## Mathematics Paper 2 Questions and Answers - Samia Joint Mock Examination 2021/2022

INSTRUCTIONS TO THE CANDIDATES

1. Write your name and index number in the spaces provided above.
2. Sign and write the date of examination in the spaces provided above.
3. This paper consists of TWO sections; Section I and II.
4. Answer ALL the questions in sections 1 and ONLY FIVE questions from section II
6. Marks may be given for correct working even if the answer is wrong.
7. Non – programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
8. Candidates should answer questions in English.

SECTION 1 (50 marks)
Answer ALL questions in this section in the spaces provided.

1. The external and internal radius of a cylindrical cement pipe are given as 20cm and 14cm respectively and its length is given as 140cm. Calculate the minimum volume of cement required to make such pipe. (Take π = 3.142) (3mks)
2. Solve for x in the equation 2sin2x + 3cosx = - 1 for 0º< x <360º (4mks)
3. A quadratic curve cuts the x – axis at points (-2, 0) and (3, 0). Find the equation of this curve in the form ax2 + by = c where a, b and c are integers. (3mks)
4. Expand (2 + 3x)6 up to the term in x3. Hence use your expansion to estimate (2.09)6 (3mks)
5. Two quantities M and N are such that M varies partly as N and partly as the square of N. Determine the relationship between M and N given that when M is 1050, N = 10 and when M = 2200, N = 20. (3mks)
6. A dealer has two types of grades of tea, A and B. Grade A costs shs. 140 per kg while grade B costs shs.160 per kg. If the dealer mixes A and B in the ratio 3:5 to make a brand of tea which he sells at shs.180 per kg, calculate his percentage profit. (3mks)
7. The position vectors of A and B are given as a = 2i -3j + 4k and b = -2i - j + 2k respectively. Find to 2 decimal places, the length of the vector AB. (3mks)
8. Calculate the quartile deviation in
18, 9, 14, 20, 23, 12, 16 (3mks)
9. Find the equation of a circle whose radius is 0.75 units and centre coordinates (0.75, -0.5). Leave your answer in the form;
x2 + y2 + ax + by + c = 0 (3mks)
10. Solve for the exact value of x in the equation. (3mks)
2log10x + log105 = 1 + 2log104
11. A plane leaves an airport A (38.50N, 37.050W) and flies due North to a point B on latitude 520N. The plane then flies due East to a point C 2400km from B. Determine the position of C. (Given that the radius of the earth is 6371km) (3mks)
12. Evaluate ɭ3-1(2x2 – 3x – 14)dx (3mks)
13. Using a ruler and a pair of compass only, construct common tangents from the point P to the circle below. (3mks)
14. Given that 3 + 3√5   3 -√5 = a + b√5. Find the values of a and b. (3mks)
3 + √5
15. In the figure below angle CBD = 370, angle BCD = 200 and ABC is a tangent to the circle at B.

Find:
1. Angle BED (2mks)
2. Angle ABE (2mks)
16. A transformation is represented by the matrix  . This transformation maps a triangle
ABC of the area 12.5cm2 onto another triangle ABC of the area 12.5cm2 onto another triangle A1B1C1. Find the area of triangle A1B1C1. (3mks)

SECTION II(50marks)

Answer ANY FIVE questions in this section in the spaces provided
17. A transformation represented by matrix maps A(0,0), B(2,0), C(2,3) and D(0,3)
Onto A1B1C1 and D1 respectively
1. Draw ABCD and its image A1B1C1D1 (4mks)
2. A transformation represented by  maps A1B1C1D1 on A11B11C11D11.
Plot A11B11C11D11on the same graph. (3mks)
3. Determine the matrix of a single transformation that maps A11B11C11D11. onto ABCD. (3mks)
18. The diagram below shows a right pyramid with a square base ABCD and vertex V. O is the centre of the base. AB = 14m, VA = 20m and N is the midpoint of BC.

Find;
1. The lengths of BO, VO and VN (3mks)
2. The angle between VO and plane VBC (3mks)
3. The angle between VB and base ABCD (2mks)
4. The angle between VDC and VBC (2mks)
19. The table below shows the distribution of marks scored in a test by standard 8 pupils in one school.
 Marks No. of pupils 30 – 34 1 35 – 39 5 40 – 44 10 45 – 49 10 50 – 54 19 55 – 59 20 60 – 64 20 65 – 69 8 70 – 74 4 75 – 79 3
Using 57 as the assumed mean mark, calculate
1. The actual mean for the grouped marks (3mks)
2. The 50th percentile (3mks)
3. The standard deviation of the marks (4mks)
20.
1. Complete the table for the function y = ½ Sin2x, where 00< x< 3600. (2mks)
 x 0º 30º 60º 90º 120º 150º 180º 210º 240º 270º 300º 330º 360º 2x 0º 60º 120º 180º 240º 300º 360º 420º 480º 540º 600º 660º 720º Sin2x 0º 0.866 0º 0.866 y = ½ Sin2x 0º 0.433 0º
2. On the grid provided, draw the graph of the function y = ½ Sin2x for 00<x<3600 using the scale 1cm for 300 on the horizontal axis and 4cm for 1 unit of y axis. (3mks)
3. Use your graph to determine the amplitude and period of the function y = ½ Sin2x.(2mks)
4. Use the graph to solve
1. ½ Sin2xº = 0 (1mk)
2. ½ Sin 2xº – 0.5 = 0 (2mks)
21. Income tax rate were charged as follows in a given year.
 Income in ksh. p.m Rate of tax in each sh. 1 – 11, 180 10% 11, 181, - 21, 714 15% 21, 715 – 32, 248 20% 32, 249 – 42, 781 25% 42, 782 and above 30%
A teacher earns a basic salary of Ksh. 48, 000. He is housed by the employer and pays a rent of Ksh. 3,000 per month. His allowances are: Commuter Ksh. 2, 500 and medical Ksh. 3,500. He is entitled to a family relief of Ksh. 1,648 per month.
Determine his:
1. Taxable income per month (2mks)
2. Net tax per month (4mks)
NHIF Ksh. 1,250
WCPS Ksh. 1,200
Co – operative shares Ksh. 3,000
Calculate the net salary. (3mks)
22.
1. The first term of an Arthmetric progression (AP) is 2. The sum of the first 8 terms of the AP is 156
1. Find the common difference of the AP (2mks)
2. Given that the sum of the first n terms of the AP is 416, find n (2mks)
2. The 3rd, 5th and 8th terms of another AP form the first three terms of a Geometric Progression (GP). If the common difference of the AP is 3, find;
1. The first term of the GP; (4mks)
2. The sum of the first 9 terms of the GP, to 4 significant figures. (2mks)
23. A married couple intends to have 3 children. They consult an expert who tells them that the probability of a male birth is 0.55.
1. Draw a tree diagram to represent this occurrence (2mks)
2. Find the probability that
1. All the three children will be female (2mks)
2. At least a male is born (2mks)
3. At least 2 will be females, giving your answer to 3 s.f. (4mks)
24. A trader is required to supply two types of sweaters, type A and type B. the total number of sweaters must not be more than 400. He has to supply more of type A than type B sweaters. However the number of type A sweaters must not be more than 300 and the number of type Bsweaters must not be less than 80. Let x be the number of type A sweaters and y be the number of type B sweaters.
1. Write down in terms of x and y all the linear inequalities representing the information above. (4mks)
2. On the grid provided, draw the inequalities and shade the unwanted regions.(4mks)
3. The profits were as follows;
Type A: sh. 600 per sweater
Type B: sh. 400 per sweater
1. Use the graph to determine the number of sweaters of each type that he should make to maximize the profit. (1mk)
2. Calculate the maximum possible profit (1mk)

## MARKING SCHEME

1. πR2H(min) - πr2h(max)
(3.142 x 19.5x 139.5) - (3.142 x 14.52 x 140.5)
= 166,666.9973 - 92815.07275
= 73851.92455
2. 2(1 - cos2x) + 3cosx - 1 = 0
2 - 2cos2x - 3cosx - 1 = 0
2cos2x - 3cosx - 1 = 0
let cosx = y
2y2 - 3y - 1 = 0
y =3 ± √9 + 8 = 3 ± √17
4                 4
= 1.780776406
- 0.280776406
cosx = 1.780776406
cosx = - 0.280776406 = 73.69º
∴ x = 180 - 73.69 = 106.31
x = 180 + 73.69 = 253.69
3. (-2,0)(3,0)
(x + 2)(x - 3) = 0
x(x - 3) +2(x - 3) = 0;
x2 - 3x + 2x - 6 = 0
x2 - x = 6
4.    26     25     24     23
(3x)2 (3x)1 (3x)2 (3x)
1      6         15     20
64 + 576x + 2160x2 + 4320x
2 + 3x = 2.09x = 0.03
64 + 576(0.03) + 2160(0.03)2 + 4320(0.03)3
= 83.34064
5. M = KN + CN2
1050 = 10K + 100C
2200 = 20K + 400C
2100 = 20K + 200C
2200 = 20K + 400C
+100 = -200C
-200      -200
C = ½
1050 = 10K + 50
K = 100
M = 100N + ½N2
6.    A          B
sh140   160
3           5
cost  price per kg of the mixture
(140 x 3) + (160 x 5) = 152.50
8
profit = 180 - 152.50 = 27.5
%p =   27.5   x 100% = 182/61%
152.50
7. AB = B - A

√(-4)2 + 22 + (-2)2 = √24 = 4.90
8. 9,12,14,16,18,20,23
Q1 = 12   Q= 20
20 - 12 = = 4
2         2
9. R = 3/4 centre (3/4 - ½)
(x - 3/)2 + (y + ½)2 =(3/4)
x263x + + y + y + ¼ =
42    16                      16
x2 = 3x + y2 + y + 9   - ¼
2                  16    16
x2 - 3x  + y2 + y = - ¼
2
4x+ 4y- 6x + 4y + 1 = 0
10. log10x2 + log5 = log1010 + log1016
log10(5x2) = log10(160)
5x2 = 160
√x= √32 = ±4√2
x =  ±4√2
11.   θ   x 2 x 22  x 6371cos52 = 2400
360          7
θ = 35.04º
37.05 - 35.04 = 2.01
C(52ºN, 2.01ºW)
12.  2x3 - 3x2  - 14x + c 3-1
3      2
(2(3)3 - 3(3) - 14(3)) - (2(-1)3 - 3(-1) - 14(-1))
3         2                        3          2
(18 - 13.5 - 42) - (-2/33/2 + 14)
- 37.5 - 115/6
= -49 ¹/
13.
14. (3(3 -√5) + 3√5(3 + √5)
(3 + √5)(3 - √5)
9 - 3√5 + 9√5 + 15
9 - 5
24 + 6√5 = 6 + 1.5√5
4
a = 6 and b = 1.5 or ½
15.
1. ∠BED = 37º; angles in alternate segments
2. ∠ABE = 57º; angles in alternate segment
16. Det = (1 x 2) - (3 x 4) = -10
Area of ΔA1B1C= 12.5 x -10 = -12.5
= 125cm2
17.
18.
1. BO= √142 + 142  = 9.899
2
VO = √202 - 9.8992 = 17.38
VN = √72 + 17.382 = 18.74

2. Sinθ =17.38
18.74
= 68.04º

3. Tanθ =17.38
9.899
= 60.34

4. Sinθ =9.899
13.12
= 48.99 x 2
= 97.98º
19.
 Marks f x x - 57d fd d2 fd2 cf 30 – 34 1 32 -25 -25 625 625 1 35 – 39 5 37 -20 -100 400 2000 6 40 – 44 10 42 -15 150 225 2250 16 45 – 49 10 47 -10 -100 100 1000 26 50 – 54 19 52 -5 -95 25 475 45 55 – 59 20 57 0 0 0 0 65 60 – 64 20 62 5 100 25 500 85 65 – 69 8 67 10 80 100 800 93. 70 – 74 4 72 15 60 225 900 97 75 – 79 3 77 20 60 400 1200 100 N= Σfd = -170 Σfd2 = 9750
1. Actual mean
X= A + Σfd
Σf
57 + (-170/100) = 55.30
2. 50th percentile
50  x 100 = 50
100
54.5 +(50 - 45)5
20
= 55.75
3. Standard deviation

= √94.61 = 9.726767192
20.
1.
 x 0º 30º 60º 90º 120º 150º 180º 210º 240º 270º 300º 330º 360º 2x 0º 60º 120º 180º 240º 300º 360º 420º 480º 540º 600º 660º 720º Sin2x 0º 0.866 0.866 0º -0.866 -0.866 0 0.866 0.866 0 -0.866 -0.866 0 y = ½ Sin2x 0º 0.433 0.433 0º -0.433 -0.433 0 0.433 0.433 0 -0.433 -0.433 0
2.

3. Amplitude = 0.5
Period = 180º
4.
1. x = 0º,90º,180º,270º,360º
2. x = 45º,225º
21.
1. Taxable income per month
115 x 48000 - 3000 + 2500 + 3500
100
= 58200
2. 11180 x  10  = 1118
100
10534 x 15  = 1580.10
100
10534 x 20  = 2106.80
100
10533 x 25  = 2633.25
100
15419 x 30  = 4625.70
100
1118 + 1580.10 + 2106.80 + 2633.25 + 4625.70
= 12063.85
12063.85 - 1648 = 10415.85
3. Total deductions
10415.85 + 1250 + 1200 + 3000
15865.85
58200 - 15865.85
= 42334.15
22.
1.
1. a = 2
S8 =n/2(2a +(n - 1)d)
156 = 8/2(4 + 7d)
7d = 35   d =5
2. 416 = n/2(4 +(n - 1)5)
832 = 4n + 5n2 - 5n
5n- n - 832 = 0
n = 1 ± √1 + 16640129 = ±129   n = 3
10                      10
n = 13 only
2.
1. a + 2d, a + 4d, a + 7d
a + 4da + 7d
a + 2d    a + 4d
a = 2d ; a = 2 x 3 = 6
a + 2d
6 + 2(3) = 6 + 6 = 12
2. a + 4d =6 + 4(3) = 1.5
a + 2d       12
S9 =a (rn - 1)
r - 1
12(1.59 - 1) = 898.640625
1.5 - 1
= 898.6
23.
1.
2. P(FFF) = 0.45 x 0.45 x 0.45
= 0.161125 or 729   = 0.908875
8000
3. P(At least 2 females)
P(MFF) or P(FMf) or P(FFM) + P(FFF)
(0.55 x 0.45 x 0.45) + (0.45 x 0.55 x 0.45) + (0.45 x 0.45 x 0.55) + (0.45 x 0.45 x 0.45)
= 0.425
24.
1. x + y ≤ 400
x >y
x ≤ 300
y ≥ 80
2.
3. 600x + 400y = P
1.   x    y   = 1
100    180
x = 300  y = 100
Type A = 300
Type B = 100
2. 600(300) + 400(100)
= 220,00