2018 WAEC GCE GENERAL MATHEMATICS OBJECTIVE & ESSAY ANSWER NOW AVAILABLE.

Maths – Obj
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11ACBACCDADA
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31ABCDABCBCA
41DCABCDABBB

1a)
sn=n/2{2a+(n-1)d}
n=10, sn=130
130=10/2{2a+(10-1)d}
130=5(2a+ad)
130=10a+45d———(eq1)

Tn=a+(n-1)d
T5=3*a=3a,n=5
3a=a+(5-1)d
3a=a+4d
3a-a-4d=0
2a-4d=0———-(eq2)

Solve equation 1 and 2 simultaneously
10a+45d=130*1
2a-4d=0*5
10a+45d=130
10a-20d=0
65d/65=130/65
d=2

(1b)
First term a = 2d = 2(2) = 4

(1c)
L = a+(n-1)d
28 = 4+(ń – 1)2
28 = 4+2n – 2
2n = 28 – 2
2n = 26
2n/2 = 26/2
n = 13
=========================

(2a)
Draw the triangle
Using Pythagoras rule
a² = b² + c²
5² = 3² + c²
25 = 9 + c²
c² = 25 – 9
c² = 16
C = √16 = 4, Cos X = 4/5
5(cosx)² – 3 = 5((4/5))² – 3
= 5(4/5 × 4/5) – 3
= 16/5 – 3/1 L. C. M = 5
= 16 – 15/5 = 1/5.

(2b)
Volume of a pyramid
=1/3b×L×h
Volume of a cone = 1/3πr²h
Vp = volume of pyramid
Vc = volume of cone
Hp = height of pyramid
Hc = height of cone
Vp = Vc; Hp = Hc
Vp ± Vc
1/3bLh = 1/3πr²h
42 × 11 = 22/7r²
21 = 1/7 × r²
r² = 21 × 7
r² = 147
r = √147
r = 12 3/25cm

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

(5a)
5 – X > 1 9 + X >_ 8
5 – 1 > X X >_ 8 – 9
6 > X X >_ -1
Range is
-1_< X < 6 OR 6 >X >_ -1

(5b)
PQR + PSR = 180(supplementary angles of a cyclic quad)
PQR + 56 = 180
PQR = 180 – 56
PQR = 124°
Next, join P to R
QRP = QPR(base angles of an isosceles)
PQR + 2QRP = 180(Sum of angles in a triangle)
124 + 2QRP = 180
2QRP = 56°
QRP = 56/2 = 28°
PRS = 90°(angle in a semi – circle)
= 28 + 90
= 118°
===============================

(4a)
(2y+x) + (6y-2x+1) + 4y = 28 …(i)
6y-2x + 1 = 4y… (II)
2y+ x +6y – 2x + 1 +4 = 28
12y + x +6 -2x + 1 + 4 = 28
12y – x + 1 = 28
12y – x = 29… (III)
6y-2x + 1 = 4y, 6y-2x – 4y = 1
2y – 2x = -1… (iv)
24y – 2x = 54
2y – 2x = 1

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22y/2w = 55/22 y = 2.5
12y – x = 27
12 (2.5) – x = 27
30 x 27 x=3

(b)
2y + x = 2 (2.5) + 3 = 5+3 = 8cm
6y – 2x + 1 = (2.5)-2 (3) + 1
= 15-6 + 1 = 10cm
4y = 4(2.5) = 10cm
=============================
(6ai)
The profit y = X²/8 + 5x
y = GHc20,000.00
Hence 20,000 = X²/8 + 5x
160,000 = X² + 40x
X² + 40x – 160,000 = 0
Since X is in thousands
X² + 40x – 160 = 0

(6aii)
Using quadratic formula
X = -b±√b² – 4ac/2a
Where; a = 1, b = 40 & c = -160
X = -40±√40² – 4(1)(-160)/2(1)
X= -40 ±√1600 + 640/2
X = -40 ±√2240/2
X = -40 ± 47.32/2
X = -40±47.32/2
= 7.32/2
X = 3.66
X ≈ 4

(6b)
Draw the diagram
Using ΔTOP
tan 28 = H/OP
OP = H/tan28

Then for ΔROP
tantita = H/2/OP
OP = H/2/tantita
Hence H/tan 28 = H/2/tantita
tantita = H/2 × tan 28/H
tantita = tan28/2
Hence Tita = 28/2 = 14•
=========================
(7a)
2
S(2x³ – 4x + 6)dx
1
= 2x³+¹/3+1 – 4x¹+¹/1+1 + 6x]2
1

=2x^4/4 – 4x²/2 +6x]2, 1
= x^4/2 – 2x² + 6x]2, 1
=(2^4/2 – 2(2²) + 6(2)) – (1^4/2 – 2(1)² + 6(1))
=(8 – 8 + 12) – (1/2 – 2 + 6)
=12 – 4½
= 7½

(7b)
Given; P^-1 = (-1 1)
(4 -3)
P = (p-1)^-1 = C^T/|p^-¹|
=(-3 -1)
(-4 -1)/3 – 4
=(-3 -1)
(-4 -1)/-1
=(3 1)
(4 1)
==============================

(8)
(I) Draw The Diagram

(II)
V = 1/3 Ah, = x r²
V = 1/3 xr²h
V = 4.158 liters
V = 4.158cm³, V = 1/3 xr²h
4158 = 1/3 x 22/7 x 21 x 21 x h
4158 x 3 = 22 x 63h
h = 4158/21 x 22 = 9cm
h = 9cm

(8b)
d = 28cm, r = d/2 = 28/2 = 14cm
V2 = 1/3 xr²h
V2 = 1/3 x 22/7 x 14 x 14 x 9
V2 = 1/2 x 22/1 x 2 X 14 x 3
= 1848cm³
V2 = 1.845 liters
=============================

(12a)
Draw the triangle

(12bi)
Using sine rule
285/sin Z = 307/sin 78
Sin Z = 285sin78/307
sin z = 285 × 0.9781/307
sin z = 278.7585/307
sin z = 0.9080
z = sin-¹(0.9080)
z = 65.23°
Bearing of X from Z = 270+(90 – 65.23)
Bearing of X from Z =270+24.77
Bearing of X from Z = 294.77
Bearing of X from Z = 295°

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(12bii)
YXZ = 180 -(78+65.23) sum of angles in a triangle
180 – 143.23
YXZ = 36.77°
Using sine rule
YZ/sin 36.77 = 307/sin78
YZ = 307sin36.77/sin78
YZ = 307 × 0.5986/0.9781
YZ = 187.88
YZ = 188m

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