ADSENSE CODE HERE
Waec 2017 General Mathematics Obj and Theory Complete Answer Now Available
==============================
Maths-Obj
1-10: CBBDCADBBC
11-20: CACBAAACAB
21-30: BBACBCCAAB
31-40: CBCACBDADB
41-50: DBDCCDBDAB
================================
SECTION A ANS ALL QUESTIONS
================================
1 a)
(y -1 ) log 4 ^ 10 = ylog 16 ^ 10
log 4 ^ 10 ( y -1 )= log 16 ^ y 10
4 ^ ( y -1 )= 16 y
4 ^ y -1 = 4 ^ 2 y
y- 1 = 2 y
-1 = 2 y= y
-1 = y
y= -y
1 b )
let the actual time for 5 km / hr be t
for 4 km /hr = 30 mint + t
4 km /hr =0 . 5 + t
distance = 4 (0 . 5 + t )
= 2 * 4 t
for 5 km /hr , time = t
distance =5 t
1 + 4 t = 5 t
t= 2 hrs
actual distance = 5 * 2 = 10 km
================================
2a)
2/3 (3x – 5) – 3/5 (2x – 3) = 3
(15) x 2/3 (3x – 5) – 3/5 (15) (2x – 3) = 3(15)
10(3x -5) – 9(2x – 3) = 45
30x – 50 – 18x + 27 = 45.
12x – 23 = 45
12x = 45+23
12x = 68
X = 68/12
X = 5.67
===============================
3a)
Tan 23.6° = h/50
Cross multiply
Tan 23.6° x h/50
h = 50 tan 23.6°
= 21.844m
aprox. 22m
================================
4 a)
T 6 =37
T 6 =a + ( 6 – 1 )d
T 6 =a + 5 d
a + 5 d = 37 – — -( eq1 )
s 6 = 147
sn = n / 2 (2 a + (n -1 ) d )
147 = 3 (2 a + 5 d )
49 = 2 a + 5 d
2 a+ 5 d = 49 — — (eq 2 )
a + 5 d = 37 – –( eq1 )
2 a+ 5 d = 49 — -( eq2 )
a =12
4 b )
S 15 = 15 /2 ( 2 (12 )+ 14 d )
S 15 = 15 /2 ( 24 + 14 d )
from(1 )
a + 5 d = 37
12 + 5 d = 37
5 d =37 -12
5 d =25
d =5
S 15 = 15 /2 ( 24 + 14 (15 )
S 15 = 15 / 2 (24 + 70 )
S 15 = 15 /2 * 94
S 15 = 15 * 42
S 15 = 630
================================
5a)
Let bag=B
Shoe= S
U=120
n(BnS)=45, n(s)=x+11, n(b)=x
n(SnB’)=x+11-45
=x-34
n(BnS’) = 45
5b)
Y – 45 + 45 + Y – 34 = 120
2Y – 34 = 120
2Y = 120 – 34
2Y = 154
Y = 154/2
Y = 77
11+x=77+11
= 88
Therefor 88 bought shoes costumer
5c)
n(bag)= 77 customers
Pr. =77/120
================================
SECTION B ANS 5 QUESTIONS ONLY
================================
8)
In Table Form / Tabular form
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 = 4x20x12/8×40
= 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
i)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)
– MAKE SURE YOU WRITE THE SUBJECT and EXAM NAME(eg: WAEC PHYSICS/ENGLISH, ETC) WHILE SUBSCRIBING..
WAEC EXAM RUNS
(GENERAL MATHEMATICS)
Direct Mobile: N800 MTN CARD.
Password/Link: N600 MTN CARD.
Forward Your Name,MTN PIN,Subject Name, Phone number to: 08143987073
Tags:www.wace 2017 mathematics qustions and answers.net
Apr 20 2017,Waec 2017 mathematics 200 to subscribe half of the answer, Google Waec 2017 mathematics question at wapsite, 2017 waec methamatic question and answer
,www.waec2017 mathematics question.com
, waec exams 2017 answers on mathematics
Where to ,How to I need mathematics answer academy hint waec 2017,waec2017mathematics free expose
,Need waec mathematics and2017 waec mathematic answer gidifans,generalbaze,Loadedexpo,expocrew crow,superbaze,flexyhubs,wakagist,wapbaze wassce may/june 2017 mathematics answers
,Www Waec Exam Solution Mathematics Com
,Get waec online 2017 maths answer,examexpo 2017 General Mathematics Nigeria ,Ghana.com
