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Past WAEC 2019 Mathematics Theory and Obj Answers:

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

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