INSTRUCTIONS TO CANDIDATES
- Write your name,class and admission number
- The paper contains two section:section I and section II
- Answer ALL questions in section I and only five questions in section II.
- All answers and working must be written on the question paper in the space provided below each question.
- Show all the steps in your calculations, giving your answers at each stage, marks may be given for correct working even if the answer is wrong.
SECTION I
- Jane mistypes (x + y)2 as x2 + y2. Find the percentage error in the evaluation of (x + y)2 when x= -2 and y= 12 (3mks)
- An arc length a cm subtends angle of 0.39c at the centre of a circle of radius 6cm. find the value of a (3mks)
- Find the semi-interquartile range of the following set of numbers. 63, 65, 76, 65, 63, 51, 52, 95, 63, 71, 83. (3mks)
- Solve the equation.
log (3x-1)=log (2x+1)-log 4 (3mks) - Rationalize the denominator (3mk)
√5 + √3
√7 - √3 -
- Write down the first 4 terms in a seconding power of x in the expansion of (1-2x)5 (2mks)
- Use your expansion to estimate the value of (0.96)5 (2mks)
- Make q the subject of the formula (3mks)
T = (b-q/q)½ - In the figure below, PT is a tangent to the circle of T. PQ = 9cm, SA =6cm, AT = 8 cm and AR = 3 cm.
Calculate the length of- AQ (2mks)
- PT (2mks)
- Solve the simultaneous equations (4mks)
x2 + xy = 4y
– x = 2 - An arithmetic progressive whose first term is 2 and the nth term 32 has the sum of n terms equal to 357. Find n. (3mks)
- PQR is a triangle of area 9cm2. If PQ is the fixed base of the triangle and is 6cm long. On the upper side of PQ.
Draw Δ P QR and describe the locus of point R (3mks) - Use matrix method to solve. (3mks)
b = 4a + 6
3a - 2b = -2 - State the amplitude and the period of the wave y =3 sin3/4θ (2mks)
- Find the centre and radius of the circle whose equation is 4x2 – 12x + 4y2 – 8y – 3 = 0 (3mks)
- Twenty men can dig a trench 300m long in 15 days. Find the number of days it would take 30 men to dig a trench 360m long.(3mks)
- Use logarithms to evaluate (3mks)
1.763√0.2876
379.5
SECTION II
Answer any FIVE questions only. - The table below shows the marks scored by students in a mathematics test.
Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 No of students 3 5 6 21 12 6 4 2 1 - From the table above, determine the 25th percentile (2mks)
- On the grid provided, draw an Ogive curve that represents the above information. (4mks)
- Using the graph above
- Determine the pass mark if 45% of the students passed. (2mks)
- If the pass mark was to be pegged at 60%. How many students passed? (2mks)
- Three qualities P, Q and R are such that P varies directly as Q inversely as the square root of R.
Given that P = 2250 when Q = 450 and R = 64- Write down an equation connecting P, Q, & R (4mks)
- If Q decreased by 16% and R increased by 44%.
Calculate the percentage change in P (3mks) - In a soccer competition the number of goals (G) scored in a penalty shoot-out is partly constant and partly varies as the skill (S) of the player. Given that when S= 1 and G= 6 when S= 2 G=4.find the value G when S =3 (3mks)
- The cost of a minibus was sh 950000. It depreciated in value by 5% per year for the first two years and by 15% per year for subsequent years.
- Calculate the value of the minibus after 5 years. (4mks)
- After 5 years, the minibus was sold through a dealer at 25% more than its value to Mr. Nyeri. If the dealer’s sale price was to be taken as its value after depreciation, calculate the average monthly rate of depreciation for the 5years. (6mks)
- A trader deals in two types of rice .Type P & Q .Type P costs ksh 1600 per bag and type Q sh 1400 per bag
- The trader mixes 30 bags of type P & 50 bags of type Q. If he sells the mixture at a profit of 20%,
Calculate the selling price of one bag of the mixture. (4mks) - The trader now mixes type P with type Q in the ration x:y respectively. If the cost of the mixture is ksh 1534 per bag, find the ration. (4mks)
- The trader mixes one bag of the mixture in part (a) with one bag of the mixture in part (b) above.
Calculate the ration of type P rice to type Q rice in this mixture. (2mks)
- The trader mixes 30 bags of type P & 50 bags of type Q. If he sells the mixture at a profit of 20%,
- A point D’ A’ and Y’ are images of triangle DAY with vertices D (4, 4) A (0,2) and Y(-2,4) respectively under a transformation given by matrix
- Determine the co-ordinates of D’, A’ and Y’ (3mks)
- Triangle D’’ A’’ Y’’ is the image of triangle D’ A’ Y’ under another transformation whose
matrix is
Determine the coordinates of D’’ A’’ Y’’ (3mks) - Find a single matrix of transformation that maps D" A" Y" onto triangle D A Y (2mks)
- A plane figure whose area is 20cm2 undergoes transformation represented by QP. Find the area of image. (2mks)
-
- Complete the table below for the function Y = 1 + 3x - 2x2 - x3
x -3 -2 -1 0 1 2 - x3 +27 0 -1 - 2x2 -8 0 -2 1 + 3x 1 7 y 1 1 - Using a scale of 2cm represent 1 unit on the x-axis and 1 cm represent 1 unit on y-axis.
Draw the graph (3mks) - By drawing a suitable line graph; solve the equation
X3 + 2x2 - 5x – 6 = 0 (3mks) - Find the co-ordinates of the turning points. (2mks)
- Complete the table below for the function Y = 1 + 3x - 2x2 - x3
- Patients A and B are to be tested for covid - 19 virus. The probability that A will be positive is 0.6 and that B will be negative is 0.2 Find the probability that
- Both will be positive (2mks)
- Neither will be positive (2mks)
- One will be positive (2mks)
- At least one will be positive (2mks)
- At least B is negative. (2mks)
-
- Complete the table below
x 0 30 60 90 120 150 180 210 240 270 300 330 360 sin x 0 0.87 0.87 0 -0.87 -1 0.5 2 sin (x+30) 1 1.73 2 1 0 -1 - On the same axes, draw the graphs of y = sin x and y = 2 sin (x+30) 0º ≤ x ≤ 360º (5mks)
- From the graph, find the roots of
2 sin (x + 30) – sin x = 0 (1mk) - Describe fully the transformation that maps the graph of
Y = sin x onto that of Y = 2 sin (x + 30º) (2mks)
- Complete the table below
MARKING SCHEME
SECTION I
- Jane mistypes (x + y)2 as x2 + y2. Find the percentage error in the evaluation of (x + y)2 when x= -2 and y= 12 (3mks)
(x + y)2 = (12 - 2) = 100
x2 + y2 = 122 +(-2)2 = 148
error = 148 - 100 = 48
% = 48/100 x 100
= 48% - An arc length a cm subtends angle of 0.39c at the centre of a circle of radius 6cm. find the value of a (3mks)
= 57.3 x 0.39
= 22.347
22.347 x 2 x 3.142 x 6
360
=2.34 cm
OR
arc length = θ
radius
0.39 = a/6
2.34cm = a - Find the semi-interquartile range of the following set of numbers. 63, 65, 76, 65, 63, 51, 52, 95, 63, 71, 83. (3mks)
51, 52, 63, 63, 65, 71, 76, 83, 95
Q1 =52 + 63 = 57.5
2
Q3 =76 + 83 = 79.5
2
Q3 - Q1 =79.5 - 57.5
2 2
22 = 11
2 - Solve the equation.
log (3x-1)=log (2x+1)-log 4 (3mks)
3x - 1= 2x + 1
4
12x - 4 = 2x + 1
10x = 5m
x = ½A - Rationalize the denominator (3mk)
√5 + √3
√7 - √3
√5 + √3 √7 + √3 M1
√7 - √3 √7 +√3
√35 + 15+ 21+ 3
7 - 3
√35 + 15+ 21+ 3
4 -
- Write down the first 4 terms in a seconding power of x in the expansion of (1-2x)5 (2mks)
1 - 5(2x) + 10(2x)2 - 10(2x)3 + ....
1 - 10x + 40x2 - 80x3 + ..... - Use your expansion to estimate the value of (0.96)5 (2mks)
1 - 2x = 0.96
0.04 = 2x
0.02 = x
1 - 10(0.02) + 40(0.0004) - 80(0.000008)
1 - 0.2 + 0.016 - 0.00064
0.81536
- Write down the first 4 terms in a seconding power of x in the expansion of (1-2x)5 (2mks)
- Make q the subject of the formula (3mks)
T = (b-q/q)½
T2 =b - q
q
T2 q = b - q
T2 q + q = b
a(T2 + b) = b
q = b
T2 + 1 - In the figure below, PT is a tangent to the circle of T. PQ = 9cm, SA =6cm, AT = 8 cm and AR = 3 cm.
Calculate the length of- AQ (2mks)
SA - AT = RA-AQ
6 x 8 = 3AQ
16cm = AQ - PT (2mks)
RP.QP = PT2
28.q = PT2
√252 = PT2
15.87cm = PT
- AQ (2mks)
- Solve the simultaneous equations (4mks)
x2 + xy = 4y
– x = 2
y = 2 + x
x2 + x(2 + x) = 4
x2 + 2x + x2 = 4
2x2 + 2x - 4 = 0
x2 + x - 2 = 0
x2 - x + 2x - 2 = 0
x(x - 1) + 2(x - 1) = 0
(x + 2)(x - 1) = 0
x = -2 or 1
when x = -2 when x = 1
y = 0 y = 3 - An arithmetic progressive whose first term is 2 and the nth term 32 has the sum of n terms equal to 357. Find n. (3mks)
a = 2
l = 32
n/2[2a + (n - 1)d] = n/2(a + l) = 357
n/2 (2 + 32) = 357
34n = 714
n = 21 - PQR is a triangle of area 9cm2. If PQ is the fixed base of the triangle and is 6cm long. On the upper side of PQ.
Draw Δ P QR and describe the locus of point R (3mks) - Use matrix method to solve. (3mks)
b = 4a + 6
3a - 2b = -2 - State the amplitude and the period of the wave y =3 sin3/4θ(2mks)
Amplitude = 3
Period = 360 x 4/3 = 480º - Find the centre and radius of the circle whose equation is 4x2 – 12x + 4y2 – 8y – 3 = 0 (3mks)
x2 - 3x + y2 - 2y - 3/4 = 0
x2 - 3x + (3/2)2 + y2 - 2y + (-2/2) = (-3/4)
(x - 3/2 )2 + (y - 1)2 = 4
Centre (1.5 , 1)
radius = 2 - Twenty men can dig a trench 300m long in 15 days. Find the number of days it would take 30 men to dig a trench 360m long.(3mks)
Men Trench Days 20 300 15 30 360
30 300 - Use logarithms to evaluate (3mks)
1.763√0.2876
379.5
No Log 1.76
0.28761/3
379.5
3.061 x 10-3
0.003060.2455
T.4588 x 1/3
3/3 + 2.4588 /3 = T.8196
0.0651 -
2.5792
3.4859
SECTION II
Answer any FIVE questions only. - The table below shows the marks scored by students in a mathematics test.
Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 No of students 3 5 6 21 12 6 4 2 1 C.F 3 8 14 35 47 53 57 59 60 - From the table above, determine the 25th percentile (2mks)
25/100 x 60 = 15
39.5+(1/21 x 10) = 39.98 - On the grid provided, draw an Ogive curve that represents the above information. (4mks)
- Using the graph above
- Determine the pass mark if 45% of the students passed. (2mks)
45/100 x 60 = 27 passed
33 failed
Pass mark = 48 + 1 = 49 - If the pass mark was to be pegged at 60%. How many students passed? (2mks)
60 - 48 = 12 students
- Determine the pass mark if 45% of the students passed. (2mks)
- From the table above, determine the 25th percentile (2mks)
- Three qualities P, Q and R are such that P varies directly as Q inversely as the square root of R.
Given that P = 2250 when Q = 450 and R = 64- Write down an equation connecting P, Q, & R (4mks)
P = kQ
√R
2250 =450k
8
k = 40
P =40Q
√R - If Q decreased by 16% and R increased by 44%.
Calculate the percentage change in P (3mks)
Q1 = 0.84Q
R1 = √1.44R
P1 =k 0.84 Q
1.2√R
=0.84 P
1.2
(0.7P - P) x 100
P
P(-0.3) x 100 = -30
P
Decrease of 30% - In a soccer competition the number of goals (G) scored in a penalty shoot-out is partly constant and partly varies as the skill (S) of the player. Given that when S= 1 and G= 6 when S= 2 G=4.find the value G when S =3 (3mks)
G = C + MS
6 = C + M
4 = C + 2M
2 = -M
M = -2
C = 8
G = 8 - 2S
G = 8 - 6
= 2
- Write down an equation connecting P, Q, & R (4mks)
- The cost of a minibus was sh 950000. It depreciated in value by 5% per year for the first two years and by 15% per year for subsequent years.
- Calculate the value of the minibus after 5 years. (4mks)
first 2yrs
950000(1 - 5/100)2
950000(0.9025)
sh 857375
subsequent 3yrs
857375(1 - 5/100)3
857375(0.614125)
= sh 526535
OR
950000 x 0.952
= 526535.42 - After 5 years, the minibus was sold through a dealer at 25% more than its value to Mr. Nyeri. If the dealer’s sale price was to be taken as its value after depreciation, calculate the average monthly rate of depreciation for the 5years. (6mks)
125/100 x 526535 = sh658168.75
658168.75 = 950000(1 - r/100)5
0.6928 = (1 - r/100)5
5√0.6928 = 1 - r/100
0.9292 = 1 - r/100
r/100 = 0.07077
r = 7.077p.a
Rate per month =7.077 = 0.58975
12
- Calculate the value of the minibus after 5 years. (4mks)
- A trader deals in two types of rice .Type P & Q .Type P costs ksh 1600 per bag and type Q sh 1400 per bag
- The trader mixes 30 bags of type P & 50 bags of type Q. If he sells the mixture at a profit of 20%,
Calculate the selling price of one bag of the mixture. (4mks)
Buying Price of mixture
(30 x 1600) + (50 x 1400)
30 + 50
118000 = sh1475
80
Selling price of mixture
120 x 1475
100
= sh1770 - The trader now mixes type P with type Q in the ration x:y respectively. If the cost of the mixture is ksh 1534 per bag, find the ration. (4mks)
1600x + 1400y = 1534
x + y
66x = 134y
x/y = 67/33
x:y = 67:33 - The trader mixes one bag of the mixture in part (a) with one bag of the mixture in part (b) above.
Calculate the ration of type P rice to type Q rice in this mixture. (2mks)
Fraction of P in (a) 30/80
in (b) 67/100
Total = 30/80 + 67/100 = 1.045
Fraction of Q in (a) 50/80
in (b) 33/100
Total = 50/80 + 33/100 = 0.955
1.045:0.955
209:191
- The trader mixes 30 bags of type P & 50 bags of type Q. If he sells the mixture at a profit of 20%,
- A point D’ A’ and Y’ are images of triangle DAY with vertices D (4, 4) A (0,2) and Y(-2,4) respectively under a transformation given by matrix
- Determine the co-ordinates of D’, A’ and Y’ (3mks)
- Triangle D’’ A’’ Y’’ is the image of triangle D’ A’ Y’ under another transformation whose matrix is
Determine the coordinates of D’’ A’’ Y’’ (3mks) - Find a single matrix of transformation that maps D" A" Y" onto triangle D A Y (2mks)
- A plane figure whose area is 20cm2 undergoes transformation represented by QP. Find the area of image. (2mks)
- Determine the co-ordinates of D’, A’ and Y’ (3mks)
-
- Complete the table below for the function Y = 1 + 3x - 2x2 - x3
x -3 -2 -1 0 1 2 - x3 +27 0 -1 - 2x2 -8 0 -2 1 + 3x 1 7 y 1 1 - Using a scale of 2cm represent 1 unit on the x-axis and 1 cm represent 1 unit on y-axis.
Draw the graph (3mks) - By drawing a suitable line graph; solve the equation
X3 + 2x2 - 5x – 6 = 0 (3mks) - Find the co-ordinates of the turning points. (2mks)
- Complete the table below for the function Y = 1 + 3x - 2x2 - x3
- Patients A and B are to be tested for covid - 19 virus. The probability that A will be positive is 0.6 and that B will be negative is 0.2 Find the probability that
- Both will be positive (2mks)
0.6 x 0.8
= 0.48 - Neither will be positive (2mks)
0.4 x 0.2 = 0.08 - One will be positive (2mks)
(0.6 x 0.2) + (0.4 x 0.8)
0.12 + 0.32 = 0.44 - At least one will be positive (2mks)
0.44 + 0.48 = 0.92 - At least B is negative. (2mks)
(0.6 x 0.2) + (0.4 x 0.2)
0.12 + 0.08
= 0.2
- Both will be positive (2mks)
-
- Complete the table below
x 0 30 60 90 120 150 180 210 240 270 300 330 360 sin x 0 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.8 0.5 0 2 sin (x+30) 1 1.73 2 1.73 1 0 -1 -1.73 -2 -1.73 -1 0 1 - On the same axes, draw the graphs of y = sin x and y = 2 sin (x+30) 0º ≤ x ≤ 360º (5mks)
- From the graph, find the roots of
2 sin (x + 30) – sin x = 0 (1mk)
x = 126º or 306º - Describe fully the transformation that maps the graph of
Y = sin x onto that of Y = 2 sin (x + 30º) (2mks)
Stretch parallel to the y axis scale factor 2, followed by translation
- Complete the table below
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