Mathematics Paper 1 Questions and Answers - Mwakican Joint Pre Mock Examination 2021

INSTRUCTIONS TO CANDIDATES

1. Write your name and Index number in the spaces provided above.
2. This paper consists of two sections: Section I and Section II.
3. Answer all questions in Section I and only Five questions from Section II.
4. Show all the steps in your calculations giving your answer at each stage in the spaces provided below each question.
5. Non-programmable silent electronic calculators and KNEC mathematical tables may be used.

SECTION I ( 50 MKS)
Attempt all questions.

1. Use tables of reciprocal only to evaluate ¹/_0.325 hence evaluate : ³ √0.000125 (3mks)
                                                                                                         0.325.
2. Two boys and a girl shared some money. The elder got 4⁄9 of it, the younger boy got 2⁄5 of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girls share. (3mks)
3. Annette has some money in two denominations only. Fifty shillings notes and twenty shilling coins. She has three times as many fifty shilling notes as twenty shilling coins. If altogether she has sh. 3,400, find the number of fifty shilling notes and 20 shilling coin. (3mks)
4. The figure below shows a rhombus PQRS with PQ= 9cm and ‹SPQ=60º . SXQ is a circular arc, centre P.
   
   Calculate the area of the shaded region correct to two decimal places (Take Pie= 22⁄7) (4mks)
5. Solve the equation 2x² + 3x=5 by completing the square method (3mks)
6. Simplify the expression 3x² – 4xy² + y (3mks)
                                         9x²—y²
7. Solve the equation 8x² + 2x – 3 =0 hence solve the equation 8Cos²y + 2Cosy – 3= 0
   For the range 0º< y <180º (4mks)
8. Show that the points P(3,4), Q(4,3) and R(1,6) are collinear. (3mks)
9. Solve the inequalities x≤ 2x + 7≤ ⅓x +14 hence represent the solution on a number line. (3mks)
10. The mean of five numbers is 20. The mean of the first three numbers is 16. The fifth number is greater than the fourth by 8. Find the fifth number. (3mks)
11. The volume of two similar solid spheres are 4752cm³ and 1408cm³. If the surface area of the small sphere is 352cm², find the surface area of the larger sphere. (3mks)
12. Solve for x in the equation ½log₂81 + log₂(x - x⁄3) =1 (3mks)
13. A farmer has a piece of land measuring 840m by 396m. He divides it into square plots of equal size. Find the maximum area of one plot. (3mks)
14.    
    1. find the inverse of the matrix  (1mk)
    2. Hence solve the simultaneous equation using the matrix method (2mks)
       4x + 3y =6
       3x + 5y +5
15. In the figure below O is the centre of the circle and <OAB=200. Find;
    
    
    1. <AOB (1mk)
    2. <ACB (2mks)
16. Each interior angle of a regular polygon is 1200 larger than the exterior angle. How many sides has the polygon? (3mks)

SECTION II (50MKS)
Choose any fie questions

17. Three business partners, Bela Joan and Trinity contributed Kshs 112,000, Ksh,128,000 and ksh,210,000 respectively to start a business. They agreed to share their profit as follows:
    30% to be shared equally
    30% to be shared in the ratio of their contributions
    40% to be retained for running the business.
    If at the end of the year, the business realized a profit of ksh 1.35 Million. Calculate:
    1. The amount of money retained for the running of the business at the end of the year. (1mk)
    2. The difference between the amounts received by Trinity and Bela (6mks)
    3. Express Joan’s share as a percentage of the total amount of money shared between the three partners. (3mks)
18. Complete the table below for the function y=x³ + 6x² + 8x for -5 ≤ x ≤1 (3mks)
    

    X	-5	-4	-3	-2	-1	0	1
    X3	-125				-1	0	8
    6X2			54		6	0
    8X	-40		-24	-16		0	8
    Y		0	3			0	15

    1. Draw the graph of the function y=x³ + 6x² + 8x for -5 ≤ x ≤1 (3mks)
       (use a scale of 1cm to represent 1 unit on the x-axis . 1cm to represent 5 units on the y-axis)
    2. Hence use your graph to estimate the roots of the equation
       X³ + 5x²+ 4x= -x² – 3x -1 (4mks)
19. Three islands P,Q,R and S are on an ocean such that island Q is 400Km on a bearing of 030º from island P. island R is 520Km and a bearing of 120º from island Q. A port S is sighted 750Km due South of island Q.
    1. Taking a scale of 1cm to represent 100Km, give a scale drawing showing the relative positions of P,Q,R and S. (4mks)
       Use the scale drawing to:
    2. Find the bearing of:
       1. Island R from island P (1mk)
       2. Port S from island R (1mk)
    3. Find the distance between island P and R (2mks)
    4. A warship T is such that it is equidistant from the islands P,S and R. by construction locate the position of T. (2mks)
20. In the figure below, E is the midpoint of AB, OD:DB=@:3 and f is the point of intersection of OE and AD. 
    Given OA= a and OB= B
    1. Express in terms of a and b
       1. AD (1mk)
       2. OE 2(mks)
    2. Given that AF= sAD and OF= tOE find the values of s and t (5mks)
    3. Show that E,F and O are collinear (2mks)
21. A bag contains 5 red, 4 white and 3 blue beads . two beads are selected at random one after another.
    1. Draw a tree diagram and show the probability space. (2mks)
    2. From the tree diagram, find the probability that;
       1. The last bead selected is red (3mks)
       2. The beads selected were of the same colour (2mks)
       3. At least one of the selected beads is blue. 3(mks)
22. The table below shows how income tax was charged on income earned in a certain year.
    

    Taxable income per year(Kenyan pounds)	Rate shilling per K£
    1-3630  3631- 7260 7261 -10890 10891 - 14520	2 3 4 5

    Mr. Gideon is an employee of a certain company and earns a salary of ksh.15,200 per month. He is housed by the company and pas a nominal rent of Ksh. 1050 per month. He is married and is entitled to a family relief of ksh. 450 per month.
    1. Calculate his taxable income in K£ p.a (2mks)
    2. Calculate his gross tax per month. (4mks)
    3. Calculate his net tax per month (2mks)
    4. Calculate his net salary per month (2mks)
23. The table below shows the distribution of mathematics marks of form 4 candindates in Mavoko Secondary school.
    

    Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
    F	4	7	12	9	15	23	21	5	4

    Use the above date to calculate:
    1. Mean using assumed mean of 65 (3mks)
    2. Median (3mks)
    3. Standard deviation (4mks)
24. Coast bus left Nairobi at 8.00am and travelled towards Mombasa at an average speed of 80Km/hr. At 8.30am, Lamu bus left Mombasa towards Nairobi at an average speed of 120 km per hour. Given that the distance between Nairobi and Mombasa is 400Km.: determine:
    1. The time Lamu bus arrived in Nairobi. (2mks)
    2. The time the two buses met. (4mks)
    3. The distance from Nairobi to the point where the two buses met. (2mks)
    4. How far coast bus is from Mombasa when Lamu bus arrives in Nairobi (3mks)

MARKING SCHEME

1.    1     =    1    x    1    
   0.325     3.25      10^-1
      1    = 0.3077 x 10 = 3.077
   0.325
   ³ √0.000125 = ³√0.000125 = x 3.077
          0.325
   ³√0.000125 = ³√125 x 10^-6 = ³√10^-6 x 5
   5 x 10^-2 = 0.05
   Hence 0.05 x 3.077
   0.15385
2. Let the amount of money shred be x 
   Elder boy → ⁴/₉ x
   Younger boy → ²/₅ × ⁵/₉ x = ²/₉x
   Girl → x - (⁴/₉x + ²/₉x) = ¹/₃x
   ²/₉x ÷ ¹/₃ x 100
   66 ²/₃%
3. Let sh 20 coins be x
   Then sh 50 notes be 3x
   20x + 3x(50) = 3400
   170 x = 3400
   x =3400= 20
         170
   Sh 50 notes = 3(20) = 60
   Sh20 coins = 20
4. Area of the rhombus = 9²sin60º = 70.1481
   Area of sector = ⁶⁰/₃₆₀ x ²²/₇ x 9² = 42.4286
   Shaded area = 70.1481 - 42.4286
   Shaded area = 277195 
   ≈27.72 (2 dp)
5. 2x² + 3x = 5
   x² + ³/₂x + (³/₄)² = ⁵/₂ + (³/₄)² 
   x + (³/₄)² = ⁴⁹/₁₆
   x + ³/₄ = ±√⁴⁹/₁₆
   x = ³/₄ ± ⁷/₄
   x = -2.5 or x =1
6. 3x² - 4xy²  +y
       9x² - y²
   3x² - 4xy + xy + y²
     (3x - y)(3x + y)
   = (3x - y)(x - y) 
     (3x - y)(3x + y)
   = x - y 
     3x + y
   
7. 8x² + 2x - 3 = 0
   8x² + 6x - 4x -3 = 0
   (2x - 1) (4x + 3) = 0
   x = ½ x = -¾
   cos y = ½ or cos y = -¾
   y = 60º or y = 138.6º
8. 
   
   PR = -2PQ shows PR is parallel to PQ and they share a common point P. 
   Therefore, P, Q and R are colinear
9. x ≤ 2x + 7 ≤ -¹/₃x + 14
   x ≤ 2x + 7
   -7 ≤ x
   x ≥  -7
   2x + 7 ≤ -¹/₃x + 14
   6x + 21 ≤ -x + 42
   x ≤ 3
   -7 ≤ x ≤ 3
   
10.    
    1. Let the nuber be a, b, c, d, e
       a + b + c + d + e = 20
                  5
       a + b + c = 16
             3
    2. 48 + d + e = 1
       d + e = 52
       e - d = 8
       2e = 60
       5th number e = 30
11. V.S.F = 4752 = 3.375
               1408
    L.S.F = ³√3.375 = 1.5
    A.S.F = 1.5²
    Area of larger container 
    =352 x 2.25
    = 792cm²
12. ½log₂81 + log₂(x² - ^x/₃) = 1
    log₂9 + log₂ (x² - ^x/₃) = log₂ 2
    9x² - 3x - 2 = 0
    (3x + 1)(3x - 2) = 0
    x = -¹/₃ or x = ²/₃
13. 840 = 2³ x 3 x 5 x 7
    396 = 2² x 3² x 11
    G.C.O = 2² x 3² = 12
    Area = 12 x 12 = 144
14.    
    1. Inverse of 
       
    2. 
       
       x = = ¹⁵/₁₁  y = ²/₁₁
       
       
15.    
    1. <AOB = 180º -(20 + 20) = 140º
    2. <ACB = ½reflex <AOB
       = ½(360 - 140)
       = ½ x 220
       = 110º
16. Exterior x Interior = 120 + x
    120 + x + x = 180
    120 + 2x = 180
    2x = 60
    x = 30º
    360 = 12 sides
     30
17. 1.     40  x 1,350,000 = 540,000
           100
    2. Shared equally = 30  x 1,350,000 x ¹/₃ = 1,350,000
                                100
       Ratio 112:128:210
       Trinity shared in the ratio
       210 x  30  x 1,350,000 = 189,000
       450   100
       Total trinity's amount
       135,000 + 189,000 = 324,000
       Bela's ratio share
       112 x  30  x 1,350,000 = 100,800
       450    100
       Total Bela's amount
       135,000 + 100,800 = 235,800
       Difference = 324,000 - 235,800 = 88,200
    3. Joan's 
       135,000 + 128  x 30  x 1,350,000 = 250,000
                       450    100
         250,000   x 100
        810, 000
       = 30⁸/₉%
18.    
    1. Table
       

       X	-5	-4	-3	-2	-1	0	1
       X3	-125	-64	-27	-8	-1	0	1
       6X2	150	96	54	24	6	0	6
       8X	-40	-32	-24	-16		0	8
       Y	-15	0	3	0		0	15

    2. Curve
    3. Roots of equation
       x³ + 5x² x² + 3x + 1 = 0
       x³ + 6x² + 7x + 1 = 0
       y = x³  + 6x² + 8x
       0 = x³ + 6x² + 7x + 1
       y = x - 1
       x = -4.2
       x = -1.8
       x = 0.2
19.    
    1. Scale drawing
    2.    
       1. Bearing of island R from island P = 82º ± 1 
       2. Bearing of port S from island R = 263º ± 1º
    3. Distance PR = 6.5 x 100 = 650km ± 10km
    4. Bisecting any two sides of PSR
       Locating T in the scale drawing 
20.    
    1. Expressing in terms of a and b, the vectors
       1. AD = AO + OD = -a + ²/₅b or ²/₅b -a
       2. OE = a + ½(-a +b)
          OE = ½a + ½b
          Alternatives
          OE = b + ½(-b +a)
          OE = ½b + ½a
    2. Values of S and T
       AF = sAD = s(²/₅ b - a) = ²/₅sb - sa
       OF = O±E = (½a + ½b) = ½±a + ½±b
       Also
       OF = OA + AF = a - sa + ²/₅b = (1-5)a + ²/₅sb
       Comparing the coefficients of a and b
       ½t = 1 - s
       s = 1 - ½t
       2x½t = ²/₅s x 2     t = ⁴/₅s
       s = 1 - ½(⁴/₅s) → s = 1 - ²/₅s
       s + ²/₅s = 1 → ⁷/₅s = 1 s = ⁵/₇
       ^{t = 4}/₅(⁵/₇) → t = ⁴/₇ 
    3. O, F and E colinear
       consider OF and OE
       OE = ½(a + b) and OF = (1 -s)a + ²/₅ b = (1 - ⁵/₇) a + ²/₅ b
       = ²/₇(a + b)
       let OE = KOF
       ½(a + b) = ²/₇(a + b)
       K = ²/₇ = ½ = ⁴/₇ hence OE = ⁴/₇ OF 
       and O is common hence O, F and E are colinear
21.    
    1. 
       
    2.       
       1. P(RR) or P(WR) or P(BR)
          (⁵/₁₂ x ⁴/₁₁) + (⁴/₁₂ x ⁵/₁₁) + (³/₁₂ x ⁵/₁₁)
          ⁵/₃₃ + ⁵/₃₃ + ⁵/₄₄ = 20 + 20 + 15 
                                              132
          = ⁵⁵/₁₃₂ = ⁵/₁₂
       2. P(RR) or P(WW) or P(BB)
          (⁵/₁₁ x ⁴/₁₁) +(⁴/₁₂ x ³/₁₁) + (³/₁₂ x ²/₁₁)
          = ²⁰/₁₃₂ + ¹²/₁₃₂ + ⁶/₁₃₂ = ³⁸/₁₃₂ = ¹⁹/₆₆
       3. P(RB) or P(WB) or P(BB) or P(BW) or P(BR)
          (⁵/₁₁ x ³/₁₁) + (⁴/₁₂ x ³/₁₁) + (³/₁₂ x ²/₁₁) + (³/₁₂ x ⁴/₁₁) +(³/₁₂ x ⁵/₁₁)
          ¹⁵/₁₃₂ + ¹²/₁₃₂ + ⁶/₁₃₂ + ¹²/₁₃₂ + ¹⁵/₁₃₂
          ⁶⁰/₁₃₂ = ⁵/₁₁ 
          
          
22.      
    1. Taxable income 
       (¹¹⁵/₁₀₀ x 15,200) - 1050
       17480 - 1050
       16430
       16430 x 12 = 9858
              20
    2. 1st £3630 → 2 x 3630 = 7260
       2nd £3630 → 3 x 3630 = 10890
       Remaining £2598 → 4 x 2598 = 10392
       Total tax = Sh28543
       Gross tax per month = 28543 = 2378050
                                            12
    3. Net tax = sh(2378.50 - 450)
       1928.50
    4. Net salary sh(15200 -(1928.50 + 1050))
       = 12221.50
23. 
    

    Marks	M.P(x)	f	CF	d=x-65	t	t2	ft	ft2
    10-20	15	4	4	-50	-5	25	-20	100
    20-30	25	7	11	-40	-4	16	-28	112
    30-40	35	12	23	-30	-3	9	-36	108
    40-50	45	9	32	-20	-2	4	-18	36
    50-60	55	15	47	-10	-1	1	-15	15
    60-70	65	23	70	0	0	0	0	0
    70-80	75	21	91	10	1	1	21	21
    80-90	85	5	96	20	2	4	10	20
    90-100	95	4	100	30	3	9	12	36
    							εft=-74	εft2=448

    Mean = A + (εft x c) A = 65
                        εf
    = 65 + (-74 x 10)
                    100
    = 65 - 7.4 = 57.6
    Median, M = L + (^N/₂ - cp)i     100 = 50
                                   f              2
    59.5 + (50 - 47)10
                    23
    60 + ³/₂₃ x 10
    60 + 6.522 
    = 66.522
    Standard deviation, s = c√εf² - (εf)²
                                           εf      εf
    s² = [448 - ( -74 )² ] x 10²
            100      100
    s² = [4.48 - 0.5476]100
    s² = 3.9324 x 100
    s² =393.24
    s = √393.24   s = 19.8303    s ≈ 19.88
24.    
    1. Time taken = 400  = 3½hrs
                           120
       Arrival time = 8.30 + 3hrs20 mins
       = 11.50am
    2. Relative speed = 120 + 80 = 200km/hr
       Distance between them at 8.30am
       = 400 - (80 x ½)
       = 400 - 40
       = 360km
       Time taken = 360  = 1.8hrs = 1hr 48 mins
                           200
       Time of meeting = 8.30am + 1hr 48mins
       = 10.18 am
    3. Distance travelled = 1.8 x 80 = 144 + 40
       184 km
    4. Lamu bus taken 3hrs 20mins
       Arrival time = 11.50am
       Coast bus → Distance covered in 3hrs 20mins
       =80 x 10 =  800 
              3           3 
       =266.67km
       Distance required = 400 -(266.67 + 40)
       = 93.33km