5a)
Area of trapezium =1/2(a+b)*h
=1/2(20+60) * 40
=40*40
=1600cm^2

5b)
<QRS=90degrees(right angle)
=======================
4a)
Sum of 3(2/3) and 2(1/5)
=3(2/3) + 2(1/5)
=11/3 + 11/5
=(55+33)/15 =88/15
=5.87
Amount=7-5.87
=1.13
=7-5(13/15)
=7-(88/15) =17/15

4b)
i)h=6+4cos(15p)degrees
h=6+4cos(15*4)
h=6+4cos60degrees
h=6+(4*0.5) h=6+2
=8m
ii)9=6+4cos(15p)degrees
4cos(15p)degrees=3
Cos15p=3/4
=0.75 15p=Cos^-1(0.75)
15p=41.4
P=41.4/15
=2.76

(6ai)
I:yes
II:NO
III:NO

(6b)
Cost(c)=h+y
c=h+y h=fixed amount
y=variable
y is directly proportional to n=>Y=kn
C1=GHc 950.00
n1=600 bricks
c2=GHc 1030 n2=1000 bricks
c=h+kn
950=h+600k=> eq(1)
1030=h+1000k=> eq(2) from eq(2) and eq(1)
substract eq(1) from eq(2)
1030-950=h-h+1000k-600k
80=400k
k=80/400
k=8/40=1/5
=0.2
substitute k=0.2 into eq(1) 950= h+ 600k
950=h +6000(0.2)
950=h+120
h=950-120,
h=830
h=830, k=0.2 c=h+kn
c=830+0.2n

(6bi)
C = h + y
C = h + kn

(6bii)
950 = h + 600k..........(I)
1030 = h + 1000k .......(II)
80= 400k
K=80/400
K= 1/5 Sub for k in I
950 = h + 600/5
950 = h + 120
H = 830 The cost of producing 500 bricks is C= 830 + 500/5
C= 830 + 100
C= 930
C=GHe 930

=======================
(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
=======================
(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

3b) 74+SPR+40=180(sum of 114degrees + SPR =180 SPR=180-114 SPR=66degrees

=======================

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= { }

=======================

(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

=======================

(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

Use this one.
(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

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