MATHEMATICS FORM THREE
MARKING SCHEME
NO WORKING MARKS REMARKS
1 Numerator
-24÷-8×5+30
3×5+30
15+30=45
Denominator
-16×4÷2+(-8)
-16×2-8
-32-8=-40
Num/Den=45/(-40)=-1 1/8
M 1
M 1
A 1
2 〖3x〗^2-3xy+xy-y^2
3x(x-y)+y(x-y)
(x-y)(3x+y) M1
M1
A1
3 100 500 dollars to shillings = 9.84×100 500
=sh. 988 920
Balance after expenses = Sh.988 920-430 897
=Sh.558 023
Amount received in Japanese Yen = (558 023 ×100)/75.12
=742 842 Yen
M 1
M 1
A 1
4
NO Standard form Log
0.7432
80.23 7.432×10^(-1)
8.023×10^1 1 ̅.8712
1.9049+
1.7756
4.125 4.125×10^0 0.6154×2
1.2308-
0.5448÷3
1.519 ←0.1816×10^0 0.1816
B 1
B1
B 1
A 1
All logarithms
Addition and subtraction
Cube root
Correct answer
5 Absolute error = 0.05
Actual Perimeter = 6.5+7.4+8.2=22.1cm
Max perimeter = 6.55+7.45+8.25=22.25cm
Min perimeter =6.45+7.35+8.15=21.95 cm
% error =1/2 ((22.25-21.95)/22.1)×100%
=0.3/44.2×100%
=0.679%
B1
M1
A1
6 Mid-point of AB = ((2+6)/2,(-4 + (-8))/2)=(4,-6)
m_1=(-8-(-4))/(6-2)=(-4)/4=-1
m_2=1
Equation of L_2⇒1/1=(y -(-6))/(x – 4)
x-4=y+6
x-y-10=0 M 1
M 1
A 1 Correct coordinates of mid-point
Gradient of line L2
Equation of L2
7 〖AB〗^2=6^2+9^2-2(6×9) cos〖132°〗
=36+81-108 cos〖132°〗
=√189.266
= 13.76 cm
Area = πR^2=22/7×(13.76/(2 sin132 ))×(13.76/(2 sin132 ))
=22/7×9.26×9.26=269.49 〖cm〗^2 M 1
A 1
M 1
A 1
8 log_10〖25-log_10〖4+log_101600 〗 〗
log_10((25×1600)/4)
log_10〖10 000〗
4 M1
M1
A1
9 3(2√5-√2)/(2√5+√2)(2√5-√2)
(6√5-3√2)/(20-2√10+2√10-2)
(6√5-3√2)/(20-2)
(6√5-3√2)/18⇒1/3 √5-1/6 √2 M 1
M 1
A 1
10 〖6a〗^2-5a=4
a^2-5/6 a=2/3
a^2-5/6 a+25/144=2/3+25/144
(a-5/12)^2=0.8403
a-5/12=√0.8403=±0.9167
a=0.9167+5/12 OR-0.9167+5/12
a=1.333 OR-0.5
M 1
M 1
A 1
11 HP =41400+(900×6)=9800/ =
Cost price 100/175×9800=5600 / =
∴Cash price=125/100×5600
=7000 /= M1
M1A1
expression
12
13 A^2=[■(1&2@4&3)][■(1&2@4&3)]=(■(9&8@16&17))
(■(9&8@16&17))=(■(1&2@4&3))+(■(a&b@c&d))
B=(■(8&6@12&14)) B 1
M 1
A 1
14 (1+r/100)^n=A/P
n log〖(1+r/100)=log〖A/P〗 〗
n=log〖A/P〗/log(1+r/100) M1
M1
A1
15 4x-3≤0.5(x+8)
4x-3≤0.5x+4
4x-0.5x≤4+3
3.5x≤7
x≤14
0.5(x+8)-2
-2<x≤14
M 1
A 1
B 1
Both inequalities solved
Compound inequality
Number line
16 a+5d=27
a+9d=43-
-4d=-16
d=4
a+20=27 ;a=7
16th term = 7+15(4)=60+7
=67
B 1
M 1
A 1
17 (21,750 + 15,000 + 8,000 ) × 12
= Kshs 537,000
Calculation of tax due1st slab = 116 160×10%=Shs.11 616
2nd slab = 109 440×15%=Shs.16 416
3rd slab = 109 440×20%=Shs.21 888
4th slab = 109 440×25%=Shs.27 360
5th slab = 92 520×30%=Shs.27 756
Tax due = 11 616+16 416+21 888+27 360+27 756
=105 036
Net tax = 105 036-12 672
=Sh.92 364
Net income = Gross income-total deductions
=(21 750+15 000+8 000)-((92 364)/12+200+4 500)
=44 750-12 397
=Shs.32 353
M1
A1
M1
M1
M1
A1
M1
A1
M1
A1
- a)
x° -90 -75 -60 -45 -30 -15 0 15 30 45 60 75 90
3 cos〖2x°〗 -3 -2.6 -1.5 0 1.5 2.6 3 2.6 1.5 0 -1.5 -2.6 -3
sin(2x+30°) -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1 0.87 0.5 0 -0.5 (i) x=-54° and x=33°
(ii) 6 cos〖2x°〗+5=0
3 cos〖2x°=-5/2=-2.5〗
x=-75° and x=75°
- B 1 Diagram
B 1 Plotting
B 1 Diagram
B 1 Coordinates A^’ (2,-1) B^’ (-1,-2) and C^’ (1,-4)
B 1 plotting
B 1 Diagram
B 1 Coordinates A^” (-1,-2) 〖B’〗^’ (-2,1) and 〖C’〗^’ (-4,-1)
(i) (2¦3)-((-1)¦(-2))=(3¦5)
T=(3¦5) B 1
(ii) B 1 coordinates
B^”’→((-2)¦1)+(3¦5)=(1¦6)=(1,6)
C^”’→((-4)¦(-1))+(3¦5)=((-1)¦4)=(-1,4)
B 1 Diagram
20 (i) x-5
(ii) [x+(x-5)]×2
4x-10
b) i) (x+20)(x+15)=15(4x+10)
x^2-25x+150=0
(x-10)(x-15)=0
x=10 or x=15
ii) 4×10-10 or 4×15-10
=30 or 50
iii) (10-5)+20 or (15-5 )+20
= 25 or 30 B1B1
B1
M1
M1
A1
M1
A1
M1
A1
Correct equation
solving
21 The completed table is shown below.
Mass in Kg Class mid-point x Frequency
f fx
41-45 43 5 215
46-50 48 10 480
51-55 53 14 742
56-60 58 8 464
61-65 63 3 189
∑f = 40 ∑f x=2090
(i) Mean mass x ̅=(∑f x)/(∑f )=2090/40
=52.25 kg
(ii) Median = 50.5+((20 – 15)/14)×5
=50.5+25/14
=52.29 kg
Frequency polygon on the attached graph.
B 1 – column for xB 1 – column for f
B 1 – column for fx
B 1 – ∑f x=2090
M 1
A 1
M 1
A 1
B 1 all points plotted
B 1 polygon drawn
22 〖DC〗^2=50^2-3^2=2500-9=2491
DC=√2491=49.91 cm
cos〖θ=3/50=0.06 θ=86.52×2=173.12°〗
ϑ=360-173.12=186.88
l=186.88/360×22/7×2×9=29.37 cm
sin〖β=3/50=0.06; β=3.4398〗
γ=2(90+3.4398)=186.88
360-186.88=173.12
l=173.12/360×2×22/7×6=18.14 cm
Total length = 49.91+29.37+49.91+18.14=147.33 cm M 1
A 1
B 1
M 1 A 1
B 1
M1 A1
M 1
A 1
23 (i) Area of curved part =22/7×4×4×0.5
= 25.14 〖cm〗^2
Area of Rectangular part = 12×6=72〖cm〗^2
Total Area of solid =25.14+72
=97.14〖cm〗^2
(ii) Curved length = 22/7×8 =12.57
Total perimeter =12.57+2+6+12+6+2
= 40.57cm
Volume of solid =97.14×900
=87.43 〖cm〗^3Density = ( 306×1000)/87.43 g/〖cm〗^3= 3.499
~3.5〖g/cm〗^3
M1
M1
A1
M1
M1 A 1M1
A1
M1
A1
24
(I) 4.4×50=220±5 km(II) 180+18=198°
B 1 – angle of depression
B 1 – construction of perpendicular bisector of PQ from M.
B 1 – 6×50=300 km
B 1
B 1
