QUESTIONS
SECTION I (50 marks)
Answer all questions in this section in the spaces provided.
 The length of two similar iron bars were given as 12.5 m and 9.23 m. Calculate the maximum possible difference in length between the two bars. (3 marks)
 The first term of an arithmetic sequence is 7 and the common difference is 3.
 List the first six terms of the sequence; (1 mark)
 Determine the sum of the first 50 terms of the sequence. (2 marks)
 In the figure below, BOD is the diameter of the circle centre O. Angle ABD =30^{0} and angle AXD = 70^{0}.
Determine the size of ; Reflex angle BOC; (2 marks)
 Angle ACO. (1 mark)
 Three quantities L, M and N are such that L, varies directly as M and inversely as the square of N. Given that L = 2 when M = 12 and N = 6, determine the equation connecting the three quantities. (3 marks)
 The table below shows the frequency distribution of marks of students scored in a test.
Marks Frequency 110 2 1120 4 2130 11 3140 5 4150 3  Determine the amplitude and period of the function, y = 2 cos (3x  45)^{0}. (2 marks)
 In a transformation, an object with n area of 5 cm^{2} is mapped onto an image whose area is 30 cm^{2}. Given the matrix of transformation is , find the value of x. (3 marks)
 Expand (3  x )^{7} up to the term containing x^{4}. Hence find the aproximate value of (2.8)^{7}. (3 marks)
 Solve for the equation;
2 log 15  log x = log 5 + log (x  4). (4 marks)  The figure below represents a cuboid PQRSTUVW.
PQ = 60 cm, QR = 11 cm, RW = 10 cm.
Calculate the angle between line PW and plane PQRS, correct to 2 decimal places. (3 marks)  Solve the simultaneous equations; (4 marks)
3x  y = 9
x^{2}  xy = 4  Muga bought a plot of land for Ksh 280 000. After 4 years, the value of plot was Ksh 495 000. Determine the rate of appreciation, per annum, correct to one decimal place. (3 marks)
 The shortest distance between two points A (40^{0}N, 20^{0}W) and B(0^{0}S, 20^{0}W) on the surface of the earth IS 8008 km. Given that the radius of the earth is 6370 km, determine the position of B.
(Take π = ^{22}/_{7} )(3 marks)  Vectors r and s are such that r = 7i + 2j  k and s =  i + j  k. Find r + s. (3 marks)
 The gradient of a curve is given by . The curve passes through the point (1, 0). Find the equation of the curve. (3 marks)
 The graph below shows the rate of cooling of a liquid with respect to time.
Determine the average rate of cooling of the liquid between the second and the eleventh minutes. (3 marks)
SECTION II (50 marks)
Answer any five questions in this section in the spaces provided.
 A paint dealer mixes three types of paint A, B and C, in the ratios A:B = 3:4 and B:C = 1:2. The mixture is to contain 168 litres of C.
 Find the ratio A:B:C. (2 mark)
 Find the required number of litres of B. (2 marks)
 The cost of A is Ksh 160, type B is Ksh 205 and type C is Ksh 100.
 Calculate the cost per litre of the mixture. (2 marks)
 Find the percentage profit if the selling price of the mixture is Ksh 182 per litre. (2 marks)
 Find the selling price of a litre of the mixture if if the dealer makes a 25% profit. (2 marks)
 In the figure below OS is the radius of the circle centre O. Chords SQ and TU are extended to meet at P and QR is perpendicular to QS at R. QS = 61 cm, PU = 50 cm, UT = 40 cm and PQ = 30 cm.
 Calculate the length of:
 QS; (2 marks)
 QR. (3 marks)
 Calculate correct to 1 decimal place:
 The size of angle ROS; (2 marks)
 The length of the minor arc QS. (3 marks)
 Calculate the length of:
 The table below shows income tax rates for a certain year.
A tax relief of Ksh 1162 per month was allowed. In a certain month, of that year, an employee's taxable income of that year in the fifth band was Ksh 2018. Calculate:
 The employees total taxable income in that month; (2 marks)
 The tax payable by the employee in that month. (5 marks)
 The employee's income included a house allowance of Ksh 15 000 per month.
The employee contributed 5% of the basic salary to a cooperative society.
Calculate the employee's net pay for that month. (3 marks)
 Calculate:
 The dimensions of a rectangular floor of a proposed building are such that:
 the length is greater than the width but at most twice the width;
the sum of the width and the length is, more than 8 metres but less than 20 metres. If x represents the width and y the length. Write the inequalities to represent the above information. (4 marks)
 Represent the inequalities in part (a) above on the grid provided. (4 marks)
 Using integral values of x and y, find the possible maximum area of the floor. (2 marks)
 Write the inequalities to represent the above information. (4 marks)
 Each morning Gataro does one of the following exercises:
Cycling, jogging and weightlifting.
He chooses the exercise to do by rolling a fair die. The faces of the die are numbered 1, 1, 2, 3, 4 and 5.
If the score is 2, 3 or 5, he goes for cycling.
If the score is 1, he goes jogging.
If the score is 4, he goes weightlifting. Find the probability that:
 On a given morning, he goes for cycling or weightlifting; (2 marks)
 On two consecutive mornings he goes for jogging. (2 mark)
 In the afternoon, Gtaro plays either football or hockey but never both games. The probability that Gataro plays hockey in the afternoon is:
1/3 if the cycled
2/5 if he jogged and
1/2 if he did weightlifting in the morning.
Complete the tree diagram below by writing the appropriate probability on each branch. (2 marks)  Find the probability that on any given day:
 Gataro plays football; (2 marks)
 Gataro neither jogs nor plays football. (2 marks)
 Find the probability that:
 In the figure below OA = a and OB = b. M is the mid point of OA and AN:NB = 2:1.
 Express in terms of a and b:
 BA; (1 mark)
 BN; (1 mark)
 ON. (2 marks)
 Given that BX = hBX and OX = kON determine the values of h and k. (6 marks)
 Express in terms of a and b:
 Figure ABCD is a scale drawing representing a square plot of side 80 metres.
 On the scale drawing, construct:
 The locus of a point P, such that it is equidistant from AD and BC. (2 marks)
 The locus of point Q such that <AQB = 60^{0}. (3 marks)

 Mark on the drawing the point Q_{1}, the intersection of the locus of Q and lines AD.
Determine the length of BQ_{1}, in metres. (1 marks)  Calculate, correct to the nearest m^{2}, the area of the region bounded by the locus of P, the locus of Q and the line BQ_{1}. (4 marks)
 Mark on the drawing the point Q_{1}, the intersection of the locus of Q and lines AD.
 On the scale drawing, construct:
 In an experiment involving two variables t and r, the following results were obtained.
 On the grid provided, draw the line of best fit for the data. (4 marks)
 The variables r and t are connected by the equation r = at + k where a and k are constants.
Determine: The values of a and k; (3 marks)
 The equation of the line of the best fit. (1 mark)
 The value of t when r = 0. (2 marks)
MARKING SCHEME
 Limits: 12.5 ± 0.05 m and 9.23 ± 0.005 m
Maximum difference
= 12.55  9.225
= 3.325 m B1 
 First 6 terms
7, 4, 1, 2, 5, 8  Sum of 1st 50 terms
S_{50} = 50 {2 × 7 + 49 × 3}
2
= 3325
 First 6 terms

 angles BAC = 70º  30º = 40º
Reflex angles BOC = 360º  80º
= 280º  angles ACO = 40º 30º = 10º
 angles BAC = 70º  30º = 40º
 L = kM
N^{2}
2 = k × 12
36
k = 6
therefore: equation L = 6M
N^{2} 
marks frequency c.f 110 2 2 1120 4 6 2130 11 17 3140 5 22 4150 3 25
= 20.5 + 12.5  6 × 10
11
= 20.5 + 5.91
= 26.41
= 26  Amplitude = 2
Period = 360 = 120º
3  Area scale factor = 30 = 6
5
4x  2x + 2 = 6
2x = 4
x = 2  (3  x)^{7} = 3^{7}  7 (3)^{6} x + 21(3)^{5} x^{2}  35(3)^{4} x^{3 }+35 (3)^{3} x^{4} + ...
= 2187  5103x + 5103x^{2}  2835x^{3} + 945x^{4}
(2.8)^{7} = (3  0.2)^{7}= 2187  5103(0.2) + 5103(0.2)^{2} 2835(0.2)^{3} + 945(0.2)^{4}= 1349.352  Log 15^{2} = log 5 (x  4)
x
15^{2} = 5 x  4
x
x^{2}  4x  45 = 0
(x  9) (x + 5) = 0
x = 9 or 5
x = 9  PR = √60^{2} + 11^{2} = 61
Tan θ = 10
61
θ = 9.31º  3x  y = 9 ................ × x
x^{2}  xy = 4
3x^{2}  xy = 9x
x^{2}  xy = 4
2x^{2} = 9x  4
2x^{2}  9x + 4 = 0
(2x  1) (x  4) = 0
x = 1 or x = 4
2
y = 3(^{1}/_{2} )  9 or 3(4)  9
=  7^{1}/_{2} or 3  (1 + r )^{4} = 495000
100 280000
1 + r = 1.153
100
r = 15.3  8008 = 40 + θ × 2 × 22 × 6370
360 7
40 + θ = 8008 × 360 × 7 = 72
2 × 22 × 6370
θ = 72º  40º
= 32º
Position of B(32º S, 20ºW) 
= 7 
 Temperature at the 2nd minute = 60º
Temperature at the 11th minute = 18º
Average rate of cooling 
 A = 3 B, C = 2B
4
= A: B: C = 3 B: B: 2B
4
= 3: 4: 8  (168 × 4)litres
8
= 84 L 
 3 × 160 + 4 × 205 + 8 × 100
3 + 4 + 8
=Ksh 140  182  140 × 100%
140
= 30%  Ksh 140 × 125
100
=Ksh 175
 3 × 160 + 4 × 205 + 8 × 100
 A = 3 B, C = 2B


 (50 + 40) (50) = 30 (30 + x)
4500 = 900 + 30 x
30 x = 3600
QS = x = 120 cm  RS = ^{1}/_{2}QS
=^{1}/_{2}(120) = 60cm
QR = √61^{2}  60^{2}=11 cm
 (50 + 40) (50) = 30 (30 + x)

 Sin θ = 60
61
θ = 79.6º  Angle at the centre
= 2 × 79.6
= 159.2º
Length of minor arc QS
= 159.2 × 2π × 61
360
= 169.5 cm
 Sin θ = 60



 38392 + 2108
= Ksh 41000  10164 × 0.1 + 9576 × 0.15 + 9576 × 0.2 + 9576 × 0.25 + 2108 × 0.3
= 1016.4 + 1436.4 + 1915.2 + 2394 + 632.4
= Ksh 7394.4
monthly income tax
= 7394.4  1162
= Ksh 6232.4
 38392 + 2108
 Amount saved in coop society
= 5 × (41000  15000)
100
= Ksh1300
Nett pay
41000  (6232.4 + 1300)
= Ksh 33467.6


 y > x
y ≤ 2x
x + y < 20
x + y > 8 

 Maximum area:
9 × 10
= 90 m^{2}

 y > x


 3 + 1
6 6
= 2
3  2 × 2
6 6
= 1
9
 3 + 1


 P(Gataro plays football)
= 1 × 2 + 1 × 3 + 1 × 1
2 3 3 5 6 2
= 37
60  P(neither jogs nor plays football)
= 1 × 1 + 1 × 1
2 3 6 2
= 1
4
 P(Gataro plays football)



 BA = a  b
 BN = 1BA = 1(a  b)
3 3  ON = b + 1(a  b)
3
=1a + 2b
3 3
 BX = hBM = h(^{1}/_{2}a  b)
OX = kON = k(^{1}/_{3}a +^{2}/_{3}b)
also
OX = OB + BX
= b + h(^{1}/_{2}a  b)
k(1a + 2b) = b + h (1a  b)
3 3 2
1ka = 1ha
3 2
1k = 1h = k = 3h.............(i)
3 2 2
2kb = b  hb
3
2k = 1  h .....................(ii)
3
Substituting k = 3 h in (ii)
2
2 (3h) = 1  h = h = 1
3 2 2
Substituting h = 1 in (i)
2
k = 3 (1) = 3
2 2 4




 locus of P
construction of 30º
identification of centre
drawing of arc
Identifying region
for radius and angle of sector


 9.2 × 10 = 92 m
 area of region bounded by locus of P, locus of Q and line BQ_{1}angle = 60º radius = 46 m
= π × 46^{2} × 60
360
= 1107.94
≈ 1108 m^{2}




 value of a
=  0.7
3.5
= 0.2
value of k = 1.7  equation: r = 0.2t + 1.7
 value of t when r = 0
therefore: 0 =0.2t + 1.7
0.2 t = 1.7
t = 1.7 = 8.5
0.2
 value of a

Download KCSE 2014 Mathematics Alternative A Paper 2 Questions with Marking Scheme.
Tap Here to Download for 50/
Get on WhatsApp for 50/
Why download?
 ✔ To read offline at any time.
 ✔ To Print at your convenience
 ✔ Share Easily with Friends / Students