Member

idexpaul1

Re- waec may/June assistance

(Apr 23 2015 at 07:34am)

Bring it out let start solving . don't waste time bro

Member

merrylarry

Re- waec may/June assistance

(Apr 23 2015 at 09:08am)

1a)
= 3 4/9 ÷ (5 1/3 - 2 3/4) + 5 9/10
= 31/9 ÷ Lcm (3 4-9/12 ) + 5 9/10
= 31/9 ÷ 3 -5/12 + 5 9/10
= 31/9 ÷ 2 7/12 + 5 9/10
= 31/9 ÷ 31/12 + 5 9/10 = 31/9 x 12/31 + 5 9/10
= 4/3 + 59/10
Find the Lcm
= 40+177/30
= 217/30
= 7 7/30 1b)
A { 2,3,4 }
B { 1,3,5 }
= Possibility are { 1,2 }, {2,3}, {3,1},{3,3},
{3,5},{4,1},
{4,3},and {4,5} = Prob! Of sum greater than 3 = 8/9
= Prob of sum less than 7 = 5/9
= Prob of sum greater than 3 and less
than 7 = 8/9
x 5/9
= 40/81 2a)
4 + 3/4 (x+2) < eqaul to 3/8x + 1
Multiply through by 8
= 8(4) + 3(2)(x+2) < eqaul to 3x + 8(1)
= 32 + 6(x+2) < eqaul to 3x + 8
= 32 + 6x + 12 < eqaul to 3x + 8 = 6x - 3x < eqaul to 8 - 44
= 3x < eqaul to -36
= x < eqaul to -36/3
= x < eqaul to -12
2b) Area of the shaded portion = total
area - area of the removed square
484 = 20(20+x) - x^2
484 = 400 + 20x - x^2
X^2 - 20x + 84 = 0
X^2 - 6x - 14x + 84 = 0
X(x-6) - 14 ( x-6 ) = 0 (X-14) , (x-6) = 0
X=14 or 6
Ie x= 6cm or 14cm
3a) The ratio of interior angle to.
Exterior angle =
6 : 2 interior angle = 5/7 x 180/1
= 900/7
= 128 4/7
Exterior angle = 2/7 x 180/1
= 360/7
= 51 3/7 Number of side = 360/exterior angle
= 360 / 51 3/7 = 6.9999 approx!! 7
The number of side of the polygon is 7
9a)
a= -8
a + 6d : a + 8d 5:8
a+6d/a+8d = 5/8
8a + 48d = 5a + 40d
8a - 5a = 40d - 48d
3a = -8d
3 (- = -8d -24 = -8d
d= 24/8
d = 3
10a)
2tan60 + cos 30/sin 60
= 2 ( square root 3/1 + square root 3/2 )/ square
root 3/2
Open the bracket
= 2 square root 3/1 + square root. 3/2 /
square
root. 3/2 = 4 square root 3 + square root 3/2 x 2/
square root
3
= 8 square root 3 + 2 square root
3/2square root 3
= 2 ( 4square root 3 + square root 3 ) / 2square
root 3
= 4 square root 3 + square root 3 /
square root 3 x
square root 3 / square root 3
= 4(3) + 3 / 3 = 15/3
= 5
10b)
Using < ACE
Tan Θ = opp/ adj
Tan 41 degree = CE / 1050 CE = 1050 (tan 41)
CE = 1050 ( 0.8693 )
CE = 912.751m
Using. < BCD
Tan 36/1 =

Member

bigDoc

Re- waec may/June assistance

(Apr 23 2015 at 09:11am)

12) Vol of. The reservoir = vol of cone + vol
of hem
= 1/3 pi r^2 h + 2/3 pi r^3 where r =x
333 1/3 = 1/3 pi x^2 6x + 2/3 pi x^3
1000/3 = 2 pi x^3 + 2/3 pi x^3
1000/3 = 2 2/3 pi x^3
1000/3 � 8/3 = pi x^3
1000/3 x 3/8 = pi x^3
7000/176 = x^3
X^3 = 39.77
X= 3sqare root 39.77
X = 3.41
Radius = 3.41m
(12bI) Vol of hemsphere = 2/3 pi r^3
Vol = 2/3 x 22/7 x ( 3.41)^3
V = 2/3 x 22/7 x 39.77
V = 1749.88/21
V = 83.33m
(12bII) Total surface area of the reservoir =
curved surface area of the cone + curve
surface area of the hempsere
TSA = 22/7 x R x L + 2 x 22/7 x r^2
L^2 = (20.46)^2 + (3.41)^2
L^2 = 418.61 + 11.63
L^2 = 430.24
L = square root ( 430.24)
L= 20.74
TSA = 1555.91/7 + 511.54/7
= 2067.55/7
= 295.36m

Member

bigDoc

Re- waec may/June assistance

(Apr 23 2015 at 09:12am)

1a)
= 3 4/9 � (5 1/3 - 2 3/4) + 5 9/10
= 31/9 � Lcm (3 4-9/12 ) + 5 9/10
= 31/9 � 3 -5/12 + 5 9/10
= 31/9 � 2 7/12 + 5 9/10
= 31/9 � 31/12 + 5 9/10
= 31/9 x 12/31 + 5 9/10
= 4/3 + 59/10
Find the Lcm
= 40+177/30
= 217/30
= 7 7/30
1b)
A { 2,3,4 }
B { 1,3,5 }
= Possibility are { 1,2 }, {2,3}, {3,1},{3,3},
{3,5},{4,1},{4,3},and {4,5}
= Prob! Of sum greater than 3 = 8/9
= Prob of sum less than 7 = 5/9
= Prob of sum greater than 3 and less than 7
= 8/9 x 5/9
= 40/81
2a)
4 + 3/4 (x+2) < eqaul to 3/8x + 1
Multiply through by 8
= 8(4) + 3(2)(x+2) < eqaul to 3x + 8(1)
= 32 + 6(x+2) < eqaul to 3x + 8
= 32 + 6x + 12 < eqaul to 3x + 8
= 6x - 3x < eqaul to 8 - 44
= 3x < eqaul to -36
= x < eqaul to -36/3
= x < eqaul to -12
2b) Area of the shaded portion = total area
- area of the removed square
484 = 20(20+x) - x^2
484 = 400 + 20x - x^2
X^2 - 20x + 84 = 0
X^2 - 6x - 14x + 84 = 0
X(x-6) - 14 ( x-6 ) = 0
(X-14) , (x-6) = 0
X=14 or 6
Ie x= 6cm or 14cm
3a) The ratio of interior angle to. Exterior
angle = 6 : 2
interior angle = 5/7 x 180/1
= 900/7
= 128 4/7
Exterior angle = 2/7 x 180/1
= 360/7
= 51 3/7
Number of side = 360/exterior angle
= 360 / 51 3/7 = 6.9999 approx!! 7
The number of side of the polygon is 7
11a)
H = mt/d(m+p) find M
= cross multiply
= mt = hd(m+p)
= mt = hdm + hdp
= mt - hdm = hdp
= m(t-hd) = hdp
= m = hdp/t-hd
or
= m = hdp/t-dh
11b)
From Angle WXM
< WXM = 90 degree ( angle at the center is
twice < at circumference )
< WMX = 180degree - (90 - 48)degrees (sum
of < in a triangles)
< WMX = 42
< WMX + < XMZ - 180 ( < on a straight lines)
< XMZ = 180 - 42 = 138
< WYZ + < XMZ = 180
< WYZ = 42 degree
9a)
a= -8
a + 6d : a + 8d
5:8
a+6d/a+8d = 5/8
8a + 48d = 5a + 40d
8a - 5a = 40d - 48d
3a = -8d
3 (-8) = -8d
-24 = -8d
d= 24/8
d = 3
9b)
30p + 100m = 2450 ................ (I)
40/100p + 30/100m = 855 .................(II)
From equation 1
3p + 10m = 245 ................ (III)
From equation 2
4/10p + 3/10m = 855
Multiply through by 10 to give 2
4p + 3m = 8550 ............... (Iv)
Multiply (III) by 4 and. (Iv) by 3
12p + 40m = 980
12p + 9m = 25
10a)
2tan60 + cos 30/sin 60
= 2 ( square root 3/1 + square root 3/2 )/
square root 3/2
Open the bracket
= 2 square root 3/1 + square root. 3/2 /
square root. 3/2
= 4 square root 3 + square root 3/2 x 2/
square root 3
= 8 square root 3 + 2 square root 3/2square
root 3
= 2 ( 4square root 3 + square root 3 ) /
2square root 3
= 4 square root 3 + square root 3 / square
root 3 x square root 3 / square root 3
= 4(3) + 3 / 3
= 15/3
= 5
10b)
Using < ACE
Tan Θ = opp/ adj
Tan 41 degree = CE / 1050
CE = 1050 (tan 41)
CE = 1050 ( 0.8693 )
CE = 912.751m
Using. < BCD
Tan 36/1 = CD/1050
CD = 1050 (tan 36)
CD = 762.8697m
(I) Height of control tower = CE - DE
= 912.751 - 762.8697
= 149.8813 Approx! 150m
(II) Using < ACE
Cos Θ = adj/hyp
Cos 41 = 1050/AC
AC = 1050/cos 41
AC = 1391.2636 aprox! 1391m
The shortest distance = AC
11a)
H = mt/d(m+p) find M
= cross multiply
= mt = hd(m+p)
= mt = hdm + hdp
= mt - hdm = hdp
= m(t-hd) = hdp
= m = hdp/t-hd
or
= m = hdp/t-dh
11b)
From Angle WXM
< WXM = 90 degree ( angle at the center is
twice < at circumference )
< WMX = 180degree - (90 - 48)degrees (sum
of < in a triangles)
< WMX = 42
< WMX + < XMZ - 180 ( < on a straight lines)
< XMZ = 180 - 42 = 138
< WYZ + < XMZ = 180
< WYZ = 42 degree
11c)
Operation Table
* | 1 | 3 | 5 | 6
1 | 4 | 6 | 1 | 2
3 | 6 | 1 | 3 | 4
5 | 1 | 3 | 5 | 6
6 | 2 | 4 | 6 | 0
I= {5}
II= { }
Nice job merrylarry

Member

merrylarry

Re- waec may/June assistance

(Apr 23 2015 at 09:30am)

(12bII) Total surface area of the
reservoir = curved surface area of the
cone + curve surface area of the
hempsere
TSA = 22/7 x R x L + 2 x 22/7 x r^2
L^2 = (20.46)^2 + (3.41)^2 L^2 = 418.61 + 11.63
L^2 = 430.24
L = square root ( 430.24)
L= 20.74
TSA = 1555.91/7 + 511.54/7
= 2067.55/7 = 295.36m

Member

idexpaul1

Re- waec may/June assistance

(Apr 23 2015 at 09:58am)

Thks

Member

merrylarry

Re- waec may/June assistance

(Apr 23 2015 at 10:37am)

(9a)
First term(a)=-8
7th term = a+6d
9th term=a+8d
(a+6d)/(a+8d)=5/8
8*(a+6d)=5*(a+8d) 8a+48d=5a+40d
8a-5a=-48d+40d
3a=-8d
3(-8)=-8d
d=3
Common difference(d)=3 (9b)
No of pawpaw=30
No of mangoes=100
Cost price(CP) for P&M=N2450
Profit=40%(pawpaw)
Profit=30%(mangoes) Overall profit= N855:00
(i)
Profit%=(Sp-Cp)/Cp
70%=Sp-2450/2450
1715=Sp-2450
Sp=1715+2450 =N4165
(ii) Cost price of a basket of pawpaw
=N4165/30
=N138.8
(iii)
=N4165/100 =N416.5