MATHS OBJ:

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/

=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(

xdegrees=128degrees+180degrees

xdegrees=308degrees

(5a)

(1,1),(1,2),(1,3),(1,4),(1,5),

(2,1),(2,2),(2,3),(2,4),(2,5),

(3,1),(3,2),(3,3),(3,4),(3,5),

(4,1),(4,2),(4,3),(4,4),(4,5),

(5,1),(5,2),(5,3),(5,4),(5,5),

(6,1),(6,2),(6,3),(6,4),(6,5),

(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)/

=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

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