1CBADBDDBDA
11DCDBACDCCC
21CACBCDCBCA
31ADBACCDCCD
41DBBBDBCAAD
(1a)
=1/2log25/4-2log4/5
+log320/125
=log(25/4)^1/2-log(4/5)^2
+log(320/125)
=log{sqroot(25/4)}-log
(16/25)+log(320/125)
=log(5/2)-log(320/125)-log(16/25)
=log[5/2*320/125/(16/25)
=log[5/2*320/125*25/16]
=log10
=1
(1b)
%Increment=20%
Grants per land=GH 15.00
The total population from 2003 to
2007=1.2*1.2*1.2*1.2*3000
=6220.8
Total grant=population * grant per head
=6220.8*15
=GH9331
Total grants=GH93312
(2a)
1/x+(1/x+3)=1/2
LCM=x(x+3)
(x+3+x)/x(x+3)=1/2
2(2x+3)=x(x+3)
4x+6=x^2+3x
x^2+3x=4x+6
x^2+3x-4x-6=0
(x^2-3x)+(2x-6)=0
x(x-3)+2(x-3)=0
(x+2)(x-3)=0
x=-2 or x=3
(2b)
Let the bag of rice be x
Let the bag of beans be y
x+y=17(eq1)
2250x+2400y=39600(eq2)
from (eq1)
x=17-y
substitute for x in eq2
2250(17-y)+2400y=39600
38250+150y=39600
y=(39600-3850)/150
y=9
therefore bags of beans=9
substitute for 9 in eq1
x+y=17
x+9=17
x=17-9
x=8
(3)
Area of garden=L^2
17=(L+2)*(L+L)
17=L^2+3L+2-17
L^2+3L+2-17=0
L^2+3L-15=0
-b+_sqroot(b^2-4ac)/2a
=-3+_sqroot(9-4*1*-15)/2*1
=-3+_sqroot69/2
=-3+_8.03/2
=11.307/2 or 5.307/2
=5.654 or 2.653
L=5.654 p=4L
p=4(5.654)
p=22.616m
(3b)
Area=L^2=5.654^2
=31.98m^2
Area of the path=L*b
=2*1
=2m^2
(4)
3^2+y^2=5^2
9+y^2=25
y^2=25-9
y^2=16
y=sqroot16
y=4
therefore (cosx+tanx)/sinx
=(4/5)+(3/4)/(3/5)
=(16+15/20)/(3/5)
=(31/20)/(3/5)
=31/20*5/3
=31/12
=2(7/12)
(4b)
From the diagram
200degrees+32degrees
+ydegrees=360degrees
(angles at a point)
ydegrees+232degrees=360degrees
ydegrees =360degrees-232degrees
y=128degrees
Ndegrees=128degrees(alternative angles)
xdegrees=128degrees+180degrees
xdegrees=308degrees
(5a)
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
(5b)
(i)Pr(sum of outcome is 8)=5/36
(ii)Pr(product of outcome 10)=17/36
(iii)Pr(outcome contain atleast a 3)
=32/36=8/9
(6a)
2basex(37basex)=75basex
(2*x^1)(3*x^1+7*x^0)=7*x+5*x^0
(2*1)(3x+7)=7x+5
2(3x+7)=7x+5
6x+14=7x+5
6x-7x=5-14
-x=-9
x=9
(6b)
let the number of boys=x no of girls=5+x
(x+5)/(x+2)=5/4
4(x+5)=5(x+20)
4x+20=5(x+2)
4x+20=5x+10
4x-5x=10-20
-x=-10
x=10
(i)No of girls=x+5
=10+5=15girls
(ii)Total No of pupils =x+x+5
=20+5=25pupils
(iii)probability of boy
=No of boy/total pupil
=10/25
=0.4
(7a)
PQ=(5-x)^2+x^2
PQ=25+x^2-10x+x^2
therefore Area of the square=2x^2-10x
+25
If the area of PQRS=3/5
2x^2-10x+25=3/5*25
2x^2-10x+25=15
2x^2-10x=15-25
2x^2-10x+10=0
divide through by 2
x^2-5x+5
Using formular==-b+_sqroot(b^2-4ac)/2a
=5+_sqroot(25-4*1*5)/2*1
=5+_sqroot(25-20)/2
=5+_sqroot4/2
=5+_2/2
=5+2/2 or 5-2/2
=7/2 or 3/2
=3.5 or 1.5
(7b)
(1+a)/(n-1)=d
1+a=dn-d
a=d(n-1)-L
2s=n(a+L)
s=n(d(n-1)+L)-L/2
s=n(dn-d+L)-L/2
(8a)
diagram
(8+x)^2 = x^2+32
64+16x+x^2= x^2 + 1024
16x=1024-64=960
therefore 960/16= 60
x=960/16
=60
therefore the radius = 60+8
=68cm
(8b)
diagram
(i)volume of a pyramid
=1/3 AH
2601= 1/3 * A * 27
A=7803/27
=289cm^3
Area of square =289
t^2= 289
t= sqr rut(289)
l=17cm
(8bii)
AC^2 = 17^2 +17^2
AC^2 = 289 +289
Ac^2 =578
AC =sqr root (578)
AC=24.04cm
for the triangele COP
CO= 1/2 AC
=1/2 * 24.04
VC^2= 27^2 + 12.07
VC^2= 929 +144.49
VC= SQR root (873.48)
=29.55cm
cos tita = ADJ/hyp
cos x= 8.5/29.55
cos x=0.2877
x=cos^-1 0.2877
=73.66 degree
(9a)
CBP=128-x(sum of angle in a triangle)
CBA=180-(128-x)
sum of angle on a straight line
CBA=52+x
ADC=180-(128-x)
=52+x
Also BCD=180-x(angle on a straight line)
DCQ=180-(180-x)
DCQ=180-180+x
DCQ=x
x+52+x+76=180
2x=180-52-76
2x/2=52/2
x=26degrees
10a)
YX/XZ=XM/MZ
W/10=8/15
15W=10*8
W=5.33cm
10bi)
q^2=p^2+r^2-2prcos tita
q^2=20^2+15-2*20*15 c0s 90
q^2=400+225-0
q^2=625
q=sqroot625
q=25km
10bii)
p/sinP=q/sinQ=r/sinR
25/sin90=15/sinR
sinR=15*1/25
sinR=0.6
R=sin^-1(0.6)
R=36.86degrees
The bearing of p from R
=90+90+90+alpha
alpha=45-36.86
=90+90+90+8.14
=278.14
=278degrees
The bearing of p from
R=278degrees
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