MATHEMATICS
PAPER 1
TERM 2 OPENER EXAM
INSTRUCTIONS
- This paper consists of two sections I and II. Answer all the questions in section 1 and any 5 questions in section 2.
- KNEC mathematical tables and electronic non- programmable calculators may be used where necessary.
SECTION I
Answer all questions in this section
- Find without using tables or a calculator the value of (3mks)
- The ratio of the size of the exterior angle to the interior angle of a regular polygon is 1:3. Determine the number of sides of the polygon and name it. (3mks)
- Given that 2x²–kx+18 is a perfect square, find k and hence solve the equation 2x²–kx+18=0 by factorization. (4mks)
- Work out using logarithms to 4 s.f (4mks)
√ (6.225log1.001)
(56.7×0.031)³ - Mr. Kanja,Miss Kanene and Mrs Nyaga have to mark a form three math contest for 160 students. They take 5mins, 4mins, and 12mins respectively to mark a script. If they all start to mark at 9.00 am non-stop, what is the shortest time they can take to complete the marking? (3mks)
- Jackie takes 5minutes to run a distance of 1km in a race. Express her speed in(2mks)
- km/hr
- m/s
- Use reciprocal tables to find the value of f given that (3mks)
- A man left ¾ of his estate in Kerugoya to his wife and to each of his two sons .The remainder was to be shared equally among his six brothers. If the estate was worth sh 3 456 000, how much did each of those people get? (3mks)
- A distance of 12km is represented by a length of 4cm on a map. Given that the scale of the map is 1:n, find the
- value of n
- actual area in hectares of a field on the map with an area of 32cm² (3mks)
- Solve the equation ⅓(x+4)–½(2x–4) =2 (2mks)
- The sides of a right angled triangle measured to the nearest cm are 5cm, 12cm and 13cm
Determine the- limits within which the measured dimensions lie (1mks)
- percentage error in the area of the triangle. (3mks)
- Form a quadratic equation in the form ax²+bx+c=0 whose roots are b and twice the negative reciprocal of b. (3mk)
- The coordinates of points A and B are A (2, 3).B (4,–5). M is the midpoint of vector AB.
Determine the coordinates of point M and the magnitude of vector BM. (3mk) - The equation of line L is y=3x–4 and is perpendicular to line H. They cross each other at the y-intercept of line L. Find the equation of line H. (3mks)
- In a circle radius 10cm, an arc PQ subtends an angle of radians at the centre of the circle. Calculate the radius of another circle whose circumference is equal to the length of arc PQ (4mks)
- Solve for a in 2187 (3mks)
SECTION II
Answer any 5 questions in this section
- Four towns are situated in such a way that town Q is 500km on a bearing of 120º from P. Town R is 240km on a bearing of 210º from town P, while town S is due north of town Q and due east of town P.
- Draw a sketch diagram showing the relative positions of P, Q, R and S. (2mks)
- Find by calculation
- the distance QR (2mks)
- the distance QS (2mks)
- the angle PRQ (2mks)
- area of triangle PQS (2mks)
-
- Represent the following inequalities graphically by shading the unwanted region
x≥0, y≥0, x+y≥5, x+y≤10, y≤7, x≤7 (6mk) - write down the coordinates of one point that is inside the wanted region (1mk)
- Name the figure formed by the unshaded region (1mk)
- measure and find the sum of all the angles in the figure formed in c) above. (2mks)
- Represent the following inequalities graphically by shading the unwanted region
- In the figure below, O is the centre of the circle and ∟EAD=40º,∟BCD=118º
Find giving reasons- ∟ADE (2mks)
- reflex ∟EOD (2mks)
- ∟EBD (2mks)
- ∟EAB (2mks)
- ∟DAB (2mks)
- The marks scored in a form three maths exam were recorded as follows
69 70 72 40 52 60 22 31 78 53 28 67 63 54 57 48 47 56 55 62
75 38 37 44 62 64 58 39 45 48 65 50 85 46 47 57 35 34 58 64
37 41 42 36 54 82 48 53 57 56 72 56 48 44 55 78 59 50 45- Make a grouped frequency table with classes 20—29,30—39,40—49,etc (2mks)
- What is the modal distribution of the test (1mk)
- Calculate the mean of the data (4mks)
- Calculate the median mark (3mks)
- The velocity(v)of a vehicle measured at intervals of time(t) were recorded as follows
t(s) 0 2 4 6 8 10 12 v(m/s) 0 20 40 40 30 8 0 - Represent this motion on a graph (3mks)
- Calculate the acceleration (2mks)
- Calculate the total distance travelled by the vehicle (3mks)
- Calculate the average velocity of the vehicle (2mks)
- A wooden stool is in the form of a frustum of a cone with slant edge 40cm,top diameter 30cm and bottom diameter 50cm.
- calculate the perpendicular height of the stool (3mks)
- calculate the total surface area of the stool in terms of (3mks)
- calculate the volume of wood used to make the stool in terms of (3mks)
- given that the density of the wood used to make the stool is 0.8g/cm³, calculate the mass of the stool in kg (1mk)
- Using ruler and compasses only,
- construct triangle ABC in which AB=5cm,BC=6cm and angle ABC=120º. (3mks)
- measure angle ACB (1mk)
- drop a perpendicular from C to cut AB produced at P. Measure CP. (2mks)
- hence calculate area of triangle ABC to 1dp (2mks)
- calculate the radius of a circle that passes through the vertices of triangle ABC (2mks)
- The distance between two towns A and B is360km.A minibus left town A at 8.15a.m and travelled towards B at an average speed of 90km/hr. A matatu left town B at 10.35a.m on the same day and travelled towards A at an average speed of 110km/hr.
-
- how far from A did they meet? (4mks)
- at what time did the two vehicles meet? (2mks)
- A motorist left his home at 10.30a.m on the same day and travelled at an average speed of 100km/hr. He arrived at B at the same time as minibus. Calculate the distance from B to his home. (4mks)
-
MARKING SCHEME
- 1.33 X 0.51 X 1000000
0.19 X 0.0017
=133 x 51 x 100
19 x 17
= 2100 - Ext: Int
1ext= 1/4 x 180 = 45º
no. of sides = 360/45 = 8 sides
Name - octagon - (-k/2)2 = 2 x 18
k2 = 36 x 4 = 144
k= ± 12
2x2 - 12x + 18 = 0
2x2 - 4x - 9x + 18=0
2x(x-2)-9(x-2)=0
(x-2)(2x-9)=0
x=2
x=4½
No log 6.225
0.0004341
56.7
0.0310.7962
+4.6376
3.4338/2
4 + 1.4338/2
2.7169
1.7536
+2.4914
0.2450
x 3
0.7350
10-3 x 9.592 = 0.009592-
LCM os 5, 4 , 12=60
in 60 mins
Kanja - 12 scripts
Kanene- 15 scripts
Nyaga - 5 scripts
All - 32scripts
60 → 32
? → 160
=300mins
=5hrs -
- 1 x 60/5 = 12km/hr
- 1000/5 x 60 = 31/3m/s
(accept 3.3333m/s
reject 10/3 m/s)
-
-
- QR=√ 5002 + 2402= 554.7km
- Sin 30= QS/500
QS=500sin30=250km - Tan PRQ=500/240=2.083
L PRQ=64.36º - A=1/2 x 500x 250sin 60
=54126.6km2
-
-
3/4 x 3456000=2592000
wife and 2 sons got a total of 2592000
3456000-2592000=864000
864000/6=144,000
each brother got 144,000
-
- 4cm:12km
1cm:3km
1cm:300000cm
1:300,000
n=300,000 - 1cm rep 3km
1cm2 rep 9km2
32cm2 rep ?
32x9=288km2
1km2 = 1,000,000m2
10,000m2 = 1 ha
288 x 1000000
10,000
=28800ha
- 4cm:12km
- 1/3 x + 4/3 - x + 2=2
-2/3x = -4/3
x= -4/3x - 3/2 = 2 -
upper limit lower limit 5.5
12.5
13.54.5
11.5
12.5- Actual area= 1/2 x 5 x 12=30cm2
max area= 1/2 x 5.5 x 12.5 = 34.375
min area= 1/2 x 4.5 x 11.5= 25.875
%age error= (34.375 - 25.875/2) x 100
30
=14.17%
- x=b
x=-1/b
(x-b)(x+1/b)=0
x2+x/b-bx-1=0
x2+(1/b-b)x-1=0 - M(2+4/2, 3+-5/2) M(3,-1)
BM=(3-1)-(4-5)=(-14)
|BM|= √-12 + 42= √17 - Grad of L=3
Grad of H= -1/3
coordinates of intersection(0,-4)
y--4 = -1
x 3
3(y+4)= -x
3y+12=-x
3y=-x-12 - Length of Arc PQ = 5/12Π/2Π x 2Π x 10
25Π/2
25Π/2= 2Πr ∴ r= 25Π/2 x 1/2Π
25/4=6.25cm - 32a+3=37
2a+3=7
2a=4
a=2 -
- x+y > 5
x 0 5 y 5 0
x+y < 10
x 0 10 y 10 0 - Any one
- Hexagon
- 135º x 4
90º x 2
Sum= 720
- x+y > 5
-
- <ADE = 50º
(<AED= 90º-<s in a semi-circle) - Reflex EOD= 360 - (40x2)
=360 -80=280º
(<at centre= 2 <on circumfrence) - <EBD=40º
(<s in same segment)
(or <s subtended by same chord/arc) - <EAB=90º
(<s in a semi-circle) - <DAB=180-118=62º
(opp.<s in a cyclic quad are supp)
- <ADE = 50º
-
-
Class f x xf cf 20-29 2 24.5 49 2 30-39 8 34.5 276 10 40-49 14 44.5 623 24 50-59 18 54.5 981 42 60-69 10 64.5 645 52 70-79 6 74.5 447 58 80-89 2 84.5 169 60 60 3190 - 50-59
- x= 3190/60 = 53.17
- M=49.5 + (30-24/18)x 10
49.5 + 10/3 = 52.83
-
-
- acc= 40-0/4-0 = 10m/s
- Areas of
A= 1/2 x 4 x 40=80
B= 40 x 2=80
C= 1/2(40+30)2=70
D=1/2(30+8)2=38
E=1/2x2x8=8
Total=276 - Av. Vel = 276/12 = 23m/s
-
15/25 = L/L+40
15L + 600 = 25l
l=60cm
H=√1002 - 252 = 96.82cm
96.82/h = 100/60
h=58.09
height=96.82-58.09=38.73- Base = Πr2 = 625Π
Top=225Π
CSA=ΠRL - Πrl
2500Π - 900Π= 1600Π
TSA=2450Π - V= 1/3 x 625Π x 96.82 - 1/3 x 225Π x 58.09
=20170.8Π - 4356.75Π=15814.05Π - M=v x d
=15814.05 x 22/7 x 0.8
1000
=39.76kg
-
- <ACB =27º
- CP=5.2cm
- A=1/2 x 5 x 5.2=13cm2
- 5/sin27 = 2r ∴r=5/sin27=5.507
-
-
- | 360 |
A B
8.15 10.35
Dist travelled by minibus when matatu leaves B 90 21/3= 90 x 7/3=210km
R.d= 360-210 =150
R.s= 90 +110=200
t=150/200 x 60 =45mins
dist minibus has travelled = 90 x 35/60
=90 x 185/60 = 277.5km
10.35
+ .45
11.20
11.20am - Time minibus reached B
=360/90 = 4hrs
8.15 + 4 =12.15pm
Time motorist took
12.15
-10.30
1.45
1hr 45 mins
d=100 x 1 45/60
100x13/4
100x7/4=175km
- | 360 |
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