KCSE 2018 Mathematics Alternative A Paper 1 with Marking Scheme

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KCSE 2018 Mathematics Alternative A Paper 2 with Marking Scheme

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QUESTIONS

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

Without using a calculator, evaluate:
(3 marks)
Given that 6^2n-3 = 7776, find the value of n. (3 marks)
The base of a right pyramid is a rectangle of length 80 cm and width 60 cm. Each slant edge of the pyramid is 130 cm. Calculate the volume of the pyramid.(3 marks)
In the figure below ABCDEF is a uniform cross section of a solid. Given that FG is one of the visible edges of the solid, complete the sketch showing the hidden edges with broken lines.
(3 marks)
The lengths of three wires were 30m, 36 m and 84 m. Pieces of wire of equal length were cut from the three wires. Calculate the least number of pieces obtained.(4 marks)
A two digit number is such that, the sum of its digits is 13. When the digits are interchanged, the original number is increased by 9. Find the original number.(4 marks)
1. Using a ruler and a pair of compasses only, construct a quadrilateral PQRS in which PQ = 5 cm, PS = 3 cm, QR = 4cm, ∠POR = 135° and ∠SPQ is a right angle. (2 marks)
2. The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in metres.(2 marks)
Given that OA = [2] and OB = [-4]. Find the mid point M of AB. (2 marks)
[3] [ 5 ]
Two towns R and S are 245 km apart. A bus travelling at an average speed of 60 km/h left towr R for town S at 8.00a.m. A truck left town S for town R at 9.00 am and met with the bus a 11.00a.m. Determine the average speed of the truck. (4 marks)
In the parallelogram WXYZ below, WX= 10 cm, XY = 5 cm and ∠WXY = 150°.

Calculate the area of the parallelogram.(3 marks)
Without using mathematical tables or a calculator, evaluate
sin 30°-sin 60° (3 marks)
tan 60°
Use matrix method to solve:
5x + 3y = 35
3x - 4y = -8. (3 marks)
Expand and simplify.
(2x +1)² + (x - 1)(x-3). (2 marks)
Use mathematical tables to find the reciprocal of 0.0247, hence evaluate

correct to 2 decimal places. (3 marks)
A Kenyan businessman intended to buy goods worth US dollar 20000 from South Africa. Calculate the value of the goods to the nearest South Africa (S.A) Rand given that, 1 US dollar - Ksh 101.9378 and 1 S.A Rand = Ksh 7.6326.(3 marks)
A photograph print measuring 24 cm by 15 cm is enclosed in a frame. A uniform space of width x cm is left in between the edges of the photograph and the frame. If the area of the space is 270 cm", find the value of x.(3 marks)

SECTION II (50 marks)
Answer any five questions from this section in the spaces provided.

A school water tank is in the shape of a frustum of a cone. The height of the tank is 7.2 m and the top and bottom radii are 6 m and 12 m respectively.
1. Calculate the area of the curved surface of the tank, correct to 2 decimal places.(4 marks)
2. Find the capacity of the tank, in litres, correct to the nearest litre.(3 marks)
3. On a certain day, the tank was filled with water. If the school has 500 students and each student uses an average of 40 litres of water per day, determine the number of days that the students would use the water.(3 marks)
Two vertices of a triangle ABC are A (3,6) and B (7,12).
1. Find the equation of line AB(3 marks)
2. Find the equation of the perpendicular bisector of line AB.(4 marks)
3. Given that AC is perpendicular to AB and the equation of line BC is y = -5x + 47, find the co-ordinates of C.(3 marks)
The distance covered by a moving particle through point O is given by the equation, s= t³ - 15t² + 63t - 10
Find:
1. distance covered when t = 2(2 marks)
2. the distance covered during the 3rd second;(3 marks)
3. the time when the particle is momentarily at rest;(3 marks)
4. the acceleration when t = 5.(2 marks)
The diagram below shows triangle ABC with vertices A(-1, -3), B(1, -1) and C(0,0), and line M.
1. Draw triangle A'B'C' the image of triangle ABC under a reflection in the line M.(2 marks)
2. Triangle A"B"C" is the image of triangle A'B'C' under a transformation represented by the matrix
  T =
  1. Draw triangle A"B"C"(3 marks)
  2. Describe fully the transformation represented by matrix T.(3 marks)
  3. Find the area of triangle A'B'C' hence find area of triangle A"B"C".(2 marks)
The figure below shows two triangles, ABC and BCD with a common base BC = 3.4 cm. AC = 7.2 cm, CD = 7.5 cm and ∠ABC = 90°. The area of triangle ABC = Area of triangle BCD.

Calculate, correct to one decimal place:
1. the area of triangle ABC;(3 marks)
2. the size of ∠BCD;(3 marks)
3. the length of BD;(2 marks)
4. the size of ∠BDC.(2 marks)
1. On the grid provided, draw the graph of.(2 marks)
2. Using trapezium rule, with 8 strips, estimate the area bounded by the curve and the x-axis.(3 marks)
3. Find the area estimated in part (b) above by integration.(3 marks)
4. Calculate the percentage error in estimating the area using trapezium rule.(2 marks)
Three business partners Abila, Bwire and Chirchir contributed Ksh 120000, Ksh 180 000 and Ksh 240 000 respectively, to boost their business. They agreed to put 20% of the profit accrued back into the business and to use 35% of the profits for running the business (official operations). The remainder was to be shared among the business partners in the ratio of their contribution. At the end of the year, a gross profit of Ksh 225 000 was realised.
1. Calculate the amount:
  1. put back into the business;(2 marks)
  2. used for official operations.(1 mark)
2. Calculate the amount of profit each partner got.(4 marks)
3. If the amount put back into the business was added to individuals's shares proportionately to their initial contribution, find the amount of Chirchir's new shares. (3 marks)
The equation of a curve is given as . Determine:
1. the value of when x = 3;(2 marks)
2. the gradient of the curve at x = 3;(3 marks)
3. the turning points of the curve and their nature.(5 marks)

MARKING SCHEME

7776 = 6⁵
6^2n-3 = 6⁵
2n-3=5
n=4
Height h = √130²-50²= 120cm
Volume = ¹/₃ x 80 x 60 x 120
= 192000cm³
30 = 3x2x5
36=2x2x3x3
84 = 2x2x3x7
G.C.D.=2x3
= 6

No of pieces obtained
= 30 + 36 + 84
6 6 6
= 25
Let the number be xy
x+y=13
(10y+x)-(10x + y)=9 or -x+y=1
x+y=13
y - x=1
2y=14
y=7
x=6

Number is 67.
2. RS=(7.8 ± 0.1)cm
  Actua x 40m
  = 312 ± 4m
Midpoint of AB
Distance covered by truck
= 245 - 60 x 3
= 65km

Time taken by truck = 11 - 9 = 2h
Average speed of truck
= 65
2
= 32.5km/hr
h = 5 sin 30°
= 2.5cm

Area = 2.5x10
= 25cm³
x=4
y = 5
(2x+1)²+(x-1)(x-3) = 4x² +4x+1+x² - 4x+3
= 5x² +4
1 = 1 x 10²
0.0247 2.47
= 0.4049x10²= 40.49

= 58.56
20000 dollars = 20000 x 101.9378
= Ksh. 2038756
In S.A.rand = 20000x101.93.78
7.6326
= 267112 rands
Area of space = 2×(15+2x)x + 2 × 24 × x
30x +4x²+48x=270
4x² + 78x-270 =0
4x² - 12x +90x-270 = 0
4x(x-3)+90(x-3)=0
4x(x-3)+90(x-3)=0
(4x +90)(x-3)=0
x= -22.5 or x=3
x=3cm
1. h = 6
  h+7.2 12
  12h-6h=6x7.2
  6h=6x7.2
  h=7.2m
  
  Curved surface area of tank.
  
  = 706.65 - 176.66
  =529.99m²
2. = 1900.0 m³
  Capacity = 1900 x 1000 litres
  = 1900000 litres
3. Amount used by students per day.
  = 40 x 500
  = 20000 litres
  No. of days = 1900000
  20000
  = 95 days
1. Gradient of AB = 12 - 6 = 6 = 3
  7 - 3 4 2
  Equation of AB:
  y-6 = 3
  x-3 2
  y = 3x + 3
  2 2
2. Midpoint of AB
  
  =(5,9)
  Equation of perpendicular bisector of AB:
3. Equations of AC: y - 6 = -2
  x - 3 3
  y = 2x + 8
  3
  At Point of intersection
  y = 2x + 8 = -5x + 47
  3
  13x=117
  x=9
  y=-5 x 9+47 = 2
  Coordinates of C is (9,2)
1. S₍₂₎=2³ -15(2)² +63(2)-10
  = 8-60+126-10
  = 64
2. S₍₃₎= 3³ -15(3)² +63(3)-10
  = 27 - 135 + 189-10
  = 71
  
  Distance in 3^rd second
  S₍₃₎-S₍₂₎= 71 - 64
  =7
4. Acceleration = dv =6t - 30
  dt
  = 6(5) - 30
  = 0
1. Correct position of the vertices of A'B'C'
  Correctly complete triangle A''B''C'' drawn
2. 1. Triangle A''B''C'' correctly drawn
  2. It's a shear,
    The x axis invariant
    point B'(-2.2) is mapped onto B"(2,2)
  3. Area of triangle A'B'C'= ¹/₂ x (3+1) x 2 - 1.5-0.5
    = 4-2
    = 2 sq units
    Area of A"B"C" = Area of A'B'C'
    =2 square units
1. AB= √7.2² -3.4²= 6.3cm
  Area of ΔABC
  =¹/₂ x 6.3 x 3.4
  = 10.7 cm²
2. Area of ΔABC = Area of ΔBCD
  ¹/₂ x 3.4 x 7.5 x sinθ =10.7
  Sin θ =10.7x2
  3.4x7.5
  θ = 57.1°
  Obtuse Angle BCD = 180 - 57.1
  = 122.9
3. BD² = 7.5² +3.4²-2x3.4 x 7.5 cos 122.9
  = 95.51
  BD = 9.8cm
4. Angle BDC:
  3. 4 = 9 .8
  Sin θ Sin 122.9
  Sin θ = 3.4 sin 122.9
  9.8
  θ = 16.9°
1. x -4 -3 -2 -1 0 1 2 3 4
  
  y 0 1.75 3 3.75 4 3.75 3 1.75 0
2. Area =
  ¹/₂ x 1 {0 + 0 + 2(1.75 + 3.75 + 3 + 4 + 3.75 + 3 + +1.75)}
  = ¹/₂ x 1 x 2 x 21
  = 21sq units
1. 1. 20 x 225000
    100
    = 45000
  2. 35 x 225000
    100
    = 78750
2. Amount for each contribution ratio contributions:
  Abiro: Bwire: Chirchir
  = 120000:180000:240000
  = 2 : 3 : 4
  Abila = 2 x 45 x 225 000
  9 100
  = 22500
  Bwire =3 x 45 x 225 000
  9 100
  = 33750
  Chirchir = 4 x 45 x 225 000
  9 100
  = 45000
3. 20 x 225000 x 4 + 240000
  100 9
  20000+ 240000
  = 260000
1. y = 1x³ - 4x + 5
  3
  When x = 3
  y=1 (3)³ - 4(2)+5
  3
  =2
2. Gradient at x=3
  dy=x²-4
  dx
  at x = 3
  dy = (3)² -4
  dx
  = 5
3. At turning points
  dy=x²-4=0
  dx
  (x-2)(x+2) =0
  x=2 or -2
  (When x=2, y = -¹/₃) and (x=2, y = 10¹/₃)
  Turning points are