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WAEC 2017 Mathematics Theory and Obj Answers

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WAEC 2017 Mathematics Theory and Obj Answers by : 12:00 am On April 20, 2017

WAEC 2017 Mathematics Theory and Obj Answers

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MATHS OBJ:
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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  1. immanuella says:

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  5. A B Y says:

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