Mathematics Paper 2 Questions and Answers - Pavement Pre-Mock Exams 2021/2022

INSTRUCTIONS TO CANDIDATES

• The paper contains two sections. Section I and II.
• Answer all the questions in section I and only any FIVE questions from section II.
• Show all steps in your calculations, giving answers at each stage.
• Marks may be given for each correct working even if the answer is wrong.
• Non-programmable silent electronic calculators and KNEC mathematical tables may be used.

SECTION I (50 Marks)
Answer ALL Questions in this Section

1. Use logarithm tables to evaluate. (4 Marks)
           32.95        
   √(5.25 × 0.069)
2. A two digit number is such that the sum of its digits is 9. If the digits are reversed the new number formed exceeds the original number by 9. Find the number. (3 Marks)
3. The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m. Calculate the largest angle of the field. (3 Marks)
4. Solve for in the logarithmic equation.   (3 Marks)
   log₁₀(3x + 4) − log₁₀(3 − x) = 1
5. Given that x = 2i −3j + k, y = 3i − 4 − k, and z = j + k and thatOR = 3x − y + z
   Find the magnitude of vector OR correct to 3 significant figures. (3 Marks)
6. Make the subject of the equation. (3 Marks)
   3y = y +     p     
                 q + ¹/_x 
7. Two letters are selected from the word ECONOMICS, find the probability that the letters are a constant and a vowel in that order (2 Marks)
8. In the circle below, OA = xcm , AB = 3cm, OD = 6cm and DC = 3cm.
   
   Find the value of x. (3 Marks)
9. Simplify and leave your answer in surd form. (3 Marks)
   1 + √5 + 1 − √5
   2 + √5     2 − √5
10. Omolo bought a new car for Ksh.800, 000. After 5 years he sold it through a second hand car dealer. The dealer charged a commission of 4% for the sale of the car. If Omolo received Ksh.480, 000, calculate the annual rate of depreciation of the car. (3 Marks)
11. Expand (1 + 3x)⁶ upto the fourth term, hence evaluate (1.3)⁶ to 2 significant figures.   (4 Marks)
12. A trader has 3 grades of tea P, Q and R. Grade P costs Shs.140 per kg, grade Q costs Shs.160 per kg and grade R costs sh.256 per kg. The trader mixes grade P and grade Q in the ratio 5:3 to make a brand of tea which he sells at Shs.180 per kg. Calculate the percentage profit he makes. (3 Marks)
13. Each interior angle of a regular polygon of n sides exceeds each exterior angle of regular polygon 3n of sides by 8°. Find the value of n. (3 Marks)
14. A quantity P is partly constant and partly varies as the square root of Q.
    1. Using constants k and m , write down an equation connecting P and Q. (1 Mark)
    2. If Q = 25 when P = 20 and Q = 49, when Q = 30; find the values of k and m. (2 Marks)
15. Use matrix method to find the point of intersection of lines 5x − 1 = 7y and 2 − 5y = 4x       (3 Marks)
16. The equation of a circle centre (a, b) is 2x² + 2y² − 12x − 20y + 60 = 0 . Find the value of a and b . (3 Marks)

SECTION II (50 Marks)
Answer any FIVE Questions in this Section.A church has a sitting capacity of 468 people with the members sitting in rows which have 3 long benches and 2 short ones. The long bench takes 2 people more than the short bench. Let the number of people sitting on the short bench be x.

17.  
    1. Form an expression in x for the number of rows of benches. (2 Marks)
    2. A new pastor finds this arrangement crowded and decides that by having one more person on each long bench, he can take out some rows and still sit the same number of people. Find an expression in x for the new number of rows of benches. (2 Marks)
    3. If one row of benches was taken out, find the original number of people per row. (6 Marks)
18. A form three class has 32 students. They are distributed between four houses as follows: -
    

    House	Boys	Girls
    Chania  Athi  Tana  Mara	6      3     4     5	2     3     6     3

    Two students are selected at random from the class. Find the probability that both students will be: -
    1. From Chania (3 Marks)
    2. Girls (2 Marks)
    3. Boys of Tana House (2 Marks)
    4. Either boys or from Mara House (3 Marks)
19. A tank is filled by tap A in 16 minutes. Tap B can empty the same tank in 24 minutes. Starting with an empty tank, both taps are turned on for 8 minutes and then tap B is turned off. Find: -
    1. The time taken to fill the tank. (6 Marks)
    2. The capacity of the tank if A delivered 18400 litres. (4 Marks)
20. In the diagram below TA and TB are tangents to the circle at A and B respectively. TBR is a straight line. The straight lines AB and QP intersect at K. ∠QAB=38°, ∠PBT=44° and ∠ATB=62°.
    
    Calculate:
    1. ∠PAB (1 Mark)
    2. ∠PBA (2Marks)
    3. ∠ABQ (2 Marks)\
    4. ∠AKP (3 Marks)
    5. If AK=KB=3cm and KQ=2cm, calculate the length QP. (3 Marks)
21. The data collected from an experiment involving two variables X and Y was recorded as shown below.
    

    X	1.1	1.2	1.3	1.4	1.5	1.6
    Y	− 0.3	0.5	1.4	2.5	3.8	5.3

    The variables are known to satisfy a relation of the form y = ax³ + b where a and b are constants.
    1. For each value of x in the table above, write down the value of x³ (2 Marks)
    2. By drawing a suitable straight line graph, estimate the value of a and b (6 Marks)
    3. Write down the relationship connecting y and x. (2 Marks)
22. The table below shows income tax rates for the year 2011.
    

    Taxable Income (Sh. p.a.)	Tax Rate (%)
    1 – 121 968 121 696 – 236 880 236 881 – 351 792 351 793 – 466 704 Above 466704	10      15      20      25      30

    In the year 2011, Ole Sayeye’s monthly earnings were as follows: -
    Basic salary                          Sh.42,800
    House allowance                  Sh.15,000
    Medical allowance                Sh.2,880
    Transport allowance             Sh.2,664
    Hardship allowance                30% of Basic salary
    Mr. Ole Sayeye was entitled to a monthly tax relief of Sh.1,162. Calulate: -
    1. His annual taxable income. (2 Marks)
    2. The monthly tax paid by Ole Sayeye. (5 Marks)
    3. Ole Sayeye got a salary increment equal to 50% of the year 2011 allowances. Calculate the percentage increase in the tax paid. (3 Marks)
23.  
    1. The first term of an arithmetic progression (AP) is 2. The sum of the first 8 terms of the AP is 156.
       1. Find the common difference of the AP. (2 Marks)
       2. Given that the sum of the first n terms of the AP is 416, find n . (4 Marks)
    2. The 3^rd, 5^th and 8^th terms of another AP form the first three terms of geometric progression (GP). If the common difference of the AP is 3, find: -
       The 1^st term of the G.P. (4 Marks)
24.  
    1. The weight of a metal rod varies jointly as the square of its length and inversely as the square root of its radius. Given that the length of the metal rod was increased by 5% and the radius decreased by 36%, what was the percentage change in weight. (6 Marks)
    2. If the weight of the metal rod is 6N when the length is 12cm and radius 25cm, find the weight of the metal rod when the length is 15cm and radius 81cm. (4 Marks)

MARKING SCHEME

1		M1   M1      M1   A1	All correct logs Multiplication by ½ Addition & subtraction
			4 marks
2	x + y = 9 .......(i) 10y + x − (10x + y ) = 9 10y − y + x − 10x = 9 9y − 9x = 9 y − x = 1 .......(ii) x + y = 9 → x = 9 − y  y − (9 − y) = 1 y − (9 + y) = 1 2y = 10 y = 4 x = 4 Number 45	M1  M1   A1	Equation (i) and (ii) Solving the two equations
			3 marks
3	120 − (40 + 21) = 59m 592 = 402 + 212 − 2(40)(21) cos θ 3481 = 1600 + 441 − 1680 cos θ   1440    = cos θ − 1680 cos1 − 0.8571 = 149°	M1  M1  A1
			3 marks
4	log10(3x + 4) − log10(3 − x) = log1010  3x + 4 = 10   3 − x  3x = 4 = 30 − 10x 13x = 26 x = 2	M1  M1  A1
			3 Marks
5	=3(2i −3j + k) − (3i − 4j − k) +  j + 3k  OR = 6i − 9j + 3k − 3i + 4j + k + j + 3k OR = 3i − 4j + 7k IORI = √(32 + (−4)2 + 72))         = √74        = 8.6023        = 8.602 3sf	M1  M1  A1
			3 marks
6	3y = y +     p                    q + 1/x   2y =     p               q + 1/x   2yq + 2y/x = p  2y/x = P − 2yp  x/2y =      1                P − 2yq x =     2y            P − 2yq	M1  M1 A1
			3 marks
7	P(Consonant) × P(Vowel) = 5/9 × 4/8 = 20/72 = 5/18	M1  A1
			2 marks
8	(x + 3)x = (6 + 3)6  x2 + 3x = 54  x2 + 3x − 54 = 0  (x2 + 9x) − 6x− 54) = 0  x(x + 9) − 6(x + 9) = 0 x = −9, x = 6    x = 6cm	M1   M1   A1
9	1 + √5 + 1 − √5 × 2 + √5 2 + √5     2 − √5    2 + √5  (1 + √5)(2 − √5) + (1 − √5)(2 + √5)    (2)2 − (√5)2               (2)2 − (√5)2   −3 + √5  + −3 − √5     −1              −1  3 − √5 + 3 + √5            = 6	M1  M1  A1	3 marks
			3 Marks
10	100 × 480000 =Sh. 500000          96  500 000 = 800 000 (1 + R/100)5   0.625 = (1 + R/100)5   5√(0.625) = 1 + R/100   0.91028 =  1 + R/100  0.91028 − 1 = R/100   0.089710 × 100 = 8.971%	M1  M1  A1
			3 marks
11	(1)6 + 6(1)5 (3x) + 15(1)4(3x)2 + 20(1)3(3x)3     1 + 18x + 135x2 + 540x3   1.3 = 1 + 3x  0.3 = 3x  0.1 = x  1 + 18(0.1) + 135(0.1)2 + 540(0.1)3   1 + 1.8 + 1.35 + 0.54 = 4.69                    = 4.7	M1  A1 M1    A1	C.A.O
			3 marks
12	140 × 5 + 160 × 3 = Sh. 147.50          5 + 3  180 − 147.50 = 32.5    32.5   × 100 = 22.03%  147.50	M1  M1  A1
			3 marks
13	360 − 360 = 8    n       3n   1080 − 360 = 8          3n  720 = 24n  30 = n	M1  M1  A1
			3 marks
14	P = K + M√Q 5m + k = 20 7m + k = 30 −2m    = −10 m = 5 25 + k = 20 k = −5	A1  M1  M1
			3 marks
15	5x − 7y = 1     −4x − 5y = − 2  x = 19/53 , y = 6/53	M1  M1  A1
			3 Marks
16	2x2 + 2y2 − 12x − 20y = − 60  x2 + y2 − 6x − 10y = − 30  x2 − 6x + C1 + y2 − 10y + C2 = − 30 + C1 + C2   C1 = 9    C2 = 25  x2 − 6x + 9 + y2 − 10y + 25 = − 30 + 9 + 25  (x− 3)2 + (y− 5)2 = 22   a = 3    b = 5	M1  M1  A1
			3 marks
17	No. of rows =  468   Of bences      5x + 6   No. of rows =    468   Of benches     5x + 9   468   −   468   = 1 5x + 6     5x + 9  (5x + 6)(5x + 9)  468   − (5x + 6)(5x + 9)   468   = (5x + 6)(5x + 9)                           5x + 6                              5x + 9  1404 = 25x2 + 75x + 54 25x2 + 75x − 1350 = 0 x = −9 or x = 6 ∴ x = 6 original no. of people per row = 5x + 6 = 5(6) + 6 = 36	M1 A1   M1 A1  M1    M1  M1  M1  A1
			10 marks
18	8/32 × 7/31 = 7/124  = 14/32 × 13/31 = 91/496  = ¼ × 4/10 × 3/9  = 1/30  18/32 × 17/31 + 8/32 × 7/31  = 306  +  56     992     992 = 181    496	M1   A1   M1   A1   M1   A1   M1   M1   A1
			10 marks
19	In 1 minute A fills 1/16  In 1 minute B empty 1/24  In 8 minutes both put 8(1/16 − 1/24) = 1/6  Part of tank to be filled = 6/6 − 1/6 = 5/6  1/16 → 1 Minute 5/6 = ? 5/6 × 16 = 131/3 Min. Total time = 8 + 131/3 = 211/3 Min 1 min → 1/16  211/3 ? ←  64/3 × 1/16 = 4/3  4/3 = 18400 1 → ? 1 × ¾ × 18400 = 13,800L	B1  B1  M1  M1  M1  A1  M1  M1   M1 A1
			10 Marks
20	∠PAB =44° ∠TBA = ∠TAB = 180 − 62 = 59°                                 2 ∠PBA =59 - 44 = 15° ABQ = 180 – (44 + 38 + 15) = 180 – 97  = 83° ∠PKB = 180 + (38 + 15) = 180 – 53 = 127° ∠AKP = 180 – 127 = 53° AK.KB = PK.KQ 3 × 3 = 2PK PK = 9/2 but QP = QK + KP = 2 + 4.5 QP = 6.5	B1   B1   B1   B1   B1     B1   B1   M1     M1   M1   A1
			10 marks
21	X   1.1   1.2   1.3   1.4   1.5   1.6   Y  − 0.3   0.5  1.4   2.5   3.8   5.3   x3   1.331   1.728   2.197   2.744   3.375   4.096  Line passes through (4.1, 5.2) and (1.7, 0.5) m = 5.2 − 0.5        4.1 − 1.7     = 4.7        2.4     ≈ 2  ∴ a = 2 b = y intercept = − 2.8 y = 2x3 − 2.8	M1 (For any two correct points) B1 B1 B1B1
			10 marks
22	42800                          15000                             2880                             2664 30/100 × 42800 = 12840+                            76184 × 12 = 914208 First Sh 121968 × 10/100 = Sh. 12196.8 Next Sh. 114912 × 15/100 = Sh. 17236.8 Next Sh. 114912 × 20/100 = Sh. 22982.4 Next Sh. 114912 × 25/100 = Sh. 28728 Remaining Sh. 447504 × 30/100 = Sh. 134251.2 Total tax = 215 395.2 Less relief ( 1162 × 12) = 13944 Tax paid = Sh. 201451.2 Salary Increment = 50/100 ×  33384                             = 16692 Tax Increment = 30/100 ×  16692                         = 5007.6 % Tax Increment =   5007.6   ×  100                                201451.2                              = 2.5%	M1 ( For sum and x by 12 A1   M1 M1 M1 M1 A1   M1 M1 A1
			10 Marks
23	S = n/2 (2a + (n − 1)d) 156 = 8/2(2(2) + (8 − 1)d) 156 = 16 + 28d 28d = 140 d = 140/28 d = 5 S = n/2 (2a + (n − 1)d) 416 = n/2 (2a + ( n − 1)d) 832 = 4n + 5n2 − 5n 5n2 − n − 832 = 0 n = 130 or − 128 ∴ n = 130   T3 = a + 2d = a + 6 Ts = a + 4d = a + 12 T8 = a = 7d = a + 21 a + 21 = a + 12 a + 12     a + 6 a = −6	M1   A1   M1 M1 M1 (Simplified equation) A1   M1(For all correct values T3, T5 & T8 M1 M1 (For equation and solving) A1
24	w α t²/√r  →  w = kt²/√r  l1 = 105/100 = 1.05 r1 = 64/100 = 0.64 w1 = (1.05)²k         √0.64       = 1.378125k % change in w = 1.378125 − k × 100                                       k = 0.378125 × 100 = 37.81% increase  w = kt²/√r   6 = 12²k         √25  6 = 144k          5 k = 30/144 = 15/72 = 5/24 → w =  5t²             12√r w = 5(15)² = 5 × 225       12√81     12 × 9 w = 10.42N	M1 (For correct formula) M1 M1 M1 M1 A1     M1 (For correct value of k) M1 (For correct equation) M1 (Correct sub) A1
			10 marks