Questions
 Evaluate and simplify without using a calculator. (3mks)
 The sum of interior angles of a polygon is 1980^{o}. Find the number of sides the polygon has. (2mks)
 Simplify as far as possible by rationalizing the denominator. (3mks)
 Use table of reciprocal only to work out (3mks)
 Solve 3x  2≤ 5x  6 < 2x + 12 and represent your solution on a number line. Hence state the integral values. (4mks)
 Evaluate without using mathematical tables. (3mks)
2 log 5  ½log 16 + 2 log 40  Given that P=2.6 cm, Q=4.0 cm and R=7.8 cm. Find the percentage error in the expression. (3mks)
P + Q
R  From a point 20m away on a level ground the angle of elevation to the lower window line is 29^{o} and the angle of elevation to the top line of the window is 32^{o}. Calculate the height of the window. (3mks)
 The size of an interior angle of a regular polygon is 156^{o}. Find the number of sides of the polygon. (2mks)
 The number 5.81 contains an integral part and a recurring decimal. Convert the number into an improper fraction and hence a mixed fraction. (3mks)
 Simplify. (3mks)
2y^{2}  xy + x^{2} 2x^{2}  2y^{2}  Given that Sin A= ^{4}/_{5} , Cos B=^{5}/_{12} A and B are acute angles. Without using tables calculate
Sin B Cos A + Sin A tan B (3mks)  A two digit number is such that the sum of the digits is 11 where the digits are reversed the number exceed the original number by 9. Calculate the original number. (3mks)
 Two boys and a girl shared some money. The elder boy ^{4}/_{5} got of it, the younger boy got ^{2}/_{5} of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girls share. (4mks)
 Solve the following simultaneous equations. (4mks)
x^{2} xy=2
x + y=3  Use the table of squares, square roots and reciprocals to evaluate to 3 decimal places the question below. (4mks)
 The diagram below shows two circles centre A and B which intersect a point P and Q. Angle PAQ=70^{o} and <PRQ=40^{o} and PA=AQ=8 cm.
Use the diagram to calculate to 2 d.p the length PQ (2mks)
 The length PB (2mks)
 Area of minor segment circle centre A. (2mks)
 Area of the shaded region (4mks)
 The following table shows the heights to the nearest centimeter of some maize plants in a research farm.
Height (cm)
8084
8589
9094
9599
100105
105109
110114
115119
Frequency
5
14
16
17
24
12
11
4
 State the modal class (1mk)
 Find to 2 d.p
 Mean height (4mks)
 The difference between the mean height and the median height. (5mk)

 Three points A(0,4) B(2,3) and C(2, 1) are vertices of a triangle. Find ;
 The gradient of AC (1mk)
 The gradient of the perpendicular bisector of line AC (1mk)
 The coordinates of the midpoint of line AC. (1mk)

 the gradient of AB (1mk)
 The gradient of the perpendicular bisector of lines AB (1mk)
 The coordinates of the midpoint of AB (1mk)

 Find the equation of the perpendicular bisector of AC (1mk)
 The equation of perpendicular bisector of AB (1mk)
 Hence find the coordinates of the circumcentre of the triangle.(2mks)
 Three points A(0,4) B(2,3) and C(2, 1) are vertices of a triangle. Find ;
 The position vectors of points A and B with respect to the origin O, are (^{8}_{5}) and (^{12}_{5}) respectively. Points M and N are the mid points of AB and OA respectively.
 Find
 The coordinates of N and M (3mks)
 The magnitude of NM. (3mks)
 Express vector NM in terms of OB (1mk)
 Point P maps onto P^{1} by a translation (^{5}_{8}) given that OP=OM +2 MN, find the coordinates of P^{1} (3mks)
 Find
 The information of a piece of land was entered in a field book as shown below.
 Sketch the map of the land (3mks)
 Find the area of the land in hectares. (7mks)

 Draw the graph of the function y=2x^{2} + 4x  3 on the graph paper provided `for 4 ≤ x ≤ 2.5 (5mks)
 Use your graph to solve the equations
 2x^{2} +4 x3=0 (2mks)
 x^{2}+ x5=0 (3mks)
 The base of an open rectangular tank is 3m by 2.5 m and its height is 4m.
 Calculate
 The capacity of the tank in litres. (3mks)
 The total surface area in m^{2} of the tank. (2mks)
 An open cylindrical tank has an equal capacity and same height as the rectangular tank in (a) above. Calculate ;(correct to one decimal places)
 The radius of the cylindrical tank. (3mks)
 The total surface area, in m^{2}, of the tank. (2mks)
 Calculate
 Transline bus left Nairobi at 8.00 and travelled Kisii at an average speed of 80km/h. Given that the distance between Nairobi and Kisii is 400km, calculate;
 The time the car arrived in Nairobi. (3mks)
 The time the two vehicles met. (3mks)
 The distance from Nairobi to the meeting point. (2mks)
 The distance of the bus from Kisii when the car arrived in Nairobi. (2mks)
Answers
 ^{16}/_{5} + (^{1}/_{4} x ^{7}/_{2})  ^{31}/_{16}
^{16}/_{5} + ^{7}/_{8}  ^{31}/_{6} = ^{131}/_{120 }√1 Numerator
^{8}/_{3}  (^{7}/_{5} x ^{3}/_{4})+^{15}/_{4}
^{8}/_{3} ^{21}/_{20} + ^{15}/_{4} = ^{322}/_{60 }√1 Denominator
^{131}/_{120} x ^{60}/_{322} = ^{131}/_{644} √1  (2n4)90 = 1980 √1
2n4=22
2n=26
n=13 √1 
 ^{3}/_{0.6735} + ^{13}/_{0.156}
3(^{1}/_{0.6735}) + 13(^{1}/_{0.156})
(3 x 6.735 x 10^{1}) + (13 x 1.56 x 10^{1}) √1
3 x 1.485 + 13(6.41) √1
4.455 + 83.3333 = 87.785 √1  3x  2 ≤ 5x  6 < 2x + 12
3x  2 ≤ 5x  6
3x  5x ≤ 6 + 2
2x ≤ 4
x ≥ 2
5x  6 < 2x + 12
5x  2x < 12 + 6
3x < 18
x < 6
2,3,4,5  2log5  ^{1}/_{2}log16 + 2log40
log 5^{2}  log 16^{1/2} + log 40^{2}
log(25x 40^{2})
4
log 1000 = 4  max quotient =
2.65 + 4.05
7.75
=0.8645
min quotient =
2.55 + 3.95
7.85
=0.8280
Working quotient =
2.6 + 4.0
7.8
=0.8462
Error =
0.8645  0.8280
2
=0.01825
%error = ^{0.01825}/_{0.8462} x 100%
=2.157%  x + y = 20tan32
y = 20tan27
x + 20tan27 = 20tan32
x = 20tan32  20tan27
x = 2.307
height of the window = 2.307m  Exterior <
= 180156
=24^{o}
n = ^{360}/_{24}
= 15 sides  let n = 5.8111.... (i)
10n = 58.1111....(ii)
100n = 581.1111....(iii)
subtract eqn (ii) from eq(iii)
90n = 523
n=^{523}/_{90}
=5^{73}/_{90}  2y^{2}  xy x^{2}
2x^{2}  2y^{2}2y^{2}  xy  x^{2} = (2y^{2}  2xy) + (xy  x^{2})
= 2y(yx) + x (yx)
= (2y+x)(yx)
2x^{2}  2y^{2} = 2 (xy)(x+y)
(2y+x)(yx) = 1(2y+x)(xy)
2(xy)(x+y) 2(xy)(x+y)
= 2yx
2(x+y)
sin A = ^{4}/_{5}
sin B = ^{12}/_{13}
cos B =^{ 5}/_{13}
cos A = ^{3}/_{5}
Tan B = ^{12}/_{5}
sinBcosA + sinAtanB
^{12}/_{13} x ^{3}/_{5} + ^{4}/_{5} x ^{12}/_{5}
^{36}/_{65} + ^{48}/_{25} =
900+3120
625
=^{4020}/_{1625}=2^{154}/_{325} x + y = 11 ... (i)
(10y + x)  (10x + y) = 9 ... (ii)
10y+x10xy=9
9y9x=9
yx = 1 ...(iii)
x+y = 1
x+y = 1
2x=10
x=5, y = 6
the original number = 56  let x be amount shared
elder boy = ^{4}/_{9}x
younger boy = ^{2}/_{5}(^{5}/_{9}x) = ^{2}/_{9}x
girl = 1  (^{4}/_{9}x + ^{2}/_{9}x)
= ^{1}/_{3}x
(^{2}/_{9}x ÷ ^{1}/_{3}x) x 100%
=66.67%  x^{2}  xy = 2
x+y = 3, y = 3x
x^{2}  x(3x) = 2
x^{2}  3x + x^{2} = 2
2x^{2}  3x  2 = 0
(2x^{2}  4x)+(x2)=0
2x(x2)+1(x2)=0
(2x+1)(x2)=0
x = ^{1}/_{2} or 2
when x = 2, y =1
when x = ^{1}/_{2}, y = 3^{1}/_{2}

 sin 35 = ^{x}/_{8}
x=8sin35
PQ = 2 x 8sin35
= 9.18cm  sin 20 = ^{4.5886}/_{PB}PB = ^{4.5886}/_{sin20} = 13.42cm
 ^{70}/_{360} x ^{22}/_{7} x 8 x 8  ½x 8 x 8sin70
=9.04cm^{2}  Area of minor segment centre B
^{40}/_{360} x ^{22}/_{7} x 13.42^{2}  ½x 13.42^{2}sin40
=5.01
Area of shaded region
(½ x 8^{2}sin70 + ½ x 13.42^{2}sin40)(9.04+5.01)
=73.90cm^{2}
 sin 35 = ^{x}/_{8}

 100104

Height f x fx cf 8084 5 82 410 5 8589 14 87 1218 19 9094 16 92 1472 35 9599 17 97 1649 52 100104 24 102 2448 76 105109 12 107 1284 88 110114 4 117 468 103 ∑f=103 ∑fx = 10181  X = ^{∑fx}/_{∑f} = ^{10181}/_{103}
=98.84  median = 94.5+(^{(103/2  35)}/_{17})5
=99.35
99.3598.84 = 0.66cm
 X = ^{∑fx}/_{∑f} = ^{10181}/_{103}


 A(0,4)
C(2,1)
4+1 = ^{5}/_{2} = 2.5
0+2  ^{2}/_{5}
 (^{02}/_{2},^{41}/_{2}) = (1,1.5)
 A(0,4)

 M =
34 =  ^{1}/_{2}
20  m = 2
 (^{0+2}/_{2},^{4+3}/_{2})
(1,3.5)
 M =

 y  1.5 = ^{2}/_{5}(x+1)
y = ^{2}/_{5}x + 1.1  y3.5 = 2(x1)
y = 2x+2.5  y = ^{2}/_{5}x + 1.1
y = 2x + 2.5
solve simultaneously
x = 0.58
y= 1.33
(0.58,1.33)
 y  1.5 = ^{2}/_{5}(x+1)




1. ^{1}/_{2} x 35 x 55 = 962.5
2. ^{1}/_{2} x 85 x 65 = 2762.5
3. ^{1}/_{2} x 50 x 60 = 1500
4. ^{1}/_{2} x 90 x 15 = 675
5. ^{1}/_{2} x 115 x 60 = 3450
6. ^{1}/_{2} x 40 x 30 = 600
Total = 9950m^{2}
^{9950}/_{10000} = 0.995ha 

 x = 0.6 or 2.6
 y = 2x2 + 4x  2
0 = 2x2 +2x 10
y = 2x + 7
x = 2 or 2.5


 V = 3 x 2.5 x 4
=30 m^{2}
=30x1000
=30,000l  (3x2.5x1)+(2.5x4x2)+(3x4x2)
=55.5
 V = 3 x 2.5 x 4

 Πr^{2} = 3 x 2.5
r = square root of ^{(3x2.5)}/_{11}
=1.545m
=1.5m  ^{22}/_{7} x 1.52 + ^{22}/_{7} x 2 x 1.5 x 4
=7.071 + 37.716
=44.787m^{2}
 Πr^{2} = 3 x 2.5


 T = ^{400}/_{80} = 5hrs
8:00AM
5:00
1:00PM  80 x ^{3}/_{2} = 120km
RS = 80 + 120 = 200km/h
t = ^{120}/_{200} = ^{3}/_{5}hrs
=36min
9:30AM
36
10:06AM  10:06
8:00
2:06
time = ^{126}/_{20}
S = 80km/h
D = ^{126}/_{60} x 80 = 168km  t = ^{400}/_{120} = 3hrs 20 min
9:30
3:20
11:50
13:00
11:50
1:10
D = ^{70}/_{60} x 80
=^{280}/_{3}
 T = ^{400}/_{80} = 5hrs
Download MATHEMATICS  FORM 3 END TERM 1 EXAMS 2020.
Tap Here to Download for 50/
Get on WhatsApp for 50/
Why download?
 ✔ To read offline at any time.
 ✔ To Print at your convenience
 ✔ Share Easily with Friends / Students
Join our whatsapp group for latest updates