# WAEC 2019 Mathematics Obj & Essay Answer Now Available

UPCOMING NEXTWAEC 2019 Economics Obj And Essay Answer Now Available

Mathematics OBJ:
11-20: BCBCBBAACD
21-30: DCCBCCABBA
31-40: AABDDCDDCC
41-50: BBCDCCBCCD

COMPLETED

Question 1.
110ₓ = 40₅
change all to base ten.
110ₓ = 40₅

1x²+1x ¹+0x⁰ = 4×5¹+0×5⁰
x² + x + 0 = 20
x² + x – 20 = 0

x² + 5x – 4x – 20 = 0
x(x+5) – 4(x+5) = 0
(x-4)(x+5) = 0
Either x-4 = 0 or x + 5 = 0
x = -5 or 4
therefore, x= 4

1b)
15/√75 + √108 + √432
= 15/√253 + √363 + √144×3
= 15/5√3 + 6√3 + 12√3
= 3/√3 + 18√3
= 3√3/3 = 18√3
= √3 + 18√3
=19√3

Question 2.
M = -3-7/2+2 = m = (y₂-y₁/-10)/4
m = -5/2

Equation line = y – y₁ = m(x-x₁)
= y – 7 = -5/2(x+2)
= 2y – 14 = -5(x+20)
= 2y – 14 = -5x – 10
5x + 2y – 14 + 10 = 0
5x + 2y – 4 = 0

2b) 5b -a /8b + 3a = 1/5
5b – a = 1 * 3
8b + 3a = 5
15a – 3b = 3
8b + 3a = 5/23b = 8
b = 8/23 = 0.35
using equation (1)
5b – a = 1
5(0.35) – a = 1
1.75 – a = 1
-a = 1 – 1.75
-a = -0.75 (- divide -)
a = 0.75
a/b = 0.75/0.35
a/b = 2.14

Question 3.
Ali : Musah : Yusif = ₦420,000
3 : 5 : 8
Sum of ratio shared;
3 + 5 + 8 = 16

therefore, Ali share = 3/16 * ₦420,000
= ₦78,750

Musah share = 5/16 * ₦420,000
= ₦131,250

Yusif share = 8/16 * ₦420,000
= ₦210,000

therefore, sum of Ali + Yusif = ₦78,750 + ₦210,000
=₦288,750

3b)
Solve: 2(1/8)^x = 32^x-1
2^1 x (8^-1) = 32^x-1
2^1 x (8^-1) = 32^x-1
2^1 x (2^3(-1))^3 = 2^5(x-1)
2^1 x (2^-3)^x = 2^5x-5
2^1 x 2^-3x = 2^5x-5
2^1+(-3x) = 2^5x-5
2^1-3x = 2^5x-5 (2 cancel 2)
1-3x = 5x -5
-3x – 5x = -5 -1
-8x/-8 = -6/-8
x = 3/4

========

(4)
Since (5a)
No of red balls = 3
No of green balls = 5
No of blue balls = x
Prob.(red ball) = no of total outcome/no of possible outcome
Pr(red) = 3/3+5+x = 1/6
3/8+x = 1/6
6(3) = 1(8+x)
18 = 8 + x
X = 18 – 8 = 10
Therefore the no of blue ball = 10

(5b)
Probability of picking a green ball
P(g) = no of green balls/no of possible outcome
P(g) = 5/3+5+10 = 5/18
=5/18

(6ai)
F α M1M2/d²
F = KM1M2/d²
Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m
20 = k(25)(10)/5²
250k = 500
k = 500/250 = 2
Expression is
F = 2M1M2/d²

(6aii)
Making d subject
d = √2M1M2/F
d = √2 ×7.5×4/30
d = √60/30 = √2
d = √2m or 1.41m

(6b)
Draw the diagram
X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)
5x + 200 = 540
5x = 540 – 200
5x = 340
X = 340/5
X = 68

(8a)
1/3x – 1/4(x+2)>_ 3x -1⅓
1/3x – 1/4(x+2)>_3x – 4/3
Multiply through by the L. C. M(12), we have
4x – 3(x + 2)>_36x – 16
4x – 3x – 6 >_ 36x – 16
-6+16 >_36x + 3x – 4x
10 >_ 35x
35x _< 10
X = 10/35
X = 2/7

(8bi)
Draw the triangle
|AB|/66 = sin35
|AB| = 66sin35 = 66×0.5736 = 37.8576

Draw the right angled triangle
|AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m
Height of tower = 48.45m

(11ai)
ar² = 1/4 ……(1)
ar^5= 1/32 …..(2)
Divide eqn (2) by eqn(1)
ar^5/ar² = 1/32÷1/4
r³ = 1/32 × 4/1
r³= 1/8
r³ = 2-³
r = 2-¹
r = 1/2
Common ratio = 1/2
Put this into eqn (1)
a(1/2)² = 1/4
a(1/4) = 1/4
a = (1/4)/(1/4) = 1
First term, a = 1

(11aii)
Seventh term, T7 = ar^6
=(1)(1/2)^6
=1/64

(11b)
Given : X = 2 and X = -3
(X – 2)(X + 3) = 0
X² + 3x – 2x – 6 , 0
X² + x – 6 = 0
Comparing with ax²+bx+c = 0
a = 1
b = 1
C = -6

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