QUESTIONS
SECTION I (40 Marks)  Answer all questions in this section.

 Use the table of cubes to evaluate 23.5^{3} (2 mark)
 Find the cube root of 3.375 using prime factor method (2marks)
 Evaluate using the table of reciprocals: 2.5 + 2 (marks)
0.0842 64.5  Simplify the expression ∛27x^{3}y^{9 } (2 marks)
x^{6}y^{3}  Use logarithm tables to evaluate: ^{2}√124.4 × 35.8(3 marks)
745  Solve for x given that 2^{x }× 8^{x }÷ 4 = 64 (3 marks)
 A bank in Kenya buys and sells foreign currencies as follows.
Currency Buying(Ksh) Selling(Kshs)
1Sterling 134.20 134.65
1US dollar 71.40 71.84
A tourist arrived in Kenya with 4500 US dollars. He converted all the dollars to Kenya shillings at the bank. While in Kenya he spent Kshs 215,000 and then converted the remaining amount to sterling pounds in the same bank. Calculate the amount he received in sterling pound. (3marks)  The length of an arc of a circle is 8.8cm. If the arc subtends an angle 144° at the centre, calculate;
 the radius of the circle (Take π=^{22}/_{7}) (2 marks)
 the area enclosed by the arc and the radii (2 marks)
 Determine the number of sides of a regular polygon whose sum of interior angles is 1440º. (2marks)
 A shopkeeper made a loss of 20% by selling a trouser at Sh. 960. What profit would he have made if he had sold it at sh.1500 (3marks)
 A student spent 2⁄7 of his pocket money on stationeries, a third on foodstuffs and 5⁄8 of the remainder on transport. If he had Ksh. 150 left, how much pocket money did he have at the beginning? (3marks)
 Two bells ring at intervals of 35 and 42 minutes respectively. The bells ring together at 8:48 a.m. Determine the time when the bells will ring together again. (3 marks)
 A twodigit number is 18 more than the number formed by reversing the digits. If the sum of the digits is 10. Find the number. (3 marks)
 Evaluate; (3 marks)
14÷^{1}/3 of 5¼  3¾ × 1^{1}/_{3}
SECTION II (30Marks)
Answer any THREE questions
 The figure below represents a piece of land consisting of a trapezoidal region and a semicircular end of radius 87.5m
 Calculate
 The perimeter of the land (3marks)
 The area of the land in hectares (3marks)
 A private developer bought this piece of land at a price of Ksh 400,000 per hectare and later sold the all land at 2.25 million shillings. Determine;
 The price at which he bought the whole piece of land (2marks)
 His percentage profit (2marks)
 Calculate
 The corner points A, B, C and D of a ranch are such that B is 8 km directly East of A and C is 6 km from B on a bearing of 30º. D is 7 km from C on a bearing of 300º.
 Using a scale of 1 cm to represent 1 km, draw a diagram to show the position of A, B, C and D. (4 marks)
 Use the scale to determine the;
 compass bearing of A from D. (1 mark)
 distance BD in kilometers. (2 marks)
 bearing of D from B (1 mark)
 perimeter of the ranch in kilometers. (2 marks)

 A straight line L_{1}, whose equation is 3y  2x = 2 meets the xaxis at R. Determine the coordinates of R. (2 marks)
 A second line L_{2} is perpendicular to L_{1} at R. Find the equation of L_{2} in the form y = mx +c, where m and c are constants. (3 marks)
 A third line L_{3} passes through (4, 1) and is parallel to L_{2}. Find:
 the equation of L_{3} in the form y = mx +c, where m and c are constants. (2 marks)
 The coordinates of a point S, at which L_{1} and L_{3} intersects (3 marks)
 A cylindrical tank of diameter 3.6m and height 2.5m internally is twothirds full of juice.
 Calculate the volume of the juice in litres. (3 marks)
 The juice is packed in small packets measuring 8cm by 5cm by 12cm. A packet retails at Kes. 40. Calculate;
 the capacity of each packet in cm^{3} (2 marks)
 the number of full packets obtained (3 marks)
 the amount of money realized from the sale of the juice (2 marks)
 Triangle PQR has vertices P(3,2), Q(1,1) and R(2,1).
 Draw PQR on the grid provided. (1mark)
 Under a rotation the vertices of P^{1}Q^{1}R^{1} are P^{1}(1,4), Q^{1}(2,0) and R^{1}(4,1). Find the centre and angle of rotation using points P and Q. (4marks)
 Triangle PQR is enlarged with scale factor 3 centre O (0,0) to give triangle P^{2}Q^{2}R^{2}. Draw triangle P^{2}Q^{2}R^{2} and state its coordinates. (2marks)
 Triangle P^{1}Q^{1}R^{1} undergoes reflection in line y = x to give triangle P^{3}Q^{3}R^{3}. Draw P^{3}Q^{3}R^{3} and state its coordinates. (3mks)
MARKING SCHEME

 23.5^{3} = (2.35 x 10^{1})^{3}
= 12.978 x 10^{3}
= 12,978  ^{3}√3.375 = ^{3}√3375 = ^{3}√3375
10^{3} 10
3375 = 3 x 3 x 3 x 5 x 5 x 5
^{3}√3375 = 3 x 5 = 15
^{3}√3.375 = ^{15}/_{10}
= 1.5
 23.5^{3} = (2.35 x 10^{1})^{3}
 2.5(8.42^{1} x 10^{2}) + 2(6.45^{1} x 10^{1})
8.42^{1} = 0.1188
6.45^{1} = 0.1550
= 2.5(0.1188 x 10^{2}) + 2(0.1550 x 10^{1})
= 29.7 + 0.0310
= 29.731  [3^{3}.x^{3}.y^{9}]
x^{6}.y^{3}
= 3.x.y^{3}
x^{2}.y
= 3y^{2} or 3xy
x No Log 124.4
x 35.82.0948
+ 1.5339÷ 745 3.6487
 2.8722√x 0.7765 x ½
0.3883
= 2.445 2^{x} x 2^{3x} ÷ 2^{2} = 2^{6}
2^{(x + 3x  2)} = 2^{6}
2^{(4x  2)} = 2^{6}
4x  2 = 6
^{4x}/_{x} = ^{8}/_{4}
x = 2  $ to Ksh
=4500 x 71.40
= Ksh 321,300
Less expenditure:
321,300  215,000
=Ksh 106,300
Ksh to £
=106,300
134.65
= £789.45 
 L = 8.8cm
θ = 144º
r = ?
8.8 = ^{144}/_{360} x 2 x ^{22}/_{7} x r
r = 8.8 x 360 x 7
144 x 2 x 22
r = 22,176
6,336
r = 3.5cm  A = ^{θ}/_{360} x πr^{2}
= 144 x ^{22}/_{7 }x 3.5 x 3.5
A = 15.4cm^{2}
 L = 8.8cm
 sum = 180(n2)
1440 = 180(n2)
180 180
8 = n2
n = 8+2
n = 10sides  80% of B.P = S.P
^{80}/_{100} x B.P = 960
B.P = 960 x ^{100}/_{80}
= Ksh1,200
If S.P = Ksh1500
Profit = 1500  1200 = 300
= Ksh 300  Let pocket money = x
Stat + food = ^{2}/_{7}x + ^{1}/_{3}x
= ^{13}/_{21}x
Trans = ^{5}/_{8} (x  ^{13}/_{21}x)
= ^{5}/_{8} x ^{8}/_{21}x
= 5/21x
Total = ^{13}/_{21}x + ^{5}/_{21}x = ^{18}/_{21}x
Rem = x  ^{18}/_{21}x = ^{3}/_{21}x
But ^{3}/_{21}x = 150
x = 150 x ^{21}/_{3}
x = Ksh1050  LCM =
2 35 42 3 35 21 5 35 7 7 7 7 1 1
= 210min
^{210}/_{60} = 3hr 30min
Time = 8:48 am + 3:30 = 11:78
= 12:18P.M  Let No be XY
(10x + y)  (10y + x) = 18
10x + y  10y  x = 18
9x  9y = 18 ...(i)
Also, x + y = 10...(ii)
Solving simultaneously;
^{1}/_{9}(9x  9y = 18)
x + y = 10
+
x  y = 2
2x + 0 = 12
6 + y = 10
x = 6
y = 4
XY is 64  14 ÷ ^{7}/_{4}  ^{15}/_{4} x ^{4}/_{3}
= 14 x ^{4}/_{7} = 8
^{15}/_{4} x ^{4}/_{3} = 5
8  5 = 3
SECTION II


 a^{2} = 175^{2} + 50^{2}
a = 33125
a = 182.002m
b = ½ x ^{22}/_{7} 87.5m
b = 137.5m
Perimeter : 182.002 + 250 + 137.5 + 200
P = 769.502m  A1 = ½ x (250 + 200) x 175
½ x 450 x 175
= 39,375m^{2}
A2 = ½ x ^{22}/_{7} x 87.5^{2}
= 12.031.25m^{2}
Total A = 39,375
+ 12,031.25m^{2}
Total A = 51,406.25
10,000
= 5.140625ha
 a^{2} = 175^{2} + 50^{2}

 B.P = 5.140625 x 400,000
Ksh 2,056,250  % profit = (2,250,000  2,056,250) x 100
2,056,250
= 9.422%
 B.P = 5.140625 x 400,000




 530ºW
 B.D = 9.2cm x 1km/cm
= 9.2km ± 0.1km  340º or N20ºW
 AD = 10Km ± 0.1km
Perimeter = 8 + 6 + 7 + 10
P = 31km ± 0.1km


 At xaxis y = 0
3y  2x = 2
0  ^{2x}/_{2} =^{ 2}/_{2}
x = 1
R is (1,0)  M_{1} = ^{2}/_{3}, m_{1}m_{2} = 1
^{2}/_{3}.m_{2} = 1
m_{2} = ^{3}/_{2}
using (1,0) (x,y)
y0 = ^{3}/_{2}
x1
y = ^{3}/_{2} (x  1)
y = ^{3}/_{2}x + ^{3}/_{2} 
 m_{2} = m_{3} = ^{3}/_{2}
Using(4, 1)(x,y)
y 1 = ^{3}/_{2}
x + 4
y  1 = ^{3}/_{2} (x + 4)
y  1= ^{3}/_{2}x  6
y = ^{3}/_{2}x 5  ^{2}/_{3}x ^{2}/_{3} = y
^{3}/_{2}x  5 = y
At intersection
^{2}/_{3}x  ^{2}/_{3} = ^{3}/_{2}x  5
^{2}/_{3}x + ^{3}/_{2}x = 5 + ^{2}/_{3}
^{6}/_{13} . ^{13}/_{6}x = ^{13}/_{3}.^{6}/_{13}
x = 2
y = ^{2}/_{3}(2)^{2}/_{3}
y = ^{4}/_{3}  ^{2}/_{3}
y = 2
5(2,2)
 m_{2} = m_{3} = ^{3}/_{2}
 At xaxis y = 0

 V = ^{2}/_{3} x ^{22}/_{7} x 1.8^{2} x 2.5
= 16.97143m^{3}
1m^{3} = 1000l
V = 16.97143 x 1000
= 16,971.43l 
 V= 8 x 5 x 12cm^{3}
=480cm^{3}  1pct = 480cm^{3 }= 480ml
1000
= 0.48l
No of pcts = 16,971.43
0.48
= 35,357.145
Full pets = 35,357  Sales = 35,357 x 40
= Ksh 1,414,280
 V= 8 x 5 x 12cm^{3}
 V = ^{2}/_{3} x ^{22}/_{7} x 1.8^{2} x 2.5

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