MATHS OBJ:
1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: ADDDBDADCA
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(1)
On February 28th 2012, value = (100-30/100) * #900,00.00
= 70/100 * #900,00
= #630,000.00

On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
= #299,000

2)
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2
2b)
The lines must be solved simultenously
3y – 2x = 21 ——- (1)
4y + 5x = 5 ——-(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——-(4)
equ (4) minus equ(3)
23x = -69
x = -69/23
x = -3
Put this into equation (1)
3y -2(-3) = 21
3y = 6 = 21
3y = 21 -6
3y = 15
y =15/3
y = 5
coordinates of Q is (-3, 5)


3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80
3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² – 3²
X²= 25 – 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x – 4tan x
5(4/5)- 4(3/4)
20/5 – 12/4
4-3= 1


4ai)
sum of angle in a D =180degree
xdegree + 90degree + 180degree – (3x+15)=180degree
xdegree + 90degree + 180degree – 3x+15=180degree
-2x=180degree – 255
+2x/2=+75/2
x=37.5
4aii)
152 degrees


8a)
Cost price for Lami= #300.00
Profit made by lami = x%
Ie selling price for lami=(100+x/100)×#300
=#3(100+x)
=#(300+3x)
Bola’s cost price = #3(100+x)
Profit made by bola =x%
Selling price for bola =(100+x/100)×#3(100+x)
=#3/100(100+x)²
James cost price =#3/100(100+x)²=300+(6x+3/4)
expanding;
3/100(10000+200+x²) = 300+3/4+6x
3(10000+200x+x²)=30000+75+600x
30000+600x+3x²=30000+75+600x
3x²=75
X² = 75/3
X² = 25
X = square root 25
X = 5
8b)
3x-2<10+x<2+5x
3x-2<10+x & 10+x<2+5x
3x-x<10+2 & 10-2<5x-x
2x<12 8<4x
X<12/2 4x>8
X<6 x>8/4
X>2
Also; 3x-2<2+5x
-4<2x
2x > -4
X > -2
Therefore; Range is -2

9a)
Draw the diagram
Angles PTR and PSR are similar
|PT|/|PS| = |TQ|/|SR|
In angle PTR
|TQ|²=|PT|²+|PQ|²-2|PT||PQ|cos30degrees
=4²+6²-2×4×6×cos30
=16+36-48×0.8660
=52-41.568
=10.432
|TQ|=√10.432 =3.22cm
4/10 = 3.22/|SR|
4|SR| = 10×3.22
|SR| = 32.2/4
|SR| = 8.05cn
Approximately 8cm(to the nearest whole number)
9b)
Atqrs = AΔPSR – AΔPTR
AΔPTR = 1/2×4×6×sin30
=2×6×0.5
=6cm²
AanglePTQ/AanglePSR = |PT|²/|PS|²
6/AanglePSR = 4²/10²
6/AanglePSR = 16/100
16×AanglePSR = 6×100
AanglePSR = 600/16 = 37.5cm2
ATQRS = 37.5 – 6
=31.5cm2
=32cm2


10a)
Using Pythagoras theorem from SPQ
|SQ|^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^-1(0.3846)
= 22.6degrees
From PRQ
Sin tita= |PR|/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12×0.3843
PR= 4.61cm
10bii)
Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^-1(0.
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12×0.60000
y= 7.2m


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(13a)
Frequency=16+x+y
16+x+y=30
x+y=30-16
x+y=14–(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560-900
30x+50y=660
divide through by 10
3x+5y=66–(eqii)
From (i)
x+y=14
x=14-y–(eqiii)
sub for x in eqii
3(14-y) +5y=66
42-3y+5y=66
2y=66-42
y=24/2
y=12
feom eqiii
x=14-12
x=2

(13b)
TABULATE
Class interval:1-10,11-20,21-30,41,50,51-60,61-70,71-80,81-90
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.5-10.5,10.5-20.5,20.5-30.5,30.5-40.5,40.5-50.5,50.5-60.5,60.5-70.5,70.5-80.5,80.5-90.5

(13C)
DRAW THE GRAPH