Mathematics Paper 1 Questions and Answers - Mumias West Pre Mocks 2022

QUESTIONS

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

1. Without using a calculator evaluate(3mks)
   5x⁶ + (-76)÷4+27÷3
         (-5)÷3 x (-4) 
2.      
   1. Express 2268 in terms of its prime factors (1mk)
   2. Hence determine the smallest positive number x such that 2268x is a perfect square. (2mks)
3. Elvis arrived in Kenya with 5000 sterling pound, he exchanged it to Kenya Shilling and spent sh. 267 100. Before jetting out of the country, he exchanged the balance into Euros. Using the exchange rates below, calculate the amount he obtained in Euros in Kenya shillings. (3mks)
   Currency             Buying    Selling
   1 Sterling pound   114.20      114.50
   1Euro                    101.20     101.30
4. Simplify the expression (3mks)
   2x²+3x-2
     x³- 4x
5. When two wires of length 179m and 234m are divided into pieces of equal lengths a remainder of 3m is left in each case. Find the least number of pieces that can be obtained. (3mks)
6. Without using calculator, solve for n in the equation 1 - (¹/₃)ⁿ = ²⁴²/₂₄₃ (3mks)
7. Solve for y in the equation 7 - y  - 9 - 2y = ½ (3mks)
                                                4         3
8. Two similar solids have surface area of 48cm² and 108cm² respectively. Find the volume of the smaller solid if the bigger solid has a volume of 162 cm³ (3 mks)
9. Use reciprocal table only to evaluate ¹/_0.325 (3mks)
   Hence, evaluate √0.25  to 1.d.p
                              0.325
10. A plot measuring 1.2m by 19.1 m is surrounded by a path 0.5m wide. Find the area of the path in square metres. (3mks)
11. The interior angle of a regular polygon is 600 more than its exterior angle, find the number of sides of the polygon. (3mks)
12. Complete the following solid given that ABC is its cross-section (3mks)
    
13. If tan x = ¹/_√3 find without using tables or calculator the value of Sin (90-x)+ cos (90-x) leaving your answer in simplified surd form (3mks)
14. A line perpendicular to the line 3y-2x=2 passes through the point (-3,2). Determine the equation of the line and write it in the form ax +by = c where a, b, and c are constant. (3mks)
15. Given that A = (⁴_-1 ³_-2) and C = (³_-1 ⁷_-2) (4mks)
    Find B such that A² + B = C^-1
16. Ali travelled a distance of 5km from village A to village B in direction of N60ºE. He then changed direction and travelled a distance of 4km in the direction of 135º to village C.
17. Using a scale of 1cm to represent 1.0 km represent the information on an accurate diagram. (2mks)
    1. Using scale drawing (a) above determine
    2. distance between A and C (1mk
    3. bearing of A from C (1mk)

SECTION B (50 MARKS):
Answer all the questions in this section in the spaces provided.

17. The figure below shows a frustrum. The top and bottom radii are 5cm and 10cm respectively, while the vertical height of the frustrum is 12cm.
    
    Find the:-
    1. Slant height of the frustum. (3marks)
    2. Curved area of the frustum. (3marks)
    3. Volume of the frustum. (4marks)
18. Bumala is a market centre 600km from Kisumu town.A bus starts from Kisumu for Bumala at 7.00am at an average speed of 80 km/h. At 8.30 am a car started from Kisumu to Bumala and moved at an average speed of 120 km/hr. Calculate
    1. The distance bus covered before the car started moving. (3marks)
    2. The relative speed for the two vehicles. (2marks)
    3. The time the car overtook the bus. (1 mark)
    4. Distance covered by the car before overtaking the bus. (2marks)
    5. Distance from Bumala to the car at the time the car was overtaking the bus. (2marks)
19. The height of 36 students in a class was recorded to the nearest centimeter as follows:-
    148 159 158 163 166 155 155 179 158
    161 160 157 165 165 175 173 172 178
    147 168 157 172 165 154 170 157 167
    155 159 173 171 168 160 172 156 167
    1. Make a frequency distribution table using a class interval of 5 and starting with the class 145 – 149. (2marks)
    2. From the table above
       1. Calculate the mean mark (3marks)
       2. Calculate the median (3marks)
    3. Draw a frequency polygon using the table in (a) above. (2 marks)
       
20. Bujumba Boys Secondary School. Intends to buy a certain number of chairs For Ksh. 16,200. The supplier agreed to offer a discount of Ksh. 60 per chair which will enable the school to get 3 chairs more.
    Taking y as the originally intended number of chairs:-
    1. Write an expression in terms of y for
       1. Original price per chair. (1mark)
       2. Price per chair after discount. (1mark)
    2. Determine
       1. The number of chair the school originally intended to buy. (4marks)
       2. Price per chair after discount. (2marks)
       3. The amount of money the school would have saved per chair of it got the intended number of chairs at a discount of 15%. (2marks)
21.      
    1. Without using a protractor, construct triangle ABC such that angle ABC = 60º, BC = 8cm and AC = 9cm.Measure AB. (3marks)
    2. Drop a perpendicular from A to BC and measure its length. (2marks)
    3. Hence calculate the area of triangle ABC. (2marks)
    4. Locate a point D on BC such that the area of triangle ABC is three times that of triangle ABD. (3marks)
22. In triangle ABC, shown below, AB = a AC = b point M lies on AB such that AM: MB = 2:3 and point N lies on AC such that AN: NC = 5:1 line BN intersects line MC at X.
    
    
    1. Express the following in terms of a and b
       1. BN (1 mark)
       2. CM (1 mark)
    2. Given that BX = kBN and CX = rCM where k and r are scalars
       1. Write two different expressions for AX in term of a, b, k and r (4marks)
       2. Find the values of k and r (4 marks)
23. A triangle ABC has vertices A(2,1), B(5,2) and C(0,4).
    1. On the grid provided plot the triangle ABC. (2 marks)
       
    2. A¹B¹C¹ is the image of ABC under a translation (²_-5). Plot A¹B¹C¹ and state its coordinates. (2 marks)
    3. Plot A¹¹B¹¹C¹¹ the image of A¹B¹C¹ after a rotation about the origin through a negative quarter turn. State its coordinates. (3 marks)
    4. A¹¹¹B¹¹¹C¹¹¹ is the image of A¹¹B¹¹C¹¹ after a reflection on the line y = 0.
       Plot A¹¹¹B¹¹¹C¹¹¹ and state its coordinates.(3 marks)
24. The table below shows income tax for a certain year
    

    Monthly Income in Kenya Shillings (Ksh)	Tax Rate
    0 - 10164	10%
    10165 - 19740	15%
    19741 - 29316	20%
    29317 - 38892	25%
    Over 38892	30%

    A tax relief of Ksh 1162 per month was allowed. In a certain month of the year, an employee’s taxable income in the fifth band was Ksh.2108.
    1. Calculate
       1. Employees total taxable income in that month (2 Marks)
       2. The tax payable by the employee in that month (5 Marks)
    2. The employee’s income include a house allowance of Ksh 15,000 per month. The employee contributed of the 5% basic salary to a co-operative society. Calculate the employee’s net pay for that month (3 Marks)

MARKING SCHEME

1. 5x⁶ +(-76) ÷ 4 +27÷3
        -15÷ 3 x (-4)
   = 30 + (-19) + 9
            -5 x-4
   = ²⁰/₂₀ = 1
2. 2268= 2² x 3⁴ x 7
   x = 2² x 34 x 7²= 7
          2² x 3⁴ x 7
3. Amt in Ksh = 5000 x 114.2
   = 571000
   Remainder = 571000-276100
   = 303900
   Amt in Euros = 303900
                            101.30
   = 3000 Euros M1
4. 2x² + 3x -2 = (2x -1)( x+ 2)
       X³-4x           x(x² -4)
   = (2x -1)( x+ 2)
     x( x -2)( x + 2)
   = 2x - 1 
     x( x- 2)
5. 179 - 3 = 176
   234 – 3 = 231
   176 = 2⁴ x 11
   234 = 3 x 7 x 11
   GCD = 11
   Number of pieces = ¹⁷⁶/₁₁ + ²³¹/₁₁
   = 16 + 21
   = 37
6. 1(¹/₃)ⁿ = ²⁴²/₂₄₃
   (¹/₃)ⁿ= 1 - ²⁴²/₂₄₃
   = ¹/₂₄₃
   (¹/₃)ⁿ = (¹/₃)⁵
   ∴ n =5
7. 4 - y - 9 - 2y= ½
     4         3 
   3(4-y) – 4(9-2y) =6
   12-3y -36 +8y = 6
   5y =30
   Y=6
8. For locus (i)
   For locus (ii)
   For locus (iii)
   For the region
9.     1     =       1       
   0.325    3.25×10^-1
   = 0.3077×10^-1
   = 3.077
   ^√0.25/_0.325 = 0.5×3.077
   = 1.5385
   = 1.5
10. External area = 20.1 × 2.2 = 44.22
    Internal area = 19.1 × 1.2 = 22.92
    Area of path = 44.22 – 22.92
    = 21.3 m²
    
    
11. x + x +60 = 180°
    2x = 120˚
    x = 60˚
    Exterior =60
    No. of sides = ³⁶⁰/₆₀
    = 6 sides
12. 
    
    Sides
    Broken lines
    Shape of solid
13. Total = 42×24 = 1008
    Total with abs = 1008-65
    = 943
    Average = ⁹⁴³/₂₃ =41
14. γ = ^2x/₃ + ²/₃
    Gradient = ^-3/₂
    γ - 2 = ^-3/₂
    x--3
    2(γ-2) = -3(x+3)
    2γ – 4 = -3x – 9
    3x + 2γ = -5
15. B = A² + C^-1
    = (⁴_-1 ³₂)(⁴_-1 ³₂) + (^-2₁ ^-7₃)
    = (¹³_-6 ¹⁸₁) + (^-2₁ ^-7₃)
    = (¹¹_-5 ¹¹₄)
16.      
    1. AC = 7.2km
    2. Bearing 273º
17.        
    1. X = 100-78
       = 22
    2. Modal class 35-44
    3. 
       

       	x	f	cf	d	fd
       15-24	19.5	6	6	-40	-240
       25-34	29.5	14	20	-30	-420
       35-44	39.5	24	44	-20	-480
       45-54	49.5	14	58	-10	-140
       55-64	59.5	22	80	0	0
       65-74	69.5	10	90	10	100
       75-84	79.5	6	96	20	120
       85-94	89.5	4	100	30	120
       		100			-940

       Median = 44.5 + 50-44 × 10
                                    14
       = 44.5 + ⁶/₁₄ × 10
       = 44.5 + 4.29
       = 48.79
    4. Mean = ^∑fd/_Σf + A
       = 59.5 - ⁹⁴⁰/₁₀₀
       = 50.1
18.  
    1.     x     = ⁵/₁₀
       x + 12 
       10x = 5x + 60
       5x = 60
       X = 12
       L = 24² + 10² = 26
       ¹/₂₆ = ⁵/₁₀
       L = 13
       Slant height 26 – 13
       = 13
    2. Curved area of the frustum
       S.A = πRL - πrl
       = π(260 – 65)
       = 612.6
    3. Volume of the frustrum
       ¹/₃π(R²H – r²h)
       =¹/₃π(2400 – 300)
       =¹/₃π(2100)
       = 2200 cm²
19.      
    1. 1½ x 80
       = 120km
    2. 120 – 50
       = 40
    3. Time taken
       ¹²⁰/₄₀
       = 3hrs
       8.300 + 3
       11.30A.m
    4. 3 x 120
       = 360km
    5. 600 – 360 = 240km
       For 240
       Total marks 10
20.      
    1. 
       

       Class	Tally	Frequency	Mid point	Fx	Cf
       145 – 149	11	2	147	294	2
       150 – 154	1	1	152	152	3
       155 – 159	1	11	157	1727	14
       160 – 164		5	162	810	19
       165 – 169	11	7	167	1169	26
       170 – 174	11	7	172	1204	33
       175 - 179	111	3	177	531	36
       				Fx2 5887

    2.    
       1. 5887
          36
          = 163.53
       2. 159.5+(³⁶/₂-14)5
                         11
          = 159.5 + 1. 8182
          = 161 . 32 
    3. Graph
       
       Plotting
       Curve
       (No hanging curve. Not drawn by use of free hand)
       Total marks 10
21.      
    1.    
       1.  16200
             Y
       2. 16 200 or 16 200 – 60
           y + 3            y
    2.    
       1.  16 200 – 16 200 = 60
             y            y + 3
          60y² + 180y – 48 600 = 0
          Y² + 3y – 810 = 0
          (y + 30) ( y – 27) = 0
          = 27 or -30
          = y = 27
       2. 16200 = Ksh 540
             30
       3.  15  x 16 200
          100       27
          = Ksh 90
22.     
    1. 
       
       Constructed locating point A
       Dropping a
       Length the 8.2 + 0.1
       For a line at an angle to BC
       For joining C to the last point
       For locating point D
       (following through)
       Total marks 10
    2. 1⁄2×8×8.2=32.8Cm²
23.      
    1.      
       1. BN = BA + AN
          = -a + ⁵/₆ b
          = ⁵/₆b – a
       2. CM = -b + ²/₅a
          = b + r (²/₅a - b)
       3. AX = b + rCM
          = b – rb + ²/₅ra
          = (1 – r) b + ²/₅ar
          Ax = a + k ( ⁵/₆b – a)
          (1 – k)a + ⁵/₆kb
       4. (1 – r) b + ²/₅ar = ( 1 -k) a + ⁵/₆kb
          ⁵/₆K = 1 – r – (i)
          ²/₅r = 1 – k -- (ii)
          K = ⁶/₅ – ⁶/₅r
          ²/₅ = 1- (⁶/₅ –⁶/₅r) = 1 – ⁶/₅ +⁶/₅r
          -4r = -1 = 7 r = ¼
          ⁵/₆K = ¾
          K = ⁹/₁₀
24.      
    1. Equating two values of AX.
       Two equations extracted
    2. A¹(4,-4) B¹(7,-3) C¹(2,-1)
    3. A¹¹(4,4) B¹¹(3,7) C¹¹(1,2)
    4. A¹¹¹(4,-4) B¹¹¹(3,-7) C¹¹¹(1,-2)
       
       For plotting
       For ∆ABC
       For ∆A¹B¹C¹
       For construction or otherwise
       For ∆A¹¹B¹¹C¹¹
       For construction or otherwise
       For ∆A¹¹¹B¹¹¹C¹¹¹
25.      
    1. S = t (t² – t – 2t + 2)
       = t³ – 3t² + 2t
       ^ds/_dt = 3t² – 6 (2) + 2
       V = 3 ( 4) - 6 ( 2) + 2
       V = 2m/s
    2. ^dv/_dt = 0
       ^dv/_dt = 6t – 6
       6t – 6 = 0
       t = 1
       = 3(1)2 – 6(1) + 2
       = -1m/s
       3t² – 6t + 2 = 0
    3. T = 6 + (-6)2 – 4 ( 3) ( 2)
       T = 6 + 3.464
       T = 1.577 or 0.4227
       a = 6 ( 3) – 6 = 12m/s²