Mathematics Paper 1 Questions and Answers - Sunrise Pre Mock Exams 2023

• This paper consists of two sections; Section I and Section II.
• Answer all the questions in Section I and any five questions from Section II
• Show all the steps in your calculations, giving your 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.

SECTION I (50MKS)

1. Simplify without using a calculator. (4mks)
     2 ⅖ ÷ 3.3 – 1.75   
   2 ¾ − 7⁵/₉ ÷ √5¹/₁₆
2. Prime numbers in between 41 and 55 were cubed and summed up. The result was multiplied by 29. Write the total value of the second digit. (3mks)
3. The table below shows the currency exchange rate table used by the bank of Baroda in Nairobi.
   

   	Selling (Ksh)	Buying (Ksh)
   10 US dollars($)	869.80	867.00
   200 South African Rand (Rn)	5590	5440

   Peterson arrived in Kenya with $ 75340. He exchanged all the dollars into Kenya shillings at a commission of 2%. While in Kenya he used Ksh 2,570,000 on charity and Ksh 1,750,000 on food and accommodation. At the end of his tour, he decided to exchange all the remaining Kenya shillings into South African Rands at a commission of 5%. Calculate to the nearest Rand, how many Rands he received from the bank. (4mks)
4. Kimani set out on a journey at 9.49a.m from Zimmaman to Ruiru town cycling at a speed of 30km/h. He cycled for a distance of 60km and then took a rest of 1.25 hours. He then walked the remaining distance at an average speed of 15km/h. If the distance between Zimmaman and Ruiru town was 75km, at what time did he arrive? (Write your answer in 24 hour  clock) (3mks)
5. Find the value of ∟ABC (3mks)
   
6. Use the reciprocal tables and square root tables to evaluate. (3mks)
     − 20.15      
   (0.007429)^½ 
7. A map is drawn to a scale of 1:200,000.  Find the area of a village whose shape is a rectangle measuring 2cm by 3.2cm on the map in km2. (2mks)
8. Use matrix method to solve the pair of simultaneous equations. (3mks)
   2x – 3y + 17 = 0
   6y + 5x + 4 = 2
9. 100 pupils running 20km a day for 30 days survive on 12 dozens of milk. How many days will 200 pupils covering 5km a day survive on 3 grosses of milk? (2mks)
10. Simplify without using a calculator. (3mks)
    8⅔ x 625^³/₂
     4 ⁵/₂ x 25^³/₂
11. Given that tan ⁡θ = ⁴/₃ , solve for (cos ⁡θ sin ⁡θ  ) without using a calculator or mathematical tables.      (2mks)
12. Vector a ̌ passes through the points (6,8) and (2,4).Vector b ̌ passes through (x,-2) and (-5,0). If a ̌ is parallel to b ̌, determine the value of x.         (3mks)
13. In the figure below AB is an arc of a circle centre O .Given that ∠AOC = 30°, OA = OB = 8cm andBC = 5cm. Calculate the area of the unshaded region to 2 decimal places  ( use π=3.142)                                                                        (3mks)
    
14. Use logarithm table to evaluate       (4mks)
    
15. The perimeter of a rectangle is 36cm. If the length of the rectangle is twice its width , find the length of the rectangle            (3mks)
    1. Sketch the net of the sketch given below. (3mks)
       
    2. Find its surface area from the net. (2mks)

SECTION II (50MKS)

16.  
    1. The line L₁ whose equation is 2 x + 1 ½ y = 8 is perpendicular to another line L2 passing through the point (−2, 3.5). Find the equation of line L₂. (4mks)
    2. Line L₂ is parallel to another line L₃, which passes through the point (−4, −5). Find the equation of L₃. Write your answer in the form of
       ax + by = c. (2mks)
    3. The line L₃ intersects another line L₄ whose equation is 2x + 5y + 33 = 0. Find the co-ordinates of the point of intersection. (4mks)
17. A, B and C contributed Ksh 2,451,300 such that A:B = 2 ⅔:5 ¾ and B:C = 3 ¼ :5¹/₉. They decided to buy a matatu with all the money. Each month the matatu delivered a profit of Ksh 90,000. The matatu was operated for one full year. The matatu driver was being paid Ksh 27,000 a month. At the end of the year, they decided that the net profit would be shared in the ration of their contributions.
    1. Determine the ratio A:B:C (3mks)
    2. Determine the net profit realized after one year. (2mks)
    3. Partner A decided to buy dairy cows with his share at the rate of Ksh 60,000 per cow. How many cows did he buy? (2mks)
    4. In the second year, the partners decided to breakup. They decided to sell the car after it had depreciated by 20% and they divided the money in the ratio of contribution. If the profit was divided equally, how much did C receive? (3mks)
18. A triangle has vertices A(1,2),B(7,2) and C(5,4).
    1. Draw triangle ABC on a cartesian plane.                                                                (1mk)
    2. Draw triangle A'B'C' the image of ABC under a rotation 90° clockwise about the origin.                        (2mks
    3. Draw ∆A''B''C'' the image of ABC under a reflection in the line y − x = 0. State the coordinates of  ∆A''B''C''.               (3mks)
    4. ∆A'''B'''C''' is the image of ∆A''B''C'' under reflection in the line y = 0. Draw ∆A'''B'''C''' and state its coordinates.       (4mks)
19. Mr Kamau a salesman is paid a commission of 6% on goods sold worth over Sh. 600,000. In addition he is paid a monthly salary of Sh.35,000
    1. Calculate his total earnings in a month when his total sales was Shs.750,000.                                    (3mks)
    2. In the following month the rate of commission was changed to x%  but his monthly salary remained the same. If Mr. Kamau received a total monthly earning of k.sh 100,000 for selling goods worth the same amount. Find the value of x                                                      (3mks)
    3. A trader made a loss of 30 % by selling an electric kettle at Sh.700. Calculate the percentage profit he would have made if he had sold it at Sh. 1150                (4mks)
20. On the grid provided, by shading the unwanted region, determine and label the region R that satisfies the three following inequalities
    y ≤ ²/₃ x+4, 4y> −3x − 24 and y+6 ≥ 3x on the graph below.                                                                                                                  (10mks)
21. The distance between towns A and B is 360km. A minibus left town A at 8.15 a.m. and traveled towards town B at an average speed of 90km/hr. A matatu left town B two and a third hours later on the same day and travelled towards A at average speed of 110km/hr.
    1. At what time did the two vehicles meet? (4mks)
    2. How far from A did the two vehicles meet? (2mks)
    3. A motorist started from his home at 10.30 a.m. on the same day as the matatu and  travelled at an average speed of 100km/h. He arrived at B at the same time as the minibus. Calculate the distance from A to his house. (4mks)
22. In the figure below E is the midpoint of BC. AD: DC =3:2 and F is the meeting point of BD and AE. AC =  and AB = 
    
    
    1. Express the following vectors in terms of b and c
       
       1.  → 
          BD                                                                                                     (1mk)
       2.  →
          AE                                                                                                      (2mks)
    2.     →      →         →        →
       If  BF = tBD and AF = nAE,  Express AF in two different ways hence find the value of t and n           (5mks)
    3. State the ratio in which F divides
       1. BD                                                                                                    (1mk)
       2. AE                                                                                                    (1mk)
23.  
    1. Solve the equation                                                                                                (4mks)
       x + 3 =    1    
          24     x − 2
    2. The length of a floor of a rectangular hall is 9m more than its width. If the area of the floor is 136 m²,
       1. Calculate the perimeter of the floor                                                                (3mks)
       2. A rectangular carpet is placed on the hall leaving an area of 64m².If the length of the carpet is twice its width, determine the width of the carpet.                        (2mks)

MARKING SCHEME

1	Numerator: 12 x 3 – 7/4 = 36 – 7/4 = 72 – 175                     5    10            50               100                                                               = -103                                                                   100 Denominator: 11 – 68 x 4/9 = 11 – 272 = 891 − 1088                         4      9              4      81           324                                                              = −197                                                                   324                                           162                     N/D= -103 x 324 = 16686 =  6836                               100     197     9850      9850                                                             50                                                               =   3418                                                                    4925	B1    B1       A1	Numerator     Denominator
2	=  433 + 473 + 533  = 332 207 x 29 = 9,634.003 Total value 600,000	M1 M1  A1	Cubing and summing up Multiplying by 29 C.A.O.
3	Amount received = 86.70 x 75 340 x 98                                            100                              = Sh. 6401,338.44 Total expenditure = 2,570,000 + 1,750,000                               = Sh. 4,320,000 The balance           = 6401338.44 – 4320000                               = Sh. 2,081,338.44 No. of  Rands        = 2,081,388.44 x   95                                        27.95            100                                = 70,744.86648                                = 70745	M1     M1   M1   A1	Multiplication     Subtraction    Multiplication/Division   C.A.O.
4	Time taken cycling  = 60/30 = 2 hrs Remaining distance = 74 – 60 = 15km Time taken walking = 15/15 = 1 hr Total time spent       = 2+1+1h 15 min                                = 3hr 15 min Arrival time             = 9.49+3hr 15min                                 = 1304hrs	M1 M1 A1	Totalling of the time Addition of the time Expressing in 24 hours
5	S = (2n – 4) 90  S = (2x5 – 4) 90 = 540⁰    = 2x + 100⁰ + 100⁰ + 2x – 40 + 2x - 10⁰ = 540                                                  6x + 150⁰ = 540⁰                                                             6x = 390⁰                                                               x = 65⁰                                          ∟ABC = 2x65 – 10 = 120⁰	M1 M1   A1	Formation of equation Collection of like terms   C.A.O.
6	0.007429 ½  = (7.429 X 10-3) ½                       = (74.29 X 10-4) ½                       = √74.29 X 10-2                      = 8.6191 X 10-2                      = 0.086191 (8.6191 X 10-2)-1 = 8.6191-1 X 102                              = 0.116                               = −21.15 X 0.116                               = −2.453	B1 B1 A1	Square root Reciprocal Correct product
7.	A.S.F. = 40000000000cm2 Area of village = 40000000000 x 2 x 3.2                         = 256 000 000 000cm2     Area in km2 = 256 000 000 000                                10000000000                          = 25.6km2	M1 M1   A1	Area of the village Conversion into km2   C.A.O.
8.	2X – 3Y = −17 5X + 6Y = −2                                       x = −4                                       y =  3	M1     M1     A1	Pre-multiplication by determinant    Simplification     Correct values of x and y
9	Pupils         Km           days             Milk 100             20             30                 144 200              5               x                  432 No.. of days = 30 x 200 x 5/20 x 432                                100             144                     = 45 days	M1   A1	Compounding the ratio   C.A.O.
10.	=  (23)⅔ x (54)3/2         (22)5/2 x (52)3/2 = 22 x 5⁶    25 x 53 = 2-3 x 53 = ⅛ x 125 = 125      8 = 15 ⅝	M1   M1     A1	Expressing in index form     Applying laws of indices   Expressing as a fraction
11.	cos θ = 3/5  and sin θ = 4/5 cos θ × sin θ = 3/5 × 4/5 = 12/25	M1 A1
12	ă=kb̆ 4 = −2k ⟹k = −2 4 = −2(x + 5) x = 14  = −7       −2	M1 M1     A1
13.	Area of triangle AOC = ½ × 8 × 13 × sin 30°=26cm²  Area of sector = 30/360 × 3.142 × 8 × 8 = 16.76cm³ Area of unshaded region = 26 − 16.76 = 9.24cm²	M1 M1 A1
14	No  Std f Log  36.152 3.615×101 1.5581×2 3.1162 0.02573 2.573×10-2     1.5266 1.938 1.938×100 0.2143 N/D   1.3123 ÷ 3 2.737           ← 0.4374	M1      M1   M1 A1
15  a)		B1           B1	Correct shape           Correct + labeling dimensions
b)	P = 2L+W W = x  and  L= 2x 2(2x+x) = 36  6x = 36 x = 6 cm Length = 2 × 6 = 12 cm	M1 A1   B1
16  a)	1 ½ y = 8 − ⅔x        y =  8    − ⅔ ÷ 1 ½ x             1½          y = 5 1/3 – 4/9x         y = -4/9x + 5 1/3        M1 = -4/9        M2 = 9/4              = y – 3.5 = 9/4                  x + 2                 4y – 14 = 9x + 18                            y = 9/4x + 8	M1 B1 M1   A1
b)	y + 5 = 9/4  x + 4 4y + 20 = 9x + 36 −9x + 4y = 16 9x – 4y = 16	M1   A1
c)	9x – 4y = −16 2x + 5y = −33 18x – 8y = −32 (18x + 45y = 297)  −53y = 265        y = −5 but 2x + 5y = −33        2x + 25= −33                2x = −8                  x = −4 the lines meet at (−4,−5)	M1   M1   M1     A1	Formation of simultaneous equations Attempting to solve for y Attempting to solve for x     C.A.O.
17  a)	A:B = ( 8/3:23/4 )  x 12 = A:B = 32:69 B:C =    (13/4:46/9 )  x 36 = B:C = 117:184 A:B =32:69        (i) x 39 B:C = 117:184   (II) X 23 A:B = 1248:2691 B:C = 2691:4232         = A:B:C = 1248:2691:4232	M1     M1 A1	Ratios express as whole numbers   Equalizing the value of B Ratios in simplified form
b)	Gross profit = 90,000 x 12                     = 1,080,000 Net profit    = 1,080,000 – 12 x 27,000                     = 756,000	M1 A1	Gross profit – expenses Net profit
c)	A share = 1248 x 756000                  8171               = Sh. 115 467.90 No. of cows = 115 467.90                            60000                     = 1.92446457                     = 1 cow	M1 A1	Division
d)	New value =  80 x 2451300                      100                  = 1,961,040      C gets  = 4232 x 1961040                         8171                   = 1,015,680 (share)       C gets = 756000                          3                    = Sh. 252,000 Amount received by C = 1015680 + 252000                                      = Sh. 1,267,680	M1       M1 A1
18.
19	(i) commission = 6100 x 150000 = Sh. 9000 Total earnings = 35000 + 9000 = Sh.44 000 (ii) Commission earned = 100,000 − 35000 = sh 65 000 x/100 × 150 000 = 65000 1500x = 65000         x = 43.3% (iii)Marked price 100/70 × 700 = 1000         Profit =1150 − 1000 = Sh. 150    % Profit =150 × 100 = 15%                        1000	M1 M1  A1 M1 M1 A1 M1 A1 M1 A1
20		B3 – 2 mks  for each inequality B3 for correct line ie bold or dotted and  must be written M2 for a good scale B2 – for region R NB Many stds dislike it hence a motivation for those who attempted
21	(i) = 7/3 x 90 = 210km Remaining distance = 360 – 210 = 150km As = 90 + 110 = 200km Time for meeting =  150km     = 0.75 hrs                               200km/hr                            = 45 mins      Meeting time = 10.35                               +  .45                                11.20 a.m	B1   B1 M1   A1
	(ii) Distance from A 210 + (0.75 x 90)                                 = 210 + 67.5                                  = 277.5 km	M1 A1
	(iii)Time minibus arrived at B  Time = D/S = 360/90 = 4 hrs          = 8.15 + 4 hrs = 12.15 p.m Time taken by the tourist to arrive B = 12.15 pm – 10.30 a.m =  1 hr 45 min  = 145/60 x 100 = 175km ∴ Home to B = 175km Home to A = 360 – 175 = 185km	M1   M1 A1   B1
22	(a) (i) →      →     →     BD = BA + AD = -b̌ + 3/5č = 3/5č − b̌      (ii) →     →     →          AE = AB + BE = b̌ + 1/2BC            =b̌ + ½ (−b̌ + č) = ½ (b̌ + č)       AF = n(½b̌ + ½č = ½nb̌ + ½nč       AF = AB + BF = b̌ + (−b̌ + 3/5č)             =(1 − t) b̌ + 3/5tč (b)comparing the co-efficients of b and c      ½n = 1−t ⟹ n = 2−2t      ½n = 3/5t ⟹ n = 6/5t     2−2t = 65t    ⟹10−10t = 6t     t = 5/8    n = 6/5 × 5/8 = ¾ (c)   (i) BD = 5:4        (ii)AE =  3:1	B1 B1 B1 B1 B1   M1 M1 A1 (both n & t) B1 B1
23	(a) x + 3 =    1    ⟹ (x+3)(x−2) = 24        24      x − 2     x2 + x − 30 = 0      x2 + 6x − 5x − 30 = 0     (x + 6) (x−5) = 0      x=5 or x=-6 (b) (i) Width = x Length = x + 9    x(x+9) = 136  x2 + 9x − 136 = 0   x2 + 17x − 8x − 136 = 0  (x+17)(x−8) = 0 x = 8 or x = −17 width = 8 cm  and Length = 17 cm P=28+17    =50 cm (ii) Remaining area = 136 − 64 = 72m2     2x2 = 72        x  = √36 = 6 cm	M1 M1 M1   A1 M1   A1 M1 A1   M1 A1