2

    1. State the modal class (1 mark)
    2. Using an assumed mean of 44.5 calculate
       1. The mean age (3 marks)
       2. The standard deviation (3 marks)
       3. The median age (3 marks)
24. A supermarket is stocked with plates which come from two suppliers A and B. They are bought in the ratio 3:5 respectively, 10% of plates from A are defective and 6% of the plates from B are defective.
    1. A plate is chosen by a buyer at randon. Find the probability that;
       1. It is from A (2 marks)
       2. It is from B and it is defective (2 marks)
       3. It is defective (2 marks)
    2. Two plates are chosen at random. Find the probability that;
       1. Both are defective (2 marks)
       2. At least one is defective (2 marks)

MARKING SCHEME

1.  0.339 + (         1           )
                  (5.041 X 10⁻³) ……M1   Correct square & cube root
   0.3309 + 3(198.4)
   0.3309 + 595.2
   (595.5309)⁻² 
   =       1        
      595.5309²         M1  Correct reciprocal
   =        1        
      35.46 ×10⁴ 
   =        1           =  0.2820 × 10⁻⁵…… M1
      3.546 × 10⁵
   = 2.82 × 10⁻⁶……..A1           4
2. 
   
   x² +2 =K …………..M1
   x² = K − 2
   x = ±√(k−2) ……………..A1      3
3. 100,000 ×12.5 × 4 = 50,000 ………M1
               100
   A=100,000 + 50,000 = 150,000
   100,000(1 + ^r/₁₀₀) = 150,000 ……….. M1
   1+^r/₁₀₀ = (1.5)^0.25 r = 10.7 (1d.p) …….A1
4.   
   
5. 1 + 4 (−2x)+6 (−2x)² + 4 (−2x)³ +(−2x)⁴
   = 1 − 8x + 24x² − 32x³ +16x⁴ ……. B1
   1− 2x = 1.02
   −2x = −0.0
   x = −0.01
   ∴1−8(−0.01)=24(− 0.01)² ………… M1
   32(−0.01)³ + 16(−0.01)⁴ …..M1 ( substitution of −0.01)
   1 + 0.08 + (0.0024) + 0.000032 + 0.00000016
   = 1.08243216
   = 1.08 (3sf) ……..A1   3\
6. tan⁡ x =   2b   
               2b√3
   =  1 
      √3  …….. M1
   x = tan⁻¹ ⁡(¹/_√3) = 30°
   sin⁡ 30° = 2b ………M1
    b = sin ⁡30
              2
   = 0.5 = 0.25 ……A1    3
        2
7. Max value = 77.5 + 23.5 = 101 = 5.3158
                        35.5 −16.5       19                        …….M1
   Min⁡ value = 76.5 + 22.5 = 4.7143
                       36.5 −15.5
   Absolute error = 5.3158 − 4.7143 …….M1
                                          2
                      = 0.30075
   % error = 0.30075 × 100 =6.015% √ A1
                        5                                3
   
8. Cos⁡ 30° = ^√3/₂ ………..B1
   
   
   = 4√3 + 6 ………A1    3
   
9. 3x²  − 8px + 12=0
   ( ^−8p/₂)² = 3×12 ……..M1
   16p² = 36
   P² = ±√(³⁶/₁₆) =± ⁶/₄
   p=±1.5 ……….A1
10. Let the pay for the craftsman be y
    ¾ (200)y+ ¼  (200) 2½y = 1485000 √ M1
                                                   18
    150y +125y = 82,500 M1
    y = sh.300
    crafts man = sh.300……
    artisan = ⁵/₂ × 300 = 750          A1    3
11. Cross-section area of the pipe = ²²/₇ × 3.5 × 3.5
                                                     = 38.5cm²
    Vol. drained per sec
    = ²²/₇ ×  3.5 ×3.5   ×1.5√ ……. M1
                100 ×100
    =        0.005775m³
       Vol. drained per day
    = 0.005775 ×8×3600√… M1
    =166.32 m³ A1
    No.of days required
    =   4158 
       166.32
    = 25 days B1     4
12. Arabica – x shs 4700
    Robusta – 1− x → shs.4200
    4700x + 4200 − 4200x=4600 √ M1
    47x − 42x=46 − 42
    5x=4
    x = ⁴/₅ √ A1
    Arabica : Robusta = 4:1 √ ….B1
13. Area scale factor = determinant = ¹¹⁰/₁₀
                                                       =11√ m1
    5x² + 6=11 ……√ m1
    5x² = 5
    x² = 1
    x = ±1 ……. A1         3
14. x² + y² 3x + 5y + (³/₂)² + (⁵/₂)² = ⁻⁷/₂ + ⁹/₄ + ²⁵/₄
    (x+³/₂)²+(y+⁵/₂)² = 5………. B1
    BO=√((^-3/₂+4) + (⁵/₂-1) )²)…….M1
    =√(²⁵/₄ + ⁹/₄)
    =√(³⁴/₄) = 2.915 units A1      3
15. (log₂⁡x)² + log_2⁡8 = log₂⁡x⁴ 
    let log₂⁡x be t
    log_2⁡8 = 3……….
    ∴ t² + 3=4t
    t² − t − 3t + 3=0……… B1
    t(t − 1) −3( t − 1)=0
    (t−3)(t−1)=0
    t=3 or t=1 ……… B1
    ∴log₂⁡x = 3 or log_2⁡x =1
    2³ = x    2¹ = x
    8  =x    B1     x = 2    B1     4
    
16. <QOP=2sin^-1(¹⁸/_19.5)    M1
    =134.76° √ B1
    Arc PQ=^134.76/₃₆₀ × ²²/₇ × 15
    =17.65m ……… A1
    Radius = √(19.5² ) c 18²
    =7.5cm √ B1
17.  
    1. Paye (p.a) = 5800x12
       = kshs.69,600/=
       Gross tax = 69,600 + 420 x 20
       = 78,000 √ B1
       Taxable income   Tax (Kshs)
       9600 x 2             19,200 √ M1
       9600 x 3             28,000 √ M1
        y x 5                  30,000 √ M1
                                  78,000
       5y=30,000
       y=K€ 6000 √ A1
       Gross annual salary
       = K€ 25,200 √ B1
    2. P (1+^r/₁₀₀)^t − 1 − ^rt/₁₀₀ =52,477.50
       P(1.15³ − 1  − 0.15 × 3) = 52,477.50√√ M1 M1
       P = 52.477.50 √ M1
              0.070875
       P=740,000 √ M1
18.  
    1. Complete the table below for y  = x³ + 4x² − 5x − 5 (2 marks)
       

       X	−5	−4	−3	−2	−1	0	1	2
       Y	−5	15	19	13	3	−5	−5	9

       √ B2
    2. On the grid provided draw the graph of y  = x³ + 4x² − 5x − 5 for -5≤ × ≤2  (3 marks)
       1. Use the graph to solve the equation
          x³ + 4x² − 5x − 5 = 0   (2 marks)
       2. By drawing a suitable straight line on the graph, solve the equation  x³ + 4x² − 5x − 5 = -4x-1 (3 marks)
          
19.  
    1.  
       1. PQ = − P + q            B1
       2. OY = ²/₃ P + ¹/₃ q     B1
       3. QX = − q + ³/₅ P       B1
    2. OT = ²/₃ pk + ¹/₃ qk …….(i) M1
       OT= q(1− h) + ³/₅ ph ……. (ii) M1
       ¹/₃ qk = (1− h)q …….. M1
       ¹/₃ k = 1− h
       K=3 – 3h
       ³/₅ ph = ²/₃ pk
       ³/₅h = ²/₃ k
       ³/₅ h = ²/₃ (3 − 3h)
       ³/₅ h = 2− 2h
       3h = 10− 10h …… B1
       h = ¹⁰/₁₃ …….. B1
       k = 3 − 3(¹⁰/₁₃)
       k = ⁹/₁₃ …….B1
    3. OT:TY=9:4 …………. B1
20.  
    1.  
       
    2.  
       
          =  46.80 √ A1
    3. S_n = ⁿ/₂ (a + l)
       Tn = a + (n-1)d
       −310 = −13 + (n −1)(−3)
       n=100 ……………√ B1
       S₁₀₀ = ¹⁰⁰/₂ {−13 + (−310)} √…………. M1
       =−16,150 √ A1
21.  DC=EF
    
    DC = EF = AX√(25² − 2²) = 24.92cm …..B1 for the length
    Sin⁡ φ = ²/₂₅ = 0.8
    ∴φ = 4.589° ………A1
    2φ = 9.178°
    Reflex Angle = 180 + 9.178 = 189.178 ………..B1
    Angle subtended by belt = 360°−189.178°
    On small circle =170.8° ….B1
    Minor arc DE = ^170.8/₃₆₀ × 2 × ²²/₇ × 6 =11.93cm….. M1 A1
    Major arc CF = ^189.2/₃₆₀ × 2 × ²²/₇ × 6 = 19.82cm ….M1 A1
    Length of belt = 2 × 24.92 + 11.93 + 19.82 …..M1
                           = 81.59cm…….. A1        10
22.  
    1.  
    2. 
         P¹ (3,4) Q¹ (5,8) R¹ (7,8) S¹ (6,6)      √ A1     (All coordinates)
    3.  
       
       P¹¹ (−2,4) Q¹¹ (−2,8) R¹¹ (−6,8) S¹¹ (−6,6) √ A1 (Must be in coordinate form)
    4.  
        √( M1 (matrix equ)
       2a + b = −2   2c +d = 4       M1 formation of 4 equations
       4a + b = −2   4c +d = 8
       2a + b = −2   2c + d = 4
             2a = 0           2c = 4
                a = 0            c = 2
                 b = −2          d = 0
       matrix = √ A1
23.  
    1. Modal class
       40 - 49   B1
    2.  
       

       Age	Frequency(f)	Midpoint x	d = x − 44.5	fd	d2	Fd2	cf
       10-19 20-29 30-39 40-49 50-59 60-69 70-79	2   4   4   8   6   3   2       εf=29	14.5  24.5  34.5  44.5  54.5  64.5  74.5	−30   −20   −10     0    10    20    30	−60  −80  −40    0   60 M1    60   60        εfd=0	900   400  100    0  100  400  900	1800  1600   400    0  600 M1   1200  1800       εfd2	2  6 10 18 24 27 29 M1

       1. x̄ = A  + εfd
                        εf
            = 44.5 + ⁰/₂₉  √m1
            = 44.5     √ A1
       2. Standard deviation
          S² = εfd² = 7400 √ m1
                    εf       29
               = 255.17 √ A1
       3. Median
          
          = 39.5 + 5.625 = 45.125     A1
24.  
    1.   
       
       
       1. P(A)= 3⁄8 √ B1
       2. P(B and D) = ⁵/₈ × ⁶/₁₀₀ √ m1
          = ³/₈₀ √ A1
       3. P(D)=P(A and D)or P(B and D)
          = (³/₈ × ¹⁰/₁₀₀) + (⁵/₈ × ⁶/₁₀₀) √ m1
          = ³⁰/₈₀₀ + ³/₈₀ 
          = ³/₄₀     √ A1
    2.  
       1. P(Two defective)=P(ADD) or P(BDD)
          = (³/₈ × ¹⁰/₁₀₀ × ¹⁰/₁₀₀) + (⁵/₈ × ⁶/₁₀₀ × ⁶/₁₀₀) √ m1
          =   3   +    180  
             800     80000
           =  3     √ A1
              500
       2. P(at least one defective) = 1− {P(AD¹D¹ or BD¹D¹ )} m1
          = 1 − {³/₈ × ⁹⁰/₁₀₀ × ⁹⁰/₁₀₀ + ⁵/₈ × ^94√/₁₀₀ × ^94√/₁₀₀}
          = 1 − 68480 = 11520  = 0.144   √ A1
                   80000     80000