2019 WAEC Mathematics Questions And Answers

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Maths Obj

Maths Obj

MATHS OBJ By CODEDβ„’
1-10: BABDDACCBC
11-20: ACBCBBAACD
21-30: DCCBCCABBA
31-40: AABDDCDDCC
41-50: BBCDCCBCCD

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(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

(8a)
1/3x – 1/4(x+2)>_ 3x -1β…“
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 = 66Γ—0.5736 = 37.8576

Draw the right angled triangle
|AD|/|AB| = Tan52
|AD| = 37.8576 Γ— Tan52Β° = 37.8576 Γ— 1.2799 = 48.45m
Height of tower = 48.45m

(8a)
1/3x – 1/4(x+2)>_ 3x -1β…“
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 = 66Γ—0.5736 = 37.8576

Draw the right angled triangle
|AD|/|AB| = Tan52
|AD| = 37.8576 Γ— Tan52Β° = 37.8576 Γ— 1.2799 = 48.45m
Height of tower = 48.45m

(11ai)
arΒ² = 1/4 ……(1)
ar^5= 1/32 …..(2)
Divide eqn (2) by eqn(1)
ar^5/arΒ² = 1/32Γ·1/4
rΒ³ = 1/32 Γ— 4/1
rΒ³= 1/8
rΒ³ = 2-Β³
r = 2-ΒΉ
r = 1/2
Common ratio = 1/2
Put this into eqn (1)
a(1/2)Β² = 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

(1b)
Given : X = 2 and X = -3
(X – 2)(X + 3) = 0
XΒ² + 3x – 2x – 6 , 0
XΒ² + x – 6 = 0
Comparing with axΒ²+bx+c = 0
a = 1
b = 1
C = -6

(4)
Since <PQR = <PRS = 90Β°
Using Pythagoras theorem
|PR|Β² = |PQ|Β² + |QR|Β²
|PR|Β² = 3Β² + 4Β²
|PR|Β² = 9 + 16
|PR|² = 25 PR = √25
|PR| = 5cm
Considering PRS
|PS|Β² = |PR|Β²+|SR|Β²
13Β² = 5Β² + |SR|Β²
169 = 25 + |SR|Β²
|SR|Β² = 169 – 25
|SR|Β² = 144
|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS
= 1/2bh + 1/2bh
= 1/2Γ—4Γ—3 + 1/2Γ—12Γ—5
= 6+30 = 36cm
-1a
1+4x/2 – 5+2x/7 < x-2
The LCM
7(1-4x)-2(5+2x)/14 < x-2/1
Cross and multiply
7-28x-10-4x+1<(x-2)
7-28x-10-4x<14x-28
CLT
7-10+28<14x+4x+28x
25<46x
Divide both sides by x
25/46<46x/46
X<25/46
(3b)
2(1/8)^x=32^x-1
2(1/2^3)^x=2^5(x-1)
2(2^-3)^x=2^5(x-1)
2^1X2^-3x=2^5(x-1)
~2~ ^1-3x= ~2~ ^5x-5
1-3x=5x-5
-3x-5x=-5-1
-8x=-6
x=-6/-8
x=3/4

(12a)
Given : siny = 8/17
Draw the right angle
From Pythagorean triple, third side is 15
Draw the right angle triangle
tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)
Amount shared = #300,000
Otobo’s share = #60,000
Ade’s share = 5/12 Γ— #(300,000-60,000)
= 5/12 Γ— #240,000
=#100,000

Adeobi’s share = #300,000 – (#60,000 + #100,000)
= 300,000 – 160,000
=#140,000

Ratio : Otobo : Ade : Adeola
60,000 : 100,000 : 140,000
60 : 100 : 140
6 : 10 : 14
3 : 5 : 7

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2019 WAEC MATHEMATICS OBJ AND THEORY QUESTIONS AND ANSWERS

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