Mathematics Paper 2 Questions and Answers - Cekana Mock Exams 2023

INSTRUCTION

• The paper consists of two sections. Section I and Section II.
• Answer ALL the questions in Section I and any FIVE questions in Section II.
• Show all the steps in your calculations, giving your answer 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.
• Candidates should answer the questions in English.

QUESTIONS

SECTION A (50MARKS)
Answer All Questions in this section

1. The area of a rectangle is 48.4cm2 and its length is 9.37cm. Calculate the percentage error in width (3marks)
2. The base of a triangle is (3x-2) cm. The height of the triangle is 5cm shorter than its base. Given that the area of the triangle is 25cm2. Find length of the base. (4marks)
3. Find the area of the triangle shown below (3marks)
4. Express the following expression in surd form and simplify by rationalising the denominator (3marks)
2/ 1-tan30
5. Solve for x in the expression (log2x)2 + log2x3=10 (3marks)
6. The table below is part of the tax for monthly income for the year 2007
 Monthly income (ksh) Rate % Under ksh 10165 10 From 10165 but under ksh19741 15 From 19741 but under ksh29317 20

In that year, John’s monthly gross tax was Ksh 2885. Calculate his monthly income (3marks)
7. A quantity E is partly constant and partly variaes as square root of F.
1. Write down an equation connecting E and F where K and C are constants respectively.
2. If F=25 where E=22 and F=49 when E=28. Find the value of K and C (3marks)
8.
1. Expand (1+ 1⁄2 x)7 in ascending power of x (1mark)
2. Using the first four times of the expansion in (a) above estimate the value of (1 1⁄30)7 (2marks)
9. Find the number of terms of the series 5+8+11+14+17+……….. that will give a sum of 2183 (3marks)
10. Find the inverse of the matrix M=  hence find the co-ordinates of the point at which
line 4x-3y=4  and 2x+y=7 intersect (3marks)
11. A trader mixed grade I, II and III of coffee in the ratio 2:3:5 respectively. Grade I cost Sh650 per kg, grade II costs sh 500 per kg and grade III costs sh 420 per kg.
1. Find the cost of one kg of the mixture (2marks)
2. If the trader sold the mixture at a profit of 20% calculate the selling price of 3kg of the mixture (2marks)
12. Solve for the equation 4cos2 x=5-4 sin x  for 0°≤ x ≤360 (3marks)
13. The figure below has a cross-section of the prism which is an isosceles triangle of side AE= 8cm, DE= 8cm and AD=6cm, where AB=20cm

1. If G is the mid- point of side BC find
1. GF (1mark)
2. AG (1mark)
2. The angle between line DF and plane ABCD (2marks)
14.
1. Construct triangle PQR in which PQ=QR=5cm and angle PQR=90° (2marks)
2. Draw the locus of S which moves in such a way that it is always on the same side of PR as Q and angle
PSR =45°(1mark)
15. Determine the amplitude and period in the function y= 1⁄2 sin(3x-60) (2marks)
16. Two towns P and Q are located on the equator such that P is due east of Q. The distance between the two towns is 1920nm. If the latitude of p is 50°E. Determine the longitude of Q(3marks)

SECTION B (50 MARKS)
Answer any five questions from this section

1. The cost of Jane’s car at the beginning of year 2000 was sh 750,000. It depreciated in value by 7% per year for the first 3 years, by 8% for the next two years and 11% per year for the subsequent years.
1. Find the value of the car at
1. The start of the year 2003 (2marks)
2. The end of year 2007 (3marks)
2. At the beginning of 2008, Jane sold the car through Mary,s dealers at 22% more than its actual depreciated value to Lucy. Taking Mary’s sale price as the car’s value after depreciation. Find the average monthly rate of depreciation for the 8years. (5marks)
2. MNP is a triangle with vertices M,(2,2),N(2,-2) and P(-1,-4). Draw the triangle MNP. (1mark)
1. If vertex N(2,-2) is mapped to N1 (5,-2) by a shear with x-axis invariant, draw triangle M1N1P1, the image of triangle MNP under the shear. (2marks)
2. Find the matrix that represent the shear in (a) above (2marks)
3. A transformation where matrix is T= (0  1) maps triangle M1N1P1 onto M11N11P11. Draw the triangle M11N11P11 (3marks)
1  1.5
4. Describe fully a single transformation that maps triangle M11N11P11 back onto triangle MNP and give its transformation matrix. (2marks)
3.
1. Complete the table below for the equation y=x3-3x2-5x+7  (2marks)

 x -3 -2 -1 0 1 2 3 4 5 y -32 8 7 0 -8 32
2. Draw the graph of  y=x3-3x2-5x+7(3marks)
3. Use your graph to estimate the roots of the equation  x3-3x2-5x+7=0 (2marks)
4. Use your graph to solve the equation  x3-3x2-10x+17=0 (3marks)
4. The table below shows marks scored by 40 students in a test
 Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 frequency 1 5 8 12 7 4 2 1

1. Calculate the upper quartile and lower quartile (4marks)
2. If 30% of students failed the test, find the pass mark (3marks)
3. The pass mark was set at 25 marks. How many students passed the test. (3marks)
5. The diagram below shows two intersecting circles with centres X and Y. HG is a tangent to the circle centre X at C. Angle GCE=70° and angle CEF=130°. Given that CB=5cm, BA=4cm, AE=12cm and radius DY=6cm.

1. Determine
1. Angle DXE (2marks)
2. Length DE (2marks)
3. Angle DYE (2marks)
2. Calculate the area of the shaded region (take π= 22⁄7) (4marks)
6.
1. Wekesa has two fair tetrahedral solids one red and other green. The faces of red solid are numbered 1to 4 while the faces of green solid are numbered 2,3,5 and 6. He tosses the two solids at the same time and recorded the number that is the bottom face of each solid. If k is the number that is at the bottom face of red solid while t is the number that is at the bottom face of green solid, find the probability that
1. 2k+t=7 (1mark)
2. (2k+t) is at most 9 (1mark)
3. 2k≥t (2marks)
2. In a group of 40 people, 10 are healthy and every person of the remaining 30 has either high blood pressure, a high level of cholesterol or both. Given that 15 have high blood pressure, 25 have high level of cholesterol and x have both. If a person is selected at random from this group. What is the probability that he/ she
1. Has high pressure only (2marks)
2. Has high level of cholesterol only (1mark)
3. Hass high blood pressure and high level of cholesterol (1mark)
4. Has either high blood pressure or high level of cholesterol (2marks)
7. A company was contracted to transport 1200 tonnes of sand. The company used type A and type B trucks to do the job. Each type A truck carries 10 tonnes of sand per trip while B curries 15 tonnes per trip. The total number of trips must not be less than 70 and type B truck must make at least twice as many trips as type A truck while the later must make not less than 10trips. Taking x to represent the number of trips made by type A trucks and y to represent the number of trips made by type B trucks
1. Write down all the inequalities representing the above information (4marks)
2. Represent the inequality graphically (4marks)
3. The company makes a profit of sh2000 per trip made by each type A truck and sh 3000 per trip made by each type B truck.
1. Using a search line, determine the number of trips each type of truck must be made to maximise the profit (1mark)
2. Hence calculate the maximum profit (1mark)
8.
1. Find the area of the region bounded by the cuver y=x3-x and the axes between the point x=-1 and x=1 (4 marks)
2. At a certain instant a body is moving in a straight line at 20cm/s. Its acceleration during the first 5seconds of the subsequent motion is (30-6t) cm/s2 where t is the time in seconds. After 5 seconds it travels with a constant speed. Find
1. Its velocity after 2seconds (2marks)
2. The greatest velocity attained (2marks)
3. The distance travelled in 10 seconds (2marks)

Marking Scheme

1.

min with = 48.35/9.375 =5.157cm
max width = 48.45/ 9.365 =5.174cm
absolute error = 5.174-5.157/ 2
=0.0085
%age error =0.0085/ (48.4 ÷ 9.37) ×100
=0.164%

M1

M1

A1

Alt

Error in area=0.05

Error in length=0.005

R.E=

%age =0.001846 error

3

2.

1⁄2 (3x-2)(3x-7)=25
9x2×27x-36=0
x2-3x-4=0
(x+1)(x-4)=0
x=-1 or 4⇒x≠-1
x=4

Length of the base 3(4)-2=10

M1

M1

A1

B1

4

3.

C = 180-(48+69)= 63°
a/ sin 69 = 8.4/ sin63
a=8.8 cm
b/ sin48 = 8.4/sin63 ⇒b=7cm
∴A1⁄3 × 8.8 × 7sin 63
=27.44cm2

M1

M1

A1

3

4.

2/ 1-tan 30 = 2/ 1- 1/√3
=2√3√3-1
=2√3(√3+1)/ (√3-1)(√3+1)
=2√3(√3+1)/ 3-1 =6+2√3/ 2
3+√3

B1

M1

A1

Returning denominator

3

5.

(log2x)2 + 3log2 x -10=0
let log 2 x=y
y2+3y-10=0
(y-2)(y+5)=0
y=2 or -5
log2 x=2⇒x=4
log2x=-5⇒x=2-5 = 1⁄32

M1

A1

B1

Attempting to solve

(y-2)(+r)=0

For

For both

3MARKS

6.

1st slab 10164×10%=1016.40

2nd slab (19740-10164)× 15⁄100=1436.40
∴2885=1016.40+1436.40+(x-19740)× 20⁄100
0.2x=4380.20
x=sh21,901
monthly income=sh21901

M1

M1

A1

For both

3MARKS

7.i)

E=K+C√F
22=K+C√25
28=K+C√49
K+5C=22
K+7C=28/ 2C=6
C=3
K=7

B1

B1

M1

A1

Attempt to solve

Both

3MARKS

8    .a)

b)

(1+1⁄2x)7=1+7⁄2x+21⁄4x2+35⁄8x3+35⁄16x4+21⁄32x5+7⁄64x6+1⁄28x

x=1⁄15
∴1+7⁄2(1⁄15)+21⁄4(1⁄15)+35⁄8(1⁄15)3
=1.25796

B1

M1

A1

CAO

3MARKS

9.

2183= n⁄2 (10+(n-1)3)
3⁄2 n2+7⁄2 n-2183=0
3n2+72 - 4366 =0
n=
n=37or-39.33
no.of terns could be-ve
∴=37

M1

M1

A1

Attempt to solve for n

C.A.O

3MARKS

10.

B1

M1

A1

3MRKS

11  .a)

b)

cost of kg =2 × 650+3 × 500+5 × 420/ 2+3+5
=sh490
selling price for kg = 490 × 1.2
=588
for 3kg =588 × 3
=1764

M1

a1

M1

A1

4MARKS

12.

4(1-sin2x)=5-4sin x
4sin2 x-4 sin x+1=0
let y = sin x
4y2 - 4y+1=0
(2y-1)(2y-1)=0
y= 1⁄2
sin x =1⁄2
x=30°, 150

m1

a1

b1

3marks

13.i)

ii)

GF=
=7.42
AG=
=20.22

tan°= 7.42/ 20.22
θ=20.15°

b1

b1

m1

a1

4marks

14.

b1

b1

b1

üconstruction

Of 90o

forü construction of DPQR

üconstruction of 450 and labelled

3marks

15.

amplitude= 1⁄2
period= 360⁄3 =120

b1

b1

Allow

2marks

16.

60θ=d
60θ=1920
θ= 1920/ 60
=32°
longitude of =(50-32)° E
=18°E

m1

a1

b1

3marks

17   .i)

ii)

750000(1 - 7⁄100)3
=750000(0.93)3
=sh603268
750000  0.93× 0.922 × 0.893
=sh359961

m1

a1

m1 m1

a1

Alt

b)

1.22 × 359961
sh43152
∴750000(1- r⁄1200)96 =439152
1- r⁄1200=
1 - r⁄1200 =0.9944
r⁄1200 =0.00556
r⁄12 0.556%
average rate per month =6.72%

m1

m1

m1

m1

a1

For 96 indicated

For root

Collecting like terms

10marks

18.a)

b1

b1

b1

B1

For  MNP üdrawn

For M1N1 P1

location of Point

üfor location of

For ü  drawn

b)

2a+2b=-1
2c-2d=5
4b=-6⇒ = -3⁄2  a=1
2c+2d=2
2c-2d=-2
4d=4
d=1
c=0
matrix

m1

a1

c)

m''(2,2)N''(-2,2)P''(-4,-1)

m1

a1

d)

reflaction in the line y=x

b1

b1

10marks

19.a)

y=-3, -7, 3

b2

B1

For allü values

For two values

s1

p1

c1

l1

For  drawn

c)

d)

1.8, 1.0, 3.8
y=x3-3x2-5x+7
0=x3-3x2-10x+17

y=5x-10
x=-2.7, 1.4, 4.4

b1 B1

b1

B1

For y=5x-10

10marks

20.a)

CF 1,6,14,26,33,37,39,40
Q1 =30.5 + (10-6/8)10
=35.5
Q2=50.5+(30-26/7)10
=56.21

m1

a1

m1

a1

b)

30% of 40 =12failed
∴30.5+(13-6/8)10
=30.5+8.75
pass mark = 39.25 marks

b1

m1

a1

(or 13 seen)

c)

Let the percentage be x
20.5+(x⁄100 × 40-1/ 5)10 =25
20.5+(0.4x-1)2 =25
x=3.25/ 0.4
=8.125% failed the exam
(100-8.128)% passed the exam
no.of student who passed
91.875/ × 100  40
=36.75
∴37 students passed the exam

m 1

m1

a1

Alternative

10marks

21.a)

i)
CDE=70∠angle in alt segment
DEC=180-130=50°
DCE=180-(50+70°)=60°
∴DXE=2(DCE)=2 × 60
=120°

ii)
∴DE=12-3=9cm

iii)
cos(YDE)=4.5/ 6
YDE=41.41
=138.59°

iv)
DYE=2(90-41.41)=97.18°
area of segment =97.18/ 360 ×  22⁄7 × 62- 1⁄2 × 62 sin97.18
=12.68cm2
radius of section DXE = 4.5/ cos30
=5.196cm
area of segment 120⁄360 × 22⁄7 × 5.1962 - 1⁄2 × 5.962 × sin120
=16.19cm2
total area of intersection =12.68+16.59
=29.27cm2

m1

a1

m1

a1

m1

a1

m1

m1

m1

a1

10marks

22a)

 2 4 2 8 2 3 5 6 4  6   8    10 5  7   9    11 7  9   11  13 8  10 12  14

i)

P(2K+t=7)= 2⁄16

b1

ii)

p(2k+t atmost p) = 9⁄16

B1

iii)

p(2k-t ≥ 0) = 12⁄16

t

2k

 2    4    6    8 2 3 4 6 0    2    4    6 -1   1    3    5 -2    0    2   4 -4    -2   0   2

B1

B1

For table

For

b)i)

x+y+z=30
x+y=15
x+z=25
=15  x=10  y=5

M1

A1

For attempting to solve all questions

ii)

p(high level cholestral only) = 15⁄40
= 1⁄4

B1

iii)

p(both) 10⁄40
= 1⁄4

B1

iv)

p(e3itherHp) = 5⁄40 + 15⁄40
= 20⁄40
= 1⁄2

M1

A1

10marks

23.a)

x+y≥70
y≥2x
10x+15y≤1200⇒2x+3y≤240
x≥10

b1

b1

b1

b1

b)

b1

b1

b1

B1

For ü and drawn

forü

forü  drawn

c)i)

ii

iii)

2000+3000y=n

For maximum point

30 of type A

60 of type B

Maximum profit

30×200+3000×60

Sh240000

b1

b1

b1

Search line drawn

both

10marks

24.a)

areaA = 01 (x3 4x)dx
=[x4/ 4 -x2/ 2]0-1
=[1⁄4 - 1⁄2]+[0]
=1⁄2 A = 1⁄2 sq units

areaB = [x4/ 4 - 42/ 2]10 =[1⁄4 - 1⁄2]
= 1⁄2
area= 1⁄2
total area = 1⁄2 + 1⁄2 =1sq units

m1

m1

m1

a1

forü integration

for substitution

of-1 and 0

for substitution

b) i)

ii

v= ∫(30-6t)dx=30t-3t2+c
20=0+0+c
c=20
v=30t-3t2+20
t=2
v=60-12+20
68cm/s

30-6t=0⇒t=5
v=30(5)-75+20
=95cm/s

m1

a1

m1

a1

(a=0)

c)

50 (30t-3t2+20)dx+5 × 95
[15t2-t3+20t]50+5 × 95
350+495

=82559 sq units

m1

a1

10marks

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