**Instructions:**

- Show all steps in your calculations

## QUESTIONS

- Use logarithms only to evaluate, correct to 4 decimal places. (4 marks)
- Without using a calculator or mathematical tables, evaluate, leaving your answer in the form a√b+c , where a, b and c are integral values (3 marks)
- Calculate the percentage error in the volume of a cone of radius 7.0 cm and height 20 cm.(3marks)
- Make Q the subject of the formula (3 marks)
- The figure below, O is the centre of the circle. P joined to Q passes through O
- Calculate the radius and centre of the circle. (2 marks)
- Express the equation of the circle in the form 〖(x-a)〗^2+〖(y-b)〗^2=r^2 (2 marks)

- Apudo wants to buy a computer on hire purchase. The cash price of the computer is Kshs. 40,000. Apudo makes a down payment of Kshs. 14, 000 followed by 16 equal monthly instalments of Kshs. 2,625. Calculate the rate of compound interest per month correct to 2 decimal places. (3 marks)
- Without using logarithms table or calculator, solve for x in;

log 5 – 2 + log (2x + 10) = log (x - 4) (3 marks) - Expand (3+x)
^{6}up to the terms in x^{3}(1 marks) - Use the expansion in (a) above to estimate (2.97)6 correct to 4 decimal places. (2marks)

- Expand (3+x)
- The value of a piece of land was Ksh. 400000 five years ago. Currently the piece of land is valued at Ksh. 587731.20. Find the annual rate of appreciation of the piece of land. (3marks)
- Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (3marks)
- Use completing of the square method to solve: 2x
^{2}-6x+4=0 (3 marks) - An arc length of 18.5 cm subtends an of angle of 1.2
^{c}at the centre of the circle. Find the diameter of the circle to one decimal place. (3 marks) - Fatima bought maize and beans from Kami. She mixed the maize and beans in the ratio 3:2 she bought the maize at sh.90 per kg and the beans at sh.150 per kg. If she was to make a profit of 30% what would be the selling price of 1kg of the mixture. (3mks)
- Solve for x given that the following is a singular matrix. (3mks)
- The figure below shows a circle centre O. AB and PQ are chords intersecting externally at a point C. AB=9cm, PQ=5cm and QC =4cm, find the length BC. (3marks)
- A body is moving along a straight line and its acceleration after t seconds is (5 – 2t) ms
^{-2.}Its initial velocity Vms^{-1}is 4ms^{-1}. Find V in terms of t. (3marks)

**SECTION II **

- The table below represents marks scored in a mathematics test.

Using an assumed mean of 44.5, Determine- Mean marks for the test (3 marks)
- Standard deviation (4 marks)
- Determine the pass mark if 30% of the students failed the exam. (3 marks)

- Taking the radius of the earth, R=6370km and π=22⁄7, calculate the

shortest distance between two cities P(600N, 290W) and Q(600N, 310E) along the parallel of latitude. (3marks) - If it is 1200hrs at P, what is the local time at Q (3marks)

- Taking the radius of the earth, R=6370km and π=22⁄7, calculate the
- An aeroplane flew due south from a point A(60°N, 45°E) to a point B, the distance covered by the aeroplane was 8000km, determine the position of B. (4marks)

- In the figure below , O is the centre of the circle.PQR is a tangent to the circle at Q. Angle PQS=28°, angle UTQ=54° and UT=TQ

Giving reasons, determine the size of- Angle STQ (2mks)
- Angle TQU (2mks)
- Angle TQS (2mks)
- Reflex angle UOQ (2mks)
- Angle RQU (2mks)

- An arithmetic progression AP has the first term a and the common difference d.
- Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (2mks)
- The AP above is increasing and the third, ninth and twenty fifth terms form the first three consecutive terms of a geometric progression (G.P).

The sum of the seventh and twice the sixth term of AP is 78. Calculate- The first term and common difference of the A.P (5mks)
- The sum of the first 5 terms of the G.P (3mks)

- complete the table for the curves y = 3sin (2x + 300) and y = Cos2x ,use the range 0≤x ≤180°
- Using the scale Horizontal axis 1cm represent 300, vertical axis 1cm represent 1 unit, draw the graphs of y = 3 Sin (2x + 30) and y = Cos 2x (4 Marks)
- Use your graph to solve the equation 3Sin (2x + 30) = Cos 2x (1 Mark)
- Determine the following from your graph
- Amplitude of y = 3Sin (2x + 30) (1 Mark)
- Period of y = 3 Sin (2x + 30) (1 Mark)
- Period of y = Cos 2x (1 Mark)

- complete the table for the curves y = 3sin (2x + 300) and y = Cos2x ,use the range 0≤x ≤180°
- The probability of Patrick passing his exam is 0.8 that of James is 0.6 while that of Lewis is 0.2.
- Draw a tree diagram to represent the above information (2 marks)
- Use your tree diagram to find the probability that;
- Exactly two students pass in the exam (2 marks)
- At most two students pass the exam (2 marks)
- Only one student passes the exam (2 marks)
- At least one passes in the exam (2 marks)

- A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = -3, When p = - 0.2, m = 3 and n= 2.
- Find
- The equation that connects p, m and n (4marks)
- The value of p when m = 10 and n = 4 (1mark)

- Express m in terms of p and n (2marks)
- If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)

- Find
- A trader bought 8 cows and 12 goats for a total of ksh.294, 000. If he had bought 1 more cow and 3 more goats, he would have spent ksh.337, 500.
- Form two equations to represent the above information. (2mks)
- Use matrix method to determine the cost of a cow and that of a goat. (3mks)
- The trader sold the animals he had bought making a profit of 40% per cow and 45% per goat.
- Calculate the total amount of money he received. (3mks)
- Determine his profit in Kenyan shillings. (2mks)

## MARKING SCHEME

- Use logarithms only to evaluate, correct to 4 decimal places. (4 marks)

∛0.4591×7.805^{2}

629.3

No Lo *g*0.4591

7.805

629.3

1.6619

0.8924×2 +

1.78481.4467 -

2.7989

2.6478

=-23+0.64783

=1.64783=0.5493= 15493 antilog

3.542×10-1=0.3542 - Without using a calculator or mathematical tables, evaluate, leaving your answer in the form a√b+c , where a, b and c are integral values (3 marks)

√3 = √3

1 -^{√3}/_{2}2-^{√3}/_{2}

= √3 x 2 2 2 - √3

ans = 6 + 4√3 - Calculate the percentage error in the volume of a cone of radius 7.0 cm and height 20 cm.(3marks)

lower radius upper

6.95 7.0 7.05

height

19.5 20cm 20.5

%error = (max-actual)+(actual-min)×100

^{2}actual

actual=^{1}/_{3}×π×7^{2}×20=1026.2536

min=^{1}/_{3}×π×6.95^{2}×19.5=986.3541

max=^{1}/_{3}×π×(7.05)^{2}×20.5=1066.9909

(1066.9909-1026.2536)+(1026.2536-986.35412)

=40.3184/1026.2536×100=3.93% - Make Q the subject of the formula (3 marks)

x=3a+2bQ

Q^{2}

Q^{2}-2^{2}=3a+2x Q=±√3a+2x

x-2b x-2b - The figure below, O is the centre of the circle. P joined to Q passes through O
- Calculate the radius and centre of the circle. (2 marks)

(-6.0)(0.8)

√(0 - - 6)^{2}+ (8 - 0)^{2}

√36 + 64 √100 = 10 cm

radius =^{10}/_{5}5 units - Express the equation of the circle in the form (x-a)
^{2}+(y-b)^{2}=r^{2}(2 marks)

- Calculate the radius and centre of the circle. (2 marks)
- Apudo wants to buy a computer on hire purchase. The cash price of the computer is Kshs. 40,000. Apudo makes a down payment of Kshs. 14, 000 followed by 16 equal monthly instalments of Kshs. 2,625. Calculate the rate of compound interest per month correct to 2 decimal places. (3 marks)

Carrying charge = 40,000-14000=sh 26000

instalments=2625×16=sh 42,000

(42000/26000)

^{1/16}=1+R/1001.6154116=1+R100

1.0304=1+R/100

R/100=0.0304 R=3.04% - Without using logarithms table or calculator, solve for x in;

log 5 – 2 + log (2x + 10) = log (x - 4) (3 marks)

log5-log100+log2x+10=logx+4

log5-log100=logx-4-log2x+10

log5100=logx-42x+10

5100=x-42x+10

52x+10100x-400

10x+50=100x-400

-90x-90=-450-90

x=5 - Expand (3+x)
^{6}up to the terms in x^{3}(1 marks)

(3+x)^{6}upto x^{3}coefficients 1,6,15,20,15,6,1

3^{6}+6(3)^{5}(x)+15(3^{4})x^{2}+20(3^{3})x^{3}729+1458x+1215x^{2}+540x^{3} - Use the expansion in (a) above to estimate (2.97)
^{6}correct to 4 decimal places. (2marks)

3+x=0.97

x=(-0.03)

729+1458(-0.03)+1215(-0.03)^{2}+540(-0.03)^{3}=686.3389

- Expand (3+x)
- The value of a piece of land was Ksh. 400000 five years ago. Currently the piece of land is valued at Ksh. 587731.20. Find the annual rate of appreciation of the piece of land.

(3marks) - Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (3marks)

cos2x

^{0}=0.8070^{0}=36.20^{0}^{0}=36.20 323.80, 396.2, 683.8^{0}=18.10^{0},161.9^{0},198.1^{0},341.9^{0} - Use completing of the square method to solve: 2x
^{2}-6x+4=0 (3 marks)

x=3/2 ± 1/2

x_{1}=3/2+1/2=4/2=2

x_{2}=3/2-1/2=2/2=1 - An arc length of 18.5 cm subtends an of angle of 1.2c at the centre of the circle. Find the diameter of the circle to one decimal place. (3 marks)

1.2

^{c}=18.5cm^{c}=15.42cm - Fatima bought maize and beans from Kami. She mixed the maize and beans in the ratio

3: 2 she bought the maize at sh.90 per kg and the beans at sh.150 per kg. If she was to make a profit of 30% what would be the selling price of 1kg of the mixture. (3mks)

price of maize=90

beans=150

ratio=3:2

(90×3)(150×2)=x(3+2)

270+300=5x

570⁄5=5x⁄5

= 114

114×130⁄100

=148.20/= - Solve for x given that the following is a singular matrix. (3mks)

det =0

x+3-2x^{2}=0

2x^{2}-x-3=0

2x^{2}-3x+2x-3=0

x(2x-3)+1(2x-3)=0

(2x-3)(x+1)=0

x=-1 or x=3⁄2 - The figure below shows a circle centre O. AB and PQ are chords intersecting externally at a point C. AB=9cm, PQ=5cm and QC =4cm, find the length BC.

(3marks)

(9 × x)x=(9 × 4)

9x+x^{2}=36

x^{2}+9x-36=0

x^{2}+12x-3x-36=0

x(x+12)-3(x+12)=0

x=3

BC=3 - A body is moving along a straight line and its acceleration after t seconds is (5 – 2t) ms-2. Its initial velocity Vms-1 is 4ms-1. Find V in terms of t. (3marks)

v=ʃa

=ʃ5-2t

=5t-t^{2}+c

4=5t-t^{2}+c

when t=0

5(0)-(t)^{2}+c=4

c=4

v=5t-t^{2}+4

**SECTION II **

- The table below represents marks scored in a mathematics test.

Marks

10-19

20-29

30-39

40-49

50-59

60-69

70-79

No. of students

2

6

7

13

6

4

2

- Mean marks for the test (3 marks)

Class

x

f

t=x-A

*Ft*t2

Ft2

cf

10-19

14.5

2

-30

-60

900

1800

2

20-29

24.5

6

-20

-120

400

2400

8

30-39

34.5

7

-10

-70

100

700

15

40-49

44.5

13

0

0

0

0

28

50-59

54.5

6

10

60

100

600

34

60-69

64.5

4

20

80

400

1600

38

70-79

74.5

2

30

60

900

1800

40

?

?f=40

Eft-50

Eft2 8900

t¯=-1.25

x ̅=44.5-1.25

=43.25 - Standard deviation (4 marks)

√(8900⁄40)-(0.25)^{2}=14.86 - Determine the pass mark if 30% of the students failed the exam. (3 marks)

30⁄100×40=12 failed

pass mark 13th std

c.f up to 3rd class

29.5+(13·8/7)^{10}29.5+7.143

=36.64

- Mean marks for the test (3 marks)
- Taking the radius of the earth, R=6370km and π=22/7, calculate the

shortest distance between two cities P(600N, 290W) and Q(600N, 310E) along the parallel of latitude. (3marks)

θ=600

time diff=4×θ - If it is 1200hrs at P, what is the local time at Q (3marks)

θ=60°

time diff=4×θ

=4×60= 240mins/60=4hrs

1200-4=08:00hrs

=08:00am

- Taking the radius of the earth, R=6370km and π=22/7, calculate the
- An aeroplane flew due south from a point A(60°N, 45°E) to a point B, the distance covered by the aeroplane was 8000km, determine the position of B.

(4marks)

D =θ/360 × 2πR

θ/360 × 2 × 22/7 × 6370 =8000

1001/9 θ =8000

θ=71.92°

71.92-60 =11.92°

B=(11.92° S45°E)

- In the figure below , O is the centre of the circle.PQR is a tangent to the circle at Q. Angle PQS=280, angle UTQ=540 and UT=TQ

Giving reasons, determine the size of- Angle STQ (2mks)

=28°Angles in alternate segement are equal - Angle TQU (2mks)

= 63°base angles of an isosceles triangle are equal - Angle TQS (2mks)

=35°angles on a straignt line addup to 180° - Reflex angle UOQ (2mks)

=-252° angles at a point add up to 360° - Angle RQU = 54°(2mks)

- Angle STQ (2mks)
- An arithmetic progression AP has the first term a and the common difference d.
- Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (2mks)

T_{3}= a+2d

T_{9}= a+8d

T_{25}=a+24 - The AP above is increasing and the third, ninth and twenty fifth terms form the first three consecutive terms of a geometric progression (G.P).

The sum of the seventh and twice the sixth term of AP is 78. Calculate- The first term and common difference of the A.P (5mks)
- The sum of the first 5 terms of the G.P (3mks)

- Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (2mks)
- complete the table for the curves y = 3sin (2x + 30°) and y = Cos2x ,use the range 0 ≤x ≤180°

X

0

15

30

45

60

75

90

105

120

135

150

165

180

y=3sin (2x+30)

1.5

3

1.5

-1.5

-2.60

-1.00

1.5

y=Cos 2x

1

0

-0.866

-0.866

-0.5

0.866

1

- Using the scale Horizontal axis 1cm represent 300, vertical axis 1cm represent 1 unit, draw the graphs of y = 3 Sin (2x + 30) and y = Cos 2x (4 Marks)
- Use your graph to solve the equation 3Sin (2x + 30) = Cos 2x (1 Mark)

85°and 176° - Determine the following from your graph
- Amplitude of y = 3Sin (2x + 30) (1 Mark)

360⁄2=180°=AMP=3 - Period of y = 3 Sin (2x + 30) (1 Mark)

360⁄2=180° - Period of y = Cos 2x (1 Mark)

360⁄2=180°

- Amplitude of y = 3Sin (2x + 30) (1 Mark)

- complete the table for the curves y = 3sin (2x + 30°) and y = Cos2x ,use the range 0 ≤x ≤180°
- The probability of Patrick passing his exam is 0.8 that of James is 0.6 while that of Lewis is 0.2.
- Draw a tree diagram to represent the above information (2 marks)
- Use your tree diagram to find the probability that;
- Exactly two students pass in the exam (2 marks)

(8⁄10×4⁄10×2⁄10)+(8⁄10×6⁄10×8⁄10)+(2⁄10×6⁄10×2⁄10)=59⁄125 - At most two students pass the exam (2 marks)

59⁄121+(8⁄10×4⁄10×8⁄10)+(2⁄10×6⁄10×8⁄10)+(2⁄10×4⁄10×2⁄10)=21⁄25 - Only one student passes the exam (2 marks)

(2⁄10×4⁄10×2⁄10)+(2⁄10×6⁄10×8⁄10)+(8⁄10×4⁄10×8⁄10)=46⁄125 - At least one passes in the exam (2 marks)

1-(2⁄10×4⁄10×8⁄10)=117⁄125

- Exactly two students pass in the exam (2 marks)

- Draw a tree diagram to represent the above information (2 marks)
- A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = -3, When p = - 0.2, m = 3 and n= 2.
- Find
- The equation that connects p, m and n (4marks)

p=xm^{2}+yn

3.8=4x-3y

-0.2=9x+2y

7.6=8x-6y

-0.6=27x+6y

7=35x

7⁄35=x

x=1⁄5=0.2

3.8=0.8-3y

i=3=-3y

-1=y

p=0.2m^{2}-n - The value of p when m = 10 and n = 4 (1mark)

p =0.2m^{2}-n

=20-4

p=16

- The equation that connects p, m and n (4marks)
- Express m in terms of p and n (2marks)

p =0.2m^{2}-n

0.2m^{2}=p+n

m^{2}=(p+n)/(0.2)

m=√((p+n)/(0.2))

m=±√((p+n)/(0.2)) - If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3mark

- Find
- A trader bought 8 cows and 12 goats for a total of ksh.294, 000. If he had bought 1 more cow and 3 more goats, he would have spent ksh.337, 500.
- Form two equations to represent the above information. (2mks)

let the price of the cow by x

goat be y

8x+12y=294000

9x+15y=337500 - Use matrix method to determine the cost of a cow and that of a goat. (3mks)
- The trader sold the animals he had bought making a profit of 40% per cow and 45% per goat.
- Calculate the total amount of money he received. (3mks)

140⁄100×30,000

=42,000×8=336000

145⁄100×4500

=6525×12=78300+/414,300 - Determine his profit in Kenyan shillings. (2mks)

414300

-294000

120,300

- Calculate the total amount of money he received. (3mks)

- Form two equations to represent the above information. (2mks)

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