Questions
INSTRUCTIONS TO CANDIDATES
 The paper contains two sections: Section I and II
 Answer ALL questions in section I and ONLY FIVE questions from section II.
 All working and answers must be written on the question paper in the spaces provided below each question.
 Marks may be awarded for correct working even if the answer is wrong.
 Negligence and slovenly work will be penalized.
 Nonprogrammable silent electronic calculators and mathematical tables are allowed for use.
SECTION 1 (50 Marks)
Answer ALL questions from this section
 Four wooden poles have lengths 280cm, 336cm, 476 and 420cm. The owner wishes to cut them into shorter pieces of equal length. Find the greatest possible length of each piece if no wood is left over. (3mks)
 simplify; ^{3}/_{5} ÷ ^{2}/_{3} ^{1}/_{6} x ^{7}/_{12} of (^{1}/_{2} + ^{4}/_{5}) (4mks)
 A rectangular lawn measures 50m by 40m. There is a concrete path of width 100cm all round it. what is the area of the path (3mks)
 The length of the major and minor arcs of a circle are 154cm and 110cm respectively. Taking π =^{22}/_{7}, find the angle subtended at the centre by the minor arc. (4mks)
 Without using mathematical tables, evaluate (2mks)
Log_{10}20 + Log_{10}40  Log_{10}8  A trader bought 360 trays of eggs at sh120 per tray. He later discovered that 10% of the eggs were bad and could not sell them. For how much must he sell the good eggs per dozen in order to make a profit of 50% (4 mks)
 Use matrices to solve the simultaneous equations (3 mks)
4x 5y=13
2y+ 3x=8  A form IV maths teacher originally worked out the mean mark of her 30 students to be 41. After the correction of the test, she added some marks to Amina, Nduku and Karimi in the ratio 2:3:4. If the new mean mark for the class is 42.5, determine how many more marks Karimi was added than Nduku (4mks)
 Find the value of K so that the expression 25x^{2} – 10x + ^{1}/_{8} +k is a perfect square (3mks)
 Find all the integers satisfying the inequalities 32x < x3 ≤4 (3mks)
 The crosssection of a head of a bolt is in the form of a regular hexagon as shown below. Determine the area of the crosssection to 2 dp (3 mks)
 A truck left Kisumu to Mombasa a distance of 1000km at 11.30am on Monday at a speed of 60km/hr. On the way it stopped for 2 ⅓hrs. At what time did it reach Mombasa (3 mks
 Solve the equation (3 mks)
^{4}/_{x}1= 3  x
x+3  The wheel of a bicycle has a diameter of 70cm and is rotating at the rate of 200 revolutions per minute. Calculate the speed of the bicycle in km/hr. (3mks)
 Given the column vectors and that p=2a  ^{1}/_{3} b +c, express p as a column vector and hence calculate its magnitude to 3 significant figures. (3mks)
 A commission agent gets 15% commission on the total value of goods sold. In one month he earned sh11,250 as commission. Calculate the total value of goods sold (2mks)
SECTION 1I (50 Marks)
Answer ANY FIVE questions from this section
 A bus left Mombasa and travelled towards Machakos at an average speed of 60km/h. After 2 ½ hrs, a car left Mombasa and travelled along the same road at an average speed of 100km/hr. If the distance between Mombasa and Machakos is 500km, determine:

 The distance of the bus from Mombasa when the car took off (2mks)
 The distance the car travelled to catch up with the bus (4mks)
 Immediately the car caught up with the bus, the car stopped for 25 minutes. Find the new average speed at which the car travelled in order to reach Machakos at the same time as the bus. (4mks)


 On the graph paper provided, plot the triangle whose coordinates are A(1,2) B(5,4) and C(2,6). (1mk)
 On the same axes
 Draw the image A^{1}B^{1}C^{1} of ABC under a rotation of 90° clockwise about the origin (2mks)
 Draw the image A^{11}B^{11}C^{11 }of A^{1}B^{1}C^{1} under a reflection in the line y=x. State the coordinates of A^{11}B^{11}C^{11} (3mks)
 A^{111}B^{111}C^{111} is the image of A^{11}B^{11}C^{11} under the reflection in the line x=o. Draw the image A^{111}B^{111}C^{111} and state its coordinates (2mks)
 Describe a single transformation that maps A^{111}B^{111}C^{111 }onto ABC (2mks)


 The gradient function for a certain curve is 3x2 4x 15. Determine the gradient of the curve at x= 5 (2mks)
 Determine the coordinates of the turning points of the curve (3mks)
 A particle moving along a straight line covers a distance of S metres in t seconds from a fixed point O on the line where S=t3 _6t2 +8t – 4. Find;
 the acceleration when t=5 seconds (3mks)
 the time when the velocity of the particle is constant (2mks)

 Four towns P,Q,R and S are such that town Q is 120km due east of town P. Town R is 160km due north of town Q. Town S is on a bearing of 330° from P and on a bearing of 300° from R. Use a ruler and compasses only for all constructions in this question.
 Taking a scale of 1cm=50km, construct a scale drawing to show the positions of towns P,Q, R and S (6mks)
 Use the scale drawing to determine
 the distance SP (1mk)
 The distance SR (1mk)
 The bearing of town S from town Q (2mks)
 Three variables P,Q and R are such that P varies directly as Q and inversely as the square of R.
 When P=18, Q=24 and R=4 . Find P when Q=30 and R=10 (4mks)
 Express P in terms of Q and R (1mk)
 If Q is increased by 20% and R is decreased by 10% find:
 A simplified expression for the change in P in terms of Q & R (3 mks)
 The percentage change in P (2 mks)
 In the figure below OA=a and OB=b. M is the midpoint of OA and AN: NB= 2:1
 Express in terms of a and b:
 BA (1 mk)
 NB (1mk)
 ON (2mks)
 Given that BX= hBM and OX=kON determine the values of h & k (6mks)
 Express in terms of a and b:
 The table below gives the marks obtained in a chemistry test by 47 students. By completing the table and using 55 as the assumed mean, calculate the measures that follow.
Marks No. of students Midpoint
(x)d = (X  A) fd fd^{2} CF 41  45 3 46  50 7 51  55 14 56  60 13 61  65 7 66  70 3  The actual mean mark (2mks)
 The median mark (2mks)
 The semiinterquartile range (3mks)
 The standard deviation (3mks)
 The diagram below represents a conical vessel which stands vertically. The vessel contains water to a depth of 30cm. The radius of the surface of water in the vessel is 21cm. (Take π =^{22}/_{7})
 Calculate the volume of the water in the vessels in cm^{3} (3mks)
 When a metal sphere is completely submerged in the water, the level of the water in the vessel rises by 6cm.
Calculate: The radius of the new water surface in the vessel (2mks)
 The volume of the metal sphere in cm3 (3mks)
 The radius of the sphere (3mks)
Marking Scheme
 Find GCD of 280cm, 336cm, 476cm and 420 cm
2 280 336 476 420 2 140 168 238 210 7 70 84 119 105 10 12 17 15  ^{3}/_{5} ÷ ^{2}/_{3}  ^{1}/_{6} x ^{7}/_{12} of (^{1}/_{2} + ^{4}/_{5})
= ^{3}/_{5} ÷ ^{2}/_{3}  ^{1}/_{6} X ^{7}/_{12} of ^{13}/_{10 }
= ^{3}/_{5} ÷ ^{2}/_{3}  ^{1}/_{6} X ^{7}/_{12} x ^{13}/_{10}
= ^{3}/_{5} x ^{3}/_{2}  ^{1}/_{6} X ^{7}/_{12} x^{ 13}/_{10}
=^{ 9}/_{10}  ^{91}/_{720}
= 648  91 = 557
720 720  Length of outer rectangle = 50 + 1+ 1= 52m
Breath of outer rectangle = 40 + 1 + 1 = 42cm
Area of lawn togther with path = 52 x 42 = 2184
Area of lawn = 50 x 40 = 2000
Area of patrh alone = 2184  2000 = 184m^{2}  Length of major arc + minor arc = circumference
2πr = 154 + 110
2 x ^{22}/_{7} xr = 264
r = 264 x ^{7}/_{22} x 1/2 = 42cm
If the angle is x , then:
^{x}/_{360} x 2πr = arc length
^{x}/_{360} x 2 x ^{22}/_{7} x 42 = 110
x = 110 x 7 x 360
2 x 22 x 42
= x = 150  Log_{10}20 + Log_{10}40  Log_{10}8 = Log_{10} (^{20 x 40}/_{8})
Log_{10}100
= 2  Good eggs: 90/100 x 360 = 324
1 tray = 30 eggs
324 x 30 = 810 dozens
12
cost of eggs = 360 x 120
Selling 150 x 360 x 120
100 810
= 80 
coeff matrix M= (4 5 3 2)
det M = (4 x 2)  (3x  5)
= (8)  (15) = 7  Original total marks = 41 x 30 = 1230
New total marks = 42.5 x 30 = 1270
∴ marks added = 1275  1230
= 45
Karimi got 4/9 x 45 = 20
Nduku got 3/9 x 45 = 15
Hence karimi got 5 more  b² = c
4a
1/3 + k = 1
(10)² = 1/3 + k
4 x 25
100/100 = 1/3 + k
1 = 1/3 + k
k = 2/3  32x < x3
3 + 3 < 3x
6< 3x
2<x
x3 ≤4
x < 7
2 < x < 7
integers {3, 4, 5, 6, 7}  Tan 60° = ^{5}/_{x}
x = 50 = 2.88675
Tan 60°
Sin 60° = ^{5}/_{L}
L = 5/Sin 60° = 5.7735
Area = (5.7735 x 10) + (1/2 x 10 x 2.88675) x 2
= 57.735 + 28.8675
= 86.6025
= 86.60  Total time taken for journey:
^{1000}/_{60} + 2^{1}/_{3} = 19 hrs
1130 + 1900
= 3030  2400
= 0630
∴ Tuesday at 6:30am  4/x  1 = 3  x
x + 3
4  x = 3  x
x x + 3
Cross multiply:
(x + 3)(4  x) = x (3  x)
4x  x^{2} + 12  3x = 3x  x^{2}
x + 12 = 3x
12 = 2x
∴ x = 6  Distance moved in 1 revolution = πd
Distance covered in 1 minute = 200πd
= 200 x 22/7 x 70cm
Distance moved in 1 hour = 200 x 22/7 x 70 x 60
100 x 100
hence speed = 26.4km/hr 
∴ P = √((3)² + (1)² + (2)²)
=√(9 + 1 + 4)
= √14
= 3.74  15% of goods sold = 11250
100% of goods sold = ?
11250 x 100
15
= 75, 000 

 2½ = 2.5 x 60 = 150km
 Time taken = 150
100  60
= 3.75  3 hrs 45 minutes
Distance = 3.75 x 100 = 375km
 Time taken by bus:
^{125}/_{60 }
= 2^{1}/_{12}
2^{1}/_{12} hrs of 2hrs 05 monutes
^{25}/_{12}  ^{25}/_{60} = 1^{2}/_{3}hrs
Spd = ^{125}/ 1^{2}/_{3} = 75km




 3 (25) + 20  15
75 + 20  15 = 80  3x^{2}  4x  15 = 0
3x^{2}  9x + 5x  15 = 0
3x (x 3)+ 5(x 3)= 0
(3x + 5)(x  3)= 0
3x + 5 = 0
x =3, x = ^{5}/_{3}
Coordinates of turning point
(^{5}/_{3} , ^{670}/_{27}) (3, 26)
 3 (25) + 20  15

 ds/dt = 3t^{2}  12t + 8
dv/dt = a = 6t  12
When t = 5
a = 30 12
a = 18m/s^{2}  At constant velocity, acceleration = 0
6t  12 = 0
6t = 12
t = 2 seconds
 ds/dt = 3t^{2}  12t + 8


 Scale: 1cm = 50km

 Distance SP = 7.9 x 50 = 395km (±5km)
 Distance SR = 7.4 x 50 = 370km (±5km)
 Bearing of S from R = 270° + 47° = 317° (±1)
 Scale: 1cm = 50km

 P α Q/R^{2}
P = kQ/R^{2}
K 24/4^{2} = k = 18 x 4²
24
K= 12
Q = 30 and r = 10
P = 12Q/R^{2}
12 x 30/10^{2} = 3.6  P = 12Q/R^{2}
 P = K ^{120}/_{1002}Q
_________
(^{90}/_{100}R)^{2}
P = K 1.2Q ⇒ P = K1.48Q/R^{2}
(0.9R)^{2}  P = 1.48 (K^{Q}/_{R2})
P increases by 48%
 P α Q/R^{2}


 BA
→ → →
BA = BO + OA
= b + a
= a  b  NB
→
NB = 1/3AB
= 1/3 b  1/3a  ON
→ → →
ON = OA + AN
= a + 2/3AB
= 1/3a + 2/3b
 BA
 →
OX = k (1/3a + 2/3b)
→
OX = 1/3ka + 2/3kb
→ → →
OX = OB + BX
= b + h (1/2a  b)
→
OX = (1  h)b + 1/2ha
1/3ka + 2/3kb = b  bh + 1/2ha
2/3k = 1  h
1/2h = 1/3k
h = 1/2
k = 3/4


Marks No. of students Midpoint
(x)d = (X  A) fd fd^{2} CF 41  45 3 43 12 36 432 3 46  50 7 48 7 49 343 10 51  55 14 53 2 28 56 24 56  60 13 58 3 39 117 37 61  65 7 63 8 56 448 44 66  70 3 68 13 39 507 47 47 21 1903
_
X = 55 + 21/47 = 55.45
Median mark
50.5 + (24  10) x 5
14
= 55.5
Semi interquartile range
Q1 = 50.5 + (12  10) x 5 = 51.21
14
Q3 = 55.5 + (36  24) x 5 = 60.12
13
S.I.Q.R = (60.12  51.21)
2
= 4.453
Standard deviation
S= √ ((1903/47)  (21/47)^{2})
S= 6.347 
 Volume of water = 1/3 x 22/7 x 21^{2} x 30
= 13, 860cm^{3} 
 h/H = r/R
30/36 = 21/R ⇒ 21 x 36 = 25.2cm
30  New volume = 1/3 x 22/7 x 25.2^{2} x 36
= 23950.08cm^{3}
Volume of sphere = 23950.08  13860
= 10090.08cm^{3}  4/3 x 22/7 x r^{3} = 10090.08
r^{3} = 10090.08 x 21
88
r = ³√(10090.08 x 21/88)
= 13.40cm
 h/H = r/R
 Volume of water = 1/3 x 22/7 x 21^{2} x 30
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