Mathematics Paper 1 Questions and Answers - Lainaku 1 Joint PreMock Exams 2023

Instructions to candidates

1. This paper consists of TWO sections I and II.
2. Answer ALL questions in section I and any five from section II.
3. All answers and working must be done on the question paper in the spaces provided below each question.
4. Show all the steps in your calculations giving your answers at each stage in the spaces below each question.
5. Marks may be given for correct working even if the answer is wrong.

                                                                                        SECTION I (50 Marks)

                                                           Answer all questions in this section in the spaces provided

1. Without using tables or calculator, evaluate the following. (3mks)
                                −8 + (−13) x 3 – (−5)
                                    −1 + (−6) ÷ 2 x 2
2. The straight line through the points D (6, 3) and E (3, -2) meets the y – axis at point F. Find the co-ordinates of F. (3 mks)
3. The circle below whose area is 18.05cm² circumscribes a triangle ABC where AB = 6.3cm,BC = 5.7cm and AC = 4.2cm. Find the area of the shaded part to 2 dp. (3 mks)
                                                            
4. A number n is such that when it is divided by 27, 30, or 45, the remainder is always 3. Find the smallest value of n. (3 mks)
5. The actual area of an estate is 3510 hectares. The estate is represented by a rectangle measuring 2.6cm by 1.5cm on the map whose scale is l: n. Find the value of n. (give your answer in standard form) (3 mks)
6. Find the obtuse angle the line y – 2x = 7 makes with the x – axis (2 mks)
7. Given the column vector 
   1. Expressas a column vector (2mks)
   2. Calculate the magnitude of vector in  (i) above correct to two decimal places. (2mks)
8. Muthoni went to a shop and bought 50 packets of milk and 25 packets of salt all for Kshs.200.00. She sold the milk at a profit of 28% and the salt at a profit of 24% thereby making a net profit of Kshs.53.50. Find the cost price of a packet of milk and a packet of salt. (3 mks)
9. The angles of elevation from two points A and B to the top of a storey building are 480 and 570 respectively. If AB = 50m and the point A and B are opposite each other; Calculate;
   1. The distance of point A to the building (3 mks)
   2. The height of the building (1 mks)
10. Find x if 3^2x+3 + 1 = 28 (2 mks)
11. Simplify as simple as possible (3 mks)
                       (4x + 2y)² − (2y − 4x)²
                         2x + y)² + (y − 2x)²
12. The cost of a camera outside Kenya is US$1000. James intends to buy one camera through an agent who deals in Japanese Yen. The agent charges him a commission of 5% on the price of the camera and further 1260 Yen as importation tax. How much in Ksh. Will he need to send to the agent to obtain the camera, given that:- (3 mks)
                        1 US$= 105.00 Yen.                   1 US$= Kshs.63.00
13. State all the integral values of x which satisfy the inequality      (3mks)
                         3x + 2 ≤ 2x + 3 ≤ 4x + 15
                              4          5              6
14. Without using a protractor, construct a triangle ABC such that angle ABC = 1350, AB = 4.6cm and BC =6.1cm. Measure AC and angle ACB (4 mks)
15. Without using mathematical tables or calculators, find the volume of a container whose base is a regular hexagon of side √3cm and height cm 2√3    (3 mks)
16. Below is a net of a model of a three dimensional figure. The lengths AB = BC = AC = 6.0cm and lengths AF = FB = BD = CD = CE = AE = 8.0cm.
                                                                                    
    
    1. Draw the solid when the net is folded by taking ABC as the base and the height 5cm. (3 mks)
    2. State the name of the figure drawn  (1 mk)

                                                                                        SECTION II        (50 Marks)

                                                        Answer only five questions in this section in the spaces provided.

17. The distance between towns A and B is 360km. A minibus left A at 8.15am and traveled towards B at an average speed of 90km/hr. A matatu left B two and a third hours later on the same day and traveled towards A at an average speed of 110km/hr.
    1.  
       1. At what time did the two vehicles meet? (4mks)
       2. How far from A did the vehicles meet? (2mks)
    2. A motorist started from his home which is between A and B at 10.30am on the same day and travelled at an average speed of 100km/hr. He arrived at B at the same time as the minibus. Calculate the distance from A to his house. (4 mks)
18. Consider the vessel below
                                                                         
    
    1. Calculate the volume of water in the vessel. (Take π = 3.142) (2mks)
    2. When a metallic hemisphere is completely submerged in the water, the level of the water rose by 6cm. Calculate:
       1. The radius of the new water surface. (2mks)
       2. The volume of the metallic hemisphere (to 2 d.p.) (3mks)
       3. The diameter of the hemisphere (to 1 d.p) (3 mks)
19. The American government hired two planes to airlift football fans to Qatar for the World cup tournament. Each plane took 10½ hours to reach the destination.
    Boeing 747 has carrying capacity of 300 people and consumes fuel at 120 litres per minute. It makes 5 trips at full capacity. Boeing 740 has carrying capacity of 140 people and consumes fuel at 200 liters per minute. It makes 8 trips at full capacity. If the government sponsored the fans one way at a cost of 800 dollars per fan, and the fans pays for the return ticket. Calculate:
    1. The total number of fans airlifted to Qatar. (1mk)
    2. The total cost of fuel used if one litre costs 0.3 dollars. (5mks)
    3. The total collection in dollars made by each plane. (2mks)
    4. The net profit made by each plane. (2mks)
20. The following data shows the length of trees grown in Mau Forest measured to the nearest cm by a research team. Use the given data to answer the given questions.
         230     240     250     253    260    253     274    238     263    260    231    284    257    260
         275     271     257     267    255    265     241    256     256    257    260    262    234    259
         263     244     254     248    281    240     247    236     256    282    242    246    277    238
         250     279     252     269    284    271     249    273
    1. Arrange the data in a frequency distribution table with a class interval of five and starting with the class of 230 – 234,… (6mks)
    2. Using the frequency distribution in (a) above and 257 as an assumed mean, find:-
       1. Mean of the data. (2mks)
       2. The standard deviation of the data. (2mks)
21. Using a ruler and a pair of compasses only, draw a triangle ABC such that AB = 5cm, BC = 8cm and angle ABC = 60°. Measure AC and angle ACB. (5mks)
    1. Locate point O in triangle ABC such that OA = OB = OC. Using O as the center and radius OA draw a circle (3mks)
    2. Construct a perpendicular from A to BC to meet BC at D. Measure AD, hence find the area of triangle ABC. (2mks)
22. Three brick layers have to lay a total of 5400 bricks . The average number of bricks they can lay in an hour are in the ratio 5:6:9.If the slowest man lays 60 brick in an hour. Calculate;
    1. How many bricks each of the other two men lay in an hour. (4mks)
    2. How many of the bricks each man will lay to complete the work if they are all employed for the same number of hours. (6mks)
23. Four towns P, Q, R and S are such that town Q is 120km due east of town P. Town R is 160km due North of town Q. Town S is on a bearing of 330° from P and on a bearing 300°from R. use a ruler and a pair of compasses only for all your constructions.
    1. Using a scale of 1cm to represent 50km, construct a scale drawing showing the positions P, Q, R and S. (6mks)
    2. Use the scale to determine
       1. The distance from town S to town P. (1mk)
       2. The distance from town S to town R. (1mk)
       3. The bearing of town S from town Q. (2mks)
24. A carpenter constructed a closed wooden box with internal measurements 1.5m long 0.8m wide and 0.4m high. The wood used in constructing the box was 1.0cm thick and had a density of 0.6g/cm³.
    1. Determine the:
       1. Volume in cm³ of the wood used in constructing the box. (4 mks)
       2. Mass of the box in kg correct to 1 d.p (2 mks)
    2. Identical cylindrical tins of diameter 10cm height 20cm with a mass of 120g each were packed in the box. Calculate the:
       1. Maximum number of tins that were packed (2 mks)
       2. Total mass of the box with the tins in kg. (to 1d.p) (2 mks)

                                                                                            MARKING SCHEME.

No.	Working	Marks	Remarks
1	−8 −39 + 5  −1 −3 x 2  = −42      −7  = 6	M1    M1       A1	Simplifying Num  Simplifying Denom
		03
2	3 + 2 = 5  6 − 3    3  y − 3 = 5  x − 6    3  3y – 9 = 5x - 30  3y = 5x −21  y = 5/3x – 7  F = (0, −7)	M1      M1    A1	For grad & eqn.     Expressing in form   y=mx +c
		03
3		M1      M1    A1	For the difference
		03
4	2 x 33 x 5 = 270                  = 270 + 3                  = 273	M1        M1     A1	✔Process of finding LCM         For addition
		03
5	Area = 35100000m2                               = 351000000000cm2                          Area = 2.6x1.5 = 3.9cm2                             Scale = 351000000000                                                   3.9                                       = 90000000000                                    n = 9.0 x1010	M1   M1       A1	For conversion.
		03
6	y = 2x + 7              tanθ = 2                   θ = tan − 12 = 63.430               Obtuse angle = 180 – 63.43                                      = 116.570	B1     B1	For ✔tan-12   Accept   63.43 seen
		02
7	(i)                     (ii)	M1     A1       M1       A1
		04
8	50x + 25y = 20    2x + y = 8…….(i)    50x +  28 + 24 x 25y = 53.50              100  100    14x + 6y = 53.50......(ii)     ∴ 2x + y = 8     14x + 6y = 53.50     ∴ 2x + y = 8    14x + 6y = 53.50     ∴ y = 2.50, x = 2.75     milk = Ksh.2.75, salt = Ksh.2.50	M1       M1     A1	✔eqn (i) and   ✔ eqn (ii)       ✔ solving using any method   ✔ both
		03
9	tan 57° = h/x ⇒ x tan57°     tan 48° = h/50 − x h = (50 − x)tan 48°     xtan57° = (50 − x)tan48     1.53986x = 55.53 − 1.1106x                  x = 20.95                     29.05m       h = xtan57 = 20.95 tan57                        = 32.26m	M1   M1       A1     B1
		04
10	32x + 3 + 1 = 28                     32x+3 = 27                     32x x 33 = 33                      27.32x = 27                           32x = 1                             2x = 0                                X = 0	M1     A1	Exp in same base
		02
11	[(4x + 2y) − (2y − 4x][(4x + 2y) + (2y − 4x)]      [(2x + y) − (y − 2x)][(2x + y) + (y − 2x)]     = 8x x 4y        4x x 2y     = 8x x 4y        4x    2y     = 2 x 2 = 4	M1   M1     A1	Simplifying num.  Simplifying den.
		03
12	Yen = 1000 x 105 x 105 + 1260                                     100           = 111510 Yen     Ksh = 111510 x 63                              105            = Ksh. 66906	M1   A1     B1
		03
13	3x + 2 ≤ 2x + 3 ≤ 4x + 15                             4             5            6                          3x + 2 ≤ 2x + 3                               4            5                          5(3x + 2) ≤ 4(2x +3)                           15x + 10 ≤ 8x + 12                                     7x ≤ 2                                       x ≤ 0.2857                           2x + 3 ≤ 4x + 15                                5             5                            6(2x + 3) ≤ 5(4x + 15)                                −8x/−8 ≤ 57/−8                                − 7.125 ≤ x                                − 7.125 ≤ x ≤ 0.28           Integral value                                  { -7, -6, -5, -4, -3, -2, -1,0}	B1       B1       B1
		03
14		B1     B1   B1   B1	Line AB BC and angle ABC   Triangle drawn.   AC measured   Angle ACB measured.
		04
15	½ x √3 x √3 sin60 x 6   Base area = 9/2 √3 or 45√3cm2   vol = 2√3 x 9/2 √3        = 27cm3	M1     M1     A1	Exp for area     ✔exp for vol     CAO
		03
16		B1   B1   B1       B1	Base drawn   Slant faces   Sketch completed and the lines dotted.
		04
17	(a) A                360km                         B      8.15am → 90km/hr                    ← 10.35am 110km/h         Minibus                                             matatu  (i) Distance traveled by minibus for 2 1/3 hr                     = 7/3 x 90                     = 210km              Remaining distance = 360 – 210 = 150km.              Relative speed = 90 + 110 = 200km/hr              Time for meeting = 150km = 0.75hrs                                             200km/hr                           or              = 45mins              Meeting time = 10.35                                      +   .45                                             11.20am (ii)  (a) Distance from A 210 + (0.75 x 90)                                 210 + 67.5                                                    = 277.5km       (b) Time minibus arrived at B              time = D = 360 = 4hrs                         S     90                      = 8.15 + 4hrs                      = 12.15pm       Time taken by the motorist to arrive at       B = 12.15pm – 10.30am = 1hr 45min       Distance = Time x speed       distance= 1 45/60 x 100       distance= 175km     ∴Home to B = 175km       Home to A = 360 – 175                                             = 185km	B1    B1      M1    A1    M1      A1          B1        M1        M1    A1	✔Distance covered by minibus for 2 1/3 hrs ✔ R.S   ✔addition CAO                       For the difference
		10
18	Volume = πr2h                       3       = 3.142 x 21 x 21 x 30              3       = 13856.22cm3                                       x = 36          21   30        x = 25.2cm      (ii)     New volume = 1/3 x 3.142 x 25.2 x 25.2 x 36                                    = 23943.55cm3               Volume change = 23943.55 – 13856.22                                    = 10087.33cm3( 2 d.p)      (iii) 2/3πr3 = 10087.33                  r3 = 10087.3 x 3/2 x 1/π                  r3 = 4815.72                    r = 3√4815.72                    r = 16.89cm       diameter = 16.89 x 2                      = 33.8cm (I d.p)	M1     A1           M1   A1     M1   A1     B1   M1       A1   B1	✔vol of hemisphere
		10
1	(a)        (300x5) + (140x8)          = 1500 + 1120          = 2620fans    (b) Cost of fuel          Boeng 747      = 120 x 10.5 x 60 x 5 x 2 x 0.3      = 226800 dollars   Boeng 740      = 200x10.5x60x8x2x0.3      = 604,800 dollars        Total cost = 226800 + 604800                        = 831600 dollars     (c) Total collection          Boeng 747                     = 300x5x800x2                     = 1,200,000x2                     = 2400000 dollars          Boeng 740                     = 140x8x800x2                     = 896,000x2                     = 1792000 dollars      (d) Net profit           Boeng 747                     = 2400000 – 226800                     = 2173200 dollars          Boeng 740                     = 1792000– 604,800                     = 1187200 dollars	B1     M1           A1   M1     A1   B1     B1   B1       B1   B1
		10
20	Class  f   x   d   fd   fd2   230-234  235-239  240-244  245-249  250-254  255-259  260-264  265-269  270-274  275-279  280-284  3  3  5  4  6  8  7  3  4  3  4    232    237    242    247    252    257    262    267    272    277    282    - 25    - 20    - 15    - 10    - 5      0      5     10     15     20     25  -75  -60  -75  -40  -30    0   35   30   60   60  100  1875  1200  1125  400  150     0  175  300  900  1200  2500    B1    ∑f   =50  B1     B1   B1   ∑fd=5  B1   ∑fd2  =9825  B1     (b) (i)                          = 257 + 5                                    50                     = 257 + 0.1                     = 257.1cm       (ii) Standard deviation	M1    A1      M1     A1	B1 For all vaues ✔ in each column   Accept equivalent
		04
21		B1 .  B1    B1    B1  B1  B1    B1  B1  B1    B1	Line AB & BC drawn   Angle ABC=60°constructed   ✔triangle ABC.     2-Perpendiculars drawn   Pt O located   Circle drawn     AC Measured   Angle ACB measured   For ✔ perpendicular & AD measured     ✔ Area
		10
2	a)            5:6 = 60 : x                 5x = 360                  X = 72                5:9 = 60:y                 5y = 540                  Y = 108 (b)          Total ratio = 5 + 6 + 9                         5/20 x 5400                                = 1350 bricks                        6/20 x 5400                                = 1620 bricks                        9/20 x 5400                                = 2430 bricks	M1  A1    M1  A1    M1  A1    M1  A1    M1  A1
		10
23	(a)   (b) (i)    7.8 x 50 = 390 km.       (ii)    7.10 x 50 = 355 km       (iii)   Bearing = 320°	B1  B1    B1    B1  B1    B2	Position of Q   Position of R   Position of S       Angle 300°   Angle 330°     Diagram drawn
		10
24	(a)    (i)   Interval volume = 150cm x 80cm x 40cm                                         = 480,000cm3               External volume = 152cm x 82cm x 42cm                                          = 523,488cm3               Volume of the wood = 523,488 – 480,000                                                = 43,488 cm3         (ii) Mass = Density x Volume                       = 0.6 x 43,488                       = 26.0928g                              1000                       = 26.0928kg                       = 26.1kg (to 1d.p) (b)    (i)    No. of tins = 8x15x2                              = 240 tins         (ii) Total mass = Mass of the box + Total mass of the tins                                = 26.1 + {(120x240)/1000}                                = 26.1+ 28.8                                = 54.9 kg	M1   M1   M1   A1     M1     A1     M1   A1     M1   A1	For the difference       Exp for mass and   Conversion to kg.
		10