# Mathematics Paper 2 Questions and Answers - Achievers Joint Mock Exams 2023

Instructions:

• Show all steps in your calculations

## QUESTIONS

1. Use logarithms only to evaluate, correct to 4 decimal places. (4 marks)
2. 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. Calculate the percentage error in the volume of a cone of radius 7.0 cm and height 20 cm.(3marks)
4. Make Q the subject of the formula (3 marks)
5. The figure below, O is the centre of the circle. P joined to Q passes through O

1. Calculate the radius and centre of the circle. (2 marks)
2. Express the equation of the circle in the form 〖(x-a)〗^2+〖(y-b)〗^2=r^2 (2 marks)
6. 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)
7. Without using logarithms table or calculator, solve for x in;
log 5 – 2 + log (2x + 10) = log (x - 4) (3 marks)
1. Expand (3+x)6 up to the terms in x3 (1 marks)
2. Use the expansion in (a) above to estimate (2.97)6 correct to 4 decimal places. (2marks)
8. 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)
9. Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (3marks)
10. Use completing of the square method to solve: 2x2-6x+4=0 (3 marks)
11. 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)
12. 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)
13. Solve for x given that the following is a singular matrix. (3mks)
14. 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)
15. 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

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

Using an assumed mean of 44.5, Determine
1. Mean marks for the test (3 marks)
2. Standard deviation (4 marks)
3. Determine the pass mark if 30% of the students failed the exam. (3 marks)
1. 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)
2. If it is 1200hrs at P, what is the local time at Q (3marks)
1. 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)
2. 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
1. Angle STQ (2mks)
2. Angle TQU (2mks)
3. Angle TQS (2mks)
4. Reflex angle UOQ (2mks)
5. Angle RQU (2mks)
3. An arithmetic progression AP has the first term a and the common difference d.
1. Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (2mks)
2. 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
1. The first term and common difference of the A.P (5mks)
2. The sum of the first 5 terms of the G.P (3mks)
1. complete the table for the curves y = 3sin (2x + 300) and y = Cos2x ,use the range 0≤x ≤180°
2. 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)
3. Use your graph to solve the equation 3Sin (2x + 30) = Cos 2x (1 Mark)
4. Determine the following from your graph
1. Amplitude of y = 3Sin (2x + 30) (1 Mark)
2. Period of y = 3 Sin (2x + 30) (1 Mark)
3. Period of y = Cos 2x (1 Mark)
4. The probability of Patrick passing his exam is 0.8 that of James is 0.6 while that of Lewis is 0.2.
1. Draw a tree diagram to represent the above information (2 marks)
2. Use your tree diagram to find the probability that;
1. Exactly two students pass in the exam (2 marks)
2. At most two students pass the exam (2 marks)
3. Only one student passes the exam (2 marks)
4. At least one passes in the exam (2 marks)
5. 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.
1. Find
1. The equation that connects p, m and n (4marks)
2. The value of p when m = 10 and n = 4 (1mark)
2. Express m in terms of p and n (2marks)
3. If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)
6. 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.
1. Form two equations to represent the above information. (2mks)
2. Use matrix method to determine the cost of a cow and that of a goat. (3mks)
3. The trader sold the animals he had bought making a profit of 40% per cow and 45% per goat.
1. Calculate the total amount of money he received. (3mks)
2. Determine his profit in Kenyan shillings. (2mks)

## MARKING SCHEME

1. Use logarithms only to evaluate, correct to 4 decimal places. (4 marks)
0.4591×7.8052
629.3
 No Log 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
2. 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
3. Calculate the percentage error in the volume of a cone of radius 7.0 cm and height 20 cm.(3marks)
6.95    7.0    7.05
height
19.5    20cm   20.5
%error = (max-actual)+(actual-min)×100
2actual
actual=1/3×π×72×20=1026.2536
min=1/3×π×6.952×19.5=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%
4. Make Q the subject of the formula (3 marks)

x=3a+2bQ2
Q2-2

Q2=3a+2x           Q=±√3a+2x
x-2b                       x-2b
5. The figure below, O is the centre of the circle. P joined to Q passes through O
1. 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
5 units
2. Express the equation of the circle in the form (x-a)2+(y-b)2=r2 (2 marks)
6. 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/100

1.6154116=1+R100

1.0304=1+R/100

R/100=0.0304 R=3.04%
7. 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
1. Expand (3+x)6 up to the terms in x3 (1 marks)
(3+x)6 upto x3
coefficients 1,6,15,20,15,6,1
36+6(3)5 (x)+15(34)x2+20(33)x3
729+1458x+1215x2+540x3
2. 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
8. 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)
9. Given that Cos 2x°=0.8070, find x to 1 decimal place when 0°≤x≤360° (3marks)

cos2x0=0.8070
cos 2x0=36.200
2x0=36.20    323.80, 396.2, 683.8
x0=18.100,161.90,198.10,341.90

10. Use completing of the square method to solve: 2x2-6x+4=0 (3 marks)

x=3/2 ± 1/2
x1=3/2+1/2=4/2=2
x2=3/2-1/2=2/2=1

11. 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)

1.2c=18.5cm
1.0=x
=18.5/1.2
1c=15.42cm

12. 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/=
13. Solve for x given that the following is a singular matrix. (3mks)
det =0
x+3-2x2=0
2x2-x-3=0
2x2-3x+2x-3=0
x(2x-3)+1(2x-3)=0
(2x-3)(x+1)=0
x=-1 or x=3⁄2
14. 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+x2=36
x2+9x-36=0
x2+12x-3x-36=0
x(x+12)-3(x+12)=0
x=3
BC=3
15. 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-t2+c
4=5t-t2+c
when t=0
5(0)-(t)2+c=4
c=4
v=5t-t2+4

SECTION II

1. 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
Using an assumed mean of 44.5, Determine
1. 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 ̅=∑ƒt⁄∑ƒ=-50⁄40
t¯=-1.25
x ̅=44.5-1.25
=43.25
2. Standard deviation (4 marks)
√(8900⁄40)-(0.25)2
=14.86
3. 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
1. 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
=4×60= 240mins/60=4hrs
1200-4=08:00hrs
=08:00am

2. 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
1. 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)
2. 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
1. Angle STQ (2mks)
=28°Angles in alternate segement are equal
2. Angle TQU (2mks)
= 63°base angles of an isosceles triangle are equal
3. Angle TQS (2mks)
=35°angles on a straignt line addup to 180°
4. Reflex angle UOQ (2mks)
=-252° angles at a point add up to 360°
5. Angle RQU = 54°(2mks)
3. An arithmetic progression AP has the first term a and the common difference d.
1. Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (2mks)
T3 = a+2d
T9 = a+8d
T25=a+24
2. 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
1. The first term and common difference of the A.P (5mks)
2. The sum of the first 5 terms of the G.P (3mks)
1. 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
2. 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)
3. Use your graph to solve the equation 3Sin (2x + 30) = Cos 2x (1 Mark)
85°and 176°
4. Determine the following from your graph
1. Amplitude of y = 3Sin (2x + 30) (1 Mark)
360⁄2=180°=AMP=3
2. Period of y = 3 Sin (2x + 30) (1 Mark)
360⁄2=180°
3. Period of y = Cos 2x (1 Mark)
360⁄2=180°
4. The probability of Patrick passing his exam is 0.8 that of James is 0.6 while that of Lewis is 0.2.
1. Draw a tree diagram to represent the above information (2 marks)
2. Use your tree diagram to find the probability that;
1. 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
2. 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
3. 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
4. At least one passes in the exam (2 marks)
1-(2⁄10×4⁄10×8⁄10)=117⁄125
5. 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.
1. Find
1. The equation that connects p, m and n (4marks)
p=xm2+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=0.2m2-n
2. The value of p when m = 10 and n = 4 (1mark)
p =0.2m2-n
=20-4
p=16
2. Express m in terms of p and n (2marks)
p =0.2m2-n
0.2m2=p+n
m2=(p+n)/(0.2)
m=√((p+n)/(0.2))
m=±√((p+n)/(0.2))
3. If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3mark
6. 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.
1. 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
2. Use matrix method to determine the cost of a cow and that of a goat. (3mks)
3. The trader sold the animals he had bought making a profit of 40% per cow and 45% per goat.
1. 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
2. Determine his profit in Kenyan shillings. (2mks)
414300
-294000
120,300

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

### Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp