Waec Real Expo: 2015/2016 Mathematics Essay / Theory and OBJ Questions and Answers Now Available
100% Correct WAEC 2015 MATHEMATICS ANSWERS
MATHEMATICS OBJ AND ESSAY
MATHEMATICS-THE ORY-ANSWERS (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} (3a)
Let p=1/x ,q=1/y
p+q=5 —->equ(1)
q-p=1 —->equ(2)
p+q=5 —->equ(1)
p+q=1 —->equ(2) Add equ(1) and equ(2)
2q=6 , q=6/2=3
Substitute q=3 into equation(1)
p+q=5
p+3=5
P=5-3 p=2
Recall p=1/x , 2=1/x, x=1/2
q=1/y, 3=1/y , y=1/3
Therefore: x=1/2 , y=1/3 (3b)
Where the road is good
72=x/0.75
x=72*0.75
x=54km
Where the road is bad 48=x/0.75
x=48*0.75
x=36km
Number of kilometers of good surface is
54km
[10:15am, 4/23/2015] Emmzy: MATHEMATICS-THEORY-ANSWERS
(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
(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= { }
=======================
(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 (-8) = -8d
-24 = -8d
d= 24/8
d = 3
======================= More Answers Here
MATHEMATICS OBJ AND ESSAY
MATHEMATICS-THE ORY-ANSWERS (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} (3a)
Let p=1/x ,q=1/y
p+q=5 —->equ(1)
q-p=1 —->equ(2)
p+q=5 —->equ(1)
p+q=1 —->equ(2) Add equ(1) and equ(2)
2q=6 , q=6/2=3
Substitute q=3 into equation(1)
p+q=5
p+3=5
P=5-3 p=2
Recall p=1/x , 2=1/x, x=1/2
q=1/y, 3=1/y , y=1/3
Therefore: x=1/2 , y=1/3 (3b)
Where the road is good
72=x/0.75
x=72*0.75
x=54km
Where the road is bad 48=x/0.75
x=48*0.75
x=36km
Number of kilometers of good surface is
54km
[10:15am, 4/23/2015] Emmzy: MATHEMATICS-THEORY-ANSWERS
(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
(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= { }
=======================
(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 (-8) = -8d
-24 = -8d
d= 24/8
d = 3
======================= More Answers Here
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