QUESTIONS
 Use logarithms to 4 decimal places to evaluate: (4 marks)
(0.7841 x √0.1356)^{1/3}
Log 84.92  A globe representing the earth has a radius of 0.5m. point A(0º, 10ºW), B (0º, 35ºE), P(60ºN, 110ºE) and Q(60ºN, 120ºW) are marked on the globe.
Find the length of arc AB, leaving your answer in term of (3mks)  A circle centre is the point C(2,3) passes through a point P(a,b). A point M(2 , ^{5}/_{2}) is the midpoint of the line CP.
 Calculate the coordinates of P. (1mk)
 Determine the equation of the circle in the form x^{2} +y^{2 }+ax + by + c = 0 (3mks)
 Make a the subject of the formula:
x = y +√(x^{2}+ a^{2}) (3marks)  Given that Sin (^{2}/_{3} x+ 20º )  Cos (^{5}/_{6}x + 10º) = 0. Without using a mathematical table or a calculator, determine tan (x+ 20º). (3 marks)
 Two fair dice one a regular tetrahedron (4 faces) and the other a cube are thrown. The scores are added together. Complete the table below to show all possible outcomes. (2 mark)
Find the probability that: The sum is 6. (1 mark)
 The sum is 6 or 9. (2 marks)
 A particle moves along a straight line such that its displacement s metres from a given point is s = t^{3} – 5t^{2} + 3t + 4 where t is time in seconds. Find:
 The displacement of the particle at t = 8. ( 2 marks )
 The velocity of the particle when t = 10. (3 marks )
 A classroom measures (x + 2) m by (x – 5)m. If the area of the classroom is 60m^{2}.
Find its length. (3 marks )
SECTION B
 Lengths of 100 mango leaves from a certain mango tree were measured t the nearest centimeter and recorded as per the table below,
Length in cm No. of leaves
10 to 12 3
13 to 15 16
16 to 18 36
19 to 21 31
22 to 24 14 On the grid provided draw a cumulative frequency graph to represent this data. (5mks)
 Use your graph to estimate
 The median length of the leaves (2mks)
 The number of leaves whose lengths lie between 13cm and 17cm. (3mks)

 Draw ΔPQR whose vertices are P (1, 1), Q (3, 2) and R (0, 3) on the grid provided.(1mk)
Find and draw the image of ΔPQR under the transformation whose matrix (^{3}_{1 }^{0}_{1}) is and label the image P¹Q¹R¹. (2 marks)  P¹Q¹R¹ is then transformed into P¹¹Q¹¹R¹¹ by the transformation with the matrix (^{1}_{1 }^{0}_{3}) .Find the coordinates of P¹¹Q¹¹R¹¹ and draw P¹¹Q¹¹R¹¹. (3 marks)
 Describe fully the single transformation which maps PQR onto P¹¹Q¹¹R¹¹ find the matrix of this transformation. (3 marks)
 Describe fully the single transformation which maps PQR onto P¹¹Q¹¹R¹¹ find the matrix of this transformation. (3 marks)
 Draw ΔPQR whose vertices are P (1, 1), Q (3, 2) and R (0, 3) on the grid provided.(1mk)
MARKING SCHEME
No Log 0.7841
0.1356½
Log 84.92 = 1.929
0.53101.8944
1.1323 = 1.5662
2 14606
1.2853
1.1753
^{3}/_{3} + 2.1753 = 1.7251
3
1.7251 Difference in longitude = 10 + 35 = 45º
length of arc AB = ^{45}/_{360} x 2 x π x 0.5
= ^{1}/_{8}π = 0.125π 
 2 + a = 2 = 2 + a = 4
2
a = 6
3 + b = 2.5 = 3 + b = 5
2
b = 8  (a, b) = (6,8)
r = √(2 6)^{2} + (38)^{2} = √185
(x2)^{2} + (y3)^{2} = 185
x^{2}  4x + 4 + y^{2}  6y + 9 = 185
x^{2} + y^{2}  4x  6y  172 = 0
 2 + a = 2 = 2 + a = 4
 x = y + √(x^{2}+ a^{2})
x – y = √(x^{2}+ a^{2})
(x – y)^{2} x^{2} + a^{2}
(x – y)^{2} x^{2} = a^{2}
± √((xy+x)(xyx) = a
± √(2xy)(y) = a
± √(y^{2} 2xy) = a  (^{2}/_{3} x+20)+^{5}/_{6} x+10º=90
^{9}/_{6} x+30=90
^{9}/_{6} x=60º
x = 40º
=Tan (x+20)
=Tan 60 M1  CUBE
1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10  P(6) = ^{4}/_{24} = ^{1}/_{6}
 P(6 or 9) = ¼
P(6) = ^{4}/_{24}
P(9) = ^{2}/_{24}
^{4}/_{24} + ^{2}/_{24}
^{6}/_{24} = 1/4

 s = 8^{3} – 5 x 8^{2} + 3 x 8 + 4
512 – 320 + 24 + 4
= 220m  V =^{ ds}/_{dt} = 3t^{2} – 10t + 3
= 3 x 10^{2} – 10 x 10 + 3
300 – 100 + 3 = 197m
 s = 8^{3} – 5 x 8^{2} + 3 x 8 + 4
 (x + 2) (x – 5 ) = 60
x^{2} – 5x + 2x – 10 = 60
x^{2} – 3x – 70 = 0
x^{2} – 10x + 7x – 70 = 0
x (x – 10) + 7(x – 10) = 0
(x – 10) (x +7) = 0
x = 10
x = 7
Length 10 + 2 = 12m  Upper class limits 12.5, 15.5, 18.5, 21.5 24.5
Cumulative frequency 3, 19 55, 86,100 
 Median = 50th 18.2 ± 0.15
 Leaves below 13 =4 leaves
leaves below 17 =35 leaves
leaves between 13 and 17 = 354=31

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