QUESTIONS
SECTION 1 (50 MARKS)
Answer all the questions in the space provided below each question
 Find the equation of a straight line passing through the points A (1,3) and B (2, 5).Express your answer in the form ax + by = c where a, b and c are integers. (3marks)
 Evaluate without using mathematical tables or calculator 10÷2+6×48×5 (3marks)
5+(12)÷3×2  Solve for x in the equation Cos(2x30)°= tan45° (3marks)
Sin(3x+10)°  Two taps P and Q together can fill a water tank in 6 minutes. Tap P alone takes 5 minutes longer than tap Q. How many minutes does it take tap P alone to fill the tank? (3marks)
 Given that, 27^{5x2y}=243 and 81^{2xy}=3, Calculate the values of x and y. (3marks)
 A point P is mapped onto P’ by a negative quarter turn about the origin. P’ is mapped onto P’’ by a translation represented by the vector (^{2}_{3}) . If P’’ has coordinates (11,5) determine the coordinates of p. (3marks)
 A metallic pipe which is 21 meters long has an internal radius of 13 cm and an external radius of 15 cm. if the density of the metal is 8620 kg/ m^{3}, find its mass. (3marks)
 Using logarithms evaluate ∛(82.73×0.2943^{2})(3marks)
613.5  A proper fraction is such that the denominator exceeds the numerator by 3. If 2 is subtracted from both the numerator and denominator, the fraction formed is ^{1}/_{8} less. Determine the original fraction. (3 marks)
 Given that OM = 2i +3j 6k and ON = 3i + 5j +k.Find the magnitude of MN to 2 decimal places. (3marks)
 Find the range of the integral values of x in the inequality 10<3(x+2)<35 , giving your answer in the form a≤x≤b (3marks)
 Simplify completely 22x ÷ x1 (3marks)
6x^{2}x12 2x3  The marked price of a recliner sofa set in a furniture store was ksh 400,000.A customer bought the recliner at 10% discount. The dealer still made a profit of 20%, Calculate the amount of money the dealer paid for the recliner. (3marks)
 Draw a line AB of length 9 cm. On one side of line AB construct the locus of a point P such that the area of triangle ABC is 13.5 cm^{2}.On this locus locate two positions of a point P1 and P2 such that
 Given that the area of an image is four times the area of the object under a transformation whose matrix is , find the possible value of x .(3 marks)
 Construct a triangle ABC in which AB = 5cm and AC = 8cm and ∠ABC=105°. Using line AC, locate point x on AB produced such that AX: XB =3: 2. (4marks)
SECTION II (50 MARKS)
Answer only five questions in this section
 The table below shows the weekly salary (k£) paid to workers in a school.
Salary (k£) 50≤x≤100 100≤x≤150 150≤x≤250 250≤x≤350 350≤x≤500 No. of Workers 25 27 30 26 24  Calculate the differences between the mean and the median. (6 marks)
 Draw a frequency polygon to illustrate the above information. (4marks)

 Complete the table of values for the equation, y=2x^(2 )+x+8. (2marks)
x 3 2 1 0 1 2 3 4 y  Use the values above to draw the graph of y=2x^{2}+x+8 . (3marks)
 Using the graph drawn above Solve the equations:
 2x^{2}=x+8 (2marks)
 2x^{2}+4x+12=0 (3marks)
 Complete the table of values for the equation, y=2x^(2 )+x+8. (2marks)
 Three towns P, Q and R are such that Q is 16 km north of P and the distance of R is 12 km from P and on a bearing of 60º from Q.
 Using a scale of 1cm to represent 4km, Make a scale drawing showing the relative positions of the three towns. (3marks)
 Using the scale drawing above, find the
 Distance of R from Q. (1mark)
 Bearing of P from Q. (1mark)
 How far town R is east of Q (1mark)
 A Passenger in an aero plane after takeoff from town R spotted town P at an angle of depression of 48º, by means of a scale drawing determine the vertical height of the plane at town R. (3marks)

 The equation of a straight line L1 is of the form 3y+2x=5.L1 is perpendicular to L2 and meets it at the point where X=2, determine the equation of L2 in the form y = mx+c where m and c are constants. (5marks)
 L3 is parallel to the line L2 and passes through the point (3,2).,find the equation of L3, leaving your answer in its double intercept form. (3marks)
 Determine the angle of inclination of L2 to the Yaxis. (2marks)
 The points P, Q, R and S, have position vectors 2p, 3p, r and 3r respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6.
 Find, in its simplest form OT, QT and TR in terms of p and r. (6 marks)
 Show that the points Q, T and R, are collinear. (3marks)
 Determine the ratio in which T divides QR. (1mark)
 In the figure below, O_{1} and O_{2} are the centers of the circles whose radii are 5 cm and 8 cm respectively. The circles intersect at A and B and angle AO_{1}O_{2} = 64˚.
Calculate the area of the: Sector
 AO_{1}B (2marks)
 AO_{2}B (3 marks)
 Intersecting region. (3marks)
 The shaded region. (2marks)
 Sector

 Find the x –intercept of the curve y = (x+2) (x1)^{2}. (1mark).
 Find the gradient function of the curve y = (x+2) (x1)^{2} (2marks)
 Find the coordinates of the turning point. Hence sketch the curve y= (x+2) (x1)^{2}. (4 marks)
 Calculate the exact area enclosed by the curve and the x  axis (3marks)
 P and Q are two points on latitude 40°N.Their longitudes are 30°E and 150°W respectively. Find to one decimal place :( Take the radius of the earth = 6370km andπ=^{22}/_{7})
 The distance in km between P and Q along the parallel of latitudes. (2marks)
 The shortest distance along the earth’s surface between P and Q in km. (3marks)
 A weather forecaster reports that the center of a cyclone at (40°N, 60°W) is moving due north at 24 knots. How long will it take to reach a point (45°N, 60°W). (2marks)
 A plane leaves P at 2.15 pm at a speed of 350 knots to town R (40°N, 65°E). Determine the time at R when the plane arrived. (3marks)
MARKING SCHEME
 Gradient =^{ 5}/_{2} = ^{3}/_{1} = ^{8}/_{3}
y  5  ^{8}/_{3}
x + 2
3(y  5) = 8(x + 2)
3y  15 = 8  16
3y + 8x = 1  Numerator = 5 + 24  40
21
den. = 5  8
= 13
^{21}/_{13}
= 1^{8}/_{13}  cos(2x  30)º = sin(3x + 10)º
cos(2x  30)º = cos 90 (3x + 10)
2x  30 = 90  (3x + 10)
2x  30 + (3x + 10) = 90
5x  20 = 90
5x = 110
x = 22  ^{1}/_{x+5} + ^{1}/_{x} = ^{1}/_{6}
6(x + x + 5) = x(x + 5)
6x + 6x + 30 = x2 + 5x
7x  30 = 0
 10x + 3x  30 = 0
x(x  10) + 3(x  10) = 0
(x + 3)(x  10) = 0
x = 3 or 10
Top P = 5 + 10
=15mins  3^{3(5x  2y)} = 3^{5}
3(5x  2y)= 5
15x  6y = 5
3^{4}(2x  y) = 3^{1}
4(2x  y) = 1
(15x  6y = 5)4
(8x  4y = 1)6
60x  24y = 20
48x  24y = 6
12x = 14
x = 1^{1}/_{6}
8x^{7}/_{6}  4y = 1
^{56}/_{6}  4y = 1
8^{1}/_{3} = 4y
2^{1}/_{12} = y
y  2 = 11
y = 13
x + 3 = 5
x = 8
P(8, 13) External volume  Internal volume
^{22}/_{7} x 15^{2} x 2100  ^{22}/_{7} x 13^{2} x 2100
^{22}/_{7} x 2100 (225  169)
^{22}/_{7} x 2100 x 56
369,600cm^{3}
0.3696m^{3}
Mass = 8620 x 0.3696
= 8620 x 3696
1000
3185.952
= 3186kg No Log 0.2943^{2}
82.73
613.5
0.2269T.4688 x 2
2.9376
1.9177 +
0.8553
2.7878
2.0675
2.0675 = 3/3 + 1.0675 = T + 0.3558
33
T.3558 ^{x}/_{y}
y  x = 3
y = x = 3
y = 3 + x
x  2 = ^{x}/_{y} ^{ 1}/_{8}
y  2
x  2 = x  ^{1}/_{8}
3+x2 3+x
(x  2) (x + 3 )8 = x (x + 1)8((x+1)(x + 3))
(x^{2} + x  6)8 = 8x^{2} + 8x  (x^{2} + 4x + 3)
8x^{2} + 8x  48 = 8x^{2} + 8x  x^{2 } 4x  3
0 = y^{2 }4x + 45
x^{2 }+ 4x  45 = 0
x^{2 }+ 9x  5x  45  0
x(x + 9) 5(x + 9) = 0
(x  5)(x + 9) = 0
x = 5 or 9  MN = √(32)^{2} + (53)^{2} + (16)^{2}
√25+ 4 + 49
√78
8.832
8.83  10<33(x + 2)
10<3x + 6
4<3x
1^{1}/_{3} < x
3x + 6 < 35
3x < 29
x < 9^{2}/_{3}
2≤x≤9  2(1x)
6x^{2}  9x + 8x  12
2(x  1)
3x(2x3)+4(2x3)
2(x1)
(3x+4)(2x3)
= 2
3x+4  100% = 400000
90% = 90 x 400000
100
= 360000
120% = 360000
100% = 100 x 360000
120
= 300000
½ x 9 x h = 13.5
h = 3cm x^{2}  ((x4)(x+8)) = 4
x^{2 } (+4x  32) = 4
x^{2} x^{2 } 4x + 32 = 4
^{ }4x = 28
x = 7 

050 f x fx cf fd 50100 25 75 1875 25 0.5 100150 27 125 3375 52 0.54 150250 30 200 6000 82 0.3 250350 26 300 7800 108 0.26 350500 24 425 10200 132 0.16 500650 Σf=132 Σfx= 29250
132
= 221.60
Median = 66th position
150 + 82
150 + (66  52) x 100
200
221.60  157
= 64.60


x 3 2 1 0 1 2 3 4 y 13 2 5 8 7 2 7 20 

 2x^{2} + x + 8 = 0
2x^{2} + x + 8 = y
0 = y
x = 1.78 or x =2.2  2x^{2} + x + 8 = y
3x  4 = y
x = 1.55
x = 3.67
 2x^{2} + x + 8 = 0




 5.3cm = 5.3 x 4 = 21.2km
21 ± 0.4  215º
214 ± 1  4.7cm = 4.7 x 4 = 18.8km
18.1 ± 0.4
 5.3cm = 5.3 x 4 = 21.2km



 3y = 2x + 5
y = ^{2}/_{3}x + ^{5}/_{3}
Gr = ^{3}/_{2}
at x = 2
y = ^{4}/_{3} + ^{5}/_{3}= ^{9}/_{3} = 3
(2, 3)
y  3 = ^{3}/_{2}
x + 2
2y  6 = 3x + 6
2y = 3x + 12
y = ^{3}/_{2}x + 6  y 2 = ^{3}/_{2}
x + 3
3(x + 3) = 2(y  2)
3x + 9 = 2y  4
3x  2y = 13
^{3x}/_{13} + ^{2y}/_{13} = 1
x + y = 1
^{13}/_{3 } ^{13}/_{3 }
x + y = 1
4^{1}/_{3} 6½
tanθ = ^{3}/_{2}
θ = 56.31º
90  56.31º = 33.69º
angle 33.69º or 146 .31º
 3y = 2x + 5

OT = ^{3}/_{7}r + ^{12}/_{7}p
QT = QO + OT
= 3p +^{3}/_{7}r + ^{12}/_{7}p
= ^{3}/_{7}r  ^{9}/_{7}p
TR = TO + OR
= ^{3}/_{7}r  ^{12}/_{7}p + r
= ^{4}/_{7}r ^{12}/_{7}p QT = ^{3}/_{7}(r  3p)
TR = ^{4}/_{7}(r  3p)
^{4}/_{7}QT = ^{3}/_{7}TR
4QT = 3TR
QT//TR  QT/TR = ¾
QT:TR = 3:4

 ^{128}/_{360} x ^{22}/_{7} x 25
27.94  ^{8}/_{sin64} =^{ 5}/_{sinx}
sin x = 5 x sin64
8
= 0.5617
= 34.18º
2x = 68.36º
68.36 x ^{22}/_{7} x 64
360
= 38.19
 ^{128}/_{360} x ^{22}/_{7} x 25
 (27.94  x 25 x sin128) + 38.19  x 64 x sin68..36
27.94  9.850 38.19  29.74
18.09 + 8.45
26.54  9.850 + 29.74  26.54
39.59  26.54
13.05


 (x+2)(x1) = 0
x = 1 or 2  y = (x+2)(x^{2}  2x + 1)
y = x^{3}  2x^{2} + 2x^{2} + x  4x
y = x^{3}  3x
^{dy}/_{dx} = 3x^{2}  3  3(x^{2} 1) = 0
3(x1)(x+1) = 0
3(x1)(x1) = 0
x = 1 or 1
when x = 1, y = 0
(1,0)
d^{2}y = 6x at x=1
dx^{2}
= 6
(1,0)minimal
when x = 1
y = (1+2)(11)^{2}
y = 4
(1,4)
d^{2}y = 6x at x = 1
dx^{2}
2^{2}y = 6
dx^{2 }
1.25  (2)
0.75
 (x+2)(x1) = 0

dif in longitude = 180
180 x ^{22}/_{7} x 2 6370 x cos40
360
= 15336.2
100 x ^{22}/_{7} x 2 x 6370
360
= 8272.7
differences in latitude = 4540
= 50
= 5 x 60 = 300NM
Time = ^{300}/_{24}
= 12.5hrs
differences in longitude = 650
= 35º
24hrs = 360º
35 x 24
360
2hrs 20mins
Distance = 35 x 60 x c0s40
= 1,608.7NM
Time = 1608.7
350
= 4.596hrs
4hrs, 36 mins
= 8.11p.m
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