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WAEC 2019 Mathematics Theory and Obj Questions and Answers

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WAEC 2019 Mathematics Theory and Obj Questions and Answers by : 12:00 am On April 14, 2019

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MATHS OBJ
1-10: CABDBACCBC
11-20: BCBCBBDACD
21-30:ACCBCCABBA
31-40:AADDDCBBCC
41-50: BBCDCCACCD
==============================
PLEASE USE THIS NUMBER 2
(2a)
The equation of the line through the points
A(-2,7) and B(2,-3)
Using the equation Y=mx+b
Where m=slope of the gradient, b=the intercept at the vertical axis
Hence slope=Change in Y/Change in X
=>Y2-Y1/X2-X1=Y-y1/X-x1
=(-3-7)/(2–2)=Y-7/X+2
=-10/4=(Y-7)/(X+2)
=4(y-2)=-10(x+2)
4y-28=-10x-20
4y=-10x+8
5x+2y=4
(2b)
(5b-a)/(8b+3a)=1/5
5(5b-a)=1(8b+3a)
25b-5a=8b+3a
25b-8b=3a+5a
17b=8a
Therefore a/b=17/8
a/b=2(1/8)
==========================================
(3a)
Ali : Masah : Yusuf = #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
Yusuf share = 8/16 × 420,000
=#210,000
Therefore
Sum of Ali + Yusuf
=#78,750 + #210,000
= #288,750
(3b)
Solve : 2(1/8)^2 = 32^x-1
2^1 × (8^-1)^x = 32^x-1
2^1 × (2^-3(-1))^x = 2^5(x – 1)
2^1 × (2^-3)^x = 2^5x – 5
2^1 × 2^-3x = 2^5x – 5
2^1+(-3x) = 2^5x – 5
2^1-3x = 2^5x – 5
1 – 3x = 5x – 5
-3x – 5x = -5 – 1
-8x/-8 = -6/-8
X = 3/4
=========================================
(4)
Since Using Pythagoras theorem
|PR|² = |PQ|² + |QR|²
|PR|² = 3² + 4²
|PR|² = 9 + 16
|PR|² = 25 PR = √25
|PR| = 5cm
Considering PRS
|PS|² = |PR|²+|SR|²
13² = 5² + |SR|²
169 = 25 + |SR|²
|SR|² = 169 – 25
|SR|² = 144
|SR| = √144 = 12cm
Hence the area of the quadrilateral = Area of triangle PQR + area of PRS
= 1/2bh + 1/2bh
= 1/2×4×3 + 1/2×12×5
= 6+30 = 36cm
================================
(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|/|AB| = Tan52 |AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m Height of tower = 48.45m ========================================== (10) 130kg of tomatoes for #52,000 Half of the tomatoes 130/2 = 65kg sold at 30% Profit = #52,000/2 = 26,000 #26,000 = 100% X = 130% X = 26000 × 130/100 = #33,800 Then 65kg was then sold at reduction of 12% per kg Recall that the initial cost price = 52000/130 =400kg 65kg sold at = 33,000/65 =#520/kg Then for 12% reduction 520 × 88/100 = 457.6/kg (a) The new selling price per kg = #457.6/kg (b) 65kg - 5kg = 60kg (60kg×457.6kg)+33,800 = #61,256.00 #profit = selling price /cost price × 1000/1 =61256/52000×100/1= 117.8 = 17.8% ==================================== (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 ====================================== (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 × #(300,000-60,000) = 5/12 × #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

Past WAEC 2019 Mathematics Theory and Obj Answers:

1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB

1a)
(y-1)log4=ylog16
log4^(y-1)=log16^y
log4^(y-1)=log4^2*y
y-1=2y
y=-1
1b)Distance =speed*time
let the time of the walk be x hours
=>4*(x 0.5)=5x
Because 30 minutes 0.5hours
4x 2=5x
5x-4x=2
x=2hours
=>5*2
=10km

2a)
2/3(3x-5)-3/5(2x-3)=3
2/3(3x-5)-3/5(2x-3)=3
lcm=15
=>(6x-10)/3-(6x-9)/9=3
Multiply both sides by 15
15(6x-10)/3-15(6x-9)5=15(3)
30x-50-3(6x-9)=45
30x-50-18x 27=45
12x-23=45
12x=45 23
x=68/12
x=5(2/3)
(2b)
=180-n-88=92-n
Also UTQ=180cm–(sum of 80 92-n 180-m=180(sum of 352-n-m=180
-n-m=180-352
-n-m=-172
-(n m)=-172
m n=172degrees

3a)
DRAW THE DIAGRAM
tanx=50/x
tan66.4=50/x
x=50/2.289
x=21.844m
x=22m
3b)
DRAW THE DIAGRAM
Area of=1/2*10*h=45
=>5h=45
h=45/5
h=9cm
Area of parallelogram=L*h
=9*6
=54cm^2
Therefore the area of QTUS=54 45
=99cm

4a)
Tnth=a (n-1)d
Snth=n/2[2a (n-1)d] T6th=37
S6th=147
T6th=>37=a (6-1)d
=>37=a 5d—–(i)
S6th=>147=6/2[2a 96-1)d
=>147=3(2a 5d)
147=6a 15d——(ii)
Therefore a 5d=37—-(i)
6a 15d=147—–(ii)
divide equation ii by 3
2a 5d=49–(ii)
multiply eq i by ii
=>a 10d=74
a 5d=49
5d/5=25/5
therefore d=5
substitute for d in eq i
a 5d=37
a 5(5)=37
a=37-25
a=12
The first term=12
4b)
Sn=n/2[2a (n-1)d] S15=15/2[2(12) (15-1)5] =15/2(24 (14*5)
=7.5(24 70)
=705

5a)
Let bag =B
shoe=S
U=120
n(BnS)=45,
n(S)=x 11
n(B)=x
n(SnB^1)=x 11-45
=x-34
n(BnS^1)=x-45
5b)
DRAW THE VENN DIAGRAM
x-45 45 x-34=120
2x-34=120
2x=120 34
2x/2=154/2
x=77
=11 x
=11 77=88
Therefore 88customers bought shoes
5c)
Pr(bag)=n(B)/U=77/120

(8a)
Tabulate:
X = 1,2,3,4,5
F = m 2, m-1, 2m-3, m 5, 3m-4 = 8m – 1‎
Fx = m 2, 2m-2, 6m-9, 4m 20, 15m-20 = 28m – 9‎
But x̄ ( this symbol (x̄) means X bar)
= 75/23‎
ΣFx / Σf = 75/23 = 28m – 9/8m-1
75/23 = 28m – 9/8m – 1
Cross multiply
75(8m-1) = 23(28m-9)
600m – 75 = 644m – 207
-75 207 = 644m – 600m
132 = 44m
M = 3
8b)
In tabular form
X = 1,2,3,4,5
F = 5,2,3,8,5
C*m Freq= 5,7,10,18,23
Q1 = (N 1/4) = (23 1/4)
= 6
Q3 = (3N 1/4) = (3*23 1/4)
= 18
Inter quarter range = Q3 – Q1
= 18-6
= 12
8bii)
Pr. (at least 4 mark)
= 8 3 2 5/23
= 18/23‎

10)
Sin x = 5/13
Using pythagoras rule
M^2 = 13^2 – 5^2 (^ means Raise to power)
M^2 = 169 – 25
M ^2 = 144
M = √144
M = 12
Hence:Cos x – 2sin x / 2tan x
12/13 – 2(5/13) / 2(5/12)
= 12/13 – 10/23 / 5/6
FIND LCM
= 12 – 10/13 / 5/6= 12/65
10b)
Draw a triangle LACB
in triangle LCB
Hyp^2 = Opp^2 Adj^2
12^2 = 9.6^2 |CB|^2
144 = 92.16 = |CB|^2
144 – 92.16 = |CB|^2
51.84 = |CB|^2
therefore, |CB| = √51.84
|CB| = 7.2m
|AC| |CB| =|AB|
|AC| 7.2m = 10m
|AC| = 10m – 7.2m
|AC| = 2.8m
In triangle LCA
Hyp^2 = Opp^2 Adj^2
|LA|^2 = |AC|^2 |LC|^2
|LA|^2 = 2.8^2 9.6^2
|LA|^2 = 7.84 92.16
|LA|^2 =100
|LA| = √100
|LA| = 10m
10bii)
in triangle LCA
sinθ = Opp/Hyp
sinθ = |LC|/|LA|
sinθ = 9.6/10
sinθ = 0.96
θ = sin^-1 (0.96)
θ = 73.74

13ai)
given
x(*)y=x y/2
3(*)2/5=3 2/5/2
=(15 2/5)*1/2
=17/5*1/2
=17/10= 1,7/10
13aii)
8(*)y=8^1/4
=8 y/2 =33/4
32 4y=66
4y=66-32
4y=34
y=34/4
y=17/2
y=8^1/2
13b)
given DABC
AB=(^-4/6) and AC =(3/^-8)
so AP =1/2(^-4/6)
AP=(^-2/3)
hence
CP = CA AP
CP= -(3/^8) (^-2/3)
CP = (^-5/11)

9ai)
DRAW THE DIAGRAM
9aii)
RTS=?
PTS=180-65(Sum of9b)
Therefore tita=180-52-32-70(sum of < in a triangle)
tita=20degrees
Therefore LWYZ=20 degrees

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