Mathematics Paper 2 Questions and Answers - Form 3 End Term 2 Exams 2023

• This paper consists of two sections: Section I and Section II.
• Answer all questions in section I and any five questions in Section II.
• 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.
• KNEC Mathematical tables may be used.

SECTION I (50 marks)

1. Without using a calculator evaluate                                                               (3 marks)
    2¹/₃ − 1²/₃ ÷ ⁵/₉  
   ⁴/₇ of 2¹/₃ − 2²/₇
2. Simplify the expression.                                                                                  (3 marks)
   18xy−18xr
      9xr−9xy
3. Solve the equation sin (⁵/₂x) = −½  for 0°≤x≤360°                                   (3 marks)
4. The sum of the interior angles of a regular polygon is 1800°. Find the size of each exterior angle   (3 marks)
5. In the figure below the shaded region is a segment of the circle with Centre O and radius r. AB=8 cm, ON = 3 cm, angle AOB =106.3°. 
   
   Find the area of the shaded part.                                                                                           (4 marks)
6. Simplify the expression    √48      leaving your answer in the form a+b√c where a, b and c are integers.    (3 marks)
                                         √5 + √3
7. By use of a quadratic formula solve the equation                                                                   (3 marks)
   2P(P+1)=4
8. Solve for x in the equation                                                                                                      (3 marks)
   32^(x−3) ÷ 8^(x−4) = 64 ÷ 2^x
9. Use logarithms correct to 4 decimal places to evaluate;        (4 marks)
   
10.  Make H the subject of the formula                                                                                           (3 marks)
    
11. In the figure below angle BAC and ADC are equal. Angle ACD is a right angle. The ratio of the sides AC:BC =4:3. Given that area of triangle ABC is 24 cm2, find the area of triangle ACD.  (3 marks)
    
12. Evaluate without using a calculator or a mathematical table,                    (3 marks)                                                   
     log (3x+8) − 2 = log (x − 4)
13. In the figure below O is the centre  .Calculate the angles marked x ,y and z                            (3 marks)
    
14. The exchange rate in January 2000 was US $ 1 = Ksh 75.60 and UK £1 = Ksh 115.80. A tourist came to Kenya with US $ 5000 and out of it spent ksh.189,000. He changed the balance in UK£. How many pounds did he receive?                                      (3 marks)
15. The length and width of a rectangle are given to the nearest 0.1 cm as 18.5 cm and 12.4 cm respectively. Calculate the percentage error in the area of the rectangle.               (3 marks)
16. Solve the following inequalities and hence state the integral values that satisfy the inequalities.              (3 marks)
    x + 1 ≤ 4x −5 < 3x+2

SECTION II (50 Marks)

17.  
    1. L₁ is a line which passes through (2, 5), (5, 16) and (−4, h). Find the equation of the line and the value of h.       (3 marks)
    2. Equation of L₂ is 3y + x =23 and it intersects line L₁ at point A.  Find the co-ordinates of A. (2 marks)
    3. B (−1, p) is on line L₁. Find p hence find the length of AB.              (2 marks)
    4. C(t, 9) is on the line L₂. Find t. It is given that the line L₃ passes through C and parallel to L₁.  Find the equation of L₃   (3 marks)
18. The table below shows the rates of taxation in a certain year.
    

    Income in K£ p.a	Rates in Ksh per K£
    1 - 3900	2
    3901- 7800	3
    7801 - 11700	4
    11701 - 15600	5
    15601 - 19500	7
    19501 and above	9

    In that year, Juma was earning a basic salary of Ksh. 41 000 per month. In  addition he was entitled to a house allowance of Ksh. 13 000 p.m. and a personal relief of Ksh. 1056 p.m.
    1. Calculate Juma’s taxable income in K£ per annum.                                   (2 marks)
    2. How much tax did he pay per month?                                                        (5 marks)
    3. His other deductions per month included: Sacco contribution of Ksh 4000 and loan repayment of Ksh 5 500. Calculate Juma’s net salary.                                (3 marks)
19. The distance between towns A and B is 360km .A minibus left town A at 8.15a.m and travelled towards town B at an average speed of 90km/hr .A matatu left town B two and  a third hours later on the same day and travelled towards town A at an average speed of 110km/hr. 
    1.  
       1. At what time did the two vehicles meet                                                       (4 marks)
       2. How far from town A did the two vehicles meet                                       (2 marks)
    2. A motorist started from his home at 10.30 a.m. on the same day as the matatu and travelled at an average speed of 100km/hr .He arrived at B at the same time as the minibus .Calculate the distance from A to his house.                                      (4 marks)
20. A triangle ABC with vertices  A(−4,2), B(−6,6) and C(-−6,2) undergoes enlargement scale factor −1 and centre (−2,6) to produce triangle A'B'C'.
    1. On the grid provided draw triangle ABC and its image A'B'C', state the co-ordinates of   ∆A'B'C'  (4 marks)
    2. Triangle ∆A'B'C'  is the reflected in the line y+x=0 to give ∆A''B''C''. Draw triangle ∆A''B''C'' and state the co-ordinates of its vertices. (3 marks)
    3. If triangle A'B'C' is mapped onto a triangle whose co-ordinates are '''(−4,−2) , B'''(−6,−6) and C'''(−6,−2) by a rotation, find the centre and angle of rotation.                                                                                                               (3 marks)
21. Three quantities R, S and T are such that R varies directly as S and inversely as the square root of T.
    1. R = 480 when S = 150 and T = 25.Write an equation connecting R, S and T.   (4 marks)
    2. Find;
       1. The value of R when S = 360 and T = 2.25.                                           (2 marks)
       2. The percentage change in R if S is increased by 5% and T decreased by 15.36%.              (4 marks)
22. The diagram below represents a lampshade in the form of an open-ended frustum of a cone. Its bottom and the top diameters are 14cm and 7cm respectively. Its height is 10cm. Use (π=3.142)
    
    
    1. Find
       1. The height of the cone from which the frustum was cut.               (2 marks)
       2. The area of the curved surface of the frustum                                (5 marks)
    2. The material used for making the lampshade is sold at Shs.800 per square metre. Find the cost of 10 lampshades if a lampshade is sold at twice the cost of the material.                                                                                       (3 marks)
23. The table below shows the mass of 60 women working in hotels
    

    Mass (Kg)	60 - 64	65 - 69	70 - 74	75 - 79	80 - 84	85 - 89
    No. of women	8	14	18	15	3	2

    1. State:
       1. The modal class          (1 mark)
       2. The median class           (1 mark)
    2. Estimate the mean mark                     (4 marks)
    3. Draw a histogram for the data                     (2 marks)
24. Two vertical walls of height h and 2h respectively stand on a level ground and the walls are 100m apart. From a point P which is y m from the shorter wall, the angles of elevation of the top of the walls are 35° and 28° respectively. From another point Q, d metres from P, the angle of elevation of the top of the taller building is 75° as shown. 
    
    Calculate;
    1. The vertical heights of the two walls.                                                     (6 marks)
    2. The distance PQ in metres correct to 1 decimal place.                                (4 marks)

MARKING SCHEME

No	Working	Remarks
1.	N⟹7/3 − (5/3 ÷ 5/9) = −2/3D ⟹ (4/7 × 7/3) − 16/7 = −20/21 N/D = −2/3 × − 21/20 = 7/10	M1M1 A1
2.	N ⟹ 18x (y−r) D ⟹ −9x(y−r) N/D ⟹  18x(y−r)  = −2            −9x(y−r)	M1  M1  A1
3.	(½) = 30°  But sine is negative in 3rd and 4th quadrants 5/2x = 210°, 330°, 570° and 690° x = 84°, 132°, 228° and 276°	B1 for sine inverse B1 B1
4.	(2n−4)90 = 1800 2n−4 = 20 n = 12 sides         ext angle = 360/12 = 30°
5.	r = 8×sin 36.85 = 5 cm        sin 106.3 Area of segment = (106.3 × 22 × 25 − (½ × 25 × sin 106.3)                                   360      7                                    = 23.19 – 12                                   = 11.192	M1  M1  M1  A1
6.	√48    × √5 − √3  √5 + √3    √5 − √3 (√48 × √5) − (√48 × √3)            5 − 3  4√15 − 12 ⇒ 2√15 − 6        2	M1 for conjugate pair  M1 simplification  A1
7.	−2 ± √36        4 p = 1 OR p = −2	M1  M1  A1
8.	25(x−3) ÷ 23(x−4) = 26−x 5x −15 − 3x + 12 = 6−x 5x − 3 + x = 15 −12 + 6. 3x = 9  x = 3	M1 M1  A1
9.		M1 All correct logs M1 Addition and Subtraction M1 Square root A1  correct answer
10.	F2 = H − 2WL             3H 3HF2 = H − 2WL H − 3HF2 = 2WL H =    2WL            1 − 3F2	M1  M1  A1
11	ASF = LSF2 (4/3)2 = 16/9 Area of triangle ACD = 16 × 24                                           9 = 422/3cm2	M1 M1  A1
12.	log (3x+8) = log x − 4            23 3x+8 = x−4    8 3x+8 = 8x−32 5x = 40  x = 8	M1 Appropriate use of logarithmic laws M1 A1
13.	<x = 90 − 26  =  64° <y = 2×26 = 52° <z = 180−64 =116°	B1 B1 B1
14.	Total Ksh received =  5000 x 75.60 = Ksh.378 000 Bal in Ksh 378000 − 189000 = Ksh.189 000 UK £ recieved =  189000   = UK£ 1,632.12                              115.80	M1 M1 A1
15.	Working area = 18.5×12.4 = 229.4cm2 Maximum area = 18.55×12.45 = 230.9475cm2 Minimum area = 18.45×12.35 = 227.8575cm2 Absolute error in area = 230.9475 − 227.8575 =1.545                                                         2 Percentage error = 1.545  × 100 = 0.6735%                                 229.4	M1 max and min area M1 absolute error  A1
16.	x+1 ≤ 4x − 5  −3x ≤ −6 x ≥ 2 4x − 5 < 3x+2 x < 7 Integral values 2,3,4,5, and 6	M1 M1 A1
	SECTION II
17.	m1 = 16 − 5 = 11/3             5 −2 11 =  y − 5   3      x  − 2 11x − 3y = 7 Substituting ( −4, h) 11 (−4) −3h = 7  −3h = 7 + 44 h = −17 At A, 11x − 7 = −x+23 12x = 30  x = 2.5  3y + 2.5 = 23  y = 6.83 Coordinates of A (2.5, 6.83)   L1 = 3y = 11x − 7  3p = 11 ( −1) −7  p = −6 Coordinates of B ( −1, −6) lABl = √(( −1 − 2.5)2 + ( −6 − 6.83)2) lABl = √(12.25 + 164.6089) =13.30 units   L2 ⟹ 3y = − x + 23 3(9) = −t + 23 t = 23 −27 = −4 Coordinates of C ( −4, 9) m3 = 11/3 11 = y − 9  3      x+4 11x − 3y + 71 = 0	M1 M1  A1 M1  A1 M1  A1 M1 M1 A1
18.	Taxable income = K£(41000+13000) × 12                                             20                           =K£32 400 Tax due 1st band = 3900×2 = Sh. 7800 2nd band = 3900×3 = Sh. 11 700 3rd  band = 3900×4 = Sh. 15 600 4th band = 3900×5 = Sh. 19 500 5th band = 3900×7 = Sh. 27 300 6th band = 12 900×9 = Sh. 116 100 Total tax due = Sh. 198 000 Total tax relief = Sh(1056×12) = 12 672 Net tax = Sh. 198 000 − 12 672 = Ksh. 15 444                                    12 Total deductions = 15 444 + 4000 + 5500                             = Ksh. 24 944 net salary = 54 000 − 24 944 = Ksh.29 056	M1  A1 M1 M1 M1 M1A1 M1 M1 A1
19.	Relative speed = 110 + 90 = 200 km/hr Distance between the cars in 21/3 hrs = 360 − (7/3×90)                                               =150 km Time taken = 150/200 = 45 mins Time of the day = 8.15 a.m. + 3 hrs. 5 mins                             = 11.20 a.m. Distance = 90 × 31/12 = 277.5 km Time taken by minibus = 360 = 4 hrs                                          90 Arrival time of minibus = 8.15 a.m. + 4 hrs = 12.15 p.m. Time difference = 12.15 p.m. – 10.30 a.m.                              = 1 hr 45 mins Distance = 360 − (1.75×100) = 185 km	M1  A1  M1  A1  M1A1  B1  B1  M1A1
20.
21.	R = kS         √T R = 150k          √25 k = 480×5  =16          150 R = 16S         √T   R = 16×360 =384          √225 R1 =  16×1.05S   = 18.26S           √0.8464T         √T R = 16S         √T % change in R = (18.26 − 16) × 100 = 14.13%                                    16	M1 M1A1  B1  M1A1  M1  M1 M1A1
22	3.5  =   x       7     x+10 7x = 3.5x + 35 x=10 cm Height of bigger cone = 10 + 10 = 20 cm Slant height of bigger cone = √(72 + 202)                                             =21.19 cm Slant height of smaller cone = √(3.52 + 102)                                               =10.59 cm CSA = πRL − πrl = 3.142((7×21.19)  − (3.5×10.59)) = 3.142×11.265 = 349.59cm2 Cost of material = 349.59 × 800 = Sh. 27.9672                               10000 Cost of 10 lamp shades = 2×10×27.9672                                        = Sh. 559.344	M1 A1 M1 M1 M1 M1A1 M1 M1 A1
23.	70 - 74      70 - 74    x  f  fx  cf  62  8  496  8  67  14  938  22  72  18  1296  40  77  15  1155  55  82  3  246  58  87  2  174  60    N = 60  Σfx = 4305   X̄ = 4305 = 71.75kg          60 Histogram	B1 B1         B1 for fx column B1 for f column             M1 A1           B1 for scale B1 for blocks
24.	tan 35 = h/y ⟹ h = y.tan 35  tan 28 =    2h    ⟹ h = (100−y) .tan 28               100−y                  2 y.tan 35= (100−y) .tan 28                       2 0.7y = 0.5317 (100−y)                             2 y = 27.52 m AD = h = 27.52 × 0.7 = 19.264 m BR = 2h = 2 × 19.264 = 38.528m QR = 100 − (27.52 + d) = 72.48 − d PR = 100 − 27.52 = 72.48m tan 28 =   BR     ⟹ BR = 72.48 × tan 28                72.48 tan 75 =      BR                    72.48−d ⟹ BR = (72.48 − d) tan 75 equating the two equations for BR, 72.48 × tan 28 = (72.48 − d) tan 75 d = 270.5 − 38.54 = 62.19 m              3.73	M1 A1 M1A1 M1A1    M1 M1 M1 A1