# Mathematics P1 Questions and Answers - Mangu High School Trial Mock Exams 2021/2022

INSTRUCTIONS TO CANDIDATES

1. Write your name and indexnumberin the spaces provided above.
2. Sign and write date of examination in the spaces provided above.
3. This paper consists of two sections; Section I and Section II.
4. Answer All questions in Section I andonly Fivequestions from section II
5. All answers and working must be written on the question paper in the spaces provided below each question.
6. Show all the steps in your calculations giving answers at each stage in the spaces provided below each question.
7. Marks may be given for correct working even if the answer is wrong.
8. Non-programmable silent electronic calculators and KNEC Mathematical tables may be used except where stated otherwise.
9. Candidates should answer questions in English.

For examiner’s use only.
Section I

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total

Section II

 17 18 19 20 21 22 23 24 Total

GRAND TOTAL __________ ## QUESTIONS

SECTION 1 (50 MARKS)

1. Without using a calculator evaluate:-  (3 marks)
-2(5 + 3) – 9 ÷ 3 +5
-3 + -16 ÷ -8 x 4
2. Wafula uses 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 8.30 a.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)
(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 [-6] and [-3] respectively.
[-2]         [ 0] 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:-
(8y)2/3 x y11/3 -6 ÷ 2y-2
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 = 135º 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:-
x4 – 2x2y2 + y4 (2 mks)
13. Solve for y given that y is acute and sin (3y – 50º) – cos (2y + 10º) = 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 ofthe 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

1. 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)
2. A sector of angle 1080 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)
3.
1. Find A-1, the inverse of matrix (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)
4. A village Q is 7 km from village P on a bearing of 045º. Village R is 5 km from village Q on a bearing of 120º and village S is 4 km from village R on a bearing of 270º.
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)
5. 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)
6.
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 .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)
7. 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 70 170 370 470 500
Calculate:
1. The mean (6 mks)
2. The median (4 mks)
8.
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 M1M1A1 NumeratorDenominator- 14  A0       Accept – 2.8    5 3 2. 1  +  1  =     2 + 1    =   36     12          12          124  x   9    =   19     12         3 1   +  1   +  1  =  2 + 1 + 4    =  76     12      3           12             121 –  7    =   5      12       1212  x 10 524 ha B1B1M1A1 Fraction for grazingFraction for unused land. 4 3. 2y + 3x = 11 y = -3x   + 11         2        2M1 = -  -3            2M2 =  2         3y + 4 =  2x + 3     33y – 2x = - 6A(3,0)              B(0, -2) B1B1B1 Check equivalents.For both coordinates of A and B. 3 4. Kitale                      400 km                      Lodwar8.00 a.m.                                                   8.30Bus                                                           Car80 km/h                                                     120 km/h30  x  80  = 4060400 – 40 = 360Relative speed 120 + 80 = 200T = 360  = 1.8 hrs      2001hr   48 min.    8.30+  1.48  10.18  10.18 am B1B1B1 Must have units 3 5. x  + x + 2 + x + 4+x + 6 > 244x + 12 > 244x > 12x > 35,  7,  9, 11 M1A1B1 All four correct 3 6. 1 Euro = ksh 112.326000 Euros =6000 x 112.32673920673920 – 260,000        = 413920Ksh 96.90 = 1 US dollarKsh. 413920 =413920 x 1    96.904271.62 M1A1M1A1 7.  M1M1A1 3 8. (8y)2/3 x y11/3 -6 ÷ 2y-2= 22xy6/3 - 6 ÷  2                        y2= 4y2 – 6 x y2                  2= 4y2 – 3y2= y2 M1M1A1 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 B1M1A1 Complete and √ labelled diagram 3 10.  QS = 4.8 ± 0.1cm B1B1B1 Construction of ∠450Completetion of PQRS.Measure of the shortest diagnol SQ. 3 11. No. of oranges sold on Saturday;           6144   = 768              8No. of oranges that remained on Friday;       768 – 560  = 208No. of oranges bought on Thursday .       208 + 750  + 750 + 240        = 1948 M1M1M1A1 √ division√ subtraction√ Addition 4 12. x4 – x2y2 – x2y2 + y4x2 (x2 – y2) – y2 (x2 – y2)   (x2 – y2) (x2 – y2) (x + y) (x – y) (x + y) (x – y) M1A1 Award marks for working by inspection 2 13. Sin (3y – 500) = cos (2y + 100)3y – 500 + 2y + 100 = 900   5y – 400 = 90   5y = 1300      y = 260 M1M1A1 3 14. S.A of solid = prl + 2pr2= 3.142 x 6 x 10 + 2 x 3.14 x 62= 188.52 + 226.224= 414.744= 414.74 cm2 (2 d.p) M1M1A1 Exp. Of areasAddition of areas 3 15. Length of an arc = θ crLet OA = x 2x   + 2  (x + 5) = 14 5        52x + 2x + 10 = 70    4x = 60    x = 15  ∴ OA= 15 cm M1M1A1 PerimeterSolving 3 16 L1 = χ ≥  0L2            (0, 4)       (5, 0) 0 - 4 = -45 - 0     5y - 0 = -4χ - 5     55y = -4χ + 205y < -4χ + 20y < -4 + 4       5 B1 3 17. Commission480,000 – 120,000 = 360,000  8   x y = 360,000100y = 360,000 x 100          8 y = 4,500,0001,000,000 + 4,500,000 5,500,000                   118   x 5500,000100  6,490,000  8    x  (6,490,000 = 1,000,000)100  8     x  5,490,000100    =  439200 70   x  6,490,0001004,543,00080    x (4,543,000 – 1,000,000)100 80   x 3,543,000100 = 283,440120, 000 + 283,440= 403,440 M1M1M1A1M1M1A1M1M1A1 Process of finding 4,500,000Summation 10 18.  108 x 22  x 20 x 20360    7 = 377.14 3.77.14 = 22 x  r  x 20                 7r = 377.14 x 7      22 x 20       r = 6 h2 + 62 = 202h = √202 – 62h = 18 18   = 612      r r = 4 Volume of smaller cone = 1 π (14)2 x 12                                          3Volume of  larger cone   =  1   π (6)2 x 18                                           31  π (6)2 x 18 – 1/3 π(4)2 x 123= 477.71 M1A1M1M1A1M1A1M1M1A1 π cals = 3.76. 99π = 3.142 = 377.04Follow through.Follow through.Expressions for the two volumes.SubtractionExpression for the 2 VolumeSubtractionπ cal = 477.52π = 3.142 = 477.584 10 19.  ∴ x = 45; y = 50(c) Total cost of loaves of bread= (45 x 240) + (50 x 100)= 10800 + 5000 = 15,800Total 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% B1B1B1M1M1A1M1M1A1  10/100 x 45 x 240 + 13/100 x50 x 100 10 20.  9.6 km ± 0.1750 ± 0.1 5.8 km ± 0.12460 ± 0.1 Area = ½ x 7 x 5 sin 1050 – ½ x 4 x 5.8 sin 154= 16.90370196 – 5.085105303= 11.81859666= 11.82 cm2 B1 B1B1B1B1B1B1M1 M1A1 Location of point QLocation of point RLocation of point S.Distance of R from PBearing of R from PDistance of P from SBearing of P from S 10 21. Length = 2x - 120Width = x – 120Volume = (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 0002x2 – 240x – 120x +14400 = 32 0002x2 – 240x – 120x = 17600x2 – 180x – 8800 = 0                          x= 180  ±  √ (-180)2 – 4 x 1 x -8800                      2 x1= 180 + √32400 + 35200                  2 = 180 + 260           2Either x = 220 or -40 x # -40x = 220 cmLength = 440 cm Area of sheet = 440 x 220 = 96 800 cm2 = 96 800 = 9.68m2   10 000Cost  = 9.68 x 1000= sh 9680Labour = 300 x 6 = sh 1800Total cost = 9680 + 1800= sh 11480S.P. = 130 x sh.11480              100= sh 14924 B1M1A1M1M1M1A1B1M1A1 10 22. PQRS drawn  Line y = 1Diagram PII  QII   RII  SII Image P111  QIII  RIII   SIII  Two ⊥ bisectorsCentre of rotation, C (7.5, 0.5) ± 0.1 for each coordinate.Angle of rotation = -900 B1B1B1B1B1B2B1B1B1 DiagramDiagram Coordinates(Line y – x = 0 can be impliedAccept 900 clockwise 10 22.23.  Time (Min)       f.         x           fx                 cf11 – 20           70       15.5       1085             7021 – 30           100     25.5       2550            17031 – 40           200     35.5       7100            37041 – 50           100     45.5        4550           47051 – 60             30      55.5       1665           500                        Σf= 500           Σfx= 16950 Mean, x    =  Σfx                      Σf = 16950      500= 33.90 Median position = 500   = 250th                               2Median class: 31 – 40Median value = 30.5 + 250 – 170 x 10                                         200 = 30.5 + 80   x 10             200 = 30.5 x 4= 34.5 B1B1B1B1M1A1B1B1M1A1 f columnx columnfx column Σf and  Σfx√ exp. For  xMedian classFor 250 – 170Expression 10 24. a = 5 – 2tV = ∫(5 - 2t)dt = 5t - t2 + CM1          C must be includedV = 5(0) + (0)² + C = 2M1 t = 0                        C = 2 V = 5t - t² + 2  A1 t = 2V = 5(2) – (2)² + 2 M1= 8m/s A1 a = 5 – 2t = 0M1          Max velocity t = 0t = 2.5 seconds A1  10

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