KCSE 2014 Mathematics Alternative B Paper 1 Questions with Marking Scheme

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KCSE 2014 Mathematics Alternative B Paper 2 Questions with Marking Scheme

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QUESTIONS

Evaluate
(2 marks)
During a prize giving day in a school, there were four times as many students as parent. The number of girls was 84 more than the number of boys. If there were 630 girls, calculate the number of parents present (3 marks)
Given that 3x + 5y = 300 and x + y = 78, find the value of 10x + 15y (3 marks)
A group of families shared 96 packets of maize meal, 84 packets of wheat flour and 36 packets of sugar
Determine:
1. the greatest number of families that shared he foodstuffs equally (2 marks)
2. the total number of packets of foodstuffs that each family received (2 marks)
Express in the simplest index form. (3 marks)
Line AB shown below is a side of a parallelogram ABCD in which AD=6cm and angle DAB=30º

Using a pair of compasses and ruler only, complete the parallelogram ABCD (3 marks)
An acute angle α is such that sin(4α)º = cos(α + 10)º. Find
1. the value of α (2 marks)
2. sin α, correct to 3 decimal places (1 marks)
Without using a calculator or mathematical table, evaluate:
0.375 ÷ 0.06 - 4.2
3.96 + 2.8 × 0.05
Three children Awino, Buko and Chebet had two types of fruits each. Awino had twice as many mangoes as Buko while Buko had three times as many mangoes as Chebet. Also, Buko had three times as many oranges as Awino while Chebet had twice as many oranges as Awino. If Buko had x mangoes and y oranges, write a simplified expression to represent the total number of fruits the three children had. (3 marks)
A solid has a circular cross-section of radius 1.4cm and a height of 4 cm.
1. Name the solid (1 marks)
2. Draw an accurate net of the solid (2 marks)
The tip of the minute hand of a clock moves through a distance of 17.6cm between 3:00pm and 3:12pm. Find the length of the minute hand. (4 marks)
In the figure below, O is the center of the circle. PQRS is a cyclic quadrilateral and RST is a straight line. Angle RPQ=21º and angle TSP=147º

Calculate the size of angle SRQ (3 marks)
Factorize 2x² + 6y - 3x - 4xy (2 marks)
The area of a rhombus is 34cm². One of the interior angles is 30º. Calculate the length of a side of the rhombus to the nearest centimeter. (3 marks)
The vertices of a triangle are P(-3,1), Q(1,3) and R(4,2). The vertices of its image under an enlargement are P'(-7,4)Q'(1,8) AND R'(7,6).
1. On the grip provided, draw triangle PQR and its image. (2 marks)
2. Determine the centre and scale factor of the enlargement. (2 marks)
A tumbler is in the shape of a frustum of a cone. The radii of the circular ends are 2.1 cm and 3.5 cm. The slant height of the tumbler is 5cm. Calculate the area of the curved surface.(4 marks)
Keya, Limo and Mumo invested some money into a business. Keya contributed Ksh 30 000, Limo contributed Ksh 50 000 and Mumo contributed 25% of the total amount contributed by Keya and Limo.
1. Calculate:
  1. the amount of money contributed by Mumo;(2 marks)
  2. the ratio in which Keya, Limo and Mumo made their contribution.(2 marks)
2. After one year, the business realised a profit of Ksh 25 000 which was shared by the partners in the ratio of their contributions. Find the amount of money Mumo got. (2 marks)
3. During the second year, Mumo added some more money to the business. The new ratio of their contributions was, Keya:Limo:Mumo = 3:5:7.
  Calculate:
  1. the total amount of money Mumo added to the business in the second year.(2 marks)
  2. Mumo's percentage contribution in the business by the end of the second year.(2 marks)
The capacity of a cylindrical container is 1.54 litres. The height of the container is 10 cm. (Take pi = ²²/₇)
1. Calculate the diameter of the container.(3 marks)
2. Along each end of the curved surface, a ribbon of width 1.5 cm is fixed with an overlap of 2 cm.
  Calculate:
  1. the total length of the ribbon used;(3 marks)
  2. the surface area of the part of container covered by the ribbon. (1 mark)
3. Given that the container is open at one end, calculate the outer surface area of the container.(3 marks)
Figure ABCD below is a scale drawing of a piece of land in which AD represents 90 m.
The square PQRS represents the homestead.
1. State, in ratio form, the scale used. Calculate:
  1. the perimeter of the homestead in metres. (2 marks)
  2. the area of the piece of land in hectares. (3 marks)
2. On the scale drawing show the shortest distance from the centre of the homestead to side CD and hence determine the distance in metres.(3 marks)
A line L, passes through points (-3,-2) and (6, 1).
1. Find the equation of L₁ in the form y = mx + c, where m and c are constants. (3 marks)
2. A line L₂, is perpendicular to L, and passes through point (-1,2). Find the equation of L₂, in the form ax + by + c = 0, where a, b and care constants. (3 marks)
3. Another line L₃, is parallel to L₂, and passes through point (1,1). Determine the co-ordinates of the x-intercept and the y-intercept of L₃. (4 marks)
Using a ruler and a pair of compasses only:
1. Construct triangle PQR such that PQ = 6.5 cm, QR = 8 cm, and angle PQR = 75°.(4 marks)
2. On triangle PQR in part (a) above, construct:
  1. the perpendicular bisector of line RP to meet line RP at M and line RQ at N. (1 mark)
  2. the bisector of angle RPQ to meet line MN at O. Measure angle POM. (2 marks)
  3. a circle centre O and radius OM, to meet line RQ at X and Y. Measure chord XY and angle XOY (3 marks)
An athlete ran two laps around a 400 metre track. He ran the first lap in 64 seconds and then increased his speed in the second lap by 6%.
1. Calculate his speed, in metres per second, during:
  1. the first lap; (2 marks)
  2. the second lap;(2 marks)
2. Calculate, to 2 decimal places:
  1. the total time taken to run the two laps;(3 marks)
  2. the average speed, in km/h, for the two laps.(3 marks)
Kerubo bought 420 bananas at Ksh 20 for every 8 bananas. For every 70 bananas she bought, she was given one extra banana. She hired a cart for Ksh 50 to transport the bananas. During transportation 14 bananas got spoilt and the remaining ones were sold.
1. Determine:
  1. the total amount of money that Kerubo spent; (2 marks)
  2. the number of bananas sold.(1 mark)
2. Kerubo made a 60% profit after selling some of the bananas at Ksh 30 for every 5 and the rest at Ksh 10 for every 3.
  Calculate:
  1. the number of bananas sold at Ksh 30 for every 5. (4 marks)
  2. the amount of money obtained from the bananas sold at Ksh 10 for every 3. (3 marks)
1. 1. On the grid provided, draw triangle RST such that R(-3,1), S(-3,-4) and T(2,-4).(1 mark)
  2. Determine the area of the triangle RST.(2 marks)
2. On the same grid:
  1. plot point U such that RSTU is a square. State the coordinates of point U; (2 marks)
  2. plot point V such that SV = 2SU and S, U and V lie on a straight line. State the coordinates of V. (2 marks)
3. Calculate the area of RSTV.(3 marks)

MARKING SCHEME

^- 8 × ⁺2 + ^-11 = -27
⁺18 ÷ ^-2 × ⁺3 = -27
= 1
Number of boys = 630 - 84
= 546
Number of students = 630 + 546
= 1176
Number of parents = 1176 ÷ 4
= 294
3(78 - y) + 5y = 300
2y = 66
y = 33
therefore: x = 78 - 48 = 45
10x + 15y = 450 + 495 = 945
1. 96 = 2⁵ × 3
  84 = 2² × 3 × 7
  36 = 2² × 3²GCD of 96, 84 and 36 = 2² × 3 = 12
2. Number of packets of foodstuffs
  = 96 + 84 + 36
  12 12 12
  = 8 + 7 + 3 = 18
128 = 2⁷
2⁵ ÷ 2⁸2^-3
= 2¹⁰
√ construction of 30°
√ construction of AD = 6 cm identifying C and completing parallelogram
4α + α + 10 = 90°
5α = 80°
α = 16°
sin α = 0.276
0.375 ÷ 0.06 - 4.2 = 6.25 - 4.2
3.96 + 2.8 × 0.05 3.96 + 0.14
= 2.05
4.1
Evidence of division and multi- plication should be seen.
= 0.5
Mangoes: 2x + x + 1x
3
= 3¹/₃ x
Oranges: ¹/₃y + y + ²/₃y = 2y
Total Fruits = 3¹/₃x + 2y
1. Cylinder
2. Two circles of radius 1.4 touching the longer sides of a rectangle 4 cm by 8.8 cm.
  *for correct circles
  *for correct rectangle
Fraction of circumference made = 12
60
22 × 2r × 12 = 17.6
7 60
r = 7 × 60 × 17.6
22 12 2
= 14
RQP = 147° or RPS = 57°
SRP = 90°
SRQ = 90 + 12 = 102° or 180 - (57 + 21) = 102°
2x² + 6y - 3x - 4xy
= 2x² - 4xy - 3x + 6y
= 2x(x - 2y) - 3(x - 2y)
= (2x - 3)(x - 2y)
x² sin 30º = 34
x = √ 34
sin 30
≈ 8 cm
1. triangle PQR
  triangle P'Q'R'
2. Centre of enlargement (1,-2)
  Scale factor of enlargement = 10 = 2
  5
L = L + 5
2.1 3.5
3.5L - 2.1L = 10.5
L = 7.5
L = 5 + 7.5 = 12.5
Curved area
= 22 × (3.5 × 12.5 - 2.1 × 7.5)
7
= 88 cm²
1. 1. Mumo’s contribution:
    = 25 × (30000 + 50000)
    100
    = 20000
  2. Ratio - Keya : Limo : Mumo
    = 30000: 50000: 20000
    = 3: 5: 2
2. Mumo’s share of profit
  = 2 × 25000
  10
  = 5000
3. 1. 20000 + x = 80000 × 7
    8
    x = 50000
  2. Mumo’s % contribution in business during 2nd year
    = 70000 × 100
    150000
    = 46²/₃%
1. 1.54l = 1540 cm³Volume = 22 × r² × 10 = 1540
  7
  r =√ 1540 × 7
  22 × 10
  = 7
  therefore: Diameter = 2 × 7 = 14 cm
2. 1. Length of ribbon
    = 2 × 22 × 14 + 2 × 2
    7
    = 88 + 4 = 92
  2. Surface area covered by ribbon
    = 88 × 1.5 = 132 cm²
3. Surface area
  = 22 × 49 + 22 × 14 × 10
  7 7
  = 154 + 440
  = 594 cm²
1. Scale used:
  9 cm represent 90 m
  therefore: scale 1:1000
2. 1. perimeter of homestead
    (2 × 10) × 4
    = 80 m
  2. Area of piece of land in ha.
    AB = 13.8 × 10 = 138;
    BC = 6 × 10 = 60
    ¹/₂(60 + 90) × 138
    10000
    = 1.035 ha
3. ⊥ distance from centre of homestead to side CD shown
  Distance, 3.6 cm, on map
  Actual distance 3.6 × 10 = 36 m
1. Gradient of L₁= 1 - -2
  6 - -3
  =¹/₃
  equation of L₁
  = y - 1 = 1
  x - 6 3
  3y - 3 = x - 6
  3y = x - 3
  y = 1 x - 1 3
2. Gradient of L₂
  = - 1
  ¹/₃=-3
  therefore: equation y - 2 =-3
  x - -1
  y =-3x - 1
  = 3x + y + 1 = 0
3. equation of L₃y - 1 =-3
  x - 1
  y - 1 =-3(x - 1)
  y =-3x + 4
  x intercept:
  when y = 0 , x = 4
  3
  therefore: coordinates of x intercepts a (⁴/₃ , 0)
  y intercept:
  when x = 0 , y = 4
  therefore: coordinates of y intercept (0, 4)
1. Lines PQ and PR angle 75° constructed completion of Δ PQR.
2. 1. ⊥ = bisector of PR
  2. angle bisector QPR
    POM 60° ± 1°
  3. circle with radius OM
    XY = 4.3 ± 0.1
    XOY 114° ± 1°
1. 1. 400m
    64s
    = 6.25 m/s
  2. speed during second lap
    6.25 × 1.06
    = 6.625 m/s
2. 1. total time for two laps
    time for 2nd lap = 400
    6.625
    ≈ 60.38 s
    total time
    = 64 + 60.38
    = 124.38 s
  2. average speed in km/h
    800 m/s
    124.38
    = 800 × 3600
    124.38 1000
    = 23.15 km/h
1. 1. amount of money spent
    = 420 × 20 + 50
    8
    = 1100
  2. number of bananas sold
    = 420 + 420 - 14
    70
    = 412
2. 1. s.p. of bananas
    = 1100 × 1.6
    = 1760
    let x be number of bananas sold at sh 30
    therefore: x × 30 + 412 - x × 10 = 1760
    5 3
    18x + 412 - 10x = 1760
    x = 145
  2. No of bananas sold at sh 10
    = 412 - 145 = 267
    Amount of money obtained
    = 267 × 10
    3
    = 890
1. 1. Δ RST √ drawn
  2. Area of Δ RST: ¹/₂ × 5²= 12.5
2. 1. Plotting point U coordinates of point U (2, 1)
  2. Plotting of point V coordinates of point V (7, 6)
3. Area of quadrilateral RSTV
  diagonals RT = √50
  and SV = √200
  therefore: Area = ¹/₂ × √50 × √200
  = ¹/₂ × 5√2 × 10√2
  = 50