2018 Waec Gce Mathematics Answers - Nitegist

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PLEASE TAKE NOTE OF THESE COMPUTER SYMBOLS

(1) / means division or divide
Examples:
2/2 means 2/2
2 whole no 3/4 means 2¾
M=wp+(1/3)yp^2 means m=wp + ¾y 2
(2) Log means logarithm
Example: (8) Log5 (base 8) means Log8 5
(3) * means multiplication
Example: 2*2 means 2×2
(4) ^ means Raise to power
Examples:
3^(-1) means 3 -1
5^(2) means 5 2
30cm^(-2) means 30cm -2
(5) Tita means θ
Example: Tan tita means tan θ
(6) Pie means π
Example: Pie R sqr h means πr 2 h
(7) Base means subscript
Example: 2 Base 3 means A 3
(8) Sqr root means √
Examples:
√3-√6+2√3 means Root 3-root6+2root 3
√(¾) means sqr root(3/4)
(9) bar means a dash on top a number of letter
Example: X bar means x̅
(10) —(1) means equation 1
(11) Proportional means ∝
Example: R is proportional to 3/4rut m means R
∝ ¾
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MATHS OBJ:
1-10: DCBBDCCADB
11-20: ACBACCDADA
21-30: DAAAABCBBB
31-40: ABCDABCDCA
41-50: DBABCBABBB
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STUDY THE MATHEMATICS SYMBOLS

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(11ai)
Completing the table:
|(+)|2|3|4|
| 2 |4|0|1|
| 3 |0|1|2|
| 4 |1|2|3|
|(x)|2|3|4|
| 2 |4|1|3|
| 3 |1|4|2|
| 4 |3|2|4|
(11aii)
(a)
4(x)e (+)2 = 1(mod 5)
4e + 2 = 6
4e = 6 - 2
4e = 4
e = 4/4 = 1
(b)
4(+)n(x)2 = 2(mod 5)
4(+)n(x)2 = 12
4(+)n(x)2 = 12
n(x)2 = 12 - 4
n(x)2 = 8
2n = 8
n = 8/2
n = 4
===============================
===============================
(8a)
(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)
When the surface diameter drops to 14cm, radius r = 7cm
Volume of water in the tank
V = 1/3 × 22/7 × 7 × 7 × 9/7
Draw the cone
h/H = r/R
h/9 = 7/21
h = 9 × 7/21 = 3cm
Hence V = 1/3×22/7×7×7×3
V = 462cm³
Volume of water drawn out
4158 - 462
= 3696cm³
===============================
(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°
(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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(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
===============================
(4a)
4y = 6y - 2x + 1
2x - 2y = 1
2x - 2y = 1 ------(1)
2y + X + 6y - 2x + 1 + 4y = 28
12y - X = 28 -----(2)
(1) × 1 2x - 2y = 1
(2) × 2 24y - 2x = 56
Add
22y = 57
y = 57/22 = 3½
Substitute for X in (1)
2x - 2(n/2) = 1
2n = 1 + 7,
2n = 8
ń = 8/2 = 4
(4b)
PQ = 2y+X=2(7/2)+4 = 11cm
QR = 4y = 4(7/2) = 14cm
PR = 6y - 2n + 1 = 6(7/2) -2(4) + 1
= 21 - 8 +1
= 14cm
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(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)
===============================
(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°
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(1a)
S = ½n[a + L]
Where S = 130
S = n/2[a+a+(n-1)d]
2S = n[2a+(n-1)d]
2(130) = 10[2a + (10 - 1)d]
260 = 10[2a + 9d]
260 = 20a + 90d .....(I)
a + 4d = 3a...... (II)
4d = 3a - a
4d = 2a
2d = a...... (III)
Substitute a = 2d into Eqn (I)
260 = 20(2d) + 90d
260 = 40d + 90d
260 = 130d
260/130 = 130d/130
d = 2
Hence common difference 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
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6)
(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•