MATHS OBJ:
1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB
MATHS THEORY ANSWERS:
SECTION A:
(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
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
==============================
(11a)
8 students finished
12 tanks in 2/3 (60) mins
= 40 mins
4 student wil finish
X tanks in 1/3 (60)min
= 20mins
X = 4x20x12/8x40
= 3tanks
(11b)
L(AB) = 200m |ON| = 12cm
r2 = (AN)2 (ON )2
r2 = (10)2 (12)2
r2 = 100 144
r2 = 244
r = Sqr 244
r = 15.6CM
(11bii)
L(AB) = 2r sin 0/2
20 = 2 (15.6) sin 0/2
20 = 31.2 sin 0/2
sin0/2 = 20/31.2
sin0/2 = 0.6410
0/2 = sin -1 (0.6410)
0/2 = 39.87
0 = 2 (39.87)
0 = 79.74
= 79.7' (1 d.p )
(11bii)
p 2r 0/360 x 2TTr
= 2 (15.6 ) 79.7/360 x 2x 3 x42x15.6
=31.2 21.7
= 52.9 cm
==================================
(12a)
3y^2-5y 2=0
y^2 - 5/3y 2/3=0
y^2-5/3y=-2/3
y^2-5/3y (^-5/6)^2=(-^5/6)^2-2/3
(y-5/6)^2=25/36-2/3
(y-5/6)^2=25/-24/36
(y-5/6)^2=1/36
(y-5/6)= sqr1/36
y=5/6 1/6
y=5 1/6 or 5-1/6
y=6/6 or 2/3
y=1 or 2/3
(12b)
given
M N = [2,3 1,4]
hence
[1,4 2,3] * [m,n x,y] =[2,3 1,4]
[m 2n, x*2y]
[4m 3n, 4x 5y] = [2,3, 1,4]
therefore
m 2n=2------(i)
4m 3n=3------(ii)
from ------(i)
m=2-2n
4(2-2n) 3n=3
8-8n 3n=3
8-5n=3
8-3=5n
5=5n
n=1
hence
m=2-2(1)
M=0
also
x 2y=1------(i)
4x 3y=4------(ii)
from ------(iii)
x=1-2y
4(1-2y) 3y=4
4-8y 3y=4
y=0
therefore x=1-2(0)
x=1
this N=[i i]
=============================
(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)
We appreciate you for reading Teeloadeders, but we think it will be better you like our facebook fanpage and also follow us on twitter below.
Title :
WAEC 2017 MATHEMATICS QUESTION AND ANSWER
Description : MATHS OBJ: 1-10: CBBCCACBBC 11-20: CACBAAAADB 21-30: BBADBCCACB 31-40: BBCACBDADA 41-50: DDACCCDDCB MATHS THEORY ANSWERS: SECTION A: (1a) (...
Rating :
5
