Mathematics Paper 2 Questions and Answers - Form 3 End Term 1 Exams 2021

SECTION A (50MARKS)

Answer all questions in this section in the spaces provided.

1. Evaluate: ⅜ of 7³/₅ - ⅓(1¼ + 3⅓) x 2²/₅ (2mks)
2. Solve the simultaneous equations (4mks)
   xy = 4
   x + y = 5 
3. A man invests Ksh.24, 000 in an account which pays 16% interest p.a. the interest is compounded quarterly. Find the amount in the account after 1½ years. (3mks)
4. Given that position vectors of points A and B are a =  and b =  and C is a point on AB such that
   AC : CB = 1 : 2. Find the coordinates of the point C. (3mks)
5. The table below shows the number of goals scored in 40 soccer matches during a certain season.
   

   No. of goals	0	1	2	3	4	5	6	7
   No. of matches	3	9	6	8	5	5	2	1

   Calculate the mean number of goals scored per match. (3mks)
6. Simplify:   (3mks)
   
7. From a point 20m away on a level ground the angle of elevation to the lower window line is 25° and the angle of elevation to the top line of the window is 34°. Calculate the height of the window. (3mks)
8. Find the distance round the figure given below. (Take Π = ²²/₇) (2mks)
   
9. Find all the integral values of x which satisfy the inequalities. (3mks)
   x + 8 > 4x - 6 ≥ 3(4 - x)
10. Agnes paid rent which was ¹/₁₀ of her net salary. She used ½ of the remaining amount to make a down payment for a plot. She gave her mother Kshs. 2,500 and did shopping worth Kshs. 7,500 for herself. She saved the remainder which was Ksh. 12,500. How much was the down payment that she made. (4mks)
11. The figure below is a velocity time graph for a car.
    
    1. Find the total distance traveled by the car. (2mks)
    2. Calculate the deceleration of the car. (2mks)
12. A chord PQ of length 15cm subtends an angle of 65º at the circumference centre O. Find the radius of the circle. (3mks)
13. The interior angles of a hexagon are 2x + 5, 4x − 5, 4x + 5, 3x, 4x −20 and 2x.
    Find the value of x. (3mks)
14. A line passes through the point (−1, 2) and has gradient −½. Write down its equation in the form
    ax + by = c  (3mks)
15. Given x = 13.4cm and y = 4.3 cm. calculate the percentage error in ^X/_Y correct to 4 d.p   (4mks)
16. The figure below is a cone whose base radius is 3.5cm and slant height 7cm. The net of the cone is a sector of a circle.
    
    1. Find the angle subtended at the centre of the sector. (2mks)
    2. Draw the net of the solid. (1mk)

SECTION II (50 MARKS)

Answer any five questions from this section in the spaces provided.

17. A straight line L₁ has a gradient ˉ½ and passes through point P (−1, 3). Another line L₂ passes through the points Q (1, -3) and R (4, 5). Find.
    1. The equation of L₁. (2mks)
    2. The gradient of L₂. (1mk)
    3. The equation of L₂. (2mks)
    4. The equation of a line passing through a point S (0, 5) and is perpendicular to L₂. (3mks)
    5. The equation of a line through R parallel to L₁. (2mks)
18.  
    1. Given that y = 7 + 3x − x², complete the table below. (2mks)
       

       x	−3	−2	−1	0	1	2	3	4	5	6
       y	−11			7						−11

    2. On the grid provided and using a suitable scale draw the graph of
       y = 7 + 3x − x². (2mks)
    3. On the same grid draw the straight line and use your graph to solve the equation
       x² - 4x – 3 = 0. (3mks)
    4. Determine the coordinates of the turning point of the curve. (2mks)
19. From a reservoir, water flows through a cylindrical pipe of diameter 0.2m at a rate of 0.35m per second.
    1. Determine the number of liters of water discharged from reservoir in one hour. (4mks)
    2. The water flows from the reservoir for 18 hours per day for 25 days per month and serves a population of 2500 families. Determine the average consumption of water per family per month giving your answer to the nearest 100 litres. (4mks)
    3. The water is charged at the rate of sh.4.50 per 100 litres. Calculate the average water bill per family per month. (2mks)
20. In the figure below PQR is a tangent to the circle at Q. TS is a diameter and TSR and QUV are straight lines. QS is parallel to TV. Angles SQR = 40° and TQV = 55°.
    
    1. Find the following angles giving reasons each case.
       1. <QTS (2mks)
       2. <QRS (2mks)
       3. <QVT (2mks)
       4. <QUT (2mks)
    2. Given that QR = 8cm, and SR = 4cm. Find the radius of the circle. (2mks)
21. Mr. Kimutai a teacher from Tuiyotich Secondary School earns K₤ 12000 per annum and lives in a house provided by the employer at a minimum rent of Ksh.2000 per month. He gets a family relief of K₤ 1320p.a. and is entitled to a relief of 10% of his insurance of K₤ 800p.a.
    1. Calculate his annual tax bill based on the table below. (6mks)
       Income slab in K₤ p.a.              Rate
          1 – 2100                               10%
       2101 – 4200                            15%
       4201 – 6300                            25%
       6301 – 8400                            35%
         Over 8400                              45%
    2. Kimutai other deductions include.
       • W.C.P.S = sh 600.00pm
       • NHIF = sh 500.00pm
         Calculate Kimutai’s net salary monthly. (4mks)
22.  
    1. The diagram represents a solid frustum with base radius of 35cm and top radius of 21cm.The frustum is 22.5cm high and is made of a metal whose density is 4g/cm³. Taking π = ²²/₇
       
       Calculate;
       1. The volume of the metal that makes the frustum. (6mks)
       2. The mass of the frustum. (2mks)
    2. The frustum is melted in down and recast into a solid cube in the process 20% of the metal is lost. Calculate to 2d.p the length of each side of the cube. (2mks)
23. Four points B, C, Q and D lie on the same plane. Point B is 42km due southwest point Q. Point C is 50 km on a bearing of S60°E from Q. Point D is equidistant from B, Q and C.
    1. Using the scale: 1cm represents 10km, construct a diagram showing the positions of B, C, Q and D. (5mks)
    2. Determine the
       1. Distance between B and C (1mk)
       2. Bearing of D from B (2mks)
    3. Find the distance and bearing of D from C. (2mks)
24.  
    1. A small field was surveyed and the measurements recorded in a surveyor’s field book as in the table below:
       
       1. Using a scale of 1cm to 10m make an accurate drawing of the map of the field. (4mks)
       2. Find the area of the field. (3mks)
       3. Assuming that the baseline in (a) runs in a northern direction give the position of D relative to A using bearing and distance. (3mks)

Marking scheme

1.  
   
   ³/₈ of ⁵⁹/₁₅ = ⁵⁹/₄₀ 
    =1¹⁹/₄₀
2. xy = 4
   x + y = 5
   x = ⁴/_y                     
   ⁴/_y + y = 5
   4 + y² = 5y               
   y² − 5y + 4 =0
   y² − y − 4y + 4 = 0
   y(y−1) − 4(y −1) = 0
   (y − 4)(y−1) = 0
   y = 4 ; y = 1              
   x = 1   x = 4             
3. A = P(1 + ^R/₁₀₀)ⁿ
      =24000(1 + ¹⁶/₁₀₀ × ¼)⁶
      =24000(1+0.04)⁶
      =Ksh. 30367.70
4. OC = OA + AC
   
   OC = OA + ¹/₃AB
     = a + ¹/₃( a + b)
    =²/₃a + ¹/₃b
   
   ∴ C(−⁴/₃, 3)
5. x̄ = Σƒx
          ∑ƒ
   (3×0) + (9×1) + (6×2) + (8×3) + (5×4) + (5×5) + (2×6) + (1×7) = 0+9+12+24+20+25+12+7
   =109
      39
   =2.79
   =3 goals
6.  
   
     x(x − y)   × y(x−y)(x+y)
   (x+y)(x+y)      (x−y) (x−y)
    xy 
   x+y
7. 
   
   ^x/₂₀ = Tan 25
   x = 20 Tan 25
   ^y/₂₀ = tan 34
   y = 20 Tan 34
    ∴ height of window
   = 20 Tan 34° − 20 Tan 25°
   =13.49 − 9.33
   =4.16m
8. Hyp² = 6² + 8²
           = 36 + 64
           =100
   Hyp = 10cm
   πD = ²²/₇ × 14 =44
   ∴ perimeter = 10+22+6+10+22+6
   =76cm
9. x + 8 > 4x − 6
   −3x > −14
   x<¹⁴/₃
   4x − 6 ≥ 12 − 3x
   7x ≥ 18
   x ≥ ¹⁸/₇
   2.57 ≤ x ≤ 4.67
   ∴ 3, 4
10. Let her salary be x
    amount after rent = ⁹/₁₀x
    ½ of ⁹/₁₀x = ⁹/₂₀x
    ⁹/₂₀x = 2500 + 7500 + 12500
    ⁹/₂₀x = 22,500
    x = 22500 × 20
                  9
       = 50,000
    ∴ Down Payment ½ of 45,000 = 22,500
    =Kshs. 22,500
11.  
    1. Area under the curve = Distance
       (½ × 4 × 80) + (16 × 80) + (½ × 4 × 80)
       =160 + 1280 + 160
       =1600m
    2. Deceleration = change in speed
                                Time taken
                         = ⁸⁰/₄
                         =20m/s² 
12. 7.5 = Sin 32.5°
      r
    r =    7.5   
         Sin 32.5
    =12.92cm
13. (2x + 5) + (4x − 5) + (4x + 5) + 3x + (4x − 20) + 2x = 90(2×6−4)
    19x +−15 = 720
    19x = 735
    x = 38.68°
14. y−2 =−½
    x+1
    2(y−2) =−1(x+1)
    2y−4 =−x−1
    2y+3=3
15. Max val of x 13.45 min 13.35
    Max val of y 4.35 min 4.25
    max value of x  13.45 = 3.1647
                        y   4.25
    min value of x  13.35 = 3.069
                       y   4.35
    actual value of x  13.4 = 3.1163
                          y   4.3
    Absolute error  3.1642  3.069 = 0.04785
                                   2
    Percentage Err. 0.04785 × 100 = 0.015355 × 100
                            3.1163
    =1.5355%
16.  
    1. Surface area of a cone = πrl
       = ²²/₇ × 3.5 × 7
       =77
       Area of a sector =  θ  πr²
                                 360
       (^θ/₃₆₀ × ²²/₇ × 7 × 7 =77)
       θ = 77 × 360
               22 × 7
       =180°
    2. 
       
17.  
    1. y−3 =½
       x+1
       y−3 =−½x −½
       y=−½x + ⁵/₂
    2. g= 5−−3
              4−1
       =⁸/₃
    3. y−5 = 8
       x−4    3
       y−5 = ⁸/₃x − ³²/₃ 
       y = ⁸/₃x − ³²/₃ +5
       y = ⁸/₃x − ¹⁷/₃
    4. y−5 × − 3
         x          8
       y = ⁻³/₈x + 5
    5. y−5= −½
       x−4
       y−5 =−½x +2
       y=−½x + 7
18.  
    1.  
       

       x	−3	−2	−1	0	1	2	3	4	5	6
       y	−11	−3	3	7	9	9	7	3	−3	−11

    2. Scale-1mk
       Curve-1mk
       
    3. y=7+3x−x²
       0= −3−4x+x²
       ∴ 

       x	0	2
       y	4	2

    4. (1.5, 9.8)
19.  
    1. Volume per second πr²x Rate
       =3.14 × 0.1 × 0.1 × 0.35
       =0.01099m³
       =0.01099 × 1000
       1099Litres per second
       In one hour = 10.99 × 3600
                         =39,564litres per hour
    2. 39564×18×25 = 7121.52litre
             2500
       =7100litres
    3. 7100 × 4.50 = Sh. 319.50
        100
20.  
    1.  
       1. <QTS = 40°
          Reason: Angles in the alternate segments are equal.
       2. <QRS = 10°
          Reason: Sum of angles in a triangle add up to 180°
       3. <QVT=35°
          Reason: Alternate angles in parallel lines are equal.
       4. <QUT = 50°
          Reason: <QST = 50°, subtended by chord TQ. TQ subtends <QUT, hence <QUT = 50°
    2. 8 × 8 = 4(4+x)
       16 + 4x =64
       4x = 48
       x = 12
       Radius = ¹²/₂ = 6cm
21.  
    1. 
       

       Income slab in K₤ p.a.	Rate	Tax in ₤
       1 – 2100	10%	210
       2101 – 4200	15%	315
       4201 – 6300	25%	523
       6301 – 8400	35%	735
       Over 8400(12600−8400−4200)  Gross tax.	45%	1890  3675

       
       Taxable Income = 12000 + 15% of 12000 − rent
       115 × 2000 − 1200
       100
       =₤12,600 p.a 
       Net tax = 3675 − 1320 − 80
       = ₤2275
    2. Net salary p.a = 12600 − (₤2275 + 360 + 300)
       =₤9665 p.a
       = 9665 × 20
                12
       =Sh. 16,108
22.  
    1. 
       
       1.    h        = 21
          h+22.5       35
          35h = 21(h + 22.5)
          35h − 21h = 472.5
          14h − 472.5
          h = 472.5
                   14
          h=33.75
          ∴ vol of frustum = ¹/₃^π(R²H − r²h)
          ¹/₃ × 3.14(35² × 56.25 − 21² × 33.75)
          ¹/₃ × 3.14(68906.25 − 14883.75) 
          =56543.55cm³ 
       2. Mass = D × V
          = 56543.55 × 4 = 22174.2g
    2. ⁸⁰/₁₀₀ × 56543.55 = 45234.84
       L×L×L = 45234.84
       =35.63cm
23.  
    1. 
       
    2.  
       1. BC = 7.4 × 10
               =74km ±1
       2. 063° ±1
    3.  Distance; DC = 4.3 × 10
                   =43km ±1
       Bearing = 282° ±1
24.  
    1. 
       
    2. A = ½×2×2 =2cm²
       B = ½×3×2.5=3.75cm²
       C= ½(3+2)4 =10cm²
       D = ½(4+2.5)4.5 = 140625cm²
       E = ½×3×5 = 7.5cm²
       F = ½×4×3.5 =7cm²
       Total Area = 44.875cm²
       Actual Area = 4487.5m²
    3. Bearing 031° ±1°
       Distance = 7.8 × 10 =78m ±1