Mathematics Paper 1 Questions and Answers - Mangu High School Mock Exams 2023

• This paper consists of two sections; Section I and Section II.
• Answer All questions in Section I  and only Five questions  from section II
• All answers and working must be written on the question paper in the spaces provided below each question.
• Show all the steps in your calculations giving answers at each stage in the spaces 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.
• Candidates should answer questions in English.

SECTION 1 (50 MARKS)

1. Without using a calculator evaluate:-       (3 marks)
   −2(5 + 3) – 9 ÷ 3 + 5
    − 3 + −16 ÷ −8 x 4
2. Wafulauses ¹/₆ of his land for planting maize, ¹/₁₂ for beans and ⁴/₉ of the remainder for grazing.  He still has 10 hectares of unused land. Find the size of Wafula’s land. (4 mks)
3. A straight line passing through point (−3,  −4) is perpendicular to the line whose equation is  2y + 3x = 11 and intersects x axis and y axis at A and B respectively. Determine the equation of the  second line and hence write down the co-ordinates of A and B. (3 mks)
4. A bus left Kitale at 8.00 a.m. and travelled towards Lodwar at an average speed of 80 km/h. At 8.30a.m a car left Lodwar towards Kitale at an average speed of 120km/h. Given that the distance between Kitale and Lodwar is 400km. Calculate the time the two vehicles met. (3 mks)
5. The sum of four consecutive odd integers is greater than 24. Determine the first four such integers.     (3 mks)
6. Wanyama on arrival in Kenya to play for Harambee Stars against Uganda Cranes converted 6000 Euros into Kenyan Shillings. During his stay in Kenya he spent Kshs. 260,000 and converted the remaining amount into US Dollars before travelling back to England. Using the exchange rates  below, find how many US Dollars he got? (4 mks)
   Currency        Buying      Selling
                          (Kshs.)     (Kshs)
   1 US Dollar      96.20       96.90
   1 Euro             112.32     112.83
7. In the diagram below, the position vector of points A and B with respect to point O are and 
   
   Given that B is a point on AC such that AB = ½ BC. Use vector method to determine the coordinates  of C. (3 mks)
8. Simplify:-
   
9. Complete the diagram below so as to make a net for a cuboid. Hence find the surface area of the cuboid. (3 mks)
   
10. Using a ruler and a pair of compasses only, construct a rhombus PQRS such that PQ = 6 cm and angle PQR = 135° hence measure the shortest diagonal.     (3mks)
11. Janice, a fruit vendor obtained a total of Kshs. 6144 from her sales of oranges on Saturday at Kshs. 8.00 each. She had bought 560 more oranges to add to what had remained on Friday where she had sold 240 more oranges than on Thursday. She had sold 750 oranges on Thursday. Calculate the total  number of oranges Janice had bought on Thursday. (4 mks)
12. Factorise Completely:-   (2 mks)
    x⁴ – 2x²y² + y⁴ 
13. Solve for y given that y is acute and sin (3y – 50°) – cos (2y + 10°) = 0 (3 mks)
14. A solid consists of a cone and a hemisphere. The common diameter of the cone and the hemisphere is 12 cm and the slanting height of the cone is 10 cm. Calculate correct to two decimal places, the surface area of the solid. (3 mks)
15. The figure below shows two sectors in which AB and CD are arcs of concentric circles centre O. Angle AOB = ²/₅ radians and AD = BC = 5 cm.
    
    Given that the perimeter of the shape ABCD is 24 cm, calculate the length of OA. (3 mks)
16. Find the inequalities that define the region R shown in the figure below. (3 marks)
    

SECTION II

Answer only five questions from this section

17. Nyongesa is a sales executive earning a salary of Kshs. 120,000 and a commission of 8% for the  sales in excess of Kshs. 1,000,000. If in January he earned a total of Kshs. 480,000 in salaries and commission.
    1. Determine the amount of sales he made in the month of January. (4 mks)
    2. If the total sales in the month of February increased by 18% and in the month of March dropped by 30% respectively;
       Calculate:-
       1. Nyongesa’s commission in the month of February. (3 mks)
       2. His total earning in the month of March. (3 mks)
18. A sector of angle 108° is cut from a circle of radius 20 cm. It is folded to form a cone.
    Calculate:
    1. The curved surface area of the cone. (2 mks)
    2. The base radius of the cone. (2 mks)
    3. The vertical height of the cone. (2 mks)
    4. If 12 cm of the cone is chopped off to form a frustrum as shown below.
       
       Calculate the volume of the frustum formed. (2 mks)
19.  
    1. Find A^-1, the inverse of matrix A =    (2 mks)
    2. Ibanda sells white and brown loaves of bread in his kiosk. On a certain day he sold 6 white loaves of bread and 5 brown ones for a total of Kshs. 520. The next day he sold 4 white loaves and 7 brown ones for a total of Kshs. 530.
       1. Form a matrix equation to represent the above information. (1 mk)
       2. Use matrix method to find the price of a white loaf of bread and that of a brown loaf of bread. (3 mks)
    3. A school canteen bought 240 white loaves of bread and 100 brown loaves of bread. A discount of 10% was allowed on each white loaf whereas a discount of 13% was allowed on each brown loaf of bread. Calculate the percentage discount on the cost of all the loaves of bread bought.       (4 mks)
20. A village Q is 7 km from village P on a bearing of 045°. Village R is 5 km from village Q on a bearing of 120° and village S is 4 km from village R on a bearing of 270°.
    1. Taking a scale of 1 m to represent 1 Km, locate the three villages. (3 mks)
    2. Use the scale drawing to find the:
       1. Distance and bearing of the village R from village P. (2 mks)
       2. Distance and bearing of village P from village S. (2 mks)
       3. Area of the polygon PQRS to the nearest 4 significant figures. (3 mks)
21. The figure below shows a rectangular sheet of metal whose length is twice its width.
    
    An open rectangular tank is made by cutting equal squares of length 60 cm from each of its four corners and folding along the dotted lines shown in the figure above. Given that the capacity of the tank so formed is 1920 litres and the width of the metal sheet used was x cm;
    1.  
       1. Express the volume of the tank formed in terms of x cm. (3 mks)
       2. Hence or otherwise obtain the length and width of the sheet of metal that was used.   (3 mks)
    2. If the cost of the metal sheet per m2 is Kshs 1000 and labour cost for making the tank is 300 per hour. Find the selling price of the tank in order to make a 30% profit if it took 6 hours to make the tank. (4 mks)
22.  
    1. On the Cartesian plane below, draw the quadrilateral PQRS with vertices   P(4,6),  Q(6,3), R(4,4), and S(2,3) (1 mk)
    2. Draw P′Q′R′S′ the image of PQRS under the transformation defined by the translation vector T =    .Write down the coordinates of P′Q′R′S′.     (2 mks)
    3. P′′Q′′R′′S′′ is the image of P′Q′R′S′ when reflected in the line y= 1. On the same plane, draw P′′Q′′R′′S′′. (2 mks)
    4. Draw P′′′Q′′′R′′′S′′′ the image P′′Q′′R′′S′′ when reflected in the line y – x = 0 (2 mks)
    5. Find by construction, the centre of the rotation that maps P′′′Q′′′R′′′S′′′ onto PQRS and hence determine the coordinates of the centre of the rotation and the angle of the rotation. (3 mks)
23. Andai recorded data on observation of time spent by a local university’s first year bachelor of  Commerce students at library as follows;-
    

    Time spent in minutes	11 - 20	21 - 30	31 - 40	41 - 50	51 - 60
    Cumulative frequency	70	170	370	470

    Calculate:
    1. The mean (6 mks)
    2. The median (4 mks)
24.  
    1. After t seconds, a particle moving along a straight line has a velocity of Vm/s and an  acceleration of (5 – 2t)m/s².  the particles initial velocity is 2m/s.
       1. Express V in terms of t. (3 marks)
       2. Determine the velocity of the particle at the beginning of the third second. (2 marks)
    2. Find the time taken by the particle to attain maximum velocity and the distance it covered to attain the maximum velocity. (5 marks)

MARKING SCHEME

	SOLUTION	MARK	COMMENTS
1.	–2 (5 +3) – 9 ÷ 3 +5 –16 – 3 +5 –19 + 5 = –14 – 3 + 2 x 4  – 3 + 8 = 5 – 14     5 = – 2 4/5	M1  M1  A1	Numerator  Denominator  –14/5 A0   Accept – 2.8
		3
2	1/6 + 1/12 = 2+1 = 3/12                     12  4/9 ×  9/12 =  1/3    1/6 ÷  1/12 +  1/3 = 2+1+4 =  7/12                                 12  1 −  7/12 =  5/12   12/5 × 10  = 24ha	B1  B1  M1  A1	Fraction for grazing  Fraction for unused land
		4
3.	2y + 3x = 11  y = −3x   + 11          2         2 M1 = −3            2 M2 =  2            3 y + 4 =  2   x + 3     3 3y – 2x = −6  A(3,0)              B(0, −2)	B1  B1  B1	Check equivalents  For both coordinates of A and B
		3
4.	Kitale                      400 km                      Lodwar    8.00 a.m.                                                   8.30 Bus                                                           Car 80 km/h                                                     120 km/h 30  x  80  = 40 60 400 – 40 = 360 Relative speed 120 + 80 = 200  T = 360  = 1.8 hrs        200 1hr   48 min.     8.30 +  1.48   10.18   = 10.18 am	B1	B1  B1  Must have units
		3
5.	x  + x + 2 + x + 4+x + 6 > 24     4x + 12 > 24      4x > 12        x > 3    5,  7,  9, 11	M1  A1  B1	All four correct
		3
6.	1 Euro = ksh 112.32 6000 Euros =  6000 x 112.32 673920 673920 – 260,000         = 413920 Ksh 96.90 = 1 US dollar Ksh. 413920 =  413920 x 1       96.90 4271.62	M1  A1  M1  A1
7.		M1  M1  A1
		3
8.	= 4y2 – 6 x y2                     2 = 4y2 – 3y2 = y2	M1  M1  A1
		3
9.	S.A = (5 x 3.5 x 2)  + (5 x 2 x 2) + (3.5 x 2 x 2)          = 35 + 20 + 14          = 69 cm2	B1  M1  A1	Complete and labelled diagram
		3
10.	QS = 4.8 ± 0,1cm	B1  B1  B1	Construction of ∠45°  Completion of PQRS  Measure of the shortest diagonal SQ.
		3
11.	No. of oranges sold on Saturday;            6144    = 768               8 No. of oranges that remained on Friday;        768 – 560  = 208 No. of oranges bought on Thursday .        208 + 750  + 750 + 240         = 1948	M1  M1  M1  A1	✓division  ✓ subtraction  ✓Addition
		4
12.	x4 – x2y2 – x2y2 + y4 x2 (x2 – y2) – y2 (x2 – y2)    (x2 – y2) (x2 – y2)  (x + y) (x – y) (x + y) (x – y)	M1  A1	Award marks for working by inspection
		2
13.	Sin (3y – 50°) = cos (2y + 10°) 3y – 50° + 2y + 10° = 90°    5y – 40° = 90    5y = 130°       y = 26°	M1  M1  A1
		3
14.	S.A of solid = πrl + 2πr2 = 3.142 x 6 x 10 + 2 x 3.14 x 62 = 188.52 + 226.224 = 414.744 = 414.74 cm2 (2 d.p)	M1M1  A1	Exp. Of areas   Addition of areas
		3
15.	Length of an arc = θ cr Let OA = x  2x   + 2  (x + 5) = 14  5        5  2x + 2x + 10 = 70                                                        4x = 60     x = 15   ∴ OA= 15 cm	M1  M1  A1	Perimeter  Solving
		3
16.	L1 = χ ≥  0 L2 (0, 4) (5, 0) 0 − 4 = −4 5 − 0      5 y − 0 = −4 x − 5     5 5y = −4χ + 20    5y < −4y + 20   y < −4 + 4          5
		3
17.	Commission 480,000 – 120,000 = 360,000     8     ×  y = 360,000   100    y = 360,000 x 100               8      y = 4,500,000    1,000,000 + 4,500,000                5,500,000   118   x 5500,000 100             6,490,000      8    x  (6,490,000 = 1,000,000)    100     8     x  5,490,000   100              =  439200   70   x  6,490,000  100   4,543,000       80      x (4,543,000 – 1,000,000)      100         80  x 3,543,000        100            = 283,440      120, 000 + 283,440            = 403,440	M1  M1  M1  A1  M1  M1  A1  M1  M1  A1	Process of finding 4,500,000  Summation
		10
18.	108  x 22  x 20 x 20 360      7   377.14 3.77.14 = 22 x  r  x 20                  7        r = 377.14 x 7                 22 x 20        r = 6 h2 + 62 = 202 h = √(202 – 62) h = 18 18   = 6 12       r   r = 4  Volume of smaller cone = 1/3π (14)2 x 12 Volume of  larger cone      1/3π (6)2 x 18 1/3 π(6)2 x 18 – 1/3 π(4)2 x 12                               = 477.71	M1  A1  M1  M1  A1  M1  A1  M1  M1  A1	π cals = 3.76. 99  π = 3.142 = 377.04  Follow through  Follow through  Expressions for the two volumes.  Subtraction Expression for the 2 Volume   Subtraction  π cal = 477.52  π = 3.142 = 477.584
		10
19.	(c) Total cost of loaves of bread        = (45 x 240) + (50 x 100)        = 10800 + 5000        = 15,800    Total cost with discount;        (45 x 240 x 90)     +     (50 x 100 x 87)               100                                100        9,720  + 4350        = 14070   % discount = 15 800 – 14070                             15800                                      = 10.94936709                         = 10.95%	B1  B1  B1  M1  A1  M1  M1  A1	Alt   22x = 990 ⇒ 45 22y = 1100 ⇒ y = 50A1 for both values  10/100 × 45 × 240 + 13/100 × 50 × 100
		10
20.	9.6km ± 0.1 75° ± 0.1 5.8km ± 0.1 246° ± 0.1 Area = ½ × 7 × 5 sin 105° − ½ × 4 × 5.8 sin 154  = 16.90370196 – 5.085105303                    = 11.81859666                     = 11.82 cm2	B1  B1  B1  B1  B1  B1  B1  M1 M1  A1	Location of point Q   Location of point R   Location of point S  Distance of R from P  Bearing of R from P  Distance of P from S  Bearing of P from S
		10
21	Length = 2x - 120  Width = x – 120 Volume = (2x – 120) (x – 120) 60               = (2x2 – 240x – 120x + 1440)60                = 120x2 – 14400x – 7200 + 864000                 = 120x2 – 21600x + 864800 . Volume = 1920 000 cm3               (2x – 120)  (x – 120) 60 = 1,920, 000               (2x – 120) (x – 120) = 32 000               2x2 – 240x – 120x +14400 = 32 000                2x2 – 240x – 120x = 17600                 x2 – 180x – 8800 = 0 x = 180  +√ (-180)2 – 4 x 1 x -8800)                              2 x1      = 180 + √(32400 + 35200)                            2       = 180 + 260                  2          Either x = 220 or -40                     x ≠ -40                     x = 220 cm            Length = 440 cm Area of sheet = 440 x 220                        = 96 800 cm2                       = 96 800 = 9.68m2                          10 000      Cost  = 9.68 x 1000               = sh 9680     Labour = 300 x 6 = sh 1800     Total cost = 9680 + 1800                       = sh 11480    S.P. = 130 x sh.11480               100          = sh 14924	B1  M1  A1  M1  M1  M1  A1  B1  M1  A1
22.	PQRS drawn   Line y = 1 Diagram PllQllRllSll Image PlllQlllRlllSlll Two ⊥ bisectors Centre of rotation, C (7.5, 0.5) + 0.1 for each coordinate. Angle of rotation = −90°	B1  B1  B1  B1  B1  B2  B1  B1  B1	Diagram  Diagram  Coordinates  (Line y − x = 0 can be implied  Accept 90° clockwise
23.	Time (min)  f  x  fx  cf  11-20  70  15.5  1085  70  21-30  100  25.5  2550  170  31-40  200  35.5  7100  370  41-50  100  45.5  4550  470  51-60  30  55.5  1665  500    = 500    = 16950   Mean x̄ = 16950                       500             = 33.90 Median position = 500   = 250th                                 2 Median class: 31 – 40 Median value = 30.5 + 250 – 170 x 10                                            200                                 = 30.5 +  80   x 10                                                200                                    = 30.5 + 4                                   = 34.5	M1  A1  B1  B1  M1  A1	√ exp. For  x̄  Median class  For 250-170  Expression
		10
24.	a = 5 – 2t V = ∫(5 − 2t)dt = 5t − t2 + C V = 5(0) + (0)² + C = 2  t = 0 C = 2 V = 5t − t² + 2 t = 2 V = 5(2) – (2)² + 2     = 8m/s  a = 5 – 2t = 0 t = 2.5 seconds  [5/2(2.5)2 − (2.5/3)3 + 292.5) − 5/2(0)2 − (10/3)3 + 2(0)] =  15.42 metres	M1  A1  M1  A1 M1  A1 M1  M1  A1	C must be included  Max velocity t = 0      Substitution of values
		10