Mathematics Paper 2 Questions with Answers - Sukellemo Joint Mock 2020/2021

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SUKELLEMO JOINT MOCK
Kenya Certificate of Secondary Education
MATHEMATICS PAPER 2
2 ½ hours

Instructions to Candidates

This paper consists of TWO sections; Section I and II
Answer ALL the questions in Section 1 and only five questions from Section II.
Show all the steps in your calculation, giving your answer at each stage
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.

SECTION I (50 MARKS).

Find the selling price of 6 kg of a mixture of maize flour and millet flour if 4 kg of maize flour costing sh. 60 per kg is mixed with 6 kg of millet flour costing sh. 45 per kg and a profit of 20% is realized. (3 marks)
If x = 9.6, y= 3.60 and z = 5, Find the percentage error in the calculation of ^(x+y)/_z, giving your answer to three significant figures. (3 marks)
Solve the equation log₂⁡ (x²-4) - log₂⁡ (x+2) = -4 (3 marks)
Form a quadratic equation whose roots are 2.5 + √3 and 2.5 - √3 giving your answer in the form ax²+bx+c=0 where a, b and c are integers (3 marks)
Make b the subject of the formula. (3 marks)
x = a
√(a-b)(a+b)
Express in surd form and simplify by rationalizing the denominator. (3 marks)
3 sin⁡ 45° - 2 cos⁡ 30°
tan⁡ 30°
Triangle ABC is such that AB = 8 cm, BC = 11 cm and AC = 15 cm. calculate correct to 2 decimal places the;
1. Angle ABC (2 marks)
2. Radius of the circum circle (2 marks)
In the figure below, the chords CD and AB intersect externally at T. DT = 4 cm, BT = 3 cm and CD = 5 cm. calculate the length AB. (3 marks)
Solve the following equation for 0⁰≤ x ≤ 360⁰ 2cos⁡x = sin² x + 2 (4 marks)
Given that A (4, -2, 6) and B (-3, 1, -2) and that a point N divides AB in the ratio -2:7. Find the vector ON in terms of i, j and k. (3 marks)
The equation of a circle is given by ²/₃ x²+ ²/₃ y²- 4x+ 2 ²/₃ y - 2 = 0. Determine the centre and the radius of the circle. (3 marks)
Expand (1+^2x/₃)⁸in ascending powers of x up to the fourth term. Hence use your expansion to evaluate (0.98)⁸ to three significant figures. (3 marks)
Find the base radius of a cylindrical hole with maximum volume which can be drilled into a cone of height 16 cm and radius 12 cm as shown below. (3 marks)
ABCD is a regular tetrahedron. AB=BC=CA=AD=BD=CD= 8cm.
1. Calculate the angle between line AD and plane ABC. (2 marks)
2. Calculate the angle between planes ABD and ABC (2 marks)
A contractor intends to transport 1,000 bags of cement using a lorry and a pick up. The lorry can carry a maximum of 80 bags while a pick up can carry a maximum of 20 bags. The pickup has to make more than twice the number of trips the lorry makes and the total number of trips has to be less than 30. The cost per trip is sh 2000 for the lorry and sh 900 for the pickup and the contractor wishes to minimize cost. Let x and y be the trips for the lorry and pickup respectively.
1. State the objective function. (1 mark)
2. Write down all the inequalities which govern the condition above. (2 marks)
The first, second and fifth terms of an arithmetic sequence are the first three consecutive terms of a geometric sequence. Find the common ratio (3 marks)

SECTION II (50 marks)

Using a ruler and a pair of compasses only for all constructions in this question.
1. Construct triangle ABC in which AB = 6 cm, BC = 7 cm and angle ABC = 75^o(3 marks)
2. Find locus X such that AX = 3 cm (1 mark)
3. On the same side of BC as A, construct the locus of P such that angle BPC = 120^o(3marks)
4. Show by shading the locus of Q inside triangle ABC such that angle BPC≥angle BQC. (1 mark)
5. On the side of AB opposite C, construct the locus of T such that the area of triangle ATB = 6 cm² (2 marks)
The data below shows the heights of students in a class.

Height 130 -139 140 -149 150-159 160-169 170-179 180-189 190-199

Frequency 8 10 11 18 14 12 7
1. Using assumed mean of 164.5, calculate the mean and the standard deviation(5 marks)
2. Draw a cumulative frequency curve on a grid and use it to estimate the median (5 marks)
P varies directly as the cube of Q and inversely as the square root of R
1. Given that P = 35 when Q = 8 and R = 144, find P when Q = 20 and R = 22 (5 marks)
2. If Q decreases by 24% and R increases by 40% find the percentage change in P.(5 marks)
The triangle ABC has vertices A (1, 2), B (2, 1) and C (2, 3). A^IB^IC^Iis the image of ABC under transformation given by the matrix
1. What are the coordinates of A^IB^IC^I? Plot ABC and A^IB^IC^I on the same axis (3 marks)
2. State the ratio of the areas of the two triangles and use the area of ABC to calculate the area of A^IIB^IIC^II (3 marks)
3. If A^IIB^IIC^II is the image of ABC under transformation given by the matrix . Find the coordinates of A^IIB^IIC^II. Plot A^IIB^IIC^II and describe the transformation fully. (4 marks)
1. An industrialist has 460 litres of a chemical which is 75% pure. She mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 78% pure. Find the amount of the 90% pure chemical used. (3 marks)
2. Three machines A, B and C are set to work together. A working alone takes 6 hours to complete the work; B takes 8 hours while C takes 12 hours. All the three machines started working at the same time. 40 minutes later machine A broke down. B and C continued for another 1 hour before B ran out of fuel and therefore stopped working for 20 minutes while C continued. If B resumed working after 20 minutes, calculate the:
  1. Fraction of the work left after machine A broke down. (2 marks)
  2. Fraction of the work done by C working alone for 20 minutes (2 marks)
  3. Total time taken for the work to be completed. (3 marks)
The table below shows taxation rates in Kenya

Monthly taxable income (kshs p.m) Tax rate %

1-9680 10

9681-18800 15

18801-27920 20

27921-37040 25

37041 and above 30

A civil servant is provided with a house and pays a nominal rent of sh 6,260 per month. In addition the government gives him taxable allowances amounting to sh 16,000 per month. He is entitled to a personal relief of sh 1,520 per month. He has a life insurance policy for which he pays sh 1,200 per month and claims insurance relief at the rate of sh 3 per k£. The civil servant’s PAYE is sh 6,900.Apart from PAYE and insurance his other monthly deductions are WCPS 2% of basic salary, HELB loan sh 4,000 and cooperative shares sh 600. Calculate his:
1. Taxable income per month. (6 marks)
2. Basic salary per month (2 marks)
3. Net monthly pay. (2 marks)
The probability of James, Tyson and David passing an examination are ⁴/₅,³/₄ and ²/₃ respectively. Find the probability that in one attempt:
1. only one passes the examination. (2 marks)
2. All the three passes the examination. (2 marks)
3. Two pass the examination. (2 marks)
4. None passes the examination. (2 marks)
5. At least one passes the examination. (2 marks)
When aero plane left town P (65º N,15º E) to another town Q (65º N,165º W) at a speed of 200 knots using the shortest route. Take π = ²²/₇ and radius of the earth R= 6370 km.
1. 1. Calculate the distance travelled in nautical miles. (2 marks)
  2. Calculate the time taken to travel from P to Q in hours. (2 marks)
2. Another plane left P at 1.30 pm local time and travelled to T (65ºN,60ºE) along a parallel of latitude. Calculate the:
  1. Distance between P and T to the nearest km (3 marks)
  2. Local time of arrival at town T if the plane flew at the speed of 470 km/h(3 marks)

MARKING SCHEME
SECTION I (50 MARKS).

Find the selling price of 6 kg of a mixture of maize flour and millet flour if 4 kg of maize flour costing sh. 60 per kg is mixed with 6 kg of millet flour costing sh. 45 per kg and a profit of 20% is realized. (3 marks)
BP per kg = (4 x 60) + (6 x 45)
4 + 6
= Sh. 51 BP of 6kg = 51 x 6 = Sh. 306
SP per kg = ¹²⁰/₁₀₀ x 306
= Sh. 367.20
If x = 9.6, y= 3.60 and z = 5, Find the percentage error in the calculation of ^(x+y)/_z, giving your answer to three significant figures. (3 marks)
maximum value = 9.65 + 3.605 = 2.9456
4.5
minimum value = 9.55 + 3.595 = 2.39
5.5
actual value = 9.6 + 3.60 = 2.64
5
absolute error = 2.9456 - 2.39 = 0.2778
2
percentage error = ^0.2778/_2.64 x 100% = 10.5%
Solve the equation log₂⁡ (x²-4) - log₂⁡ (x+2) = -4 (3 marks)
log₂ x² -4 = Log₂2^-4
(x+2)
(x+2)(x-2) = 2^-4
(x+2)
x-2 = ¹/₁₆
x = 2 + ¹/₁₆
x= 2¹/₁₆
Form a quadratic equation whose roots are 2.5 + √3 and 2.5 - √3 giving your answer in the form ax²+bx+c=0 where a, b and c are integers (3 marks)
x = 2.5 + √3 , x = 2.5 - √3
(x - 2.5 - √3)(x - 2.5 + √3) = 0
x² - 2.5x +x√3 - 2.5x + 6.25 - 2.5√3 - x√3 + 2.5√3 -3 =0
= 4x² - 20x + 13 = 0
Make b the subject of the formula. (3 marks)
x =   a
√(a-b)(a+b)
x =   a
  √(a² -b²)
x² =     a²
a² -b²
a² = a²x² - b²x²
b²x² = a²x² - a²
b² = a²x² - a²/x²
b² = a² - a²
x²
b = ±√( a² - a²/_x²)
Express in surd form and simplify by rationalizing the denominator. (3 marks)
3 sin⁡ 45° - 2 cos⁡ 30°
tan⁡ 30°
sin 45º = ¹/_√2 × ^√2/_√2 = ^√2/_√2
Cos 30º = ^√3/₂
Tan 30º = ¹/_√3 × ^√3/_√3 = ^√3/₃
(3 ^√2/₂ - 2 ^√3/₂)/^√3/₃
3 √2 - 2√3
2^√3/₃
(9√2 - 6√3) √3
2√3 (√3)
9√6 - 18
6
3√6 - 6
2
or 3√6 - 3
2
Triangle ABC is such that AB = 8 cm, BC = 11 cm and AC = 15 cm. calculate correct to 2 decimal places the;
1. Angle ABC (2 marks)
  b² = a² + c² - 2ac Cos B
  15² = 11² + 8² - 2(11 x 8)Cos B
  225 = 185 - 176 Cos B
  176 Cos B = - 40
  Cos B = - 0.2273
  B = 103.14º
2. Radius of the circum circle (2 marks)
  15 = 2r
  sin 103.14º
  R = 7.70cm
In the figure below, the chords CD and AB intersect externally at T. DT = 4 cm, BT = 3 cm and CD = 5 cm. calculate the length AB. (3 marks)

AT.BT = CT.DT
(AB + 3).3 = 9 x 4
AB + 3 = ³⁶/₃AB = 12 - 3
9cm
Solve the following equation for 0⁰≤ x ≤ 360⁰ 2cos⁡x = sin² x + 2 (4 marks)
2 cos x = 1 - cos²x + 2
cos²x + 2cosx - 3 = 0
p = cos x
p² + 2p -3 = 0
p² + 3p -p -3 =0
p(p+3) - 1(p+3) =0
p = 1 or -3
cos x = 1
x=0º, 360º
Given that A (4, -2, 6) and B (-3, 1, -2) and that a point N divides AB in the ratio -2:7. Find the vector ON in terms of i, j and k. (3 marks)
The equation of a circle is given by ²/₃ x²+ ²/₃ y²- 4x+ 2 ²/₃ y - 2 = 0. Determine the centre and the radius of the circle. (3 marks)
³/₂ x ( ²/₃x² + ²/₃y² - 4x + 2 ²/₃y - 2=0)
= x² + y² - 6x + 4y -3 = 0
x² - 6x + 9 + y² + 4y + 4 = 3 + 9 + 4
(x - 3)² ( y +2)² = 4²
centre = (3,-2)
Radius = 4 units
Expand (1+^2x/₃)⁸in ascending powers of x up to the fourth term. Hence use your expansion to evaluate (0.98)⁸ to three significant figures. (3 marks)
1 + 8(^2x/₃) + 28(^2x/₃)2 + 56(^2x/₃)³
= 1 + ^16x/₃ + 112x²/9 + 448x³/27
(1 + ^2x/₃) = [ 1 + (-0.02)]
^2x/₃ = -0.02
x = -0.03
0.988
1 + ¹⁶/₃(-0.03) + ¹¹²/₉(-0.03)²
+ ⁴⁴⁸/₂₇( -0.02)³
1 - 0.16 + 0.0112 - 0.000448
= 0.851
Find the base radius of a cylindrical hole with maximum volume which can be drilled into a cone of height 16 cm and radius 12 cm as shown below. (3 marks)

^r/₁₂ = ^16-h/₁₆
16r = 12(16 -h)
⁴/₃r 16 -h
h = 16 - ⁴/₃ r
V = πr²h
=πr²( 16 - ⁴/₃ r)
= (16πr² - ⁴/₃πr³)
^dv/_dr = 32πr - 4πr²
at max, ^dv/_dr = 0
4πr(8 -r) = 0
r=0 or 8
r=8cm
ABCD is a regular tetrahedron. AB=BC=CA=AD=BD=CD= 8cm.
1. Calculate the angle between line AD and plane ABC. (2 marks)
  Sin 60º = ⁴/_AO
  AO = ⁴/_Sin60º
  =4.619cm
  
  Cosθ = ^4.619/₈
  θ = 54.73º
2. Calculate the angle between planes ABD and ABC (2 marks)
  
  h = √(8² - 4²)
  = 6.928cm
  
  Cos θ = ^2.309/_6.928
  θ = 70.53º
A contractor intends to transport 1,000 bags of cement using a lorry and a pick up. The lorry can carry a maximum of 80 bags while a pick up can carry a maximum of 20 bags. The pickup has to make more than twice the number of trips the lorry makes and the total number of trips has to be less than 30. The cost per trip is sh 2000 for the lorry and sh 900 for the pickup and the contractor wishes to minimize cost. Let x and y be the trips for the lorry and pickup respectively.
1. State the objective function. (1 mark)
  2000x + 900y
2. Write down all the inequalities which govern the condition above. (2 marks)
  y > 2x
  x + y <30
  80x + 20y ≥ 1000, 4x + y ≥ 50
The first, second and fifth terms of an arithmetic sequence are the first three consecutive terms of a geometric sequence. Find the common ratio (3 marks)
a, a+d, a+ 4d
^{a + d}/_a = ^{a + 4d}/_a+d
a(a + 4d) = (a+d)²
a² + 4ad = a² +2ad+d²
2ad = d²
2a=d
common ratio (r)
^a+2d/_a =^a+2a/_a
=^3a/_a =3

SECTION II (50 marks)

Using a ruler and a pair of compasses only for all constructions in this question.
1. Construct triangle ABC in which AB = 6 cm, BC = 7 cm and angle ABC = 75^o(3 marks)
2. Find locus X such that AX = 3 cm (1 mark)
3. On the same side of BC as A, construct the locus of P such that angle BPC = 120^o(3marks)
4. Show by shading the locus of Q inside triangle ABC such that angle BPC≥angle BQC. (1 mark)
5. On the side of AB opposite C, construct the locus of T such that the area of triangle ATB = 6 cm² (2 marks)

The data below shows the heights of students in a class.

Height	130 -139	140 -149	150-159	160-169	170-179	180-189	190-199
Frequency	8	10	11	18	14	12	7

Using assumed mean of 164.5, calculate the mean and the standard deviation(5 marks)

Class	x	d=x-A	f	fd	fd²	cf
130-139	134.5	-30	8	-240	7200	8
140-149	144.5	-20	10	-200	4000	18
150-159	154.5	-10	11	-110	1100	29
160-169	164.5	0	18	0	0	47
170-179	174.5	10	14	140	1400	61
180-189	184.5	20	12	240	4800	73
190-199	194.5	30	7	210	6300	80

∑f = 80
∑fd=40
∑fd²= 24,800
mean = A + ∑fd
∑f
164.5 + 40/80 =165
standard dev = √(∑fd²/∑f) - (^∑fd/_∑f)²=√(²⁴⁸⁰⁰/₈₀) - (⁴⁰/₈₀)²
= 17.160

Draw a cumulative frequency curve on a grid and use it to estimate the median (5 marks)

median = ½ x 80 =40
From cumulative frequency
curve, median = 165.5

P varies directly as the cube of Q and inversely as the square root of R
1. Given that P = 35 when Q = 8 and R = 144, find P when Q = 20 and R = 22 (5 marks)
  P=KQ³
  √R
  35 = K x 8 x 8 x 8
  √144
  35=^512k/₁₂
  K= ¹⁰⁵/₁₂₈
  P = 105Q³
  128√R
  P= 105 x 20³
  128 x √225
  = 105 x 8000m
  128 x 15
  =437.5
2. If Q decreases by 24% and R increases by 40% find the percentage change in P.(5 marks)
  Original value of P =KQ³
    √R
  new value = K x (0.76Q)³
    √1.4R
  = 0.438976KQ³
  1.1832√R
  = 0.3710 KQ³
  √R
  change = (0.3710 -1) KQ³
    √R
  0.6290KQ³
  √R
  %change = -0.6290 x 100
  -62.90%
  P has decreased by 62.90%
The triangle ABC has vertices A (1, 2), B (2, 1) and C (2, 3). A^IB^IC^Iis the image of ABC under transformation given by the matrix
1. What are the coordinates of A^IB^IC^I? Plot ABC and A^IB^IC^I on the same axis #(3 marks)
2. State the ratio of the areas of the two triangles and use the area of ABC to calculate the area of A^IIB^IIC^II (3 marks)
  Ratio of the 2 triangles = determinant of matrix
  (2 x 3) = (0 x 0) =6
  thus ratio = 6:1
  Area of ABC = ½ x 2 x 1= 1 square unit
  Area of A^IB^IC^I= 1 x 6= 6 square units
3. If A^IIB^IIC^II is the image of ABC under transformation given by the matrix . Find the coordinates of A^IIB^IIC^II. Plot A^IIB^IIC^II and describe the transformation fully. (4 marks)
  
  A^II (-3,2) B^II (0,1)C^II(-4,3)
  The transformation is a shear parallel to the x axis with the x axis invariant shear factor. The image of A(1,2) is A¹(-3,2)
1. An industrialist has 460 litres of a chemical which is 75% pure. She mixes it with a chemical of the same type but 90% pure so as to obtain a mixture which is 78% pure. Find the amount of the 90% pure chemical used. (3 marks)
  ⁷⁵/₁₀₀ x 460 = 345
  345 + 0.9x = 358.8 + 0.78x
  Let 90% pure chemical use x 0.12x=13.8
  345 + ⁹⁰/₁₀₀x =⁷⁵/₁₀₀(460 + x)
  x = 115 litres
2. Three machines A, B and C are set to work together. A working alone takes 6 hours to complete the work; B takes 8 hours while C takes 12 hours. All the three machines started working at the same time. 40 minutes later machine A broke down. B and C continued for another 1 hour before B ran out of fuel and therefore stopped working for 20 minutes while C continued. If B resumed working after 20 minutes, calculate the:
  1. Fraction of the work left after machine A broke down. (2 marks)
    ¹/₆ + ¹/₈ + ¹/₁₂ = ³/₈
    ⁴⁰/₆₀ x ³/₈ = ¹/₄
    = 1 - ¹/₄ = ³/₄
  2. Fraction of the work done by C working alone for 20 minutes (2 marks)
    1hr = ¹/₁₂
    20min
    ^20/₆₀ x ¹/₁₂ = ¹/₃₆
  3. Total time taken for the work to be completed. (3 marks)
    B & C = ¹/₈ + ¹/₁₂= ⁵/₂₄
    ³/₄ - ⁵/₂₄ = ¹³/₂₄, ¹³/₂₄ - ¹/₃₆ =³⁷/₇₂
    ⁵/₂₄ = 1hr
    ³⁷/₇₂ = ³⁷/₇₂ x ²⁴/₅ = 2⁷/₁₅
    time= 40minutes + 1hr 20 mins + 2hrs 28mins
    =4hrs 28min
The table below shows taxation rates in Kenya

Monthly taxable income (kshs p.m) Tax rate %

1-9680 10

9681-18800 15

18801-27920 20

27921-37040 25

37041 and above 30

A civil servant is provided with a house and pays a nominal rent of sh 6,260 per month. In addition the government gives him taxable allowances amounting to sh 16,000 per month. He is entitled to a personal relief of sh 1,520 per month. He has a life insurance policy for which he pays sh 1,200 per month and claims insurance relief at the rate of sh 3 per k£. The civil servant’s PAYE is sh 6,900.Apart from PAYE and insurance his other monthly deductions are WCPS 2% of basic salary, HELB loan sh 4,000 and cooperative shares sh 600. Calculate his:
1. Taxable income per month. (6 marks)
  Tax= net tax + relief = 6900 + 1520 + (3/20x1200) = Sh.8600
  8600 - 6440=2160
  9680 x 0.1 = 968
  9120 x 0.15=1368
  9120 x 0.20=1824
  9120 x 0.25=2280
  6440
  y x 0.30 = 2160
  y = 7200
  taxable income = 7200 +(9120 x 3)+ 9680
  = 44,240
2. Basic salary per month (2 marks)
  Taxable income = (1.15 of BS) + allowance
  44240 = 1.15x + 16000 - 6260
  1.15x = 34500
  x= 30,000
3. Net monthly pay. (2 marks)
  Total deductions = 6900 + 1200 (¹²/₁₀₀ x 30,000) + 4000 + 600
  =13,300
  Net monthly pay = 44,240 - (0.15 x 30,000) - 13,300
  =26,440
The probability of James, Tyson and David passing an examination are ⁴/₅,³/₄ and ²/₃ respectively. Find the probability that in one attempt:
1. only one passes the examination. (2 marks)
  (¹/₃ x ⁴/₅ x ¹/₄) + (³/₄ x ¹/₅ x ¹/₃) + (²/₃ x ¹/₅ x ¹/₄)
  = ¹/₁₅ + ¹/₂₀ + ¹/₃₀= ³/₂₀
2. All the three passes the examination. (2 marks)
  ⁴/₅ x ³/₄ x ²/₃ = ²/₅
3. Two pass the examination. (2 marks)
  (⁴/₅ x ³/₄ x¹/₃) + (³/₄ x ²/₃ x ¹/₅) + (⁴/₅ x ²/₃ x ¹/₄)
  = ¹/₅ + ¹/₁₀ +²/₁₅
  =¹³/₂₀
4. None passes the examination. (2 marks)
  ¹/₅ x ¹/₄ x¹/₃ = ¹/₆₀
5. At least one passes the examination. (2 marks)
  1 - ¹/₆₀ = ⁵⁹/₆₀
When aero plane left town P (65º N,15º E) to another town Q (65º N,165º W) at a speed of 200 knots using the shortest route. Take π = ²²/₇ and radius of the earth R= 6370 km.
1. 1. Calculate the distance travelled in nautical miles. (2 marks)
    θ = 180º - (65 + 65) =50º
    D= 60x50 = 3000nm
  2. Calculate the time taken to travel from P to Q in hours. (2 marks)
    Time = ^d/_s = ^3000nm/_200knots = 15hours
2. Another plane left P at 1.30 pm local time and travelled to T (65ºN,60ºE) along a parallel of latitude. Calculate the:
  1. Distance between P and T to the nearest km (3 marks)
    Longitude difference = 60-15 = 45º
    Distance(km) = ^θ/₃₆₀2πRcos α
    ⁴⁵/₃₆₀ x 2 x ²²/₇ x 6370 x cos65
    =2115km
  2. Local time of arrival at town T if the plane flew at the speed of 470 km/h(3 marks)
    Time = ^2115.2/₄₇₀ = 4hrs 30 minutes
    ^{45 x 4}/₆₀ = 3hrs
    time at T = 1:30 + 4¹/₂ + 3hrs
    =9:00pm