QUESTIONS
SECTION 1 (50 MARKS)
Answer all the questions in the space provided below each question
 Evaluate without using mathematical tables or calculator 12÷46×3+5×2 (3marks)
8+(5)÷2×(4)  A threedigit number is such that twice the hundreds digit is more than the tens digit by 2.the unit digit is thrice the hundred digit. When the digits are reversed, the number is increased by 594.Find the number. (4marks)
 Simplify x + 4  5x+20 (3marks)
x  4 x^{2} 16  The mass of maize flour to the nearest 10 grams is 8. 67kg.Determine the percentage error in this measurement.(3marks)
 Without using a calculator, solve for x in the equation 0.5^{x} × 0.125^{1x} =32. (3marks)
 The velocity V m/s of a particle in motion is given by V=3t^{2}2t+5.Calculate the distance travelled by the particle between t= 2 seconds and t = 6 seconds.(3 marks)
 Given that, the coordinates of two points P and Q are (2, 3, 5) and (6, k1, 15) respectively and that their position vectors are parallel, Calculate the value of PQ. (4marks)
 A dealer sells a mobile phone at a profit of 25%. The customer sells it to a friend at ksh 60,000, making a profit of 20 %. Find the cost prize of the mobile phone. (3 marks)
 A wall clock that gains 20 seconds after every hour was set to read the correct time on Tuesday at 03 25. Determine the time the wall clock will read on Thursday 03 25 h. (3 marks)
 A cylinder of radius 15 cm and height 24 cm is filled with water. A solid hemisphere of radius 7cm is submerged into the cylinder and removed. Find the change in height of water level in the cylinder(4 marks)
 The average rate of depreciation in value of a laptop is 10% per annum. After three complete years its value was ksh 35,000. Determine its value at the start of the threeyear period. (3marks)
 If 3x+2y∶ 7xy=3:2,calculate the ratio x:y (3marks)
 State the inequalities that satisfy the region defined by R. (3marks)
 Solve for θ in the equation sin(2θ+30)° = tan45°(3marks)
cos(3θ40)°  The figure below shows a circle PQRS Centre O with SR produced to T.PQ//SR and ∠QSR=55°.Calculate the size of∠QRT.(3 marks)
 The scale of a map is 1: 200.Calculate the actual area of a triangular coffee field whose sides are 6cm, 8cm and 10cm on the map(3 marks)
SECTION II (50 MARKS)
Answer only five questions from this section.
 A tower is on a bearing of 030º from a point P and a distance of 100m.From P, the angle of elevation of the top of the tower is 15º and the angle of depression of the foot of the tower is 1º.
 Calculate the height of the tower. (4 marks)
 A point Q is on the same horizontal plane as point P. The tower is on a bearing of 330ºfrom Q and a distance of 70 m. Calculate:
 The distance from P to Q. (3 marks)
 The bearing of P from Q. (3 marks)

 The diagram below shows a bucket in the shape of a frustum of a cone with diameters 36 cm and 24 cm and a vertical height of 28 cm. The bucket contains water such that the diameter of the water surface is 30cm. Calculate the volume of the bucket. (6 marks)
 If the bucket above has a hole and 1.1 cm^{3} of water leaks out every 5 seconds and collects in a cylindrical can of base radius and height 10 cm and 25 cm respectively. Calculate how long it takes to fill the cylindrical can. (4 marks)
 The diagram below shows a bucket in the shape of a frustum of a cone with diameters 36 cm and 24 cm and a vertical height of 28 cm. The bucket contains water such that the diameter of the water surface is 30cm. Calculate the volume of the bucket. (6 marks)
 A piece of wire, 18 cm long is cut into two parts. The first part is bent to form the four sides of a rectangle having length x cm and breath 1 cm.
 State two expressions in terms of x only for the perimeter of the square and the rectangle (2 marks)
 If the sum of the areas of the square and the rectangle is A cm^{2}, show that A =16  3x + ^{x2}/_{4} (2 marks)
 If A =8cm^{2}, Solve the equation in (b) above for x, hence find the possible dimensions of the two pieces of wire. (6 marks)
 The distance between two towns P and Q is 300 km. A bus started at P at 10.30 am and travelled towards town Q at 80 km/h. After 45 minutes a car started at Q and travelled to town P at x km/h. The car met the bus after 1hour 20 minutes.
 Determine the value of x. (3 marks)
 Find the distance from P where the car met the bus. (2 marks)
 At what time did the car meet the bus? (2 marks)
 If t a shuttle started at P, 1hour after the car left Q for P. Calculate the speed to the nearest km/h at which the shuttle should be driven in order to arrive at Q at the same time with the bus. (3 marks)

 Complete the table of values for the equation y = x^{2}+ 3x  6, given that 6 ≤ x ≤4.(2marks)
x 6 5 4 3 2 1 0 1 2 3 4 y 12 6 22  Using a scale of 1cm to represent 2 units in both axes, draw the graph of y= x^{2 }+ 3x  6 (3 marks)
 Using the graph drawn above Solve the equation
 x^{2}+3x6=0 (2 marks)
 x^{2}+3x2=0 (3 marks)
 Complete the table of values for the equation y = x^{2}+ 3x  6, given that 6 ≤ x ≤4.(2marks)
 A Business man is paid a commission of 5% on sales of goods worth over ksh 100,000. He is paid a monthly salary of ksh 15,000 but 2% of his total earning is remitted as tax. In a certain month he sold goods worthy khs 500,000
 Calculate his monthly net earnings that month.(5marks)
 The following month, his monthly salary increased by 20%. His commission was increased to 10% but on goods worth over ksh 200,000. If his total earnings that month was ksh 64,800, Calculate the money received from the sale of goods. (5marks)
 The coordinates of two points A and B are (2, 3) and (4, 5) and R is the midpoint of AB.
 Determine the coordinates of R. (2 marks)
 Find the equation of a straight line joining A and B, expressing it in the form y= mx + c where m and c are constants (3 marks)
 The straight line L_{1 }which is a perpendicular bisector of AB meets the Xaxis at T. Find the coordinates of T.(3 marks)
 If the straight line L_{1 }is parallel to a line that passes through the point (1, 6) and (a, 8), find a. (2 marks)
 Two inlet taps P and Q opened at the same time can fill a tank in 2 1/2 h. The two taps were opened together at the same time and after 1 hour 10 minutes tap Q was closed and P continued alone and filled the tank after a further 4 hours. Find:
 The fraction of the tank filled by both taps for 1 hour.(1 marks)
 The fraction of the tank filled by tap P after Q was closed.(2 marks)
 The time which each tap working alone would have taken to fill the tank. (7 marks)
MARKING SCHEME
 Num 3 18 + 10 = 11
Deno 5  ^{5}/_{8} x 4
8 + 10 = 18
^{11}/_{18}  xyz = 100x + 10y + z
2x = y + 2
z = 3x
zyx = 100z + 10y + x
100x + 10y + z = 100z + 10y + x  594
99x  99z = 594
x  z = 6
z = 3x
2y = y + 2
2x = y + 2
x  3x = 6
2x = 6
x = 3
z = 9
6 = y + 2
4 = y
= 349  (x+4)(x+4)5x20
(x4)(x+4)
x^{2} + 8x + 16  5x  20
(x  4)(x + 4)
x^{2} + 3x  4
(x4)(x+4)
x(x+4)1(x+4)
(x4)(x+4)
x  1
x  4  Error = ½ x10 = 4gm
= 0.005kg
%error = 0.005 x 100
8.67
=0.05767%  (½)x + (^{1}/_{8})1x = 25
2x + 23(1x) = 25
x + 3 + 3x = 5
2x  3 = 5
2x = 8
x = 4  ∫^{6}_{2}(3t^{2}  2t + 5)dt
(t^{3}  t^{2} + 5t)^{6}_{2}
(216  36 + 30)  (8  4 + 10)
210  14 = 196  2 = 6x
^{1}/_{3} = x
3 = ^{1}/_{3}(k1)
9 = k1
10 = k
ΙPQI = √(62)^{2} + (93)^{2} + (155)^{2}
= √16 + 36 + 100
= 12.33  120% → 60000
100% → 100 x 60000
120
125% → 50000
100% →100 x 50000
125
= 40000  1hr → 20seconds
48hrs → 48 x 20
960 seconds
^{960}/_{60}
= 16minutes
325 + 16 = 0341hrs  V = ^{22}/_{7} x 15^{2} x 24
= 16971
^{4}/_{3} x ½ x ^{22}/_{7} x 49 x 7
718.7
16971  718.7 = 16252.3
^{22}/_{7} x 15^{2} x h = 16252.3
h = 22.98
change in height
= 24  22.98
= 1.017  x(1  ^{10}/_{100})^{3} = 35000
(0.9)^{3}x = 35000
x = 35000
0.729
48010.98 ≈48011  3x + 2y = ^{3}/_{2}
7x  y
2(3x + 2y) = 3(7x  y)
6x + 4y = 21x  3y
7y = 15x
15x = 7y
^{x}/_{y} = ^{7}/_{15}
x:y = 7:15  ^{x}/_{3} + ^{y}/_{4} = 1
^{x}/_{3} + ^{y}/_{4} ≥1
4x + 3y ≥ 12
^{x}/_{2} +^{y}/_{4} = 1
2x + y = 4
2x + y < 4
^{x}/_{3} +^{ y}/1 = 1
x  3y ≤ 3  Sin(2θ + 30)= cos(3θ  40)
2θ + 30 = 90(3θ 40)
2θ + 30 = 90  3θ + 40
5θ = 100
θ = 20º
<PQS = <QSR alternate angles are equal
<PSQ = 90º diameter subtends 90º at circumference
<QPS = 9055 = 35º
<QPS = <QRT = 35º exterior angle is equal to interior opposite angle in a cyclic quadrilateral 1cm rep 2m
1cm^{2} rep 4m^{2}
½ x 6 x 8 = 24cm^{2}
1cm^{2} rep 4m^{2}
24cm^{2} rep 24 x 4
1
= 96 
tan15 = ^{x}/_{100}
x = 26.79m
tan1º = ^{y}/_{100}
y = 1.746m
26.79 + 1.746
= 28.536
= 28.54
PQ = √70^{2} + 100^{2}  2 x 70 x 100cos60
= √7900
= 88.88 100 = 88.88
sinθ sin60
sinθ = 100 x sin60
88.88
sinθ = 0.9744
θ = 77.00º
360  (30 + 77)
360  107
253º

28 + x = ^{18}/_{12}
x
28 + x = ^{1.5}/_{1}
x
1.5x = 28 + x
0.5x = 28
x = 56
^{1}/_{3} x^{ 22}/_{7} x 18^{2} x 84  ^{1}/_{3} x ^{ 22}/_{7} x 12^{2} x 56
28512  8448
20,064 ^{ 22}/_{7} x 10^{2} x 25
7857
5seconds → 1.1cm^{3}
?→ 7857
7857 x 5
1.1
35713.6seconds
35713.6
3600
9.920
9hrs 55minutes

P = (4  ½x)4
P = 16  2x
P = 2x + 2 A = (4  ½x)(4  ½x) + x
A = 16  4 x ¼x^{2} + x
A = 16  3x + ¼x^{2}  16  3x + ¼x^{2} = 8
8  3x + ¼x^{2} = 0
x2  12x + 32 = 0
x = 12 ± √144  128
2
x = 12 ± 4
2
x = 8 or 4
x = 4 8impossible
16  8 = 8
2 x 4 + 2 = 10

 ^{45}/_{60} x 80
60km
RS = x + 80
240 = ^{4}/_{3}
x + 60
720 = 4x + 240
4x = 480
x = 120  ^{4}/_{3} x 120 = 160lm
300  160
= 140km  10:30
+ 45
11.15 +
1:20
= 12:35pm  ^{45}/_{60} x 80 + 1 x 80
60 + 80
= 140
160 = 300
80 y
160y = 300 x 80
y = 300 x 80
160
= 150km/h
 ^{45}/_{60} x 80


x 6 5 4 3 2 1 0 1 2 3 4 y 12 4 2 8 8 6 2 4 12 22 

 x = 1.4 or 4.4
 x^{2} + 3x  6 = y
4 = y
at point of interception
x = 3.5 or 0.6


 ^{5}/_{100} x 400000 = 25000
total earnings = 15000 + 25000
= 40000
tan x = ^{2}/_{100} x 40000
= 800
net earnings = 40000  800
= 39200  ^{120}/_{100} x 15000 = 18000
^{10}/_{100} x x = 0.1x
2/100(10/100x + 18000)
0.002x + 360
(0.1x + 18000)  (0.002x + 360)
= 64800
0.098x + 17640 = 64800
0.098x = 47160
x = 481224
total = 481224 + 200000
= 681224
 ^{5}/_{100} x 400000 = 25000

 (2 + 4 , 3 + 5)
2 2
(1, 1)  grad = 5 3 = ^{8}/_{6} = ^{4}/_{3}
4 2
y  5 = ^{4}/_{3}
x + 4
3(y  5) = 4(x + 4)
3y  15 = 4x  16
3y = 4x + 1
y = ^{4}/_{3}x + ^{1}/_{3}  3(x + 1) = 4(y  1)
3x + 3 = 4y  4
3x + 7 = 4y
4y = 3x + 7
T(k,0)
4y = 7
y = ^{7}/_{4}
y = 1¾
T(1¾ , 0)  G = ¾
¾ = 8  6
a + 1
¾ = 2
a+1
8 = 3(a + 1)
8 = 3a + 3
5 = 3a
a =^{ 5}/_{3}
= 1^{2}/_{3}
 (2 + 4 , 3 + 5)

 ^{5}/_{2}hrs  full tank
1hr  1 x ^{1}/_{1} ÷ ^{5}/_{2} = ^{2}/_{5}  1hr  ^{2}/_{5}
11/6hrs  ^{2}/_{5} x 7/6 x 1 = 7/15
remaining 1  7/15 = 8/15  4hrs ^{ 8}/_{15}
1hr  ^{8}/_{15} x 1 x = ^{2}/_{15}
1hr  ^{2}/_{15}
?  ^{15}/_{15}
15 x 1 ÷ ^{2}/_{15}
^{15}/_{15} x 1 x ^{15}/_{2} = 7.5hrs
let rate of working
^{2}/_{15} + y = ^{2}/_{5}
y = ^{2}/_{5}  ^{2}/_{15}
y = 6  2
15
y = ^{4}/_{15}
^{15}/_{15} x 1 x ^{15}/_{4}
^{15}/_{4} = 3hrs
 ^{5}/_{2}hrs  full tank
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