2017 WAEC / MATHEMATICS ANSWERS IS NOW AVAILABLE
MATHS OBJ:
1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB
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MATHS THEORY 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
========================================
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(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
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SECTION B:
(8)
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
Cum 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
==============================
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SECTION B:
(8)
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
Cum 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
==============================
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(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)
(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)
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