Mathematics Paper 1 Questions and Answers - Form 3 Mid Term 3 Exams 2023

INSTRUCTIONS TO CANDIDATES

• This paper consists of two sections; Section I and Section II.
• Answer ALL the questions in Section I and ONLY FIVE questions in Section II.
• Show all the steps in your calculations, giving your answer at each stage in the space provided below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise.

SECTION I (50 Marks)

Answer all the questions in the spaces provided.

1. Without using tables or a calculator evaluate; (3 marks)
   −24 ÷ −8 × 5 −(−30)  
     −16 × 4 ÷ 2 + (−8)
2. Factorize completely 3x² − 2xy − y² (2 marks)
3. A Kenyan bank buys and sells foreign currencies as shown below.
   

   Currency	Buying (Ksh)	Selling (Ksh)
   1 Hong Kong dollar 100 Japanese yen	9.84  76.08	9.87  76.12

   A tourist arrived in Kenya with 100 500 Hong Kong Dollars and changed the whole amount to Kenyan shillings. While in Kenya he spent Ksh. 430 897 and changed the balance to Japanese yen before leaving for Tokyo. Calculate the amount in Japanese Yen that he received. (3 marks)
4. Use logarithm tables to evaluate (4 marks)
   
5. The sides of a triangle were measured to 1 decimal as 6.5cm, 7.4cm and 8.2cm respectively. Calculate the percentage error in its perimeter
   (4 marks)
6. Line L₁ passes through points A(2,−4) and B(6, −8). Find the equation of the line L2, the perpendicular bisector of AB leaving your answer in the form ax + by + c = 0 . (3 marks)
7. In the figure below, O is the centre of the circle. AC = 6 cm, BC = 9 cm and ∠ACB = 132°
   
   1. Calculate length AB correct to 2 decimal places. (2 marks)
   2. Calculate the area of the circle. (2 marks)
8. Evalute, without using mathematical tables or the calculator, the expression. (3 marks)
   2 log₁₀5 − ½log₁₀16 + 2log₁₀40
9. Rationalize the denominator and leave your answer in surd form. (3 marks)
         3      
   2√5 + √2
10. Solve the quadratic equation 6a² = 5a + 4 using completing the square method. (3 marks)
11. The hire purchase terms of a cupboard is a deposit of Ksh 4,400 and six monthly installments of Ksh 900 each. The hire purchase price is 175% of the cost price while the cash price is 25% more than the cost price .Calculate the cash price of the cupboard. (3 marks)
12. Construct triangle PQR in which PQ=QR=5 cm and ∠ PRQ=35°. Measure PR. (3 marks)
13. A and B are two matrices. If A =  find B given that A² = A + B. (3 marks)
14. Make n the subject of the formulae in A = P(1 + ^r/₁₀₀)ⁿ (3 marks)
15. Solve the following inequality and show your solution on a number line. (3 marks)
    4x − 3 ≤ ½(x + 8) < x + 5
16. The sixth term of an arithmetic progression is 27 and the tenth term is 43. Find the 16^th term. (3 marks)

SECTION II (50 Marks)

Answer ONLY five the questions in the spaces provided.

17. Mr. Omwega is employed. His basic salary is Kshs. 21, 750 and is entitled to a house allowance of Kshs 15, 000 and a travelling allowance of Kshs 8, 000 per month. He also claims a family monthly relief of Kshs 1, 056 per month. Other deductions are;
    • Union dues Kshs 200 and
    • Co-operative shares Kshs 4, 500 per month.
      The table below shows the tax rates for the year.
      

      Income (Kshs per annum)	Tax rates
      1 - 116,600  116,161 - 225, 600  225, 601 - 335, 040  335,041 - 444, 480  Over 444, 480	10%   15%   20%   25%   30%

      Calculate;
      1. Mr. Omwega’s annual taxable income. (2 marks)
      2. The tax paid by Mr. Omwega in the year. (6 marks)
      3. Mr. Omwega’s net income per month. (2 marks)
18.  
    1. Complete the table below giving your values correct to 2 decimal places. (2 marks)
       

       x°	−90	−75	−60	−45	−30	−15	0	15	30	45	60	75	90
       3 cos 2x°	−3	−2.6		0	1.5		3	2.6		0	−1.5		−3
       Sin (2x + 30)°	−0.5		−1	−0.87		0	0.5		1		0.5	0	−0.5

    2. On the grid provided, draw on the same axes the graph of y=3cos 2x° and y=sin 2x+30° for interval -90°≤x≤90°. Take the scale 1 cm represent 15° on x – axis and 2 cm represent 1 unit on the y – axis. (4 marks)
    3. Use the graph in (b) above to solve the equation;
       1. 3cos 2x° =sin 2x+30° (2 marks)
       2. 6cos 2x° +5=0 (2 marks)
19. Triangle ABC has coordinates A (−4,1, B (−1,2) and C (−3,4).
    1. On the grid provided, draw triangle ABC. (1 mark)
    2. Triangle A'B'C' is the image of triangle ABC under an enlargement with scale factor -1 about the origin. On the same grid draw triangle A'B'C' and state its coordinates. (3 marks)
    3. Triangle A''B''C'' is the image of triangle A'B'C' under a rotation of −90° about the origin. On the same grid draw triangle A''B''C'' and state its coordinates. (3 marks)
    4. Under a certain translation T, the image of points A''B''C'' are mapped onto triangle A'''B'''C''' such that point A'' is mapped onto A'''2, 3.
       1. Find the translation T. (1 mark)
       2. Find the coordinates of the image points B''' and C''' and plot triangle A'''B'''C''' on the same grid. (2 marks)
20. A parent has two children whose age difference is 5 years. Twice the sum of the ages of the two children is equal to the age of the parent.
    1. Taking x to be the age of the elder child, write an expression for
       1. The age of the younger child. (1 mark)
       2. The age of the parent (1 mark)
    2. In twenty years’ time, the product of the children’s age will be 15 times the age of their parent
       1. Form an equation in x and hence determine the present possible ages of the older child. (4 marks)
       2. Find the present possible ages of the parent. (2 marks)
       3. Determine the possible ages of the younger child in twenty years’ time. (2 marks)
21. The masses of 40 students were measured to the nearest kilogram and recorded as shown below.
    52             58               54            51              59            53              56              51
    43             41               53            58              54            65              58              59
    49             63               49            49              47            45              46              52
    52             55               52            55              49            57              53              63
    42             45               46            48              60            49              48              53
    1. Use this data to complete the table below. (4 marks)
       

       Mass in Kg	Class mid point x	Frequency f	fx
       41 - 45
       46 - 50	48	10	480
       51 - 55
       56 - 60
       61 - 65	63	3	189
       		Σf = 40	Σfx =

    2. Calculate ;
       1. The mean mass. (2 marks)
       2. The median mass (2 marks)
    3. Draw a frequency polygon for the distribution. (2 marks)
22. The figure below shows two pulleys with centres A and B of radii 9 cm and 6 cm respectively. C and D are contact points of the belt with the pulleys. The distance between the centres of the two pulleys is 50 cm. a belt is tied around the two pulleys as shown.
    
    Calculate,
    1. Length DC (2 marks)
    2. The length of arc DE (3 marks)
    3. The length of arc CF (3 marks)
    4. The total length of the belt. (2 marks)
23. The figure below shows a rectangle attached to a semi-circle
    
    1. Calculate;
       1. the area of the figure (3marks)
       2. the perimeter of the figure (3 marks)
    2. If the figure represents a cross-section of a steel girder of length 9m, determine the volume of the steel girder in cubic centimetres (2 marks)
    3. If the mass of the steel girder is 306kg, calculate its density in g/cm3. (2 marks)
24. Five towns P,Q, R S and T are situated such that Q is 200 km east of P. R is 300 km from Q on a bearing of 150° from Q. S is 350 km on a bearing of 240° from R. T is 250 km from S and its bearing from P is 160°.
    1. Using a scale of 1 cm to represent 50 km, draw a diagram representing the position of the towns. (5 marks)
    2. From the diagram, determine:
       1. The distance in km from P to T. (1 mark)
       2. The bearing of S from Q. (1 mark)
    3. A plane heading to town R takes off from town P and flies upward at a constant angle which is less than 90°. After flying a distance of 100 km in the air, it sees a town R at angle of depression of 50°. By scale drawing, find the horizontal distance of the plane from R at this point.
       (3 marks)

MARKING SCHEME

NO	WORKING	MARKS	REMARKS
1	Numerator −24÷−8×5+30 3×5+30 15+30=45 Denominator −16×4÷2+(−8) −16×2−8 −32−8=−40 Num =   45   = −11/8 Den     −40	M1 M1 A1
2.	3x2 − 3xy + xy − y2 3xx − y + yx − y (x−y)(3x+y)	M1  M1  A1
3.	100 500 dollars to shillings = 9.84×100 500                                             = sh. 988 920 Balance after expenses = Sh. 988 920-430 897                                       = Sh. 558 023 Amount received in Japanese Yen = 558 023 ×100                                                                      75.12                                      = 742 842 Yen	M1  M1  A1
4.		B1  B1  B1    A1	All logarithms  Addition and subtraction  Cube root  Correct answer
5.	Absolute error = 0.05 Actual Perimeter = 6.5 + 7.4 + 8.2 = 22.1cm Max perimeter = 6.55 + 7.45 + 8.25 = 22.25cm Min perimeter = 6.45 + 7.35 + 8.15 = 21.95 cm % error =½(22.25 − 21.95) × 100%                           22.1 = 0.34  × 100%    44.2 =0.679%	B1  M1  A1
6.	Mid-point of AB = (2+6,−4 + (−8)) = (4, −6)                                 2          2    m1 = −8−(−4) = −4 = −1            6 − 2        4 m2 = 1 Equation of L2 ⇒ 1/1 = y −(−6)                                      x − 4 x − 4 = y + 6 x − y − 10 = 0	M1  M1  A1	Correct coordinates of mid-point  Gradient of line L2  Equation of L2
7.	AB2 = 62 + 92 − 2(6 × 9) cos 132°         = 36 + 81 − 108 cos 132°         = √189.266          = 13.76 cm Area = πR2          = 22/7 ×    (13.76)   ×  (13.76)                          2sin 132      2sin 132          = 22/7 × 9.26 × 9.26 = 269.49 cm2	M1    A1  M1  A1
8.	log1025 − log104 + log101600 log10 (25 × 1600)                  4 log10 10 000    =   4	M1  M1  A1
9.	3(2√ 5 − 2√          (2√5 + √2) (2√5 − √2)          6√5 − 3√ 2          20 − 2√10 + 2√10 − 2 6√5 − 3√2       20−2 6√5 − 3√2  ⇒ 1/3√5 − 1/6√2        18	M1  M1  A1
10.	6a2 − 5a = 4 a2 − 5/6a = 23 a2 − 5/6a + 25/144 = 2/3 + 25/144 (a − 5/12)2 = 0.8403 a − 5/12 = 0.8403 = ±0.9167 a = 0.9167 + 5/12 OR − 0.9167 + 5/12 a = 1.333 OR −0.5	M1  M1  A1
11.	HP = 41400 + (900×6) = 9800/ = Cost price 100/175 × 9800 = 5600 / = ∴ Cash price = 125/100 × 5600 = 7000 /=	M1  M1  A1
12.	PR = 8.3 ± 0.1cm	B1  B1    B1	Lines PQ = QR = 5cm  ∠ R = 35°, ∠Q = 110° and complete triangle
13		B1  M1  A1
14.	(1 + r/100)n = A/P n log (1 + r/100)n = Log A/P n =    log A/P              log(1+r/100)	M1  M1  A1
15.	4x − 3 ≤ 0.5x + 8 4x − 3 ≤ 0.5x + 4 4x − 0.5x ≤ 4 + 3 3.5x ≤ 7 x ≤ 14 0.5(x+8) < x+5 0.5x+4 < x+5 0.5x−x < 5−4 x >−2 −2 < x ≤ 14	M1  A1  B1	Both inequalities solved  Compound inequality  Number line
16.	a + 5d = 27 a + 9d = 43−        −4d = −16 d = 4 a + 20 = 27 ;a = 7 16th term = 7 + 15(4) = 60 + 7                 = 67	B1    M1  A1
17.	(21,750 + 15,000 + 8,000 ) × 12 = Kshs 537,000 Calculation of tax due 1st slab = 116 160 × 10% = Shs. 11 616 2nd slab = 109 440 × 15% = Shs. 16 416 3rd slab = 109 440 × 20% = Shs. 21 888 4th slab = 109 440 × 25% = Shs. 27 360 5th slab = 92 520 × 30% = Shs. 27 756 Tax due = 11 616 + 16 416 + 21 888 + 27 360 + 27 756               = 105 036 Net tax = 105 036 − 12 672             = Sh. 92 364 Net income = Gross income − total deductions = (21 750+15 000+8 000) − (92 364 + 200 + 4 500)                                                    12 = 44 750 − 12 397 = Shs. 32 353	M1   A1  M1  M1  M1  A1  M1  A1  M1  A1
18.	x°  −90  −75  −60  −45  −30  −15   0  15  30  45  60  75  90  3 cos 2x°  −3  −2.6  −1.5  0  1.5  2.6  3  2.6  1.5   0  −1.5  −2.6  −3  Sin (2x + 30)°  −0.5  −0.87  −1  −0.87  −0.5  0  0.5  0.87  1  0.87  0.5  0  −0.5                                                                    x = − 54° and x = 33° 6cos 2x° + 5 = 0 3cos 2x° = −52 = −2.5 x = −75° and x = 75°
19.	B1 - DIagram B1 Plotting B1   Diagram B 1 Coordinates A'(2, −1) B'(−1, −2) and C'(1, −4)B 1 plotting B 1 Diagram B 1 Coordinates A''(−1, −2) B''(−2, 1) and C''(−4, −1)
20.	x − 5 [x + (x−5)] × 2 4x − 10   x + 20x + 15 = 15(4x+10) x2 − 25x + 150 = 0 (x −10) (x − 15) = 0 x = 10 or x = 15 4 × 10 − 10 or 4 × 15 − 10    = 30 or 50 (10 − 5) + 20 or (15 − 5) + 20 = 25 or 30	B1   B1   B1  M1  M1  A1  M1  A1  M1  A1	Correct equation solving
21.	a). The completed table is show below.  Mass in Kg  Class mid point x   Frequency f   fx  41 - 45  43  5  215  46 - 50  48  10  480  51 - 55  53  14  742  56 - 60  58  8  464  61 - 65  63  3  189      Σf = 40  Σfx = 2090 b)  (i)Mean mass x̄ = ∑f x                            ∑f                       = 2090                              40                       = 52.25 kg (ii) Median = 50.5 + (20 – 15) × 5                                      14                   = 50.5 + 25/14                   = 52.29 kg c) Frequency polygon on the attached graph.	B 1 – column for x B 1 – column for f B 1 – column for fx B 1 – Σf x = 2090 M1 A1  M1  A1 B1 all points plotted                        B1 polygon drawn
22	DC2 = 502 − 32 = 2500 − 9 = 2491 DC = √2491 = 49.91 cm cos θ = 3/50 = 0.06  θ = 86.52 × 2 = 173.12° θ =360 − 173.12 = 186.88 l = 186.88 × 22/7 × 2 × 9 = 29.37 cm          360 sin β = 3/50 = 0.06; β = 3.4398     γ = 2(90+3.4398) = 186.88 360 − 186.88 = 173.12 l = 173.12 × 2 × 22/7 × 6 = 18.14 cm          360 Total length = 49.91 + 29.37 + 49.91 + 18.14                      =147.33 cm	M1  A1  B1  M1A1  B1  M1A1  A1
23.	Area of curved part = 22/7 × 4 × 4 × 0.5                                = 25.14 cm2 Area of Rectangular part = 12×6 = 72cm2    Total Area of solid = 25.14 + 72                                 = 97.14cm2 Curved length = 22/7 × 8 =12.57 Total perimeter = 12.57 + 2 + 6 + 12 + 6 + 2 = 40.57cm Volume of solid = 97.14 × 900                          = 87.43 cm3 Density = 306×1000 g/cm3 = 3.499                     87.43                ~3.5g/cm3	M1  M1  A1  M1  M1  A1  M1  A1  M1  A1
24.	4.4 × 50 = 220 ± 5km 180 + 18 = 198° B1 - angle of depression B1 - Construction of perpendicular bisector of PQ from M B1 - 6 × 50 = 300km	B1  B1