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COMPLETE MATHS OBJ:
1-10: BACBCCCBCD
11-20: CBBACACBBC
21-30: BBCCDACDCD
31-40: ACAABADAAA
41-50: DAABADCBAD.
=================================
9a)
x 62 63 64 65 66 67 68
tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i ii
freq 1 5 12 14 10 6 2 total = 50
fx 62 315 768 910 660 402 136 tot = 3253
x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94
(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436 f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot = 92.82
9bi) mean = EFX/EF = 3253/50 =65.06
9bii) Standard deviation = Square root of EF(x-x)2/Ef
= Square root of 92.82/50
= square root of 1.8564
= 1.36
No9)
================================
5a)
Area of triangle=1/2bh where h=6:. 1/2*6/1*b/1=366b/2=36/16b=72B=72/6Base of angle PSR=12cmSince |TS| //PQQR=PR-PQQR=12-8QR=4cm5b)Draw d triangleFrom angle ABC, to get |BC|Tan60/1=10cm/|BC||BC|=10.65/tan60|BC|=10.65/1.7321=6.1486 =6.15mHence |BC|=|DE|From angle AED, to get |AE|Draw d diagram|AB|=6.15tan456.15*1=6.15mThere4, th height of the tree =10.65-6.15=4.50m
=============
8a)
13a)(x1,y1) =(2,-3)for 2x + y =6y=-2x +6compare y = mx +6m2 = -2for parallel lines, m1=m2: y-y1/x-x1 = my--3/x-2 =-2/1y + 3 = -2(x -2)y + 3 = -2x + 4y = -2x + 4 -3y = -2x +1
==================
11a)3p+4q/3p-4q* 2/1 find p:q1(rp+4q)= 2(3p-4q)3p+4q=6p-8qCollect like terms3p-6p=-8q-4q-3p=-12qTher4 p:q = -3p/-3=-12q/-3P=4q, p=4, q=1P:q= 4:1
=========
6a)Sn=n/2(2a+(n-1)d)A=1, d=2, n=nSn=n/2(2*1+(n-1)2)N/2(2+2,-2)=n/2*2=n^2
(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
No6b)
(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
= (4a)
Pr(3)=x/total
total=25+30+x+28+40+32
=155+x
0.225=x/155+ x
=34.875+0.225x=x
x-0.225x=34.875
=0.775x=34.875 degree
x=34.875/0.775
=45
(4b)
Pr(even)=30+28+40+32
=90
Pr(even)=90/200=9/20
Pr(prime no)=25=30+45+40
=140
Pr(prime no)=140/200=14/20
Pr(even or prime)=9/20 + 14/20
(5a)
Area PSR=1/2*PR*SM
PR=8+6=14cm
36=1/2 * 14* 6
36=1/2 * 14* 6 *QR
QR=8.6cm
(5b)
diagram
sin45 degree= 10.65/h
h=10.65/sin 45 degree
=10.65/0.7071
h=15.06cm
(3b)
h=7cm
T=462cm^2
T=pie sqr + 2 pie rH
462= pie( r ^2 +2*7*r)
462/3.142
=r sqr= =14r
r sqr+ 14r=147
r sqr +14r- 147=0
r= 14+-( (sqr(14))-4*1*147)/2*1
r=-(14 +-(196-4))/2
r=(-14+_12)/2
r=(-14+12)/2 or
r=2/2
=1cm
(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)x-2/4 :x+2/2xCross multiply2x(x-2)=4(x+2)Opening the bracket2x^2-4x = 4x+8Collect like terms together2x^2-4x-4x-8=02x^2-8x-8=0Solving with completing the square method2x^2-8x=8Divide through with the coefficient of x^22x^2/2-8x/2=8/2X^2-4x=4Half of coefficient of x(-4 x 1/2)^2 = (-2)^2X^2-4x+(-2)^2=4+(-2)^2X^2-4x+(-2)^2=4+4(x-2)^2=8X-2:square root 8X=2 + or - squar root 8X=2+ square root 8 or x=2- square root 8X=2+2.828=4.828X=2-2.828= -0.828X=4:83 or - 0.83.
9a)
x 62 63 64 65 66 67 68tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i iifreq 1 5 12 14 10 6 2 total = 50fx 62 315 768 910 660 402 136 tot = 3253x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot = 92.82bi) mean = EFX/EF = 3253/50 =65.06bii) Standard deviation = Square root of EF(x-x)2/Ef= Square root of 92.82/50= square root of 1.8564= 1.36================
=========
11a)
3p+4q/3p-4q* 2/1 find p:q
1(rp+4q)= 2(3p-4q)
3p+4q=6p-8q
Collect like terms
3p-6p=-8q-4q
-3p=-12q
Ther4 p:q = -3p/-3=-12q/-3
P=4q, p=4, q=1
P:q= 4:1
11b)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
11bi.
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
=2pie(pq-4)
Perimeter=pq+4+4+2+2+2pie(pq-4)
34=pq+12+2pie(pq-4)
34-12=pq+2piePQ-8pie
22=PQ(1+2pie)-8pie
22=pq(1(2*22/7)-8*22/7
22=pq(1+44/7)-176/7
22=PQ(7+44/7)-176/7
22+176/7 =pq(51/7)
154+176/7=Pq(51/7)
330/7=pq(51/7)
330/51=pq
Pq=6.4706
11bii.
TS=PQ-4=6.4706-4=2.4706m
Area of semi circle TS=pieD
=7.7626m^2
Area of rectangle PqRU=|pu|*|Pq|
4**6.4706=25.8824
Area of the cross section=area of rectangle-area of semi circle
=25.8824-7.7626
=18.1198m^2
================================
================================
(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 15.00The total population from 2003 to2007=1.2*1.2*1.2*1.2*3000=6220.8Total grant=population * grant per head=6220.8*15=GH9331Total grants=GH93312(2a)1/x+(1/x+3)=1/2LCM=x(x+3)(x+3+x)/x(x+3)=1/22(2x+3)=x(x+3)4x+6=x^2+3xx^2+3x=4x+6x^2+3x-4x-6=0(x^2-3x)+(2x-6)=0x(x-3)+2(x-3)=0(x+2)(x-3)=0x=-2 or x=3(2b)Let the bag of rice be xLet the bag of beans be yx+y=17(eq1)2250x+2400y=39600(eq2)from (eq1)x=17-ysubstitute for x in eq22250(17-y)+2400y=3960038250+150y=39600y=(39600-3850)/150y=9therefore bags of beans=9substitute for 9 in eq1x+y=17x+9=17x=17-9x=8(3)Area of garden=L^217=(L+2)*(L+L)17=L^2+3L+2-17L^2+3L+2-17=0L^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 2.653L=5.654 p=4Lp=4(5.654)p=22.616m(3b)Area=L^2=5.654^2=31.98m^2Area of the path=L*b=2*1=2m^2(4)3^2+y^2=5^29+y^2=25y^2=25-9y^2=16y=sqroot16y=4therefore (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 diagram200degrees+32degrees+ydegrees=360degrees(angles at a point)ydegrees+232degrees=360degreesydegrees =360degrees-232degreesy=128degreesNdegrees=128degrees(alternative angles)xdegrees=128degrees+180degreesxdegrees=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+52(3x+7)=7x+56x+14=7x+56x-7x=5-14-x=-9x=9(6b)let the number of boys=x no of girls=5+x(x+5)/(x+2)=5/44(x+5)=5(x+20)4x+20=5(x+2)4x+20=5x+104x-5x=10-20-x=-10x=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^2PQ=25+x^2-10x+x^2therefore Area of the square=2x^2-10x+25If the area of PQRS=3/52x^2-10x+25=3/5*252x^2-10x+25=152x^2-10x=15-252x^2-10x+10=0divide through by 2x^2-5x+5Using 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)=d1+a=dn-da=d(n-1)-L2s=n(a+L)s=n(d(n-1)+L)-L/2s=n(dn-d+L)-L/2(8a)diagram(8+x)^2 = x^2+3264+16x+x^2= x^2 + 102416x=1024-64=960therefore 960/16= 60x=960/16=60therefore the radius = 60+8=68cm(8b)diagram(i)volume of a pyramid=1/3 AH2601= 1/3 * A * 27A=7803/27=289cm^3Area of square =289t^2= 289t= sqr rut(289)l=17cm(8bii)AC^2 = 17^2 +17^2AC^2 = 289 +289Ac^2 =578AC =sqr root (578)AC=24.04cmfor the triangele COPCO= 1/2 AC=1/2 * 24.04VC^2= 27^2 + 12.07VC^2= 929 +144.49VC= SQR root (873.48)=29.55cmcos tita = ADJ/hypcos x= 8.5/29.55cos x=0.2877x=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 lineCBA=52+xADC=180-(128-x)=52+xAlso BCD=180-x(angle on a straight line)DCQ=180-(180-x)DCQ=180-180+xDCQ=xx+52+x+76=1802x=180-52-762x/2=52/2x=26degrees10a)YX/XZ=XM/MZW/10=8/1515W=10*8W=5.33cm10bi)q^2=p^2+r^2-2prcos titaq^2=20^2+15-2*20*15 c0s 90q^2=400+225-0q^2=625q=sqroot625q=25km10bii)p/sinP=q/sinQ=r/sinR25/sin90=15/sinRsinR=15*1/25sinR=0.6R=sin^-1(0.6)R=36.86degreesThe bearing of p from R=90+90+90+alphaalpha=45-36.86=90+90+90+8.14=278.14=278degreesThe bearing of p fromR=278degrees
Wace Maths Answers
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COMPLETE MATHS OBJ:
1-10: BACBCCCBCD
11-20: CBBACACBBC
21-30: BBCCDACDCD
31-40: ACAABADAAA
41-50: DAABADCBAD.
=================================
9a)
x 62 63 64 65 66 67 68
tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i ii
freq 1 5 12 14 10 6 2 total = 50
fx 62 315 768 910 660 402 136 tot = 3253
x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94
(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436 f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot = 92.82
9bi) mean = EFX/EF = 3253/50 =65.06
9bii) Standard deviation = Square root of EF(x-x)2/Ef
= Square root of 92.82/50
= square root of 1.8564
= 1.36
No9)
================================
5a)
Area of triangle=1/2bh where h=6:. 1/2*6/1*b/1=366b/2=36/16b=72B=72/6Base of angle PSR=12cmSince |TS| //PQQR=PR-PQQR=12-8QR=4cm5b)Draw d triangleFrom angle ABC, to get |BC|Tan60/1=10cm/|BC||BC|=10.65/tan60|BC|=10.65/1.7321=6.1486 =6.15mHence |BC|=|DE|From angle AED, to get |AE|Draw d diagram|AB|=6.15tan456.15*1=6.15mThere4, th height of the tree =10.65-6.15=4.50m
=============
8a)
13a)(x1,y1) =(2,-3)for 2x + y =6y=-2x +6compare y = mx +6m2 = -2for parallel lines, m1=m2: y-y1/x-x1 = my--3/x-2 =-2/1y + 3 = -2(x -2)y + 3 = -2x + 4y = -2x + 4 -3y = -2x +1
==================
11a)3p+4q/3p-4q* 2/1 find p:q1(rp+4q)= 2(3p-4q)3p+4q=6p-8qCollect like terms3p-6p=-8q-4q-3p=-12qTher4 p:q = -3p/-3=-12q/-3P=4q, p=4, q=1P:q= 4:1
=========
6a)Sn=n/2(2a+(n-1)d)A=1, d=2, n=nSn=n/2(2*1+(n-1)2)N/2(2+2,-2)=n/2*2=n^2
(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
No6b)
(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
= (4a)
Pr(3)=x/total
total=25+30+x+28+40+32
=155+x
0.225=x/155+ x
=34.875+0.225x=x
x-0.225x=34.875
=0.775x=34.875 degree
x=34.875/0.775
=45
(4b)
Pr(even)=30+28+40+32
=90
Pr(even)=90/200=9/20
Pr(prime no)=25=30+45+40
=140
Pr(prime no)=140/200=14/20
Pr(even or prime)=9/20 + 14/20
(5a)
Area PSR=1/2*PR*SM
PR=8+6=14cm
36=1/2 * 14* 6
36=1/2 * 14* 6 *QR
QR=8.6cm
(5b)
diagram
sin45 degree= 10.65/h
h=10.65/sin 45 degree
=10.65/0.7071
h=15.06cm
(3b)
h=7cm
T=462cm^2
T=pie sqr + 2 pie rH
462= pie( r ^2 +2*7*r)
462/3.142
=r sqr= =14r
r sqr+ 14r=147
r sqr +14r- 147=0
r= 14+-( (sqr(14))-4*1*147)/2*1
r=-(14 +-(196-4))/2
r=(-14+_12)/2
r=(-14+12)/2 or
r=2/2
=1cm
(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)x-2/4 :x+2/2xCross multiply2x(x-2)=4(x+2)Opening the bracket2x^2-4x = 4x+8Collect like terms together2x^2-4x-4x-8=02x^2-8x-8=0Solving with completing the square method2x^2-8x=8Divide through with the coefficient of x^22x^2/2-8x/2=8/2X^2-4x=4Half of coefficient of x(-4 x 1/2)^2 = (-2)^2X^2-4x+(-2)^2=4+(-2)^2X^2-4x+(-2)^2=4+4(x-2)^2=8X-2:square root 8X=2 + or - squar root 8X=2+ square root 8 or x=2- square root 8X=2+2.828=4.828X=2-2.828= -0.828X=4:83 or - 0.83.
9a)
x 62 63 64 65 66 67 68tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i iifreq 1 5 12 14 10 6 2 total = 50fx 62 315 768 910 660 402 136 tot = 3253x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot = 92.82bi) mean = EFX/EF = 3253/50 =65.06bii) Standard deviation = Square root of EF(x-x)2/Ef= Square root of 92.82/50= square root of 1.8564= 1.36================
=========
11a)
3p+4q/3p-4q* 2/1 find p:q
1(rp+4q)= 2(3p-4q)
3p+4q=6p-8q
Collect like terms
3p-6p=-8q-4q
-3p=-12q
Ther4 p:q = -3p/-3=-12q/-3
P=4q, p=4, q=1
P:q= 4:1
11b)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
11bi.
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
=2pie(pq-4)
Perimeter=pq+4+4+2+2+2pie(pq-4)
34=pq+12+2pie(pq-4)
34-12=pq+2piePQ-8pie
22=PQ(1+2pie)-8pie
22=pq(1(2*22/7)-8*22/7
22=pq(1+44/7)-176/7
22=PQ(7+44/7)-176/7
22+176/7 =pq(51/7)
154+176/7=Pq(51/7)
330/7=pq(51/7)
330/51=pq
Pq=6.4706
11bii.
TS=PQ-4=6.4706-4=2.4706m
Area of semi circle TS=pieD
=7.7626m^2
Area of rectangle PqRU=|pu|*|Pq|
4**6.4706=25.8824
Area of the cross section=area of rectangle-area of semi circle
=25.8824-7.7626
=18.1198m^2
================================
================================
(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 15.00The total population from 2003 to2007=1.2*1.2*1.2*1.2*3000=6220.8Total grant=population * grant per head=6220.8*15=GH9331Total grants=GH93312(2a)1/x+(1/x+3)=1/2LCM=x(x+3)(x+3+x)/x(x+3)=1/22(2x+3)=x(x+3)4x+6=x^2+3xx^2+3x=4x+6x^2+3x-4x-6=0(x^2-3x)+(2x-6)=0x(x-3)+2(x-3)=0(x+2)(x-3)=0x=-2 or x=3(2b)Let the bag of rice be xLet the bag of beans be yx+y=17(eq1)2250x+2400y=39600(eq2)from (eq1)x=17-ysubstitute for x in eq22250(17-y)+2400y=3960038250+150y=39600y=(39600-3850)/150y=9therefore bags of beans=9substitute for 9 in eq1x+y=17x+9=17x=17-9x=8(3)Area of garden=L^217=(L+2)*(L+L)17=L^2+3L+2-17L^2+3L+2-17=0L^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 2.653L=5.654 p=4Lp=4(5.654)p=22.616m(3b)Area=L^2=5.654^2=31.98m^2Area of the path=L*b=2*1=2m^2(4)3^2+y^2=5^29+y^2=25y^2=25-9y^2=16y=sqroot16y=4therefore (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 diagram200degrees+32degrees+ydegrees=360degrees(angles at a point)ydegrees+232degrees=360degreesydegrees =360degrees-232degreesy=128degreesNdegrees=128degrees(alternative angles)xdegrees=128degrees+180degreesxdegrees=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+52(3x+7)=7x+56x+14=7x+56x-7x=5-14-x=-9x=9(6b)let the number of boys=x no of girls=5+x(x+5)/(x+2)=5/44(x+5)=5(x+20)4x+20=5(x+2)4x+20=5x+104x-5x=10-20-x=-10x=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^2PQ=25+x^2-10x+x^2therefore Area of the square=2x^2-10x+25If the area of PQRS=3/52x^2-10x+25=3/5*252x^2-10x+25=152x^2-10x=15-252x^2-10x+10=0divide through by 2x^2-5x+5Using 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)=d1+a=dn-da=d(n-1)-L2s=n(a+L)s=n(d(n-1)+L)-L/2s=n(dn-d+L)-L/2(8a)diagram(8+x)^2 = x^2+3264+16x+x^2= x^2 + 102416x=1024-64=960therefore 960/16= 60x=960/16=60therefore the radius = 60+8=68cm(8b)diagram(i)volume of a pyramid=1/3 AH2601= 1/3 * A * 27A=7803/27=289cm^3Area of square =289t^2= 289t= sqr rut(289)l=17cm(8bii)AC^2 = 17^2 +17^2AC^2 = 289 +289Ac^2 =578AC =sqr root (578)AC=24.04cmfor the triangele COPCO= 1/2 AC=1/2 * 24.04VC^2= 27^2 + 12.07VC^2= 929 +144.49VC= SQR root (873.48)=29.55cmcos tita = ADJ/hypcos x= 8.5/29.55cos x=0.2877x=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 lineCBA=52+xADC=180-(128-x)=52+xAlso BCD=180-x(angle on a straight line)DCQ=180-(180-x)DCQ=180-180+xDCQ=xx+52+x+76=1802x=180-52-762x/2=52/2x=26degrees10a)YX/XZ=XM/MZW/10=8/1515W=10*8W=5.33cm10bi)q^2=p^2+r^2-2prcos titaq^2=20^2+15-2*20*15 c0s 90q^2=400+225-0q^2=625q=sqroot625q=25km10bii)p/sinP=q/sinQ=r/sinR25/sin90=15/sinRsinR=15*1/25sinR=0.6R=sin^-1(0.6)R=36.86degreesThe bearing of p from R=90+90+90+alphaalpha=45-36.86=90+90+90+8.14=278.14=278degreesThe bearing of p fromR=278degrees
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